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https://keiseronlineuniversity.com/regrouping-of-numbers-from-ones-to-tens-and-tens-to-ones/ | 1,701,918,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00399.warc.gz | 376,127,639 | 85,985 | Thursday, December 7, 2023
HomeMathRegrouping of Numbers From Ones to Tens and Tens to Ones
# Regrouping of Numbers From Ones to Tens and Tens to Ones
Regrouping of Numbers from Ones to Tens:
Writing a quantity in several methods known as regrouping.
Allow us to contemplate the quantity 18. We are able to consider it in these two methods:
18 ones could be regrouped as 1 ten 8 ones.
Now, contemplate the quantity 24.
Once more, 2 tens 15 ones is identical as 3 tens 5 ones.
Extra Examples:
18 ones = 1 ten + 8 ones
14 ones = 1 ten + 4 ones
3 tens + 12 ones = 4 tens + 2 ones
5 tens + 19 ones = 6 tens + 9 ones
1 ten +18 ones = 2 tens + 8 ones
Regrouping of Numbers from Tens to Ones:
Allow us to contemplate the quantity 16. We are able to consider it in two methods.
1 ten 6 ones could be regrouped as 16 ones.
Now, contemplate the quantity 23.
2 tens 3 ones could be regrouped as 1 ten 13 ones.
Once more, 3 tens 2 ones is identical as 2 tens 12 ones
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Original Equation: $\int$ $(2x-3)(4x^{2}+1)dx$ on the interval $[2,0]$ First, we have to expand the equation to make it easy to find the integral. We do this using the FOIL method. We see that after FOILing, the final equation becomes $\int$ $(8x^{3}-12x^{2}+2x-3)dx$ To solve this integral, we first need to find the anti-derivative. The anti-derivative of $x^{n}$ is found through the equation $\frac{x^{n+1}}{n+1}$. by applying this formula to each term in the equation, we see that the final anti-derivative is $(8x^{4}/4-12x^{3}/3+2x^{2}/2-3x)$ which can be simplified to $(2x^{4}-4x^{3}+x^{2}-3x)$. Now that we have this equation, we simply subtract the bottom range from the upper range. Our range is $[2,0]$, so we plug 2 and 0 into the anti derivative and the difference of the two is our final answer. $(2(2)^{4}-4(2)^{3}+(2)^{2}-3(2))$ - $(2(0)^{4}-4(0)^{3}+(0)^{2}-3(0))$ $= -2$ | 332 | 938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-34 | longest | en | 0.819225 |
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Chebyshev's_sum_inequality | 1,369,226,524,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701670866/warc/CC-MAIN-20130516105430-00053-ip-10-60-113-184.ec2.internal.warc.gz | 317,230,444 | 4,385 | # All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
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# Chebyshev's sum inequality
Another article treats Chebyshev's inequality in probability theory.
In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if
$a_1 \geq a_2 \geq \cdots \geq a_n$
and
$b_1 \geq b_2 \geq \cdots \geq b_n,$
then
$n \sum_{k=1}^n a_kb_k \geq \left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n b_k\right).$
Chebyshev's sum inequality follows from the rearrangement inequality.
Last updated: 06-13-2005 18:35:27
03-10-2013 05:06:04 | 272 | 918 | {"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.602001 |
https://www.citehr.com/330638-leave-calculation-el.html?status=closed | 1,580,231,965,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251779833.86/warc/CC-MAIN-20200128153713-20200128183713-00370.warc.gz | 794,111,264 | 11,027 | Leave Calculation (EL) - CiteHR
Pon1965
Construction
Yaasmin
Cite.Co is a repository of information created by your industry peers and experienced seniors sharing their experience and insights.
The XYZ company have a system of 21 EL and 4 CL in a year.
Now the case of discussion is:
If an employee remains absence from work ( treated under loss of pay) so then does than his leave of 1.75 days stands cancel for that month, and he will not be eligible for any leave accumalation.
Suppose every month he is absent any one day due to unavoidadable reoson like sickness or so, so end of the year is he not eligible for leave.
Yes He was on Leave without pay at all times, but what is prorata basis and how is it calculated
Average working days of your company - X days No. of working days by the said employee - Y days Allowed EL as per your Company Policy - 21 days EL on prorata = (21 days/X days) multiplied by Y days.
This discussion thread is closed. If you want to continue this discussion or have a follow up question, please post it on the network.
Add the url of this thread if you want to cite this discussion. | 263 | 1,123 | {"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} | 2.546875 | 3 | CC-MAIN-2020-05 | latest | en | 0.961134 |
https://www.sanfoundry.com/mathematics-questions-answers-finding-increase-decrease-percentage/ | 1,716,069,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00138.warc.gz | 870,947,442 | 20,100 | # Class 8 Maths MCQ – Finding Increase and Decrease of Percentage
This set of Class 8 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Finding Increase and Decrease of Percentage”.
1. The price of bike is Rs. 75000 last year. The price increased by 20% this year. What is the price of bike this year?
a) Rs.50000
b) Rs.60000
c) Rs.80000
d) Rs.90000
Explanation: Price of bike last year was Rs. 75000. The price increased by 20%.
Increase in price = $$\frac{20}{100}$$×75000=15000
New price = Old price + Increase in price = 75000 + 15000 = 90000.
2. The price of bike is Rs. 75000 last year. The price decreased by 20% this year. What is the price of bike this year?
a) Rs.50000
b) Rs.60000
c) Rs.80000
d) Rs.90000
Explanation: Price of bike last year was Rs. 75000. The price decreased by 20%.
Decrease in price = $$\frac{20}{100}$$×75000=15000
New price = Old price + Decrease in price = 75000 – 15000 = 60000.
3. An item is marked as Rs. 6000. There is a 25% discount on the item. What is the amount of discount received on the item?
a) Rs.6000
b) Rs.1000
c) Rs.1500
d) Rs.2500
Explanation: Marked price of item Rs. 6000. Discount of item is 25%.
Discount = $$\frac{25}{100}$$×6000=1500.
4. An item is marked as Rs. 6000. There is a 25% discount on the item. What is the discounted price of item?
a) Rs.6000
b) Rs.4500
c) Rs.1500
d) Rs.2500
Explanation: Marked price of item is Rs. 6000. Discount on item is 25%.
Discount = $$\frac{25}{100}$$×6000=1500
Discounted price = Marked price – Discount = 6000 – 1500 = 4500.
5. An item was marked for Rs. 5000 but sold for Rs. 4500. What is the % discount offered on the product?
a) 5
b) 10
c) 15
d) 20
Explanation: Marked price of item is Rs. 5000. Discounted price of item is Rs. 4500.
Discount = Marked price – Selling price = 5000 – 4500 = 500
% discount=$$\frac{discount}{marked price}=\frac{500}{5000}$$=10.
6. The list price of shirt is Rs. 2500. There is 30% discount on the item. What is the sell price?
a) Rs.1750
b) Rs.4500
c) Rs.1500
d) Rs.2500
Explanation: List price of shirt is Rs. 2500. Discount on item is 30%.
Discount = $$\frac{30}{100}$$×2500=750
Discounted price = Marked price – Discount = 2500 – 750 = 1750.
7. The list price of shirt is Rs. 2500. There is 30% discount on the item. What is the discount offered on the shirt?
a) Rs.1750
b) Rs.4500
c) Rs.1500
d) Rs.750
Explanation: List price of shirt is Rs. 2500. Discount on item is 30%.
Discount = $$\frac{30}{100}$$×2500=750.
8. Zayn bought tickets to concert for Rs. 1000. He wants to sell them at a premium of 20%. What is the selling price of tickets?
a) Rs.1750
b) Rs.1200
c) Rs.1500
d) Rs.800
Explanation: Buying price of tickets is Rs. 1000. Premium on tickets is 20%
Premium = $$\frac{20}{100}$$×1000=200
Selling price = Buying price + Premium = 1000 + 200 = 1200.
9. Zayn bought tickets to concert for Rs. 2700. He wants to sell them at a discount of 15%. What is the selling price of tickets?
a) Rs.2295
b) Rs.1200
c) Rs.2700
d) Rs.3105
Explanation: Buying price of tickets is Rs. 2700. Discount on tickets is 15%
Discount = $$\frac{15}{100}$$×2700=405
Selling price = Buying price – Discount = 2700 – 405 = 2295.
10. Zayn bought tickets to concert for Rs. 3500. He wants to sell them at a discount of 15%. What is the discount in Rs.?
a) Rs.1525
b) Rs.350
c) Rs.525
d) Rs.1050
Explanation: Buying price of tickets is Rs. 3500. Discount on tickets is 15%
Discount = $$\frac{15}{100}$$×3500=525.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,261 | 3,781 | {"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} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.918007 |
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Discussion in 'C++' started by Gary Wessle, Aug 18, 2006.
1. ### Gary WessleGuest
Hi
I have 2 valarray<double> a and b, how can I make a = 1/b, that is
each element of a is 1 / each element of b?
I have been reading around and could not understand how it is done.
let a and b have the same size.
transform(a.begin(), a.end(), &b[0], operation)
or
b = a /=(1);
not sure.
thanks
Gary Wessle, Aug 18, 2006
2. ### Guest
Gary Wessle wrote:
> Hi
>
> I have 2 valarray<double> a and b, how can I make a = 1/b, that is
> each element of a is 1 / each element of b?
>
> I have been reading around and could not understand how it is done.
> let a and b have the same size.
> transform(a.begin(), a.end(), &b[0], operation)
> or
> b = a /=(1);
>
> not sure.
>
> thanks
Well, it is a better idea to switch to vectors - as valarrays don't
have iterators, and thren transform is not usefull. An example then
would be:
#include <iostream>
#include <algorithm>
#include <vector>
double reciproke(const double& d)
{
return 1.0/d;
}
int main()
{
std::vector<double> a(2), b(2);
a[0] = 2.0;
a[1] = 4.0;
std::transform(a.begin(), a.end(), b.begin(), reciproke);
for(std::vector<double>::iterator it = a.begin(); it !=
a.end();++it)
{
std::cout << *it << std::endl;
}
for(std::vector<double>::iterator it = b.begin(); it !=
b.end();++it)
{
std::cout << *it << std::endl;
}
return 0;
}
, Aug 18, 2006
3. ### Pierre Barbier de ReuilleGuest
Gary Wessle wrote:
> Hi
>
> I have 2 valarray<double> a and b, how can I make a = 1/b, that is
> each element of a is 1 / each element of b?
>
> I have been reading around and could not understand how it is done.
> let a and b have the same size.
> transform(a.begin(), a.end(), &b[0], operation)
> or
> b = a /=(1);
>
> not sure.
>
> thanks
And why not simply :
a = 1.0 / b;
??
This operation is defined in the standard ...
Pierre
Pierre Barbier de Reuille, Aug 18, 2006
4. ### Stefan NaeweGuest
Gary Wessle schrieb:
> Hi
>
> I have 2 valarray<double> a and b, how can I make a = 1/b, that is
> each element of a is 1 / each element of b?
What's wrong with
valarray<double> b(2.0, 10);
valarray<double> a(1.0/b);
??
/S
--
Stefan Naewe
stefan_DOT_naewe_AT_atlas_DOT_de
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PRIMER DILUTION
2 replies to this topic
#1 THIRUMURUGAN
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Posted 27 June 2011 - 10:59 PM
HI TO EVERY ONE...
THE QUERY WHICH AM POSTING HERE WOULD HELP EVERYONE, WHO DEALING WITH PCR PRIMER--
THE QUERY IS ;;;
How to dilute a STOCK solution of original primer 100microMolar concentration of 1ML..(SIGMA PRIMER)
to
a> 0.36 micromolar
b> 1 micromolar
c> 10 picomoles
in a 50 microlitre PCR mix..
kindly brief the protocol and the principle behind it,,,,
I will be so grateful to all who rectify my query..
Thanks
Regards
Thiru Murugan. N
#2 scwvin
scwvin
member
• Active Members
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Neutral
Posted 27 June 2011 - 11:33 PM
Dear Ms Thiru Murugan,
You can use the M1V1=M2V2 formula.
M1 = Molarity of the stock solution (100uM)
M2 = Molarity of the final solution (a. 0.36uM, b. 1.00uM and c. 10pM)
V1 = Initial concentration (???)
V2 = Final concentration (50ul)
Example:
M1:100uM; V1:?; M2:0.36uM; V2:50ul)
M1V1 = M2V2
(100uM)(V1) = (0.36uM)(50ul)
(100uM)(V1) = 18
V1 = (18uM/ul)/(100uM)
V1 = 0.18ul
*** This is applicable if your lab have a 0.1-2.5ul pipette. If you don't have, then dilute the stock primer further to 50uM.***
Best Regards,
Vin
HI TO EVERY ONE...
THE QUERY WHICH AM POSTING HERE WOULD HELP EVERYONE, WHO DEALING WITH PCR PRIMER--
THE QUERY IS ;;;
How to dilute a STOCK solution of original primer 100microMolar concentration of 1ML..(SIGMA PRIMER)
to
a> 0.36 micromolar
b> 1 micromolar
c> 10 picomoles
in a 50 microlitre PCR mix..
kindly brief the protocol and the principle behind it,,,,
I will be so grateful to all who rectify my query..
Thanks
Regards
Thiru Murugan. N
#3 almost a doctor
almost a doctor
Veteran
• Active Members
• 246 posts
16
Good
Posted 28 June 2011 - 12:46 AM
Just a minor correction: V1 and V2 refer to VOLUME and not concentration.
The formula can also be noted as: C1V1 = C2V2 where C is concentration and V is volume just make sure to keep units consistent. | 695 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.790203 |
https://www.4-hbrandnetwork.org/commercial-property-depreciation-calculator/ | 1,569,237,257,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576355.92/warc/CC-MAIN-20190923105314-20190923131314-00540.warc.gz | 723,323,867 | 8,021 | # Commercial Property Depreciation Calculator
How is Commercial Property Depreciated? Commercial property is depreciated, generally, in one of two ways – straight line depreciation and cost segregation depreciation. The method of straight line depreciation is a simple, three-step process. You calculate the total cost basis of your commercial property. Then, divide that value by 39 to.
This calculator is geared towards residential rental property depreciation, but you can still use it to show the depreciation of commercial real estate for one or more years. However, using the general macrs method, commercial property typically has a useful life of 39 years, and the calculator only shows depreciation for up to 27.5 years.
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As mentioned earlier, commercial property owners can claim depreciation on any assets they own within the property, and tenants can claim depreciation on any assets they installed during the fit-out. If the asset is worth less than \$300, you can claim an immediate deduction in the income year that you bought it.
This managerial financial model reflects, how we measure, analyze and discuss financial results by segregating commercial performance. In the same period our real estate portfolio decreased.
An electing real property trade or business (as defined in section 163(j)(7)(B)) and electing farming business (as defined in section 163(j)(7)(C)) are required to use the alternative depreciation system for certain property to figure depreciation under MACRS for tax years beginning after 2017. Recovery period for residential rental property.
In addition, as our cost structure is based in local currency, we benefited from the strong currency depreciation in Argentina. adding — currently building very significant commercial real estate.
Calculate Term Of Loan Based On Payment Personal Loan Calculator – The Calculator Site – This loan calculator compounds interest on a monthly basis (the compound interest calculator has multiple options for compounding). What is a balloon payment? A balloon payment is a large, lump-sum payment made at the end of a long-term loan.
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How to Calculate Depreciation on a Rental Property. By Cathie. Real estate depreciation is a complex subject. BGC Partners (NASDAQ:BGCP) is a combined interdealer broker/commercial real estate enterprise that since 2014 has grown. usually the following description accompanies the "Adjustments made to. | 678 | 3,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-39 | longest | en | 0.917513 |
https://www.doubtnut.com/question-answer/find-the-areas-of-rectangles-with-the-following-pairs-of-monomials-as-their-length-and-breadth-respe-1534127 | 1,632,301,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00374.warc.gz | 751,619,249 | 84,643 | Home
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# Find the areas of rectangles with the following pairs of monomials as their length and breadth respectively: (x , y) (ii) (10, 5y) (iii) (2x^2,\ 5y^2) (4a ,\ 3a^2) (v) (3m n ,\ 4n p)
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1:25 | 445 | 1,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.509399 |
http://stackoverflow.com/questions/12763829/naive-bayse-classifier-for-multiclass-getting-same-error-rate?answertab=votes | 1,394,629,721,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021767149/warc/CC-MAIN-20140305121607-00014-ip-10-183-142-35.ec2.internal.warc.gz | 174,066,925 | 15,617 | # Naive Bayse Classifier for Multiclass: Getting Same Error Rate
I have implemented the Naive Bayse Classifier for multiclass but problem is my error rate is same while I increase the training data set. I was debugging this over an over but wasn't able to figure why its happening. So I thought I ll post it here to find if I am doing anything wrong.
``````%Naive Bayse Classifier
%This function split data to 80:20 as data and test, then from 80
%We use incremental 5,10,15,20,30 as the test data to understand the error
%rate.
