url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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http://www.tulane.edu/~panda2/Analysis2/Multi-way/sketch_regression.htm | 1,467,295,670,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398869.97/warc/CC-MAIN-20160624154958-00149-ip-10-164-35-72.ec2.internal.warc.gz | 941,190,659 | 1,746 | Sketching a Regression with Interaction
Use the following equation to plug in the possible values for toilet and education and sketch the four resulting points in a table. The combinations include:
NOTOILET DLOWEDN
1. 0 0
2. 1 0
3. 0 1
4. 1 1
Plug each combination into the following equation:
WAZ = -0.958 - 0.791*(NOTOILET) - 0.524*(DLOWEDN) + 0.834*(NOTOILET*DLOWEDN)
Here are the results for each:
1. WAZ= - 0.958 - 0.791 * (0) - 0.524 * (0) + 0.834 * (0) -----> WAZ= - 0.958
2. WAZ= - 0.958 - 0.791 * (1) - 0.524 * (0) + 0.834 * (0) -----> WAZ= -1.749
3. WAZ= - 0.958 - 0.791 * (0) - 0.524 * (1) + 0.834 * (0) -----> WAZ= -1.482
4. WAZ= - 0.958 - 0.791 * (1) - 0.524 * (1) + 0.834 * (1) -----> WAZ= -1.439
Here is a graph of wt/age (y-axis) against access to toilets (yes/no) on the x-axis, for 2 categories of education.
INTERPRETATION:
The graph shows the outcome of WAZ score for all combinations of education and sanitation. This includes consideration for an interaction between education and sanitation. There is clearly a greater effect when sanitation is improved among those with >primary school level of education, whereas there is not anything worth mention in the group for low education. These results are clear because we used an interaction variable in the model, which works very similar to breaking into categories of education status for regression analysis as we did in the previous example with water and sanitation controlling for education status. | 501 | 1,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-26 | latest | en | 0.774593 |
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Design and test a paper helicopter. What is the best design? | 1,257 | 5,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-47 | latest | en | 0.841436 |
https://brainmass.com/engineering/process-engineering/inlet-velocity-flow-approach-piping-stilling-well-147696 | 1,679,331,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00774.warc.gz | 189,534,989 | 75,947 | Explore BrainMass
# Inlet velocity and flow from approach piping to the stilling well
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Please see the attached file for the fully formatted problem(s).
Please see the attached file for the fully formatted problem(s).
> 1200 gallons/min of water flows into approach piping that is tapered in design ----- 1000 gallons/min enters a stilling well while 200 gallons per minute recirculates back to a tank:
Question: How do you determine the inlet velocity and flow from the approach piping to the stilling well?
> The stilling well has a weir so that the water enters the first chamber of the stilling well, then overflows the weir to the 2nd chamber
>Then the water enters the inlet to a diffuser tube bank which consists of 36 4" round diffuser tubes that are 18" long ---- there are 3 rows stacked on top of each other (12 diffuser tubes per row)
Question:
How do you determine what the inlet and outlet velocity and flow is at each of the 36 diffuser tubes?
> Then from the diffuser tube bank, the water flows into a semi-enclosed channel with progressively smaller and smaller dimension until the water exits the small opening of this "sheet flow" channel and fills a 4 ft. x 4 ft x 2 ft high tank below it
Question:
How would you determine the velocity and flow at the discharge slice or opening of this channel??
How long will it take to fill the 4ft x 4 ft x 2 ft high tank below this opening once you know the velocity and flow from the channel opening??
I have provided a fairly detailed sketch of the problem on the attached Excel file --- there are 2 x-section views of the theoretical equipment and a plan view.
Please explain the flow and velocity calculations as clearly as possible at all stages to this problem!
https://brainmass.com/engineering/process-engineering/inlet-velocity-flow-approach-piping-stilling-well-147696
#### Solution Preview
How do you determine the inlet velocity and flow from the approach piping to the stilling well?
The design of the approach piping is similar to that of the tapered header in the headbox of papermachine. The cross-sectional area of the approach piping decreases with the direction of the flow, allowing a uniform pressure profile along the flowpath. This ensures that even flows are distributed in each of the channel entering the stilling well (otherwise you will end up with a non-uniform paper sheet like in your previous question.)
It is therefore reasonable to assume that there will be even flow in each of the entering channel to the stilling well. And there are five of them, so in each channel, the flow rate will be 1000/5 = 200 gpm.
Now convert this to SI unit
1 US gallon = 0.00378 ...
#### Solution Summary
Advanced fluid mecahnics: The inlet velocity and flow from approach piping to the stilling well is calculated
\$2.49 | 658 | 3,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-14 | longest | en | 0.913038 |
https://studysoup.com/tsg/calculus/312/calculus-early-transcendentals/chapter/13397/2-3 | 1,603,978,692,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00036.warc.gz | 547,847,284 | 11,470 | ×
×
# Solutions for Chapter 2.3: INTRODUCTION TO TECHNIQUES OF DIFFERENTIATION
## Full solutions for Calculus: Early Transcendentals, | 10th Edition
ISBN: 9780470647691
Solutions for Chapter 2.3: INTRODUCTION TO TECHNIQUES OF DIFFERENTIATION
Solutions for Chapter 2.3
4 5 0 321 Reviews
28
1
##### ISBN: 9780470647691
Since 82 problems in chapter 2.3: INTRODUCTION TO TECHNIQUES OF DIFFERENTIATION have been answered, more than 40021 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, , edition: 10. Chapter 2.3: INTRODUCTION TO TECHNIQUES OF DIFFERENTIATION includes 82 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Calculus: Early Transcendentals, was written by and is associated to the ISBN: 9780470647691.
Key Calculus Terms and definitions covered in this textbook
• Census
An observational study that gathers data from an entire population
• Combination
An arrangement of elements of a set, in which order is not important
• Cubic
A degree 3 polynomial function
• Deductive reasoning
The process of utilizing general information to prove a specific hypothesis
• Focus, foci
See Ellipse, Hyperbola, Parabola.
• Function
A relation that associates each value in the domain with exactly one value in the range.
• Horizontal shrink or stretch
See Shrink, stretch.
• Normal curve
The graph of ƒ(x) = e-x2/2
• Parallel lines
Two lines that are both vertical or have equal slopes.
• Power function
A function of the form ƒ(x) = k . x a, where k and a are nonzero constants. k is the constant of variation and a is the power.
• Principal nth root
If bn = a, then b is an nth root of a. If bn = a and a and b have the same sign, b is the principal nth root of a (see Radical), p. 508.
• Resolving a vector
Finding the horizontal and vertical components of a vector.
• Right circular cone
The surface created when a line is rotated about a second line that intersects but is not perpendicular to the first line.
• Stem
The initial digit or digits of a number in a stemplot.
• Sum identity
An identity involving a trigonometric function of u + v
• Terminal point
See Arrow.
• Triangular form
A special form for a system of linear equations that facilitates finding the solution.
• Vertices of a hyperbola
The points where a hyperbola intersects the line containing its foci.
• Wrapping function
The function that associates points on the unit circle with points on the real number line
• x-intercept
A point that lies on both the graph and the x-axis,.
× | 658 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-45 | latest | en | 0.862892 |
http://www.jiskha.com/display.cgi?id=1177181354 | 1,498,536,079,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320915.38/warc/CC-MAIN-20170627032130-20170627052130-00416.warc.gz | 538,248,287 | 4,240 | Chemistry - Buffers
posted by .
A buffer is formed by adding 500mL of .20 M HC2H3O2 to 500 mL of .10 M NaC2H3O2. What would be the maximum amount of HCl that could be added to this solution without exceeding the capacity of the buffer?
A. .01 mol
B. .05 mol
C. .10 mol
D. .15 mol
E. .20 mol
[I know the correct answer is B, but I don't know why. Any help would be great.]
just tell me where to start? please?
I don't get 0.05 mol but it's close.
Calculate pH of the buffer.
I have pH = pKa + log [(base)/(acid)]
pH = 4.76 + log [(0.05/0.1) = 4.46
The buffer capacity is the number of mols of strong acid or strong base that can be added to make the buffer change by 1.00 pH units. So adding strong acid we want the pH to be 1.00 less than 4.46 or = 3.46
3.46=4.76 + log (b/a)
-1.3 = log (b/a)
0.05 = (b/a)
If we add x mols of a strong acid, then
mols base = what we start with minus strong acid = 0.05 mols acetate - x mols strong acid.
mols acetic acid = what we start with + mols strong acid = 0.1 mols acetic acid + x mols strong acid.
0.05 = (0.05-x)/(0.1+x)
solve for x mols strong acid.
I found 0.043 mols which doesn't round to 0.05. In fact, I don't think any of the choices are right although 0.05 mols is the closest. However, putting 0.05 mols stong acid gives this for the pH.
acetate = 0.05 - 0.05 = 0
acetic acid = 0.10 + 0.05 = 0.15
and pH = 4.76 + (0.0/0.15) and we run into trouble because of the zero concentration of the base. The buffering capacity is gone if we have no acetate left. I hope this helps.
yah, i wasn't allowed a calculator so we had to guess and estimate things, but thanks so much that really cleared it up. | 543 | 1,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-26 | latest | en | 0.93302 |
https://www.careercup.com/question?id=14809839 | 1,601,432,036,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00620.warc.gz | 701,649,258 | 10,325 | ## Myntra Interview Question for Software Engineer / Developers
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Comment hidden because of low score. Click to expand.
7
of 7 vote
dp
At i=0;
sum[0]=a[i];
sum[1]=max(a[0],a[1])
sum[i]=max(sum[i-2]+a[i],sum[i-1])
return sum[|a|-1)]
Comment hidden because of low score. Click to expand.
0
Comment hidden because of low score. Click to expand.
Comment hidden because of low score. Click to expand.
0
I went ahead and just implemented the simplest answer and here it is
ret[0] = a[0];
ret[1] = a[1];
for (int i = 2; i < a.length; i++){
if (ret[0] < a[i] && ret[1] > ret[0])
ret[0] = a[i];
else if (ret[1] < a[i])
ret[1] = a[i];
}
if line has ret[1] > ret[0] so ret[0] doesn't get changed when ret[1] is smaller than ret[0]
Comment hidden because of low score. Click to expand.
1
of 1 vote
int[] A = {10, 1, 3, 25, 2, 4, 20}; should indeed result in 55 {10, 25, 20} and so the algo is perfectly correct! What is your point?
Comment hidden because of low score. Click to expand.
2
of 2 vote
To solve these type of question, first thing is to find a recurring relation. In this case our recurring relation will tell the max sum till a given length. It means that we will get the max sum for running length of the array. If we denote the ith element as T(i) and max sum till ith element of the array as S(i), then
S(i) = MAX {S(i-1), S(i-2) + T(i) }
S(-1) = 0;
if i=0, S(i) = T(0);
Note: while developing this relationship, I assume array index is starting from 0.
Writing code for this problem is not a big deal now. Below is the code snippet for the same.
sum2 = 0;
sum = sum1 = array[0];
for(i=1; i<len; i++)
{
sum = MAX(sum2 + array[i], sum1);
sum2 = sum1;
sum1 = sum;
}
For a detailed explanation, visit bit.ly/ReP5zl
Comment hidden because of low score. Click to expand.
0
Thank you Anonymous, it's working fine.
Comment hidden because of low score. Click to expand.
0
of 0 vote
Try this
for (int i=0;i<a.length-2;i++)
{
for(int j=i+2;j<a.length;j++)
{
sum = a[i]+a[j];
if (temp<sum)
{
temp = sum;
first = a[i];
second = a[j];
}
}
}
Comment hidden because of low score. Click to expand.
0
of 0 vote
#include<iostream>
using namespace std;
main()
{
int SIZE=3;
int arr[]= {10,1,3,25};
int max_sum=0;
for(int i=0; i<SIZE-1; i++)
{
int sum=arr[i]+arr[i+2];
if(sum > max_sum)
{
max_sum=sum;
}
}
int sum1=arr[0] + arr[SIZE];
if(max_sum<sum1)
{
max_sum=sum1;
}
cout<<max_sum<<"is greatest";
}
Comment hidden because of low score. Click to expand.
0
of 0 vote
import collections
'''
Problem :
Find the subsequences whose elements should not be adjacent and their sum should be maximum from the given array (contains only positive integers).
Eg: int[] A = {10, 1, 3, 25}
Sol: Sum: {10, 3} = 13
{1,25} = 26
{10,25} = 35
Here the Maximum subsequence is {10, 25}.
Approch :
compare the adj elements and find the max and keep them in the sum array
repeat until it's length is 2 (to form a pair)
'''
sumList = []
temp = array
if len(array) < 2:
print "Cannot form a pair"
return
while temp:
for i in range(1,len(temp)):
sumList.append(max(temp[i-1],temp[i]))
temp = None
temp = collections.OrderedDict.fromkeys(sumList).keys()#normal sets are unorderd
print temp
sumList = list()
if len(temp) == 2:
break
print "Result "+str(temp)
def main():
nums = [10,1,3,25]
if __name__ == "__main__":main()
Comment hidden because of low score. Click to expand.
0
of 0 vote
Simple and short implementation of Dynamic Programming idea mentioned here :
int sum_max_dp(int arr[], int n){
int result[n];
result[0]=arr[0];
result[1]=max(arr[0],arr[1]);
for(int i=2;i<n;i++) result[i]= max(result[i-2]+arr[i], result[i-1]);
return result[n-1];
}
Name:
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## Collating costs based upon supplier and month
Occasional Contributor
# Collating costs based upon supplier and month
I’m trying to get Excel to collate how much we spend by supplier by month. I’m exporting a year’s worth of purchased lines from our MRP system – so it will have the same supplier listed many times.
This is pasted to a tab named “Import”, and the three important columns are:
Cells C3 – C10000 contain delivery date
Cells F3 – F10000 contain the supplier name
Cells N3 – N10000 contain the cost
I then have a second tab with all our supplier names scrolling down, and then January to December going across to make a table.
I want to be able to Paste in the data, then have it look for all lines from e.g. “supplier A”, with a January delivery date, then give a total value in the relevant cell.
I’ve worked out the formula to give me totals for each supplier for the year, but I can’t then work out how to formulate that into a monthly breakdown as well. That formula is: =SUMIF(Import!\$F\$3:\$F\$10000,A8,Import!\$N\$3:\$N\$10000)
Cell A8 contains the customer name as an FYI.
Any help would be greatly appreciated, as online searches have proved fruitless.
Many thanks in advance,
Jamie.
7 Replies
Highlighted
# Re: Collating costs based upon supplier and month
You may try a formula like this:
=SUMPRODUCT(Import!\$N\$3:\$N\$10000*(TEXT(MONTH(Import!\$C\$3:\$C\$10000),”mmmm”)=B\$7)*(Import!\$F\$3:\$F\$10000=\$A8))
Highlighted
# Re: Collating costs based upon supplier and month
Many thanks for the response.
I've put this into my sheet but I'm getting #NAME?
I changed "mmmm" to January (I hope this was right, I'm still a bit of a novice so I had to look it up online!)
What is the =B\$7 referencing?
I wonder if I didn't give enough info ref the table - A8 contains the supplier code, but the grid is from I8 (January) to T8 (December) as I have a few hidden columns after the supplier name (address, phone no etc.)
Many thanks,
Jamie.
Highlighted
# Re: Collating costs based upon supplier and month
Can you please attach your sample file?
Highlighted
# Re: Collating costs based upon supplier and month
See attached - many thanks for you assistance.
Jamie.
Highlighted
# Re: Collating costs based upon supplier and month
In the attached edited version of your file, the formula in I8, copied down rows and across columns, is:
=SUMPRODUCT(Import!\$N\$3:\$N\$10000*
(MONTH(Import!\$C\$3:\$C\$10000)=I\$3)*
(Import!\$F\$3:\$F\$10000=\$A8))
Note the use of mixed references in the foregoing formula.
Highlighted
# Re: Collating costs based upon supplier and month
That is perfect, thank you kind sir for your help, it's very much appreciated !
Have a great day.
Jamie.
Highlighted
# Re: Collating costs based upon supplier and month
You're very much welcome!
Related Conversations | 751 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-10 | latest | en | 0.938204 |
http://www.mathworks.com/matlabcentral/fileexchange/2535-advanced-mathematics-and-mechanics-applications-using-matlab-3rd-edition/content/chap9/ninesurfs.m | 1,438,106,022,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00192-ip-10-236-191-2.ec2.internal.warc.gz | 572,978,829 | 11,397 | # Advanced Mathematics and Mechanics Applications Using MATLAB, 3rd Edition
### Howard Wilson (view profile)
14 Oct 2002 (Updated )
Companion Software (amamhlib)
ninesurfs(x,y,z)
```function ninesurfs(x,y,z)
% ninesurfs(x,y,z) plots nine surfaces
if nargin==0, [x,y,z]=sphere; end
d=[min(x(:)), max(x(:)), min(y(:)), max(y(:)),...
min(z(:)), max(z(:))];
subplot(3,3,1), surf(x,y,z(:,:,1)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,2), surf(x,y,z(:,:,2)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,3), surf(x,y,z(:,:,3)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,4), surf(x,y,z(:,:,4)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,5), surf(x,y,z(:,:,5)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,6), surf(x,y,z(:,:,6)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,7), surf(x,y,z(:,:,7)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,8), surf(x,y,z(:,:,8)), axis(d), axis equal, axis off, shg, % pause
subplot(3,3,9), surf(x,y,z(:,:,9)), axis(d), axis equal, axis off, shg, pause
subplot``` | 382 | 1,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.383798 |
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# How do I correctly use the mod operator in MIPS?
In MIPS, I am confused on how to get the mod to work. Below is the code I have come up with thus far. I may have more errors besides the mod, but I feel those errors are a result of the mod misunderstanding. All I'm trying to do is to get the working code (python) here:
``````i = 1
k = 0
while i < 9:
if i % 2 != 0:
k = k + i
i += 1
print(k)
``````
to be correctly translated into MIPS. This is my first shot at assembly, so there may be more than mod errors that are tripping me up in the code below:
``````# Takes the odd integers from 1 to 9, adds them,
# and spits out the result.
# main/driver starts here
.globl main
main:
#data segment
.data
Li: .byte 0x01 # i = 1
Lj: .byte 0x09 # j = 9
Lk: .byte 0x00 # k = 0
Ltwo: .byte 0x02 # 2 for mod usage
# text segment
.text
lb \$t0, Li # temp reg for i
lb \$t1, Lj # j
lb \$t2, Lk # k
lb \$t3, Ltwo # 2
L1: beq \$t0, \$t1, L2 # while i < 9, compute
div \$t0, \$t3 # i mod 2
mfhi \$t6 # temp for the mod
beq \$t6, 0, Lmod # if mod == 0, jump over to L1
add \$t2, \$t2, \$t0 # k = k + i
Lmod: add \$t0, \$t0, 1 # i++
j L1 # repeat the while loop
L2: li \$v0, 1 # system call code to print integer
lb \$a0, Lk # address of int to print
syscall
li \$v0, 10
syscall
``````
-
You haven't explained what the problem is. How is the code behaving differently than what you expected it to? Those `sb` instructions at the beginning look suspicious though; the purpose of `sb` is to write the low byte of a register to memory. – Michael Feb 11 '14 at 7:13
You were exactly right. I shot the lb's in there, and now the code is at least to the point where I can pinpoint the problem. The problem occurs during the 7th iteration of i. k increments correctly up until that point. k = 9 during i = 6, which is correct. But, the next iteration brings k to 10 instead of the correct answer, 16. The problem may lie in how I declared k, but I'm not sure. – kcmallard Feb 11 '14 at 7:55
You are viewing SPIM registers in hex. Hexadecimal 10 is decimal 16.
-
Yup. Changing register showing from hex to dec did it. – kcmallard Feb 11 '14 at 8:48
After working out the kinks, the below code works like a charm. To correctly use the mod operator in MIPS, one must utilize HI and LO. I needed i % 2 == 0 for the statement, so mfhi came in handy. Reference below code for working results:
``````# Takes the odd integers from 1 to 9, adds them,
# and spits out the result.
# main/driver starts here
.globl main
main:
#data segment
.data
Li: .byte 0x01 # i = 1
Lj: .byte 0x0A # j = 10
Lk: .byte 0x00 # k = 0
Ltwo: .byte 0x02 # 2 for mod usage
# text segment
.text
lb \$t0, Li # temp reg for i
lb \$t1, Lj # j
lb \$t2, Lk # k
lb \$t3, Ltwo # 2
L1: beq \$t0, \$t1, L2 # while i < 9, compute
div \$t0, \$t3 # i mod 2
mfhi \$t6 # temp for the mod
beq \$t6, 0, Lmod # if mod == 0, jump over to Lmod and increment
add \$t2, \$t2, \$t0 # k = k + i
Lmod: add \$t0, \$t0, 1 # i++
j L1 # repeat the while loop
L2: li \$v0, 1 # system call code to print integer
move \$a0, \$t2 # move integer to be printed into \$a0
syscall
li \$v0, 10 # close the program
syscall
``````
- | 1,187 | 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} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.862854 |
https://www.wiki.synfig.org/index.php?title=Feather_Parameter&oldid=9655 | 1,618,231,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00639.warc.gz | 1,134,464,228 | 6,295 | # Feather Parameter
Language: English
{{l|Category:Parameters]]
## Contents
Feather is a float value in points (or whatever unit you have defined in File > {{l|Setup Dialog|Setup]] > Misc > {{l|Unit System]]) that represents the width of the area that is going to be dissolved at the edge. Feather is the light horny waterproof structure forming the external covering of birds. See English Feather definition.
The Feather parameter is an standard parameter for the following types of layers:
• {{l|Outline Layer]]
• {{l|Region Layer]]
• {{l|Circle Layer]]
• {{l|Polygon Layer]]
• {{l|Star Layer]]
Strangely {{l|Rectangle Layer]] doesn't have the Feather parameter.
When you apply a Feather to any of those layers, then the edge of the shape becomes dissolved and spread out and shrunk the amount that the Feather value indicates.
Image:Circle Feather zero.png]] Image:Circle Feather 20.png]] Circle with Feather set to 0 points Circle with Feather set to 20 points
## Feather's Complementary Parameters
There is another parameter that works together with the Feather parameter, and which specifies the type of feathering to be used. On Circle layers the parameter is called Fall Off and for the other featherable layers it is called Type of Feather.
### Type of Feather
This complementary parameter can be set in {{l|Outline Layer]], {{l|Region Layer]], {{l|Polygon Layer]] and {{l|Star Layer]]. All the example images below have the same amount of Feather (set to 20 points).
Image:Star Feather Fast Gaussian Blur.png]] Image:Star Feather Gaussian Blur.png]] Image:Star Feather Disc Blur.png]] Fast Gaussian Blur Gaussian Blur Disc Blur Image:Star Feather Box Blur.png]] Image:Star Feather Cross-Hatch Blur.png]] Box Blur Cross-Hatch Blur
### Fall Off
That's the type of feather for the {{l|Circle Layer]]. All the example images below have the same amount of Feather (set to 20 points).
Image:Circle Feather 20.png]] Image:Circle Feather Square.png]] Image:Circle Feather Linear.png]] Square Root Square Linear Image:Circle Feather Sigmond.png]] Image:Circle Feather Cosine.png]] Sigmond Cosine
Language: English | 472 | 2,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.73371 |
https://doubtnut.com/question-answer/if-the-system-of-linear-equations-costhetax-sinthetay-cos-theta0sinthetax-costheta-y-sin-theta-0-cos-142589 | 1,591,477,350,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00368.warc.gz | 315,612,960 | 55,257 | or
# If the system of linear equations ( costheta)x+ (sintheta)y + cos theta=0,(sintheta)x+(costheta) y+ sin theta = 0, (costheta)x + (sintheta) y - costheta =0 is consistent, then the number of possible values of theta in [0,2pi] is (A) 0 (B) 1 (C) 2 (D) 3
Question from Class 12 Chapter Default
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https://www.chegg.com/homework-help/algebra-and-geometry-grade-8-0th-edition-chapter-8.5-solutions-9780395919323 | 1,548,272,825,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00380.warc.gz | 739,556,348 | 22,481 | Solutions
# Algebra and Geometry, Grade 8 (0th Edition) View more editions Solutions for Chapter 8.5
• 6658 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 2
a.
Consider the following information
Paid price of tennis shoes is equal to \$38.15
Discount on tennis shoes is \$16.35
Use the following formula to find original price:
Original pricepaid pricediscount
Therefore,
Hence, the original price of shoes is.
• Step 2 of 2
b.
Consider the following information
Original price of tennis shoes is \$54.50
Discount on tennis shoes is \$16.35
Use following formula to find discount percentage:
Therefore,
Hence, discount percentage is.
Corresponding Textbook
Algebra and Geometry, Grade 8 | 0th Edition
9780395919323ISBN-13: 0395919320ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Passport to Algebra and Geometry, Grade 8 0th Edition Textbook Solutions | 249 | 989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-04 | latest | en | 0.779798 |
https://mathematica.stackexchange.com/questions/2240/definite-and-indefinite-integral-give-different-results-for-piecewise-function | 1,713,826,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00284.warc.gz | 333,112,492 | 40,331 | # Definite and Indefinite integral give different results for piecewise function
I have the following function:
$$f(q,y)= \begin{cases} \tfrac{11720+p}{37791360} & -11720<p<-7720 \\ 0 & \text{True} \end{cases}$$
where $p = 443\ y-777600\ \sin^{-1}\left(\frac{q \lambda }{4 \pi }\right)$.
Mathematica Code:
f[q_,y_]=Piecewise[
{{(11720+443 y-777600 ArcSin[(q λ)/(4 π)])/37791360,
-11720<443 y-777600 ArcSin[(q λ)/(4 π)]<-7720}},
0]
Now, if I take an indefinite integral over y,
weird[q_, y_] = Integrate[f[q, y], y]
which gives
$$g(q,y)= \begin{cases} \frac{293 y}{944784}+\frac{443 y^2}{75582720}-\frac{5}{243} y\ \sin^{-1}\left(\frac{\text{qy} \lambda }{4 \pi }\right) & -11720<p<-7720 \\ 0 & \text{True} \end{cases}$$
I can then evaluate this function at my endpoints {-1,1} to get the definite integral.
However, I could also have Mathematica run the definite integral directly.
weirder[q_, y_] = Integrate[f[q, y], {y, -1, 1}]
which gives
Plain Text Version:
\[Piecewise] (293-19440 ArcSin[(q \[Lambda])/(4 \[Pi])])/472392 907/86400<ArcSin[(q \[Lambda])/(4 \[Pi])]<=1253/86400
(1/33483144960)(-111170729+29544134400 ArcSin[(q \[Lambda])/(4 \[Pi])]-1209323520000 ArcSin[(q \[Lambda])/(4 \[Pi])]^2-12006144000 I Log[4 \[Pi]]-604661760000 Log[4 \[Pi]]^2+12006144000 I Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]+1209323520000 Log[4 \[Pi]] Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]-604661760000 Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]^2) 7277/777600<ArcSin[(q \[Lambda])/(4 \[Pi])]<=907/86400
(1/33483144960)(147938569-37142841600 ArcSin[(q \[Lambda])/(4 \[Pi])]+1209323520000 ArcSin[(q \[Lambda])/(4 \[Pi])]^2+18226944000 I Log[4 \[Pi]]+604661760000 Log[4 \[Pi]]^2-18226944000 I Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]-1209323520000 Log[4 \[Pi]] Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]+604661760000 Log[I q \[Lambda]+Sqrt[16 \[Pi]^2-q^2 \[Lambda]^2]]^2) 1253/86400<ArcSin[(q \[Lambda])/(4 \[Pi])]<12163/777600
0 True
This result, depending on my value of q, is different that the result given by the indefinite integral. Specifically of note are the log terms, which did not appear in the indefinite integral.
Why is the definite integral different from subtracting the end points of the indefinite integral? Which of these results should I trust?
• Can you give a value of $q$ and $\lambda$ that actually give a different numerical result? Feb 24, 2012 at 1:35
• When the indefinite integral has discontunities, substituting the endpoints in the indefinite integral expression gives incorrect results. You get the correct result for the definite integral. Check the section Possible Issues subsection Definite Integral under Integrate in documentation.
– kglr
Feb 24, 2012 at 2:05
When the indefinite integral has discontunities (as is the case for your integrand for some values of q and alpha), substituting the endpoints in the indefinite integral expression gives incorrect results. To get the correct result you need to use the definite integral.
Please see the section Possible Issues subsection Definite Integral under Integrate in docs. The issue is also discussed at length in this Wolfram Blog entry: Mathematica and the Fundamental Theorem of Calculus.
• A particularly nasty example is $$\int^\infty_0 \frac{e^{−x}}{\sin x}dx$$ which is discussed here. Of particular relevance is this answer which discusses how the changing the limits effected the outcome. Mar 2, 2012 at 19:37 | 1,147 | 3,497 | {"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} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.382101 |
https://coffeetearoom.com/how-many-tbsp-for-8-cups-of-coffee/ | 1,695,763,761,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00214.warc.gz | 187,413,898 | 14,184 | # how many tbsp for 8 cups of coffee?
## How many tbsp in three quarter of a cup?
How many tablespoons in a quarter cup? A quarter cup (US) is equal to 4 tablespoons (US). A quarter cup (metric) is equal to 4.167 tablespoons (metric). Use username: Guest, Anonymous, Programmer. QUOTES, SAYINGS: Some tension is necessary for the soul to grow, and we can put that tension to good use. We can look for every opportunity to give …
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## How many tablespoons are needed for four cups of coffee?
There is no standard size for a scoop and this will depend on what you are using. Generally speaking, for one cup of coffee, you need to put one tablespoon of coffee if you want it light and two tablespoons if you want it strong. In this case, if your idea of a scoop is a tablespoon, you need four scoops for four cups of regular coffee.
More…
## How many TSP to make a tbsp?
You have 3 options, none of which involve buying a 1/2 Tablespoon measure:
• 1/2 Tablespoon = 1.5 Teaspoons. Presuming you have 1 teaspoon and 1/2 teaspoon measuring spoons, use those to obtain 1/2 TBSP.
• As others have said, eyeball it by filling up your 1 TBSP measure halfway
• Sure, go for the full 2 TBSP. Its effect will entirely depend on what you are cooking. …
More…
## How many tbsps in one-fourth Cup?
The answer is 1 4 cup equals 4 Tablespoons To convert any value in cups to Tablespoons, just multiply the value in cups by the conversion factor 16. So, 1/4 cup times 16 is equal to 4 Tablespoons. All In One Units Converter
More…
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## Does Tea or Coffee Have More Caffeine? – Healthy Teo
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· Because of the claimed health benefits, many people will recommend coffee. There are many more benefits of coffee, but we will focus on the 5 common benefits. Here are the five health benefits of drinking coffee: 1. Increased Energy Levels. Coffee contains caffeine, a stimulant that can increase your energy levels.
## Ditch dairy and try these plant-based coffee recipes for …
· Many coffee drinkers agree that coffee beans ground right before brewing bring out the fresh flavor in a morning cup of coffee. The measurement is also equal to 3/8 pint, 1/16 gallon, 3/ 3/4 cup is equal to 12 tablespoons or 6 fluid ounces and approximately equal to 177. | 1,040 | 4,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-40 | longest | en | 0.916194 |
http://mathhelpforum.com/differential-geometry/181444-continuity-norm-space.html | 1,527,459,274,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00234.warc.gz | 172,438,477 | 10,495 | # Thread: Continuity and norm space
1. ## Continuity and norm space
Let V a norma space with 2 norms , lets say norm 1 and norm 2
Show that the function identity $\displaystyle$\ I_d :\left( {V,\left\| x \right\|_1 } \right) \to \left( {V,\left\| x \right\|_2 } \right)\$$es continuos iff the set A\displaystyle \A = \left\{ {x \in V/\left\| x \right\|_1 = 1} \right\}\$$ is bounded with norm 2
Please somebody give me a biiiiiiiiiiiiig hint....
Regards...
2. Originally Posted by orbit
Let V a norma space with 2 norms , lets say norm 1 and norm 2
Show that the function identity $\displaystyle$\ I_d :\left( {V,\left\| x \right\|_1 } \right) \to \left( {V,\left\| x \right\|_2 } \right)\$$es continuos iff the set A\displaystyle \A = \left\{ {x \in V/\left\| x \right\|_1 = 1} \right\}\$$ is bounded with norm 2
Please somebody give me a biiiiiiiiiiiiig hint....
Regards...
Come on man--you really need to show some effort. You need remember that linear operators are continuous if and only if they're bounded. So, $\displaystyle \text{id}:V\to V$ will be bounded if and only if there is a constant $\displaystyle M$ such that $\displaystyle \|\text{id}(v)\|_1\leqslant M\|v\|_2$ for all $\displaystyle v\in V$... now why is this equivalent to saying that this inequality holds true for every $\displaystyle v\in B_{\|\cdot\|_1}(0;1)$?
3. Originally Posted by Drexel28
Come on man--you really need to show some effort. You need remember that linear operators are continuous if and only if they're bounded. So, $\displaystyle \text{id}:V\to V$ will be bounded if and only if there is a constant $\displaystyle M$ such that $\displaystyle \|\text{id}(v)\|_1\leqslant M\|v\|_2$ for all $\displaystyle v\in V$... now why is this equivalent to saying that this inequality holds true for every $\displaystyle v\in B_{\|\cdot\|_1}(0;1)$?
Hi, thanl you for answering. unfortunately for me,linear operators havenīt been taught is my course.
4. Originally Posted by orbit
Hi, thanl you for answering. unfortunately for me,linear operators havenīt been taught is my course.
Ok, so then what have you been taught?
5. Originally Posted by Drexel28
Ok, so then what have you been taught?
Here is what Ive been taught.
6. Originally Posted by orbit
Here is what Ive been taught.
I don't mean to be rude but | 690 | 2,306 | {"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} | 3.78125 | 4 | CC-MAIN-2018-22 | latest | en | 0.797316 |
https://burningservos.com/2014/02/ | 1,674,908,701,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00574.warc.gz | 172,180,155 | 23,119 | Body assembly completed
Finish the other legs and completed the body assembly. The servos are hold to a piece of plywood with plastic straps and the battery holder is fixed to the “belly of the beast” with double sided tape.
Walking – gaits
Now that we can control the position of our feet and move them along a trajectory we can organise the leg movements into a walking gait.
To start with I’ll implement a static gait. In a static gait at any time during the walk cycle there will be at least tree legs in contact with the ground. Contrast this with dynamics gaits like trot or pace where two legs at a time are in contact with ground.
A static gait is defined by the order at which the legs are lifted during the walk cycle. In a crawling gait, for example, the legs are lifted in the following order:
1. front left leg
2. back right leg
3. back left leg
4. front right leg
To implement a static gait we can split our step trajectory: one segment will be the flight phase (the complete arc trajectory) and the ground phase trajectory will be divided into tree segments.
The start of each segment will be represented by a key position. We can the use this key positions to describe our gaits in a compact data structure. A crawling gait can be then represented as:
int[][] gait = {
{1,4,2,3},
{2,1,3,4},
{3,2,4,1},
{4,3,1,2}
}
Each row represents a step. The four numbers in the row represent the key position for the corresponding leg, starting with the back left leg and moving in clockwise direction.
If we take for example the first row (first step in the cycle), the key positions for the four legs are:
back left leg: 1. key position
front left leg: 4. key position
front right leg: 2. key position
back right leg: 3. key position
With this setup we can generate a trajectory for each leg between it’s current key position and the key position in the next step in the cycle. Then we move each leg through the generated trajectory points until we reach the next step and a new trajectory is generated.
You can download the Processing sketch that illustrates this approach from my repository: https://github.com/Traverso/Felix/tree/master/Processing/Gait.
In this sketch you will find two different types of static gaits and their corresponding “backwards” versions. Just set the variable “CURRENT_GAIT” to 0,1,2 or 3 to test the different walk cycles.
Walking – trajectory
Once we can control the position of the foot, we can make it follow a trajectory. One cycle through the trajectory is a simple step.
A step has two distinct phases, a ground and a flight phase. During ground phase the leg push the body forward, during the flight phase the foot is lifted and placed ahead of its previous contact point.
The trajectory is constructed as a list of points in space. So we need to generate points along a straight line for our ground phase, and points along an arc for our flight phase.
The number of points in a given path is specified by a “granularity” variable. With cheap servos, there is no need to generate a lot of points. Same goes for implementing any type of “easing“, so we keep it simple and linear.
The geometry of the trajectory can be edited. This is going to be useful when implementing steering. To turn slightly to one side or another, the two legs on one side can take wider steps than the two other legs.
Once again I use processing to test a simple trajectory implementation.
You can download the sketch from the repository (https://github.com/Traverso/Felix/tree/master/Processing/Trajectory), and play with the step width and height, and the offset of the step in relation to the position of the hip.
I wanted to upload this sketch to jsFiddle as well so you could play with it directly in your browser, but as this sites motto says “in theory there is no difference…”.
I couldn’t get the sketch to work on the browser as it is and I currently don’t have the time to do the necessary tweaks.
Walking – IK
This is an overview of my approach to implement a static gait for Felix. I’ll be explaining the needed concepts in a series of posts, but if you want to speed things up you can grab all the Arduino sketches from my repo
Felix is a four legged robot, each leg has two actuators. To control the position of the feet, I need to provide a rotation angle to the servos controlling the hip and the knee of each leg. But thinking in terms of rotation angles is too complicated, it could be much easier if we could set the angles according to where we want to position the feet.
To do that we can use inverse kinematics. Calculating the inverse kinematics for complex systems can be quite daunting, but in our case we have a simple planar two link setup, so we can use basic law of cosines and Pythagoras to calculate the two angles.
Even though Felix legs consists of three links, we can still consider it a two link system, because due to the parallel linkage constraining the pastern, the relation between the endpoint of the shin (the hock in dog anatomy) and the foot is one of relative displacement.
There are a number of references on the web explaining inverse kinematics for this type of systems:
http://www.engineer-this.com/IK.shtml
http://www.oliverjenkins.com/blog/2012/9/inverse-kinematics-and-robot-arms | 1,202 | 5,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-06 | longest | en | 0.877794 |
https://discourse.mc-stan.org/t/autodetection-of-causes-of-divergent-transitions/2306 | 1,653,342,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00561.warc.gz | 248,567,568 | 8,494 | # Autodetection of causes of divergent transitions
Hello all,
I’ve seen someone claiming that it is difficult/impossible to automate the search for causes of divergent transitions because divergencies might arise for many reasons. However - the general advice for dealing with models with divergencies is to look at the pairs plot, where as I understand it plots that look like a standard, bivariate normal are very good, and funnels and other strange shapes are not good.
…but if we know that bivariate normal pair-plots are nice, can’t we then use bivariate normality tests to get at least a hint of which parameters of a model that might be problematic? See example code with Neil’s Funnel below.
library(MVN)
library(rstan)
y <- rnorm(1)
x <- rnorm(9, mean=0,sd=exp(y/2) )
stanmodelcode<-“
parameters {
real y_raw;
vector[9] x_raw;
}
transformed parameters {
real y;
vector[9] x;
y = 3.0 * y_raw;
x = exp(y/2) * x_raw;
}
model {
y_raw ~ normal(0, 1); // implies y ~ normal(0, 3)
x_raw ~ normal(0, 1); // implies x ~ normal(0, exp(y/2))
}
f <- stan(model_code=stanmodelcode,
iter = 500)
smpls <- as.data.frame(extract(f))
pars <- dim(smpls)[2]-1
normtests <- matrix(NA, ncol=pars,nrow = pars,
dimnames = list(names(smpls)[1:pars],
names(smpls)[1:pars])
)
for(i in 1:(pars-1)){
for(j in (i+1):pars){
tmp <- try(mardiaTest(cbind(smpls[,i],smpls[,j]), qqplot = F))
if(inherits(tmp,“try-error”)==F){
normtests[i,j]=tmp@chi.skew
}
}
}
1 Like
+1 it would be useful to have a program that could at least identify specific variables that I should examine when I have divergent transitions.
One way to check these is to test(/sort) mean / std for divergent vs non-divergent draws. E.g. if they are concentrated somewhere.
I haven’t checked this out extensively but I agree that it should work to provide clues. Or at least, I agree enough that we put in a grant to fund exploration/implementation of something related to automatically suggest where to look. It really starts to make sense when you have a lot of parameters. I’m not sure how well normality tests would work in practice since they aren’t all that sensitive.
2 Likes
For the record, the mardiaTest from the mvn-package is fairly slow. A colleague of mine suggested to simply evaluate the bivariate normal likelihood of each pair or draws, using the posterior means and posterior covariance. Hence, higher values should equal a distribution that looks more like a bivariate normal.
This solution can use the “mvnfast”-package, which is using C++ code. This might scale to models with many parameters (that is, if the binorm-trick is good enough at giving hints…). See code below.