%Goal is to compare the plots in stanford paper
%http://ai.stanford.edu/~ang/papers/nips01-discriminativegenerative.pdf
function[tPercent] = naivebayes(file, iter, percent)
for i=1:iter
%Getting the index common to test and train data
idx = randperm(size(dm.data,1))
%Using same idx for data and labels
shuffledMatrix_data = dm.data(idx,:);
shuffledMatrix_label = dm.labels(idx,:);
percent_data_80 = round((0.8) * length(shuffledMatrix_data));
%Doing 80-20 split
train = shuffledMatrix_data(1:percent_data_80,:);
test = shuffledMatrix_data(percent_data_80+1:length(shuffledMatrix_data),:);
%Getting the label data from the 80:20 split
train_labels = shuffledMatrix_label(1:percent_data_80,:);
test_labels = shuffledMatrix_label(percent_data_80+1:length(shuffledMatrix_data),:);
%Getting the array of percents [5 10 15..]
percent_tracker = zeros(length(percent), 2);
for pRows = 1:length(percent)
percentOfRows = round((percent(pRows)/100) * length(train));
new_train = train(1:percentOfRows,:);
new_train_label = train_labels(1:percentOfRows);
%get unique labels in training
numClasses = size(unique(new_train_label),1);
classMean = zeros(numClasses,size(new_train,2));
classStd = zeros(numClasses, size(new_train,2));
priorClass = zeros(numClasses, size(2,1));
% Doing the K class mean and std with prior
for kclass=1:numClasses
classMean(kclass,:) = mean(new_train(new_train_label == kclass,:));
classStd(kclass, :) = std(new_train(new_train_label == kclass,:));
priorClass(kclass, :) = length(new_train(new_train_label == kclass))/length(new_train);
end
error = 0;
p = zeros(numClasses,1);
% Calculating the posterior for each test row for each k class
for testRow=1:length(test)
c=0; k=0;
for class=1:numClasses
temp_p = normpdf(test(testRow,:),classMean(class,:), classStd(class,:));
p(class, 1) = sum(log(temp_p)) + (log(priorClass(class)));
end
%Take the max of posterior
[c,k] = max(p(1,:));
if test_labels(testRow) ~= k
error = error + 1;
end
end
avgError = error/length(test);
percent_tracker(pRows,:) = [avgError percent(pRows)];
tPercent = percent_tracker;
plot(percent_tracker)
end
end
end
``````
Here is the dimentionality of my data
``````x =
data: [768x8 double]
labels: [768x1 double]
``````
I am using Pima data set from UCI
-
And btw: it is Bayes, not Bayse. – Thomas Jungblut Oct 7 '12 at 10:48 | 796 | 2,868 | {"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} | 2.84375 | 3 | CC-MAIN-2014-10 | latest | en | 0.706393 |
http://www.exampleproblems.com/wiki/index.php/Supremum | 1,371,531,205,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706934574/warc/CC-MAIN-20130516122214-00040-ip-10-60-113-184.ec2.internal.warc.gz | 362,869,902 | 9,461 | # Supremum
In mathematics, the supremum of an ordered set S is the least element that is greater than or equal to each element of S. Consequently, it is also referred to as the least upper bound (also lub and LUB). The supremum may, or may not, belong to the set S. If S contains a greatest element, then that element is the supremum; and if not, then the supremum does not belong to the set.
Suprema are often considered for subsets of real numbers, rational numbers, or any other well-known mathematical structures for which it is immediately clear what it means for an element to be "greater-or-equal" than another element. Nonetheless, the definition generalizes easily to the more abstract setting of order theory where one considers arbitrary partially ordered sets.
In any case, suprema must not be confused with minimal upper bounds, or with maximal or greatest elements. Some notes on these issues follow below.
## Supremum of a set of real numbers
In analysis the supremum or least upper bound of a set S of real numbers is denoted by sup(S) and is defined to be the smallest real number that is greater than or equal to every number in S. An important property of the real numbers is its completeness: every nonempty set of real numbers that is bounded above has a supremum. If, in addition, we define sup(S) = −∞ when S is empty and sup(S) = +∞ when S is not bounded above, then every set of real numbers has a supremum (see extended real number line).
Examples:
sup { 1, 2, 3 } = 3
sup { xR : 0 < x < 1 } = sup { xR : 0 ≤ x ≤ 1 } = 1
sup { xQ : x2 < 2 } = √2
sup { (−1)n − 1/n : n = 1, 2, 3, ...} = 1
sup Z = +∞
sup { a + b : aA and bB} = sup(A) + sup(B)
The supremum of S may or may not belong to S. In particular, note the third example where the supremum of a set of rationals is irrational (which means that the rationals are incomplete). However, if the supremum value belongs to the set then it is the greatest element in the set. The term maximal element is also synonymous as long as one deals with real numbers or any other totally ordered set.
Since sup(S) is the least upper bound, to show that sup(S) ≤ a, one only has to show that a itself is an upper bound for S, i.e. one only has to show that xa for all x in S. Showing that sup(S) ≥ a is a bit harder: for any b < a, we must find an x in S with xb.
In functional analysis, one often considers the supremum norm of a bounded function f : X -> R (or C); it is defined as
$\|f\|_{\infty}=\mbox{ sup }\{\|f(x)\|:x \in X\}$
and gives rise to several important Banach spaces.
## Suprema within partially ordered sets
Least upper bounds are important concepts in order theory, where they are also called joins (especially in lattice theory). As in the special case treated above, a supremum of a given set is just the least element of the set of its upper bounds, provided that such an element exists.
Formally, we have: For subsets S of arbitrary partially ordered sets (P, ≤), a supremum or least upper bound of S is an element u in P such that
1. xu for all x in S, and
2. for any v in P such that xv for all x in S it holds that uv.
It can easily be shown that, if S has a supremum, then the supremum is unique: if u1 and u2 are both suprema of S then it follows that u1u2 and u2u1, and since ≤ is antisymmetric, one finds that u1 = u2. The dual concept of supremum, the greatest lower bound, is called infimum and is also known as meet.
If the supremum of a set S exists, it can be denoted as sup(S) or, which is more common in order theory, by $\wedge$S. Likewise, infima are denoted by inf(S) or $\vee$S.
Subsets of a partially ordered set may well fail to have a supremum, even if they have upper bounds. Some discussion on this is provided in the sections below, where the difference between suprema, maximal elements, and minimal upper bounds is stressed. As a consequence of the possible absence of suprema, classes of partially ordered sets for which certain types of subsets are guaranteed to have least upper bound become especially interesting. This leads to the consideration of so-called completeness properties and to numerous definitions of special partially ordered sets.
## Comparison with other order theoretical notions
### Greatest elements
The difference between the supremum of a set and the greatest element of a set may not be immediately obvious. The difference is exemplified by the set of negative real numbers. Since 0 is not a negative number, this set has no greatest element: for every element of the set, there is another, larger element. For instance, for any negative real number x, there is a negative real number x/2, which is greater. On the other hand, the upper bounds of the set of negative reals as a subset of the real numbers obviously constitute of all real numbers greater than or equal to 0. Hence, 0 is the least upper bound of the negative reals.
In general, this situation occurs for all subsets that do not contain a greatest element. In contrast, if a set does contain a greatest element, then it also has a supremum given by the greatest element.
### Maximal elements
For an example where there are no greatest but still some maximal elements, consider the set of all subsets of the set of natural numbers (the powerset). We take the usual subset inclusion as an ordering, i.e. a set is greater than another set if it contains all elements of the other set. Now consider the set S of all sets that contain at most ten natural numbers. The set S has many maximal elements, i.e. elements for which there is no greater element. In fact, all sets with ten elements are maximal. However, the supremum of S is the (only and therefore least) set which contains all natural numbers. One can compute least upper bounds of an element of a powerset (i.e. a set of sets) by just taking the union of its elements.
### Minimal upper bounds
Finally, a set may have many minimal upper bounds without having a least upper bound. Minimal upper bounds are those upper bounds for which there is no strictly smaller element that also is an upper bound. This does not say that each minimal upper bound is smaller than all other upper bounds, it merely is not greater. Of course this is only possible when the given order is not a total one (like the real numbers above).
As an example, let S be the set of all finite subsets of natural numbers and consider the partially ordered set obtained by taking all sets from S together with the set of integers Z and the set of positive real numbers R+, ordered by subset inclusion as above. Then clearly both Z and R+ are greater than all finite sets of natural numbers. Yet, neither is R+ smaller than Z nor is the converse true: both sets are minimal upper bounds but none is a supremum.
## Least-upper-bound property
The least-upper-bound property is an example of the aforementioned completeness properties which is typical for the set of real numbers.
If an ordered set S has the property that every nonempty subset of S having an upper bound also has a least upper bound, then S is said to have the least-upper-bound property. As noted above, the set R of all real numbers has the least-upper-bound property. Similarly, the set Z of integers has the least-upper-bound property; if S is a nonempty subset of Z and there is some number n such that every element s of S is less than or equal to n, then there is a least upper bound u for S, an integer that is an upper bound for S and is less than or equal to every other upper bound for S.
An example of a set that lacks the least-upper-bound property is Q, the set of rational numbers. Let S be the set of all rational numbers q such that q2 < 2. Then S has an upper bound (1000, for example, or 6) but no least upper bound in Q. For suppose pQ is an upper bound for S, so p2 > 2. Then q = (2p+2)/(p + 2) is also an upper bound for S, and q < p. (To see this, note that q = p − (p2 − 2)/(p + 2), and that q2 − 2 is positive.)
There is a corresponding 'greatest-lower-bound property'; an ordered set possesses the greatest-lower-bound property if and only if it also possesses the least-upper-bound property; the least-upper-bound of the set of lower bounds of a set is the greatest-lower-bound, and the greatest-lower-bound of the set of upper bounds of a set is the least-upper-bound of the set. | 1,991 | 8,333 | {"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2013-20 | latest | en | 0.905548 |
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## Find the velocity, acceleration, and speed of a particle with the given position function Sketch th
Leave a tip for good service: https://paypal.me/jjthetutor Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of t. r(t)=(t^2, 1/t^2), t=1 Student Solution Manuals: https://amzn.to/2WZrFnD More help via http://jjthetutor.com Download my eBooks via…
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algebra tutor – trigonometry tutor – precalculus tutor – calculus tutor – differential calculus tutor – integral calculus tutor – linear algebra tutor – differential equations tutor – physics tutor – mechanics physics tutor – electromagnetism physics, tutor | 3,997 | 19,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-24 | latest | en | 0.705929 |
https://www.physicsforums.com/threads/dividing-large-numbers.301470/ | 1,566,449,665,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00409.warc.gz | 934,163,144 | 17,086 | # Dividing large numbers
#### Stratosphere
How would you Divide very large numbers without using a calculator?
EX. $$\frac{125000}{299000000}$$
#### Helios
Long ago, before calculators, logarithms were used and invented for this purpose. You'd divide by subtracting logarithms and antilog the result to get the answer.
#### Santa1
One should usually first take out the obvious powers of ten, then factorize.
e.g.
$$\frac{125000}{299000000} = \frac{125}{299000}=\frac{5^3}{299\cdot 10^3} = \frac{5^3}{299\cdot (2\cdot 5)^3} = \frac{1}{299\cdot 2^3}$$
And $$299\cdot 8 = 3 \cdot 10^2 \cdot 8 - 8 = 24 \cdot 10^2 - 8 = 2400 - 8 = 2392$$,
so that
$$\frac{125000}{299000000} = \frac{1}{2392}$$
Which by hand is good enough for me.
(This might be wrong tho, it is kinda late here)
#### csprof2000
"How would you Divide very large numbers without using a calculator? "
Long division is a correct algorithm. Are you asking whether or not there exists a faster way?
#### Stratosphere
"How would you Divide very large numbers without using a calculator? "
Long division is a correct algorithm. Are you asking whether or not there exists a faster way?
Yes I am asking for a faster way.
#### Count Iblis
You could use Newton-Raphson. Computing x = 1/y for given y amounts to solving the equation:
1/x - y = 0
Then, Newton-Raphson yields the following recursion for the nth approximation
x_{n+1} = x_n - (1/x_n - y)/(-1/x_n^2) =
x_n +x_n -y x_n^2 =
2 x_n - y x_n^2
The iteration doesn't involve any divisions, so it is a true division algorithm. The number of correct digits doubles after each iteration, while with long division you only get one decimal at a time, so it is much faster than long division.
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datetime in microseconds
Hi I have a time in microseconds, for example 0x8C905CBA7F84AF4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Aug 20 '07 #1
5 5336
On Aug 20, 6:52 am, mroelo...@gmail.com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84AF4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Just truncate the value so that it's less than 1000000. A slight
difference of a couple seconds shouldn't matter for the output format
Mike
Aug 20 '07 #2
On Aug 20, 9:52 pm, mroelo...@gmail.com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84AF4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Your question can be interpreted in two possible ways:
1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>>microsecs = 0x8C905CBA7F84AF4
secs = microsecs // 1000000 # or round to nearest if you prefer
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
days, hrs, mins, secs
(7326893L, 11L, 1L, 16L)
>>>
2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84AF4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>>datetime.datetime.fromordinal(1) + datetime.timedelta(microseconds=microsecs
)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
>># Whoops!
years_approx = days / 365.25
years_approx
20059.939767282682
>>>
Hmmm, one of us seems to be missing something ...
Aug 20 '07 #3
On Aug 20, 3:15 pm, John Machin <sjmac...@lexicon.netwrote:
On Aug 20, 9:52 pm, mroelo...@gmail.com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84AF4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Your question can be interpreted in two possible ways:
1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>microsecs = 0x8C905CBA7F84AF4
secs = microsecs // 1000000 # or round to nearest if you prefer
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
days, hrs, mins, secs
(7326893L, 11L, 1L, 16L)
2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84AF4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>datetime.datetime.fromordinal(1) + datetime.timedelta(microseconds=microsecs
)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
># Whoops!
years_approx = days / 365.25
years_approx
20059.939767282682
Hmmm, one of us seems to be missing something ...
Sorry, sorry, sorry it was the wrong value, it should be
0xE0E6FAC3FF3AB2.
Aug 20 '07 #4
On Aug 20, 4:17 pm, mroelo...@gmail.com wrote:
On Aug 20, 3:15 pm, John Machin <sjmac...@lexicon.netwrote:
On Aug 20, 9:52 pm, mroelo...@gmail.com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84AF4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Your question can be interpreted in two possible ways:
1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>>microsecs = 0x8C905CBA7F84AF4
>>secs = microsecs // 1000000 # or round to nearest if you prefer
>>mins, secs = divmod(secs, 60)
>>hrs, mins = divmod(mins, 60)
>>days, hrs = divmod(hrs, 24)
>>days, hrs, mins, secs
(7326893L, 11L, 1L, 16L)
2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84AF4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>>datetime.datetime.fromordinal(1) + datetime.timedelta(microseconds=microsecs
)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
>># Whoops!
>>years_approx = days / 365.25
>>years_approx
20059.939767282682
Hmmm, one of us seems to be missing something ...
Sorry, sorry, sorry it was the wrong value, it should be
0xE0E6FAC3FF3AB2.
The solution I made, with thanks to John. Maybe someone a better one??
def DecodeDateTime(self,dateTime):
dateTime = self.Rotate(dateTime)
microsecs = int(hexlify(dateTime),16)
microsecs -= 31536000000000 # -1 Year
microsecs -= 1123200000000 # -13 Days (magic?)
secs = microsecs // 1000000
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
timed = datetime.datetime.fromordinal(1) +
datetime.timedelta(days)
return "%02d-%02d-%02d %02d:%02d:%02d"%(timed.day,
timed.month, timed.year, hrs, mins, secs)
Aug 20 '07 #5
Robert Dailey wrote:
A small off topic question. Why use divmod() instead of the modulus
operator?
Because he needed both the quotient and the remainder. % only gives you the
remainder.
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco
Aug 20 '07 #6 | 1,830 | 5,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | latest | en | 0.845958 |
http://nanobukva.ru/b/sior/Brezinski_C__Numerical_Analysis_2000%5Bc%5D_Interpolation_and_Extrapolation__Vol__2_(2000)(en)(355s)_5.html | 1,537,642,822,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158633.40/warc/CC-MAIN-20180922182020-20180922202420-00372.warc.gz | 181,444,521 | 9,771 | << Ïðåäûäóùàÿ ñòð. 5(èç 83 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>
and to new results. He also discussed applications to linear systems. In fact, as showed by Sidi [133],
and Jbilou and Sadok [75], vector sequence transformations are closely related to projection methods
for the solution of systems of equations. In particular, the RPA, a vector sequence transformation
deÿned by Brezinski [20] was extensively studied by Messaoudi who showed its connections to
direct and iterative methods for solving systems of linear equations [98,99].
Vector sequence transformations lead to new methods for the solution of systems of nonlinear
equations. They also have other applications. First of all, it is quite important to accelerate the con-
vergence of iterative methods for the solution of systems of linear equations, see [32,33,36]. Special
vector extrapolation techniques were designed for the regularization of ill-posed linear systems in
[43] and the idea of extrapolation was used in [35] to obtain estimates of the norm of the error
when solving a system of linear equations by an arbitrary method, direct or iterative.
General theoretical results similar to those obtained in the scalar case are still lacking in the vector
case although some partial results have been obtained. Relevant results on quasilinear transformations
are in the papers by Sadok [123] and Benazzouz [8]. The present author proposed a mechanism for
vector sequence transformations in [45,34].
4. Conclusions and perspectives
In this paper, I have tried to give a survey of the development of convergence acceleration
methods for scalar and vector sequences in the 20th century. These methods are based on the idea
of extrapolation. Since a universal algorithm for accelerating the convergence of all sequences cannot
exist (and this is even true for some restricted classes of sequences), it was necessary to deÿne and
study a large variety of algorithms, each of them being appropriate for some special subsets of
sequences.
It is, of course, always possible to construct other convergence acceleration methods for scalar
sequences. However, to be of interest, such new processes must provide a major improvement
over existing ones. For scalar sequence transformations, the emphasis must be placed on the theory
rather than on special devices (unless a quite powerful one is found) and on the application of new
16 C. Brezinski / Journal of Computational and Applied Mathematics 122 (2000) 1–21
methods to particular algorithms in numerical analysis and to various domains of applied sciences. In
particular, the connection between convergence acceleration algorithms and continuous and discrete
integrable systems brings a di erent and fresh look to both domains and could be of beneÿt to them.