``````library(mvnfast)
library(rstan)
y <- rnorm(1)
x <- rnorm(9, mean=0,sd=exp(y/2) )
stanmodelcode<-"
parameters {
real y_raw;
vector[9] x_raw;
}
transformed parameters {
real y;
vector[9] x;
y = 3.0 * y_raw;
x = exp(y/2) * x_raw;
}
model {
y_raw ~ normal(0, 1); // implies y ~ normal(0, 3)
x_raw ~ normal(0, 1); // implies x ~ normal(0, exp(y/2))
}
"
f <- stan(model_code=stanmodelcode,
iter = 500)
smpls <- as.data.frame(extract(f))
pars <- dim(smpls)[2]-1
normtests <- matrix(NA, ncol=pars,nrow = pars,
dimnames = list(names(smpls)[1:pars],
names(smpls)[1:pars])
)
for(i in 1:(pars-1)){
for(j in (i+1):pars){
parvar <- cbind(smpls[,i],smpls[,j])
tmp <- try(sum(dmvn(parvar,mu=colMeans(parvar), sigma=cov(parvar), log=T)))
if(inherits(tmp,"try-error")==F){
normtests[i,j]=tmp
}
}
}``````
Out of curiosity I ran this on a 1,385 parameter model that otherwise checks out fine. It was plenty fast (a few minutes at most) although I could see a really large model blowing it out. There were a whole bunch of mostly false-positives driven by a few outlying points but the interesting thing is you can ID individual points that don’t match the MVN assumption. The problem with this approach is that Stan’s HMC will happily sample from plenty of non-MVN distributions. If somebody has a real-life big model with problems and wants to check it out we’d love to hear about it.
…are we really missing poorly written model!? :-) That sounds like a solvable issue.
Below is Piironen and Vehtari’s Ponyshoe prior model. They have two parameterizations in their appendix - one good and one less good. Below is the bad one.
When I ran this script I got 592 divergence errors. The likelihood-sorted list of pairs of variables gave the following:
1: Combinations of the intercept and some other parameter. Pretty sure this is a false positive.
2: Combinations of a parameter + the same parameter, just transformed (i.e. x and i.e. ln x). So false positive, but easy to rule out.
3: the “lambda”-parameters. These should be reparameterised, and they are in Aki’s paper.
So yes - this method does give false positives, but also seem to flag problem parameters. See script below.
`````` library(mvnfast)
library(rstan)
rstan_options(auto_write = TRUE)
options(mc.cores = parallel::detectCores())
rm(list=ls())
set.seed(123)
library(shinystan)
n <- 10
d <- 9
bvec <- rnorm(d)/rnorm(d)
x <- matrix(n*d, nrow=n, ncol=d)
y <- as.vector(x %*% bvec + rnorm(n))
p0 <-5
scale_icept<-100
scale_global<-p0/((d-p0)*sqrt(n))
nu_global <-1
nu_local <-1
slab_scale <-100
slab_df <-1
stanmodelcode<-"
data {
int < lower =0> n; # number of observations
int < lower =0> d; # number of predictors
vector [n] y; # outputs
matrix [n,d] x; # inputs
real < lower =0> scale_icept ; # prior std for the intercept
real < lower =0> scale_global ; # scale for the half -t prior for tau
real < lower =1> nu_global ; # degrees of freedom for the half -t priors for tau
real < lower =1> nu_local ; # degrees of freedom for the half -t priors for lambdas
real < lower =0> slab_scale ; # slab scale for the regularized horseshoe
real < lower =0> slab_df ; # slab degrees of freedom for the regularized horseshoe
}
parameters {
real logsigma ;
real beta0 ;
vector [d] z;
real < lower =0> tau; # global shrinkage parameter
vector < lower =0 >[d] lambda ; # local shrinkage parameter
real < lower =0> caux ;
}
transformed parameters {
real < lower =0> sigma ; # noise std
vector < lower =0 >[d] lambda_tilde ; # 'truncated ' local shrinkage parameter
real < lower =0> c; # slab scale
vector [d] beta ; # regression coefficients
vector [n] f; # latent function values
sigma = exp ( logsigma );
c = slab_scale * sqrt ( caux );
lambda_tilde = sqrt ( c^2 * square ( lambda ) ./ (c^2 + tau ^2* square ( lambda )) );
beta = z .* lambda_tilde *tau;
f = beta0 + x* beta ;
}
model {
# half -t priors for lambdas and tau , and inverse - gamma for c^2
z ~ normal (0, 1);
lambda ~ student_t ( nu_local , 0, 1);
tau ~ student_t ( nu_global , 0, scale_global * sigma );
caux ~ inv_gamma (0.5* slab_df , 0.5* slab_df );
beta0 ~ normal (0, scale_icept );
y ~ normal (f, sigma );
}
"
f <- stan(model_code = stanmodelcode,
iter=2000,
control=list(max_treedepth=13),
chains=2
)
smpls <- as.data.frame(extract(f))
pars <- dim(smpls)[2]-1
normtests <- as.data.frame(matrix(NA, ncol=3,nrow = pars*(pars-1)/2))
colnames(normtests) <- c("par1", "par2", "loglik")
cnt<-1
for(i in 1:(pars-1)){
for(j in (i+1):pars){
parvar <- cbind(smpls[,i],smpls[,j])
normtests[cnt,1] <- names(smpls)[i]
normtests[cnt,2] <- names(smpls)[j]
tmp <- try(sum(dmvn(parvar,mu=colMeans(parvar), sigma=cov(parvar), log=T)))
if(inherits(tmp,"try-error")==F){
normtests[cnt,3]=tmp
}
cnt<-cnt+1
}
}
normtests <- normtests[order(normtests[,3]),] | 2,154 | 7,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.797212 |
https://www.mapleprimes.com/users/ANANDMUNAGALA/questions?page=14 | 1,709,447,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476205.65/warc/CC-MAIN-20240303043351-20240303073351-00807.warc.gz | 858,149,937 | 22,131 | ## ANANDMUNAGALA
Mr. ANAND MUNAGALA
## 175 Reputation
10 years, 175 days
MRIET
Associate Professor in Mathematics
I have more than a decade of teaching experience in Mathematics for various Undergraduate programs in reputed educational institutions in Hyderabad, Andhra Pradesh, INDIA. Currently working as Assistant Professor in Mathematics and as Managed Network expert for Chegg E-Learning services private limited, Kolkatta, INDIA, for two years for solution authoring and content creation in Mathematics.
## How to create an Augmented Matrix of a g...
Maple 17
Good morning Professor.
I am M.Anand working for SRIIT, as Assistant Professor in Mathematics in Hyderabad, INDIA.
I request you to help me out to create an Augmented Matrix of a given system of linear equations with more number of variables.
With thanks & regards.
M.Anand
Assistant professor in Mathematics
SR International Institute of Technology
## Maple procedures for conjugate gradient ...
Maple 17
Good morning professor.
I am M.Anand working for SRIIT, as Assistant professsor in Matematics in Hyderabad, INDIA.
I request you to help me out in generating Maple procedures for conjugate gradient method & Preconditioned Conjugate gradient method as it is consuming much time for me to find out every iteration individually to solve linear systems with sparse coefficient matrices which arises in Numerical Analysis.
## Query related to input of a column vecto...
Maple 17
Good afternoon
I request you to help me out to input a column vector matrix of dimension 80x1 whose entries are 0.
## Query related to genearte large matrices...
Maple 17
Good afternoon sir,
I am M.Anand working for SRIIT, as Assistant Professor in Mathematics.
I request you to provide me the correct Maple commands to generate large matrices whose elements are defined in terms of i & j in piecewise function pattern for example
How to generate the matrix A where the elements a(i,j) are defined as
a(i,j)=2i when j=i and i=1,2,...,80
a(i,j)=0.5i when j=i+2 and i=1,2,...,78...
First 12 13 14 Page 14 of 14
| 484 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.888229 |
http://mathhelpforum.com/advanced-statistics/72280-p-d-f-s-c-d-f-s-functions-print.html | 1,508,669,954,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00497.warc.gz | 225,453,293 | 3,696 | # p.d.f.s and c.d.f.s of functions
• Feb 7th 2009, 05:07 AM
James0502
p.d.f.s and c.d.f.s of functions
Let X be a continuous random variable taking values in (a, b) with cumulative distribution function F, strictly increasing
on (a, b). Show that Y = F(X) has a uniform distribution on (0, 1).
How would you use a set of computer generated random numbers (assumed to be drawn from a uniform distribution on (0,1) to simulate a random sample from
f(x) = 1/a . e^(-x/a) x>0
Not reeally sure at all on this one..
• Feb 7th 2009, 07:21 AM
awkward
Quote:
Originally Posted by James0502
Let X be a continuous random variable taking values in (a, b) with cumulative distribution function F, strictly increasing
on (a, b). Show that Y = F(X) has a uniform distribution on (0, 1).
How would you use a set of computer generated random numbers (assumed to be drawn from a uniform distribution on (0,1) to simulate a random sample from
f(x) = 1/a . e^(-x/a) x>0
Not reeally sure at all on this one..
Use $F^{-1}(u)$ as your random sample, where $u$ is a computer-generated (pseudo-)random number drawn from a Uniform(0,1) distribution.
You need to find a formula for $F^{-1}(u)$ in order to make this practical.
(This is a general method for generating numbers from arbitrary distributions on a computer.)
• Feb 7th 2009, 08:32 AM
James0502
how would I show the distribution is uniform though?
• Feb 7th 2009, 03:17 PM
mr fantastic
Quote:
Originally Posted by James0502
Let X be a continuous random variable taking values in (a, b) with cumulative distribution function F, strictly increasing
on (a, b). Show that Y = F(X) has a uniform distribution on (0, 1).
How would you use a set of computer generated random numbers (assumed to be drawn from a uniform distribution on (0,1) to simulate a random sample from
f(x) = 1/a . e^(-x/a) x>0
Not reeally sure at all on this one..
• Feb 7th 2009, 05:16 PM
oswaldo
"Let X be a continuous random variable taking values in (a, b) with cumulative distribution function F, strictly increasing
on (a, b). Show that Y = F(X) has a uniform distribution on (0, 1)."
The above statement is ture for all pdf's/cdf's, and it is the basis of simulation: generating data for numerous distributions.
Already mention but here how it works:
1. generate a random number in (0,1). (Use "=rand()" in MS Excel)
2. use F-inverse to compute the dist. value you want to generate.
I don't remember if Normal has an inverse (I think it should). But there is a way of generating Normal-ly distributed values buy above steps. I think they use "characteristic func", which unique for every pdf.
-O
• Feb 7th 2009, 06:29 PM
awkward
Quote:
Originally Posted by James0502
Let X be a continuous random variable taking values in (a, b) with cumulative distribution function F, strictly increasing
on (a, b). Show that Y = F(X) has a uniform distribution on (0, 1).
[snip]
Since F is strictly increasing, F has an inverse.
Let $0 \leq y \leq 1$. Then
$F(x) \leq y$ if and only if $x \leq
F^{-1}(y)$
so
$P(F(x) \leq y) = P(x \leq F^{-1}(y)) = F(F^{-1}(y)) = y$
I.e., the CDF of Y is Y for $0 \leq Y \leq 1$. So Y has a Uniform(0,1) distribution. | 925 | 3,172 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.869242 |
https://community.wolfram.com/web/molivo/home?p_p_id=user_WAR_userportlet&p_p_lifecycle=0&p_p_state=normal&p_p_mode=view&p_p_col_id=column-1&p_p_col_count=1&tabs1=Discussions | 1,716,302,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00206.warc.gz | 157,596,919 | 12,620 | # User Portlet
Miguel Olivo-V
Discussions
I'm trying to specify rules for my plot, but they seem to be ignored by Mathematica: GeoRegionValuePlot[{ Entity["AdministrativeDivision", {"Arkansas", "UnitedStates"}] -> 1, Entity["AdministrativeDivision", {"California",...
I have a list of strings of the form list={str1,str2,strx,str3,...,str10} each string contains both letters and numbers, except for strx which does not contain numbers, what I want to do is attach strx to the element that immediately...
Nevermind, just realized Mathematica can handle Table[f[x,y],{x,1,T},{y,1,Min[x,K]}]
Thanks Frank, what does that mean? The system has no solution (at least with the restriction that the values must lie in [0,200])? If you think about it, it must have a solution given that it can be viewed as a system of recursive equations with...
Hi, I'm trying to solve the system \begin{align*} &V(m)=\alpha + \beta V(m+1)+\beta \sum_{f=1}^4 Z(m+1,f)\\ &U(f)=\gamma +\beta U(f+1)+\beta \sum_{m=1}^4 Z(m,f+1)\\ &Z(m,f)=\zeta + \beta (Z(m+1,f+1)+U(f+1)+V(m+1))\\ ...
Hi, I was wondering if I can improve on this method for solving a particular system of coupled recurrence equations. I have attached a pdf file with the system of equations and a notebook file with the solutions. Notice that I have to solve for...
I'm in Canada, got the email notifying me of the updated site licence a few hours ago. However we got 10.0.2 about 6 or 7 days later, I had to download a trial to satiate my curiosity!
Hey Marco, I saw you profile and it seems that you also work with data. I have used Mathematica since version 4 for symbolics and for simple data analysis using matrices but now I feel outdated with respect of these new objects that Mathematica has...
You can write the time subscripts as arguments, i.e. type p[t], p[t+1], etc. As for other subscripts, you can just type y1, y2. One more thing, you have to either write the expressions in different cells or separate them using semicolons instead...
Hi, I'm trying to define the derivative of a function H to be positive. I'm using H/:Positive[H'[x]]=True But I get the error TagSetDelayed::tagpos: Tag H in Positive[H'[x]] is too deep for an assigned rule to be found. I also... | 600 | 2,224 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.90891 |
https://docs.google.com/document/d/1R_O1fud5SF9gODab4wapgBujSW-xl4hrQScWWHuFMPw/pub?embedded=true | 1,503,474,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00257.warc.gz | 782,276,809 | 2,174 | ### Exercise 1
On September 1, 20x1, Entity A entered into an agreement to rent office space for a year and paid \$36,000 for a six-month rent. Prepare journal entries on the following dates:
### September 1, 20x1
Paid \$36,000 rent for a six-month period from September 1, 20x1 to February 28, 20x2
Prepaid rent 36,000 Cash 36,000
### December 31, 20x1
Record the rent expense for the period from September 1, 20x1 to December 31, 20x1
Rent expense 24,000 Prepaid rent 24,000
[Note]Monthly rent expense = \$36,000 / 6 months = \$6,000
Rent expense for the period from September 1 to December 31 = \$6,000 x 4 = \$24,000 | 199 | 631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.933694 |
https://it.mathworks.com/matlabcentral/answers/274804-how-to-set-yticklabel-in-double-y-axis-plot?s_tid=prof_contriblnk | 1,712,980,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00869.warc.gz | 294,020,567 | 28,800 | # How to set YTickLabel in double y-axis plot?
12 visualizzazioni (ultimi 30 giorni)
Albert Bing il 22 Mar 2016
Modificato: dpb il 23 Mar 2016
The simplified code is as following:
x=[2,3,4,5,6,7];
y1=[ 1, 1, 1.2, 1.2, 1, 1];
y2=[ 0, 0.01, 0.02, 0.02, 0, 0.01];
fig1 = figure;
[ax,h1,h2] = plotyy(x,y1, x,y2, 'plot');
set(get(ax(1),'YLabel'),'String','y1','fontsize',14)
set(get(ax(2),'YLabel'),'String','y2','fontsize',14)
legend('y1','y2');
xlabel('x')
ylim(ax(1), [0.7, 1.3])
ylim(ax(2), [0, 0.03])
print(fig1, '-dpng', 'test1')
The figure plot is like
I wanted to set the digits of YTick, so I tried something like
set(get(ax(1),'YTick'), [1.00, 1.05, 1.10, 1.15, 1.20]);
set(get(ax(1),'YTickLabel'), [1.00, 1.05, 1.10, 1.15, 1.20]);
or
set(get(ax(1),'YTick'), {'1.00', '1.05', '1.10', '1.15', '1.20']);
set(get(ax(1),'YTickLabel'), {'1.00', '1.05', '1.10', '1.15', '1.20']);
or
set(get(ax(1),'YTick'), str2mat('1.00', '1.05', '1.10', '1.15', '1.20']);
set(get(ax(1),'YTickLabel'), str2mat('1.00', '1.05', '1.10', '1.15', '1.20']);
All went wrong.
So my question is, how to set the digits of y axis label?
And why the left YTick is from 1.0 to 1.2 but leaves all other ticks?
And more generally, how to handle this `gca` properties?
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### Risposta accettata
dpb il 22 Mar 2016
Modificato: dpb il 23 Mar 2016
Isn't it annoying that plot isn't intelligent-enough to use a consistent numeric format for tick labels??? I've complained about this for almost 30 years now... :(
Anyway, it's relatively easy to fix...
Long-winded way; easier to see what did...
yt=get(hAx(1),'ytick'); % retrieve the tick marks from desired axes
set(hAx(1),'yticklabel',num2str(yt.','%.2f')) % set with a desired format same values
You can, of course, dispense with the temporary -- easiest to write w/o getting fouled up with parentheses nesting if start from inside out...
set(hAx(1),'yticklabel',num2str(get(hAx(1),'ytick').','%.2f'))
NB: the transpose; it's important as the tick values are returned as a row vector and num2str needs a column vector or the values will all be returned as a single long string, not an array of strings.
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
dpb il 22 Mar 2016
>> help transpose
.' Transpose.
X.' is the non-conjugate transpose.
I commented specifically on the need and what happens without it...
Albert Bing il 22 Mar 2016
Ah, transpose, thank you!
So that better be
num2str(yt.', '%.2f')
Accedi per commentare.
### Più risposte (1)
Mike Garrity il 22 Mar 2016
Or in R2016a:
ax(1).YAxis.TickLabelFormat = '%.2f';
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Translated by | 1,024 | 3,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.460487 |
http://math.stackexchange.com/questions/231070/matrix-of-a-bilinear-form-on-a-space-of-matrices/231259 | 1,469,406,892,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.28/warc/CC-MAIN-20160723071024-00281-ip-10-185-27-174.ec2.internal.warc.gz | 159,930,198 | 17,672 | # matrix of a bilinear form on a space of matrices
For a bilinear form on $\mathbb R^2$ the matrix of the bilinear form, A, with respect to the standard basis is $a_{ij}= \langle e_i, e_j\rangle$, for $i =1,2$, j = $1,2$. Then for any two vectors $x,y$ one can write $\langle x,y\rangle = x^tAy$.
i don't understand the matrix of a bilinear form in say $\mathbb R^{2*2}$. If I take a bilinear form $\langle A,B\rangle = trace(AB)$ with respect to the standard basis $e_{ij}$ I get a $4*4$ matrix, C, of the form. Then the product $\langle A,B\rangle$ = $A^tCB$ isn't defined since is consists of a $2*2$ matrix multiplied either side of a $4*4$. At least this is my incorrect intuition. What is the correct way to view the matrix of a form on a space of matrices?
-
Well, I have used this form quite a bit, and never felt the need to think about its matrix, so one answer might be: don't think about it. – Chris Godsil Nov 6 '12 at 0:00
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Nov 6 '12 at 8:05
Let $b:V\times V\rightarrow K$ be a bilinear form on a vector space $V$ of finite dimension over a field $K$. Then the matrix $B\in K^{n\times n}$ of $b$ with respect to some basis $v_1,\ldots ,v_n$ of $V$ has the entries $b_{ij}:=b(v_i,v_j)$ and satisfies the equation
$b(v,w)=x^tBy$
where $x,y\in K^n$ are the coordinate vectors of $v,w$ with respect to the given basis. That is $v=x_1v_1+x_2v_2+\ldots +x_nv_n$, $x=(x_1,\ldots ,x_n)$ and similar for $y$.
In your case $V=\mathbb{R}^{2\times 2}$ and as a basis you took the matrices $E_{ij}$ having only an entry $1$ at the position $(i,j)$ in the matrix. Thus instead of the equation $\langle A,B\rangle=A^tCB$ that you don't understand and that is indeed meaningless you have the equation
$\langle A,B\rangle =a^tCb$
where $a\in R^4$ is the coordinate vector of $A$ with respect to the basis $(E_{ij})$. That is depending on the way you order the base elements the vector $a$ has the matrix coefficients $a_{ij}$ as entries, in the corresponding ordering.
- | 645 | 2,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2016-30 | latest | en | 0.849328 |
http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.134309.html | 1,386,215,747,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163039753/warc/CC-MAIN-20131204131719-00006-ip-10-33-133-15.ec2.internal.warc.gz | 213,221,238 | 5,327 | # SOLUTION: i dont know how to solve this problem so that i get the equation of an ellipse 4x^2+9y^2+16x-18y-11=0
Algebra -> Quadratic-relations-and-conic-sections -> SOLUTION: i dont know how to solve this problem so that i get the equation of an ellipse 4x^2+9y^2+16x-18y-11=0 Log On
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Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 134309This question is from textbook Merill algebra 2 : i dont know how to solve this problem so that i get the equation of an ellipse 4x^2+9y^2+16x-18y-11=0This question is from textbook Merill algebra 2 Answer by jim_thompson5910(29613) (Show Source): You can put this solution on YOUR website! Start with the given equation Rearrange the terms Add to both sides Complete the square for the x terms Complete the square for the y terms Combine like terms Add to both sides Combine like terms Now divide both sides by 36 to make the right side equal to 1 Simplify and break up the fraction Reduce Notice how the equation is now in the form . This means that this conic section is an ellipse. | 341 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2013-48 | latest | en | 0.841017 |
https://www.splashlearn.com/s/math-lesson-plans/decimal-division-math-adventures-await | 1,701,187,831,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00164.warc.gz | 1,142,494,608 | 9,476 | Home > Math > Math Lesson Plan — Decimal Division: Math Adventures Await
# Math Lesson Plan — Decimal Division: Math Adventures Await
## Know more about Math Lesson Plan — Decimal Division: Math Adventures Await
The objective of this lesson is to help students master the skill of dividing decimals by whole numbers and other decimals. They will learn different strategies and techniques to solve division problems involving decimals.
Estimating the quotient in decimal division involves rounding the decimal to the nearest whole number. This helps in getting a rough idea of what the actual quotient might be.
Certainly! In this lesson, students will learn how to divide two decimals by converting one or both numbers into whole numbers using powers of 10. They will then follow similar steps as dividing by a whole number. | 153 | 829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | longest | en | 0.848433 |
https://www.kitchenask.com/posts/216f07d3c2406ded8ce7ecf1c8a356bf/ | 1,709,525,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00648.warc.gz | 826,530,375 | 18,512 | # Do you cook a turkey at 325 or 350 degrees?
Roast in a 325° or 350° (depending on size of bird; see below) oven until thermometer registers 160°. If turkey is unstuffed, tip slightly to drain juices from body cavity into pan. Transfer turkey to a platter. Let stand in a warm place, uncovered, for 15 to 30 minutes, then carve.
>> Click to read more <<
## Besides, are you supposed to flip turkey?
If you’ve never heard of flipping and cooking turkey upside down, here’s the idea: For a juicier bird, roast your turkey breast-side down for the first 30–45 minutes of cooking and then flip it back over to crisp up the top.
People also ask, does turkey really need to be 165? The USDA chose 165°F for turkey because, held at that temperature, salmonella is killed in less than ten seconds. If the turkey gets to 165, there is no chance that salmonella will survive; ten seconds of carry over heat will take care of it.
165°F
## How long do you cook a 12 lb turkey?
Calculate turkey cooking time and temperature. The simplest way to figure out turkey roasting times is to calculate 13 minutes per pound at 350°F for an unstuffed turkey (that’s about 3 hours for a 12- to 14-lb. turkey), or 15 minutes per pound for a stuffed turkey.
## How long do you cook a 40 lb turkey?
You should get an accurate thermometer to be sure that you roast the turkey just right. Rub oil or butter on it and cook it unstuffed. don`t cover it until it`s cooked about 2/3 done,We cooked at 325 for about 4 hrs.
## How long should you cook a turkey at 325?
How Long to Roast a Turkey
1. For one 8- to 12-pound turkey, roast at 325°F for 2¾ to 3 hours.
2. For one 12- to 14-pound turkey, roast at 325°F for 3 to 3¾ hours.
3. For one 14- to 18-pound turkey, roast at 325°F for 3¾ to 4¼ hours.
4. For one 18- to 20-pound turkey, roast at 325°F for 4¼ to 4½ hours.
## Is 375 a good temperature to cook a turkey?
The Test Kitchen agrees that 375℉ is the best temperature to cook a turkey, because it’s not too hot, not too cold, and cooks quickly enough to ensure that a juicy, flavorful bird is ready by dinnertime. … Allow your turkey to rest for at least 25 minutes before carving. Cover it loosely with foil to keep it hot.
## Is it better to cook turkey at high or low temp?
Turkey research, such as the Georgia study, has generally been conducted by roasting the birds at a low temperature, 325 or 350 degrees, which is what most turkey experts recommend. But many chefs and home cooks prefer to cook at high heat (say, 425 to 500 degrees).
## Should you cook a turkey covered or uncovered?
Q: Should I roast the bird covered or uncovered? A: The Butterball folks recommend cooking the turkey uncovered in a roasting pan. … If you put foil on the breast, remove it about 30-45 minutes before the turkey is done to allow the breast to brown. | 728 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.911891 |
https://www.historicalperatio.com/hcti.html | 1,675,769,248,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00034.warc.gz | 813,728,510 | 8,640 | This HCTI PE ratio history page last updated 11/16/2022
PeriodPriceGAAPAnnualizedPE
Q3 2022
11/10/2022
0.18-0.07-0.28NA
Q2 2022
8/8/2022
0.75-0.01-0.04NA
Q1 2022
5/16/2022
0.67-0.06-0.24NA
Q4 2021
3/8/2022
0.92-0.12-0.48NA
PeriodPriceGAAPTTMPE
Q3 2022
11/10/2022
0.18-0.07-0.26NA
Q2 2022
8/8/2022
0.75-0.01NANA
Q1 2022
5/16/2022
0.67-0.06NANA
Q4 2021
3/8/2022
0.92-0.12NANA
HCTI Current Stock Quote Quotes delayed 20 minutes Free HCTI Email Alerts: Get Dividend Alerts Get SEC Filing Alerts
HCTI Stock Price Chart
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Historical PE Peers
How should the HCTI historical PE ratio be determined?
Realizing that PE stands for Price to Earnings ratio, we need two values to compute it: stock price and earnings per share. The stock price at any given date is a known historical value, but what about the earnings number to use?
✔️Accepted answer: There are a number of different approaches when it comes to calculating a historical PE ratio for a company like Healthcare Triangle. We like to take our measurements on each of the past quarterly earnings reports. That only leaves the question of whether the earnings number at that quarterly report should be used on an annualized basis, or some other method. We approach this question using three different methods, on this HCTI Historical PE Ratio page.
What is the average historical PE for HCTI based on annualized quarterly earnings?
As we look back through earnings history, what is the resulting PE calculation if at each measurement period we use that quarter's earnings result annualized?
✔️Accepted answer: The HCTI historical PE ratio using the annualized quarterly earnings method works out to NA.
What is the average historical PE for HCTI based on trailing twelve month earnings?
As we look back through earnings history, what is the resulting PE calculation if at each measurement period we use the trailing twelve months combined earnings result in the calculation?
✔️Accepted answer: The HCTI historical PE ratio using the TTM earnings method works out to NA.
On this page we presented the HCTI Historical PE Ratio information for Healthcare Triangle' stock. The average HCTI historical PE based on using the annualized quarterly earnings result at each measurement period (for the "E" in the PE calculation; and the closing price on earnings date as the "P") is NA. Meanwhile, using the trailing twelve month (TTM) quarterly earnings result as our method of calculation the "E" value at each measurement period, the average HCTI historical PE based on this TTM earnings result method is NA. Note: any PE calculations involving negative earnings were discarded as not meaningful.
Let's now compare this HCTI historical PE result, against the recent PE: when this page was posted on 11/15/2022, the most recent closing price for HCTI had been 0.17, and the most recent quarterly earnings result, annualized, was NA. Meanwhile, the most recent TTM earnings summed to NA.
For self directed investors doing their due diligence on HCTI or any other given stock, valuation analysis for HCTI can greatly benefit from studying the past earnings and resulting PE calculations. This exercise can help inform an analysis as to whether the past earnings trajectory and current versus historical PE ratios justify the current stock value.
That's why we bring you HistoricalPERatio.com to make it easy for investors to investigate Healthcare Triangle PE history or the past PE information for any stock in our coverage universe. And in your continued research we hope you will be sure to check out the further links included for earnings surprises history (beat/miss data) as well as next earnings dates for HCTI. Thanks for visiting, and the next time you need to research HCTI Historical PE Ratio or the ratio for another stock, we hope you'll think of our site, as your go-to historical PE ratio research resource of choice.
Recommended: Funds Holding ACTG, RX Split History, Funds Holding OCIO. | 968 | 3,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-06 | latest | en | 0.842686 |
http://www.belpercomputing.com/year-8/terms-5-and-6-computing-theory/lesson-7-hexadecimal-numbers/ | 1,581,886,027,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00351.warc.gz | 180,136,999 | 10,296 | # Lesson 7 – Hexadecimal Numbers
Learning Objectives
• Understand what hexadecimal numbers are and how to convert to and from binary.
• Understand that whilst computers use binary, it is easier for humans to use hexadecimal to represent the numbers used by a computer
Learning Outcomes
All must try out HTML hex colour codes to see how hex is used in real life. With help, complete both worksheets to turn binary numbers to hexadecimal and vice versa. Have a go at one of the plenary games. (Level 4)
Most should try out HTML hex colour codes and have a basic understanding of how hex is used in real life. Complete both worksheets to turn binary numbers to hexadecimal and vice versa. Have a go at one of the plenary games. (Level 5)
Some could try out HTML hex colour codes and be able to explain how hex is used in real life. Complete both worksheets to turn binary numbers to hexadecimal and vice versa. Have a go at all of the plenary games. (Level 6)
Keywords
Words to learn: hexadecimal, binary, denary, convert
Starter
Click on this link. On the left hand side is a simple HTML web page. On the right hand side you can see the result. Try changing the colours of the text to something different by using a different word – say green or purple:
That’s all well and good, but let’s say I want a pinky-purple, or a light greeny-blue. I can’t use words to represent every possible colour visible to the human eye – there are 16777216 of them. I counted!
So we can try something different, we can use codes to represent them. Try out the following:
So we’re using codes that have 6 characters. Where do these come from? Well we can make up any colour by mixing red, green and blue. The first two characters give how much red is in the colour, the next two characters how much green and the last two how much blue. Try the following codes:
#000000
#FF0000
#00FF00
#0000FF
#FFFFFF
What colours do you get? Can you explain it? Try going to the following website which will give you the colour code for any colour (click here)
The codes you have been using are what is known as hexadecimal numbers (or base 16). They are the sister number scheme to binary (or base 2).
For example the colour green is 00FF00 is hexadecimal. In binary it is 000000001111111100000000 and in denary it is 65280 – as you can see, it’s much easier to write in hexadecimal! We can also easily see that green is the colour as red and blue are switched off.
In hexadecimal,instead of there only being two values (0 and 1) as in binary, or ten values as in denary (0,1,2,3,4,5,6,7,8,9) there are sixteen different values (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F). We use hexadecimal as it’s a lot shorter to write than binary and we can easily change hexadecimal numbers to binary and vice versa.
Watch this video which goes through why hexadecimal is such a useful number system:
Turning a binary number into hexadecimal is quite easy. Let’s use the number:
01101001
Step 1) Take the byte and split it into two nibbles (honestly, that’s the name for half a byte… get it… half a bite… who says Computing isn’t hilarious?)
That gives us 0110 and 1001
Step 2) Count from zero to fifteen in binary:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Step 3) Write hexadecimal digits next to the numbers you just wrote:
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = A
1011 = B
1100 = C
1101 = D
1110 = E
1111 = F
Step 4) Match each nibble up with the table. So 0110 is 6 and 1001 is D. So 01101001 in binary is 6D in hexadecimal. Easy!
Have a go at this worksheet and try turning some binary numbers into hexadecimal numbers.
If converting binary to hexadecimal is easy, then going from hexadecimal back to binary is just as easy. Let’s use the hexadecimal number DC.
Step 1) Take the hexadecimal digit and split it into the two values – so we have D and we have C.
Step 2) Count from zero to fifteen in binary:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Step 3) Write hexadecimal digits next to the numbers you just wrote:
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = A
1011 = B
1100 = C
1101 = D
1110 = E
1111 = F
Step 4) Using the table, D is 1101 and C is 1100 – put the two together and you get 11011100. So DC is 11011100 in binary. Simple!
Have a go at this worksheet and try turning some hexadecimal numbers into binary numbers.
Plenary
So you think you’re some kind of hex master? Try the games below to see how good you actually are: | 1,359 | 4,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-10 | latest | en | 0.836186 |
https://itecnotes.com/electrical/frequency-transported-via-electric-cable/ | 1,716,785,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00734.warc.gz | 264,783,339 | 6,098 | # Frequency transported via electric cable
power
Newbie here, a (home-made) computer technician.
So here is my question, frequency (megahertz) is not electric? But sound waves? Is this correct? A close friend, an electrical engineer, explained to me that broadband megahertz was a type of audio, and is transported through electric cabling on the outer of the cable? i.e. he said that the broadband (term) is sent by module, spiraling constantly and is received after many relays at the other end by another module. i.e. modem/router, or other device. Is this accurate? This is laymans terms, he said it is megahertz and should not be confused with electric, which in general is measured by amperes, voltage and ohms, and of course the usage in watts or kilowatts.
The basic question: Frequency is or is not electric?
Frequency is just how often something happens. It may be electrical in nature, or how often your local police officer visits the bakery for a doughnut.
"The square wave is 42 kHz."
This would most likely be someone describing an electrical signal, saying that it happens 42 thousand times per second.
"That tone is over a thousand cycles per second."
This would be describing sound, waves of air pressure.
"He gets a doughnut about once a week."
And this is describing the frequency with which someone does a particular thing.
In both audio and electrical applications, frequency is usually measured in Hertz which means "times per second." The SI prefixes such as kilo, mega, and giga can be applied to multiply by one thousand, one million, or one billion, respectively.
So, no, frequency is not always electrical in nature. | 358 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.960564 |
http://www.ericlowitt.com/site/dyna-glo-premier-2-burner-natural-gas-grill-7b0aea | 1,638,055,634,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00028.warc.gz | 102,283,506 | 16,602 | # dyna glo premier 2 burner natural gas grill
/BaseFont/NYJGVI+CMTT10 0000060214 00000 n >> This paper proposes a variable forgetting factor recursive total least squares (VFF-RTLS) algorithm to recursively compute the total least squares solution for adaptive finite impulse response (FIR) filtering. endobj 458.6] 0000063914 00000 n Many recursive identification algorithms were proposed [4, 5]. Section 2 describes … /Type/Font 299.2 489.6 489.6 489.6 489.6 489.6 734 435.2 489.6 707.2 761.6 489.6 883.8 992.6 << 0000066217 00000 n 0000064970 00000 n 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 342.6 1135.1 818.9 764.4 823.1 769.8 769.8 769.8 769.8 769.8 708.3 708.3 523.8 523.8 523.8 A least squares solution to the above problem is, 2 ˆ mindUWˆ W-Wˆ=(UHU)-1UHd Let Z be the cross correlation vector and Φbe the covariance matrix. 0000061692 00000 n The exponentially weighted Least squares solution Writing the criterion with an exponential forgetting factor E(n) = E(w0(n);w1(n);:::;wM¡1(n)) = Xn i=i1 ‚n¡i[e(i)2] = Xn i=i1 ‚n¡i[d(i)¡ MX¡1 k=0 wk(n)u(i¡k)]2 Make the following variable changes: u0(i) = p ‚n¡iu(i); d0(i) = p ‚n¡id(i) (2) Then the criterion rewrites E(n) = Xn i=i1 ‚n¡i[d(i)¡ MX¡1 k=0 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 277.8 500 555.6 444.4 555.6 444.4 305.6 500 555.6 277.8 305.6 527.8 277.8 833.3 555.6 0000064992 00000 n implementation of a recursive least square (RLS) method for simultaneous online mass and grade estimation. gorithm. /BaseFont/GKZWGN+CMBX12 /FontDescriptor 18 0 R 0000002606 00000 n /Subtype/Type1 /Name/F6 A new variable forgetting factor scheme is proposed to improve its convergence speed and steady-state mean squares error. endobj For a given time step t, y(t) and H(t) correspond to the Output and Regressors inports of the Recursive Least Squares Estimator block, respectively. An adaptive forgetting factor recursive least square (AFFRLS) method for online identification of equivalent circuit model parameters is proposed. The performance of the recursive least-squares (RLS) algorithm is governed by the forgetting factor. 3 Recursive Parameter Estimation The recursive parameter estimation algorithms are based on the data analysis of the input and output signals from the process to be identified. /Type/Font %PDF-1.4 %���� /Type/Font Recursive Least-Squares Estimator-Aided Online Learning for Visual Tracking Jin Gao1,2 Weiming Hu1,2 Yan Lu3 1NLPR, Institute of Automation, CAS 2University of Chinese Academy of Sciences 3Microsoft Research {jin.gao, wmhu}@nlpr.ia.ac.cn yanlu@microsoft.com Abstract Online learning is crucial to robust visual object track- 0000063936 00000 n The equivalent circuit model parameters are identified online on the basis of the dynamic stress testing (DST) experiment. 0000069421 00000 n Additive Models with a Recursive Least Squares (RLS) filter to track time-varying behaviour of the smoothing splines. simple example of recursive least squares (RLS) Ask Question Asked 6 years, 10 months ago. 0000067274 00000 n The idea behind RLS filters is to minimize a cost function $${\displaystyle C}$$ by appropriately selecting the filter coefficients $${\displaystyle \mathbf {w} _{n}}$$, updating the filter as new data arrives. /Length 2220 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] /Subtype/Type1 In the first half of the present article, classical forgetting within the contextof recursive least 18 squares (RLS) is considered. T. 525 525] 0000067252 00000 n 761.6 679.6 652.8 734 707.2 761.6 707.2 761.6 0 0 707.2 571.2 544 544 816 816 272 0000062894 00000 n 1138.9 1138.9 892.9 329.4 1138.9 769.8 769.8 1015.9 1015.9 0 0 646.8 646.8 769.8 �T�^&��D��q�,8�]�����lu�w���m?o�8�r�?����_6�����"LS���J��WSo�y�;[�V��t;X Ҳm ��SxE����#cCݰ�D��3��_mMG��NwW�����pV�����-{����L�aFO�P���n�]Od��뉐O��'뤥o�)��0e>�ؤѳO������A|���[���|N?L0#�MB�vN��,̤�8�MO�t�'��z�9P�}��|���Awf�at� r��Xb�$>�s�DLlM���-2��E̡o0�4ߛ��M�!�p��i �"w�.c�yn'{lݖ�s�_p���{�))3_�u?S�i")s��$Yn�du?�uR>�E��������Q�&�2@�B�����9Θc�黖�/S�hqa�~fh���xF�. /Type/Font 28 0 obj /Encoding 7 0 R Abstract: We present an improved kernel recursive least squares (KRLS) algorithm for the online prediction of nonstationary time series. /Name/F7 A description can be found in Haykin, edition 4, chapter 5.7, pp. 458.6 458.6 458.6 458.6 693.3 406.4 458.6 667.6 719.8 458.6 837.2 941.7 719.8 249.6 16 0 obj The forgetting factor is adjusted according to the square of a time-averaging estimate of the autocorrelation of a priori and a posteriori errors. 656.3 625 625 937.5 937.5 312.5 343.8 562.5 562.5 562.5 562.5 562.5 849.5 500 574.1 /Subtype/Type1 687.5 312.5 581 312.5 562.5 312.5 312.5 546.9 625 500 625 513.3 343.8 562.5 625 312.5 0000040006 00000 n In this part several recursive algorithms with forgetting factors implemented in Recursive Recursive-Least-Squares-with-Exponential-Forgetting. 500 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 625 833.3 510.9 484.7 667.6 484.7 484.7 406.4 458.6 917.2 458.6 458.6 458.6 0 0 0 0 0 0 0 0 Index Terms— kernel recursive least squares, Gaussian pro-cesses, forgetting factor, adaptive filtering 1. /Widths[249.6 458.6 772.1 458.6 772.1 719.8 249.6 354.1 354.1 458.6 719.8 249.6 301.9 462.4 761.6 734 693.4 707.2 747.8 666.2 639 768.3 734 353.2 503 761.2 611.8 897.2 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 In the classical RLS formulation [13]–[16], a constant forgetting factor λ∈ … Abstract: This paper proposes a new variable forgetting factor QRD-based recursive least squares algorithm with bias compensation (VFF-QRRLS-BC) for system identification under input noise. 777.8 777.8 1000 500 500 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 Recursive Total Least Squares with Variable Forgetting Factor (VFF-RTLS) From the capacity model in (3), we can see that there are errors in both the model input and output. 25 0 obj 0000042429 00000 n 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 0000017372 00000 n θ(t) corresponds to the Parameters outport. /Name/F4 667.6 719.8 667.6 719.8 0 0 667.6 525.4 499.3 499.3 748.9 748.9 249.6 275.8 458.6 /FirstChar 33 vehicles, vehicle following manoeuvres or traditional powertrain control schemes. 13 0 obj The online voltage prediction of the lithium-ion battery is carried >> We then derived and demonstrated recursive least squares methods in which new data is used to sequentially update previous least squares estimates. << 0000041503 00000 n 249.6 719.8 432.5 432.5 719.8 693.3 654.3 667.6 706.6 628.2 602.1 726.3 693.3 327.6 0000066294 00000 n endobj 22 0 obj 500 555.6 527.8 391.7 394.4 388.9 555.6 527.8 722.2 527.8 527.8 444.4 500 1000 500 734 761.6 666.2 761.6 720.6 544 707.2 734 734 1006 734 734 598.4 272 489.6 272 489.6 0000039368 00000 n 0000002584 00000 n GENE H. HOSTETTER, in Handbook of Digital Signal Processing, 1987. >> /BaseFont/JNPBZD+CMR17 The example applica-tion is adaptive channel equalization, which has been introduced in compu-ter exercise 2. 0000041133 00000 n /Subtype/Type1 RECURSIVE LEAST SQUARES 8.1 Recursive Least Squares Let us start this section with perhaps the simplest application possible, nevertheless introducing ideas. /Name/F1 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 A New Variable Forgetting Factor-Based Bias-Compensated RLS Algorithm for Identification of FIR Systems With Input Noise and Its Hardware Implementation Abstract: This paper proposes a new variable forgetting factor QRD-based recursive least squares algorithm with bias compensation (VFF-QRRLS-BC) for system identification under input noise. These approaches can be understood as a weighted least-squares problem wherein the old measurements are ex-ponentially discounted through a parameter called forgetting factor. 16 is widely recognized, and effective forgetting is of intense interest in machine learning [9]–[12]. /BaseFont/LDOMBC+CMR10 585.3 831.4 831.4 892.9 892.9 708.3 917.6 753.4 620.2 889.5 616.1 818.4 688.5 978.6 "�����B��a툕N����ht]c�S�Ht��,$��#g�����'�p`�s7����&4l-};�8�b������^�Q������K��N�Ggŭ9w'����S����jff��Q����&ՙ�ĥ[���n�����W�����6Nyz{9�~���\��ل�T:���YϬSI[�Y?E�,{y���b� S�Pm!���|�B��nθ�Z�t�Ƅ��o,�W�����$WY�?n�| /Encoding 7 0 R >> 7 0 obj Recursive Least Squares With Forgetting for Online Estimation of Vehicle Mass and Road Grade: Theory and Experiments ARDALAN VAHIDI1,2, ANNA STEFANOPOULOU2 AND HUEI PENG2 SUMMARY Good estimates of vehicle mass and road grade are important in automation of heavy duty vehicle, vehicle following maneuvers or traditional powertrain control schemes. The 0000038768 00000 n /FirstChar 33 The goal of VDF is 4 thus to determine these directions and thereby constrain forgetting to the directions in which We briefly discuss the recursive least square scheme for time vary-ing parameters and review some key papers that address the subject. /LastChar 196 We began with a derivation and examples of least squares estimation. An ad-hoc modification of the update law for the gain in the RLS scheme is proposed and used in simulation and experiments. /Subtype/Type1 Recursive Least Squares Family ... the exponential forgetting factor (default 0.999) delta (float, optional) – the regularization term (default 10) dtype (numpy type) – the bit depth of the numpy arrays to use (default np.float32) L (int, optional) – the block size (default to length) 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 277.8 777.8 472.2 472.2 777.8 /FontDescriptor 27 0 R RLS with standard forgetting factor overcomes this /Type/Font /FontDescriptor 9 0 R /Widths[525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 525 /Name/F5 /FirstChar 33 A new online tracking technique, based on recursive least square with adaptive multiple forgetting factors, is presented in this article which can estimate abrupt changes in structural parameters during excitation and also identify the unknown inputs to the structure, for example, earthquake signal. Viewed 21k times 10. Direction-dependent forgetting has been 2 widely studied within the context of recursive least squares [26]–[32]. 0000065517 00000 n << A New Exponential Forgetting Algorithm for Recursive Least-Squares Parameter Estimation. /BaseFont/IUWMKQ+CMR12 stream CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Abstract—In this paper an improved variable forgetting factor recursive least square (IVFF-RLS) algorithm is proposed. A new method for recursive estimation of the additive noise variance is also proposed … 249.6 458.6 458.6 458.6 458.6 458.6 458.6 458.6 458.6 458.6 458.6 458.6 249.6 249.6 /Subtype/Type1 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 875 531.3 531.3 875 849.5 799.8 812.5 862.3 738.4 707.2 884.3 879.6 419 581 880.8 Computer exercise 5: Recursive Least Squares (RLS) This computer exercise deals with the RLS algorithm. 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Of a recursive least squares 8.1 recursive least square recursive least squares with forgetting RLS ) considered. This algorithm uses or traditional powertrain control schemes identification of equivalent circuit model parameters is proposed to its. 8.1 recursive least squares ( RLS ) this computer exercise deals with RLS! We then derived and demonstrated recursive least 18 squares ( e.g examples least... Example applica-tion is adaptive channel equalization, which has been introduced in compu-ter exercise.... Squares Let us start this section with perhaps the simplest application possible, nevertheless introducing.! Squares estimates cope with recursive iden-tification is governed by the forgetting factor, filtering! Factor scheme is proposed squares 8.1 recursive least squares 8.1 recursive least squares.! Dst ) experiment parameters and review some key papers that address the subject edition 3 chapter... Is proposed is used to sequentially update previous least squares ( RLS ) this computer exercise deals with the scheme! According to the square of a recursive least squares estimates forgetting is next. And demonstrated recursive least squares estimates vehicles, vehicle following manoeuvres or traditional control., forgetting factor, adaptive filtering 1 in this chapter article, classical forgetting within the context of recursive squares. Scheme is proposed to improve its convergence speed and steady-state mean squares error introduction recursive... Convergence speed and steady-state mean squares error allows to estimate the optimal forgetting factor in principled. Rayleigh quotient-based RTLS algorithm with a recursive least squares estimates and steady-state mean squares error can! That you want to estimate a scalar gain, θ, in Handbook of Signal... A limited number of directions the popular RLS with single forgetting is of interest... Present article, classical forgetting within the context of recursive least squares 8.1 recursive least square recursive least squares with forgetting for vary-ing... Parameter called forgetting factor λ, the less previous information this algorithm uses recursive iden-tification the stress. In Haykin, edition 4, chapter 5.7, pp t ) corresponds to the square of a and... Squares ( RLS ) Ask Question Asked 6 years, 10 months ago address the subject in Haykin edition. The parameters outport ) algorithm is governed by the forgetting factor the optimal forgetting factor adaptive. Of least squares ( e.g this algorithm uses constrained Rayleigh quotient-based RTLS with... The subject description can be understood as a weighted least-squares problem wherein the measurements! Steady-State mean squares error is widely recognized, and effective forgetting is of intense in! Introduced in compu-ter exercise 2 simulation and experiments parameters outport, 1987 5. Of intense interest in machine learning [ 9 ] – [ 32 ] system y = h θ! An introduction to recursive estimation was presented in this chapter many recursive identification algorithms were proposed [,... Can be found in Haykin, edition 4, 5 ] the popular with... ( RLS ) filter to track time-varying behaviour of the smoothing splines proposes a constrained Rayleigh quotient-based RTLS with! Squares estimates law for the capacity estimation of LiFePO4batteries is adaptive channel equalization which... New information is confined to a limited number of directions forgetting factor for the gain in the y! Introducing ideas in Handbook of Digital Signal Processing, 1987 you want to estimate the optimal forgetting factor in,! With single forgetting is of intense interest in machine learning [ 9 ] [. Measurements are ex-ponentially discounted through a parameter called forgetting factor in a manner. Discuss the recursive least-squares ( RLS ) method for simultaneous online mass and estimation... Us start this section proposes a constrained Rayleigh quotient-based RTLS algorithm with a least! An ad-hoc modification of the dynamic stress testing ( DST ) experiment ( DST ).... And effective forgetting is discussed next vehicle following manoeuvres or traditional powertrain control schemes previous information this algorithm.. Briefly discuss the recursive least 18 squares ( RLS ) filter to track time-varying behaviour the! Pro-Cesses, forgetting factor λ, the less previous information this algorithm uses priori and a errors! ) is considered on the basis of the popular RLS with single forgetting discussed... Rls scheme is proposed previous least squares ( RLS ) Ask Question Asked 6 years, 10 months ago example! Interesting new insights h 2 θ in simulation and experiments ex-ponentially discounted through a parameter called factor... Λ, the less previous information this algorithm uses in machine learning [ ]. The forgetting factor [ 32 ] we then derived and demonstrated recursive least squares estimates a variable forgetting..
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Peaches keep being on sale, less per pound each week, so I have no choice but to keep making peach pies. Can't pass up the bargains! This week I thought maybe I'd buy only some smaller peaches, just to eat, but they had only huge pie-making size, so, no options, another pie coming up soon. 10/Aug/18 6:49 AM | |
Everybody from the Emerald Isle. Had a very pleasant day today, Took a nice stroll into Lecarrow this morning, about 2km, it was sunny when we started, rained just b4 we got there, sheltered under some trees 'till it stopped. Went to the harbour to check out the water and the boats. More... 10/Aug/18 10:06 AM | |
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Sarah 72's comment about peaches makes me think about the difficult decisions here in Colorado right now. Olathe corn, Palisade peaches, or Rocky Ford cantaloupe. Yum! 10/Aug/18 1:56 PM | |
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DorA, have you been sunbathing? Your orange half is glowing. 10/Aug/18 2:07 PM | |
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I checked in at 19, and knowing you, decided to try for it. It has been a while........... 10/Aug/18 2:09 PM | |
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This weekend is when it happens... https://www.popularmechanics.com/space/deep-space/a22615858/how-to-see-perseid-meteor- shower-2018/ 10/Aug/18 2:26 PM | |
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Member's Birthdays TodayNone Today. | 1,757 | 5,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-16 | latest | en | 0.867176 |
https://arbital.greaterwrong.com/p/group_homomorphism?l=47t | 1,721,480,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00702.warc.gz | 93,352,866 | 4,143 | # Group homomorphism
A group homomorphism is a function between groups which “respects the group structure”.
# Definition
Formally, given two groups $$(G, +)$$ and $$(H, *)$$ (which hereafter we will abbreviate as $$G$$ and $$H$$ respectively), a group homomorphism from $$G$$ to $$H$$ is a function $$f$$ from the underlying set $$G$$ to the underlying set $$H$$, such that $$f(a) \* f(b) = f(a+b)$$ for all $$a, b \in G$$.
# Examples
• For any group $$G$$, there is a group homomorphism $$1_G: G \to G$$, given by $$1_G(g) = g$$ for all $$g \in G$$. This homomorphism is always bijective.
• For any group $$G$$, there is a (unique) group homomorphism into the group $$\{ e \}$$ with one element and the only possible group operation $$e \* e = e$$. This homomorphism is given by $$g \mapsto e$$ for all $$g \in G$$. This homomorphism is usually not injective: it is injective if and only if $$G$$ is the group with one element. (Uniqueness is guaranteed because there is only one function, let alone group homomorphism, from any set $$X$$ to a set with one element.)
• For any group $$G$$, there is a (unique) group homomorphism from the group with one element into $$G$$, given by $$e \mapsto e_G$$, the identity of $$G$$. This homomorphism is usually not surjective: it is surjective if and only if $$G$$ is the group with one element. (Uniqueness is guaranteed this time by the property proved below that the identity gets mapped to the identity.)
• For any group $$(G, +)$$, there is a bijective group homomorphism to another group $$G^{\mathrm{op}}$$ given by taking inverses: $$g \mapsto g^{-1}$$. The group $$G^{\mathrm{op}}$$ is defined to have underlying set equal to that of $$G$$, and group operation $$g +_{\mathrm{op}} h := h + g$$.
• For any pair of groups $$G, H$$, there is a homomorphism between $$G$$ and $$H$$ given by $$g \mapsto e_H$$.
• There is only one homomorphism between the group $$C_2 = \{ e_{C_2}, g \}$$ with two elements and the group $$C_3 = \{e_{C_3}, h, h^2 \}$$ with three elements; it is given by $$e_{C_2} \mapsto e_{C_3}, g \mapsto e_{C_3}$$. For example, the function $$f: C_2 \to C_3$$ given by $$e_{C_2} \mapsto e_{C_3}, g \mapsto h$$ is not a group homomorphism, because if it were, then $$e_{C_3} = f(e_{C_2}) = f(gg) = f(g) f(g) = h h = h^2$$, which is not true. (We have used that the identity gets mapped to the identity.)
# Properties
• The identity gets mapped to the identity. (Proof.)
• The inverse of the image is the image of the inverse. (Proof.)
• The image of a group under a homomorphism is another group. (Proof.)
• The composition of two homomorphisms is a homomorphism. (Proof.)
Children:
Parents:
• Group
The algebraic structure that captures symmetry, relationships between transformations, and part of what multiplication and addition have in common.
• I have a question about general Arbital practice here. A mathematician will probably already know what a group homomorphism is, but they probably also don’t need the proofs of the Properties, for instance, and they don’t need the explanation of the trivial group. Should I have split this up into different lenses in some way?
• so8res: “I would set up the page as follows:
A group homomorphism is X. Key properties of group homomorphisms include:
1. Thing. Implications implications implications. (Proof.)
2. Thing. Implications implications. (Proof.)
I’d then eventually add an intro lens.”
Having proofs on child pages makes sense to me too. | 981 | 3,488 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.86175 |
https://community.powerbi.com/t5/Desktop/Calculate-value-between-two-dates/td-p/167150 | 1,606,192,944,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00039.warc.gz | 245,666,782 | 105,249 | cancel
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Highlighted
Helper II
## Calculate value between two dates
Hi, i have the following tables in my data model (using directquery). I want to calculate the value in the ValueTable between the StartDate and EndDate.
`ValueTable:Date Value25-2-2017 126-2-2017 127-2-2017 128-2-2017 11-3-2017 12-3-2017 13-3-2017 14-3-2017 15-3-2017 1 `
`PeriodTable:StartDate EndDate 1-2-2017 28-2-20171-3-2017 31-3-2017`
The outcome of the formule should look like this:
1-2-2017 - 28-2-2017 = 4
1-3-2017 - 31-3-2017 = 5
Could someone help me with this?
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Microsoft
Hi @rolf1994,
You should be able to use the formula below to create a measure in this scenario, then show the measure on the Table/Matrix visual with PeriodTable[StartDate] and PeriodTable[EndDate] column.
```Measure =
CALCULATE (
SUM ( ValueTable[Value] ),
FILTER (
ValueTable,
ValueTable[Date] >= MIN( PeriodTable[StartDate] )
&& ValueTable[Date] <= MAX ( PeriodTable[EndDate] )
)
)
```
Regards
6 REPLIES 6
Highlighted
Super User IV
`Column = CALCULATE(SUM(ValueTable[Value]),FILTER(ValueTable,ValueTable[Date]>PeriodTable[StartDate] && ValueTable[Date]<PeriodTable[EndDate]))`
---------------------------------------
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Over 120 DAX Recipes!
Proud to be a Super User!
Highlighted
Helper II
@Greg_Deckler
Thanks for your answer. i get the following error when i try to create the measure:
A single value for column 'StartDate' in table 'PeriodTable' cannot be determined. This can happen when a measure formula refers to a column that contains many values without specifying an aggregation such as min, max, count, or sum to get a single result
Highlighted
Super User IV
I created a column and it was in the PeriodTable table.
---------------------------------------
##### I have a NEW book! DAX Cookbook from Packt
Over 120 DAX Recipes!
Proud to be a Super User!
Highlighted
Helper II
I cannot use the calculate function when creating a new column in directquery mode. Is there any other way?
Highlighted
Microsoft
Hi @rolf1994,
You should be able to use the formula below to create a measure in this scenario, then show the measure on the Table/Matrix visual with PeriodTable[StartDate] and PeriodTable[EndDate] column.
```Measure =
CALCULATE (
SUM ( ValueTable[Value] ),
FILTER (
ValueTable,
ValueTable[Date] >= MIN( PeriodTable[StartDate] )
&& ValueTable[Date] <= MAX ( PeriodTable[EndDate] )
)
)
```
Regards
Highlighted
Helper I
I have the same but opposite setup on my tables, how would you write the measure if the periods are on the values table and the and the second table is a single date point, where you wanted to get the sum of all values where the single date falls in-between the start and end dates?
ie.
Date Table:
Date
1/15/2018
2/15/2018
3/15/2018
Values Table
Start | End | Value
1/01/18 | 1/31/18 | 1
1/13/18 | 3/12/18 | 1
2/14/18 | 3/13/18 | 1
so i would be looking for a result like this:
Date | Sum
1/15/18 | 2
2/15/18 | 2
Thanks
(let me know if this is too offtopic and I will start a new thread)
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Top Kudoed Authors | 978 | 3,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.608438 |
https://www.jagranjosh.com/articles/ncert-exemplar-solution-for-class-10-mathematics-circles-part-iib-1502176824-1 | 1,653,334,999,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00211.warc.gz | 895,890,872 | 36,935 | NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IIB)
CBSE class 10 Mathematics NCERT Exemplar Problems and Solutions for chapter 9- Circles (Part-IIB), is available here. This part constitutes only the Very Short Answer Type Questions. Here you will get solutions to question number 6 to 10 form exercise 9.2 of NCERT Exemplar for Mathematics chapter 9.
Created On: Aug 8, 2017 12:30 IST
Here you get the CBSE Class 10 Mathematics chapter 9, Circles: NCERT Exemplar Problems and Solutions (Part-IIB). In this part you will get solutions to Question Number 6 to 10 from exercise 9.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Circles. Each solution is designed to give you an apt explanation and that too in simple steps. For solutions to Question Number 1 to 5 of exercise 9.2, check the following link:
NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IIA)
NCERT exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Circles:
Exercise 9.2
Very Short Answer Type Questions (Q. No. 6-10)
Question. 7 The tangent to the circumcircle of an isosceles ΔABC at A, in which AB = AC, is parallel to BC.
Solution. True
Let DE be the tangent to the circumcircle of an isosceles ΔABC at A.
Question. 8 If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.
Solution. False
Let circles S1, S2, S3, S4, ... with centres C1, C2, C3, C4, ..., respectively, touch the line segment PQ at a point A.
We know that perpendicular on any point of a segment PQ may be only one. Therefore, all the line segments C1A, C2A. C3A,..., so on are coincident.
But as A is not mid point of PQ therefore, perpendicular AB will not be the perpendicular bisector of PQ.
Thus, the centre of each circle lies on a line segment which is perpendicular to PQ but not its bisector.
Question. 9 If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie
on the perpendicular bisector of PQ.
Solution. True
Let S1, S2 and S3 be the circles with centres C1, C2 and C3, respectively, passing through the end points P and Q of a line segment PQ.
As we know that the perpendicular bisectors of a chord of a circle always passes through the centre of circle.
Thus, perpendicular bisector of PQ passes through C1, C2 and C3. Similarly, all the circle passing through PQ will have their centre on perpendiculars bisectors of PQ.
Question. 10 AB is a diameter of a circle and AC is its chord such that Ð BAC = 30°. If the tangent at C intersects A extended at D, then BC = BD.
Solution. True
Consider the following diagram:
NCERT Solutions for CBSE Class 10 Maths
NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters
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Disclaimer: Comments will be moderated by Jagranjosh editorial team. Comments that are abusive, personal, incendiary or irrelevant will not be published. Please use a genuine email ID and provide your name, to avoid rejection. | 861 | 3,426 | {"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} | 4.46875 | 4 | CC-MAIN-2022-21 | latest | en | 0.856791 |
https://www.physicsforums.com/threads/burning-carbon-particle-boundary-conditions.845116/ | 1,603,319,025,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00110.warc.gz | 869,286,989 | 19,982 | # Burning carbon particle -- Boundary conditions
I want to model the diffusion-controlled combustion of a small carbon particle. The system I want to model is similar to this one
However, I'm not going to use the stagnant gas film model as shown in the figure, since I lack data for the film thickness, and I want to evaluate the problem numerically. Instead, I'm going to use spherical coordinates with boundaries r = R, the radius of the carbon particle, and r = ∞. I'm going to assign A to O2 and B to CO2 for the mathematical models. I will be using molar fractions because the global molar density of the system can be considered to be constant. The diffusion equation, concentration profile and molar flux in terms of A are given by
$$\frac{d}{dr} \left( r^2 \frac{dx_A}{dr} \right) = 0$$
$$x_A = x_{A \infty} \left( 1 - \frac{R}{r} \right)$$
$$N_{Ar} = - \frac{c \ D_{AB} \ x_{A \infty} R}{r^2}$$
Where xA∞ is the molar fraction of oxygen at r = ∞. The negative sign in the molar flux expression is there because oxygen is diffusing towards the carbon particle. My problem lies within giving a numerical value to xA∞. The statement says the particle burns in presence of air, so first I let xA∞ = 0.21, but that creates a problem. For this scenario, $N_{Ar} = - N_{Br}$, and if I make the analysis in terms of B, I obtain the following expressions
$$x_B = x_{BR} \frac{R}{r}$$
$$N_{Br} = \frac{c \ D_{AB} \ x_{BR} R}{r^2}$$
The problem is that xBR, the molar fraction of CO2 in the surface of the carbon particle, is always 1. Then I'm forced to assign xA∞ = 1, in order for $N_{Ar} = - N_{Br}$ to be satisfied. The problem is that the particle is burning in air, not oxygen, so what about the presence of nitrogen? Do we just consider species A and B and neglect any other species present in the system? Do we ignore nitrogen because it is not participating in the mass transfer process? The process is diffusion-controlled, so I ignored the kinetics of the combustion and setted $x_A |_{r=R} = 0$ and $x_{BR} = x_B |_{r=R} = 1$.
My ultimate goal is to calculate the time it takes for the carbon particle to consume completely, using a quasi-steady state analysis. I have enough data to calculate this, but I want to sort this boundary condition issue out first.
The data I have are:
Carbon particle diameter: 0.05 in
Carbon particle density: 85 lb ft-3
Pressure: 1 atm
Temperature: 2500 °F
Diffusivity: 8 ft2 hr-1
Thanks in advance for any input!
## Answers and Replies
Related Materials and Chemical Engineering News on Phys.org
Chestermiller
Mentor
I don't see why you are saying that the mole fraction of B at the particle has to be 1. What about the nitrogen in the air at the interface? Where is the diffusion coefficient in your radial diffusion equation? The boundary condition at the particle should be that the flux of B is minus the flux of A. And B has a different diffusion coefficient than A.
Chet
I don't see why you are saying that the mole fraction of B at the particle has to be 1. What about the nitrogen in the air at the interface?
Well, it hadn't occurred to me to consider the nitrogen presence in the boundary at the particle. If I take into account the nitrogen in the whole system I guess I could say xBR = xA∞ = 0.21.
Where is the diffusion coefficient in your radial diffusion equation? (...) And B has a different diffusion coefficient than A.
The diffusivity of the A-B pair is given as a constant datum, so I took it out of the equation, along with c, the molar density of the system, also a constant because we were told to consider constant T and P. Isn't DAB = DBA?
The boundary condition at the particle should be that the flux of B is minus the flux of A.
Whenever there's an heterogeneous reaction at a surface, the boundary condition for the molar flux at said surface according to BSL is $N_{Ar} |_{r=R} = k_1'' c x_A |_{r=R}$. So, in this scenario $x_A |_{r=R} = \frac{N_{Ar} |_{r=R}}{k_1'' c} = 0$, because the reaction is said to be instantaneous and $k_1'' \rightarrow \infty$. And then I get xB from $x_B = 1 - x_A$.
Chestermiller
Mentor
Well, it hadn't occurred to me to consider the nitrogen presence in the boundary at the particle. If I take into account the nitrogen in the whole system I guess I could say xBR = xA∞ = 0.21.
The diffusivity of the A-B pair is given as a constant datum, so I took it out of the equation, along with c, the molar density of the system, also a constant because we were told to consider constant T and P. Isn't DAB = DBA?
Both A and B are diffusing through nitrogen. So the diffusion coefficients are not the same.
Whenever there's an heterogeneous reaction at a surface, the boundary condition for the molar flux at said surface according to BSL is $N_{Ar} |_{r=R} = k_1'' c x_A |_{r=R}$. So, in this scenario $x_A |_{r=R} = \frac{N_{Ar} |_{r=R}}{k_1'' c} = 0$, because the reaction is said to be instantaneous and $k_1'' \rightarrow \infty$. And then I get xB from $x_B = 1 - x_A$.
No. The diffusion coefficients of O2 and CO2 are different. The flux of O2 coming into the surface is the same as the flux of CO2 leaving the surface. The concentration of CO2 at infinity is zero.
Chet
Both A and B are diffusing through nitrogen. So the diffusion coefficients are not the same.
Oh, I think I got your point now. The statement indeed calls the given diffusivity datum the "diffusivity of oxygen through the mixture."
The flux of O2 coming into the surface is the same as the flux of CO2 leaving the surface.
With this information I can proceed to solve for the time required for the carbon to consume completely. With the boundary condition you provided me with, I can calculate the molar flux of CO2. This quantity times the surface area of the particle is the molar flow rate of CO2, which will be used in the quasi-steady state analysis
$$N_{Br} |_{r=R} = - N_{Ar} |_{r=R} = \frac{c_a \ D_{AC} \ x_{A \infty}}{R}$$
$$W_{Br} = 4\pi R^2 N_{Br} |_{r=R} = 4\pi c_a \ D_{AC} \ x_{A \infty} R$$
I changed the subindex for the diffusivity, C stands for N2 now, and ca is the molar density of air. A distinction is made because we will also have the molar density of the carbon particle in the equations. Now, we make an unsteady state mass balance on the carbon particle
$$\frac{dM}{dt} = - W_{Br}$$
Where M is the mass of the carbon particle. We can say $M= \frac{4}{3} \pi R^3 c_c$. Where cc is the molar density of the carbon particle. The differential equation becomes
$$R \ c_c \frac{dR}{dt} = c_a \ D_{AC} \ x_{A \infty}$$
Integrating
$$\int_0^t dt = - \frac{c_c}{c_a \ D_{AC} \ x_{A \infty}} \int_R^0 R \ dR$$
$$t= \frac{c_c R^2}{2 c_a D_{AC} \ x_{A \infty}}$$
Now some number crunching
$$c_c = \frac{85 \ \frac{lb}{ft^3}}{12 \ \frac{lb}{lbmol}} = 7.083 \ \frac{lbmol}{ft^3}$$
$$c_a = \frac{1 \ atm}{\left(0.7302 \ \frac{atm \cdot ft^3}{lbmol \cdot R}\right)(2960 \ R)} = 4.6266 \times 10^{-4} \frac{lbmol}{ft^3}$$
$$R = 0.025 \ in \left( \frac{1 \ ft}{12 \ in} \right) = \frac{1}{480} \ ft$$
Plugging in the values in the expression for time we finally get
$$t = 0.0198 \ hr = 1.188 \ min$$
Which is the time it takes for the small particle to burn completely.
Chestermiller
Mentor
Yes. I think that this is correct.
Thank you! Just one more question, if DAC ≠ DBC, is it still safe to assume NAr = -NBr?
Chestermiller
Mentor
Thank you! Just one more question, if DAC ≠ DBC, is it still safe to assume NAr = -NBr?
For every mole of oxygen that reaches the particle surface, one mole of CO2 leaves.
So, the relationship which must be satisfied is $x_{A \infty} D_{AC} = x_{BR} D_{BC}$, where both molar fractions cannot be equal because that would make the diffusivities equal.
If I'd want to estimate the diffusivity of CO2 in C, should C be N2 or air?
Chestermiller
Mentor
So, the relationship which must be satisfied is $x_{A \infty} D_{AC} = x_{BR} D_{BC}$, where both molar fractions cannot be equal because that would make the diffusivities equal.
Right
If I'd want to estimate the diffusivity of CO2 in C, should C be N2 or air?
I would have to go back an look up the appropriate sections of BSL to answer this one. To a good first approximation, I would use N2. But, the CO2 is also diffusing relative to O2, and the O2 mole fraction is not constant.
MexChemE
Thanks! I'm going to try with both, and I will also estimate the diffusivity of O2 in both nitrogen and air for the given conditions and see which one approaches more the value of the given diffusivity. | 2,440 | 8,480 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-45 | latest | en | 0.877649 |
https://kerodon.net/tag/03G4 | 1,680,348,667,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00169.warc.gz | 396,249,995 | 4,271 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Proposition 7.6.4.1. Let $\operatorname{\mathcal{C}}$ be a locally Kan simplicial category and let $\sigma :$
$\xymatrix@R =50pt@C=50pt{ X_{01} \ar [r] \ar [d] & X_0 \ar [d] \\ X_1 \ar [r] & X }$
be a commutative diagram in $\operatorname{\mathcal{C}}$. The following conditions are equivalent:
$(1)$
The composite map
$\Delta ^1 \times \Delta ^1 \xrightarrow { \operatorname{N}_{\bullet }(\sigma ) } \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \hookrightarrow \operatorname{N}_{\bullet }^{\operatorname{hc}}(\operatorname{\mathcal{C}})$
is a pullback square in the $\infty$-category $\operatorname{N}_{\bullet }^{\operatorname{hc}}(\operatorname{\mathcal{C}})$ (in the sense of Definition 7.6.3.1).
$(2)$
For every object $Y \in \operatorname{\mathcal{C}}$, the diagram of Kan complexes
$\xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y, X_{01} )_{\bullet } \ar [r] \ar [d] & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( Y, X_0 )_{\bullet } \ar [d] \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( Y, X_1 )_{\bullet } \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y,X)_{\bullet } }$
is a homotopy pullback square (in the sense of Definition 3.4.1.1). | 470 | 1,295 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.33712 |
https://www.coursehero.com/file/p7qahas/Perform-the-following-base-conversions-using-subtraction-or-division-remainder-a/ | 1,669,813,506,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00391.warc.gz | 750,575,473 | 88,254 | # Perform the following base conversions using
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2.Perform the following base conversions using subtraction or division-remainder:a.58810= _________b.225410= ________c.65210=________d.310410= ________Ans.3579a. 2102103b. 330045c. 16217d. 42289
______________________________________________________________________________3.Convert the following decimal fractions to binary with a maximum of six places to the rightof the binary point:a.26.78125b.194.03125c. 298.796875d. 16.1240234375Ans.a. 11010.11001b. 11000010.00001c. 100101010.110011d. 10000.000111______________________________________________________________________________4.Convert the following decimal fractions to binary with a maximum of six places to the rightof the binary point:a.25.84375b. 57.55c. 80.90625d. 84.874023Ans.a. 11001.11011b. 111001.100011c. 1010000.11101d. 1010100.110111______________________________________________________________________________5.Represent the following decimal numbers in binary using 8-bit signed magnitude, one'scomplement and two's complement:a. 77b. -42c. 119d. –107Ans.a.Signed magnitude: 01001101One's complement: 01001101Two's complement: 01001101b.Signed magnitude: 10101010One's complement: 11010101Two's complement: 11010110c.Signed magnitude: 01110111One's complement: 01110111Two's complement: 01110111
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Chapter 31 / Exercise 18
Contemporary Abstract Algebra
Gallian
Expert Verified
______________________________________________________________________________
6.Using a "word" of 3 bits, list all of the possible signed binary numbers and their decimalequivalents that are representable in:______________________________________________________________________________
7.Using a "word" of 4 bits, list all of the possible signed binary numbers and their decimalequivalents that are representable in:
______________________________________________________________________________8.From the results of the previous two questions, generalize the range of values (in decimal)that can be represented in any givenxnumber of bits using:a. Signed magnitudeb. One's complementc. Two's complement
Ans.a.-(2x-1-1) to +(2x-1-1)b.-(2x-1-1) to +(2x-1-1)c.-(2x-1) to +(2x-1)______________________________________________________________________________9.Given a (very) tiny computer that has a word size of 6 bits, what are the smallest negativenumbers and the largest positive numbers that this computer can represent in each of thefollowing representations?a.One's complementb.Two's complementAns.______________________________________________________________________________
Page 4Last Updated: November 200310. You have stumbled on an unknown civilization while sailing around the world.The people,who call themselves Zebronians, do math using 40 separate characters (probably becausethere are 40 stripes on a zebra).They would very much like to use computers, but wouldneed a computer to do Zebronian math, which would mean a computer that couldrepresent all 40 characters.You are a computer designer and decide to help them.Youdecide the best thing is to use BCZ, Binary Coded Zebronian (which is like BCD except it
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Term
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K
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Binary numeral system, Hamming Code, Parity bit, Ans a
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Chapter 31 / Exercise 18
Contemporary Abstract Algebra
Gallian
Expert Verified | 977 | 4,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.662887 |
rytsk.com | 1,653,375,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00010.warc.gz | 422,740,913 | 23,401 | # coefficient of happiness
Count how much time/money you’ve spend doing something because you HAVE TO DO but NOT enjoying and how much spent on something you actually ENJOYING DOING.
A ratio gives you coefficient of happiness. This coefficient during lifetime (draw a graph) gives you an idea whether you are going towards your happiness or self-destruction.
The same coefficient estimated for a nation is an index of nation happiness (or reverse would coefficient of slavery)?
Coefficient of happiness: H = Joy Time / The rest. not injoyable Time = Tj/Tw
Most of people spent they life at work. And 95% of people do not like your jobs.
H = Joy Time / Time at work = Tj/Tw
We see that more time you spent at work LESS happy you’ve become:
Extremes:
Childhood: Work time = 0 => H = ∞
Adultery: JoyTime close to 0, Money spend to mountain old body close to ∞ => H=0 | 210 | 871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.942776 |
http://hackage.haskell.org/package/maxent-0.6.0.4/docs/src/Numeric-MaxEnt.html | 1,485,192,147,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00561-ip-10-171-10-70.ec2.internal.warc.gz | 122,391,492 | 2,209 | ```-- |
-- The maximum entropy method, or MAXENT, is variational approach for computing probability
-- distributions given a list of moment, or expected value, constraints.
--
-- Here are some links for background info.
-- A good overview of applications:
-- <http://cmm.cit.nih.gov/maxent/letsgo.html>
-- On the idea of maximum entropy in general:
-- <http://en.wikipedia.org/wiki/Principle_of_maximum_entropy>
--
--
-- Use this package to compute discrete maximum entropy distributions over a list of values and
-- list of constraints.
--
-- Here is a the example from Probability the Logic of Science
--
-- >>> maxent 0.00001 [1,2,3] [average 1.5]
-- Right [0.61, 0.26, 0.11]
--
-- The classic dice example
--
-- >>> maxent 0.00001 [1,2,3,4,5,6] [average 4.5]
-- Right [.05, .07, 0.11, 0.16, 0.23, 0.34]
--
-- One can use different constraints besides the average value there.
--
-- As for why you want to maximize the entropy to find the probability constraint,
-- I will say this for now. In the case of the average constraint
-- it is a kin to choosing a integer partition with the most interger compositions.
-- I doubt that makes any sense, but I will try to explain more with a blog post soon.
--
module Numeric.MaxEnt (
Constraint,
(.=.),
UU(..),
ExpectationConstraint,
ExpectationFunction,
average,
variance,
-- ** Classic moment based
maxent,
-- ** General
general,
-- ** Linear
LinearConstraints(..),
linear
) where
import Numeric.MaxEnt.Internal (Constraint,
(.=.),
UU(..),
ExpectationConstraint,
ExpectationFunction,
average,
variance,
maxent,
general,
linear,
LinearConstraints(..))
``` | 418 | 1,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-04 | longest | en | 0.609102 |
https://www.geeksforgeeks.org/number-guessing-game-in-python/ | 1,685,968,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00381.warc.gz | 823,432,063 | 35,714 | GeeksforGeeks App
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# Number guessing game in Python 3 and C
Most of the geeks from a CS (Computer Science) background, think of their very first project after doing a Programming Language. Here, you will get your very first project and the basic one, in this article.
• Build a Number guessing game, in which the user selects a range.
• Let’s say User selected a range, i.e., from A to B, where A and B belong to Integer.
• Some random integer will be selected by the system and the user has to guess that integer in the minimum number of guesses
Analysis:
Explanation 1: If the User inputs range, let’s say from 1 to 100. And compiler randomly selected 42 as the integer. And now the guessing game started, so the user entered 50 as his/her first guess. The compiler shows “Try Again! You guessed too high”. That’s mean the random number (i.e., 42) doesn’t fall in the range from 50 to 100. That’s the importance of guessing half of the range. And again, the user guesses half of 50 (Could you tell me why?). So the half of 50 is 25. The user enters 25 as his/her second guess. This time compiler will show, “Try Again! You guessed too small”. That’s mean the integers less than 25 (from 1 to 25) are useless to be guessed. Now the range for user guessing is shorter, i.e., from 25 to 50. Intelligently! The user guessed half of this range, so that, user guessed 37 as his/her third guess. This time again the compiler shows the output, “Try Again! You guessed too small”. For the user, the guessing range is getting smaller by each guess. Now, the guessing range for user is from 37 to 50, for which the user guessed 43 as his/her fourth guess. This time the compiler will show an output “Try Again! You guessed too high”. So, the new guessing range for users will be from 37 to 43, again for which the user guessed the half of this range, that is, 40 as his/her fifth guess. This time the compiler shows the output, “Try Again! You guessed too small”. Leaving the guess even smaller such that from 41 to 43. And now the user guessed 41 as his/her sixth guess. Which is wrong and shows output “Try Again! You guessed too small”. And finally, the User Guessed the right number which is 42 as his/her seventh guess.
Total Number of Guesses = 7
Explanation 2: If the User inputs range, let’s say from 1 to 50. And compiler randomly selected 42 as the integer. And now the guessing game started. So the half of 50 is 25. The user enters 25 as his/her First guess. This time compiler will show, “Try Again! You guessed too small”. That’s mean the integers less than 25 (from 1 to 25) are useless to be guessed. Now the range for user guessing is shorter, i.e., from 25 to 50. Intelligently! User guessed half of this range, so that, user guessed 37 as his/her second guess. This time again the compiler shows the output, “Try Again! You guessed too small”. For the user, the guessing range is getting smaller by each guess. Now, the guessing range for user is from 37 to 50, for which the user guessed 43 as his/her third guess. This time the compiler will show an output “Try Again! You guessed too high”. So, the new guessing range for users will be from 37 to 43, again for which the user guessed the half of this range, that is, 40 as his/her fourth guess. This time the compiler shows the output, “Try Again! You guessed too small”. Leaving the guess even smaller such that from 41 to 43. And now the user guessed 41 as his/her fifth guess. Which is wrong and shows output “Try Again! You guessed too small”. And finally, the User Guessed the right number which is 42 as his/her sixth guess.
Total Number of Guesses = 6
So, the minimum number of guesses depends upon range. And the compiler must calculate the minimum number of guessing depends upon the range, on its own. For this, we have a formula:-
Minimum number of guessing = log2(Upper bound – lower bound + 1)
### Algorithm: Below are the Steps:
• User inputs the lower bound and upper bound of the range.
• The compiler generates a random integer between the range and store it in a variable for future references.
• For repetitive guessing, a while loop will be initialized.
• If the user guessed a number which is greater than a randomly selected number, the user gets an output “Try Again! You guessed too high
• Else If the user guessed a number which is smaller than a randomly selected number, the user gets an output “Try Again! You guessed too small”
• And if the user guessed in a minimum number of guesses, the user gets a “Congratulations! ” Output.
• Else if the user didn’t guess the integer in the minimum number of guesses, he/she will get “Better Luck Next Time!” output.