An important problem in numerical analysis is the solution of large, sparse systems of linear
equations. Most of the methods used nowadays are projection methods. Often the iterates obtained
in such problems must be subject to acceleration techniques. However, many of the known vector
convergence acceleration algorithms require the storage of too many vectors to be useful. New
and cheaper acceleration algorithms are required. This di cult project, in my opinion, o ers many
opportunities for future research.
In this paper, I only brie y mentioned the con uent algorithms whose aim is the computation of
the limit of a function when the variable tends to inÿnity (the continuous analog of the problem
of convergence acceleration for a sequence). This subject and its applications will provide fertile
ground for new discoveries.
Acknowledgements
I would like to thank Jet Wimp for his careful reading of the paper. He corrected my English
in many places, he asked me to provide more explanations when needed, and suggested many
improvements in the presentation. I am also indebted to Naoki Osada for his informations about
Takakazu Seki.
References
[1] A.C. Aitken, On Bernoulli’s numerical solution of algebraic equations, Proc. Roy. Soc. Edinburgh 46 (1926)
289–305.
[2] H. Andoyer, Interpolation, in: J. Molk (Ed.), Encyclopà die des Sciences Mathà matiques Pures et Appliquà es, Tome
e e e
I, Vol. 4, Fasc. 1, I–21, Gauthier–Villars, Paris, 1904 –1912, pp.127–160; (reprint by Editions Gabay, Paris, 1993).
[3] F.L. Bauer, La mà thode d’intà gration numà rique de Romberg, in: Colloque sur l’Analyse Numà rique, Librairie
e e e e
Universitaire, Louvain, 1961, pp. 119 –129.
[4] B. Beckermann, A connection between the E-algorithm and the epsilon-algorithm, in: C. Brezinski (Ed.), Numerical
and Applied Mathematics, Baltzer, Basel, 1989, pp. 443–446.
[5] M.D. Benchiboun, Etude de Certaines Gà nà ralisations du 2 d’Aitken et Comparaison de Procà dà s d’Accà là ration
ee ee ee
de la Convergence, Th se 3 me cycle, Università de Lille I, 1987.
e e e
[6] M.D. Benchiboun, Extension of Henrici’s method to matrix sequences, J. Comput. Appl. Math. 75 (1996) 1–21.
[7] A. Benazzouz, Quasilinear sequence transformations, Numer. Algorithms 15 (1997) 275–285.
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<< Ïðåäûäóùàÿ ñòð. 5(èç 83 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >> | 6,133 | 19,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-39 | longest | en | 0.953195 |
https://www.tutorialspoint.com/p-find-m-if-m-frac-4-5-mplus12-p | 1,701,648,178,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00522.warc.gz | 1,167,609,380 | 21,328 | # Find $m$, if $m=\frac{4}{5}(m+12)$.
Given:
$m=\frac{4}{5}(m+12)$.
To do:
We have to find the value of $m$.
Solution:
$m=\frac{4}{5}(m+12)$
$\Rightarrow 5(m)=(4)(m+12)$ (On cross multiplication)
$\Rightarrow 5m=4m+4(12)$
$\Rightarrow 5m-4m=48$
$\Rightarrow m=48$
The value of $m$ is $48$.
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Functional Programming
Long viewed as an important theoretical idea, functional programming finally became truly convenient and practical with the introduction of Mathematica's symbolic language. Treating expressions like f[x] as both symbolic data and the application of a function f provides a uniquely powerful way to integrate structure and function—and an efficient, elegant representation of many common computations.
Function (&) — specify a pure function (e.g. (#+1)&)
#, ## slots for variables in a pure function
Map (/@) — map across a list: f/@{x, y, z}{f[x], f[y], f[z]}
Apply (@@, @@@) — apply to a list: f@@{x, y, z}f[x, y, z]
MapIndexed map with index information: {f[x, {1}], f[y, {2}], f[z, {3}]}
Nest, NestList nest a function: f[f[f[x]]] etc.
Fold, FoldList fold in a list of values: f[f[f[x, 1], 2], 3] etc.
FixedPoint, FixedPointList repeatedly nest until a fixed point
List-Oriented Functions
Select select from a list according to a function
Array create an array from a function
Sort, Split sort, split according to a function
Functional Composition Operations
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http://teachhighschoolmath.blogspot.com/2011/03/you-take-left-side-and-ill-take-right.html | 1,490,590,153,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189403.13/warc/CC-MAIN-20170322212949-00498-ip-10-233-31-227.ec2.internal.warc.gz | 353,807,427 | 13,975 | ## Sunday, March 20, 2011
### You take the Left side and I'll take the Right side
As a math teacher, I give a lot of problems to my students that are equations. This provides a great opportunity for students to work together to solve a problem. Put one student on the left side of the problem and another student on the right side of the problem. Then they can only do things on their "side" of the problem. It was amazing. Basically, it forced my students to work in tandem and at the same speed. Even better was the fact that they had to talk the math out. So one person added something to her side of the equation. That meant that their partner had to add something to their side as well. They would each discuss what had to be done next. If one student tried to do something to one side, the other person would not do anything to their side until things were explained. It was a magical moment. So here is how I designed it.
I did a podcast on this very topic at www.teachingwithsmartboard.com . Check out Podcast #81.
We were doing 2 step equations in Introduction to Algebra
1. Have the students working in pairs. Have them move their desks next to each other.
2. Have one sheet of paper per pair of students.
3. Tell the student in the left desk that they are only allowed to write on the left side of the equation. Also, the person in the right desk are only allowed to write on the right side of the equation.
4. Give a question to your students that is an equation.
5. Watch your students work together to solve the problem.
6. Monitor your students to see when they might be getting done.
7. Pick a student at random to come to the board with their partner and explain the problem. The pair should bring up the paper they were working on to guide them.
8. The pair should "JUSTIFY" what they do. I like to have both students alternate talking.
9. Enjoy having your students engaged!
Seema Goghar's MSU Blog said...
Hi, thank you so much for this idea. I have been teaching Algebra 1 for several years now and am constantly looking for ways to have students work with each other collaboratively. Solving 2 step equations is an essential algebraic skill, and too often, students struggle with it. I particularly liked your idea of having the students "JUSTIFY" what they are doing. I will definitely try this out.
Sarah said...
This is such a great idea! I have been working on solving multi-step equations with my Pre-Algebra class, and when I have them do partner work, I notice one student doing the writing and the other student watching. This would allow them both to be involved in the problem. I have done worksheets before where each partner has a different problem, but both have the same answer. This works for anything from simplifying expressions, to solve equations, to slope, but I like your take on the idea. I may have to try this tomorrow!
Thanks!
DerivativeDiary said...
This seems like good idea and I believe that it could be very effective... but do you ever worry that maybe only one of the students is doing the work? That one student is just repeating what the other is doing on their side? | 685 | 3,142 | {"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} | 3.015625 | 3 | CC-MAIN-2017-13 | longest | en | 0.983897 |
https://community.oracle.com/thread/1661367?tstart=210 | 1,500,652,715,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00153.warc.gz | 625,168,577 | 29,423 | 13 Replies Latest reply on Jun 28, 2010 4:04 PM by 796262
Rotation...?
I'm making a game in java, basically like a Tower Defense game.
I've been working on this for 4 days now and its going pretty nicely.
I'm currently stuck with rotation, the rotation works but theres a issue...
The issue is that once the angle reaches 180 degrees and try's to increase more it'll instead go to -180.
I've been searching the internet for hourrrrrrss (a good 3 hours +)
Heres my code for rotation and setting rotation on a object:
`````` //Setting of the rotations is only called when a new direction is required:
level.creep.setRequiredRotation(getDirectionDegree(level.creep[i].getMoveY()[currentTileMove + 1], level.creep[i].getY(), level.creep[i].getMoveX()[currentTileMove + 1], level.creep[i].getX()));
//Rotate the current rotation until we are at the required angle.
if(level.creep[i].getRotation() - level.creep[i].getRequiredRotation() > 10) {
level.creep[i].setRotation(level.creep[i].getRotation() - 10);
} else if(level.creep[i].getRotation() - level.creep[i].getRequiredRotation() < -10) {
level.creep[i].setRotation(level.creep[i].getRotation() + 10);
} else {
level.creep[i].setRotation(level.creep[i].getRequiredRotation());
}
public double getDirectionDegree(int y1, int y2, int x1, int x2) {
double direction = Math.toDegrees(Math.atan2(y1 - y2, x1 - x2));
return direction;
}
Heres a cleaner code that is used by towers:for(int i2 = 0; i2 < level.creep.length; i2++) {
if(level.creep[i2] != null) {
Rectangle towerRangeRectangle = new Rectangle((level.tower[i].getX() / Tile.TILE_SIZE - level.tower[i].getRange() / 2) * Tile.TILE_SIZE, (level.tower[i].getY() / Tile.TILE_SIZE - level.tower[i].getRange() / 2) * Tile.TILE_SIZE, level.tower[i].getRange() * Tile.TILE_SIZE, level.tower[i].getRange() * Tile.TILE_SIZE);
Rectangle creepRectangle = new Rectangle(level.creep[i2].getX(), level.creep[i2].getY(), Tile.TILE_SIZE, Tile.TILE_SIZE);
if(towerRangeRectangle.intersects(creepRectangle)) {
target = i2;
level.tower[i].setRequiredRotation(getDirectionDegree(level.creep[i2].getY(), level.tower[i].getY(), level.creep[i2].getX(), level.tower[i].getX()));
}
}
}
//Rotate the turret.
if(System.currentTimeMillis() - level.tower[i].lastTurretMoveTime >= 75) {
if(level.tower[i].getRotation() - level.tower[i].getRequiredRotation() > 10) {
level.tower[i].setRotation(level.tower[i].getRotation() - 10);
} else if(level.tower[i].getRotation() - level.tower[i].getRequiredRotation() < -10) {
level.tower[i].setRotation(level.tower[i].getRotation() + 10);
} else {
level.tower[i].setRotation(level.tower[i].getRequiredRotation());
}
level.tower[i].lastTurretMoveTime = System.currentTimeMillis();
}
Uses the same method to convert the radians to degrees as the top.
Does also the same basic thing as the top.
Help would be very appreciated. Thanks.
Edited by: steve4448 on Jun 21, 2010 7:22 PM ``````
• 1. Re: Rotation...?
For better help sooner, post an [SSCCE |http://sscce.org] that demonstrates the problem. A single shape rotating (or whatever your problem looks like) in place would probably be sufficient.
• 2. Re: Rotation...?
(Hmm, can't edit my top post.. weird)
Okay heres a quickly made up SSCCE: (I made it into a 'game' so it shows the problem completely, hope this helps.)
``````import javax.swing.*;
import java.awt.*;
import java.awt.event.KeyEvent;
import java.awt.event.KeyListener;
import java.awt.geom.AffineTransform;
public class Main extends JFrame {
//Okay, I made this SSCCE, all you have to do is start it and move
//around the center rectangle, once you completed a trip you'll notice it spins to -180 and to 180.
public static void main(String[] args) {
new Main();
}
public Main() {
super("My rotate example");
RotateSSCCE rotateSSCCE = new RotateSSCCE();
setResizable(false);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setSize(800, 600);
setVisible(true);
}
}
class RotateSSCCE extends Component implements KeyListener, Runnable {
//Keys are WASD to move your rectangle
Rectangle rectangleDimension = new Rectangle(400, 300, 40, 20);
Rectangle rectangle2Dimension = new Rectangle(100, 300, 40, 20);
double requiredRectangleRotation;
double currentRectangleRotation;
public RotateSSCCE() {
}
@Override
public void paint(Graphics g) {
Graphics2D g2 = (Graphics2D) g;
//Work our rotation up until we meet the required rotation
//This is where the bug is, the bug will make the rotation 'reset' once it hits 180/-180 degrees
if(currentRectangleRotation - requiredRectangleRotation >= 5) {
currentRectangleRotation -= 5;
} else if(currentRectangleRotation - requiredRectangleRotation <= -5) {
currentRectangleRotation += 5;
}
AffineTransform aT = new AffineTransform();
aT.rotate(Math.toRadians(currentRectangleRotation), rectangleDimension.x + rectangleDimension.width / 2, rectangleDimension.y + rectangleDimension.height / 2);
g2.setTransform(aT);
g2.fillRect(rectangleDimension.x, rectangleDimension.y, rectangleDimension.width, rectangleDimension.height);
AffineTransform aT2 = new AffineTransform();
aT2.rotate(0.0); //Reset the rotation so it doesn't glitch this one, any other way to do this btw?
g2.setTransform(aT2);
g2.fillRect(rectangle2Dimension.x, rectangle2Dimension.y, rectangle2Dimension.width, rectangle2Dimension.height);
}
@Override
public void keyPressed(KeyEvent kE) {
if(kE.getKeyCode() == KeyEvent.VK_W) {
rectangle2Dimension.y -= 10;
} else if(kE.getKeyCode() == KeyEvent.VK_S) {
rectangle2Dimension.y += 10;
} else if(kE.getKeyCode() == KeyEvent.VK_A) {
rectangle2Dimension.x -= 10;
} else if(kE.getKeyCode() == KeyEvent.VK_D) {
rectangle2Dimension.x += 10;
}
requiredRectangleRotation = Math.toDegrees(Math.atan2(rectangle2Dimension.y - rectangleDimension.y, rectangle2Dimension.x - rectangleDimension.x));
}
@Override
public void run() {
while(true) {
repaint();
try {
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
@Override
public void keyReleased(KeyEvent kE) {}
@Override
public void keyTyped(KeyEvent kE) {}
}``````
• 3. Re: Rotation...?
Maybe work through that, and see if there is anything you can apply to your code.
• 4. Re: Rotation...?
Sadly this isn't my problem I'm having, if you read my post I said I needed a rotation to gradually move up until its designated rotation.
So say if the rotation for the rectangle was 0, and the required rotation was 45, It would gradually plus until its reached its rotation.
In my post I mentioned I can do this fine (and if you used my SSCCE I posted, you'll notice what I mean)
but the issue is that once it hits -180/180 it practically 'resets' because of how the java rotation works.
• 5. Re: Rotation...?
steve4448 wrote:
In my post I mentioned I can do this fine (and if you used my SSCCE I posted, you'll notice what I mean)
but the issue is that once it hits -180/180 it practically 'resets' because of how the java rotation works.
No. It "resets" because of how your program logic works. Particularly these lines:
``````if(currentRectangleRotation - requiredRectangleRotation >= 5) {
currentRectangleRotation -= 5;
} else if(currentRectangleRotation - requiredRectangleRotation <= -5) {
currentRectangleRotation += 5;
}``````
Reexamine that logic.
Also, putting that stuff in your paint method (not to mention using paint instead of paintComponent) and calling repaint from a Thread is pretty ghetto. Try setting up a Timer instead.
• 6. Re: Rotation...?
What is wrong with my 'logic' in that code?
Also since this is a SSCCE I wasn't looking for having everything set-up like a real application, its just a demo application of whats wrong.
Theres nothing wrong with my rotation only that it "resets".
• 7. Re: Rotation...?
steve4448 wrote:
What is wrong with my 'logic' in that code?
If you don't understand what's happening, try adding some print lines to figure out what's going on.
Also since this is a SSCCE I wasn't looking for having everything set-up like a real application, its just a demo application of whats wrong.
That's fine, as long as you understand why the two things I listed are bad ideas.
Theres nothing wrong with my rotation only that it "resets".
Wouldn't that mean there's something wrong with it then?
• 8. Re: Rotation...?
kevinaworkman wrote:
steve4448 wrote:
What is wrong with my 'logic' in that code?
If you don't understand what's happening, try adding some print lines to figure out what's going on.
What exactly am I printing to the console?
I print out the operation (currentRectangleRotation - requiredRectangleRotation)
and I honestly don't see anything wrong with this.
Also since this is a SSCCE I wasn't looking for having everything set-up like a real application, its just a demo application of whats wrong.
That's fine, as long as you understand why the two things I listed are bad ideas.
Actually honestly I've never used paintComponent or a timer for doing this type of releated things.
Although I do my calculations in another void instead of the repaint method.
Theres nothing wrong with my rotation only that it "resets".
Wouldn't that mean there's something wrong with it then?
"Only that it "resets"", yes it does have a issue but I was stating that everything else was fine.
Thanks for posting btw, hopefully we can fix my fail of a problem :P.
• 9. Re: Rotation...?
steve4448 wrote:
You're welcome
Sadly this isn't my problem I'm having, if you read my post I said I needed a rotation to gradually move up until its designated rotation.
So say if the rotation for the rectangle was 0, and the required rotation was 45, It would gradually plus until its reached its rotation.
Well actually if you play around with this example you will see that he has solved your "reset" issue. Drag the mouse up and down past the 180/-180 lines and see that the objects continue to rotate in the same direction.
In my post I mentioned I can do this fine (and if you used my SSCCE I posted, you'll notice what I mean)
but the issue is that once it hits -180/180 it practically 'resets' because of how the java rotation works.
Your assumption that I didn't run your SSCCE is incorrect, but I'm not going to tell you what to change to fix it, I've given you a pointer to example code that has already solved your problem. It's up to you to understand that code, and the underlying issue, and to solve your problem based on that. If you still can't solve it, then at least show us that you have moved on a little bit, and ask further questions.
This isn't StackOverflow.com so you will find it less likely that people will give you complete solutions, they are more likely to give you hints/tips on how to solve your problem. I personally subscribe to this, as it gets you thinking critically, as opposed to just asking for the solution every time you hit a wall.
• 10. Re: Rotation...?