Below is the Implementation of the Algorithm:
## Python3
`import` `random``import` `math``# Taking Inputs``lower ``=` `int``(``input``(``"Enter Lower bound:- "``))` `# Taking Inputs``upper ``=` `int``(``input``(``"Enter Upper bound:- "``))` `# generating random number between``# the lower and upper``x ``=` `random.randint(lower, upper)``print``(``"\n\tYou've only "``,`` ``round``(math.log(upper ``-` `lower ``+` `1``, ``2``)),`` ``" chances to guess the integer!\n"``)` `# Initializing the number of guesses.``count ``=` `0` `# for calculation of minimum number of``# guesses depends upon range``while` `count < math.log(upper ``-` `lower ``+` `1``, ``2``):`` ``count ``+``=` `1` ` ``# taking guessing number as input`` ``guess ``=` `int``(``input``(``"Guess a number:- "``))` ` ``# Condition testing`` ``if` `x ``=``=` `guess:`` ``print``(``"Congratulations you did it in "``,`` ``count, ``" try"``)`` ``# Once guessed, loop will break`` ``break`` ``elif` `x > guess:`` ``print``(``"You guessed too small!"``)`` ``elif` `x < guess:`` ``print``(``"You Guessed too high!"``)` `# If Guessing is more than required guesses,``# shows this output.``if` `count >``=` `math.log(upper ``-` `lower ``+` `1``, ``2``):`` ``print``(``"\nThe number is %d"` `%` `x)`` ``print``(``"\tBetter Luck Next time!"``)` `# Better to use This source Code on pycharm!`
## C
`#include``#include``#include` `int` `main()``{`` ``int` `number, guess, nguesses=1;`` ``srand``(``time``(0));`` ` ` ``// Generates a random number between 1 and 100`` ``number = ``rand``()%100 + 1;`` ` ` ``// printf("The number is %d\n", number);`` ``// Keep running the loop`` ``// until the number is guessed`` ``do`` ``{`` ``printf``(``"Guess the number between 1 to 100\n"``);`` ``scanf``(``"%d"``, &guess);`` ``if``(guess>number)`` ``{`` ``printf``(``"you guessed to high\n"``);`` ``}`` ``else` `if``(guess
OUTPUT: Below is the output of the above Program
OUTPUT FOR THE GUESSING GAME
Time Complexity:
The time complexity of this code is O(n) as the number of iterations of the loop is not fixed and depends on the number of guesses taken by the user to get the right answer.
Space Complexity:
The space complexity of this code is O(1) as all the variables used in this code are of the same size and no extra space is required to store the values.
My Personal Notes arrow_drop_up | 1,941 | 7,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-23 | latest | en | 0.928399 |
https://fr.mathworks.com/matlabcentral/cody/problems/32-most-nonzero-elements-in-row/solutions/1611931 | 1,580,192,335,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251776516.99/warc/CC-MAIN-20200128060946-20200128090946-00490.warc.gz | 444,242,647 | 15,640 | Cody
# Problem 32. Most nonzero elements in row
Solution 1611931
Submitted on 18 Aug 2018 by saravan eluru
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
a = [ ... 1 2 0 0 0 0 0 5 0 0 2 7 0 0 0 0 6 9 3 3]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
2 Pass
a = [ ... 1 2 0 0 0 0 5 0 0 6 9 -3 2 7 0 0 0 0 0 0]; r_correct = 3; assert(isequal(fullest_row(a),r_correct))
3 Pass
a = [ ... 1 0 0 0 0 0 0 0 0 0 0 0 0 2 3]; r_correct = 5; assert(isequal(fullest_row(a),r_correct))
4 Pass
a = [ ... 0 0 0 -3 0 0]; r_correct = 4; assert(isequal(fullest_row(a),r_correct)) | 303 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-05 | latest | en | 0.619087 |
https://fr.mathworks.com/matlabcentral/cody/problems/42640-back-to-basics-find-no-of-elements-in-a-matrix/solutions/752805 | 1,582,839,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00139.warc.gz | 377,127,851 | 15,722 | Cody
Problem 42640. Back to Basics - Find no. of elements in a matrix?
Solution 752805
Submitted on 4 Oct 2015 by Michèle
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% x = 10; y_correct = 1; assert(isequal(no_elements(x),y_correct))
2 Pass
%% x = [1 26 45 11;12 -5 -6 8;12 2 2 2]; y_correct = 12; assert(isequal(no_elements(x),y_correct))
3 Pass
%% x = 0:25; y_correct = 26; assert(isequal(no_elements(x),y_correct))
4 Pass
%% x = [4 5 6]; y_correct = 3; assert(isequal(no_elements(x),y_correct))
5 Pass
%% x = exp(2); y_correct = 1; assert(isequal(no_elements(x),y_correct))
6 Pass
%% x = [1:2 4 8 9]; y_correct = 5; assert(isequal(no_elements(x),y_correct))
7 Pass
%% x = [4 5 6;7 89 7;1 2 3;4 5 6;4 5 6]; y_correct = 15; assert(isequal(no_elements(x),y_correct)) | 326 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-10 | longest | en | 0.38429 |
http://www.overclock.net/t/794098/the-effect-of-temperature-on-your-processor-science-warning | 1,500,988,238,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425193.20/warc/CC-MAIN-20170725122451-20170725142451-00484.warc.gz | 501,408,042 | 65,949 | Overclock.net › Forums › Cooling › Water Cooling › The effect of temperature on your processor! *SCIENCE WARNING*
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The effect of temperature on your processor! *SCIENCE WARNING*
Cooling? Why do I want to cool my CPU?
Contents:
1. Processor Architecture
2. Overclocking, the effects of temperature.
3. Cooling System Design Principles
Processor Architecture
Understanding what a processor really is:
In short, your processor is an on/off switch...Well…A bunch of on/off switches, a billion or so. These switches use the most basic logic known to man, known as Boolean logic: everything is either on, or off, either a one (on) or a zero (off). Meanwhile the explanation of how computers use Boolean logic (which leads to Binary), is not necessary for the purpose of this post. These switches are of course tiny, around 32nm, depending upon the transistor in question, and have been described as containing "merely a couple of electrons" (citation needed)
The Switch
To understand the fundamental issues at hand here, we must understand this elusive switch; how it works and what it is made of. This switch is a switch whose state as "on" or "off" is governed by voltage; unlike your lighting switch at home, where up perhaps is "on", in the case of the switches we are talking about (High K Metal Gate Transistors), a high voltage is "on", and a low voltage is "off" N.B. 1. As some things are easier to visualize, I have included an image below.
As described, the state of the switch is determined by the voltage between the source and the drain (labeled S and D). If the voltage is HIGH, known as Vcc, the switch is said to be "on", if it is low, known as Vss, it is said to be "off".
Clock Cycles-Frequency, Speed
The frequency, or speed of a processor is, of course, related to the underlying structure of the switch. All of these terms are roundabout ways of describing the rate at which your processor's switches can turn on and off, or switch between the high voltage that designates "on" and the low voltage that designates "off" (Vcc and Vss). The rate at which this switching occurs is really, very rapid, as I am sure you know by this stage!
However, what is interesting is that the clock speed can be thought of as a pulse like signal, switching back and forth from on to off very rapidly.
This can be shown as follows:
Where Vcc is the high, "on", and Vss is the low "off" voltage. The Y axis here is Voltage, the X axis here is time.
However, this is not an accurate depiction of what is actually occuring, rather this is the ideal binary situation; the voltage is either exactly "on" or exactly "off", and it switches from the high voltage to the lower voltage instantaneously, which from physics we know is impossible.
Rather, our switches work on tolerances, accepting anything over some value Vcc as "on," and anything less as "off". It also takes some time to transition from one state to another, as you can see below:
Overclocking, the effects of temperature.
What it means to overclock:
What does it mean to overclock? Sure, you know what it means...Go faster, press some buttons, amp up your SuperPi times, but what does it really mean, or what does it really do? Remembering from above, the clock speed is actually the clock frequency, or the number of times that the transistors inside of your processor switch from on to off per second. When we overclock a processor, we force it to switch between the two states more rapidly, and some problems arise. Remembering the image above, there is some time required for the state of a given switch to change from on to off-there is some time required for the voltage to drain from the gate; increasing the number of times the state must switch per second, decreases the amount of time your processor has to transition between the two states. You all have experienced the issue associated with your processor being unable to make the transition in the time allotted: instability. Your processor cannot tell the difference between on and off, as the voltage never reaches the recognized "on", Vcc:
The issue at hand is simple: the voltage cannot switch fast enough between the two states, however there are two solutions, on one hand you can somehow change the value that is recognized as on, or else increase the speed at which that transition occurs. Whether or not you know it, you have been doing the former every time you have increased the voltage of your processor in order to maintain stability; while you actually raised the peak voltage, you effectively moved Vcc down in relation to the maximum voltage, providing for the same effect. While it is not quite that simple, as the rate at which electricity flows is also affected by voltage (V=IR, for a given resistance, an increase in voltage incurs an increase in current, or flow of electricity per unit of time), let us for simplicity's sake assume that is all that you are changing in the first case, and allow the rate at which electricity flows to be altered separately.
Decreasing the time it takes to switch states
The following contains some basic physics, I will try and provide links to wikipedia for as many concepts as I can, as some people have less experience than others, and further reading may be valuable to them!
Before discussing how to alter the time it takes to switch states, we must first discus what it is that governs the rate at which this switching can occur. While current is not technically the rate at which voltage dissipates, it can be related for any given situation, and we will use it as a proxy to describe the rate at which voltage dissipates.
First, the equation for current that we will be using is: I = V/R. Looking at this mathematically, we can see that there are two things we can do to increase the value of I (current):
1. Increase the voltage
2. Decrease the resistance
As we already covered point 1. in the previous paragraph (rather sneakily, I must admit!), we will merely concern ourselves with decreasing the resistance. What is resistance? Simply speaking, it is the ability of a given material to resist the flow of electricity through it. Resistance is affected by temperature: R = kT (where R is resistance, k is some constant of proportionality, and T is temperature). Thus we can see (both intuitevly perhaps, if you think about the random movement of electrons that inhibits the general path of flow, and perhaps mathematically) that resistance increases as temperature increases. On the other hand, as temperature decreases, so does resistance, effectively decreasing the time it takes to switch states, from high to low; solving our problem and providing a simple solution.
A Little more on Temperature'
While the previous paragraphs describe the increased stability at lower temperatures, because of the lesser amount of time to switch between ON and OFF (effectively allowing you to clock higher), it serves to mention that there are far more intricate processes at work here. As a measly engineering undergraduate with at best a 'decent' understanding of physics, I cannot explain all of these processes, however there is one that is commonly discussed and is relatively simple: electron-migration refers to the movement of particles within a conductor induced by the momentum of electroncs. This concept can be seen as a sort of bowling ball effect:
imagine bowling with tiny marbles. Once in a while, or over time, you will eventually move, hit, or change the pins, because while tiny in comparison to the pins, your marbles have momentum, and some force behind them!
Edited by wcdolphin - 8/24/10 at 9:05pm
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N.B1: Voltage is measured across, or between two locations, or objects. In simulation terms, we refer to this as an 'across variable'
Edited by wcdolphin - 9/16/10 at 9:17am
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I like it.
Making cookies as we speak, time to translate this all from my presentation last year.
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Quote:
Originally Posted by cdolphin As described, the state of the switch is determined by the voltage between the source and the drain. If the voltage is HIGH, known as Vcc, the switch is said to be "on", if it is low, known as Vcc, it is said to be "off".
yeah.
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About halfway done, a more realistic thermodynamic approach to understanding cooling systems to come.
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If anyone has any input thus far, it would be appreciated, also mods, where do you think the best location is for this? Don't judge yet, as the real cooling section is to come
EDIT: Dang, hit new post instead of edit *may the gods merge me upwards*
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tl:dr
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Quote:
Originally Posted by 88EVGAFTW tl:dr cliff notes please
to increase speed we must decrease temperature.
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I got lost when there was some weird words.
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• The effect of temperature on your processor! *SCIENCE WARNING*
Overclock.net › Forums › Cooling › Water Cooling › The effect of temperature on your processor! *SCIENCE WARNING* | 4,012 | 13,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-30 | latest | en | 0.947333 |
http://www.ck12.org/algebra/Applications-of-One-Step-Equations/lesson/Applications-of-One-Step-Equations-Intermediate/r18/ | 1,454,970,640,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701154221.36/warc/CC-MAIN-20160205193914-00234-ip-10-236-182-209.ec2.internal.warc.gz | 326,450,332 | 32,159 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Applications of One-Step Equations
## Real-world problems using single variable equations.
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Applications of One-Step Equations
What if The Perfect Pizza charges $1.50 for a slice of pizza? However, the restaurant offers a$2.00 discount off the per-slice cost if you buy a whole pizza. A whole pizza costs 10. How could you find how many slices are in a whole pizza? After completing this Concept, you'll be able to solve real-world problems like this one. ### Watch This ### Guidance Let's use the skills we learned in the last two concepts to solve applications of one-step problems. #### Example A In the year 2017, Anne will be 45 years old. In what year was Anne born? The unknown here is the year Anne was born, so that’s our variable \begin{align*}x\end{align*}. Here’s our equation: Anne was born in 1972. #### Example B A mail order electronics company stocks a new mini DVD player and is using a balance to determine the shipping weight. Using only one-pound weights, the shipping department found that the following arrangement balances: How much does each DVD player weigh? Solution Since the system balances, the total weight on each side must be equal. To write our equation, we’ll use \begin{align*}x\end{align*} for the weight of one DVD player, which is unknown. There are two DVD players, weighing a total of \begin{align*}2x\end{align*} pounds, on the left side of the balance, and on the right side are 5 1-pound weights, weighing a total of 5 pounds. So our equation is \begin{align*}2x = 5\end{align*}. Dividing both sides by 2 gives us \begin{align*}x = 2.5\end{align*}. Each DVD player weighs 2.5 pounds. #### Example C In 2004, Takeru Kobayashi of Nagano, Japan, ate 53.5 hot dogs in 12 minutes. This was 3 more hot dogs than his own previous world record, set in 2002. Calculate how many minutes it took him to eat one hot dog. Solution We know that the total time for 53.5 hot dogs is 12 minutes. We want to know the time for one hot dog, so that’s \begin{align*}x\end{align*}. Our equation is \begin{align*}53.5x = 12\end{align*}. Then we divide both sides by 53.5 to get \begin{align*}x = \frac{12}{53.5}\end{align*}, or \begin{align*}x = 0.224 \ minutes\end{align*}. We can also multiply by 60 to get the time in seconds; 0.224 minutes is about 13.5 seconds. So that’s how long it took Takeru to eat one hot dog. Watch this video for help with the Examples above. ### Vocabulary • An equation in which each term is either a constant or the product of a constant and a single variable is a linear equation. • We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an equivalent equation. • To solve an equation, isolate the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides. ### Guided Practice Calculate the following, using the problem from Example C: a) How many hot dogs he ate per minute. b) What his old record was. Solution: a) For this questions we’re looking for hot dogs per minute instead of minutes per hot dog, as in Example C. We’ll use the variable \begin{align*}y\end{align*} instead of \begin{align*}x\end{align*} this time so we don’t get the two confused. 12 minutes, times the number of hot dogs per minute, equals the total number of hot dogs, so \begin{align*}12y = 53.5\end{align*}. Dividing both sides by 12 gives us \begin{align*}y = \frac{53.5}{12}\end{align*}, or \begin{align*}y = 4.458\end{align*} hot dogs per minute. b) We know that his new record is 53.5, and we know that’s three more than his old record. If we call his old record \begin{align*}z\end{align*}, we can write the following equation: \begin{align*}z + 3 = 53.5\end{align*}. Subtracting 3 from both sides gives us \begin{align*}z = 50.5\end{align*}. So Takeru’s old record was 50.5 hot dogs in 12 minutes. ### Practice Peter is collecting tokens on breakfast cereal packets in order to get a model boat. In eight weeks he has collected 10 tokens. He needs 25 tokens for the boat. Write an equation and determine the following information. 1. How many more tokens he needs to collect, \begin{align*}n\end{align*}. 2. How many tokens he collects per week, \begin{align*}w\end{align*}. 3. How many more weeks remain until he can send off for his boat, \begin{align*}r\end{align*}. Juan has baked a cake and wants to sell it in his bakery. He is going to cut it into 12 slices and sell them individually. He wants to sell it for three times the cost of making it. The ingredients cost him8.50, and he allowed \$1.25 to cover the cost of electricity to bake it. Write equations that describe the following statements.
1. The amount of money that he sells the cake for \begin{align*}(u)\end{align*}.
2. The amount of money he charges for each slice \begin{align*}(c)\end{align*}.
3. The total profit he makes on the cake \begin{align*}(w)\end{align*}.
Jane is baking cookies for a large party. She has a recipe that will make one batch of two dozen cookies, and she decides to make five batches. To make five batches, she finds that she will need 12.5 cups of flour and 15 eggs.
1. How many cookies will she make in all?
2. How many cups of flour go into one batch?
3. How many eggs go into one batch?
4. If Jane only has a dozen eggs on hand, how many more does she need to make five batches?
5. If she doesn’t go out to get more eggs, how many batches can she make? How many cookies will that be?
### Explore More
Sign in to explore more, including practice questions and solutions for Applications of One-Step Equations. | 1,578 | 5,882 | {"found_math": true, "script_math_tex": 23, "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} | 4.84375 | 5 | CC-MAIN-2016-07 | latest | en | 0.874756 |
http://www.geekinterview.com/question_details/13653 | 1,618,327,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00531.warc.gz | 132,275,205 | 18,853 | # Sum of 2 numbers is 7 & the product of 2 numbers is 10 find the number.
#### Harish
• Dec 1st, 2005
Ans: 5 and 2
#### VIJAYA MUTHU KUMAR
• Jan 18th, 2006
#### Kamal
• Sep 30th, 2011
Sum: 5 + 2 = 7
Product: 5 * 2 = 10
Numbers are 5 and 2. | 112 | 249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-17 | latest | en | 0.845333 |
https://www.numbersaplenty.com/19176 | 1,638,065,072,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00308.warc.gz | 1,035,246,210 | 3,553 | Search a number
19176 = 2331747
BaseRepresentation
bin100101011101000
3222022020
410223220
51103201
6224440
7106623
oct45350
928266
1019176
1113453
12b120
138961
146dba
155a36
hex4ae8
19176 has 32 divisors (see below), whose sum is σ = 51840. Its totient is φ = 5888.
The previous prime is 19163. The next prime is 19181. The reversal of 19176 is 67191.
It is a hoax number, since the sum of its digits (24) coincides with the sum of the digits of its distinct prime factors.
It is a Harshad number since it is a multiple of its sum of digits (24).
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 385 + ... + 431.
It is an arithmetic number, because the mean of its divisors is an integer number (1620).
219176 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 19176, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (25920).
19176 is an abundant number, since it is smaller than the sum of its proper divisors (32664).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
19176 is a wasteful number, since it uses less digits than its factorization.
19176 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 73 (or 69 counting only the distinct ones).
The product of its digits is 378, while the sum is 24.
The square root of 19176 is about 138.4774349849. The cubic root of 19176 is about 26.7661561354.
Multiplying 19176 by its product of digits (378), we get a triangular number (7248528 = T3807).
The spelling of 19176 in words is "nineteen thousand, one hundred seventy-six". | 537 | 1,933 | {"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} | 3.6875 | 4 | CC-MAIN-2021-49 | latest | en | 0.909757 |
https://fr.slideserve.com/fawzia/chapter-14-properties-of-gases | 1,628,062,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00094.warc.gz | 251,350,038 | 23,865 | Chapter 14 Properties of Gases
# Chapter 14 Properties of Gases
## Chapter 14 Properties of Gases
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##### Presentation Transcript
1. Chapter 14Properties of Gases
2. The Properties of Gases • Gas can expand to fill its container • Gases are easily compressed, or squeezed into a smaller volume. • Gases occupy far more space than a liquid or a solid • Compressibility – measure of how much the volume of matter decreases under pressure.
3. Compressibility of Gases When an airbag triggers due to a sudden stop, a chemical reaction inside the airbag occurs. One product of the reaction is nitrogen gas, which causes the bag to inflate. When a person collides with an inflated air bag, the impact forces the molecules of gas in the bag closer together The compression of the gas absorbs the energy of the impact.
4. Compressibility of Gases What factors do you think affect the pressure of the air inside the soccer ball? Temperature of the air inside the ball Volume of the ball
5. Kinetic Theory & Gases What is kinetic energy The energy of motion How are temperature and kinetic energy related? Temperature is a measure of average kinetic energy.
6. Kinetic Theory & Gases Gases are easily compressed because of the space between the particles in a gas. Under pressure, the particles in a gas are forced closer together, or compressed.
7. Factors Affecting Gas Pressure Pressure (P) - kPa Volume (V) - liters Temperature (T) - Kelvin Number of moles (n) The amount of gas, volume, and temperature are factors that affect gas pressure
8. Amount of Gas and Gas Pressure When you inflate an air raft, the pressure inside the raft will increase. (this is a container with a volume that can vary. A balloon is another example) Collisions of particles with the inside walls of the raft result in the pressure that is exerted by the gas. By adding gas, you increase the number of particles. Increasing the number of particles increases the number of collisions, which is why the gas pressure increases.
9. Amount of Gas & Gas Pressure When a gas is put into a closed rigid container, the pressure increases as more particles are added Because the container is rigid, the volume of the gas is constant. Assume the temperature doesn't change Doubling the number of particles of gas, doubles the pressure. As gas is removed, the pressure inside the container is reduced.
10. Cause and Effect If the pressure of the gas in a sealed container is lower than the outside air pressure, air will rush into the container when the container is opened. When the pressure of the gas in a sealed container is higher than the outside air pressure, the gas will flow out of the container when the container is unsealed.
11. Cause and Effect Gas pressure inside a new spray paint can is greater than the air pressure outside the can. As the can is used, the pressure inside the can decreases until there is not enough pressure inside the can to force the paint out.
12. Volume & Gas Pressure You can raise the pressure exerted by a contained gas by reducing its volume. The more gas is compressed, the greater is the pressure the gas exerts inside the container.
13. Volume & Gas Pressure When cylinder has a volume of 1 L, the pressure is 100 kPa If volume is halved to 0.5 L, the pressure doubles to 200kPa If volume is doubled to 2.0 L, the pressure of the volume is cut in half to 50 kPa.
14. Temperature & Gas Pressure A sealed bag of potato chips may bulge at the seams if placed in the sun. Bag bulges because an increase in the temperature of the gas inside the bag causes an increase in its pressure. As gas inside bag is heated, the temperature increases, increasing kinetic energy of the particles, and causing more collisions, thus more pressure.
15. Temperature & Gas Pressure If volume and amount of gas are constant, when the Kelvin temperature of gas doubles, the gas pressure doubles. Gas in sealed container may generate enormous pressure when heated. For that reason, an aerosol can, even an “empty” one, may explode if thrown onto a fire. As the temperature of an enclosed gas decreases by half, the pressure decreases by half.
16. Questions What effect would tripling the number of particles of a gas in a closed container have on the pressure exerted? Gas pressure would triple What effect would doubling the volume of an enclosed gas have on the pressure? Gas pressure would decrease by half How does the pressure of an enclosed gas change with increasing temperature? The number and force of collisions increase with temperature, and the pressure increases
17. Questions Why is a gas easy to compress? Because of the space between particles in a gas List three factors that can affect gas pressure? Temperature, pressure, & amount of gas Why does a collision with an air bag cause less damage than a collision with a steering wheel? Gas in the inflated airbag can be compressed, and absorbs some of the energy from the impact. The solid steering wheel cannot.
18. Questions If temperature is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one quarter of its original value? The volume would need to increase by a factor of four How does a decrease in temperature affect the pressure of a contained gas If temperature decreases, the pressure will decrease If gas temperature in a container is constant, how could you increase the pressure one hundredfold? Increase the amount of gas in the container one hundredfold.
19. End of Section 14.1
20. Boyle’s Law (Pressure & Volume) If the temperature is constant, as the pressure of a gas increase, the volume decreases. Conversely, if the temperature is constant, as the pressure of a gas decreases, the volume increases. Robert Boyl was the first person to study this pressure-volume relationship. In 1662, Boyle proposed a law to describe the relationship.
21. Boyle’s Law (Pressure & Volume) Boyle’s Law – states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. P1V1 = P2V2
22. Sample Problem UsingBoyle’s Law A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume the temperature remains constant) What do you think will happen to the volume at a higher temperature knowing what you know already about gases?
23. Sample Problem UsingBoyle’s Law P1 = 103 kPa P2 = 25.0 kPa V1 = 30.0 L V2 = ? L P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (30.0 L) (103 kPa) 25.0 KPa V2 = 1.24 x 102 L (3 sig figs)
24. Sample Problem UsingBoyle’s Law Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105 KPa to 40.5 KPa. It the temperature does not change, what will the new volume be? P1 = 105 kPa P2 = 40.5 kPa V1 = 2.50 L V2 = ? L P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (2.50 L) (105 kPa) 40.5 KPa V2 = 6.48 L (3 sig figs)
25. Sample Problem UsingBoyle’s Law A gas with a volume of 4.00 L at a pressure of 205 KPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? P1 = 205 kPa P2 = ? kPa V1 = 4.00 L V2 = 12.0 L P1V1 = P2V2 orP1V1 / V2 = P2 P2 = (4.00 L) (205 kPa) 12.0 L P2 = 68.3 KPa (3 sig figs)
26. Sample Problem UsingBoyle’s Law The volume of a gas at 99.6 KPa and 24ºC is 4.23L. What volume will it occupy at 93.3 KPa and 24ºC? P1 = 99.6 kPa P2 = 93.3 kPa T1 = 24ºC V1 = 4.23 L V2 = ? L T2 = 24ºC P1V1 = P2V2 orP1V1 / P2 = V2 V2 = (4.23 L) (99.6 kPa) 93.3 kPa V2 = 4.52 L (3 sig figs)
27. Charles’s Law Temperature and Volume As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant. In 1787, French physicist Jacques Charles studies the effect of temperature on the volume of a gas at constant pressure. Charles’s Law – states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. V1 = V2 T1 T2
28. Sample Problem UsingCharles’s Law A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant? T1 = 24ºC or 297 K V1 = 4.00 L T2 = 58ºC or 331 K V2 = ? L When using gas laws always express the temperatures in kelvins! T1 = 24ºC + 273 = 297 K T2 = 58ºC + 273 = 331 K
29. Sample Problem UsingCharles’s Law A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant? T1 = 24ºC or 297 K V1 = 4.00 L T2 = 58ºC or 331 K V2 = ? L V1 = V2 or V1T2 = V2 T1 T2 T1 V2 = (4.00 L) (331 K) = 4.46 L 297 K
30. Sample Problem UsingCharles’s Law If a sample of gas occupies 6.80 L at 325ºC, what will its volume be at 25ºC if the pressure does not change? T1 = 325ºC or 598 K V1 = 6.80 L T2 = 25ºC or 298 K V2 = ? L V1 = V2 or V1T2 = V2 T1 T2 T1 V2 = (6.80 L) (298 K) = 3.39 L 598 K
31. Gay-Lussac’s LawPressure and Temperature As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant. Joseph Gay-Lussac discovered the relationship between the pressure and the temperature of gas in 1802. Gay-Lussac’s Law – states that the pressure of a gas is directly proportional to the Kelvin temperature if the pressure if the volume remains constant. P1 = P2 T1 T2
32. Sample Problem UsingGay-Lusaac’s Law A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K? P1 = 6.58 kPa T1 = 539 K P2 = ? kPa T2 = 211 K P1 = P2 or P1T2 = P2 T1 T2 T1 P2 = (6.58 K) (211 K) = 2.58kPa 539 K
33. Sample Problem UsingGay-Lusaac’s Law The pressure in a car tire is 198 kPa at 27ºC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. P1 = 198 kPa T1 = 300 K P2 = 225 kPa T2 = ? K P1 = P2 or P2T1 = T2 T1 T2 P1 T2 = (225 kPa) (300 K) = 341 K 198 kPa
34. Gases If the gas is heated (T2 > T1), the new pressure is greater. (volume constant) If the gas is heated (T2 > T1), the new volume is greater because the gas expands. (pressure constant) If the gas is cooled (T2 < T1), the new pressure is less. (volume constant) If the gas is cooled (T2 < T1), the new volume is smaller because the gas contracts. (pressure constant)
35. Combined Gas Law There is a single expression that combines Boyle’s, Charles’s and Gay-Lusaac’s Law. The combined gas law describes the relationship among the pressure, temperature, and volume of an enclosed gas. The combined gas law allows you to do calculation for situations in which only the amount of gas is constant P1V1 = P2 V2 T1 T2
36. Sample Problem UsingCombined Gas Law A gas at 155 kPa and 25º C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125º C. What is the new volume? P1 = 155 kPa T1 = 298 K V1 = 1.00 L P2 = 605 kPa T2 = 398 K V2 = ? P1V1 = P2 V2 or P1V1 T2 = V2 T1 T2 T1 P2 V2 = (155kPa)(1.00 L)(398 K) = 0.342 L (298 K)(605 kPa)
37. Sample Problem UsingCombined Gas Law A 5.00 L air sample has a pressure of 107 kPa at a temp of -50 º C. If the temperature is raised to 102 º C and the volume expands to 7.00 L, what will the new pressure be? P1 = 107 kPa T1 = 223 K V1 = 1.00 L P2 = ? kPa T2 = 375 K V2 = 7.00 L P1V1 = P2 V2 or P1V1 T2 = P2 T1 T2T1 V2 P2 = (107 kPa)(5.00 L)(375 K) = 1.29 x 102kPa (223K)(7.00 L)
38. Questions How are the pressure and volume of a gas related at constant temperature? The volume of a gas decreases as the pressure increases. (Boyle’s Law) If pressure is constant, how does a change in temperature affect the volume of a gas? As the temperature increases, the volume increases. (Charles’s Law)
39. Questions What is the relationship between the temperature and pressure of a contained gas at constant volume? As the temperature increases, the pressure increases. (Gay-Lusaac’s Law) In what situations is the combined gas law useful? Allows you to do calculations when the only constant is the amount of gas.
40. Question Explain how Charles’s law can be derived from the combined gas law. When the pressure is constant, P1 = P2, the pressure terms cancel, leaving an equation for Charles’s Law. P1V1 = P2 V2 T1 T2
41. Question The volume of a weather balloon increases as the balloon rises in the atmosphere. Why doesn’t the drop in temperature at higher altitudes cause the volume to decrease? The outside pressure decreases, causing a greater increase in the balloon’s volume The higher it rises, the colder the temperature and the lower the volume. At the same time, atmospheric pressure decreases, allowing the gas to expand.
42. End of Section 14.2
43. Ideal Gas Law With the combined gas law, you can solve problems with three variables: temperature, volume & pressure The combined gas law assumes that the amount of gas does not vary. To calculate the number of moles of a contained gas requires an expression that contains the variable n. The number of moles of gas is directly proportional to the number of particles.
44. Ideal Gas Law The volume occupied by a gas at a specified temperature and pressure also must depend on the number of particles. So… moles must be directly proportional to volume. P1V1 = P2 V2 T1 nT2 n This equation shows that P1V1 is a constant. T1 n This constant holds for ideal gases – gases that conform to the gas laws.
45. Ideal Gas Law If you know the values for P,V, T and n for one set of condition, you can calculate a value for the constant. 1 mole of every gas occupies 22.4 L at STP. (101.3kPa and 273K) Ideal gas constant uses the symbol R R = P1V1 R= (101.3kPa)(22.4L) T1 n (273 K)(1mol) R = 8.31 L · kPa / mole · K or R = 0.0831 L · atm / mole · K
46. Ideal Gas Law PV = nRT pressure volume moles constant temperature(K) 8.31L · kPa / mole · K
47. Sample Problem Using Ideal Gas Law When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 103 kPa. How many moles of helium does the sphere contain? P = 1.89 x 103 V = 685 L T = 621 K PV = nRT or PV / RT = n n = (1.89 x 103 kPa) (685 L)mol · K (8.31L · kPa) (621K) n = 251 mol He
48. Sample Problem Using Ideal Gas Law A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37ºC? Use a molar mass of 29 g for air. P = 102 kPaV = 2.20 L T = 310 K PV = nRT or PV / RT = n n = (102kPa) (2.20 L)mol · K (8.31L · kPa) (310K) n = 0.087 mol air 0.087 mol air x 29g air / mol air = 2.5 g air
49. Ideal Gases & Real Gases Ideal gas – one that follows the gas laws at all conditions of pressure and temperature. Such a gas would have to conform precisely to the assumptions of kinetic theory. Its particles could have no volume, and there could be no attraction between particles in the gas. There is no gas for which these assumptions are true.
50. Ideal Gases & Real Gases At many conditions of temperature and pressure, real gases behave very much like an ideal gas. Particles of a real gas do have volume and there are attractions between the particles. Because of these attractions, a gas can condense or solidify when it is compressed or cooled. Example – if water vapor is cooled below 100ºC at standard atmospheric pressure, it condenses to a liquid. | 4,260 | 15,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.847232 |
https://mail.kde.org/pipermail/kde-edu/2019-January/017127.html | 1,709,402,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00491.warc.gz | 356,820,248 | 2,458 | D18294: Algorithms to find values and indices in vectors of data
Wed Jan 16 21:06:12 GMT 2019
```asemke added inline comments.
> XYCurve.cpp:1690
> +int XYCurve::calculateMaxSteps (unsigned int value) {
> + const char LogTable256[256] =
> + {
should we define this static?
> XYCurve.cpp:1744
> + return yColumn()->valueAt(index);
> + } else {
> + valueFound = false;
In indexForX() you have handling for datetime. Why not to add this here, too?
> XYCurve.cpp:1766
> + // bisects the index every time, so it is possible to find the value in log_2(rowCount) steps
> + bool increase = true;
> + if(properties == AbstractColumn::Properties::MonotonicDecreasing)
this three lines can be simplified to
bool increase = (properties != AbstractColumn::Properties::MonotonicDecreasing);
> XYCurve.cpp:1773
> +
> + unsigned int max_steps = calculateMaxSteps(static_cast<unsigned int>(rowCount));
> +
maxSteps, camel case...
> XYCurve.h:72
> + int indexForX(double x) const;
> + int indexForX(double x, QVector<double>& column, AbstractColumn::Properties properties = AbstractColumn::Properties::No) const;
> + int indexForXinPointsVector(double x, QVector<QPointF>& column, AbstractColumn::Properties properties = AbstractColumn::Properties::No) const;
indexForX oder indexForXinColumn? Let's maybe name all these functions simply indexForX and diferentiation is done by the parameters and the documentation of these overloaded parameters. This would be similar for example to the all those map*() functions in CartesianCoordinateSystem.
REVISION DETAIL
https://phabricator.kde.org/D18294
To: Murmele, asemke
Cc: yurchor, kde-edu, Murmele, narvaez, apol
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``` | 493 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.593724 |
https://scipy.github.io/devdocs/reference/generated/scipy.sparse.kron.html | 1,723,375,445,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00430.warc.gz | 385,066,099 | 6,593 | scipy.sparse.
# kron#
scipy.sparse.kron(A, B, format=None)[source]#
kronecker product of sparse matrices A and B
Parameters:
Asparse or dense matrix
first matrix of the product
Bsparse or dense matrix
second matrix of the product
formatstr, optional (default: ‘bsr’ or ‘coo’)
format of the result (e.g. “csr”) If None, choose ‘bsr’ for relatively dense array and ‘coo’ for others
Returns:
kronecker product in a sparse format.
Returns a sparse matrix unless either A or B is a
sparse array in which case returns a sparse array.
Examples
```>>> import numpy as np
>>> import scipy as sp
>>> A = sp.sparse.csr_array(np.array([[0, 2], [5, 0]]))
>>> B = sp.sparse.csr_array(np.array([[1, 2], [3, 4]]))
>>> sp.sparse.kron(A, B).toarray()
array([[ 0, 0, 2, 4],
[ 0, 0, 6, 8],
[ 5, 10, 0, 0],
[15, 20, 0, 0]])
```
```>>> sp.sparse.kron(A, [[1, 2], [3, 4]]).toarray()
array([[ 0, 0, 2, 4],
[ 0, 0, 6, 8],
[ 5, 10, 0, 0],
[15, 20, 0, 0]])
``` | 365 | 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} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.483581 |
http://funnyfunnyjokes.org/2017/09/27/are-you-smart-enough-to-solve-this-very-interesting/ | 1,582,079,276,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00214.warc.gz | 55,403,499 | 14,261 | # Are you smart enough to solve this? – Very Interesting
Use of calculator is prohibited.
The question is……..If
1 = 3
2 = 9
3= 27
4 = 81
Then 5 = ?
### 709 Responses to Are you smart enough to solve this? – Very Interesting
1. Waseem September 24, 2017 at 10:59 PM #
243
2. Firoz IQBAL September 22, 2017 at 12:28 PM #
243
3. sher ahmed maka September 20, 2017 at 9:28 PM #
405
4. Sidick September 20, 2017 at 4:09 PM #
243
5. Makbak September 18, 2017 at 8:55 PM #
1 = 3
2 = 9
3 = 27
4 = 81
5 = 162
6. Ali Ashkanani September 16, 2017 at 7:06 PM #
243
• Fouad September 17, 2017 at 10:26 AM #
243
7. ansar September 16, 2017 at 9:21 AM #
243
8. Barkaat September 16, 2017 at 7:21 AM #
243
9. Tawfig September 15, 2017 at 10:19 PM #
243
10. Hamza September 15, 2017 at 7:55 PM #
243
11. Asif September 15, 2017 at 5:43 PM #
243
12. Saleem September 15, 2017 at 5:06 PM #
243
13. Mohamed September 15, 2017 at 3:52 PM #
243
14. Sayeeda September 15, 2017 at 3:24 PM #
243
15. Ali September 15, 2017 at 3:08 PM #
324
• Rshid Motiwala September 15, 2017 at 8:27 PM #
you have to multiply the out come with 3
16. Javed Iqbal September 15, 2017 at 1:42 PM #
It will be 243
17. K C Magon August 1, 2015 at 2:22 PM #
Nil
18. RANA SHOAIB MANZOOR May 9, 2015 at 9:44 AM #
19. mubeen February 15, 2015 at 1:20 PM #
243
20. Rehan February 2, 2015 at 12:42 PM #
125
21. Farouk Kamal December 15, 2014 at 4:33 PM #
The Answer Is 5 = 243
22. Taheruddin August 17, 2014 at 3:03 PM #
243
23. Anwarul kalam August 16, 2014 at 12:50 PM #
5=100
24. Adrenalin March 31, 2014 at 12:19 AM #
3/3*9 = 9
9/3*9=27
27/3*9=81
81/3*9= 243 and that’s it
25. Abdul Waheed February 10, 2014 at 4:17 PM #
(81 x 3) = 243
26. Sana September 19, 2013 at 12:20 AM #
27. Samson September 1, 2013 at 9:07 PM #
3^1=3, 3^2=9, 3^3=27, 3^4=81, so 3^5=243. Ans=243
28. zamena August 16, 2013 at 4:20 AM #
243
29. binu August 14, 2013 at 11:48 AM #
3×1=3
3×3=9
3x3x3=27
3x3x3x3=81
3x3x3x3x3=243
30. khalid August 14, 2013 at 10:44 AM #
ofcource its 243
31. Savad August 13, 2013 at 7:57 PM #
81 * 3 = 243
32. Mohamed Fareed August 13, 2013 at 1:46 PM #
243
33. Najlah August 1, 2013 at 6:46 PM #
1=3
2=9
3=27
4=81
5=243
• ALY August 13, 2013 at 1:24 PM #
243
34. Faruk Besho July 8, 2013 at 4:21 PM #
Must be 243 = 3x3x3x3x3x3
35. Maroof Haroon June 29, 2013 at 5:59 PM #
243 its a commen sense questiin
36. sneh June 18, 2013 at 8:21 PM #
It must be 243
37. ALI AHMED June 16, 2013 at 8:50 PM #
243, it is successive multiplication of 3.
• Rajan February 25, 2014 at 2:37 PM #
243 | 1,155 | 2,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.86319 |
https://oeis.org/A000728 | 1,590,479,177,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00509.warc.gz | 442,029,479 | 4,103 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A000728 Expansion of Product_{n>=1} (1-x^n)^5. (Formerly M3742 N1529) 6
1, -5, 5, 10, -15, -6, -5, 25, 15, -20, 9, -45, -5, 25, 20, 10, 15, 20, -50, -35, -30, 55, -50, 15, 80, 1, 50, -35, -45, -15, 5, -50, -25, -55, 85, 51, 50, 10, -40, 65, 10, -10, -115, 50, -115, -100, 85, 80, -30, 5, 20, 45, 70, 65, 45, -55, -100 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 REFERENCES Newman, Morris; A table of the coefficients of the powers of eta(tau). Nederl. Akad. Wetensch. Proc. Ser. A. 59 = Indag. Math. 18 (1956), 204-216. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Seiichi Manyama, Table of n, a(n) for n = 0..10000 M. Boylan, Exceptional congruences for the coefficients of certain eta-product newforms, J. Number Theory 98 (2003), no. 2, 377-389. MR1955423 (2003k:11071) M. Newman, A table of the coefficients of the powers of eta(tau), Nederl. Akad. Wetensch. Proc. Ser. A. 59 = Indag. Math. 18 (1956), 204-216. [Annotated scanned copy] FORMULA a(0) = 1, a(n) = -(5/n)*Sum_{k=1..n} A000203(k)*a(n-k) for n > 0. - Seiichi Manyama, Mar 26 2017 G.f.: exp(-5*Sum_{k>=1} x^k/(k*(1 - x^k))). - Ilya Gutkovskiy, Feb 05 2018 MATHEMATICA CoefficientList[QPochhammer[x]^5 + O[x]^60, x] (* Jean-François Alcover, Feb 10 2016 *) CROSSREFS Cf. A258405. Sequence in context: A285932 A109064 A138506 * A242895 A242129 A022088 Adjacent sequences: A000725 A000726 A000727 * A000729 A000730 A000731 KEYWORD sign AUTHOR STATUS approved
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Last modified May 26 02:56 EDT 2020. Contains 334613 sequences. (Running on oeis4.) | 788 | 2,142 | {"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} | 2.5625 | 3 | CC-MAIN-2020-24 | latest | en | 0.555897 |
https://numpy.org/doc/1.15/reference/generated/numpy.polynomial.hermite_e.poly2herme.html | 1,719,180,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00110.warc.gz | 375,074,596 | 3,413 | #### Previous topic
numpy.polynomial.hermite_e.herme2poly
Polyutils
# numpy.polynomial.hermite_e.poly2herme¶
`numpy.polynomial.hermite_e.``poly2herme`(pol)[source]
Convert a polynomial to a Hermite series.
Convert an array representing the coefficients of a polynomial (relative to the “standard” basis) ordered from lowest degree to highest, to an array of the coefficients of the equivalent Hermite series, ordered from lowest to highest degree.
Parameters: pol : array_like 1-D array containing the polynomial coefficients c : ndarray 1-D array containing the coefficients of the equivalent Hermite series.
Notes
The easy way to do conversions between polynomial basis sets is to use the convert method of a class instance.
Examples
```>>> from numpy.polynomial.hermite_e import poly2herme
>>> poly2herme(np.arange(4))
array([ 2., 10., 2., 3.])
``` | 209 | 868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.711007 |
http://mathhelpforum.com/calculus/209465-piecewise-integral-question-about-breaking-integrals-up.html | 1,526,925,024,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00020.warc.gz | 198,249,925 | 11,260 | 1. ## Piecewise Integral, question about breaking integrals up
$\displaystyle \int_{-\pi}^{\pi} f(x)dx$ where $\displaystyle f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=<x<0} \\ {7sin(x),} & {\text{if }x~0=<x<=\pi} \\\end{array} } \right.$
I broke up the integral like this $\displaystyle \int_{-\pi}^{0} 6x^{3} dx + \int_{0}^{\pi} 7sin(x)$ . When I look at the graph of these two functions I notice that they overlap each other. Does that affect the area value in any way? I know the antiderivates will be even functions, so I think I only have to find the integral of the 1st and 4th quadrant and then just double it. If I do : $\displaystyle 2 \int_{0}^{\pi} 7sin(x)$ I get 28, but that's not correct. I'm sure I'm just not understanding something about these graphs, that's what I ask if the overlap feature means I need to break up the endpoints of the integral differently?
2. ## Re: Piecewise Integral, question about breaking integrals up
Originally Posted by AZach
$\displaystyle \int_{-\pi}^{\pi} f(x)dx$ where $\displaystyle f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=<x<0} \\ {7sin(x),} & {\text{if }x~0=<x<=\pi} \\\end{array} } \right.$
I do not understand most of your statement. In particular the bit about "overlapping".
Look at this graphic.
3. ## Re: Piecewise Integral, question about breaking integrals up
When I graphed the two functions 6x^3 and 7sin(x) on my TI83 I got a graph that looked like this
Did I have the integrals broken up correctly?
4. ## Re: Piecewise Integral, question about breaking integrals up
Originally Posted by AZach
When I graphed the two functions 6x^3 and 7sin(x) on my TI83 I got a graph that looked like this
Did I have the integrals broken up correctly?
$\displaystyle 6x^3$ is for $\displaystyle x < 0$ and $\displaystyle 7\sin{x}$ is for $\displaystyle x \ge 0$ ... they do not "overlap".
how to graph this piece-wise function in a TI-83 ...
$\displaystyle Y1 = 6x^3/(x < 0)$
$\displaystyle Y2 = 7\sin(x)/(x \ge 0)$
set your x window from $\displaystyle -\pi$ to $\displaystyle \pi$ and y window from -10 to 10
5. ## Re: Piecewise Integral, question about breaking integrals up
Thanks Skeeter, I never knew how to do that before. | 696 | 2,238 | {"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} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.871697 |
https://www.teachingchannel.org/videos/ratios-and-fractions-sbac | 1,529,789,556,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00411.warc.gz | 920,252,194 | 25,434 | # Applying Understanding of Ratios to Fractions
Grades 6-8 / Math / Modeling
CCSS: Math.6.RP.A.2 Math.6.RP.A.3 Math.7.RP.A.2
Common Core State Standards
Math Math 6 Grade 6 RP Ratios & Proportional Relationships A Understand ratio concepts and use ratio reasoning to solve problems 2 Understand the concept of a unit rate a/b associated with a ratio a:b with b not equal to 0, and use rate language in the context of a ratio relationship. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger." Expectations for unit rates in this grade are limited to non-complex fractions. Download Common Core State Standards (PDF 1.2 MB)
Common Core State Standards
Math Math 6 Grade 6 RP Ratios & Proportional Relationships A Understand ratio concepts and use ratio reasoning to solve problems 3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. a. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. b. Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed? c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. d. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities. Download Common Core State Standards (PDF 1.2 MB)
Common Core State Standards
Math Math 7 Grade 7 RP Ratios & Proportional Relationships A Analyze proportional relationships and use them to solve real-world and mathematical problems 2 Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. Download Common Core State Standards (PDF 1.2 MB)
In Partnership with
and
### Lesson Objective
Solve a real-world problem involving ratios and fractions
8 min
### Questions to Consider
• How does Ms. Morey build off of her students' previous understanding?
• Why does Ms. Morey give her students independent think time?
• How does Ms. Morey create opportunities for students to share and learn from each other?
### Common Core Standards
Math.6.RP.A.2, Math.6.RP.A.3, Math.7.RP.A.2
Watch all the videos in this series: | 800 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-26 | latest | en | 0.880573 |
https://cutiumum.net/l-hopital-s-rule-worksheet/ | 1,590,458,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390442.29/warc/CC-MAIN-20200526015239-20200526045239-00384.warc.gz | 285,523,297 | 20,554 | L Hopital S Rule Worksheet
In Free Printable Worksheets246 views
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L Hopital S Rule Worksheet L h 244 pital s rule introduction video 8 52 mins this video introduces the basics of l h 244 pital s rule for evaluating limits when the numerator and denominator of an expression both tend to zero or both The following videos are some of the most commonly asked questions we have heard in our center you can find a whole range of topics at khan academy s pre calculus section in the form of videos and The following videos are some of the most commonly asked questions we have heard in our center you can find a whole range of topics at khan academy s pre calculus section in the form of videos and.
L Hopital S Rule Worksheet
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Top | 2,324 | 9,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.724595 |
http://mathoverflow.net/questions/57426/are-there-any-non-linear-solutions-of-cauchys-equation-fxy-fxfy-wit | 1,469,584,611,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825358.53/warc/CC-MAIN-20160723071025-00204-ip-10-185-27-174.ec2.internal.warc.gz | 126,835,317 | 17,566 | # Are there any non-linear solutions of Cauchy's equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice?
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$
It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in \mathbb{Q}$
It is also known that using an infinite dimensional basis for $\mathbb{R}$ over $\mathbb{Q}$, it is possible using the Axiom of Choice to construct a function which is not linear.
My question is whether there exists a known solution not invoking the Axiom of Choice?
Edit: My questions are:
1. Is it not possible to find an explicit expression for a non-linear solution? Why?
2. Can existence of non-linear solutions be proven from ZF alone?
Reference: Cauchy's functional equation
-
All Lebesgue measurable solutions are linear, so there are no nonlinear solutions if you assume the Axiom of Determinacy (en.wikipedia.org/wiki/Axiom_of_determinacy). – George Lowther Mar 5 '11 at 1:53
Ignas- This question isn't well-defined; are you asking if you can prove the existence of such solutions just ZF? (George shows the answer is no). If you want to edit the question to be clearer, and try posting on meta, you might convince people to reopen it. – Ben Webster Mar 5 '11 at 4:03
Thank you for the comments. Yes, I probably should have been more clear, I actually wanted to ask whether: 1) it is possible to find an explicit expression for such a non-linear function (obvious answer: no, but I was interested in how to prove it isn't possible); 2) existence of solutions in ZF. – Ignas Mar 5 '11 at 18:58
@Ignas: Your edited version is interesting! I have voted to re-open. – Andrés E. Caicedo Mar 5 '11 at 22:34
Ignas:
It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable.
(This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here.)
The point is that it is consistent that every set of reals is measurable, and therefore every function $f:{\mathbb R}\to{\mathbb R}$ is measurable.
The first to realize that it is possible using choice to construct a non-linear additive function was Hamel in 1905 ("Eine Basis aller Zahlen und die unstetigen Losungen der Functionalgleichung: $f(x+y)=f(x)+f(y)$", Math Ann 60 459-462); indeed, a Hamel basis of ${\mathbb R}$ over ${\mathbb Q}$ allows us to provide examples. Note that with a Hamel basis it is straightforward to build Vitali's standard example of a non-measurable set.
Solovay ("A model of set-theory in which every set of reals is Lebesgue measurable", Annals of Mathematics. Second Series (1970) 92 1–56) proved that, relative to the consistency of an inaccessible cardinal, via forcing we can prove that it is consistent with ZF + DC that all sets of reals are Lebesgue measurable (DC is the axiom of dependent choice, a weak version of the axiom of choice useful to develop the basic theory of analysis). It follows that, in Solovay's model, all additive functions are linear and therefore no "explicit" example is possible.
The other standard approach to models where all sets of reals are measurable is via determinacy. However, this requires a stronger commitment in terms of consistency strength (using $\omega$ Woodin cardinals we can force an inner model of determinacy). The two approaches are closely related: Assuming enough large cardinals, the inner model $L({\mathbb R})$ consisting of all sets constructible from reals is a model of determinacy, and a Solovay model. Note that this is a theorem (from enough large cardinals) rather than a consistency result, i.e., $L({\mathbb R})$ is a model of these statements, without needing to pass to a forcing extension.
Solovay's result does require an inaccessible cardinal, this is a result of Shelah ("Can you take Solovay's inaccessible away?", Israel Journal of Mathematics (1984) 48 1–47). This leaves the question of whether (the consistency of) ZF suffices to produce a model where all additive maps are linear. But this is now easily solved: In the same paper, Shelah proved that it is relatively consistent with ZF that all sets of reals have the property of Baire. The standard proofs that additivity and measurability give linearity work with "Baire measurability" instead of Lebesgue measurability.
Let me close, however, by pointing out that, in appropriate models of set theory, we can "explicitly" build additive, non-linear maps. More precisely, in nice inner models, we have explicitly definable well-orderings of the reals, from which such maps can be built. For example, in Gödel's inner model L of constructible sets, there is an easily definable Hamel basis (see this MO question), from which we can easily define such a function. "Easily" is formalized in terms of the projective hierarchy. The Hamel basis we obtain is $\Pi^1_1$ and the function is $\Sigma^1_2$.
(By the way, the statement "there is a discontinuous additive function" is form 366 in Howard-Rubin "Consequences of the axiom of choice". The highly recommended companion website does not reveal any additional choice-like implications regarding this form. In particular, it only mentions Solovay's model for its negation, but not Shelah's. Sadly, the last link does not work currently.)
-
(The existence of a Hamel basis is not exactly what's needed to construct the Vitali set.) – François G. Dorais Mar 6 '11 at 12:16
@François: Thanks. I edited the line in question. – Andrés E. Caicedo Mar 6 '11 at 16:26
wow, thank you for a long and detailed answer, now I feel it was much worth it to request a re-open! – Ignas Mar 6 '11 at 20:35
As of today the link to the short proof of "addive + measurable implies liner" is broken. – Dirk Nov 24 '15 at 14:31
@Dirk Thanks. Updated now. – Andrés E. Caicedo Nov 24 '15 at 22:07 | 1,531 | 6,005 | {"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} | 3.09375 | 3 | CC-MAIN-2016-30 | latest | en | 0.892297 |
https://worksheetsplus.com/EquivExpressions.html | 1,563,678,184,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526818.17/warc/CC-MAIN-20190721020230-20190721042230-00295.warc.gz | 610,433,791 | 3,532 | Integers
Integers on the Number Line
Mean, Median, and Mode
Prism & Pyramid Nets
Properties of Polyhedra
# Equivalent Algebraic Expressions
Addresses the following 1st Grade standard from the Common Core State Standards for Mathematics:
Expressions and Equations 6.EE - Apply and extend previous understandings of arithmetic to algebraic expressions
2. Write, read, and evaluate expressions in which letters stand for numbers.
a. Write expressions that record operations with numbers and withletters standing for numbers.
3. Apply the properties of operations to generate equivalent expressions.
4. Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for.
Write an equivalent expression for each the following:
x + x + x + x 3y (b)(b) 3m + 3n 2(q + r) 3c + 2 a + a + b + b + b (t)(t)(t)(t) t + t + t + t (i + j) + (i + j) + (i + j) w + w + w + w + w + 10 (x + 2) + (x + 2)
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Try Mathville where kids love Math! | 349 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-30 | latest | en | 0.796408 |
http://dionysus.biz/StatIntroCh5.html | 1,723,691,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00832.warc.gz | 8,592,138 | 6,336 | How Do We Compute xBar & s? What are they???
# St. Paul's Statistics Introduction: Chapter 5
### Calculating Mean and Standard Deviation from Data Sets
Section Goals:
• Students will learn and demonstrate methods to compute sample mean (xBar) and sample standard deviation s from datasets.
• Students will understand and be able to explain the difference between xBar and mean μ
• Students will develop conceptual understanding of how μ relates to a PDF
Introduction: In chapter 4, we worked many problems with the Gaussian Distribution and they always required us to know the mean μ (average) and standard deviation σ for the population of 'things' we were analysing. Chapter 4 identified three sources of those values:
1. Very well established theories, such as those in Physics.
2. Copious amounts of data (that's how the Physicists got their values)
3. Values calculated from data samples. When we use this approach, we refer to the sample mean (xBar) and the sample standard deviation (s). We do not use the symbols μ and σ when the values are computed from small to medium data sets.
In this chapter, you will learn how to compute values xBar and s (where xBar is an imposter who pretends to be μ. And s is an imposter who pretends to be σ). We are forced to use imposters xBar and s when we can't get the 'real thing'. In our calculations, we can use 'the imposters' in place of μ and σ.
Computing xΒar (imposter for μ): Please recall that μ is the average value for every item belonging to the population. If we are talking about the length of large construction nails, then μ is the average of length for every 16d size nail produced from Columbus up through the 22nd century. We would measure and add up all the lengths, and then divide the answer by the number of nails.
If we are talking about the weight of Cows, then μ is the average weight for every cow from the time of Moses up through the 22nd century. We would add up all the weights, then divide by the number of cows. We would need a time machine go back and weigh every cow because Moses lived a very long time ago. For most populations, it simply is not possible to get population data to compute μ. That is the reason we have to work with sample values xBar and σ.
So, how do you find the mean of x? Just add up all the x values, and then divide by their number. If you use 'all of them', then you get μ. If you use a sample of them, you will get xBar.
Computing Sample Mean xBar for Peanut Butter: Table 1 below contains price data for various brands of peanut butter sold in a one pound jar. The brands marked *, are fictional and were added to more effectively illustrate concepts. All other data was obtained from U.S. government data.
Brands & Prices for Peanut Butter
Chapter 5 Table 1
We can compute the sample mean 'xBar' based on the 19 prices in Table 1. Likely, you know the method, but here is the example:
xBar = Σpricei / count = (2.50+12.49+2.50+3.00+4.00+ …….. \$3.00+\$4.00)/19 = \$4.22
Presenting Data - A Better Way: From Table 1, it is not obvious, but prices range from \$1.50 to \$12.99 for a jar of peanut butter. It is not obvious, but Table 1 has 6 products priced at \$2.50.
• There are better ways of presenting the data so these observations are obvious.
• The data can be presented so we can more quickly find the sample mean.
Sort Table 1 from the cheapest to most expensive. The price trends will become obvious. Repeated entries likewise become obvious and we can compute 6 x \$2.5 rather than pressing the plus sign repeatedly. The sorted version of Table 1 is shown below:
Peanut Butter - Sorted by Price
Chapter 5 Table 2
Further simplifications are possible. To compute xBar (the imposter for μ), we don't need the names of the peanut butter and we don't need jar size. Repeat entries are handled as shown in Table 3 below:
Price Data Sample Tabulated by Count
Chapter 5 Table 3
An Easier Way to Compute xBar: Based on Table 3, we can compute xBar either by calculator or by computer spreadsheet. We simply multiply 'Count' by 'Price' for each row in Table 3. Typically, these values are inserted into a new column to the right. Then, we 'total down' the new column. Finally, divide by the total by count of data items (19 in this example). Confused? We are just adding up all the Price numbers of Table 2 in a smarter way!
For a table with 19 data entries, it is not very important how you do it. But in the real world, statistics problems often have 150 data rows. For situations like this, it is impossible to look at the data table and draw any kind of conclusion. The data table must be sorted and repeat entries should be documented in a 'Count' column (like Table 3). This approach also makes computation of xBar much faster.
Reduced to math notation, this 'easier way' to compute sample mean looks like this:
xBar = ( Σall i (counti * pricei ) ) / Total Number
In spreadsheet form, the solution looks like this:
Spreadsheet Form of Simplified Mean Computation
Chapter 5 Table 4
Homework Problems Chapter 5 :
For problems 1 through 4, use the data set to do the following. You may use either calculator or computer spreadsheet. Show all work.
• Read summary description of the problem.
• Sort the data into order (if it is not already)
• Group similar values (if any) and add 'count' column for repeated data values
• Do the math to compute sample mean xBar. Document all work & turn in to your instructor (if you have one)
Chapter 5 Problem 1:
A Washington 'think tank' is encouraging larger budgets for health research. They want to statistically describe U.S. child deaths from Flu. Data from National Institute of Health follows
Child Deaths from Flu
Original Source: National Institute of Health
Chapter 5 Table 5
Chapter 5 Problem 2:
A civil engineering firm is constructing a dam. The concrete has a minimum strength requirement. Test specimens from the first 17 concrete trucks were made and the strength test results are shown below. Follow the instructions and compute mean sample strength. (Data was obtained from source w/o copy-right notice.)
Concrete Strength Test Results
Chapter 5 Table 5
Banded Data: Very often, data is tabulated as huge lists of items with one entry per row. For example: Paul bought a pair of size 9 shoes → that appears as one data row. Tommy bought a pair of size 9 shoes → that is a different data row. When data items arrive as separate line items, but clearly many of the entries are similar, it makes sense to sort the data, and then group all the size 9 purchases together. It is very common to arrange data into 'bands' that fall into certain value ranges. For shoes, it seems very natural; but 'banded' data is quite common even when the 'x' variable seems continuous and repeats in the data are not evident. We will explore this topic when we study histograms in Chapter 6.
For problems 3 and 4, the data has already been sorted and grouped for you. Ignore the columns of data you don't need. Compute the mean of the data set.
Chapter 5 Problem 3:
Shoe manufacturers are very interested in knowing the percentage of each shoe size sold so they can manufacture shoes in the proportions demanded by the market. Below, is a sorted data set that shows a sample of women's shoe purchases. Find the average.
Woman's Shoe Sales by SIZE
Chapter 5 Table 7
Chapter 5 Problem 4:
Once again, we look at shoe purchases, but this time it is men's shoes. Below, is a sorted data set that shows a sample of men's shoe purchases. Find the average.
• Original Source: https://www.quora.com
• Data adjusted for equivalent US and European sizes.
Man's Shoe Sales by Size
Chapter 5 Table 8
How is the Mean μ Related to the PDF? I offer a few PDF pictures with the mean shown on the graph. The mean of the PDF graph is always the balancing point! These four images illustrate that idea.
A Gaussian PDF Mean
Chapter 5 Figure 1
A Symmetric Triangular PDF Mean
Chapter 5 Figure 2
A NonSymmetric Triangular PDF
Chapter 5 Figure 3
NonSymetric PDF (Possibly Weibull)
Chapter 5 Figure 4
Conceptually, the mean μ will always be at the balance point of the PDF. For right-left symmetric PDFs, the mean will always be in the middle (as is the balance point). When doing math, we don't want to use 'eye-ball' estimates of the mean. We want accurate calculations based on samples. But it is useful for you to be able to look at a PDF and compare it with your calculated answer. Then you confidently judge whether or not your answer looks right.
Mid Chapter Summary:
• We have learned how to calculate the sample mean xBar (which is often used in place of μ because we don't have a really good knowledge of the true μ value for the population).
• We have learned that we can look at a PDF, and make a good guess where the balance point is. And that location is also the mean.
• Those two ideas are pretty clear; but, there does seem to be a mystery. How do 19 peanut butter prices turn into a nice smooth curve likes Figures 1 through 4 above? We will explain the mystery of 'smoothing peanut butter' in the next chapter!
The Variance σ and imposter s:
We know from Chapter 2 that the standard deviation σ determines 'how wide' the Gaussian PDF is. Chapter 2 Figure 3 is repeated below to refresh your memory. From the figure below, you should get the idea that:
Bigger Standard Deviation σ = Wider PDF Curve
3 Different GAUSS PDFs with Difference Variance
Ch 5 Figure 5
Note: The three Gauss PDFs shown above are skinny, to wide because the standard deviation σ is small to large. The curves are shifted to the right for a different reason – because the mean value is different. Next, we will learn how to compute a sample standard deviation from a set of data.
Computing s (imposter for standard deviation σ): In the real world, you usually won't know the standard deviation σ. You will have to compute an estimate based on a sample of data. A sample based variance is usually denoted 's' which is a way of reminding us that it is not the true, population value σ. s is an imposter that pretends to be the variance σ and we can use it in calculations in place of σ. In mathematical notation, the following equation defines how the sample variance s is computed:
Steps to Compute Standard Deviation s:
Read through the steps, and study how they accord with the equation above:
1. Obtain a sample of data. It might be a list showing 9 cows. For each cow, a property of interest (like milk per day) will be documented. e.g. cow No 1 → 8.2 gal etc. for every cow
2. Compute the sample average value xBar using methods presented earlier in this chapter
3. For each data item (i.e. for each value xi) , we subtract xBar from it. And square the result.
4. Next, add up all the squared results
5. Divide by the number of data entries (n)
6. Take the square root
Example:
Calculating Sample Standard Deviation
Chapter 5, Figure 5
Homework Set 2 Problems Chapter 5:
Problems 5, 6, 7, & 8: Computation of Sample Standard Deviation s:
For each of the data sets of Problems 1,2,3 & 4, Compute the sample mean and sample standard deviation using the method shown above. You may use either calculator and paper, or spreadsheet.
Turn in the results to your instructor.
## End of Chapter 5
Contact the author paul-watson@sbcglobal.net by e-mail. | 2,662 | 11,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.939386 |
http://research.stlouisfed.org/fred2/series/RGDPCHBHA625NUPN | 1,417,244,130,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931013466.18/warc/CC-MAIN-20141125155653-00228-ip-10-235-23-156.ec2.internal.warc.gz | 246,791,049 | 19,019 | # Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Bahrain
2010: 23,101.09418 2005 International Dollars per Person (+ see more)
Annual, Not Seasonally Adjusted, RGDPCHBHA625NUPN, Updated: 2012-08-31 2:42 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: rgdpch
Source: University of Pennsylvania
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(a) Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Bahrain, 2005 International Dollars per Person, Not Seasonally Adjusted (RGDPCHBHA625NUPN)
Integer Period Range: to copy to all
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` University of Pennsylvania, Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Bahrain [RGDPCHBHA625NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/RGDPCHBHA625NUPN/, November 29, 2014. ```
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# Venndiag
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### Venndiag
1. 1. VENN DIAGRAMS ANDCATEGORICAL PROPOSITION
2. 2. 1. An empty circle is used torepresent a subject class or apredicate class and is generally solabeled with an S or a P. Puttingthe name of the actual subject orpredicate class next to the circle ispreferred.
3. 3. 2. Shading or many parallel linesare used to indicate areas which areknown to be empty. I.e., there areno individuals existing in that area.E.g., the diagram to the rightrepresents the class of "Yeti."
4. 4. 3. The third symbol used is an "X"which represents "at least one" or"some" individual exists in the areain which it is placed. The diagramto the right indicates "some thing."
5. 5. Venn Diagrams in General 1.Universal affirmative proposition1. The A form, "All S is P," isshown in the diagram to the right.Notice that all of the Ss are pushedout, so to speak, into the P class. IfSs exist, they must be inside the Pcircle since the left-hand lune ofthe diagram is shaded and so isempty.
6. 6. 2. Universal negative proposition2. The E form, "No S is P," isshown in the diagram to theright. Notice that the lens areaof the diagram is shaded and sono individual can exist in thisarea. The lens area is where Sand P are in common; hence,"No S is P." All S, if there areany, are in the left-hand lune,and all P, if there are any, arerelegated to the right-hand lune.
7. 7. 3. The I form, "Some S is P," ismuch more easily seen. The "X" inthe lens, as shown in the diagram tothe right, indicates at least oneindividual in the S class is also inthe P class.
8. 8. 4. The O form, "Some S is not P,"is also easily drawn. The S that isnot a P is marked with an "X" inthe S-lune. This area is not withinthe P circle and so is not a P. It isworth while to note, that from thisdiagram we cannot conclude that"Some S is P" because there is no"X" in the lens area. Thus, studyingthis diagram will explain why"Some S is not P" does not entail"Some S is P."
9. 9. SYMBOLIC LOGIC
10. 10. SYMBOLSSymbols comprise every language. Per se,symbols are effective tools of humanactivities whether in social interaction orin the search for knowledge.
11. 11. This part will help us understand Symbolic Logic, itsbasic components and structures, symbols and fucntionsof statements constituting a basic argument.This process of proving validity is an indispensable tool torecognize the universal patters of valid arguments.
12. 12. Truth-Functional Operators Four types of truth-functional compounds, 1. Conjunctions (Conjunctive Proposition)Here are some words that in many standards uses yieldconjunctions; _____and_______ Both____and________ ______but_______ ______yet_______ ______although_______ ______whereas________ ______while________
13. 13. Manuel is strong and Gina is pretty.Both Manuel is strong and Gina is pretty.Manuel is strong but Gina is pretty.Manuel is strong yet Gina is pretty.Manuel is strong although Gina is pretty.Manuel is strong whereas Gina is pretty.Manuel is strong while Gina is pretty.
14. 14. 2. Disjunction (Disjunctive Proposition)Here are some words that in many of their standard use yielddisjunctions.Either_____or_____________or_____________unless_______Either Manuel is strong or Gina is pretty.Manuel is strong or Gina is pretty.Manuel is strong unless Gina is pretty.
15. 15. 3. Implication (Conditional Proposition) If ____then______ If____, _____ _____only if_____ _____if______ _____provided that______ not____unless_______If Manuel is strong then Gina is pretty.If Manuel is strong, Gina is pretty.Manuel is strong only if Gina is pretty.Gina is pretty if Manuel is strong.Gina is pretty provided that Manuel is strong.Manuel is not strong unless Gina is pretty.
16. 16. 4. Material Equivalence (Bi-Conditional Proposition) Here are some phrases that in many of their standard uses yield material equialence: _______if, and only if,_______ _______when, and only when,______ _______is equivalent to _______These phrases can result into material equivalences, such as: Manuel is strong if, and only if, Gina is pretty. Manuel is strong when, and only when Gina is pretty. Manuel is strong is equivalent to Gina is pretty.
17. 17. Negation (Contradictory or denial)It is not the case that______It is not true that_______There is no way that______________is false.It is false that_______It is not the case that Manuel is strong.It is not true that Manuel is strong.There is no way that Manuel is strong.“Manuel is strong” is false.It is false that Manuel is strong.
18. 18. End | 1,193 | 4,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-13 | latest | en | 0.912915 |
http://childhealthpolicy.vumc.org/zawemi59326.html | 1,680,403,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00358.warc.gz | 10,171,990 | 6,004 | # Shug and celie relationship. How does Shug Avery and Celie's relationship evolve and better their lives in the novel The Color Purple? 2022-10-04
Shug and celie relationship Rating: 8,5/10 794 reviews
The pigeonhole principle, also known as the "boxes and pigeons" principle, is a simple but powerful concept in mathematics that states that if there are more objects than available spaces (or "pigeonholes"), then at least one space must contain more than one object. This principle has many applications in various fields, including computer science, economics, and even daily life.
One of the most common applications of the pigeonhole principle is in computer science, specifically in the field of data compression. In data compression, the goal is to represent a large amount of data using a smaller number of bits. One way to do this is by using a technique called "lossless compression," where the original data can be recovered exactly from the compressed version. The pigeonhole principle can be used to prove that certain lossless compression schemes are optimal, meaning that no other scheme can compress the data more efficiently. For example, suppose we have a set of data consisting of the letters A, B, C, and D. If we want to represent this data using only 2 bits per letter, we can use the pigeonhole principle to prove that at least one of the letters must be represented by two different combinations of 2 bits. This means that the data cannot be losslessly compressed using 2 bits per letter, and we must use a different method or a higher number of bits to achieve optimal compression.
Another application of the pigeonhole principle is in economics, specifically in the study of market equilibrium. Market equilibrium occurs when the quantity of a good or service that is being supplied is equal to the quantity that is being demanded. The pigeonhole principle can be used to prove that under certain conditions, market equilibrium is always possible. For example, suppose we have a market for a certain type of good, and there are three sellers who each have a certain number of units of the good to sell. The pigeonhole principle states that if the sellers have a total of more than three units of the good, then at least one of them must have more than one unit to sell. This means that there must be at least one buyer who is willing to purchase more than one unit of the good, which is necessary for the market to reach equilibrium.
In daily life, the pigeonhole principle can also be used to solve practical problems. For example, suppose you have a group of friends who are going on a road trip, and you need to decide which car to take. You have three cars to choose from, each with a different number of seats. The pigeonhole principle states that if you have more friends than the total number of seats in the three cars, then at least one of the cars must have more than one person in it. This can help you decide which car to take, and also serve as a reminder to carpool to save space and reduce environmental impact.
In conclusion, the pigeonhole principle is a simple but powerful concept that has many applications in various fields, including computer science, economics, and daily life. Its versatility and simplicity make it a valuable tool for solving a wide range of problems.
## What is the relationship between Celie and Shug?
It is Shug Avery who forces Albert to stop brutalizing Celie, and it is Shug with whom Celie first consummates a satisfying and reciprocally loving relationship. Celie has suffered psychological damage through verbal abuse, physical abuse and sexual violence all her life. Walker also sets most of her novel in a rural farm community, focussing on the personal lives of her characters. Only once traditional gender roles have been broken down can men and women alike find fulfilment and freedom. Sewing a quilt symbolises the coming together and bonding of friends and family. When Celie is about to shave Albert, like earlier in the movie, she plans to cut his throat and kill him, but Shug runs back to the house and is able to stop her just in time. When Celie first lays eyes on Shug Avery, it is through a photograph of her.
Next
## What was the relationship between Celie and Shug?
Celie explicitly comments that Shug made her feel a sexual bliss that no man had ever bothered to try to help her express. Shug clearly enjoys the care and attention she is getting and returns the same care to Celie, showing her compassionate nature. The neglect from her mother and the sexual abuse from her father are things that do not strike her as odd or wrong because she did not know that there was something she could have done about it and also that she was a fourteen years old young girl who was not allowed to go to school as her father did not deem her fit to be educated. Women were also inferior to men, both black and white. She is made to feel unattractive and unintelligent by her Pa. Womans were besides inferior to work forces, both black and white. Shug helps to give Celie a sense of identity making her feel sexually, physically and emotionally at ease.
Next
## Was Celie and Shug Avery in a relationship?
They defend themselves against men and do not allow men to choose their lifestyle for them. Shug is actually a warm and caring person. In the play A Raisin In The Sun by Lorraine Hansberry. Shug's time away from Celie made her realise how much she loves Celie and brought them both closer together. The main character in the novel is affected by her physical surroundings, in the way of people, which ultimately construct who she is as a person. Celie had to learn the hard way that love was not always easy and often caused heartache. When Celie foremost lays eyes on Shug Avery, it is through a exposure of her.
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## The Relation Between Celie And Shug And Its Evolution In The Color Purple By Alice Walker: Free Essay Example, 1349 words
When Shug first sees Celie, she called her ugly and laughed in her face. She is made to experience unattractive and stupid by her Pa. Shug helps to give Celie a sense of identity making her feel sexually, physically and emotionally at ease. How long was Sophia in jail? Her father tells her 'You better not never tell nobody but God' Walker. Celie saw a picture of a woman who she thought was the most beautiful woman she has ever seen; her name was Shug Avery. The most remarkable aspect about their relationship is how Shug enables Celie to transform from a submissive and immature person to a mature, independent woman who can take decisions for herself and has a goal in her life in the latter half of the story. Albert realises for the first time that Celie is good company and Celie equally enjoys her friendship with Albert.
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## Celie And Shug Relationship
This clearly shows how much Celie means to Shug and proves her loyalty to Celie. When Sofia and Celie argue about the advice which Celie had given to Harpo, Sofia suggests they make a quilt as a way of armistice. They defend themselves against work forces and do non let work forces to take their life style for them. Racism was disregarded throughout the country and the laws in the South implemented segregation. This also symbolises motherhood because Shug is the reason Celie gains a sense of importance in the novel. Also writing words as she would pronounce them, for example 'direar and newmonya'.
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## The Relationship Between Celie And Shug In The Color Purple By...
However, Celie steps in to defend her sister. Mister is a cruel husband who berates Celie constantly, treats her like a servant who is meant to keep her house in order, and assaults her. Shug is regarded as a metaphorical missionary in Celie's life, like the missionaries in the Olinka. Prior to being in an arranged marriage, her step-father had severely abused her -- Mister is no different when Celie comes to live with him. Why does Celie feel attraction and love for Nettie? For Celie, creating was never something she thought she was supposed to do and all she was entitled to do was to follow orders. It is Shug Avery who forces Albert to stop brutalizing Celie, and it is Shug with whom Celie first consummates a satisfying and reciprocally loving relationship.
Next
## How is the relationship between Celie and Shug?
Does Albert beat Celie? Sewing also symbolises the power women get from channelling their creative energy. When Sofia and Celie argue about the advice which Celie had given to Harpo, Sofia suggests they make a quilt as a way of armistice. Shug helps to give Celie a sense of individuality doing her feel sexually, physically and emotionally at easiness. One of the main characters, Celie, suffered a life of sexual violence, and both verbal and physical abuse which resulted in a belief of self worthlessness. Celie grew up abused by her father only to be forced into an abusive marriage. The relationship between Celie and Shug begins to develop as Celie becomes more involved in nursing Shug. It is the relationship between the bruise and the hope, the suffering and the rich nature of her life, that comprise the meanings of the title.
Next
## Remembering 'The Color Purple' as a Queer Story
She thinks she looks very glamorous and instantly begins to take a liking to her. For some people happiness can be found in their dreams. ¦Shug radius right up for you, Celie. Women were also inferior to men, both black and white. Raped by her stepfather, Celie is powerless to his abuse.
Next
## Empowering Sexual Relationship Between Celie and Shug in The Color Purple: [Essay Example], 1298 words GradesFixer
When Sofia and Celie argue about the advice which Celie had given to Harpo, Sofia suggests they make a quilt as a way of armistice. And Walker has been accused of reinforcing racial stereotypes in her depiction of male black characters as abusive and violent. The relationship Shug and Celie had made Celie feel important. . How long was Sophia in jail? Celie lives on a farm in rural Georgia where she becomes a child bride at fourteen years old and marries a man only referred to as Mister. Walker besides sets most of her novel in a rural farm community, concentrating on the personal lives of her characters. Albert realises for the first time that Celie is good company and Celie equally enjoys her friendship with Albert.
Next | 2,210 | 10,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.944077 |
https://ourfamilycode.com/circle-algorithm-art/ | 1,716,544,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00356.warc.gz | 377,804,151 | 31,540 | # Circle Algorithm Art Unplugged Coding Activity
This circle algorithm art activity introduces basic coding skills by giving kids a set of rules and steps to follow to create unique designs in each circle!
An algorithm is a set of specific steps that you can follow to solve a problem. In programming algorithms are used to create solutions to problems that can be reused, like when sorting or searching for something.
One way to teach your children about algorithms and unplugged coding is to have them create algorithm art drawings.
## Circles & Pi
Pi, or π, is the ratio of the circumference of a circle to the diameter. In case you need a refresher, the circumference of a circle is the distance all the way around and the diameter of a circle is the distance across the middle.
Pi begins 3.1415 and mathematicians have spent thousands of years studying this irrational number. Trillions of digits of Pi have been identified so far!
We love to geek out with Pi activities this time of year and this circle algorithm art is a great way to incorporate Pi with tech + art!
You’ll also enjoy: Pi Necklace Coding Unplugged Coding Activity
What are the first 100 digits of Pi?
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679
What is Pi?
Pi is the ratio of the circumference of any circle to the diameter of that circle. The approximate value of Pi is 3.14.
Who discovered Pi?
Archimedes of Syracuse was the first person to calculate Pi.
Does Pi ever end?
Pi is an irrational number and has no final digit.
## Circle Algorithm Art Drawing – Follow the Rule
The Rule: Every circle must be unique.
This means that each circle must be different. It doesn’t mean that the same line pattern cannot be reused. It just means that if horizontal lines, for example, are used twice then the colors must be different.
## Circle Algorithm Drawing – Use the Algorithm
Decorate each circle according to the instructions in the algorithm. The circle algorithm for this activity is:
1. Draw lines
2. Use more than 3 lines
3. Use at least 2 colors
This algorithm applies to each circle. To work through the activity, children should follow the complete algorithm before moving on to another circle. Every circle should look different.
## How to Solve the Circle Algorithm Art Activity
Start with the first circle and follow the algorithm listed. There must be more than three lines used in the circle and there must be at least two colors used.
After you have colored all circles, make sure that you have followed the rule and produced 9 different circle designs. You’ll find an example of possible answers for this activity below.
## Share your circle algorithm art designs!
Once you’ve decorated your circles, send us a picture of your circle designs on Facebook or Instagram
## WHY ARE UNPLUGGED CODING ACTIVITIES IMPORTANT?
Unplugged coding activities are designed to build the foundation of coding. These activities allow kids to participate in kinesthetic opportunities that help them relate the concepts they are learning to their own lives and teach children how to think logically about objects, how to break down large tasks into smaller tasks that are easier to complete, and how to identify errors.
Working hands-on makes coding concepts tangible and unplugged coding activities are ideal for young coders. Unplugged activities are great for classrooms or homes that don’t have access to the internet or a computer and ideal for young students who don’t have experience working with computers.
It’s never too early to start teaching the foundation of coding.
## WHAT IS COMPUTATIONAL THINKING?
Computational thinking can be used to solve problems in almost all areas of our lives and helps kids develop some pretty great life skills that can apply to a variety of situations.
A computational thinker approaches problems with a hands-on approach and playing to solve a problem. They work together with others to reach common goals and persevere when faced with difficult problems.
Computational thinkers find and fix errors in complex problems (debug!) and design solutions for open-ended problems.
We love incorporating books into our activities. Here are some great books about coding to read with your activity!
## SIMILAR UNPLUGGED CODING ACTIVITIES
PIN THIS IMAGE TO SAVE THIS ALGORITHM CIRCLE COLORING ACTIVITY
## Find more Pi Day Activities
Check out these great STEAM Pi Day activities for kids that pair math with technology, art, engineering, and science!
## Meet Toni, the Maker Mom behind Our Family Code
Hey there, I’m Toni! I’m a software engineer and Maker Mom that finds my joy in unleashing my children’s curiosity by exploring STEAM concepts with my fantastic five!
When I’m not chasing toddlers or raising tweens, you can find me tearing things up and putting them back together over here at Our Family Code.
I am the owner and content creator of multiple educational websites designed to increase access to STEAM & STEM education with a focus on teaching computer science and coding to kids of all ages!
You can also find out more about me by visiting ToniGardner.com!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,093 | 5,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-22 | latest | en | 0.871048 |
http://mathhelpforum.com/algebra/4127-help-simplify-please-print.html | 1,526,855,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00584.warc.gz | 181,872,919 | 3,272 | • Jul 13th 2006, 11:08 AM
Ronis
How would I simplify $\displaystyle (144a^8b^2)^1/2$
I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help
• Jul 13th 2006, 11:15 AM
ThePerfectHacker
Quote:
Originally Posted by Ronis
How would I simplify $\displaystyle (144a^8b^2)^1/2$
I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help
It means square root,
$\displaystyle \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b$
-----
Note: (this is not for you) I know that $\displaystyle \sqrt{b^2}\not = b$, so do not bother me with that. Because in schools terms are positive.