I never asked anyone to give me the complete fix.
I asked for a idea on what I'm doing wrong.
I can tell thats what you guys are trying to do but I'm not getting it...
I've tried to change my AfflineTransforms to what they have but it messed up my rotation, I don't see anything else different in my code other from than that though.
• 11. Re: Rotation...?
Your entire problem is in the if statements I pointed out in a previous post. Printing out the value of requiredRectangleRotation every time it's calculated helped me visualize what's going on. I think you already figured that part out, so what's stopping you from changing that if statement?
An easy first step towards a fix might be to figure out two separate cases instead: what do you want to do if the requiredRectangleRotation is negative? What if it's positive?
When you get that working, you can try to combine it into a single step.
Edit- Actually man, you've been a pretty good sport (better than many adults in your position) and I'm feeling generous today:
``````if(currentRectangleRotation < 0 ){
currentRectangleRotation += 360;
}
if(requiredRectangleRotation < 0 ){
requiredRectangleRotation += 360;
}
double deltaRotation = currentRectangleRotation - requiredRectangleRotation;
if(deltaRotation > 180){
deltaRotation -= 360;
}
else if(deltaRotation < -180){
deltaRotation += 360;
}
if(deltaRotation >= 5) {
currentRectangleRotation -= 5;
}
else if(deltaRotation <= -5) {
currentRectangleRotation += 5;
}``````
That seems to work. Enjoy.
• 12. Re: Rotation...?
Yay!
Thank you everyone for your time.
The code you gave me didn't work still, it basically flipped it so the other side was glitchy.
But than I noticed:
`````` if(currentRectangleRotation < 0 ){
currentRectangleRotation += 360;
}
if(requiredRectangleRotation < 0 ){
requiredRectangleRotation += 360;
}``````
and thought "Hey, shouldn't there be a clause for if its above 360 too?"
`````` if(currentRectangleRotation > 360){
currentRectangleRotation -= 360;
}
if(requiredRectangleRotation > 360){
requiredRectangleRotation -= 360;
}``````
and my rotation works 100% now. | 3,419 | 20,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-30 | latest | en | 0.595394 |
http://www.gurufocus.com/term/InventoryTurnover/ELLI/Inventory%2BTurnover/Ellie%2BMae%2BInc | 1,493,517,103,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123635.74/warc/CC-MAIN-20170423031203-00237-ip-10-145-167-34.ec2.internal.warc.gz | 561,766,807 | 28,499 | Switch to:
GuruFocus has detected 3 Warning Signs with Ellie Mae Inc \$ELLI.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Ellie Mae Inc (NYSE:ELLI)
Inventory Turnover
0.00 (As of Dec. 2016)
Inventory turnover measures how fast the company turns over its inventory within a year. It is calculated as cost of goods sold divided by average inventory. Ellie Mae Inc's cost of goods sold for the three months ended in Dec. 2016 was \$32.8 Mil. Ellie Mae Inc's average inventory for the quarter that ended in Dec. 2016 was \$0.0 Mil.
Days inventory indicates the number of days of goods in sales that a company has in the inventory. Ellie Mae Inc's days inventory for the three months ended in Dec. 2016 was 0.00.
Inventory can be measured by Days Sales of Inventory (DSI). Ellie Mae Inc's days sales of inventory (DSI) for the three months ended in Dec. 2016 was 0.00.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Ellie Mae Inc's inventory to revenue ratio for the quarter that ended in Dec. 2016 was 0.00.
Definition
Ellie Mae Inc's Inventory Turnover for the fiscal year that ended in Dec. 2016 is calculated as
Inventory Turnover (A: Dec. 2016 ) = Cost of Goods Sold / Average Inventory = Cost of Goods Sold (A: Dec. 2016 ) / ( (Inventory (A: Dec. 2015 ) + Inventory (A: Dec. 2016 )) / 2 ) = 120.145 / ( (0 + 0) / 2 ) = 120.145 / 0 = N/A
Ellie Mae Inc's Inventory Turnover for the quarter that ended in Dec. 2016 is calculated as
Inventory Turnover (Q: Dec. 2016 ) = Cost of Goods Sold / Average Inventory = Cost of Goods Sold (Q: Dec. 2016 ) / ( (Inventory (Q: Sep. 2016 ) + Inventory (Q: Dec. 2016 )) / 2 ) = 32.843 / ( (0 + 0) / 2 ) = 32.843 / 0 = N/A
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Inventory Turnover measures how fast the company turns over its inventory within a year. A higher inventory turnover means the company has light inventory. Therefore the company spends less money on storage, write downs, and obsolete inventory. If the inventory is too light, it may affect sales because the company may not have enough to meet demand.
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Ellie Mae Inc's Days Inventory for the three months ended in Dec. 2016 is calculated as:
Days Inventory = Average Inventory (Q: Dec. 2016 ) / Cost of Goods Sold (Q: Dec. 2016 ) * Days in Period = 0 / 32.843 * 365 / 4 = 0.00
2. Inventory can be measured by Days Sales of Inventory (DSI).
Ellie Mae Inc's Days Sales of Inventory for the three months ended in Dec. 2016 is calculated as:
Days Sales of Inventory (DSI) = Average Inventory (Q: Dec. 2016 ) / Revenue (Q: Dec. 2016 ) * Days in Period = 0 / 96.181 * 365 / 4 = 0.00
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Ellie Mae Inc's Inventory to Revenue for the quarter that ended in Dec. 2016 is calculated as
Inventory to Revenue = Average Inventory (Q: Dec. 2016 ) / Revenue (Q: Dec. 2016 ) = 0 / 96.181 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
Usually retailers pile up their inventories at holiday seasons to meet the stronger demand. Therefore, the inventory of a particular quarter of a year should not be used to calculate inventory turnover. An average inventory is a better indication.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Ellie Mae Inc Annual Data
Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Inventory Turnover 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Ellie Mae Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Inventory Turnover 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,190 | 4,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-17 | latest | en | 0.953889 |
https://curatedsql.com/2019/08/14/local-randomness-and-r/ | 1,566,803,766,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00070.warc.gz | 425,540,873 | 9,786 | # Local Randomness and R
2019-08-14
Let’s say we have a deterministic (non-random) problem for which one of the solutions involves randomness. One very common example of such problem is a function minimization on certain interval: it can be solved non-randomly (like in most methods of optim()), or randomly (the simplest approach being to generate random set of points on interval and to choose the one with the lowest function value).
What is a “clean” way of writing a function to solve the problem? The issue with direct usage of randomness inside a function is that it affects the state of outer random number generation:
Click through for a solution which uses random numbers but doesn’t change the outside world’s random number generation after it’s done.
## From Excel to R: Three Examples
2019-08-21
Abdul Majed Raja has a few examples of things which are easy to do in Excel and how you can do them in R: Create a difference variable between the current value and the next valueThis is also known as lead and lag – especially in a time series dataset this varaible becomes very important in feature engineering. In […]
## Calculating AUC in R
2019-08-20
Andrew Treadway shows how you can calculate Area Under the Curve in R: AUC is an important metric in machine learning for classification. It is often used as a measure of a model’s performance. In effect, AUC is a measure between 0 and 1 of a model’s performance that rank-orders predictions from a model. For […]
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August 2019
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262728293031 | 383 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-35 | longest | en | 0.905059 |
https://sites.google.com/site/barrypsgisclass/lab/lab-02-coordinate-data-projections-transformations | 1,656,779,867,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00300.warc.gz | 600,387,260 | 10,578 | Lab >
## Introduction
As part of Lab 02, students participating in WATS 6920 were given total station data from a stream habitat survey in a stretch of the Bear Valley Creek. The survey data are used to assess the habitat of salmonids in the Columbia River Basin. The data was very precise but when the points are added to ArcGIS they are not given a specified coordinate system. Completion of this lab consisted of: 1) show the unprojected points on a map using ArcGIS; 2) use the Columbia Habitat Monitoring Program (CHaMP) Transformation Tool to transform the unprojected data into projected data at the study site; 3) share the transformed data using an interactive map.
## Methods
This section describes what was done with the survey data and how it was transformed to fit the study site.
The first task was to import the unprojected data provided by the instructor and create a map that showed the data. The data was very accurate so grids were added to the map. Labels were also added to the map to help understand the different survey points.
The second task was to transform the unprojected data using the Columbia Habitat Monitoring Program (CHaMP). This tool helps move, or translate, and rotate the data to a correct position and location.
The third task was to create an interactive map that allows for users to explore the study site and the surrounding area. A Google map was used using a kmz file that showed the study site.
## Results
ArcGIS was used to take the data and plot them on a assumed Cartesian coordinate system. A Cartesian coordinate system is a way of plotting data using two points. In math Cartesian coordinates commonly use x and y coordinates, in maps it is more common to use Northing and Easting coordinates. Figure 1 shows the unprojected survey points on the assumed coordinate system in the map on the left side of the figure. The map shows that it is an assumed coordinate system because the values shown for the grids are not indicative of the location. While the survey is very accurate, the map of the unprojected points needed some work. The units are unknown, the direction of north is not known, and the data need to be transformed to better represent the true location of the study site. A pdf of the map can be downloaded here.
Figure 1. Unprojected and projected data in Bear Valley Creek, Idaho.
To transform the survey data to the correct coordinates the CHaMP Transformation Tool was used. This tool helped transform the survey data from an unprojected assumed coordinate system to a projected coordinate system. The tool also helps rotate, or orient the data, so directions are known (. To transform the unprojected data to the right coordinates the data was moved, or translated to the correct coordinates. Then the data was rotated so the North arrow was pointing vertically. The transformation that was used to do this is called the Affine Transformation. This transformation uses Northing and Easting coordinates and a minimum of three control control points. Even with the accurate data provided by the survey, error is still present. Figure 2 shows the error associated with using the different possible transformations. The transformation that was chosen was 3/GPS 2, because this choice had the least amount of error associated with the transformation. The resulting map is shown on the right side of Figure 1.
Figure 2. CHaMP Transformation Tool hinges and errors.
Other transformations include Similarity and Projective (. Similarity transformations require at least two control points but three points are recommended. This was not chosen because the Similarity transformation requires a scale be set and this did not need to be changed. The projective transformation uses a complex formula which takes transforms data from aerial photography. This was very complex so this transformation was not used.
The interactive map was created using Google maps and is shown below. Users can utilize this map and use the zoom and pan controls to verify that the site is in the correct location and projection.
## Conclusions
As part of the WATS 6920 course students took survey data that was very accurate, however this data needed to be transformed to correctly display the true placement and projection. Transformation is used when data is imported and no projection is associated with the data. In cases where data has been either previously transformed or taken with correct projections, ArcGIS can automatically transform and project data in their correct position. | 898 | 4,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-27 | latest | en | 0.930573 |
http://slideplayer.com/slide/3449524/ | 1,495,896,767,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608954.74/warc/CC-MAIN-20170527133144-20170527153144-00311.warc.gz | 408,077,818 | 88,377 | # Theoretical Foundation of Computation
## Presentation on theme: "Theoretical Foundation of Computation"— Presentation transcript:
Theoretical Foundation of Computation
Instructor: Bin Fu Textbook: Introduction to the theory of computation, by Michael Siper Class Time: 5:45-8:25pm Tuesday
Contents Computational models and languages
finite automata, push automata regular languages, context free languages Computability theory decidable problem, un-decidable problem Complexity theory time, space, P, NP, PSpace
Theory of Computation What is the computation?
What problems are computable by computer in finite steps? What problems are computable by small number of steps?
Mathematical Model Real computers are very complicate
Develop simple mathematical model to define computation The mathematical models are “equivalent to” real computer under some transformation
Un-computable problems
Important Boundary Computability theory boundary Un-computable problems Computable problems
Hierarchy inside computable area
Basic Concepts Set: a set is a group of objects represented as a unit.
{7,21,57} Element: A object of a set is called an element. Subset: A is a subset of B if every member of A is also a member of B.
Some sets Natural numbers set: N={1,2,3,…} Integers set
Z={…, -2,-1,0,1,2,…} Empty set: it is a set with no elements
Set operations Union: A={2,4,9} B={1,2,5} ={1,2,4,5,9} Intersection:
={2}
Tuple Sequence: a list of objects in some order (7, 21, 57)
Tuple: finite sequence (7, 21,57) Cartesian product: A x B is the set of all pairs with first element from A and second element from B A={1,2} B={x,y,z} A x B= {(1,x),(1,y), (1,z), (2,x), (2,y), (2,z) }
Power Set Let A be a set. The power set of A is the set of subsets of A. For A={a,b}, its power set P(A) is { , {a}, {b}, {a,b}}
Cartesian product
Function A function f is mapping from one set D to another set R
f: DR for every a in the set D, there is another element b in R such that a is mapped to b by f f(a)=b Domain: D Range: R
Function example Function f: {0,1,2,3,4}{0,1,2,3,4} n f(n) 0 1 1 2
Relation For two sets A and B, a binary relation R is a subset AxB
Example: A={scissor, paper, stone} scissor beats paper paper beats stone stone beats scissor R= {(scissor, paper), (paper, stone), (stone, scissor)}
Represent binary relation
If (a,b) in the binary relation R, we often use aRb to represent them. Example A={1,2,3}. The relation = is the set {(1,1),(2,2),(3,3)} 1=1 2=2 3=3
Relation For two sets A and B, a binary relation R is a subset AxB
Example: A={scissor, paper, stone} scissor beats paper paper beats stone stone beats scissor R= {(scissor, paper), (paper, stone), (stone, scissor)}
Represent binary relation
If (a,b) in the binary relation R, we often use aRb to represent them. Example A={1,2,3}. The relation = is the set {(1,1),(2,2),(3,3)} 1=1 2=2 3=3
Equivalence relation A binary relation R is equivalence relation if
Reflexive: xRx for every x Symmetric: if xRy, then yRx Transitive: if xRy and yRz, then xRz Example 1: = on {1,2,3}x{1,2,3} Example 2: = on NxN
Example for equivalence
Relation For two integers x and y, x y if (x-y) is a multiple of 7 In other words, there is another integer z such that (x-y)=7z.
Graph Graph: A set of nodes V A set of edges E from V x V V={ } E={ }
Path Graph G=(V,E) A path is a series of edges linked one by one Loop:
Tree A graph is connected if every two nodes have a path to connect them A tree is a connected graph without loop
Connected Graph Tree Every connected graph can be converted into tree by removing some edges Removing one edge on a loop does not damage the connectivity.
A tree is a minimal connected graph
Removing any edge on a tree damages the connectivity Proof. Tree T=(V,E). Let (v1, v2) be removed from T. T T’=(V, E-{(v1,v2)}). If T’ is still connected, T has a loop containing v1 and v2 . Contradiction!
Graph Graph: A set of nodes V A set of edges E from V x V V={ } E={ }
Node degree: The number of edges connecting to the node
v2 v1 v3 v4
Mathematical approach
Definition Mathematical statement Express some object with certain property Theorem A statement proved to be true Proof
Mathematical proof Convincing logical argument that a statement is true Usually consists a series logical statements There is a small logical gap between the current logic statement with previous statements.
3 Styles of Mathematical Proofs
Proof by construction Proof by contradiction Proof by induction
Sum of node degrees For every graph, the sum of the degrees of all nodes in G is an even number Sum=2+2+2= sum= =14
Proof Let v1, v2, …, vn are the n nodes of the graph.
deg(vi) is the degree of node vi The sum of node degrees is sum=deg(v1)+deg(v2)+…+deg(vn) For each edge e=(vi, vj), it makes one contribution to both deg(vi) and deg(vj). If there are k edges, the sum is 2k.
Regular graph A graph is k-regular if every node has degree equal to k
Theorem: For each even number n>2, there exists a 3-regular graph with n nodes.
Proof by construction n/2 nodes
Let every point at top half connect to a point in the bottom half
Proof Let be the n nodes of the graph.
Add edges for i=0, 1, …,(n/2)-1 , and Add edges for i=0,1, …,(n/2)-1
Assume the theorem is false Lead to an obviously false consequence Example: Jill just came in from outdoor and is complete dry Try to Prove: No rain Proof: Assume it were raining Jill would be wet. A contradiction!
No 3-regular graph with odd nodes
Theorem: There is no 3-regular graph with odd number of nodes Proof (by contradiction) If the graph G is a 3-regular graph with 2m+1 nodes, where m is an integer at least 0. The sum of the degrees of nodes is sum=3+3+….+3=3(2m+1)=6m+3=2(3m+1)+1 It is a contradiction to our previous theorem.