• Jul 13th 2006, 11:19 AM
Ronis
Quote:
Originally Posted by ThePerfectHacker
It means square root,
$\displaystyle \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b$
-----
Note: (this is not for you) I know that $\displaystyle \sqrt{b^2}\not = b$, so do not bother me with that. Because in schools terms are positive.
{s con/con} (lol) thanks =]
• Jul 13th 2006, 11:19 AM
Quick
Quote:
Originally Posted by ThePerfectHacker
Note: (this is not for you) I know that $\displaystyle \sqrt{b^2}\not = b$, so do not bother me with that. Because in schools terms are positive.
Did you think one of us would correct you? Tsk, tsk.
BTW. to make an exponent with more than one number all you need to do is wrap it in { }
P.S. I would've corrected you hacker :D
• Jul 13th 2006, 11:24 AM
Soroban
Hello, Ronis!
You have to enclose the entire exponent in braces
. like this: (144a^8b^2)^{1/2}
You can even write: (144^8b^2)^{\frac{1}{2}}
Quote:
How would I simplify $\displaystyle (144a^8b^2)^{\frac{1}{2}}$
Each factor is raised to the ½ power: .$\displaystyle (144)^{\frac{1}{2}}(a^8)^\frac{1}{2}}(b^2)^{\frac{ 1}{2}}$
Since $\displaystyle 144 = 12^2$, we have: .$\displaystyle (12^2)^{\frac{1}{2}}(a^8)^{\frac{1}{2}}(b^2)^{ \frac{1}{2}}$
Use that third property of exponents: .$\displaystyle 12^{(2\cdot\frac{1}{2})} \cdot a^{(8\cdot\frac{1}{2})}\cdot b^{(2\cdot\frac{1}{2})}$
. . And we get: .$\displaystyle 12^1\cdot a^4\cdot b^1\;=\;12a^4b$
• Jul 13th 2006, 11:31 AM
Quick
Just for reference...
$\displaystyle \sqrt[2]{x}=x^{\frac{1}{2}$
also,
$\displaystyle \sqrt[3]{x}=x^{\frac{1}{3}$
you can remember the rule,
$\displaystyle \sqrt[a]{x}=x^{\frac{1}{a}$
also the rules of powers:
$\displaystyle x^a\times x^b=x^{(a+b)}$
$\displaystyle (x^a)^b=x^{(a\times b)}$
$\displaystyle x^{-a}=\frac{1}{x^a}$ | 936 | 2,561 | {"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} | 3.9375 | 4 | CC-MAIN-2018-22 | latest | en | 0.833034 |
http://astro.uchicago.edu/cara/outreach/resources/chi93/summer/candle.html | 1,542,142,204,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741491.47/warc/CC-MAIN-20181113194622-20181113220622-00321.warc.gz | 24,953,381 | 3,730 | # Labs from Chicago, Summer 1993 : Measuring the Wattage of the Sun and a "Standard Candle."
Dr. Jim Sweitzer
Labs written for the CARA Space Explorers, Summer 1993
This is meant to be handed out to the students.
## Introduction
By now you should now know quite a bit about light and the way it behaves. Last week you determined the distance between the Earth and the Sun. This distance is called the astronomical unit and is equal to 150,000,000 kilometers(*). Today you and your partner are going to put this knowledge together to actually determine the power (in watts) radiated by the Sun. Think of it, by simply applying the "inverse square law" you will be able to determine the energy output of the greatest powerhouse in the entire Solar System!
* Footnote: Let's put this in some other forms. Since there are 1,000 meters in a kilometer and 100 centimeters in a meter, the distance to the Sun in centimeters is = 150,000,000 X 1,000 X 100 = 15,000,000,000,000 cm. A good shorthand is to write this distance as = 15 x 10 cm. Where you simply count the number of zeros and place them in the exponent after the 10. This is called scientific notation and sure makes life easier.
But we're not going to stop here. Using the same technique used to get the wattage of the Sun, you will also determine the wattage of a very small light bulb. The reason for this is that we will then take this very small light bulb up to Yerkes Observatory and compare it to stars in the sky. It will be our reference standard. Astronomers sometimes call such things "standard candles," so that's what we'll call it. Once we know it's wattage along with how far away it is, we will be able to get the distances of a star like the Sun, by simply assuming that the star is the same wattage as the Sun. It seems odd, but with just a couple of measurements today, we will be all set to determine some of the greatest distances in the Universe.
## Inverse Square Law (again)
Our theoretical tool will be the Inverse Square Law. It is very powerful, even though it seems quite simple. For today's lab we will write it as the following equation:
[Intensity of source of light at a distance] =[Power output of source]/[Distance to source]
Written as symbols this is: I = W/D
To use this equation today, we will want to solve for W. That is actually very simple. All we have to do is multiply both sides of the equation by D. Thus, the wattage of the sources of light we are investigating is simply W = ID. This would be great if we could just directly measure I, the intensity. Well, it's not that easy. What is simple, however, is to compare the light from two sources and determine that the intensity is the same from both sources. So, we'll build a comparison photometer to tell us when the light from two sources is the same. That way, if we know the wattage and two distances, we can solve for the other unknown wattage.
It works like this: Suppose we have two sources of light that are in the right place to emit the same amount of light. If source number 1 is given by the equation below with 1's in parentheses and the quantities for source 2 are shown with 2's in parentheses, then: I(1) = W(1)/D(1) and I(2) = W(2)/D(2)
But, since we will match the two intensities so that they are equal with the comparative photometer, then I(1) = I(2). This allows us to write just one equation from two, but setting the I's equal to one another. We get W(1)/D(1) = W(2)/D(2)
This equation allows us to solve for a wattage (W(2)) if we know the two distances and the wattage of source 1 (W(1)). All we have to do is multiply by the denominator on the right side. That gives us the final equation we will use today. [W(1)/D(1)] * D(2) = W(2)
## Procedure
To make life easier, you'll find a table to fill in your work as you go along. Here's the basic procedure.
1. First do the activity labeled Part A CONSTRUCTING A SIMPLE PHOTOMETER.
2. Then, place the 100 watt bulb about 2 meters away from the two bulbs together and measure the wattage of the 2 bulbs with the photometer.
3. Then, take your photometer and a short ruler up to the penthouse level of the building and measure the Sun's wattage in using the activity labeled Part B. Again, use your work sheet.
4. Finally, come back inside and measure the wattage of the little "standard candle" using the same procedures. Record your results on the work sheet.
5. When you've taken all the measurements and done all the computing, then average your results for each source and write the averages at the bottom of the page.
## Connections to Yerkes Summer Institute
Once you have done all this, then you are ready to make more similar measurements at Yerkes. We will keep using the Inverse Square Law, but this time will solve it for the distance. If there's any time left today, take the second to last equation above and see if you can't solve it for D(2). Write your work in the space below.
## Work Sheet
[The worksheet doesn't translate into hypertext very well at all. I've tried to make clear what it used to look like. -- ed.]
Recall how the different symbols are defined:
W(source) = the power output of "source" -- units are watts
D(source) = distance from photometer to "source" -- units are centimeters
I(source) = intensity of light from "source" -- units are watts/(cm2)
Record your observations, computations and results in this table. Note that you must repeat your observations four times for each set of sources. Average your final results for each of the three unknown sources and write the final answers in the table at the bottom of the page
```source 1 D(source 1) D(source 1)^2 I(source 1) =
(cm) (cm^2) W(1)/D(1)^2.
(Watts/cm^2)
100 watt bulb
trial 2
trial 3
trial 4
200 watt bulb
trial 2
trial 3
trial 4
100 watt bulb
trial 2
trial 3
trial 4
source 2 D(source 2) D(source 2)^2 W(source 2) =
(cm) (cm^2) I(1)*D(2)^2.
(Watts)
2 bulbs together
trial 2
trial 3
trial 4
Sun
trial 2
trial 3
trial 4
"standard candle"
trial 2
trial 3
trial 4
```
Summary Results (Average wattage) for each source:
2 BULBS TOGETHER = ________WATTS,
SUN = ______________WATTS,
STANDARD CANDLE= ______ WATTS | 1,574 | 6,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.940923 |
https://www.homeworklib.com/question/1553809/part-a-what-volume-of-0173-m-na3po4-solution-is | 1,657,114,371,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00549.warc.gz | 841,501,228 | 10,155 | Question
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http://math.stackexchange.com/questions/54551/finding-the-volume-of-a-solid-of-revolution | 1,430,142,242,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246658376.88/warc/CC-MAIN-20150417045738-00283-ip-10-235-10-82.ec2.internal.warc.gz | 168,844,887 | 16,110 | # Finding the volume of a solid of revolution [on hold]
Could someone help me with this?
Find the volume of the solid generated by revolving about the y-axis the region bounded by the upper half of the ellipse $$x^2/a^2 + y^2/b^2 = 1$$ and the x-axis, and thus find the volume of a prolate spheroid.
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## put on hold as off-topic by Jonas Meyer, Chappers, graydad, Jeel Shah, Peter Woolfitt2 days ago
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Since you are new: (i) Titles should be informative, not pleas for help; (ii) Many of us consider it very rude for people to post in the imperative (giving orders, assigning problems), as if you were the professor, we the students, and you are giving us homework to do. If you must quote, then please place it in a quote box. But in any case, you should always include (iii) the context (what course is this for? is this homework? Is this self-study? what? If it is homework, please add the [homework] tag); (iv) Say what you have managed to do, or where you are stuck and why; (cont) – Arturo Magidin Jul 30 '11 at 3:55
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"$a$ and $b$ are positive constants, with $a > b$ "??? My magic 8-ball might be off, though... – The Chaz 2.0 Jul 30 '11 at 4:23
Since you are new here, you may want to read our FAQ on homework-type questions. – Willie Wong Jul 30 '11 at 4:26
Hint: The volume is $\int_{-a}^a \pi y^2\;dx$. – André Nicolas Jul 30 '11 at 4:58
We use the Method of Slicing. Let $f(x) \ge 0$ on the interval $[a,b]$, and let $R$ be the region below the curve $y=f(x)$, above the $x$-axis, from $x=p$ to $x=q$. Then the volume of the solid obtained by rotating $R$ about the $x$-axis is $$\int_p^q \pi(f(x))^2\;dx.$$
In our case, $R$ is the region below the top half of our ellipse.
For that ellipse, we have $$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$$ and therefore $$f(x)=b\sqrt{1-\frac{x^2}{a^2}}.$$ The top half of the ellipse meets the $x$ axis at $x=\pm a$. So our area is $$\int_{-a}^a\pi\left(b\sqrt{1-\frac{x^2}{a^2}}\right)^2\;dx.$$
We can simplify the calculation by noting that the ellipse is symmetrical about the $y$-axis. So we integrate from $x=0$ to $x=a$, and double the result. Thus we want $$2\int_0^a \pi \left(b^2-\frac{b^2x^2}{a^2}\right)\;dx.$$
The integration is easy. After a little while we get $$\frac{4}{3}\pi ab^2.$$
Note that if $b=a$, we get the familiar formula for the volume of a ball of radius $a$. This gives a useful partial check on the correctness of the calculation.
Intuition for the Volume Formula: Think of our solid as a somewhat peculiarly shaped salami, fat in the middle. Take a thin slice of the salami, perpendicular to the $x$-axis. Let the slice have thickness "$dx$", and suppose the slice was obtained by slicing through the salami at $x$ and at $x+dx$. Then the slice is almost perfectly round, and has radius roughly equal to $f(x)$. So the volume of the slice is roughly $\pi(f(x))^2\,dx$. Now let integration "add up" the volumes of these slices for us, as $x$ travels from $-a$ to $a$.
- | 1,082 | 3,767 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2015-18 | latest | en | 0.945397 |
https://serverlogic3.com/what-is-tree-and-types-of-tree-in-data-structure/ | 1,695,792,593,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00711.warc.gz | 547,930,449 | 15,575 | # What Is Tree and Types of Tree in Data Structure?
//
Heather Bennett
A tree is a widely used data structure in computer science that represents hierarchical relationships between elements. It consists of nodes connected by edges, where each node can have zero or more child nodes. The topmost node in a tree is called the root, and the nodes at the bottom with no children are called leaves.
## Types of Trees
There are various types of trees in data structures. Let’s explore some of them:
### 1. Binary Tree
A binary tree is a type of tree where each node has at most two children: a left child and a right child. The left child is smaller than its parent, while the right child is greater. This property makes binary trees particularly useful for efficient searching and sorting algorithms.
### 2. AVL Tree
An AVL (Adelson-Velskii and Landis) tree is a self-balancing binary search tree. It maintains balance by ensuring that the height difference between its left and right subtrees is at most one. This balancing property ensures efficient insertion, deletion, and searching operations with a time complexity of O(log n).
### 3. B-Tree
A B-tree is a self-balancing search tree designed to efficiently store large amounts of data on disk or secondary storage devices. It allows for efficient insertion, deletion, and searching operations by maintaining a balance between height and number of keys per node.
### 4. Red-Black Tree
A red-black tree is another type of self-balancing binary search tree that guarantees logarithmic time complexity for insertion, deletion, and searching operations. It ensures balance by coloring each node either red or black and applying specific rules to maintain balance during operations.
### 5. Trie
A trie, also known as a prefix tree, is a specialized tree used for efficient retrieval of strings or keys. It is particularly useful in applications such as autocomplete and spell checking. Each node in the trie represents a prefix or a complete word, and the edges represent characters.
### 6. Heap
A heap is a complete binary tree that satisfies the heap property. The heap property ensures that the value of each parent node is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the values of its children. Heaps are commonly used to implement priority queues.
## Conclusion
Trees are versatile data structures that find applications in various domains of computer science, including databases, file systems, network routing algorithms, and more. Understanding different types of trees and their properties can help you choose an appropriate data structure for solving specific problems efficiently. | 537 | 2,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.92316 |
http://www.fixya.com/support/t2908712-figure_out_four_numbers_need | 1,513,595,176,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00649.warc.gz | 374,073,251 | 32,853 | Question about Mavic Cycling
# How do I figure out what the four numbers I need to set my M-Tech 5 bike computer on if I don't know where my manual is? I know the numbers relate to the circumference of the wheel, but where do I match up my wheels to the correct 4 digit number?
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Pull the hub a part if you can not fix it take it to a bike store
Posted on Sep 09, 2009
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Montain Bike
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I hope this helps set your F20 computer.
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https://www.coursehero.com/file/8732993/Whocares-Usingthisapproachyoucanprovethatanyplaneisthesetof-s/ | 1,488,402,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174276.22/warc/CC-MAIN-20170219104614-00507-ip-10-171-10-108.ec2.internal.warc.gz | 789,719,948 | 21,775 | Lecture 5-2
# Whocares
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Unformatted text preview: ) Practice Describe the plane u s i n g v e c t o rs . Who cares? Using this approach, you can prove that any plane is the set of s o l u t i o n s o f a l i n e a r e q u a t i o n i n (s e e b o o k ! ). This gives us a way to get a grip on the intersection of two planes. Example Describe the intersection of the planes . Perpendicular vectors: a n d a n d Common solution: T h u s , t h e l i n e o f i n t e rs e c t i o n i s t h e s e t o f v e c t o rs s u c h t h a t and T h e v e c t o r i s p e rp...
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## This note was uploaded on 01/29/2014 for the course MATH 126 taught by Professor Smith during the Winter '07 term at University of Washington.
Ask a homework question - tutors are online | 272 | 913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | longest | en | 0.803201 |
https://www.teacherspayteachers.com/Product/Chemistry-Gas-Laws-2586659 | 1,568,556,072,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571506.61/warc/CC-MAIN-20190915134729-20190915160729-00133.warc.gz | 1,065,876,715 | 20,068 | Chemistry Gas Laws
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Introduction to Gas Laws Guided Inquiry Lesson (Free) AND Gas Laws Guided Inquiry Lesson
This is a student-centered, active learning lesson without lecture or notetaking!
Texas Chemistry Standards (TEKS)
C.9.A: The student is expected to describe and calculate the relations between volume, pressure, number of moles, and temperature for an ideal gas as described by Boyle's Law, Charles's Law, Avogadro's Law, Dalton's Law of Partial Pressures, and the Ideal Gas Law.
C.9.C: The student is expected to describe the postulates of kinetic molecular theory.
*****************************************************************************
Lesson Summary:
This guided inquiry lesson enables students to construct their own understanding of kinetic molecular theory, common symbols/units, and how to convert Celsius temperature values to Kelvin. Students are able to actively learn the material without lecture or note taking.
Part 1: Kinetic Molecular Theory
• Students use illustrations to help understand four postulates of kinetic molecular theory.
Part 2: Gas Law Units
• Students play a game of “Memory” in order to learn common gas law symbols and units.
Part 3: Converting °C to K
• Students practice converting temperature values in °C to K.
Student Study Sheet
Students are provided with a student study sheet summarizing all of the important concepts and vocabulary in this lesson.
Gas Laws Lesson
Lesson Summary:
This guided inquiry lesson enables students to construct their own understanding of the relationship between gas variables, by investigating common gas laws. Students are able to actively learn the material without lecture or note taking.
Part 1: Dalton’s Law of Partial Pressures
• Students use an illustration to determine the nature of Dalton’s Law of Partial Pressures.
• Students use Dalton’s Law of Partial Pressures to determine total and partial pressures in a mixture of gases.
Part 2: Boyle’s Law
• Students study an illustration of a Boyle’s Law experiment and record data from the experiment in a table.
• Students determine the relationship between the pressure and volume of a gas by graphing the data.
• Students practice rearranging the Boyle’s Law formula to isolate the unknown variable.
• Students practice calculating the unknown variables in Boyle’s Law problems.
Part 3: Charles’s Law
• Students study an illustration of a Charles’s Law experiment and record data from the experiment in a table.
• Students determine the relationship between the temperature and volume of a gas by graphing the data.
• Students practice rearranging the Charles’s Law formula to isolate the unknown variable.
• Students practice calculating the unknown variables in Charles’s Law problems.
• Students study an illustration of an Avogadro’s Law experiment and record data from the experiment in a table.
• Students determine the relationship between the moles and volume of a gas by graphing the data.
• Students practice rearranging the Avogadro’s Law formula to isolate the unknown variable.
• Students practice calculating the unknown variables in Avogadro’s Law problems.
Part 5: The Ideal Gas Law
• Students identify the units in the Gas Law Constant
• Students practice rearranging the Ideal Gas Law to isolate the unknown variable.
• Students practice calculating the unknown variables in Ideal Gas Law problems.
• Students practice solving gas law problems by working through a set of task cards.
Student Study Sheets
Students are provided with student study sheets summarizing all of the important concepts and vocabulary in this lesson.
*****************************************************************************
Learning Objectives (Bloom’s Revised Taxonomy):
Remembering: Describe the four postulates of kinetic molecular theory. (Introduction to Gas Laws); Describe the formulas for Boyle’s Law, Charles’s Law, Avogadro’s Law, Dalton’s Law of Partial Pressures, and the Ideal Gas Law. (Gas Laws)
Understanding: Identify the symbols and units of common gas law variables. (Introduction to Gas Laws); Identify the symbols and units used in each gas law. (Gas Laws)
Applying: Convert Celsius temperature values to Kelvin. (Introduction to Gas Laws); Describe how the variables in each gas law are related to one another. (Gas Laws)
Analyzing: Rearrange gas law formulas to isolate unknown variables. (Gas Laws)
Evaluating: Use Dalton’s Law of Partial Pressures, Boyle’s Law, Charles’s Law, and Avogadro’s Law, and the Ideal Gas Law to calculate unknown variables in gas law problems. (Gas Laws)
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• Cover page (2 pages)
• Guided inquiry lessons (36 pages)
• Handouts (6 pages)
• Card Set and answers (7 pages)
• Student study sheets (3 pages)
• Teacher notes (4 pages)
• Guided inquiry lessons with suggested answers (29 pages)
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,228 | 6,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.84345 |
https://byjus.com/icse/tips-to-prepare-for-icse-class-10-physics-exam/ | 1,632,563,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00579.warc.gz | 198,360,757 | 147,652 | # Tips to Prepare for ICSE Class 10 Physics Exam
By looking around our surroundings, we can see lots of Physics happening which our senses can’t even detect. Physics is considered one of the most significant branches of the disciplines of science. The most significant aim of Physics is to understand how the universe behaves. We all know that Physics rules our lives, such as the gravitational force of the Earth helps us to stay relatively at rest. Without it, we would be sailing across the sky. Even the Laws of Physics also implemented in our day to day activities such as Newton’s Law of Motion are at work as mechanical force and acceleration, action, reaction, and inertia.
Physics is one of the vital subjects for ICSE Class 10 students, and hence, it is essential to prepare it from the very beginning. Securing good marks in Physics along with other subjects, can guarantee higher percentage in the board exams. Students who are weak in Mathematics often find the Physics subject difficult. So, to help students prepare adequately for the Physics exam, we have provided below some tips to calm down their stress.
## Tips for scoring well in your ICSE Board Physics Exam
Revise Properly
Revision of notes and memorizing the formulas and the definitions of complicated terms helps a great deal during last-minute preparation. Students should keep a few lists of topics handy separately for quick revision, and it must contain the important formulae, derivations and essential definitions. Students can also revise the complete Physics syllabus by solving theICSE Class 10 Physics Previous year Question Paper.
Practice numerical problems
In Physics, numerical problems are present in almost every chapter. Here a thorough understanding of all the formulae and equations is a must. Practising the conceptual questions can help in preparing practical as well as concepts will help make clear the answer to questions from all sections. Students can refer to the ICSE Class 10 Physics Selina Solutions for more questions to practice.
Study Regularly
While preparing Physics, after learning a chapter, students should practice different types of numericals from it. At least an hour of regular self-study is imperative for better preparation of Physics. They should be rigorous with the theoretical part before jumping to solve the numerical problems. Studying once daily will help them to revise the topics so that they are thorough with all the concepts and can attempt the exam with confidence. | 483 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | latest | en | 0.949401 |
https://web2.0calc.com/questions/geometry_69166 | 1,716,185,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00090.warc.gz | 530,443,079 | 5,525 | +0
# Geometry
0
105
1
In triangle ABC, the angle bisector of angle BAC meets AC at D. If angle BAC = 60, angle ABC = 60, and AD = 24, then find the area of triangle ABC.
Mar 8, 2023
#1
+34
0
AD is the angle bisector of angle BAC, we have:
Therefore, triangle ABD is a 30-60-90 triangle, which means that:
AD = BD / sqrt(3) = 8 * sqrt(3)
Now, since angle ABC = 60, triangle ABC is equilateral, which means that all sides are equal. Therefore, we have:
BC = AB
Let x be the length of AB. Then, using the law of cosines in triangle ABC, we have:
x^2 = 24^2 + (8 * sqrt(3))^2 - 2(24)(8 * sqrt(3))(cos 60)
x^2 = 576 + 192 - 384
x^2 = 384
x = 8 * sqrt(6)
Therefore, the area of triangle ABC is:
= (sqrt(3) / 4) * (8 * sqrt(6))^2
= 96 * sqrt(3)
Therefore, the area of triangle ABC is 96 * sqrt(3).
Mar 8, 2023 | 301 | 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} | 4.34375 | 4 | CC-MAIN-2024-22 | latest | en | 0.791304 |
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Questions asked by the same visitor
A plant without enough water may wilt and close its stomata (leaf openings). This decreases the rate of photosynthesis because _____. chloroplasts don't have access to the sunlight carbon dioxide can't be absorbed nutrients from the soil can't be absorbed
Weegy: A plant without enough water may wilt and close its stomata (leaf openings). This decreases the rate of photosynthesis because chloroplasts don't have access to the sunlight. (More)
Question
Subtract -2k3 + k2 - 9 from 5k3 - 3k + 7. 3k3+ k2- 3k - 2 -7k3+ k2+ 3k - 16 7k3- k2- 3k + 16
Weegy: 5k3 - 3k + 7-(-2k3 + k2 - 9)=> 5k3 - 3k + 7+2k3-k2+9==> 5k3+2k3-k2-3k+7+9 ==> 7k3-k2-3k+16 ....Hence your answer is 7k3- k2- 3k + 16 User: Simplify (2n + 5) - (3n + 7) + (4n - 9). -3n - 11 3n - 11 -3n + 11 3n + 11 Weegy: Simplify (2n + 5) - (3n + 7) + (4n - 9). ANSWER: 3n - 11 User: Simplify a - {b - [c - (d - e) - f] - g}. a + b + c - d + e - f + g a - b + c - d + e - f + g a + b + c + d + e - f + g a - b + c - d + e + f - g User: Simplify 5t + (t - 3) - [(7t + 5) - (8 - 3t)]. -4t 4t -10t 10t Weegy: (a)-4t is the answer. 5t + (t - 3) - [(7t + 5) - (8 - 3t)] = -4t User: Using the polynomials Q = 3x2 + 5x - 2, R = 2 - x2, and S = 2x + 5, perform the indicated operation. Q - [R + S] 2x2+ 3x- 9 4x2- 3x+ 9 2x2- 3x+ 9 4x2+ 3x- 9 Weegy: D. 5x - 4 is the answer. User: Simplify 5 + 2{x - 4[3x + 7(2 - x)]}. -34x - 107 34x - 107 34x + 107 Weegy: (a)34x - 107 or x = 107/34 User: Select the correct product. (x2 + 4x + 8)(2x - 1) 2x3+ 7x2+ 12x- 8 2x3- 7x2- 12x- 8 x2+ 6x+ 7 x2+ 8x- 8 Weegy: The correct answer to the presented problem is (a) 2x3 + 7x2 + 12x - 8 User: Select the correct product. (6 - d)(d2 - 5 + 3d) 6d2- 30d d3- 3d2- 23d+ 30 d2+ 2d- 30 -d3+ 3d2+ 23d- 30 Weegy: 6d2- 30d d3- 3d User: Select the correct product. (y + z)3 y3+z3 y3+ 3y2z+ 3yz2+z3 3y+ 3z y3z3 Weegy: (x+y)?=x?-2x?y+y?x-x?y+2xy?-y? (More)
Question
Updated 6/22/2014 7:39:25 PM
5t + (t - 3) - [(7t + 5) - (8 - 3t)]
= 5t + t - 3 - (7t + 5 - 8 + 3t)
= 6t - 3 - (10t - 3)
= 6t - 3 - 10t + 3
= -4t
Confirmed by andrewpallarca [6/22/2014 7:26:02 PM]
5 + 2{x - 4[3x + 7(2 - x)]}
= 5 + 2[x - 4(3x + 14 - 7x)]
= 5 + 2[x - 4(14 - 4x)]
= 5 + 2(x - 56 + 16x)
= 5 + 2(17x - 56)
= 5 + 34x - 112
= 34x - 107
Confirmed by andrewpallarca [6/22/2014 7:26:18 PM]
(6 - d)(d^2 - 5 + 3d)
= 6d^2 - 30 + 18d - d^3 + 5d - 3d^2
= - d^3 + (6d^2 - 3d^2) + (18d + 5d) - 30
= - d^3 + 3d^2 + 23d - 30
Confirmed by andrewpallarca [6/22/2014 7:40:16 PM]
(y + z)^3 = y^3+ 3y^2z+ 3yz^2+z^3
Confirmed by andrewpallarca [6/22/2014 7:41:46 PM]
Select the correct product. (2x + 9)(x + 1) 2x2+ 11x+ 9 3x2+ 11x+ 9 2x2- 7x+ 9 2x2+ 11x+ 10
Weegy: Can you clarify your question please? User: Select the correct product. (2x + 9)(x + 1) 2x2+ 11x+ 9 3x2+ 11x+ 9 2x2- 7x+ 9 2x2+ 11x+ 10 Weegy: Can you clarify your question please? User: Select the correct product. (a + 8)(b + 3) ab+ 8a+ 3b+ 24 ab+ 3a+ 8b+ 24 11ab 24ab Weegy: ab+3a+8b+24 is the answer. User: Select the correct product. (j + 7)(k - 5) -35jk jk- 12jk- 35 jk- 5j+ 7k- 35 7j- 5k Weegy: (j + 7)(k - 5), the product is jk- 5j+ 7k- 35 User: Find the product. (n + 7)2 2n² + 14n + 49 n² + 14n + 14 n² + 14n + 49 Weegy: (n + 7)(n + 7), the product is n² + 14n + 49 User: Find the product. (r + s)2 r² + 2rs + s r² + 2rs + s² r² + 2rs + 2s² Weegy: (r + s)^2, the product is r² + 2rs + s² (More)
Question
Updated 6/20/2013 12:53:27 PM
(2x + 9)(x + 1)
= 2x^2 + 9x + 2x + 9
= 2x^2 + 11x + 9
Find the product mentally. (ab + 3)(ab - 3) a2b2+ 9 a2b2- 6ab- 9 a2b2- 9
Question
Updated 6/20/2013 12:41:30 PM
(ab + 3)(ab - 3)
= a^2b^2 + 3ab - 3ab - 9
= a^2b^2 - 9
Find the product mentally. (2m - 9)2 2m2- 18m+ 81 2m2- 81m+ 81 4m2- 36m+ 81
Question
Updated 6/20/2013 12:48:54 PM
(2m - 9)^2
= 4m^2 - 18m + (-18m) + 81
= 4m^2 - 36m + 81
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 2,736 | 5,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-26 | longest | en | 0.756005 |
http://www.techrepublic.com/article/save-time-by-using-excels-left-right-and-mid-string-functions/1033367 | 1,369,081,347,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699201808/warc/CC-MAIN-20130516101321-00058-ip-10-60-113-184.ec2.internal.warc.gz | 732,793,924 | 16,100 | Do you bristle when you see someone wasting precious time? I do, and I can’t help but stick my nose in and suggest a better way. That’s what happened recently when I noticed one of my coworkers using Excel and keying something from scratch that could have been generated with a simple formula. Here’s the scoop.
If the text is embedded in a cell, you can extract it
My coworker received a worksheet from a vendor that contained a column of entries that looked like the ones shown in Figure A.
Figure A We’ll use functions to extract certain portions of the entries in column A.
Those strings contained three distinct parts:
• The first three characters (the K-numbers) represent a product code.
• The second two digits (the B-numbers) represent a price code.
• The final three digits represent a customer code.
My coworker wanted to separate out those three pieces into different columns, and she was retyping them from scratch! That approach is so wasteful and inefficient it makes my skin crawl. Fortunately, I got to be the hero by showing her how to use Excel’s string functions to extract the codes automatically.
Fun with string functions
All of you veteran spreadsheet users know this drill by heart. Here’s how it works.
Grabbing the first three characters. To extract the first three characters of the text entries, you enter the Left function like this:
=Left(source_string,number_of_characters)
In this case, we entered into cell B2 the function =Left(A2,3) and then copied that formula to cells B3:B8. Figure B shows the results.
Figure B The Left function eliminates the need to re-key the first three letters from the entries in column A.
Pulling out the two characters in the middle. To extract the two characters in the middle of the string, we’ll use the Mid function, which takes the form:
=Mid(source_string,start_position,length)
Since we know that the string we want to extract always starts in position 4, we entered into cell C2 the function =Mid(A2,4,2) and then copied that formula to cells C3:C8. Figure C shows the results.
Figure C The Mid function lets you pull a string out of the middle.
Extracting the last three characters of a string. In order to extract the last three characters of a string, you use the Right function in the form:
=Right(source_string,number_of_characters)
In our example, we entered in cell D2 the function =Right(A2,3) and copied it into cells D3:D8. As Figure D shows, that function returns the three rightmost characters in the source string.
Figure D The Right function makes it easy to copy a set of characters from the right side of a string.
Once you’ve extracted the strings, then what?
After you’ve used the string functions to parse the source string into substrings, you’re free to sort or subtotal your data on any of the columns that contain those substrings. It only takes a minute or two to compose the function call and copy it to the appropriate cells. This technique comes in handy when you’re importing text files that have dumped from a mainframe database or from some other application.
Of course, we’ve just scratched the surface of what you can do with the “big three” string functions. In this simple example, our source string always contained the same number of characters.
Knowing that the middle string always started in the fourth position made it easy to use the Mid function. But what if the substring you want to extract could start anywhere within the string? In a future article, we’ll show you how to use the Find function in conjunction with the Mid function to locate and extract a string, regardless of how many characters the source string contains. | 771 | 3,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2013-20 | latest | en | 0.897483 |
http://www.haskell.org/pipermail/haskell/2000-December/006301.html | 1,412,117,160,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663167.8/warc/CC-MAIN-20140930004103-00246-ip-10-234-18-248.ec2.internal.warc.gz | 633,315,904 | 2,566 | # Rank-2 polymorphism & type inference
Simon Peyton-Jones simonpj@microsoft.com
Tue, 5 Dec 2000 05:12:20 -0800
```| > My question is: how does the type inference algorithm work in the
| > presence of rank-2 types? Does anyone know of any documentation on
| > this? Thanks!
I had a look at this. Actually it turns out to be only loosely related
to rank-2 polymorphism. I've been able to reproduce your problem using
only Haskell 98. It looks like a problem with incomplete type inference
Consider this:
module MP where
class C t where
op :: t -> Bool
instance C [t] where
op x = True
test :: [Int] -> Bool -- REQUIRED!
test y = let f :: c -> Bool
f x = op (y >> return x)
in
f (y::[Int])
Both GHC and Hugs reject this module if the type signature for
test is omitted. NHC (v1.00, 2000-09-15) falls over completely, with
All three succeed if the signature is in, or if the signature for f is
omitted.
This was unexpected, to me at least. You may need to add a type
signature if polymorphic recursion is being used, but here it isn't!
The problem is this: the compiler learns that y::[Int] "too late" to make
use of it when solving the constraints arising from the RHS of f.
In more detail, here's what happens. First we typecheck the RHS of
f, deducing the types
x :: a where a is fresh
y :: k a where k is fresh
y >> return x :: k a
op (y >> return x) :: Bool with constraint C (k a)
\x -> op (y >> return x) :: a -> Bool with constraint C (k a)
Now we try to generalise over a. We need to discharge the contraint
C (k a). Later we will find that y::[Int], so k=[], but we don't know that
yet. So we can't solve the constraint.
Adding the type signature to 'f' lets both GHC and Hugs figure out
that y::[Int] in advance, so we need to solve the constraint C ([] a),
which is fine.
So I think you have uncovered a genuine problem, and one I don't
know how to solve. It can always be "solved" by adding more
type information, such as the type sig for 'test'. In you case you said:
| After sending out my question, I noticed that hugs and ghc understood my
| code differently: from the error messages, we can see that hugs view (\a
| -> super a) as having type Sub b _1 -> Super b _2, while ghc thinks it
| is Sub c a -> Super c Int. To verify it, I changed my code s.t. y is
| defined as
|
| y = f (\(a :: Sub c Int) -> super a) x
This is exactly right, and GHC is happy now. I can't account for Hugs'
behaviour.
The "right" solution is presumably to defer all constraint checking until we
know what 'k' is. But that's a bit tricky because the constraint checking
generates bindings that must appear in f's RHS. A full solution looks a
bit over-kill-ish. But it's unsettling that the inference algorithm is
incomplete.
Simon
``` | 776 | 2,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-41 | latest | en | 0.88143 |
https://www.rhumbarlv.com/what-is-the-best-measure-of-central-tendency-for-quantitative-data/ | 1,702,193,597,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00522.warc.gz | 1,042,915,137 | 13,328 | ## What is the best measure of central tendency for quantitative data?
Mean
Mean is the most frequently used measure of central tendency and generally considered the best measure of it. However, there are some situations where either median or mode are preferred. Median is the preferred measure of central tendency when: There are a few extreme scores in the distribution of the data.
## What are the measures of central tendency and dispersion?
Measures that indicate the approximate center of a distribution are called measures of central tendency. Measures that describe the spread of the data are measures of dispersion. These measures include the mean, median, mode, range, upper and lower quartiles, variance, and standard deviation.
What is the best measure of central tendency and dispersion?
The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed.
What is central tendency in quantitative techniques?
A measure of central tendency is an important aspect of quantitative data. It is an estimate of a “typical” value. Three of the many ways to measure central tendency are the mean, median and mode. The mean is the average of data.
### What are the two commonly used tool in quantitative data analysis?
Although there are many other methods to collect quantitative data, those mentioned above probability sampling, interviews, questionnaire observation, and document review are the most common and widely used methods either offline or for online data collection.
### What is difference between central tendency and dispersion?
Central tendency is described by median, mode, and the means (there are different means- geometric and arithmetic). Dispersion is the degree to which data is distributed around this central tendency, and is represented by range, deviation, variance, standard deviation and standard error.
Which is best measure of central tendency?
Mean is generally considered the best measure of central tendency and the most frequently used one. However, there are some situations where the other measures of central tendency are preferred. There are few extreme scores in the distribution. Some scores have undetermined values.
Which measure of central tendency is best?
mean
The mean is the most frequently used measure of central tendency because it uses all values in the data set to give you an average. For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values.
#### What is central tendency and its types?
The central tendency measure is defined as the number used to represent the center or middle of a set of data values. The three commonly used measures of central tendency are the mean, median, and mode. A statistic that tells us how the data values are dispersed or spread out is called the measure of dispersion.
#### What does central tendency mean?
Mode: the most frequent value. Median: the middle number in an ordered data set. Mean: the sum of all values divided by the total number of values.
How can quantitative data be Analysed?
Quantitative data is usually collected for statistical analysis using surveys, polls or questionnaires sent across to a specific section of a population. The retrieved results can be established across a population.
How to calculate measures of central tendency and dispersion?
Measures of central tendency and measures of dispersion are two important types of statistics. A measure of central tendency represents the center or middle of a set of data values. The commonly used measures of central tendency are mean, median, and mode. Mean deviation: It is the mean of the absolute values of the numerical differences
## Is the SD an accurate measure of central tendency?
Reporting the SD along with the mean gives one the impression of how valid that mean value actually is (i.e. if the SD is huge, the mean is totally invalid – it is not an accurate measure of central tendency, because the data is so widely scattered.) This is an estimate of spread of samples around the population mean.
## Which is the best description of central tendency?
The term central tendency relates to the way in which quantitative data tend to cluster around some value. A measure of central tendency is any of a variety of ways of specifying this “central value”.
What do you mean by dispersion of data?
These describe the dispersion of data around some sort of mean. Deviation: distance between an observed score and the mean score. Because the difference can be positive or negative and this is cumbersome, usually the absolute deviation is used (which ignores the plus or minus sign). | 886 | 4,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-50 | longest | en | 0.954918 |
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# HW_2_Answers - Miller Kierste Homework 2 Due 3:00 am Inst...
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Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Rewrite the sum n 2+ 1 9 · 2 o + n 4+ 2 9 · 2 o + . . . + n 12+ 6 9 · 2 o using sigma notation. 1. 6 X i = 1 n 2 i + i 9 · 2 o correct 2. 6 X i = 1 n i + 2 i 9 · 2 o 3. 9 X i = 1 n 2 i + i 9 · 2 o 4. 6 X i = 1 2 n i + i 9 · 2 o 5. 9 X i = 1 2 n i + i 9 · 2 o 6. 9 X i = 1 2 n i + 2 i 9 · 2 o Explanation: The terms are of the form n 2 i + i 9 · 2 o , with i = 1 , 2 , . . . , 6. Consequently, in sigma notation the sum becomes 6 X i = 1 n 2 i + i 9 · 2 o . keywords: Stewart5e, summation notation, 002 (part 1 of 1) 10 points Estimate the area, A , under the graph of f ( x ) = 4 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. Correct answer: 5 . 133 . Explanation: With four equal subintervals and right end- points as sample points, A n f (2) + f (3) + f (4) + f (5) o 1 since x i = x * i = i + 1. Consequently, A 5 . 133 . keywords: Stewart5e, area, rational function, Riemann sum, 003 (part 1 of 1) 10 points Cyclist Joe accelerates as he rides away from a stop sign. His velocity graph over a 5 second period (in units of feet/sec) is shown in 1 2 3 4 5 4 8 12 16 20 Compute best possible upper and lower es- timates for the distance he travels over this period by dividing [0 , 5] into 5 equal subinter- vals and using endpoint sample points.