3 Styles of Mathematical Proofs
Proof by construction Proof by contradiction Proof by induction
Proof by induction Try to prove the statement P(k) is true for all nature numbers k in N={1, 2, 3,…} Basis: Prove P(1) is true Induction step: Prove if P(i) is true, then so is P(i+1)
Example for induction Theorem: For every natural number n,
Proof Basis: when n=1. The left side is 1 The right is Induction step:
Assume
Proof
Computation models Finite automata Pushdown automata Turing machine
3 Styles of Mathematical Proofs
Proof by construction Proof by contradiction Proof by induction
Proof by induction Try to prove the statement P(k) is true for all nature numbers k in N={1, 2, 3,…} Basis: Prove P(1) is true Induction step: Prove if P(i) is true, then so is P(i+1)
Example for induction Theorem: For every natural number n,
Proof Basis: when n=1. The left side is 1 The right is Induction step:
Assume
Proof
Computation models Finite automata Pushdown automata Turing machine
Finite automata Supermarket entrance
Front: A person is coming from the front. Rear: …… Both: front and rear Neither: neither front nor rear closed open
Formal definition of automata
A finite automata has Q is a finite set called the states is a finite set called the alphabet is the transition function is the start state, and is the set of accept states
Automata example q1 q3 q2
Acceptance Any sequence with at least one 1 and even number of 0s following the last 1
Notions for automata start state: q1
accept state: q2. In other words, F={q2} The set The transition function is q1 q1 q2 q2 q3 q2 q3 q2 q2 Q={q1,q2,q3}
Run the automata Start from the start state
Follow the state transition based on the current state and symbol accepts if it enters accept state, rejects otherwise
Run the machine at inputs
q1q2 q1q1q2 q1q2q3q2q3q2q3q2q3q2 q1q2 q1q2q3 101000 q1q2q3q2q3q3q3
Language and Machine Let A be a set of strings. Let M be a machine
If A is the set of strings that M accepts, A is the language of M, denoted by L(M)=A We also say “M recognizes A” or “M accepts A”
Example for language acceptance
A={ w| w has at least one 1 and an even number of 0s follow the last 1} Let M be the automata example with 3 states A=L(M)
Formal definition of automata
A finite automata has Q is a finite set called the states is a finite set called the alphabet is the transition function is the start state, and is the set of accept states
Formal definition of computation
A finite automata String w= be a string on the alphabet M accepts w if there is a sequence of states with following conditions: (1) (2) (3)
Automata What is the language accepted by the automata?
Answer L(M)={w| w ends in a 1}
Automata What is the language accepted by the automata?
Answer L(M)={w| w is an empty string or ends in a 0}
Automata What is the language accepted by the automata?
Answer L(M)={w| w starts and ends at the same symbol}
Designing Automata Language A consists all {0,1} strings with even number of 1s. Problem: design an automata M with L(M)=A.
Automata States:
Designing Automata Language A consists all {0,1} strings with 001 as substring. Problem: design an automata M with L(M)=A.
Automata States:
Regular Language M accepts language A if A={w| M accepts w}
A language is regular if some finite automata accepts it. Example: A1={w| w is {0,1} string and ends in a 1} A2={w| w is a {a,b} string that starts and ends with the same symbol}
Language and Machine Let A be a set of strings. Let M be a machine
If A is the set of strings that M accepts, A is the language of M, denoted by L(M)=A We also say “M recognizes A” or “M accepts A”
Example for language acceptance
A={ w| w has at least one 1 and an even number of 0s follow the last 1} Let M be the automata example with 3 states A=L(M)
Formal definition of computation
A finite automata String w= be a string on the alphabet M accepts w if there is a sequence of states with following conditions: (1) (2) (3)
Non-determinism blind monkey
Symbol It represents the empty symbol.
If used , one state moves to the next without consuming any symbol q1 q2
Automata example q2 q1 q3 q3 q4 q2
Acceptance of NFA The nondeterministic machine accepts if at least one of the computation branches ends in an accept state.
Language recognized Let A be a set of strings that contain either 101 or 11 as a substring. E.G , 10111
Automata example q2 q1 q3 q3 q4 q2
Language recognized Let A be a set of strings containing a 1 in the third position from the end E.G
Automata example q2 q3 q1 q3 q2 q3
Problem: What language does it accept?
q2 q3 q2 q1 q2 q5
Language recognized Accept all strings , where k is a multiple of 2 and 3.
Some notations For a set Q, P(Q) is the collection of all subsets of Q
Example, Q={q1,q2} P(Q)={empty, {q1}, {q2}, {q1,q2}} For alphabet , write
Formal definition of automata
A finite automata has Q is a finite set called the states is a finite set called the alphabet is the transition function is the start state, and is the set of accept states
Equivalence between NFA and DFA
Theorem: Every NFA has an equivalent DFA Proof. Given an NFA We will construct an to accept the same language.
Proof Q’=P(Q), which is the set of all subsets of Q
The transition function or and for some The start state F’={R in Q’| R has accept state in N}
Example Given NFA, convert it into DFA 2 1
Example Let Q={1,2}. P(Q)={empty, {1},{2}, {1,2}} 1 2 1,2
Example Let Q={1,2}. P(Q)={empty, {1},{2}, {1,2}} 2 1 1,2
Problem: What language does it accept?
q2 q3 q2 q1 q2 q5
Language Operations Let A and B be two languages Union: Concatenation:
Star:
Closure under Union Theorem: If A and B are regular languages, then
is also regular language Proof. Let A be accepted by finite automata N1, and B be accepted by finite automata N2 Find another finite automata N to accept
Construct N N1: N2:
Construct N N accepts iff one of N1 and N2 accepts
Closure under Catenation
Theorem: If A and B are regular languages, then is also a regular language Proof. Let A be accepted by finite automata N1, and B be accepted by finite automata N2 Find another finite automata N to accept
Construct N N1: N2:
Construct N N2 is linked to the accept state of N1
Closure under Concatenation
Theorem: If A is regular language, then is also a regular language Proof. Let A be accepted by finite automata N1, Find another finite automata N to accept
Construct N N1:
Construct N N1:
Exercise 1. Given NFA, convert it into DFA
2. Let A be the language recognized by the NFA. Design the automata to recognize A* 1 2
Regular Operations R is a regular expression if R is A for some a in
The special string The empty set ,where R1 and R2 are regular expressions , where R1 and R2 are regular expressions ,where R1 is a regular expression
Regular Operations Examples
0*10* 3) 4) 5) 6)
Regular Automata Theorem: Every regular expression is regular language Proof. Let R be a regular expression. Construct a DFA to accept R
6 Cases a ,where R1 and R2 are regular expressions
,where R1 is a regular expression
Case 1 a
Case 2
Case 3
Case 4 DFA N1 accepts R1 DFA N2 accepts R2
Construct N N accepts
Case 5 DFA N1 accepts R1 DFA N2 accepts R2
Construct N N accepts
Construct N N1 accepts R
Construct N N accepts R*
Example
Regular Operations R is a regular expression if R is A for some a in
The special string The empty set ,where R1 and R2 are regular expressions , where R1 and R2 are regular expressions ,where R1 is a regular expression
Regular Operations Examples
0*10* 3) 4) 5) 6)
Automata Regular expression
If a language is regular, then it is described by a regular expression Proof. Let A be recognized by a DFA A, find a regular expression R for A.
Some observation States transition 1 2 3
Some observation States transition 1 3
Some observation States transition 1 2 3
Some observations States transition 1 3
Some observation States transition 1 2 3
Some observations States transition 1 3
Some observations States transition 1 3
Proof Idea Convert DFA by shrinking it step by step
Replace the symbols by regular expressions on the state transition
Proof Idea Given DFA, covert it into regular expression
Proof Idea Add start state s and accept state a
Convert DFA by removing those old state one by one Replace the symbols by regular expressions on the state transition
Example States transition 1 2
Example States transition 1 2 s a
Example States transition 1 s a
Example States transition s a
Regular Operations For and for and
Remove a state Convert left to right
Verify The old and new machines accept the same language
Construct Regular expression
Add one start state and one accept state:
Pumping Lemma Lemma: If A is regular language, there is a number p such that if s is in A and of length at least p, s may be divided into s=xyz, satisfying 1) for each 2) 3)
Some notations For a string s, is the length of s (the number of letters) For example, |adb|=3, |a|=1, |afdsaf|=6 For a string s and integer i, is a string to repeats i times For example, if s=dgh then
Proof Idea a q9 q1 q13
Analysis Run the input string on a automata
Input string: s1 s2 s3 s4 s5 s6 … sn State q1 q2 q3 q9 q5 q9… q13 Let p be the number of states If n>p, two of the states must be equal, say q9 Repeat the substring between the two q9s reach the same accept state q13
Proof Let p be the number of the states in the Automata M for A
Let (n>=p) be a string in A Let be the state transition sequence There are two equal states (j<k) Let x takes M from r1 to rj, y takes M from rj to rj z takes M from rj to r(n+1) M must accepts for all
Application of Pumping Lemma
is not a regular language Proof: Assume it is a regular language. There is a automata M to accept it. Let p be the number from the pumping lemma Consider the string
Proof Let s= It is in the language L By the pumping lemma, s can be divided into s=xyz that satisfies 1), 2) and 3) in the pumping lemma. By 3), |xy| is at most p. So, y contains only 0s. By 1) of the pumping, is also in the language L. The number of 0s in the string is more than the number of 1s. This is a contradiction.
Application of Pumping Lemma
The language L={w | w has an equal number of 0s and 1s} is not a regular language Proof: Assume it is a regular language. There is a automata M to accept it. Let p be the number from the pumping lemma Consider the string
Proof Let s= It is in the language L By the pumping lemma, s can be divided into s=xyz that satisfies 1), 2) and 3) in the pumping lemma. By 3), |xy| is at most p. y contains only 0s. By 1) of the pumping, the string is also in the language L. The number of 0s in the string is more than the number of 1s. This is a contradiction.
Application of Pumping Lemma
The language {ww | w is a {0,1} string} is not a regular language Proof: Assume it is a regular language. There is a automata M to accept it. Let p be the number from the pumping lemma Consider the string
Proof Let s= It is in the language L By the pumping lemma, s can be divided into s=xyz that satisfies 1), 2) and 3) in the pumping lemma. By 3), |xy| is at most p. So, y contains only 0s. By 1) of the pumping, is also in the language L. The number of 0s in the first 0s area is more than the number of 0s in the second 0s area. This is a contradiction.
Application of Pumping Lemma
The language is not a regular language Proof: Assume it is a regular language. There is a automata M to accept it. Let p be the number from the pumping lemma Consider the string
Proof Let s= It is in the language L By the pumping lemma, s can be divided into s=xyz that satisfies 1), 2) and 3) in the pumping lemma. By 3), |xy| is at most p. So, y contains only 1s. By 1) of the pumping, is also in the language. The number of 1s in the string is at least and at most This is a contradiction since the language L does not contain any string with the number of 1s between and .
Problem Prove that is not a regular language.
Two ways for language A language can be accepted by a machine
A language can be also generated by some rules
Context Free Grammar Rules for generating a language: A0A1 A#
Variables: A Terminals: 0,1,# Start variable: A Rules: A0A1, A#
Conversion AXY X0 YAZ Z1 A#
Context Free Grammar Rules for generating a language: A0A1 AB B#
Variables: A, B Terminals: 0,1,# Start variable: A Rules: A0A1, AB, B#
Derivation A0A10B10#1 A0A100A11000A111000B111000#111
Context free grammar A context free grammar is
V is a finite set called the variables is a finite set of terminals R is a finite set of rultes is the start variable
Language of the grammar
If Aw is a rule, u and v are strings, then , called uAv yields uwv if for some strings The language of the grammar is
Context Free Grammar Rules for generating a language: A0A1 AB
Variables: A, B Terminals: 0,1, Start variable: A The language of this grammar is
Convert DFA to grammar Automata accepts all strings that end in 1 A B
Grammar A0A A1B B1B B0A (because B represents the accept state)
Start variable A, Variables A,B Terminals: 0,1,
DFA to grammar conversion
For every transition , add a rule For every accept state , add rule
Theorem Every regular language can be generated by a grammar with the rules like AaB, and
Chomsky normal form A context-free grammar is in chomsky normal form if every rule is of the form ABC Aa where “a” is a terminal and A,B,and C are variables. In addition, we permit if S is the start variable.
Removing empty Every context free grammar G has another equivalent context free grammar G’, which has no rule for any non-start variable A
Proof Add a new start symbol and rule
For any rule that A is not a start symbol, 1) replace every RuAv by Ruv, every RuAvAw by Ruvw, etc. 2) replace RA by Repeat the steps above until with non-start variable A does not appear
Remove Unit Rule Every context free grammar G has another equivalent context free grammar G’, which has no rule
Proof For every A B, replace every Bu by Au
Repeat the step above until there is no AB
Chomsky Normal Form Every context free grammar G has an equivalent chomsky normal form
Proof Remove the for non-start symbol A Remove the unit rule AB
For each at k>2, replace it by rules
Proof For each at k>1, replace each terminal by the new symbol and add new rule
Example A0A1 Conversion:
Remove the empty
Remove the empty
Arithmetic expression
Grammar
Example Generate
Generate the expression
Steps
Arithmetic expression
Grammar Start variable:<EXPR> Variable: <EXPR> Terminals: a,+,x,(,)
Example Generate
Example Generate
Leftmost derivation A derivation of a string w is leftmost derivation if the left most variable is replaced at every step <EXPR><EXPR> x <EXPR> <EXPR>+<EXPR> x <EXPR> a+<EXPR> x <EXPR> a+a x <EXPR> a+a x a
Ambiguity A string w is derived ambiguously in context-free grammar G if it has more than one leftmost derivation Grammar G is ambiguous if it generates some string ambiguously Ambiguity gives different interpretations by computer program
Example Design the context free grammar for
Grammar Partition the problem into two parts. One part is for
The second part is Let and S1 is used to get via S2 is used to get via
Pushdown Automata State control aaaabbbbbabaaabb a b
Example The grammar generates the language
It can be accepted by pushdown automata
Remove input 0, push 0 into stack
State control
Remove input 0, push 0 into stack
State control
Match stack 0 with input 1. Then remove both of them
State control
Match stack 0 with input 1. Then remove both of them, accept
State control
Pushdown Automata Let M be is the set of states is the input alphabet
is the stack alphabet is the state transition function is the start state, is the set of accept states q2 q1 q3 q4
States Explanation Q1 is the start state Q2 is used to push 0 symbol
Q3 is used to match 0,1 pairs between stack and input tape Q4 is the accept state
Transition a,bc: Used when the current input symbol a and the stack top symbol b, Remove the top symbol on the stack and push c on it The input is moved to the next after a
Difference between Pushdown and finite state automata
Pushdown automata has unlimited memory, which is last in, first out. Nondeterministic finite state automata is a special case of pushdown automata that has no memory
Pushdown Automata 6-tuple Q is the finite set of states
is the input alphabet (finite) is the stack alphabet (finite) is the start state, is the set of accept states
A computation of pushdown automata
Input string: State transition sequence: For i=0,…,m-1, where and Final state accepts
Theorem Every context free language can be accepted by a non-deterministic pushdown automata
Example The grammar generates the language
It generates the string 0011
Pushdown Automata State control
Replace S by 0S1 at stack State control
Match the stack top symbol with the input symbol, remove both if matched
State control
Replace S by 0S1 State control
Remove after matching State control
Replace S by empty State control
Remove after matching State control
Remove after matching, Accept
State control
Proof Idea Push \$ and start symbol to the stack in the beginning.
Repeat the three steps below Replace the top variable A on the stack with a the right side of a rule Match the top terminal on the stack with the input symbol, reject if not matched When stack has \$ on the top and all input has been read, accepts
How to replace the variable
If the current stack has the top element s, replace it with the right of the rule State q a…………….. s .
New States The states are new states, specially added for the rule
State moves from q to r after the variable is replaced on the top of the stack
Pumping Lemma Lemma: If A is context-free language, there is a number p such that if s is in A and of length at least p, s may be divided into s=uvxyz, satisfying 1) for each 2) 3)
Proof Idea Two variables are the same on a path
Proof Idea A path from root to a leaf has all variables except the last one, which is a terminal When the path is too long, same variable has to happen twice. Repeat the part of the two equal variables area.
Analysis Let V be the set of variables in a grammar.
|V| is the number of variables in V If a path has at least |V|+1 variables, two of them will be equal. A path from root to a leaf has least |V|+1 variables if its length is at least |V|+2 A tree of depth |V|+2 has leaves, where b is the maximal length of right side among all rules.
Proof Let If s has length , the parse tree T has height
It has a path has at least |V|+1 variables. Two of the variables on the path are equal (=R). We have So,
Proof The tree T is the parsing tree for s and has least size
It is impossible that is empty otherwise, T is not least. Select the bottom |V|+2 symbols on the longest path of T, this makes the
Application The language is not context free
Proof. Assume it is a context free language. It can be generated by a context free grammar G. Apply pumping lemma, to derive some strings not in the language can be also derived by G.
Proof The language is not context free Proof. Let
By Pumping lemma, s can be expressed as s=uvxyz s. t And Case 1. v and y are in the area of the same symbol, say a. Contradiction for having more a than b. Case 2. v or y contains more than one symbol, contradiction for the incorrect order of symbols in
Regular Pumping The language is not regular
Proof. Assume it is a regular language. It can be accepted by automata M. Let p be the length of the pumping lemma. Consider s can be expressed s=xyz, where The string xz has the number of 0s no more than 1s (Pump it down to get the contradiction)
Application The language is not context free
Proof. Assume it is a context free language. It can be generated by a context free grammar G. Apply pumping lemma, to derive some strings not in the language can be also derived by G.
Proof The language is not context free Proof. Let
By Pumping lemma, s can be expressed as s=uvxyz s. t And Case 1. vy have intersection with the a area. Contradiction for having more a or b than c (pumping up). Case 2. vy does not contain symbol a, contradiction for having more a than b or c (pumping down).
Application The language is not context free
Proof. Assume it is a context free language. It can be generated by a context free grammar G. Apply pumping lemma, to derive some strings not in the language can be also derived by G.
Proof The language is not context free Proof. Let
By Pumping lemma, s can be expressed as s=uvxyz s. t And Case 1. vy is in the first half. Contradiction for moving 1 to the right area(pumping up). Case 2. vy is in the second half. Contradiction for moving 0 to the left area(pumping up).
Proof The language is not context free
Proof. Case 3. vy crosses the middle line. Contradiction for reducing 0 or 1 in the middle area (pumping down). The left half and right half have different number of 1s or 0s.
Problem Write a context-free grammar for the following language
Problem Using pumping lemma to disprove the following language is context-free:
Algorithm An algorithm is a collection of simple instructions for carrying out some task
Church-Turing Thesis Intuitive Algorithm is equal to Turing machine algorithms
Unlimited Register Machine
Statement1: xx+1 Statement2: xy Statement3: x0 Statement4: if (x==y) jump m Each register can save an unbounded integer
Turing Machine Proposed in 1936
An accurate model for the general purpose computer.
Turing Machine Write on the tape and read from it
Head can move left and right Tape is infinite Rejecting and accepting states Control a b a b
Language ww Design Turing machine for L={w#w|w is in {0,1}*}
For example: 011#011 is in L 10011#10011 is in L
Movement on the tape Move the head back and forth to match all pairs
Movement on the tape Move the head
Turing Machine 7-tuple Q is the finite set of states
is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
State transition function
For
State transition function
For
Configuration Current state: q7
Current head position on the tape: 4th cell Current tape content: abab q7 a b a b
Configuration A configuration is represented by
Where is the left part of the tape content, is the right part of the tape content, a is the symbol at the head position, q is the current state
Configuration Transition
For
Configuration Transition
For
Configuration Start configuration: , where w is the input
Accepting configuration: a configuration with state Rejecting configuration: a configuration with state
Accept Computation A Turing machine M accepts input w if a sequence of configurations exists where is the start configuration of M on input w, 2. each yields , and is an accepting configuration
Language recognized by TM
For a Turing machine M, L(M) denotes the set of all strings accepted by M. A language is Turing recognizable if some Turing machine recognizes it.
a q1 q3 q2 q4 q5 q7 q6 q9 q8 q11 q10 q12 q13 q_accept q14
Multi-tape Turing Machine
1 1 Each tape has its own head Initially the input is at tape 1 and other tapes are blank Reading, writing and moving at all tapes Control a a a b a
Multi-tape Turing Machine
Transition for multi-tape Turing machine
Simulate Multitape Turing machine
Theorem: Every multitape Turing machine has an equivalent single tape Turing machine q7 # 1 1 # a a a # b a #
Simulation Initially, For one move simulation, scan the first #, then the second #, etc. It one tape goes out of its #, move the rest of tape in order to get one space for it. (stupid, slow)
Another simulation Assume there are k tapes
Let the positions i, k+i, 2k+i, …., mk+i,… be used for simulating the tape i. q7 a b 1 a a a
Simulate 3 tapes q7 a b 1 a a a
Simulate 3 tapes q7 a b 1 a a a
K tapes Use two consecutive cells for one symbol, one is for storing the symbol and the other one is for storing the dot mark The tape i’s first j-cell information is at 2(j-1)k+2i-1 and 2(j-1)k+2i
Midterm October 18, 2010 Class Time Close book Chapter 0-Chapter 3
Problem 5 Problem 5 (20) Give a description of a Turing machine that decides the following language over the alphabet {0,1} {w|w contains an equal number of 0s and 1s}.
Turing Machine Write on the tape and read from it
Head can move left and right Tape is infinite Rejecting and accepting states Control a b a b
Language L Design Turing machine for L={w|w is in {0,1}* and the same number 1s as the same number of 0s} For example: is in L is not in L
Movement on the tape Move the head back and forth to pair up 0 and 1.
Movement on the tape Move the head
Turing Machine Write on the tape and read from it
Head can move left and right Tape is infinite Rejecting and accepting states Control 1 1
Language L Design Turing machine for L={w|w is in {0,1}* and the same number 0s is two times the number of 1s} For example: is in L is not in L
Movement on the tape Move the head back and forth to pair up 0 and 1.
Movement on the tape Move the head
Problem 6 Problem 6 (10) If A and B are languages, define
and Prove that if A and B are regular languages, then is a context free language.
Problem 6 DFA N1 accepts A DFA N2 accepts B
Construct N N accepts
Problem 7 Problem 7 (10) If A and B are languages, define
and and |x|=|y|} Prove that if A and B are regular languages, then is a context free language. You will get 10 more points for midterm.
Deterministic Turing Machine
7-tuple Q is the finite set of states is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
Nondeterministic Turing Machine
7-tuple Q is the finite set of states is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
Nondeterministic Turing Machine
One configuration can have multiple choice for entering the next configuration because of
Non-determinism blind monkey
Automata example q2 q1 q3 q3 q4 q2
Acceptance of NFA The nondeterministic machine accepts if at least one of the computation branches ends in an accept state.
Acceptance of NTM The nondeterministic Turing machine accepts if at least one of the computation branches ends in an accept state.
Simulate NTM Theorem: Every Nondeterministic Turing machine has an equivalent deterministic Turing machine.
Simulate NTM with DTM Each dot is one configuration
Find an accepting path
Simulate NTM Think about the problem from the Java programming point of view Convert the strategy into deterministic Turing machine.
Simulate NTM Key points: search accepting path with width first until it finds the first one. Search from the left to right Search from small level 0, to level 1, to level 2,…
Simulate NTM 1: first branch, 2: second branch, etc
1 1 1: first branch, 2: second branch, etc 121: take the first branch at level 2, then take the second at level 3, then take the first branch at level 4. Control a a a 1 2 1
Turing machine and Computer
The computational power of Turing machine is equivalent to regular computer with unlimited memory
Successor Funtion f(x)=x+1
Successor Funtion f(x)=x+1
Successor Turing machine is able to simulate the operation xx+1 1 1 1
1 1 1 1 1
Comparison Turing machine is able to check if x=y 1 1 1 1 1 1 1 1 1 1
Transfer Turing machine is able to simulate yx 1 1 1 1 1
Transfer Turing machine is able to simulate xy 1 1 1 1 1 1 1 1 1 1
Set zero Turing machine is able to simulate x0 1 1 1 1 1
Program statements Statement1: xx+1 Statement2: xy Statement3: x0
Statement4: if (x==y) jump m
Compute Algorithm: Let x be added by 1 y times Program for x+y: 1: z0
2: If (z==y) jump 6 3: xx+1 4: zz+1 5: If (z==z) jump 2 6:
Jump TM program state Program q1 1: I1 q2 2: I2 q3 3: I3 …. ……
… …… qk k: jump 3 The jump can be achieved via state transition qkq3
Compute Algorithm: Let x be added by original x y times
Program for x*y: 1: z0 2: p0 3: If (z==y) jump 7 4: pp+x 5: zz+1 6: If (z==z) jump 3 7:
Compute Algorithm: Let x be multiplied by x y times Program for :
1: z0 2: p (via p0 and pp+1) 3: If (z==y) jump 7 4: pp*x 5: zz+1 6: If (z==z) jump 3
Algorithm An algorithm is a collection of simple instructions for carrying out some task
Church-Turing Thesis Intuitive Algorithm is equal to Turing machine algorithms
Unlimited Register Machine
Statement1: xx+1 Statement2: xy Statement3: x0 Statement4: if (x==y) jump m Each register can save an unbounded integer
Other computation model
Lambda calculus (Church) Recursion (Godel and Kleene) Post and Markov’s model
Evidences Many approaches led to the same algorithmic computable class
No one has found an algorithm that is accepted in informal sense, but it can not be implemented in Turing model
Hilbert 10th Problem Find an algorithm to decide if a polynomial has integer root Input: a polynomial e.x. (it has root x=5,y=3,z=0) Output: yes or no
Midterm Problem 1 (20) Give the state diagram of a DFA recognizing {w|w is 0,1-string with at least five 0s}. Problem 2 (20) Prove that is not regular language with the pumping lemma.
Midterm Problem 3 (20) Prove the following facts:
a) If A and B are context free languages, then so is their union . b) If A and B are context free languages, then so is their concatenation.
Midterm Problem 4 (20) a) Design a context-free grammar to recognize the language b) Prove that is not a context-free language by using pumping lemma.
Midterm Problem 5 (20) Give a description of a Turing machine that decides the following language over the alphabet {0,1} {w|w contains more 0s than 1s}.
Decidability A language L is Turing decidable if there is a deterministic Turing machine M such that If x is in L, then M accepts x in finite number of steps If x is not in L, then M rejects x in finite number of steps Example: {w#w| w is in {0,1}*} is Turing decidable
DFA ={<B,w>| B is a DFA that accepts input string w}
Theorem: is a decidable language.
Proof The input is <B,w>, where B is a DFA , and w is a string.
Simulate B on the input w: Start from the state of B and leftmost symbol w. Follow to transit the state and move the input symbols one by one Accepts if ends at an accept state; rejects otherwise.
NFA ={<B,w>| B is a NFA that accepts input string w}
Theorem: is a decidable language.
Proof The input is <B,w>, where B is a NFA and w is a string.
Convert B into a DFA C via our previous algorithm Use our last TM to decide if C accepts w Accepts if it accepts; rejects otherwise.
={<A>| L(A) is not empty}
Theorem: is a decidable language.
Proof The input is <A>, where A is a DFA. Mark the start state
Repeat Mark a new state that has a transition to it from a marked one. Until no new state can be added Accept if an accept state is marked; otherwise reject
={<A,B>| A and B are DFA and L(A)=L(B)}
Theorem: is a decidable language.
Proof The input is <A,B>, where A and B are DFA Check if
is empty
Problem Define L={ <A,B,C>: L(A) is the union of L(B) and L(C)}, where A, B, and C are DFAs. Prove that L is decidable.
One- one and onto Let A and B be two sets.
For function f: AB, if whenever then f is called one-one. For function f: A B, say f is onto if f hits every element of B(In other words, for very b in B, there is a in A such that b=f(a))
Correspondence Let A and B be two sets.
A and B are of the same size if there is a one-one and onto function f: AB For function f: AB, if it is both one-one and onto, then f is called correspondence.
Examples {1,2,3,…} and {2,4,6,…} are of the same size via f(x)=2x.
Countable A set is countable if it is finite or it has the same size as N={1,2,3,…} Theorem: The positive rational numbers set is coutable
Proof Every positive rational number is in the table below
Proof List all of them by and avoid repetition
Examples For two positive rational numbers , ( p and q have no common divisor>1, and p’ and q’ have no common divisor >1), The number is listed before if p+q<p’+q’, or p+q=p’+q’ and q<q’
(0,1) is not countable Theorem: (0,1) is not countable
Assume N={1,2,3,…} and (0,1) have the same size. There is one-one and onto map f: N (0,1)
Proof. Select such that and Since
Infinite binary strings set is not countable
Let be the set of all infinite binary strings Theorem: is not countable
Assume N={1,2,3,…} and have the same size. There is one-one and onto map f: N
Proof. Select such that and Since
The set of finite binary strings is countable
Correspondence N= {0, 1 , 2, 3, 4, 5, 6, 7, , , 10,…}
Language of Binary Strings
Every set A of binary strings is countable Proof. Let A be a set of binary strings. Its elements can be listed according to their orders in A={ s1, s2, s3, s4, ….} It is easy to see the correspondence between N and A
An infinite binary string uniquely determines a language
The language with positive even number of 1s strings
Correspondence For each infinite binary string B, it uniquely determines a binary language L(B). If B1 and B2 are different binary strings, then L(B1) and L(B2) are different language Each binary language uniquely determines an infinite binary string.
Correspondence There is a correspondence between the set of all infinite binary strings and the set of all binary languages. The set of all binary languages is not countable
Turing machine to binary string
Each Turing machine can be encoded into a binary string. The set of all Turing machines can be encoded into a set of binary strings. The set of Turing machines is countable. The set of binary languages recognized by TM is also countable
Un-computable language by TM
There is a binary language that is not Turing recognizable Proof. The set of binary languages is not countable, but the set of language recognized by Turing machine is countable
Undecidable Problem ={<M,w>| M is a Turing machine and M accepts w} Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Every entry of the table can be obtained in finite steps
accept reject accept reject accept accept accept accept reject reject reject reject accept accept reject reject ……
Proof Assume that is decidable There is a TM H such that
H(<M,w>) accepts if M accepts w H(<M,w>) rejects if M rejects w. Consider a TM D D(<M>) accepts if M rejects <M> D(<M>) rejects if M accepts <M>
Proof Since D(<M>) accepts if M rejects <M>
D(<M>) rejects if M accepts <M> We have D(<D>) accepts if D rejects <D> D(<D>) rejects if D accepts <D> A contradiction
Every entry of the table can be obtained in finite steps
accept reject accept reject accept accept accept accept reject reject reject reject accept accept reject reject ……
Every diagonal entry of the table can be obtained in finite steps
accept reject accept reject accept accept accept accept reject reject reject reject accept accept reject reject ……
Diagonal method reject reject accept accept accept reject ……
Call H in R If TM H exists, the TM R also exits via using H
(Software R uses an existing software H)
Decidability A language L is Turing decidable if there is a deterministic Turing machine M such that If x is in L, then M accepts x in finite number of steps If x is not in L, then M rejects x in finite number of steps Example: {w#w| w is in {0,1}*} is Turing decidable
Problem Is there any one-one and onto map from the set of integers in [1,10] and the set of odd integers in [1,10]? Why? Prove that there is a one-one and onto map from the set of all integers and the set of all odd integers.
Reduction Solution for Problem 1 Solution for Problem 2 Help
Example: Map and Direction
Reduction Problem 2 Solution for Problem 1 Question Answer
Reduction Software for Problem 2: Software for Problem 1 call Return
Reduction & Undecidability
It is known that P2 is undecidable We want to prove P1 is undecidable Proof by contradiction Assume P1 is decidable. There is a software solving P1 Design a software for P2 by calling the software for P1 Contradiction!!!!
Halting Problem P1: ={<M,w>| M is a Turing machine and M halts on input w} P2: . ={<M,w>| M is a Turing machine and M accepts input w} Undecidable
P2 Input <M,w> If M does not stop on w, reject it
If M stops, simulate M. If M acceps, accpets Otherwise
Halting Problem ={<M,w>| M is a Turing machine and M halts on
input w} Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Undecidable Problem ={<M,w>| M is a Turing machine and M accepts w} Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Proof Assume the Turing machine R decides We can use R to decides
For input <M, w> Run R on <M,w> if R rejects, “reject” if R accepts, simulate M until it stops if M accepts, “accept” else “reject”
Empty Problem ={<M>| M is a Turing machine and }
Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Proof Assume the Turing machine R decides
For input <M,w>, design another TM If , reject simulate M on input w, accepts if M accepts w
Proof If M accepts w, then w belongs to L( ) Otherwise, L( ) is empty
Proof Use R to decide Input <M,w> Make from <M,w>
Run R on the input If R rejects (it means L( ) is not empty), accepts I R accepts (it means L( ) is empty), rejects
Problem ={<M>| M is a Turing machine and L(M) is regular}
Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Proof Assume the Turing machine R decides
For input <M,w>, design another TM (x) If x has format , accept simulate M on input w, accepts if M accepts x
Proof If M accepts w, then Otherwise,
Proof Use R to decide Input <M,w> Make from <M,w>
Run R on the input If R rejects (it means L( ) is not regular), accepts I R accepts (it means L( ) is regular), rejects
Turing machine equivalence Problem
={ | M1 and M2 are Turing machines and L(M1)=L(M2) } Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Proof For input <M,w> for TM M1: rejects any input
TM M2: accepts any input if M accepts w. It is easy to see L(M1) not equal L(M2) iff M accepts w So,
Mapping Reducibility Convert instances of problem to instances of problem B A B
Computable function Assume that is a set of finite number of symbols
is the set of all finite strings with symbols from The function is computable if some Turing machine M, on every input w, halts with just f(w) on the tape. Example: +,x,/ are all computable functions
Mapping Reducibility Language A is mapping reducible to language B, if there is a computable function such that for every w
Example A={1,3,5,….} B={0,2,4,…} f(x)=x+1 via the funtion f
Theorem If , and B is decidable, then A is also decidable A B
Proof Let f be the reduction from A to B since
Let M be the decider for B. Decider N: Input w, Compute f(w) Run M on f(w) and accept iff M accepts
Corollary If , and A is undecidable, then B is also undecidable A B
Proof Proof by contradiction. If B is decidable, then so is A
Halting Problem P1: ={<M,w>| M is a Turing machine and M halts on input w} P2: . ={<M,w>| M is a Turing machine and M accepts input w} Undecidable
A reduction from to B: ={<M,w>| M is a Turing machine and M halts on input w} A: . ={<M,w>| M is a Turing machine and M accepts input w} Undecidable
Reduction On input <M,w> Construct Turing machine M’ Run M on w
If M accepts, accept If M rejects, enter an infinite loop Output <M’,w>
Empty Problem ={<M>| M is a Turing machine and }
Theorem: is an undecidable language. Proof by contraction. Assume it is decidable.
Reduction from to For <M,w>, design another TM M’ Input x
If x is not equal to w, reject else simulate M on input w, accepts if M accepts w
Reduction from to <M,w> is in L(M’) is not empty
<M,w> is not in L(M’) is empty
Reduction from to <M,w> is in M’ is in
<M,w> is not in M’ is not in
Problem Let ={M| M is a Turing machine and M prints 1 for some input w} Show that L is undecidable.
Turing Machine Write on the tape and read from it
Head can move left and right Tape is infinite Rejecting and accepting states Control a b a b
Turing Machine 7-tuple Q is the finite set of states
is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
Configuration Current state: q7
Current head position on the tape: 4th cell Current tape content: abab q7 a b a b
Configuration A configuration is represented by
Where is the left part of the tape content, is the right part of the tape content, a is the symbol at the head position, q is the current state
Configuration Transition
For
Configuration Transition
For
Configuration Start configuration: , where w is the input
Accepting configuration: a configuration with state Rejecting configuration: a configuration with state
Accept Computation A Turing machine M accepts input w if a sequence of configurations exists where is the start configuration of M on input w, 2. each yields , and is an accepting configuration
Language recognized by TM
For a Turing machine M, L(M) denotes the set of all strings accepted by M. A language is Turing recognizable if some Turing machine recognizes it.