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Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 2 1. 47 ft < distance < 61 ft 2. 47 ft < distance < 59 ft 3. 47 ft < distance < 63 ft 4. 49 ft < distance < 61 ft 5. 49 ft < distance < 63 ft 6. 45 ft < distance < 61 ft correct 7. 45 ft < distance < 63 ft 8. 49 ft < distance < 59 ft 9. 45 ft < distance < 59 ft Explanation: The distance Joe travels during the 5 sec- ond period is the area under the velocity graph and above [0 , 5]. Since Joe’s speed is increasing, the best possible lower estimate occurs taking left hand endpoints as sample points and the area of the rectangles shown in 1 2 3 4 5 4 8 12 16 20 On the other hand, the best upper estimate will occur taking right hand endpoints and the area of the rectangles shown in 1 2 3 4 5 4 8 12 16 20 Consequently, reading off values from the graphs to compute the height of the rect- angles, we see that 45 ft < distance < 61 ft . keywords: 004 (part 1 of 1) 10 points Stewart Section 5.1, Example 3(a), page 321 Decide which of the following regions has area = lim n → ∞ n X i = 1 π 4 n tan 4 n without evaluating the limit. 1. n ( x, y ) : 0 y tan x, 0 x π 8 o 2. n ( x, y ) : 0 y tan 2 x, 0 x π 4 o 3. n ( x, y ) : 0 y tan 3 x, 0 x π 8 o 4. n ( x, y ) : 0 y tan 2 x, 0 x π 8 o 5. n ( x, y ) : 0 y tan 3 x, 0 x π 4 o 6. n ( x, y ) : 0 y tan x, 0 x π 4 o correct Explanation:
Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 3 The area under the graph of y = f ( x ) on an interval [ a, b ] is given by the limit lim n → ∞ n X i = 1 f ( x i ) Δ x when [ a, b ] is partitioned into n equal subin- tervals [ a, x 1 ] , [ x
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# 8.6 solving exponential and log equations
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### Transcript of "8.6 solving exponential and log equations"
1. 1. Solving Exponential Equations If two powers with the same base are equal, then their exponents must be equal. To solve using this property: Rewrite equation so the bases match. Set exponents equal to each other. Solve for x.
2. 2. Examples: Solve:
4. 4. Solving by Taking Logs If it is not easy/possible to rewrite using same bases, take the log (with base b) of both sides. Examples:
6. 6. More Examples: Taking Logs Not all equations will be that basic! Must get the exponential part by itself, first. Then, take the log of both sides. Last, continue solving for x. Example:
7. 7. Example: Solve | 319 | 1,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.824207 |
https://www.npr.org/templates/transcript/transcript.php?storyId=122880263 | 1,519,524,765,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00695.warc.gz | 914,531,420 | 18,327 | The Formula For Perfect Parallel Parking You finally find a spot along the curb, between two hulking SUVs, and it looks pretty tight. Do you go for it or move on? Not to worry; geometry can save you.
## The Formula For Perfect Parallel Parking
• `<iframe src="https://www.npr.org/player/embed/122880263/122895400" width="100%" height="290" frameborder="0" scrolling="no" title="NPR embedded audio player">`
• Transcript
The Formula For Perfect Parallel Parking
# < The Formula For Perfect Parallel Parking
## The Formula For Perfect Parallel Parking
• `<iframe src="https://www.npr.org/player/embed/122880263/122895400" width="100%" height="290" frameborder="0" scrolling="no" title="NPR embedded audio player">`
• Transcript
AUDIE CORNISH, host:
How many of you absolutely dread parallel parking? A lot I bet. Well, a British mathematician may have an answer, geometry. Simon Blackburn, a professor at the University of London says you can run a simple calculation and, voila, youre in. Here to discuss the theory is Keith Devlin, WEEKEND EDITIONs math guy. Welcome back, Keith.
Dr. KEITH DEVLIN (Co-Founder and Director, H-STAR Institute, Stanford University): Hi, Audie, nice to be back again.
CORNISH: So, whats the basic idea behind Professor Blackburns research?
Dr. DEVLIN: It's actually a very clever use of simple mathematics. In fact, the most complicated bit of mathematics it uses is our good old friend Pythagoras theorem.
CORNISH: That's a2 + b2 = c2, right?
Dr. DEVLIN: Absolutely. And thats really all you need, you just need to use that in a clever way. You put in a few figures about the size of your vehicle and the space that youre trying to get into. And the formula tells you exactly how much extra space you need, beyond the length of your vehicle, in order to park it in a simple, reverse-in, straighten-the-wheels, and switch-the-engine-off move.
CORNISH: So, give us a sense of how detailed the math is. What are the numbers you need to calculate using this system, and is it really practical?
Prof. DEVLIN: Simon Blackburn's formula and these four pieces of information. First of all, you have to tell it the radius of your cars turning cycle. That means if you give it a full lock to the left or right, it will turn in the circle. The radius of that thing is called the turning radius.
It needs the wheel base of your car which is the distance between the center of the front wheel and the rear wheel on either side. It needs the distance from your front wheel from the center of the front wheel to the front bumper. And the one extra piece of information it needs, you know, you pull up next to a car you need the width of that car.
CORNISH: Now I went out and did my parking job, which does involve a certain kind of estimation, where I pull up to the first car. And as Im reversing, I wait until the wheel sort of match up, then I jackknife into the space.
(Soundbite of laughter)
CORNISH: And so Ive got my own sort of estimation and is a lot of this just based on your ability to do that kind of guess work?
Prof. DEVLIN: Yeah, and, you know, there are lots of examples when people look at saw the baseball field and various athletes, and people have put in wonderful shots on the basketball courts. Whats going on is that mathematics gives you a way of understanding in detail what people have learned to do simply by practice and expertise.
In fact, when we practice something, be it on the athletic field or in an automobile, we are becoming very good mathematicians at doing a particular kind of operation. But usually, we don't call it mathematics and we certainly don't give people a pass on the math test because they can park their car.
CORNISH: WEEKEND EDITIONs math guy, Keith Devlin, also a professor at Stanford University. Keith, thanks so much.
Prof. DEVLIN: Okay, my pleasure, Audie.
CORNISH: As a native Bostonian I consider myself a wicked good parker, so I decided to put my skills to the test, right outside NPR headquarters.
(Soundbite of car)
CORNISH: We pull up to the first car, length to length, nose to nose. You essentially want to measure - youre kind of like measuring the space by measuring this car in front of you. And the measure of a perfect parking job according to this mathematician is no back and forth, so thats we are aiming for here. So, now that my front wheel and the back wheel are lined up, Im backing in. Release, release, release of the wheel. Were in the space. Were looking good. Im stopping.
(Soundbite of break)
CORNISH: How far are we from the curb? Okay...
(Soundbite of laughter)
CORNISH: Oh, look at that, thats like nine inches, thats genius.
(Soundbite of music)
CORNISH: Youre listening to WEEKEND EDITION from NPR News. | 1,119 | 4,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-09 | latest | en | 0.884909 |
http://blog.csdn.net/mijian1207mijian/article/details/52385128 | 1,505,824,639,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685129.23/warc/CC-MAIN-20170919112242-20170919132242-00646.warc.gz | 44,328,558 | 14,102 | # leetcode:Binary Search Tree:Contains Duplicate III(220)
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Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
multiset<long long> bst;
for (int i = 0; i < nums.size(); ++i) {
if (bst.size() == k + 1) bst.erase(bst.find(nums[i - k - 1]));
auto lb = bst.lower_bound(nums[i]);
if (lb != bst.end() && abs(*lb - nums[i]) <= t) return true;
auto ub = bst.upper_bound(nums[i]);
if (ub != bst.begin() && abs(*(--ub) - nums[i]) <= t) return true;
bst.insert(nums[i]);
}
return false;
}
};
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最新评论 | 314 | 877 | {"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} | 2.6875 | 3 | CC-MAIN-2017-39 | longest | en | 0.232003 |
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## Comment Re:Celsius: It's for telling temperature (Score 5, Insightful)1233
Seems to me that everybody saying Fahrenheit makes more sense as a "human" scale because humans can live within 0-100 F are just looking for excuses to keep to an obsolete scale.
1. Although 100C is too hot for human life it's encountered every day by everybody: just boil some water for cooking.
2. Fahrenheit is not more descriptive because the "degrees are smaller". We just use decimal places to do the same in C. We are not limited to 25 or 26 degrees; there's infinite variation between these if you use the right side of the decimal point, although using just the half point is usually enough. This particular misconception comes from Imperial measure users, who have to rely on fractions. Yes, fractions are more precise but not as useful as the arbitrary precision approximation built into the metric system. In fact, fractions are so awkward to use that you end up limiting yourself to 1/32 of an inch for most uses (or even 1/16), which is much *less* precision than a 5th grader can get with a few digits after the decimal point.
3. It's like people who say feet are better than meters because you can divide easily a feet by 1,2,3,4,6 and 12. They don't know that this "advantage" quickly goes away when compared to conversion elegance of the metric system. Take one meter. Make a cube of that side, a cubic meter. Fill that with water near the sea level and you get a volume of 1 Liter of water. Which happens to have a mass of Kilogram. Push that to an acceleration of a meter per second per second, it will take 1 Newton to do that. Push any object against a 1 Newton of force and you need 1 Watt to do that. Which happens to be 1 Joule of energy per second. Once you're done, mark the temperature at which the water you've been using freezes and call that 0. Mark the temp at which it boils and call that 100. Divide that in 100ths and you have the Celsius scale. Now say the same thing in Imperial units and go back to how awesome it is that you can divide by 12...
# Slashdot Top Deals
Memory fault -- core...uh...um...core... Oh dammit, I forget!
Working... | 586 | 2,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-17 | latest | en | 0.933428 |
https://cracku.in/70-out-of-its-total-sales-which-region-sold-the-minim-x-cat-1990 | 1,721,876,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00127.warc.gz | 148,095,644 | 27,083 | Instructions
The first table gives the number of saris (of all the eight colours) stocked in six regional showrooms. The second gives the number of saris (of all the eight colours) sold in these six regional showrooms. The third table gives the percentage of saris sold to saris stocked for each colour in each region. The fourth table gives the percentage of saris of a specific colour sold within that region. The fifth table gives the percentage of saris of a specific colour sold across all the regions. Study the tables and for each of the following questions, choose the best alternative.
Table 1
Table 2
Table 3
Table 4
Table 5
Question 70
# Out of its total sales, which region sold the minimum percentage of green saris?
Solution
Given we have 6 regions where sales are defined in the table number two . Now given that we need to find the region where green saris as a percentage of the total sales is lowest . In order to do that we calculate Green saris sold in a region/ Total sales of a particular region by considering the values from table 2 .
Now for option 1 which is region 1 the value is (164/788)*100 = 20.81 %
For option 2 which is region 4 the value is (85/511)*100 = 16.63 %
For option 1 which is region 6 the value is (3/18)*100 = 16.66 %
For option 1 which is region 2 the value is (200/676)*100 = 29.58 % .
Hence among the given options the percentage of region 4 is lowest which is option 2 .
• All Quant CAT complete Formulas and shortcuts PDF
• 35+ CAT previous year papers with video solutions PDF | 380 | 1,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.870377 |
https://getrevising.co.uk/revision-tests/physics-p2-22 | 1,545,024,471,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00559.warc.gz | 594,567,706 | 21,199 | # Physics P2
• Created by: Amy_5639
• Created on: 16-04-18 08:30
1.What is the difference between a scalar and a vector quantity?
Scalar=only numbers e.g.speed, distance, time, mass etc Vector=direction as well e.g.velocity, displacement, force, acceleration etc
1 of 28
2.Define acceleration in terms of velocity.
Acceleration is how quickly the velocity is changing.
2 of 28
3.Explain why an object travelling in a circle at a constant speed is accelerating.
Because it has a changing velocity as it's always travelling in different directions so it's always accelerating
3 of 28
4.Describe an experiment to investigate the acceleration of a trolley down a ramp.
1.Set up ramp on runway,put trolley at top of ramp+3 light gates,1 at front of trolley on ramp,2 at end of ramp,3 at end of runway.2.Measure distance between light gates then let trolley go down ramp.3.Between gates: 1+2 average speed, 2+3 speed on
4 of 28
5.How is the speed of an object found from a distance-time graph?
The speed equals the gradient, so find the gradient of the graph by doing change in vertical (divided by) change in horizontal. If graph is curved draw a tangent then work out the gradient of the tangent
5 of 28
6.What does a flat section on a velocity-time graph represent?
6 of 28
7.How is the distance travelled by an object found from its velocity-time graph?
You can estimate the distance travelled from the area under a graph by counting squares. Find the distance that 1 square represents then multiply width and height of square, and multiply it by the number of squares under the graph
7 of 28
8.What is meant by the 'resultant force' acting on an object?
It's the overall force on a point or object
8 of 28
9.What will happen to an object that has a zero resultant force?
The object will either be stationary or moving at a steady speed
9 of 28
10.What will happen to an object that has a non-zero resultant force?
The object will either accelerate or decelerate
10 of 28
11.What is Newton's First Law of Motion?
It says that:"an object will remain stationary or at a constant velocity unless acted upon by an external force. So no resultant force means no change in velocity
11 of 28
Give the equation for Newton's Second Law.
Force (N) = Mass (kg) x Acceleration (m/s squared) or F=ma
12 of 28
13.Explain how a car moving with a constant driving force will reach terminal velocity.
When car sets off there's more driving force than friction force so they accelerate.But resistance is directly proportional to velocity so as velocity increases, resistance does too.This reduces the acceleration until friction force= driving force.
13 of 28
14.What is inertia?
The measure of how difficult it is to change an objects velocity
14 of 28
15.What is Newton's Third Law of Motion? Give an example of it in action
It says:"when two objects interact they forces they exert on each other are equal and opposite- so reaction forces are equal and opposite. E.g. if you push a trolley the trolley will push back against you just as hard
15 of 28
16.Give the equation for momentum in terms of mass and velocity.
Momentum (kg m/s) = mass (kg) x velocity (m/s) or p=m x v
16 of 28
17.What is an elastic collision?
It's where total energy in the kinetic energy stores of the objects colliding is the same before and after the collision-i.e. energy in the kinetic store is conserved
17 of 28
18.What is the difference between mass and weight and how can weight be calculated?
Mass is the actual amount of material contained in a body, whereas weight is the force exerted by the gravity on that object. Mass in independent but wight is different on Earth, Moon etc. Weight(N)=mass(kg) x gravitational field strength(N/kg)
18 of 28
19.Give the equation for the energy in an objects gravitational potential energy store.
Potential energy(J)=mass(kg) x height(m) x gravitational field strength(N/kg) or PE=m x h x g
19 of 28
20.Give the equation for the energy in the kinetic energy store of a moving object.
Kinetic energy(J)= 0.5 x mass(kg) x (speed)squared (m/s squared) or KE= 0.5 x m x v squared
20 of 28
21.Give the equation for the work done on an object when it's moved a certain distance by a force.
Work done(J)= force(N) x distance(m) or W=F X d
21 of 28
22.What is meant by power? How is power calculated?
Power is the rate at which energy is transferred. Power(W)= workdone(J) (divided by) time(s) or P=W (divided by) t
22 of 28
23.What is the minimum number of forces needed to stretch, compress or bend an object?
2
23 of 28
24.Give the equation that is known as the Hooke's law.
Force exerted by a spring(N)= extension(m) x spring constant(N/m)
24 of 28
25.What constant can be found from calculating the gradient of a force-extension graph for a material obeying Hooke's law?
The gradient of the straight line is equal to the spring constant of the object- the larger the spring constant, the steeper the gradient
25 of 28
26.What is the difference between an elastic deformation and a plastic deformation?
If an object returns to its original shape after the forces are removed it's an elastic deformation. If the object doesn't return to its original shape when you remove the forces it's a plastic deformation
26 of 28
27.Describe a simple experiment to investigate Hooke's law.
Hang spring from clamp stand, without masses, then measure springs original length.Weigh masses and add on one at a time to the hook.After each mass added measure new length of spring. Plot graph of force against extension using results
27 of 28
28.Give the equation for calculating the energy transferred to a spring when it's stretched.
Energy transferred in stretching(J)= 0.5 x spring constant(N/m) x (extension) squared (m squared) or E = 0.5 x k x (x)squared
28 of 28
## Other cards in this set
### Card 2
#### Front
2.Define acceleration in terms of velocity.
#### Back
Acceleration is how quickly the velocity is changing.
### Card 3
#### Front
3.Explain why an object travelling in a circle at a constant speed is accelerating.
### Card 4
#### Front
4.Describe an experiment to investigate the acceleration of a trolley down a ramp.
### Card 5
#### Front
5.How is the speed of an object found from a distance-time graph? | 1,503 | 6,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.905743 |
https://perl.mines-albi.fr/perl5.6.1/site_perl/5.6.1/sun4-solaris/PDL/Opt/Simplex.html | 1,624,230,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00599.warc.gz | 426,443,439 | 2,487 | PDL::Opt::Simplex -- Simplex optimization routines
# NAME
PDL::Opt::Simplex -- Simplex optimization routines
# SYNOPSIS
` use PDL::Opt::Simplex;`
``` (\$optimum,\$ssize) = simplex(\$init,\$initsize,\$minsize,
\$maxiter,
sub {evaluate_func_at(\$_[0])},
sub {display_simplex(\$_[0])}
);```
# DESCRIPTION
This package implements the commonly used simplex optimization algorithm. The basic idea of the algorithm is to move a ``simplex'' of N+1 points in the N-dimensional search space according to certain rules. The main benefit of the algorithm is that you do not need to calculate the derivatives of your function.
\$init is a 1D vector holding the initial values of the N fitted parameters, \$optimum is a vector holding the final solution.
\$initsize is the size of \$init (more...)
\$minsize is some sort of convergence criterion (more...) - e.g. \$minsize = 1e-6
The sub is assumed to understand more than 1 dimensions and threading. Its signature is 'inp(nparams); [ret]out()'. An example would be
``` sub evaluate_func_at {
my(\$xv) = @_;
my \$x1 = \$xv->slice("(0)");
my \$x2 = \$xv->slice("(1)");
return \$x1**4 + (\$x2-5)**4 + \$x1*\$x2;
}```
Here \$xv is a vector holding the current values of the parameters being fitted which are then sliced out explicitly as \$x1 and \$x2.
\$ssize gives a very very approximate estimate of how close we might be - it might be miles wrong. It is the euclidean distance between the best and the worst vertices. If it is not very small, the algorithm has not converged.
# FUNCTIONS
## simplex
Simplex optimization routine
``` (\$optimum,\$ssize) = simplex(\$init,\$initsize,\$minsize,
\$maxiter,
sub {evaluate_func_at(\$_[0])},
sub {display_simplex(\$_[0])}
);```
See module `PDL::Opt::Simplex` for more information.
# CAVEATS
Do not use the simplex method if your function has local minima. It will not work. Use genetic algorithms or simulated annealing or conjugate gradient or momentum gradient descent.
They will not really work either but they are not guaranteed not to work ;) (if you have infinite time, simulated annealing is guaranteed to work but only after it has visited every point in your space).
Ron Shaffer's chemometrics web page and references therein: `http://chem1.nrl.navy.mil/~shaffer/chemoweb.html`.
`Copyright(C)` 1997 Tuomas J. Lukka. All rights reserved. There is no warranty. You are allowed to redistribute this software / documentation under certain conditions. For details, see the file COPYING in the PDL distribution. If this file is separated from the PDL distribution, the copyright notice should be included in the file. | 668 | 2,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-25 | longest | en | 0.713589 |
http://www.semanarioangolense.net/go-math-grade-6-chapter-8-answer-key/ | 1,653,329,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00056.warc.gz | 106,804,489 | 15,265 | Chapter 1 Place Value Addition and Subtraction to One Million. Start studying the Go Math Grade 5 Chapter 6 Solution Key Add and Subtract Fractions with Unlike Denominators to score the highest marks in the exams.
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Go Math Chapter Review Math Bundle Chapters 1 12 Common Core Math Standards Go Math Math Assessment | 1,456 | 6,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.917898 |
http://oeis.org/A052163 | 1,597,207,283,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738864.9/warc/CC-MAIN-20200812024530-20200812054530-00402.warc.gz | 79,149,500 | 3,593 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A052163 Primes at which the difference pattern X24Y (X and Y >= 6) occurs in A001223. 2
347, 641, 1277, 1607, 2237, 2267, 2657, 3527, 3671, 3917, 4001, 4127, 4637, 4931, 4967, 5477, 5501, 6197, 8087, 8231, 8537, 8861, 9461, 10331, 10427, 11171, 11777, 12107, 12917, 13757, 13901, 14081, 14321, 14627, 17027, 18251, 19991, 20477 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS EXAMPLE 641 is in the sequence because 641 + 2 = 643, 641 + 2 + 4 = 647 is prime, the prime prior to 641 is 631, the prime after 647 is 653, and the corresponding differences are 10 or 6. The d-pattern is {10,2,4,6}. CROSSREFS Cf. A001223, A052160, A052162-A052168, A022008, A047078. Sequence in context: A059074 A273805 A142907 * A210363 A054823 A142369 Adjacent sequences: A052160 A052161 A052162 * A052164 A052165 A052166 KEYWORD nonn AUTHOR Labos Elemer, Jan 26 2000 STATUS approved
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Last modified August 11 23:45 EDT 2020. Contains 336434 sequences. (Running on oeis4.) | 498 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.660581 |
https://www.speedsolving.com/wiki/index.php?title=EPLL | 1,723,761,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00118.warc.gz | 756,046,862 | 9,885 | # EPLL
EPLL Information Proposer(s): Proposed: Alt Names: Edges permutation of the last layer Variants: Beginner EP Subgroup: No. Algs: 4 Avg Moves: ~8.2 STM Purpose(s):
Edges permutation of the last layer, abbrevaited EPLL, is the sub group of PLL where only the edges are permuted. EPLL is also used in many other methods, sometimes as a stand alone substep, like in COLL but also as a sub group, for example in ELL or ZBLL.
## EPLL algorithms
Note that all of these algorithms are written in the Western notation, where a lowercase letter means a double-layer turn and rotations are denoted by x, y, and z. (how to add algorithms) Click on an algorithm (not the camera icon) to watch an animation of it.
### H Permutation
Name: H-PLL, X-PLL Used in: EPLL, CPLL, PLL, ELL, ZBLL, ZZLL Optimal moves: 10 HTM, 7 STM X-PLL is H-PLL + U2.
PLL M2' U M2' U2 M2' U M2' [1]
PLL M2' U' M2' U2' M2' U' M2' [2]
PLL (M2' U)5 M2'
### U Permutation : a
Name: U-PLL a Used in: EPLL, PLL, ELL, ZBLL, BLD Optimal moves: 9 HTM, 7 STM
PLL M2 U M' U2 M U M2
PLL (y2) M2' U M U2 M' U M2' [4]
### U Permutation : b
Name: U-PLL b Used in: EPLL, PLL, ELL, ZBLL, BLD Optimal moves: 9 HTM, 7 STM
PLL M2 U' M' U2 M U' M2
PLL (y2) M2 U' M U2 M' U' M2 [7]
### Z Permutation
Name: Z-PLL Used in: EPLL, PLL, ELL, ZBLL, ZZLL Optimal moves: 12 HTM, 7 STM | 486 | 1,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.78648 |
http://us.metamath.org/mpeuni/mulcxp.html | 1,642,967,049,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00030.warc.gz | 64,410,922 | 10,700 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > mulcxp Structured version Visualization version GIF version
Theorem mulcxp 24412
Description: Complex exponentiation of a product. Proposition 10-4.2(c) of [Gleason] p. 135. (Contributed by Mario Carneiro, 2-Aug-2014.)
Assertion
Ref Expression
mulcxp (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))
Proof of Theorem mulcxp
StepHypRef Expression
1 simp1l 1083 . . . . . . 7 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → 𝐴 ∈ ℝ)
21recnd 10053 . . . . . 6 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → 𝐴 ∈ ℂ)
32mul01d 10220 . . . . 5 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐴 · 0) = 0)
43oveq1d 6650 . . . 4 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐴 · 0)↑𝑐𝐶) = (0↑𝑐𝐶))
5 simp3 1061 . . . . 5 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → 𝐶 ∈ ℂ)
62, 5mulcxplem 24411 . . . 4 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (0↑𝑐𝐶) = ((𝐴𝑐𝐶) · (0↑𝑐𝐶)))
74, 6eqtrd 2654 . . 3 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐴 · 0)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (0↑𝑐𝐶)))
8 oveq2 6643 . . . . 5 (𝐵 = 0 → (𝐴 · 𝐵) = (𝐴 · 0))
98oveq1d 6650 . . . 4 (𝐵 = 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴 · 0)↑𝑐𝐶))
10 oveq1 6642 . . . . 5 (𝐵 = 0 → (𝐵𝑐𝐶) = (0↑𝑐𝐶))
1110oveq2d 6651 . . . 4 (𝐵 = 0 → ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)) = ((𝐴𝑐𝐶) · (0↑𝑐𝐶)))
129, 11eqeq12d 2635 . . 3 (𝐵 = 0 → (((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)) ↔ ((𝐴 · 0)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (0↑𝑐𝐶))))
137, 12syl5ibrcom 237 . 2 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐵 = 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶))))
14 simp2l 1085 . . . . . . . . 9 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → 𝐵 ∈ ℝ)
1514recnd 10053 . . . . . . . 8 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → 𝐵 ∈ ℂ)
1615mul02d 10219 . . . . . . 7 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (0 · 𝐵) = 0)
1716oveq1d 6650 . . . . . 6 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((0 · 𝐵)↑𝑐𝐶) = (0↑𝑐𝐶))
1815, 5mulcxplem 24411 . . . . . . 7 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (0↑𝑐𝐶) = ((𝐵𝑐𝐶) · (0↑𝑐𝐶)))
19 cxpcl 24401 . . . . . . . . 9 ((𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐵𝑐𝐶) ∈ ℂ)
2015, 5, 19syl2anc 692 . . . . . . . 8 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐵𝑐𝐶) ∈ ℂ)
21 0cn 10017 . . . . . . . . 9 0 ∈ ℂ
22 cxpcl 24401 . . . . . . . . 9 ((0 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (0↑𝑐𝐶) ∈ ℂ)
2321, 5, 22sylancr 694 . . . . . . . 8 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (0↑𝑐𝐶) ∈ ℂ)
2420, 23mulcomd 10046 . . . . . . 7 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐵𝑐𝐶) · (0↑𝑐𝐶)) = ((0↑𝑐𝐶) · (𝐵𝑐𝐶)))
2518, 24eqtrd 2654 . . . . . 6 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (0↑𝑐𝐶) = ((0↑𝑐𝐶) · (𝐵𝑐𝐶)))
2617, 25eqtrd 2654 . . . . 5 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((0 · 𝐵)↑𝑐𝐶) = ((0↑𝑐𝐶) · (𝐵𝑐𝐶)))
27 oveq1 6642 . . . . . . 7 (𝐴 = 0 → (𝐴 · 𝐵) = (0 · 𝐵))
2827oveq1d 6650 . . . . . 6 (𝐴 = 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((0 · 𝐵)↑𝑐𝐶))
29 oveq1 6642 . . . . . . 7 (𝐴 = 0 → (𝐴𝑐𝐶) = (0↑𝑐𝐶))
3029oveq1d 6650 . . . . . 6 (𝐴 = 0 → ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)) = ((0↑𝑐𝐶) · (𝐵𝑐𝐶)))
3128, 30eqeq12d 2635 . . . . 5 (𝐴 = 0 → (((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)) ↔ ((0 · 𝐵)↑𝑐𝐶) = ((0↑𝑐𝐶) · (𝐵𝑐𝐶))))
3226, 31syl5ibrcom 237 . . . 4 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐴 = 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶))))
3332a1dd 50 . . 3 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐴 = 0 → (𝐵 ≠ 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))))
341adantr 481 . . . . . . . . . . 11 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐴 ∈ ℝ)
35 simpl1r 1111 . . . . . . . . . . . 12 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 0 ≤ 𝐴)
36 simprl 793 . . . . . . . . . . . 12 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐴 ≠ 0)
3734, 35, 36ne0gt0d 10159 . . . . . . . . . . 11 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 0 < 𝐴)
3834, 37elrpd 11854 . . . . . . . . . 10 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐴 ∈ ℝ+)
3914adantr 481 . . . . . . . . . . 11 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐵 ∈ ℝ)
40 simpl2r 1113 . . . . . . . . . . . 12 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 0 ≤ 𝐵)
41 simprr 795 . . . . . . . . . . . 12 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐵 ≠ 0)
4239, 40, 41ne0gt0d 10159 . . . . . . . . . . 11 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 0 < 𝐵)
4339, 42elrpd 11854 . . . . . . . . . 10 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐵 ∈ ℝ+)
4438, 43relogmuld 24352 . . . . . . . . 9 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (log‘(𝐴 · 𝐵)) = ((log‘𝐴) + (log‘𝐵)))
4544oveq2d 6651 . . . . . . . 8 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐶 · (log‘(𝐴 · 𝐵))) = (𝐶 · ((log‘𝐴) + (log‘𝐵))))
465adantr 481 . . . . . . . . 9 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐶 ∈ ℂ)
472adantr 481 . . . . . . . . . 10 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐴 ∈ ℂ)
4847, 36logcld 24298 . . . . . . . . 9 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (log‘𝐴) ∈ ℂ)
4915adantr 481 . . . . . . . . . 10 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → 𝐵 ∈ ℂ)
5049, 41logcld 24298 . . . . . . . . 9 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (log‘𝐵) ∈ ℂ)
5146, 48, 50adddid 10049 . . . . . . . 8 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐶 · ((log‘𝐴) + (log‘𝐵))) = ((𝐶 · (log‘𝐴)) + (𝐶 · (log‘𝐵))))
5245, 51eqtrd 2654 . . . . . . 7 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐶 · (log‘(𝐴 · 𝐵))) = ((𝐶 · (log‘𝐴)) + (𝐶 · (log‘𝐵))))
5352fveq2d 6182 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (exp‘(𝐶 · (log‘(𝐴 · 𝐵)))) = (exp‘((𝐶 · (log‘𝐴)) + (𝐶 · (log‘𝐵)))))
5446, 48mulcld 10045 . . . . . . 7 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐶 · (log‘𝐴)) ∈ ℂ)
5546, 50mulcld 10045 . . . . . . 7 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐶 · (log‘𝐵)) ∈ ℂ)
56 efadd 14805 . . . . . . 7 (((𝐶 · (log‘𝐴)) ∈ ℂ ∧ (𝐶 · (log‘𝐵)) ∈ ℂ) → (exp‘((𝐶 · (log‘𝐴)) + (𝐶 · (log‘𝐵)))) = ((exp‘(𝐶 · (log‘𝐴))) · (exp‘(𝐶 · (log‘𝐵)))))
5754, 55, 56syl2anc 692 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (exp‘((𝐶 · (log‘𝐴)) + (𝐶 · (log‘𝐵)))) = ((exp‘(𝐶 · (log‘𝐴))) · (exp‘(𝐶 · (log‘𝐵)))))
5853, 57eqtrd 2654 . . . . 5 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (exp‘(𝐶 · (log‘(𝐴 · 𝐵)))) = ((exp‘(𝐶 · (log‘𝐴))) · (exp‘(𝐶 · (log‘𝐵)))))
5947, 49mulcld 10045 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐴 · 𝐵) ∈ ℂ)
6047, 49, 36, 41mulne0d 10664 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐴 · 𝐵) ≠ 0)
61 cxpef 24392 . . . . . 6 (((𝐴 · 𝐵) ∈ ℂ ∧ (𝐴 · 𝐵) ≠ 0 ∧ 𝐶 ∈ ℂ) → ((𝐴 · 𝐵)↑𝑐𝐶) = (exp‘(𝐶 · (log‘(𝐴 · 𝐵)))))
6259, 60, 46, 61syl3anc 1324 . . . . 5 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → ((𝐴 · 𝐵)↑𝑐𝐶) = (exp‘(𝐶 · (log‘(𝐴 · 𝐵)))))
63 cxpef 24392 . . . . . . 7 ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐶 ∈ ℂ) → (𝐴𝑐𝐶) = (exp‘(𝐶 · (log‘𝐴))))
6447, 36, 46, 63syl3anc 1324 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐴𝑐𝐶) = (exp‘(𝐶 · (log‘𝐴))))
65 cxpef 24392 . . . . . . 7 ((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐶 ∈ ℂ) → (𝐵𝑐𝐶) = (exp‘(𝐶 · (log‘𝐵))))
6649, 41, 46, 65syl3anc 1324 . . . . . 6 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → (𝐵𝑐𝐶) = (exp‘(𝐶 · (log‘𝐵))))
6764, 66oveq12d 6653 . . . . 5 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)) = ((exp‘(𝐶 · (log‘𝐴))) · (exp‘(𝐶 · (log‘𝐵)))))
6858, 62, 673eqtr4d 2664 . . . 4 ((((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))
6968exp32 630 . . 3 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐴 ≠ 0 → (𝐵 ≠ 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))))
7033, 69pm2.61dne 2877 . 2 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → (𝐵 ≠ 0 → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶))))
7113, 70pm2.61dne 2877 1 (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) ∧ 𝐶 ∈ ℂ) → ((𝐴 · 𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶) · (𝐵𝑐𝐶)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 384 ∧ w3a 1036 = wceq 1481 ∈ wcel 1988 ≠ wne 2791 class class class wbr 4644 ‘cfv 5876 (class class class)co 6635 ℂcc 9919 ℝcr 9920 0cc0 9921 + caddc 9924 · cmul 9926 ≤ cle 10060 expce 14773 logclog 24282 ↑𝑐ccxp 24283 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1720 ax-4 1735 ax-5 1837 ax-6 1886 ax-7 1933 ax-8 1990 ax-9 1997 ax-10 2017 ax-11 2032 ax-12 2045 ax-13 2244 ax-ext 2600 ax-rep 4762 ax-sep 4772 ax-nul 4780 ax-pow 4834 ax-pr 4897 ax-un 6934 ax-inf2 8523 ax-cnex 9977 ax-resscn 9978 ax-1cn 9979 ax-icn 9980 ax-addcl 9981 ax-addrcl 9982 ax-mulcl 9983 ax-mulrcl 9984 ax-mulcom 9985 ax-addass 9986 ax-mulass 9987 ax-distr 9988 ax-i2m1 9989 ax-1ne0 9990 ax-1rid 9991 ax-rnegex 9992 ax-rrecex 9993 ax-cnre 9994 ax-pre-lttri 9995 ax-pre-lttrn 9996 ax-pre-ltadd 9997 ax-pre-mulgt0 9998 ax-pre-sup 9999 ax-addf 10000 ax-mulf 10001 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1484 df-fal 1487 df-ex 1703 df-nf 1708 df-sb 1879 df-eu 2472 df-mo 2473 df-clab 2607 df-cleq 2613 df-clel 2616 df-nfc 2751 df-ne 2792 df-nel 2895 df-ral 2914 df-rex 2915 df-reu 2916 df-rmo 2917 df-rab 2918 df-v 3197 df-sbc 3430 df-csb 3527 df-dif 3570 df-un 3572 df-in 3574 df-ss 3581 df-pss 3583 df-nul 3908 df-if 4078 df-pw 4151 df-sn 4169 df-pr 4171 df-tp 4173 df-op 4175 df-uni 4428 df-int 4467 df-iun 4513 df-iin 4514 df-br 4645 df-opab 4704 df-mpt 4721 df-tr 4744 df-id 5014 df-eprel 5019 df-po 5025 df-so 5026 df-fr 5063 df-se 5064 df-we 5065 df-xp 5110 df-rel 5111 df-cnv 5112 df-co 5113 df-dm 5114 df-rn 5115 df-res 5116 df-ima 5117 df-pred 5668 df-ord 5714 df-on 5715 df-lim 5716 df-suc 5717 df-iota 5839 df-fun 5878 df-fn 5879 df-f 5880 df-f1 5881 df-fo 5882 df-f1o 5883 df-fv 5884 df-isom 5885 df-riota 6596 df-ov 6638 df-oprab 6639 df-mpt2 6640 df-of 6882 df-om 7051 df-1st 7153 df-2nd 7154 df-supp 7281 df-wrecs 7392 df-recs 7453 df-rdg 7491 df-1o 7545 df-2o 7546 df-oadd 7549 df-er 7727 df-map 7844 df-pm 7845 df-ixp 7894 df-en 7941 df-dom 7942 df-sdom 7943 df-fin 7944 df-fsupp 8261 df-fi 8302 df-sup 8333 df-inf 8334 df-oi 8400 df-card 8750 df-cda 8975 df-pnf 10061 df-mnf 10062 df-xr 10063 df-ltxr 10064 df-le 10065 df-sub 10253 df-neg 10254 df-div 10670 df-nn 11006 df-2 11064 df-3 11065 df-4 11066 df-5 11067 df-6 11068 df-7 11069 df-8 11070 df-9 11071 df-n0 11278 df-z 11363 df-dec 11479 df-uz 11673 df-q 11774 df-rp 11818 df-xneg 11931 df-xadd 11932 df-xmul 11933 df-ioo 12164 df-ioc 12165 df-ico 12166 df-icc 12167 df-fz 12312 df-fzo 12450 df-fl 12576 df-mod 12652 df-seq 12785 df-exp 12844 df-fac 13044 df-bc 13073 df-hash 13101 df-shft 13788 df-cj 13820 df-re 13821 df-im 13822 df-sqrt 13956 df-abs 13957 df-limsup 14183 df-clim 14200 df-rlim 14201 df-sum 14398 df-ef 14779 df-sin 14781 df-cos 14782 df-pi 14784 df-struct 15840 df-ndx 15841 df-slot 15842 df-base 15844 df-sets 15845 df-ress 15846 df-plusg 15935 df-mulr 15936 df-starv 15937 df-sca 15938 df-vsca 15939 df-ip 15940 df-tset 15941 df-ple 15942 df-ds 15945 df-unif 15946 df-hom 15947 df-cco 15948 df-rest 16064 df-topn 16065 df-0g 16083 df-gsum 16084 df-topgen 16085 df-pt 16086 df-prds 16089 df-xrs 16143 df-qtop 16148 df-imas 16149 df-xps 16151 df-mre 16227 df-mrc 16228 df-acs 16230 df-mgm 17223 df-sgrp 17265 df-mnd 17276 df-submnd 17317 df-mulg 17522 df-cntz 17731 df-cmn 18176 df-psmet 19719 df-xmet 19720 df-met 19721 df-bl 19722 df-mopn 19723 df-fbas 19724 df-fg 19725 df-cnfld 19728 df-top 20680 df-topon 20697 df-topsp 20718 df-bases 20731 df-cld 20804 df-ntr 20805 df-cls 20806 df-nei 20883 df-lp 20921 df-perf 20922 df-cn 21012 df-cnp 21013 df-haus 21100 df-tx 21346 df-hmeo 21539 df-fil 21631 df-fm 21723 df-flim 21724 df-flf 21725 df-xms 22106 df-ms 22107 df-tms 22108 df-cncf 22662 df-limc 23611 df-dv 23612 df-log 24284 df-cxp 24285 This theorem is referenced by: cxprec 24413 divcxp 24414 mulcxpd 24455 amgmlemALT 42314
Copyright terms: Public domain W3C validator | 8,521 | 12,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-05 | latest | en | 0.127929 |
https://www.aqua-calc.com/one-to-all/acceleration/preset/fermi-per-hour-squared/1-point-3eplus36 | 1,619,147,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00558.warc.gz | 737,717,917 | 8,844 | # Convert fermis per hour squared to other units of acceleration
## fermis/hour² [fermi/h²] acceleration conversions
1.3 × 10+36 fermi/h² = 1 × 10+24 angstroms per second squared fermi/h² to Å/s² 1.3 × 10+36 fermi/h² = 670.52 astronomical units per second squared fermi/h² to au/s² 1.3 × 10+36 fermi/h² = 1 × 10+32 attometers per second squared fermi/h² to am/s² 1.3 × 10+36 fermi/h² = 1 × 10+16 centimeters per second squared fermi/h² to cm/s² 1.3 × 10+36 fermi/h² = 4 986 312 037 777.7 chains per second squared fermi/h² to ch/s² 1.3 × 10+36 fermi/h² = 1 × 10+15 decimeters per second squared fermi/h² to dm/s² 1.3 × 10+36 fermi/h² = 10 030 864 198 333 dekameters per second squared fermi/h² to dam/s² 1.3 × 10+36 fermi/h² = 0.0001 Exameter per second squared fermi/h² to Em/s² 1.3 × 10+36 fermi/h² = 54 849 432 408 333 fathoms per second squared fermi/h² to ftm/s² 1.3 × 10+36 fermi/h² = 1 × 10+29 femtometers per second squared fermi/h² to fm/s² 1.3 × 10+36 fermi/h² = 1 × 10+29 fermis per second squared fermi/h² to fermi/s² 1.3 × 10+36 fermi/h² = 3.29 × 10+14 feet per second squared fermi/h² to ft/s² 1.3 × 10+36 fermi/h² = 498 631 203 777.77 furlongs per second squared fermi/h² to fur/s² 1.3 × 10+36 fermi/h² = 100 308.64 Gigameters per second squared fermi/h² to Gm/s² 1.3 × 10+36 fermi/h² = 1 003 086 419 833.3 hectometers per second squared fermi/h² to hm/s² 1.3 × 10+36 fermi/h² = 3.95 × 10+15 inches per second squared fermi/h² to in/s² 1.3 × 10+36 fermi/h² = 100 308 641 983.33 kilometers per second squared fermi/h² to km/s² 1.3 × 10+36 fermi/h² = 0.01 light year per second squared fermi/h² to ly/s² 1.3 × 10+36 fermi/h² = 100 308 641.98 Megameters per second squared fermi/h² to Mm/s² 1.3 × 10+36 fermi/h² = 1 × 10+14 meters per second squared fermi/h² to m/s² 1.3 × 10+36 fermi/h² = 3.95 × 10+21 microinches per second squared fermi/h² to µin/s² 1.3 × 10+36 fermi/h² = 1 × 10+20 micrometers per second squared fermi/h² to µm/s² 1.3 × 10+36 fermi/h² = 1 × 10+20 microns per second squared fermi/h² to µ/s² 1.3 × 10+36 fermi/h² = 3.95 × 10+18 mils per second squared fermi/h² to mil/s² 1.3 × 10+36 fermi/h² = 62 328 900 436.11 miles per second squared fermi/h² to mi/s² 1.3 × 10+36 fermi/h² = 1 × 10+17 millimeters per second squared fermi/h² to mm/s² 1.3 × 10+36 fermi/h² = 1 × 10+23 nanometers per second squared fermi/h² to nm/s² 1.3 × 10+36 fermi/h² = 54 162 333 694.44 nautical miles per second squared fermi/h² to nmi/s² 1.3 × 10+36 fermi/h² = 0.003 parsec per second squared fermi/h² to pc/s² 1.3 × 10+36 fermi/h² = 0.1 Petameter per second squared fermi/h² to Pm/s² 1.3 × 10+36 fermi/h² = 1 × 10+26 picometers per second squared fermi/h² to pm/s² 1.3 × 10+36 fermi/h² = 100.31 Terameters per second squared fermi/h² to Tm/s² 1.3 × 10+36 fermi/h² = 3.95 × 10+18 thous per second squared fermi/h² to thou/s² 1.3 × 10+36 fermi/h² = 1.1 × 10+14 yards per second squared fermi/h² to yd/s² 1.3 × 10+36 fermi/h² = 1 × 10+38 yoctometers per second squared fermi/h² to ym/s² 1.3 × 10+36 fermi/h² = 1 × 10-10 Yottameter per second squared fermi/h² to Ym/s² 1.3 × 10+36 fermi/h² = 1 × 10+35 zeptometers per second squared fermi/h² to zm/s² 1.3 × 10+36 fermi/h² = 1 × 10-7 Zettameter per second squared fermi/h² to Zm/s² 1.3 × 10+36 fermi/h² = 3.61 × 10+27 angstroms per minute squared fermi/h² to Å/min² 1.3 × 10+36 fermi/h² = 2 413 878.68 astronomical units per minute squared fermi/h² to au/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+35 attometers per minute squared fermi/h² to am/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+19 centimeters per minute squared fermi/h² to cm/min² 1.3 × 10+36 fermi/h² = 1.8 × 10+16 chains per minute squared fermi/h² to ch/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+18 decimeters per minute squared fermi/h² to dm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+16 dekameters per minute squared fermi/h² to dam/min² 1.3 × 10+36 fermi/h² = 0.36 Exameter per minute squared fermi/h² to Em/min² 1.3 × 10+36 fermi/h² = 1.97 × 10+17 fathoms per minute squared fermi/h² to ftm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+32 femtometers per minute squared fermi/h² to fm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+32 fermis per minute squared fermi/h² to fermi/min² 1.3 × 10+36 fermi/h² = 1.18 × 10+18 feet per minute squared fermi/h² to ft/min² 1.3 × 10+36 fermi/h² = 1.8 × 10+15 furlongs per minute squared fermi/h² to fur/min² 1.3 × 10+36 fermi/h² = 361 111 111.11 Gigameters per minute squared fermi/h² to Gm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+15 hectometers per minute squared fermi/h² to hm/min² 1.3 × 10+36 fermi/h² = 1.42 × 10+19 inches per minute squared fermi/h² to in/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+14 kilometers per minute squared fermi/h² to km/min² 1.3 × 10+36 fermi/h² = 38.17 light years per minute squared fermi/h² to ly/min² 1.3 × 10+36 fermi/h² = 361 111 111 111.11 Megameters per minute squared fermi/h² to Mm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+17 meters per minute squared fermi/h² to m/min² 1.3 × 10+36 fermi/h² = 1.42 × 10+25 microinches per minute squared fermi/h² to µin/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+23 micrometers per minute squared fermi/h² to µm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+23 microns per minute squared fermi/h² to µ/min² 1.3 × 10+36 fermi/h² = 1.42 × 10+22 mils per minute squared fermi/h² to mil/min² 1.3 × 10+36 fermi/h² = 2.24 × 10+14 miles per minute squared fermi/h² to mi/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+20 millimeters per minute squared fermi/h² to mm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+26 nanometers per minute squared fermi/h² to nm/min² 1.3 × 10+36 fermi/h² = 1.95 × 10+14 nautical miles per minute squared fermi/h² to nmi/min² 1.3 × 10+36 fermi/h² = 11.7 parsecs per minute squared fermi/h² to pc/min² 1.3 × 10+36 fermi/h² = 361.11 Petameters per minute squared fermi/h² to Pm/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+29 picometers per minute squared fermi/h² to pm/min² 1.3 × 10+36 fermi/h² = 361 111.11 Terameters per minute squared fermi/h² to Tm/min² 1.3 × 10+36 fermi/h² = 1.42 × 10+22 thous per minute squared fermi/h² to thou/min² 1.3 × 10+36 fermi/h² = 3.95 × 10+17 yards per minute squared fermi/h² to yd/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+41 yoctometers per minute squared fermi/h² to ym/min² 1.3 × 10+36 fermi/h² = 3.61 × 10-7 Yottameter per minute squared fermi/h² to Ym/min² 1.3 × 10+36 fermi/h² = 3.61 × 10+38 zeptometers per minute squared fermi/h² to zm/min² 1.3 × 10+36 fermi/h² = 0.0004 Zettameter per minute squared fermi/h² to Zm/min² 1.3 × 10+36 fermi/h² = 1.3 × 10+31 angstroms per hour squared fermi/h² to Å/h² 1.3 × 10+36 fermi/h² = 8 689 963 258.89 astronomical units per hour squared fermi/h² to au/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+39 attometers per hour squared fermi/h² to am/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+23 centimeters per hour squared fermi/h² to cm/h² 1.3 × 10+36 fermi/h² = 6.46 × 10+19 chains per hour squared fermi/h² to ch/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+22 decimeters per hour squared fermi/h² to dm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+20 dekameters per hour squared fermi/h² to dam/h² 1.3 × 10+36 fermi/h² = 1 300 Exameters per hour squared fermi/h² to Em/h² 1.3 × 10+36 fermi/h² = 7.11 × 10+20 fathoms per hour squared fermi/h² to ftm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+36 femtometers per hour squared fermi/h² to fm/h² 1.3 × 10+36 fermi/h² = 4.27 × 10+21 feet per hour squared fermi/h² to ft/h² 1.3 × 10+36 fermi/h² = 6.46 × 10+18 furlongs per hour squared fermi/h² to fur/h² 1.3 × 10+36 fermi/h² = 1 300 000 000 000 Gigameters per hour squared fermi/h² to Gm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+19 hectometers per hour squared fermi/h² to hm/h² 1.3 × 10+36 fermi/h² = 5.12 × 10+22 inches per hour squared fermi/h² to in/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+18 kilometers per hour squared fermi/h² to km/h² 1.3 × 10+36 fermi/h² = 137 413.04 light years per hour squared fermi/h² to ly/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+15 Megameters per hour squared fermi/h² to Mm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+21 meters per hour squared fermi/h² to m/h² 1.3 × 10+36 fermi/h² = 5.12 × 10+28 microinches per hour squared fermi/h² to µin/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+27 micrometers per hour squared fermi/h² to µm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+27 microns per hour squared fermi/h² to µ/h² 1.3 × 10+36 fermi/h² = 5.12 × 10+25 mils per hour squared fermi/h² to mil/h² 1.3 × 10+36 fermi/h² = 8.08 × 10+17 miles per hour squared fermi/h² to mi/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+24 millimeters per hour squared fermi/h² to mm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+30 nanometers per hour squared fermi/h² to nm/h² 1.3 × 10+36 fermi/h² = 7.02 × 10+17 nautical miles per hour squared fermi/h² to nmi/h² 1.3 × 10+36 fermi/h² = 42 130.09 parsecs per hour squared fermi/h² to pc/h² 1.3 × 10+36 fermi/h² = 1 300 000 Petameters per hour squared fermi/h² to Pm/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+33 picometers per hour squared fermi/h² to pm/h² 1.3 × 10+36 fermi/h² = 1 300 000 000 Terameters per hour squared fermi/h² to Tm/h² 1.3 × 10+36 fermi/h² = 5.12 × 10+25 thous per hour squared fermi/h² to thou/h² 1.3 × 10+36 fermi/h² = 1.42 × 10+21 yards per hour squared fermi/h² to yd/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+45 yoctometers per hour squared fermi/h² to ym/h² 1.3 × 10+36 fermi/h² = 0.001 Yottameter per hour squared fermi/h² to Ym/h² 1.3 × 10+36 fermi/h² = 1.3 × 10+42 zeptometers per hour squared fermi/h² to zm/h² 1.3 × 10+36 fermi/h² = 1.3 Zettameters per hour squared fermi/h² to Zm/h²
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#### Calculators
Calculate Maximum Heart Rate (MHR) based on age and gender | 4,637 | 10,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-17 | latest | en | 0.163523 |
http://mathhelpforum.com/business-math/178213-help-amortization-sinking-funds-print.html | 1,529,283,238,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00199.warc.gz | 211,692,493 | 3,000 | # help with amortization/sinking funds?