Turing Recognizable Turing machine M recognizes language L
Decidability A language L is Turing decidable if there is a deterministic Turing machine M such that If x is in L, then M accepts x in finite number of steps If x is not in L, then M rejects x in finite number of steps Example: {w#w| w is in {0,1}*} is Turing decidable
Turing Decidable Turing machine M decides language L
Observation If L is Turing decidable, then L is Turing recognizable
A Turing recognizable problem
={<M,w>| M is a Turing machine and M accepts w} Turing machine R recognizable For input <M,w> Simulate M on w R accepts <M,w> if M accepts w.
Theorem Theorem: If L is a language in {0,1}*, there is a Turing machine to print out all elements in L
Proof Let M recognize L Step i(i=1,2,3…)
Simulate M each of the first i strings i steps If a string is accepted, print it out.
Theorem If , and B is Turing recognizable, so is A A B
Proof Let f be the reduction from A to B since Let M recognize B.
TM N: Input w, Compute f(w) Run M on f(w) and accept if M accepts
Theorem If , and A is not Turing recognizable, B is not Turing recognizable A B
Complement set Set and its complement
Theorem If A and are Turing recognizable, then A is decidable
Proof Let Turing machine M1 recognize A
For input x, Run M1 and M2 on x in parallel If M1 accepts, accept If M2 accepts, reject.
={<A,B>| A and B are DFA and L(A)=L(B)}
.
={<A,B>| A and B are DFA and L(A)=L(B)}
Theorem: is not Turing decidable language.
. Theorem: Neither nor its complement is Turing recognizable
Proof We first prove For input <M,w> for
TM M1: rejects any input TM M2: accepts any input if M accepts w. It is easy to see L(M1) not equal L(M2) iff M accepts w So,
Proof We prove For input <M,w> for TM M1: accepts any input
TM M2: accepts any input if M accepts w. It is easy to see L(M1)=L(M2) iff M accepts w So,
Problem Show that the language F={M| L(M) contains infinite elements} is not Turing recognizable}.
Problem 4.2 The equivalence of a DFA and a regular expression is decidable
Problem 4.5 A is a DFA and L(A) is infinite} Show is decidable
Problem 4.7 The set of all infinite binary strings is not countable
Problem 4.8 N={1,2,3,…} NxNxN is countable
Problem 5.9 All Turing recognizable problems are mapping reducible to
Problem 5.12 S={<M>| M is a TM that accepts whenever it accepts w}. Show S is undecidable.
Problem 5.13 Test if a Turing machine has useless state, which never enters.
Language Design Turing machine to recognize
Complexity Let M be a deterministic Turing machine. The running time of M is the function f:N N such that f(n) is the maximum number of steps that M uses on input of length n.
Big O-notation Let f and g be functions Say f(n)=O(g(n)) if integers c and exist so that for all
Examples Let
Small o-notation Let f and g be functions Say f(n)=o(g(n)) if
One tape Turing machine time
One tape Turing machine can recognize In steps
Two tapes Turing machine time
Two tapes Turing machine can recognize In O(n) steps
Multi-tape Turing Machine
1 1 Each tape has its own head Initially the input is at tape 1 and other tapes are blank Reading, writing and moving at all tapes Control a a a b a
Multi-tape Turing Machine
Transition for multi-tape Turing machine
Simulate Multitape Turing machine
Theorem: Every multitape Turing machine has an equivalent single tape Turing machine q7 # 1 1 # a a a # b a #
Simulation Initially, For one move simulation, scan the first #, then the second #, etc. It one tape goes out of its #, move the rest of tape in order to get one space for it. (stupid, slow)
Another simulation Assume there are k tapes
Let the positions i, k+i, 2k+i, …., mk+i,… be used for simulating the tape i. q7 a b 1 a a a
Simulate 3 tapes q7 a b 1 a a a
Simulate 3 tapes q7 a b 1 a a a
K tapes Use two consecutive cells for one symbol, one is for storing the symbol and the other one is for storing the dot mark The tape i’s first j-cell information is at 2(j-1)k+2i-1 and 2(j-1)k+2i
Simulate multi-tape by one tape
One tape Turing machine can simulate t(n) time multi-tape Turing machine in time
Time Complexity Class Let .
is the class of languages decided by O(t(n) time Turing machines. In other words is a language decided by O(t(n)) time Turing machine}
One tape Turing machine time
One tape Turing machine can decide In steps
Complexity Class P P is the class of languages decided by single tape Turing machine in polynomial time
Directed Path Problem G is a directed graph that has a directed path from s to t} a t c s d b
Algorithm Input mark “s” Repeat
For each edge (a,b), if “a” is marked, the mark “b” until no node is marked If (“t” is marked) then accept else reject
Greatest common divisor
Divisor: For two integers b and c, if b=c*z for some integer z, c is a divisor of b. Greatest common divisor: Given two integers a and b, gcd(a,b) is the greatest positive integer c such that c is the divisor for both a and b. Examples: gcd(10,4)=2, gcd(16,100)=4 Problem: How to find gcd(a,b)?
Relatively Prime Problem
Two integers x and y are relatively prime if gcd(x,y)=1. RELPRIME={<x,y>| x and y are relatively prime} Theorem:
Euclid algorithm For two integer b and a a=q*b+c with
gcd(a,b)=gcd(b,c) a and b are relatively prime iff b and c are relatively prime
Euclid algorithm Assume a1 and a2 are two positive integers
If , then and Therefore, always true
Algorithm for gcd(x,y) Input Repeat rx mod y xy yr until y=0
output x
Modular Assume a and b are two positive integers
This is a recursive equation since the second item goes down
Example Find gcd(1970,1066)
Speed of Euclid algorithm
Assume a1 and a2 are two positive integers If , we have In another words,
Euclid algorithm Assume a1 and a2 are two positive integers
Time For input <a,b> The total time of steps is O(log a+log b)
Non-determinism blind monkey
Automata example q2 q1 q3 q3 q4 q2
Deterministic Turing Machine
7-tuple Q is the finite set of states is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
Nondeterministic Turing Machine
7-tuple Q is the finite set of states is the input alphabet not containing special blank is the tape alphabet is the start state, is the accept state is the reject state, where
Nondeterministic Turing Machine
One configuration can have multiple choice for entering the next configuration because of
Non-determinism blind monkey
Acceptance of NFA The nondeterministic machine accepts if at least one of the computation branches ends in an accept state.
Acceptance of NTM The nondeterministic Turing machine accepts if at least one of the computation branches ends in an accept state.
Simulate NTM Theorem: Every Nondeterministic Turing machine has an equivalent deterministic Turing machine.
Simulate NTM with DTM Each dot is one configuration
Find an accepting path
Class NP NP is the class of languages that can be recognized by nondeterministic Turing machine in polynomial time
Hamiltonian Path Hamiltonian path goes through each node exactly once
HAMPATH={<G,s,t>| G is a directed graph with a Hamiltonian path from s to t}
Composite A natural number is composite if it is the product of two integers >1 10=2* =2*9 Composite={x| x=pq, for integers p,q>1}
Verifier A verifier for a language L is an algorithm V,
L={w| V accepts <w,c> for some string c} For the verifier V for L, c is a certificate of w if V accepts <w,c> If the verifier V for the language L runs in polynomial time, V is the polynomial time verifier for L.
Verifier for Hamiltonian Path
For <G,s,t>, a certificate is a list of nodes of G: Verifier: check if m is the number of nodes of G check if and check if each is a directed edge of G for i=1,…,m-1 If all pass, accept . Otherwise, reject.
Verifier for Composite
For integer x, a certificate is two integers p,q: Verifier: check if p>1 and q>1 check if x=pq If all pass, accept . Otherwise, reject.
Class NP NP is the class of languages that have polynomial time verifiers. Examples: COMPOSITE is in NP HAMPATH is in NP
Theorem A language has polynomial verifier iff it can be recognized by polynomial time nondeterministic Turing machine Proof: Assume L has verifier V, … Assume L has NTM M,…
Proof Assume that L has polynomial time verifier V, which runs in time, where k is a constant NTM M Nondeterministically select string c of length Run V on <w,c> If V accepts, accept; Otherwise, reject.
Proof Assume that L is recognized by polynomial time NTM M, which runs in time, where k is a constant Verifier <w,c> Let c determine a computation path of M Simulate M on the path c If M accepts, accept. Otherwise, reject.
Clique Problem Given undirected graph G, a clique is a set of nodes of G such that every two nodes are connected by an edge. A k-clique is a clique with k nodes
Clique Problem CLIQUE={<G,k>| G iss an undirected graph with k-clique} CLIQUE is in NP.
Subset Sum Problem SUBSET-SUM={<S,t>| S= and for some , we have
Polynomial Time Computable
A function is a polynomial time computable function if some polynomial time Turing machine M exists that outputs on the tape for input w.
Polynomial Time Reduction
Assume that A and B are two languages on A is polynomial time mapping reducible to A if a polynomial time computable function exists such that
Boolean Formula A literal is either a boolean variable or its negation: A clause is the disjunction of several literals Conjunctive normal form is the conjunction of several clauses
SAT A boolean formula is satisfiable if there exists assignments to its variables to make the formula true SAT={ | is satisfiable boolean formula}
3SAT A 3nd conjunctive normal formula (3nd-formula) is a conjunction form with at most 3 literals at each clause 3SAT={ | is satisfiable 3nd-formula}
3SAT to CLIQUE Example:
NP-completeness A language B is NP-complete if B is in NP, and
Every A in NP is polynomial time reducible to B Theorem. If B is NP-complete and B is in P, then P=NP.
Cook-Leving Theorem Theorem: SAT is NP-complete Proof.
1. SAT is in NP. 2. For every problem A in NP,
Proof The start configuration is legal The final state is accept.
The movement is legal. Each cell takes one legal symbol.
Proof 1 if The cell[i,j] holds symbol s; 0 otherwise
Time bound for the NTM M with constant k. The movement is legal. NTM M for accepting A.
Nondeterministic Turing Machine
Q is the finite set of states is the tape alphabet is the start state, is the accept state.
Configuration Transition
For
Configuration Transition
For
Configuration Start configuration: , where w is the input
Accepting configuration: a configuration with state Rejecting configuration: a configuration with state
Accept Computation A Turing machine M accepts input w if a sequence of configurations exists where is the start configuration of M on input w, 2. each yields , and is an accepting configuration
Language recognized by TM
For a Turing machine M, L(M) denotes the set of all strings accepted by M. A language is Turing recognizable if some Turing machine recognizes it.
Proof Each cell has only one symbol The symbol is selected from C:
Only one symbol is selected: It is true for all cell at all configuration:
Proof The start configuration is
Proof Accept computation has reached.
It makes sure the accept state will appear among the configuration transitions.
Proof Characterize the legal move
The whole move is legal if all windows are legal. Characterize one window is legal
Proof The state transition
Logic Demorgan Law:
Truth table for y y x2
Convert to CNF Conversion:
Convert to CNF Conversion:
Boolean Formula A literal is either a boolean variable or its negation: A clause is the disjunction of several literals Conjunctive normal form is the conjunction of several clauses
Prepare for the Final Regular language and automata
Context free language Decidability Undecidability Complexity theory
Regular Language Concepts: Automata, regular expression
Skills: Design automata to accept a regular language Disprove a language is a regular
Context-free Language
Concepts: Context-free grammar, parsing tree Skills: Design automata to accept a context-free language Disprove a language is context-free
Decidability Concepts: Turing machine, algorithm, Church-Turing Thesis, Turing recognizable, Turing Decidable Skills: Prove a language is decidable (design algorithm) Prove a language is Turing recognizable
Undecidability Concepts: Countable, Turing undecidable, reduction
Skills: Diagonal method: Prove is undeciable Use reduction to prove a language is undecidable
Complexity Concepts: Time on Turing machine PTIME(t(n))
NP-completeness Polynomial time reduction Polynomial time verifier
Complexity Skill: Prove a problem is in P Prove a problem is in NP
Use reduction to prove a problem is NP-complete.
Grade A:… B:… C: Miss exam or homework
SAT’ A conjunctive normal form is a conjunction of some clauses
SAT’={ | is satisfiable conjunctive normal form}
Cook-Leving Theorem’ Theorem: SAT’ is NP-complete
Proof. Same as that for SAT is NP-complete
3SAT A 3nd conjunctive normal formula (3nd-formula) is a conjunction form with at most 3 literals at each clause 3SAT={ | is satisfiable 3nd-formula}
3SAT is NP-complete Theorem: There is polynomial time reduction from SAT’ to 3SAT.
3SAT is NP-complete is satisfiable if and only if the following is satisfiable
3SAT is NP-complete is satisfiable if and only if the following is satisfiable
3SAT is NP-complete Convert every clause into 3cnf:
3SAT is NP-complete Conjunctive normal form
Each clause is convert into is satisfiable if and only if the following is satisfiable
Problem Convert the formula F into 3SAT formula F’ such that F is satisfiable iff and F’ is satisfiable.
Approximation Algorithms
Outline and Reading Approximation Algorithms for NP-Complete Problems
Approximation ratios Polynomial-Time Approximation Schemes 2-Approximation for Vertex Cover Approximate Scheme for Subset Sum 2-Approximation for TSP special case Log n-Approximation for Set Cover
Approximation Ratios Optimization Problems
We have some problem instance x that has many feasible “solutions”. We are trying to minimize (or maximize) some cost function c(S) for a “solution” S to x. For example, Finding a minimum spanning tree of a graph Finding a smallest vertex cover of a graph Finding a smallest traveling salesperson tour in a graph
Approximation Ratios An approximation produces a solution T
T is a k-approximation to the optimal solution OPT if c(T)/c(OPT) < k (assuming a min. prob.; a maximization approximation would be the reverse)
Polynomial-Time Approximation Schemes
A problem L has a polynomial-time approximation scheme (PTAS) if it has a polynomial-time (1+)-approximation algorithm, for any fixed >0 (this value can appear in the running time). Subset Sum has a PTAS.
Vertex Cover A vertex cover of graph G=(V,E) is a subset W of V, such that, for every (a,b) in E, a is in W or b is in W. OPT-VERTEX-COVER: Given an graph G, find a vertex cover of G with smallest size. OPT-VERTEX-COVER is NP-hard.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A 2-Approximation for Vertex Cover
Every chosen edge e has both ends in C But e must be covered by an optimal cover; hence, one end of e must be in OPT Thus, there is at most twice as many vertices in C as in OPT. That is, C is a 2-approx. of OPT Running time: O(m) Algorithm VertexCoverApprox(G) Input graph G Output a vertex cover C for G C empty set H G while H has edges e H.removeEdge(H.anEdge()) v H.origin(e) w H.destination(e) C.add(v) C.add(w) for each f incident to v or w H.removeEdge(f) return C
Problem 1 Show that T={(i,j)| both I and j are positive rational numbers} is countable.
Problem 2. Find a correspondence between (0,1) and [0,1]
Solution Let for n=1, 2, … Define the correspondence f(x) below:
Problem 3 3. Let A={<R,S> | R and S are regular expressions and L(R ) is a subset of L( S)}. Show that A is decidable.
R S Solution For two regular expression R and S, L( R) is a subset of L(S) if and only if the L(R ) has no intersection with . Find two automata M1 and M2 to accept R and S respectively. Construct automata M3 to accept Construct automata M4 to accept Check if L(M4) is empty
Problem 4 I={x| x is an irrational number in (0,1)}. Show that I is not countable.
Homework 2009
Problem 4.2 The equivalence of a DFA and a regular expression is decidable
Solution Proof. Given a regular language L and DFA M.
We can construct a DFA N to accept L. The problem is converted into checking if two DFAs are equivalent. By Theorem 4.5, there is an algorithm to decide if L(M)=L(N).
Problem 4.5 A is a DFA and L(A) is infinite} Show is decidable
Solution Assume M is a DFA with n states.
L(M) is infinite iff it has path from the start state to accept with a loop on it, and the loop length is no more than n. L(M) is infinite iff M accepts a string of length between n and 2n. For every string x with length between n and 2n Run M on x If M accepts x, then output “L(M) is infinite” else output “no”
Problem 4.7 The set of all infinite binary strings is not countable
Assume N={1,2,3,…} and have the same size. There is one-one and onto map f: N
Proof. Select such that and Since
Problem 1 N={1,2,3,…} NxNxN is countable
Proof List all of them by
Proof Use the similar method like the last slide to prove the NxN is countable Let f: NxNN be one-one and onto We have g: NxNxNN is one-one and onto, where g(i,j,k)=f(f(i,j),k)
Problem 2. Find a correspondence between (0,1) and [0,1]
Solution Let for n=1, 2, … Define the correspondence f(x) below:
Problem 3 3. Let A={<R,S> | R and S are regular expression and L(R ) = L( S)}. Show that A is decidable.
Solution For two regular expression R and S, L( R)=L(S) if and only if their symmetric difference is empty. Find two automata M1 and M2 to accept R and S respectively. Construct automata M3 to accept Construct automata M4 to accept Construct automata M5 to accept Check if L(M5) is empty
Problem 4 S={<M>| M is a TM that accepts whenever it accepts w}. Show S is undecidable.
Proof We are going to design a reduction from to S. For <M,w>
Design the TM N as follows Input x if x is not 01 or 10, rejects. if x=01, accepts if x=10, run M on w, if M accepts w, accepts <M,w> is in iff N accepts both 01 and 10.
Problem 5 5. Let S={<M> | M is a TM and L(M) = {<M>}}. Show that neither S nor its complement is Turing-recognizable.
Solution We first show that can be reducible to S.