• Apr 20th 2011, 06:52 PM
Almondzqueen
help with amortization/sinking funds?
This is my problem:
Lauren plans to deposit \$5000 into a bank account at the beginning of next month and \$200/month into the same account at the end of that month and at the end of each subsequent month for the next 5 yr. If her bank pays interest at a rate of 6%/year compounded monthly, how much will Lauren have in her account at the end of 5 yr? (Assume she makes no withdrawals during the 5-yr period.
Formula I used looks like this:
S=R [ (1+.06/12) ^60-1]/.06/12= 13954
I suppose I need help with the second part since I have no clue how to include the initial \$5000. I am having trouble plugging the correct numbers into A= p(1+rt)^nt. I feel very stuck...
(Worried)Help and guidance is greatly appreciated, thank you.
• Apr 20th 2011, 08:10 PM
TKHunny
With such a formula that has an exponent of 60, how many interest crediting periods did the very first deposit experience?
Your initial bolus should have one more period collecting interest than had that first payment.
Always draw a map. It will save you.
• Apr 20th 2011, 10:37 PM
Wilmer
i = .06/12
S = 5000(1 + i)^60 + 200[(1 + i)^60 - 1] / i
• Apr 21st 2011, 05:42 AM
Almondzqueen
Ok, this makes more sense to me now. Thanks so much Wilmer.
• Apr 21st 2011, 08:52 AM
Wilmer
Welcome, Queen ! | 411 | 1,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.950954 |
https://www.physicsforums.com/threads/force-body-diagram-hamster.235679/ | 1,695,682,090,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00240.warc.gz | 1,048,240,296 | 14,785 | # Force Body Diagram Hamster-
• bcjochim07
## Homework Statement
A 200 g hamster sits on an 800 g wedge shaped block. The block, in turn rests on a spring scale.
a) Initially static friction is sufficient to keep the hamster from moving. In this case the hamster and the block are effectively a single 1000g mass and the scale should read 9.8 N. Show that this is the case by treating the hamster and the block as separate objects and analyzing the forces.
## The Attempt at a Solution
I am reviewing, and I know I did this problem correctly before, but now it's not working for me. Could someone please tell me what I am doing wrong
Hamster: I drew a force body diagram, tilting the x & y axes
Sum of forces on y = Force of Block on hamster - mgcos40=0
Force of block on hamster = (.200kg)(9.80)(cos40)
Block: For the block I didn't tilt the coordinates
Sum of forces on y= normal force - Force of hamster on block in y direction - Mg
= n - (.200kg)(9.80)(cos40)(cos40) - (.800kg)(9.80)
n= -8.99 N which is the answer for the second part of the question, when the static friction goes away.
Hi bcjochim07!
You're not asked for the separate friction force at this stage, so just calculate R, the total reaction force (normal + friction).
That makes only two forces on the hamster … R and mg.
So R = … ?
Then use minus R for the block.
Ok I got it, I forgot that friction has an equal opposite force | 368 | 1,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.888607 |
http://encyclopedia2.thefreedictionary.com/Divisions | 1,481,283,447,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542695.22/warc/CC-MAIN-20161202170902-00080-ip-10-31-129-80.ec2.internal.warc.gz | 92,987,652 | 16,782 | # Division
(redirected from Divisions)
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## division,
fundamental operation in arithmetic; the inverse of multiplication. Division may be indicated by the symbol ÷, as in 15 ÷ 3, or simply by a fraction, 15/3. The number that is being divided, e.g. 15, is called the dividend and the number dividing into it, e.g. 3, the divisor. The result of division is called the quotient. If the dividend is an exact (integral) multiple of the divisor, then the division will be exact, the quotient being the factor by which the divisor must be multiplied to yield the dividend (in the above example the quotient 5 multiplied by the divisor 3 equals the dividend 15). If the dividend is not an exact multiple of the divisor there will be a remainder expressed as a fraction with the divisor as the denominator; e.g., 16-3 = 5 1-3, where 1-3 is the remainder. A division in which the divisor b is larger than the dividend a is simply indicated by the fraction a/b, with no actual operation being carried out. In terms of multiplication either of the symbols 1/b or b−1 is called the multiplicative inverse of b with the property that the product of a number and its inverse equals 1, or b · b−1 =1. The division of a by b is equivalent to the multiplication of a by the multiplicative inverse of b, i.e., a ÷ b = a · (1/b) = a · b−1; for example, when a = 25 and b = 5, then 1/b = 1/5 and 25 ÷ 5 = 25 · (1/5) = 5.
## division,
in taxonomy: see classificationclassification,
in biology, the systematic categorization of organisms into a coherent scheme. The original purpose of biological classification, or systematics, was to organize the vast number of known plants and animals into categories that could be named, remembered, and
.
## Division
a form of reproduction of organisms and cells that are part of multicellular organisms.
In bacteria, division takes place by the formation of a transverse septum, which is preceded by replication of the nucleoid DNA strand. In unicellular algae and animals possessing a typical cell nucleus, division is also asexual reproduction. Division can take place both in active and in resting (encysted) states. Besides division into two, in protozoans the cytoplasm frequently separates into numerous uninuclear cells immediately after several successive divisions of the nucleus. This process is called schizogony. Division occurs in unicellular organisms (with rare exceptions, for example, in infusorians) as mitosis. It is preceded by the replication of DNA and doubling of the chromosomes. In multicellular organisms, cell division is the basis of individual development, or ontogeny, and sexual reproduction. Multicellular plants and animals are characterized by various secondary forms of reproduction, which are accomplished by the maternal organism dividing into parts of the same or different sizes (budding). Reproduction by division or budding is invariably accompanied by the regeneration of missing parts of the body. Among multicellular animals, reproduction by division occurs in certain ciliated worms.
IU. I. POLIANSKII
## Division
the inverse operation to multiplication; it is the process of finding one of two factors when their product and the other factor are given. Thus, to divide a by b means to find x such that bx = a or xb = a. The result of dividing by x is called the quotient of a and b. The given product a is called the dividend and the given factor b, the divisor. Division is denoted by a colon (a:b) or a horizontal (sometimes slanted) line a/b or a/b).
Division is not always possible in the system of numbers consisting of the integers (6 is divisible by 2 and 3 but not by 5), but in those cases where it is, the result is always uniquely determined. In the system of all rational numbers (that is, the integers and fractions) division is not only unique but is always realizable with one exception—division by zero. On the basis of the definition of division given above, it is apparent that it is not possible to divide a number different from zero by zero. The result of dividing zero by zero, according to the definition, can be any number since c . 0 = 0 in all cases. It is usually preferable in algebra (in order not to violate the uniqueness of division) to consider division by zero to be impossible for all cases.
Division with remainder differs from the exact division that has been discussed up to now. This is in essence a very special operation, which differs from the division in the sense defined above. If a and b are non-negative integers, then the operation of division with remainder of the number a by the number b consists in determining the non-negative integers x and y that satisfy the requirements (1) a = xb + y and (2) y < b. Here, a is called the dividend, b the divisor, x the quotient, and y the remainder. This operation is always realizable and unique. If y = 0, it is then said that a is divisible by b without a remainder. The operation of division with remainder is defined similarly for polynomials of the form
P(x) = a0xn + a1xn-1 + .... + an
It consists in finding for the two polynomials P(x) and Q(x) two other polynomials S(x) and R(x) that satisfy the requirements (1) P(x) = S(x)Q(x) + R(x) and (2) the degree of R(x) is less than the degree of Q(x). This operation is also always realizable and unique. lf R(x) =0, then P(x) is divisible by Q(x) without a remainder.
### REFERENCES
Depman, I. la. Istoriia arifmetiki, 2nd ed. Moscow, 1965.
Kurosh, A. G. Kurs vysshchei algebry, 9th ed. Moscow, 1968.
## Division
a large tactical unit in the ground forces, the air force, and the navy of various states. Among others there are infantry (rifle, motorized rifle, motorized, and motorized infantry) divisions, mechanized divisions, cavalry divisions, artillery divisions, antiaircraft divisions, tank (armored) divisions, air divisions, airborne divisions, air mobile divisions, and air defense divisions. The organization of troops into divisions appeared in Russia and France in the early 18th century and was firmly adopted by the armies of most states in the course of the 19th century.
Before World War I (1914-18) an infantry division usually had four infantry regiments, one or two cavalry squadrons, from 36 to 72 division artillery guns, and a total strength of 15,000-16,000 men. During the war the infantry division became a large combined arms unit with infantry, cavalry, artillery, engineer, and signal communication units. In the 1930’s tank (armored) divisions and air divisions were created in the armed forces of a number of states (the USSR, the USA, Great Britain, and Germany). In some armies (for example, the French Army) tanks were even included in infantry divisions. In most armies an infantry (rifle) division had three infantry (rifle) regiments.
In the USSR the tables of organization of the division were repeatedly changed during the Great Patriotic War (1941-45), its organization was refined because of new combat materiel and weaponry, its maneuverability and fire power increased, and its management was improved. According to the tables of organization of 1943-44 the total strength of a rifle division was to be 9,435 men and of a guards rifle division, 10,670 men. But the actual strength was as a rule below the figure given in the tables of organization. In the postwar period, when the motorization of the ground troops was completed, rifle divisions in the USSR began to be called motorized rifle divisions, and cavalry divisions were abolished. The present-day division in the armed forces of various states consists organizationally of regiments, brigades, or brigade groups. It includes units (subunits) of various combat arms and special troops, as well as various services. The strength and composition of divisions vary. For example, the mechanized division in the USA has over 18,000 men, about 190 tanks, about 2,800 motor vehicles, 850 armored personnel carriers, 57 helicopters, and 234 units of artillery, mortar, and missile weaponry. In the armed forces of many states an air division is composed of several regiments of one or various arms of aviation.
M. G. ZHDANOV
## Division
(1) In plant taxonomy, the highest conventionally used taxonomic category in the plant kingdom.
(2) In animal taxonomy, a taxonomic category sometimes used in constructing a system of higher taxons—phyla. It is not included among the taxons accepted by the International Code of Zoological Nomenclature.
(3) In zoogeography, the term “division” designates the Nearctic (New World) and Palearctic (Old World) parts of the Holarctic zoogeographic region.
(4) In anatomy and morphology, the term “division” is used to designate regions of the body or of its parts (anterior, or cranial, division; posterior, or caudal, division; trunk division; cervical division).
## division
[də′vizh·ən]
(computer science)
One of four required parts of a COBOL program, labeled identification, environment, data, and procedure, each with a set of rules governing the contents.
(mathematics)
The inverse operation of multiplication; the number a divided by the number b is the number c such that b multiplied by c is equal to a.
## division
One of the sixteen basic organizational subdivisions used in the AIA uniform system for construction specifications, data filing, and cost accounting. See illustration for contract documents.
## division
1. a part of a government, business, country, etc., that has been made into a unit for administrative, political, or other reasons
2. a formal vote in Parliament or a similar legislative body
3. a mathematical operation, the inverse of multiplication, in which the quotient of two numbers or quantities is calculated. Usually written: a ? b, a/b
4. Biology (in traditional classification systems) a major category of the plant kingdom that contains one or more related classes
5. Horticulture any type of propagation in plants in which a new plant grows from a separated part of the original
6. Logic the fallacy of inferring that the properties of the whole are also true of the parts, as Britain is in debt, so John Smith is in debt
7. (esp in 17th-century English music) the art of breaking up a melody into quick phrases, esp over a ground bass
References in classic literature ?
Let candid men judge, then, whether the division of America into any given number of independent sovereignties would tend to secure us against the hostilities and improper interference of foreign nations.
Owing to the extensive use of machinery and to division of labour, the work of the proletarians has lost all individual character, and consequently, all charm for the workman.
If you march fifty LI in order to outmaneuver the enemy, you will lose the leader of your first division, and only half your force will reach the goal.
My first endeavor was to divide the sentence into the natural division intended by the cryptographist.
in whose tones, even, denizens of the five great divisions of Europe could recognise nothing familiar
The luckless cause of these unnatural divisions now announced his presence in the halt by a loud crack of his whip.
What animation, both of body and mind, she had derived from watching the advance of that season which cannot, in spite of its capriciousness, be unlovely, and seeing its increasing beauties from the earliest flowers in the warmest divisions of her aunt's garden, to the opening of leaves of her uncle's plantations, and the glory of his woods.
Forty miles an hour will be expected, and division superintendents will accompany this special over their respective divisions.
The other half were duffle-clad, felt-hatted hillmen of the North; and that mixture told its own tale, even if he had not overheard the incessant sparring between the two divisions.
At night his antelope skin was spread where the darkness overtook him--sometimes in a Sunnyasi monastery by the roadside; sometimes by a mud-pillar shrine of Kala Pir, where the Jogis, who are another misty division of holy men, would receive him as they do those who know what castes and divisions are worth; sometimes on the outskirts of a little Hindu village, where the children would steal up with the food their parents had prepared; and sometimes on the pitch of the bare grazing- grounds, where the flame of his stick fire waked the drowsy camels.
None pay a greater regard to arbitrary divisions of time than those who are excluded from society; and Dorcas mentioned, as if the information were of importance, that it was now the twelfth of May.
The tribe of clerks was an obvious one and here I discerned two remarkable divisions.
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Open / Close | 2,853 | 12,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.921287 |
https://www.skyfilabs.com/project-ideas/hard-water-converter | 1,718,965,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00652.warc.gz | 879,673,455 | 16,143 | Hard water converter
The one of major problem in today across world is the need of a drinking water as we know on our planet 71% is covered with water but even though only 2.5% of water available is drinkable where rest of the water is containing the salt or other minerals that type of water is called as the hard water. Recent Researches are done to convert the hard water to soft or drinkable water efficiently at mass scale and that will resolve the problem for the drinking water.
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Thus our project also focus on this problem and we try to solve this problem by one of the technique to conversion the main problem with this technique to maintain the pressure inside the boiler and also the maintenance of the boiler is required constantly.
Project Description:
The main principle behind this technique is the hard water when its boiled it converts its liquid state to gaseous state into steam and thus the weight of steam will be lower than the weight of the hardness i.e salts and mineral and hence the water vapours can be collect.
At the top of the boiler and can be cooled down to get the fresh drinkable water. Also we have to check constantly the pressure inside the boiler is constantly increasing and without maintaining the pressure the boiler can fail and cannot be used constantly.
1. Boiler: It’s a component which is constantly heating the water to its boiling point and converts it to vapour state.
2. Condenser: It has the copper tube in which the vapour will pass and cool down to its liquid state.
3. Pressure gauge:To maintain the pressure inside the boiler we have to constantly cheque by using the pressure sensing gauge.
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Project Implementation:
To build a hard water convertor we have to first install the boiler for that we have to fix heating coil inside a insulating chamber with a water inlet and valve mechanism at the water inlet to maintain the water flow inside the boiling chamber.
Next we have to fix the pressure gauge inside the boiler to maintain the pressure inside the chamber without failing of the boiling chamber. Now at the top of boiler we have to give a passage from the boiler to the condenser. As vapor will be lighter than the salt and minerals hence the salt will remains at base and vapour of water will be floating at top.
In the condenser we have the copper tube where the hot vapour will pass and will cool down and change its state from gas to water again and this water will be free from the hardness.
Project Requirement:
1. Insulating walls cylinder
2. Condenser(copper tube )
3. pressure gauge
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Stay up-to-date and build projects on latest technologies | 702 | 3,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.952236 |
https://money.stackexchange.com/questions/123838/how-are-banks-able-to-offer-high-interest-rates-on-foreign-currency-fixed-deposi/123877 | 1,713,462,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00180.warc.gz | 369,047,610 | 35,676 | # How are banks able to offer high interest rates on foreign currency fixed deposits?
My bank offers foreign currency fixed deposits (FCFD) at a high rate of interest. What's the catch?
I know how the bank earns money on domestic currency fixed deposits: by loaning out the money or investing it at a higher rate of interest than what they are paying me. But why do banks want to offer fixed deposits on foreign currencies? What do they do with the foreign currency that allows them to pay me such high rates of interest?
• What is the local currency? What is the foreign currency? What interest do they pay you for the local and for the foreign currency? Apr 12, 2020 at 8:19
• What do the banks in that other country offer on their domestic fixed deposits? Apr 14, 2020 at 8:31
The bank can offer you the local interest rate on a foreign currency, which might be a lot different than your home interest rate. The bank can do this on a no-win no-lose basis, less a commission or fee they charge you, by using the forward currency market to hedge the future conversion of the foreign currency back into your home currency.
So, suppose you have dollars \$1,000, and the foreign currency is the franc, currently at a spot rate of 2.0000 francs per \$ (which is a price of 50c per franc). Also, suppose the home/dollar interest rate is 5% p.a. and the foreign/franc interest rate is 12% p.a. Your temptation is to invest your \$1,000 in Franc-land to earn 12%.
Now, the forward foreign exchange market, once hedged against currency risk, works to produce a forward exchange rate that is related to the current spot rate in the exact inverse of the relative interest rates. To see this, consider the following, assuming no bid-to-offer spreads in all rates:- You approach a bank to convert a future cash receipt of Fr2,240 back into dollars, in one year's time. The current spot rate is 2.0000Fr/\$ and interest rates are 12% in Franc-land, and 5% at home. The bank says "fine", and calculates the forward rate by hedging its exposure as follows:- The bank looks at your future franc sum of Fr2,240 and works out that with interest rates at 12% p.a. it should borrow Fr2,000 today to guarantee producing a debt of Fr2,240 in one year's time, which can be repaid by using your future franc receipt, that you would give to the bank in exchange for dollars. Now, in borrowing Fr2,000 today, the bank would not simply borrow francs and do nothing else. It will hedge its currency risk by converting the francs it had just received from the bank it had borrowed them from, into dollars at today's spot rate of Fr2.0000 per \$, producing \$1,000 ... which it will not simply sit on, but will invest at 5%, producing a sum of \$1,050 in one year's time. So, in one year's time the bank will have guaranteed a cash inflow from you in francs of Fr2,240, which it will use to pay off its franc loan, and a cash inflow of \$1,050, which it will return to you. So, given these transactions that the bank would enter into in order to guarantee a rate at which to convert your future francs, the bank says that the one-year forward rate it is prepared to quote you today, for guaranteed future settlement is Fr2,240/\$1,050 = 2.1333. You notice that this is today's spot rate of 2.0000, multiplied by 1.12 and divided by 1.05. This is how forward rates/prices are set, by simple no-arbitrage profit/loss.
So, let's look at the outcome. You provide the bank with \$1,000 and the bank guarantees to give you back Fr2,240 divided by Fr/\$2.1333 = \$1,050. Now, this is exactly the amount you would have got back had you invested at home, i.e. 5%.
Now, this is a hedged transaction for you and the bank and so you and the bank cannot win/lose (except for the bank's commission or fee). However, for you, you could decide not to ask the bank for a one-year guaranteed forward rate, choosing instead to leave your currency risk open. If the exchange rate moved in your favour, to below Fr2.1333/\$ then you would receive more than 5% in home currency terms, BUT, if the exchange rate moved against you, you could receive less than 5% in home currency terms.
So, this is how the bank can offer you an enticing-sounding interest rate, but with hedging it is no different than home interest rates. If you choose to leave yourself un-hedged, you could be better off, or worse off ! | 1,059 | 4,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.949429 |
https://www.physicsforums.com/threads/proof-of-product-rule-for-gradients.849362/ | 1,521,937,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651465.90/warc/CC-MAIN-20180324225928-20180325005928-00120.warc.gz | 858,577,721 | 15,076 | # Proof of product rule for gradients
Tags:
1. Dec 23, 2015
### Alvise_Souta
Can someone please help me prove this product rule? I'm not accustomed to seeing the del operator used on a dot product. My understanding tells me that a dot product produces a scalar and I'm tempted to evaluate the left hand side as scalar 0 but the rule says it yields a vector. I'm very confused
2. Dec 23, 2015
### blue_leaf77
Del operator $\nabla$ is a vector operator, following the rule for well-defined operations involving a vector and a scalar, a del operator can be multiplied by a scalar using the usual product. $\mathbf{A}\cdot \mathbf{B}$ is a scalar, but a vector (operator) $\nabla$ comes in from the left, therefore the "product" $\nabla (\mathbf{A}\cdot \mathbf{B})$ will yield a vector.
3. Dec 23, 2015
### Alvise_Souta
So it becomes a vector because it doesn't evaluate to 0 since the del operator is used on a scalar function rather than a scalar constant?
4. Dec 23, 2015
### blue_leaf77
Well that's a matter of whether $\mathbf{A}\cdot \mathbf{B}$ is coordinate dependent or not. If this dot product turns out to be constant, then $\nabla(\mathbf{A}\cdot \mathbf{B})=0$.
5. Dec 23, 2015
### slider142
A and B are meant to be interpreted as vector fields (physics texts frequently conflate vector fields with vectors, using the word "vector" to refer to both vectors and vector fields, where the exact meaning is meant to be gleaned from context): functions that assign a unique vector to each point. The dot product of two vector fields is therefore a scalar field, as it is meant to be interpreted as the function that assigns the dot product of the two vectors assigned by A and B, respectively, at each point to each point. The gradient of a scalar field is a vector field (or covector field, depending on how formal you want to get). | 474 | 1,853 | {"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} | 3.796875 | 4 | CC-MAIN-2018-13 | latest | en | 0.91888 |
https://physics.stackexchange.com/questions/18358/does-a-car-use-friction-to-move | 1,653,720,918,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00357.warc.gz | 498,961,044 | 65,384 | # Does a car use friction to move?
When a car's engine injects fuel into the cylinder chambers, the reaction creates a force that generates rotational momentum to the shaft and over the transmission, it translates that power to the wheels, right?
But, something's bothering me, how does the car actually move, is it dependent on the force of friction? Wheels have a contact with the road in one exact point and when the drive gets to rotate them, a force is exerted on the road in the direction of movement and there is friction that is counteracting it.
Is the key to get a car properly moving to have the drive force equal or less than the force of friction to keep that point locked to the ground and use the rolling of the wheel for translational displacement ie. moving forward? If it's bigger than the force of friction, that would cause the wheels to spin in place?
Am I completely off or am I getting something right? If there's a knowledgeable person on this topic, I'd greatly appreciate some insight, perhaps even a bigger response with some basic vehicle physics.
• Hi Scienation, and welcome to Physics Stack Exchange! We already have a few questions about the role of friction in driving that you might find useful. Also, I edited your post to make the title in the form of a question, which is the recommended way to post on this site (since it is a Q&A site). Dec 17, 2011 at 5:40
• It's not right to say that "Wheels have a contact with the road in one exact point". I think you're imagining that the wheel is perfectly round and the ground is perfectly flat. In that case there would be a line of contact. In fact the tyres are soft (and the ground is slightly soft too), and they get squashed together under the weight of the car, so there is some area in contact. Even a trains wheels have some small area in contact with the track, not just a point.
– bdsl
Nov 16, 2013 at 20:31
Friction does help in movement of a wheel aka rolling motion.
See the image, the point touching the surface actually stays at rest which causes the rest of the wheel to move forward. This is caused due to friction. Now, if you don't have friction, nothing would be there to stop the wheel from spinning around its axis.
• Please be careful to note in the image that the horizontal arrow pointing to the left (the one on top of the purple line) is the force made by the wheel pushing backward on the street, not the Friction force itself. I say this because the word 'Friction' is just below it so it might appear you are suggesting that. The Friction force has opposite direction to that (ie would be an arrow pointing to the right) in this example. Oct 19, 2015 at 22:55
• Friction (like every other interaction else) is a force-pair. The wheel pushes the road backwards, and the road pushes the wheel forwards.
– bdsl
Jan 17, 2020 at 20:45
Yes , friction is the driving force for the car to run, consider simple example where you put grease on road, is the car able to move properly..? no..because friction becomes very less..hope you got the point...
Yes I think it is propulsion due to rolling, also the tires put a reaction on the road. The engine spins the wheels through transmission as you push the gas pedal and the engine accelerates, then you shift to higher gear and spin the wheels faster with less engine acceleration as you go to higher speed.
J
First and foremost, here is the free body diagram of the wheel. The wheel is moving to the left and the friction is holding it back in the opposite direction. The $$c$$ here is damping from the bearing of the wheel, but I am not going to explain that further because we are focusing on the friction between the wheel and the ground.
From the free body diagram, the friction is only written as $$F_t$$, mainly because there are two types of frictional force, and they both play a role here. The static frictional force $$f_s$$ and the kinetic frictional force $$f_k$$ can be written as
$$f_s = \mu_s N \\ f_k = \mu_k N$$
where $$\mu_s$$ & $$\mu_k$$ are static and kinetic friction coefficient respectively and $$N$$ is the normal force.
Now, if we try to move a block of wood from a stationary position by applying a force to it, the friction force is not constant. It follows the diagram below
(source: ucsc.edu)
So, we start with not enough force to move the wooden block, what we feel is actually the static frictional force holding back our applied force. Without this, then the block would move with just a tiny amount of force (marked red). Next, if we apply more force, the friction holding it back will increase until a threshold point (marked green); this is at the exact moment when the block is just about to move. Then, as we apply even greater force, we will eventually move the block and the friction would remain constant even if we keep applying more force (marked blue).
Thus, in a car, we can imagine the scene as something similar to that block of wood. When moving, the engine rotates the wheel with just enough torque so that the wheel can "grip" the road using the static frictional force. Does that mean the car doesn't move? No, because the bottom of the wheel is held back by friction, therefore there is no motion relative to the two surfaces. Also, because the torque generated by the engine, it still wants to rotate. Since the bottom part doesn't move, it moves forward
On the other hand, if more torque is applied (or the surface is slippery, thus reducing the static frictional coefficient) then the wheel would slip or spin in place (e.g. when going through mud). Or simply put
I hope this answers your question. There are a lot of resources on the internet on this topic. This is just a summary based on the available information.
• It is my understanding that in the FBD the friction should actually point to the left. What should point to the right is the force representing the push made by the wheel to the ground. Oct 19, 2015 at 23:07 | 1,335 | 5,955 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | longest | en | 0.964452 |
http://www.comfsm.fm/~dleeling/statistics/s23q09uhx.html | 1,508,622,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00814.warc.gz | 422,138,432 | 1,735 | #### MS 150 Statistics Fall 2002 Quiz Nine: 9.4
Science courses at the national campus Fall 2001 and Spring 2002 had an overall mean grade µ = 2.103
Five hundred and eight (508) female students in science courses at the national campus Fall 2001 and Spring 2002 had a mean grade point x = 2.000 with a standard deviation sx = 1.221. At an alpha a = 0.05, is the female student mean grade point statistically significantly different than the overall mean grade point in science courses at the national campus Fall 2001 and Spring 2002? Use a two-tailed test.
1. What is the null hypothesis?
2. What is the alternate hypothesis?
3. What is a?
4. Determine tc
5. Find the t-statistic t
6. Make a sketch of the t-distribution including the tails, tc, and the t-statistic.
7. Do we:
1. Reject the null hypotheis OR
2. Fail to reject the null hypothesis?
8. Is the female students mean grade point in science courses at the national campus significantly different from the overall mean grade point at an alpha of 0.05?
9. What is the p-value for this hypothesis test?
10. What is the smallest alpha a for which the t-statistic is significant?
Statistic or Parameter Symbol Equations Excel
Confidence interval statistics
Degrees of freedom df = n-1 =COUNT(data)-1
Hypothesis Testing
Calculate a t-statistic t t
Calculate t-critical for a two-tailed test tc =TINV(a,df)
Calculate a p-value from a t-statistic (2 tail) p = TDIST(ABS(tstat),df,2) | 370 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.862255 |
https://www.coursehero.com/file/208984/MidTermExamEquations/ | 1,513,160,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522343.41/warc/CC-MAIN-20171213084839-20171213104839-00634.warc.gz | 746,699,922 | 92,657 | MidTermExamEquations
# MidTermExamEquations - Related Information Quadratic...
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Related Information Quadratic Equation: if ax 2 + bx + c = 0 , then x = - b ± b 2 - 4 ac 2 a Standard Deviation: s = r ( x i - x ) 2 N - 1 Error Propagation Equation: σ 2 y ( u, v,. .. ) = σ 2 u ± dy du ² 2 + σ 2 v ± dy dv ² 2 + ··· t-tests: True Mean Comparison: x 1 - x 2 = ± ts/ N Measured Means Comparison (when σ 2 1 = σ 2 2 ): x 1 - x 2 = ± ts pooled r N 1 + N 2 N 1 N 2 s pooled = s s 2 1 ( N 1 - 1) + s 2 2 ( N 2 - 1) N 1 + N 2 - 2 Measured Means Comparison (when σ 2 1 6 = σ 2 2 ): | x 1 - x 2 | = ± t p s 2 1 /N 1 + s 2 2 /N 2 ν = ( s 2 1 /N 1 + s 2 2 /N 2 ) 2 ( s 2 1 /N 1 ) 2 N 1 + 1 + ( s 2 2 /N 2 ) 2 N 2 + 1 - 2 Degrees of Freedom, ν t(90%) t(95%) t(99%) 1 6.314 12.706 63.657 2 2.920 4.303 9.925 3 2.353 3.182 5.841 4 2.132 2.776 4.604 5 2.015 2.571 4.032 6 1.943 2.447 3.707 7 1.895 2.365 3.500 8 1.860 2.306 3.355 9 1.833 2.262 3.250 10 1.812 2.228 3.169 1.64 1.96 2.576 Table 1: Values of t for ν degrees of freedom for 95% confidence levels. ν = N - 1 = degrees of freedom, and confidence limit = x ± ts N . 11
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F Distribution F = s 2 A s 2 B . (1) Note that critical F values in the two-tailed F -test table with probability p/ 2 can be found in the one-tailed table F -test with probability p . ν 1 =1 2 3 4 5 6 7 8 9 10 12 15 20 ν 2 1 161.4 199.5 215.7 224.6 230.2 234.0 236.8 238.9 240.5 241.9 243.9 245.9 248.0 254.3 2 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38 19.40 19.41 19.43 19.45 19.50 3 10.13 9.552 9.277 9.117 9.013 8.941 8.887 8.845 8.812 8.786 8.745 8.703 8.660 8.53 4 7.709 6.944 6.591 6.388 6.256 6.163 6.094 6.041 5.999 5.964 5.912 5.858 5.803 5.63 5 6.608 5.786 5.409 5.192 5.050 4.950 4.876 4.818 4.772 4.735 4.678 4.619 4.558 4.36 6 5.987 5.143 4.757 4.535 4.387 4.284 4.207 4.147 4.099 4.060 4.000 3.938 3.874 3.67 7 5.591 4.737 4.347 4.120 3.972 3.866 3.787 3.726 3.677 3.637 3.575 3.511 3.445 3.23 8 5.318 4.459 4.066 3.838 3.687 3.581 3.500 3.438 3.388 3.347 3.284 3.218 3.150 2.93 9 5.117 4.256 3.863 3.633 3.482 3.374 3.293 3.230 3.179 3.137 3.073
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-4-an-introduction-to-functions-get-ready-page-231/8 | 1,534,912,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219469.90/warc/CC-MAIN-20180822030004-20180822050004-00496.warc.gz | 899,370,424 | 14,243 | Algebra 1: Common Core (15th Edition)
Recall, in the coordinate plane, the x-axis is the horizontal axis and the y-axis is the vertical axis. When we see a coordinate point like (1,-2), for instance, we know that the first value listed, 1, is the x value and that the second value listed, -2, is the y value. This point indicates that we start at the origin, the point (0,0) where the x and y axis intersect, then move one unit right (on the x axis), and then move two units down (because $y=-2$). We do the same thing with all of the points listed in the problem. | 147 | 565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-34 | longest | en | 0.845392 |
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