Let (M, w) be an input for Let Turing machine M1 be constructed below: M1 (x, Y) Input x and another TM Y If x is not equal <Y> then reject If x=<Y> then accepts if M(w) accepts.
Solution For an input (M, w), derive M1(x, M1).
Solution We show that can be reducible to . Let (M, w) be an input for
Let Turing machine M2 be constructed below: M1 Input x If x =<M1> then accept If x is not equal <M1> then accept if M(w) accepts.
Solution For an input (M, w), derive M2(x, M2).
Problem 6 Show that A is a Turing recognizable if and only if
Proof Part a) Let A be a Turing recognizable language.
There is a Turing machine M such that if w is in A iff M accepts w Let f be the reduction f(w) =<M,w>. It is easy to see that
Proof Part b) Assume that via computable function g.
It is easy to see that is Turing recognizable. Let M1 be a Turing machine that recognize We have Turing machine M2 to recognize A: M2: Input w Compute g(w) Run M1 on input g(w), accepts w if M1 accepts g(w).
Proof We are going to design a reduction from to it For <M,w>
Design the TM N as follows Input x if x=0, enters all states except if x=1 run M on w, if M accepts w, enter <M,w> is in iff N enters all states.
Problem 4 c) The language is not a context free language
Proof: Assume it is a context free language. Let p be the number from the pumping lemma Consider the string
Final Exam December 13, 5:45-7:30pm, Monday
Homework 4, 20
Problem 1 1.Let T1={<M>| M is a Turing machines and accepts infinite number of 0,1-strings of finite length}. Prove that T1 is undecidable.
Proof We are going to design a reduction from to T1.
For input Turing machine <M,w> for Design the TM N as follows Input x run M on w, if M accepts w, acceptx <M,w> is in iff N acccepts each input x.
Problem 2 Let T2={<M>| M is a Turing machines and accepts five of 0,1-strings of finite length}. Prove that T2 is undecidable.
Proof We are going to design a reduction from to T2. For <M,w>
Design the TM N as follows Input x if x is not 0,1, 00, 01 or 10, rejects. run M on w if M accepts w, accept x <M,w> is in iff N accepts 0,1,00,01, and 10. Otherwise, N accepts no string
Problem 3 S={<M>| M is a TM that accepts whenever it accepts w}. Show S is undecidable.
Proof We are going to design a reduction from to S. For <M,w>
Design the TM N as follows Input x if x is not 01 or 10, rejects. if x=01, accepts if x=10, run M on w, if M accepts w, accepts <M,w> is in iff N accepts both 01 and 10.
Problem 4 Show that A is a Turing recognizable if and only if
Proof Part a) Let A be a Turing recognizable language.
There is a Turing machine M such that if w is in A iff M accepts w Let f be the reduction f(w) =<M,w>. It is easy to see that
Proof Part b) Assume that via computable function g.
It is easy to see that is Turing recognizable. Let M1 be a Turing machine that recognize We have Turing machine M2 to recognize A: M2: Input w Compute g(w) Run M1 on input g(w), accepts w if M1 accepts g(w).
Problem 5 5. Let S={<M> | M is a TM and L(M) = {<M>}}. Show that neither S nor its complement is Turing-recognizable.
Solution We first show that can be reducible to S.
Let (M, w) be an input for Let Turing machine M1 be constructed below: M1 (x, Y) Input x and another TM Y If x is not equal <Y> then reject If x=<Y> then accepts if M(w) accepts.
Solution For an input (M, w), derive M1(x, M1).
Solution We show that can be reducible to . Let (M, w) be an input for
Let Turing machine M2 be constructed below: M1 Input x If x =<M1> then accept If x is not equal <M1> then accept if M(w) accepts.
Solution For an input (M, w), derive M2(x, M2).
Homework 5, 20
Problem 1 Call graphs G and H isomorphic if the nodes of G may be reordered so that it is identical to H. Let ISO={<G,H> | G and H are isomorphic graphs}. Show that ISO is in NP.
Solution for Problem 1 We design a polynomial time verification algorithm. A mapping f from the vertices of G to those of H is a witness. Check if 1) f is one-one, 2) f is onto, 3) (v,u) is an edge of G iff (f(u),f(v)) is an edge of H. It is easy to see that the verification takes polynomial time.
Problem 2 Let MAX-CIQUE={<G, k>| the largest clique of G has k vertices}. Whether MAX-CLIQUE is in NP is unknown. Show that if P=NP, then MAX-CLIQUE is in P, and a polynomial time algorithm exists that, for a graph G, finds one of its largest cliques.
Solution of Problem 2 Fist step is to find the largest k with (G,k) is in Clique. Try k from 1,2,… Check if (G,k) is in Clique Select the largest k. Assume the largest k for (G,k) in Clique is obtained.
Solution of Problem 2 Assume the largest k for (G,k) in Clique is obtained. Formulate the problem: (H, k,G) Determine if there is a clique of size k in G and contains all veritces in H. The problem is in NP. Extend H one by one until its size reaches k.
Problem 3 Let G represent an undirected graph and let SPATH={<G, a,b,k> | G contains a simple path of length at most k from a to b} and LPATH={<G, a,b,k} | G contains a simple path of length at least k from a to b}. Show that SPATH is in P. Show that LPATH is NP-complete.
Solution Problem 3 Show that SPATH is in P.
Directed Path Problem G is a directed graph that has a directed path from s to t} a t c s d b
Algorithm Input mark “s” Repeat
For each edge (a,b), if “a” is marked, the mark “b” until no node is marked If (“t” is marked) then accept else reject
Solution Problem 3 Show that LPATH is NP-complete.
Hamiltonian path problem is NP-complete. There is an easy reduction from Hamiltonian path to it.
Prepare for the Final Regular language and automata
Context free language Decidability Undecidability Complexity theory
Regular Language Concepts: Automata, regular expression
Skills: Design automata to accept a regular language Disprove a language is a regular
Context-free Language
Concepts: Context-free grammar, parsing tree Skills: Design automata to accept a context-free language Disprove a language is context-free
Decidability Concepts: Turing machine, algorithm, Church-Turing Thesis, Turing recognizable, Turing Decidable Skills: Prove a language is decidable (design algorithm) Prove a language is Turing recognizable
Undecidability Concepts: Countable, Turing undecidable, reduction
Skills: Diagonal method: Prove is undeciable Use reduction to prove a language is undecidable
Complexity Concepts: Time on Turing machine PTIME(t(n))
NP-completeness Polynomial time reduction Polynomial time verifier
Complexity Skill: Prove a problem is in P Prove a problem is in NP
Use reduction to prove a problem is NP-complete.
Grade A:… B:… C: Miss exam or homework
Motivation Complexity classes correspond to bounds on resources
One such resource is space: the number of tape cells a TM uses when solving a problem Complexity ©D.Moshkovits
Introduction Objectives: To define space complexity classes Overview:
Low space classes: L, NL Savitch’s Theorem Immerman’s Theorem TQBF Complexity ©D.Moshkovits
Space Complexity Classes
For any function f:NN, we define: SPACE(f(n))={ L : L is decidable by a deterministic O(f(n)) space TM} NSPACE(f(n))={ L : L is decidable by a non-deterministic O(f(n)) space TM} Complexity ©D.Moshkovits
Low Space Classes Definitions (logarithmic space classes):
L = SPACE(logn) NL = NSPACE(logn) Complexity ©D.Moshkovits
Problem! How can a TM use only logn space if the input itself takes n cells?! !? Complexity ©D.Moshkovits
3Tape Machines a b _ b _ b a _ input . . . work output . . . . . .
read-only Only the size of the work tape is counted for complexity purposes read/ write b _ . . . write-only b a _ . . . Complexity ©D.Moshkovits
Example Question: How much space would a TM that decides {anbn | n>0} require? Note: to count up to n, we need logn bits Complexity ©D.Moshkovits
Graph Connectivity CONN
An undirected version is also worth considering CONN Instance: a directed graph G=(V,E) and two vertices s,tV Problem: To decide if there is a path from s to t in G? Complexity ©D.Moshkovits
Graph Connectivity t s Complexity ©D.Moshkovits
CONN is in NL Start at s For i = 1, .., |V| {
Non-deterministically choose a neighbor and jump to it Accept if you get to t } If you got here – reject! Counting up to |V| requires log|V| space Storing the current position requires log|V| space Complexity ©D.Moshkovits
Configurations Which objects determine the configuration of a TM of the new type? The content of the work tape The machine’s state The head position on the input tape The head position on the work tape The head position on the output tape If the TM uses logarithmic space, there are polynomially many configurations Complexity ©D.Moshkovits
Log-Space Reductions Definition:
A is log-space reducible to B, written ALB, if there exists a log space TM M that, given input w, outputs f(w) s.t. wA iff f(w)B the reduction Complexity ©D.Moshkovits
Do Log-Space Reductions Imply what they should?
Suppose A1 ≤L A2 and A2L; how to construct a log space TM which decides A1? Wrong Solution: w Too Large! f(w) Use the TM for A2 to decide if f(w)A2 Complexity ©D.Moshkovits
Log-Space reductions Claim: if Then, A1 is in L
A1 ≤L A2 – f is the log-space reduction A2 L – M is a log-space machine for A2 Then, A1 is in L Proof: on input x, in or not-in A1: Simulate M and whenever M reads the ith symbol of its input tape run f on x and wait for the ith bit to be outputted Complexity ©D.Moshkovits
NL Completeness Definition: A language B is NL-Complete if BNL
For every ANL, ALB. If (2) holds, B is NL-hard Complexity ©D.Moshkovits
Savitch’s Theorem Theorem: S(n) ≥ log(n) NSPACE(S(n)) SPACE(S(n)2)
Proof: First we’ll prove NLSPACE(log2n) then, show this implies the general case Complexity ©D.Moshkovits
Savitch’s Theorem Theorem: NSPACE(logn) SPACE(log2n) Proof:
First prove CONN is NL-complete (under log-space reductions) Then show an algorithm for CONN that uses log2n space Complexity ©D.Moshkovits
CONN is NL-Complete Theorem: CONN is NL-Complete
Proof: by the following reduction: s L t “Is there a path from s to t?” “Does M accept x?” Complexity ©D.Moshkovits
Technicality Observation:
Without loss of generality, we can assume all NTM’s have exactly one accepting configuration. Complexity ©D.Moshkovits
Configurations Graph A Computation of a NTM M on an input x can be described by a graph GM,x: A vertex per configuration the start configuration s t the accepting configuration (u,v)E if M can move from u to v in one step Complexity ©D.Moshkovits
Correctness Claim: For every non-deterministic log-space Turing machine M and every input x, M accepts x iff there is a path from s to t in GM,x Complexity ©D.Moshkovits
CONN is NL-Complete Corollary: CONN is NL-Complete
Proof: We’ve shown CONN is in NL. We’ve also presented a reduction from any NL language to CONN which is computable in log space (Why?) Complexity ©D.Moshkovits
A Byproduct Claim: NLP Proof:
Any NL language is log-space reducible to CONN Thus, any NL language is poly-time reducible to CONN CONN is in P Thus any NL language is in P. Complexity ©D.Moshkovits
What Next? We need to show CONN can be decided by a deterministic TM in O(log2n) space. Complexity ©D.Moshkovits
The Trick “Is there a path from u to v of length d?”
“Is there a vertex z, so there is a path from u to z of size d/2 and one from z to v of size d/2?” d/2 d/2 u z v d Complexity ©D.Moshkovits
Recycling Space The two recursive invocations can use the same space
The Algorithm Boolean PATH(a,b,d) {
if there is an edge from a to b then return TRUE else { if d=1 return FALSE for every vertex v { if PATH(a,v, d/2) and PATH(v,b, d/2) then return TRUE } return FALSE Complexity ©D.Moshkovits
Example of Savitch’s algorithm
boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE boolean PATH(a,b,d) { if there is an edge from a to b then return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then } return FALSE 2 3 1 4 (a,b,c)=Is there a path from a to b, that takes no more than c steps. (1,4,3)(1,3,2)(2,3,1)TRUE (1,4,3)(1,2,2)TRUE (1,4,3)(1,2,2) (1,4,3)(2,4,1) (1,4,3)(2,4,1)FALSE (1,4,3)(1,3,2)TRUE (1,4,3) (1,4,3) TRUE (1,4,3)(3,4,1)TRUE (1,4,3)(1,3,2)(1,2,1) (1,4,3)(1,3,2)(2,3,1) (1,4,3)(3,4,1) (1,4,3)(1,3,2) (1,4,3)(1,3,2)(1,2,1)TRUE Complexity ©D.Moshkovits 3Log2(d)
O(log2n) Space DTM Claim: There is a deterministic TM which decides CONN in O(log2n) space. Proof: To solve CONN, we invoke PATH(s,t,|V|) The space complexity: S(n)=S(n/2)+O(logn)=O(log2n) Complexity ©D.Moshkovits
Conclusion Theorem: NSPACE(logn) SPACE(log2n)
The Padding Argument Motivation: Scaling-Up Complexity Claims
We have: can be simulated by… space space + non-determinism + determinism We want: can be simulated by… space space + non-determinism + determinism Complexity ©D.Moshkovits
Formally NSPACE(s1(f(n))) SPACE(s2(f(n)))
si(n) can be computed with space si(n) Claim: For any two space constructible functions s1(n),s2(n)logn, f(n)n: NSPACE(s1(n)) SPACE(s2(n)) NSPACE(s1(f(n))) SPACE(s2(f(n))) simulation overhead E.g NSPACE(n)SPACE(n2) NSPACE(n2)SPACE(n4) Complexity ©D.Moshkovits
Idea NTM DTM n n . . . f(n) . space: s1(.) in the size of its input
space: O(s2(f(n))) NTM n n . . . space: O(s1(f(n))) f(n) . Complexity ©D.Moshkovits
Padding argument Let LNPSPACE(s1(f(n))) There is a 3-Tape-NTM ML: |x|
Input babba Work O(s1(f(|x|))) Complexity ©D.Moshkovits
Padding argument Let L’ = { x0f(|x|)-|x| | xL }
We’ll show a NTM ML’ which decides L’ in the same number of cells as ML. f(|x|) babba# Input Work O(s1(f(|x|)) Complexity ©D.Moshkovits
Padding argument – ML’ In O(log(f(|x|)) space
Count backwards the number of 0’s and check there are f(|x|)-|x| such. 2. Run ML on x. in O(s1(f(|x|))) space f(|x|) Input babba Work O(s1(f(|x|))) Complexity ©D.Moshkovits
Padding argument Total space: O(s1(f(|x|))) f(|x|) Input
babba Work O(s1(f(|x|))) Complexity ©D.Moshkovits
Padding Argument We started with LNSPACE(s1(f(n)))
We showed: L’NSPACE(s1(n)) Thus, L’SPACE(s2(n)) Using the DTM for L’ we’ll construct a DTM for L, which will work in O(s2(f(n))) space. Complexity ©D.Moshkovits
Padding Argument The DTM for L will simulate the DTM for L’ when working on its input concatenated with zeros Input babba Complexity ©D.Moshkovits
Padding Argument When the input head leaves the input part, just pretend it encounters 0s. maintaining the simulated position (on the imaginary part of the tape) takes O(log(f(|x|))) space. Thus our machine uses O(s2(f(|x|))) space. NSPACE(s1(f(n)))SPACE(s2(f(n))) Complexity ©D.Moshkovits
Savitch: Generalized Version
Theorem (Savitch): S(n) ≥ log(n) NSPACE(S(n)) SPACE(S(n)2) Proof: We proved NLSPACE(log2n). The theorem follows from the padding argument. Complexity ©D.Moshkovits
Corollary Corollary: PSPACE = NPSPACE Proof: Clearly, PSPACENPSPACE.
By Savitch’s theorem, NPSPACEPSPACE. Complexity ©D.Moshkovits
Space Vs. Time We’ve seen space complexity probably doesn’t resemble time complexity: Non-determinism doesn’t decrease the space complexity drastically (Savitch’s theorem). We’ll next see another difference: Non-deterministic space complexity classes are closed under completion (Immerman’s theorem). Complexity ©D.Moshkovits
NON-CONN NON-CONN Instance: A directed graph G and two vertices s,tV.
Problem: To decide if there is no path from s to t. Complexity ©D.Moshkovits
NON-CONN Clearly, NON-CONN is coNL-Complete.
(Because CONN is NL-Complete. See the coNP lecture) If we’ll show it is also in NL, then NL=coNL. (Again, see the coNP lecture) Complexity ©D.Moshkovits
An Algorithm for NON-CONN
We’ll see a log space algorithm for counting reachability Count how many vertices are reachable from s. Take out t and count again. Accept if the two numbers are the same. Complexity ©D.Moshkovits
N.D. Algorithm for reachs(v, l)
1. length = l; u = s 2. while (length > 0) { 3. if u = v return ‘YES’ 4. else, for all (u’ V) { 5. if (u, u’) E nondeterministic switch: 5.1 u = u’; --length; break 5.2 continue } } 6. return ‘NO’ Takes up logarithmic space This N.D. algorithm might never stop Complexity ©D.Moshkovits
N.D. Algorithm for CRs CRs ( d ) 1. count = 0 2. for all uV {
3. countd-1 = 0 4. for all vV { 5. nondeterministic switch: 5.1 if reach(v, d - 1) then ++countd-1 else fail if (v,u) E then ++count; break 5.2 continue } 6. if countd-1 < CRs (d-1) fail 7.return count Assume (v,v) E Recursive call! Complexity ©D.Moshkovits
N.D. Algorithm for CRs parameter , C) CRs ( d 1. count = 0
2. for all uV { 3. countd-1 = 0 4. for all vV { 5. nondeterministic switch: 5.1 if reach(v, d - 1) then ++countd-1 else fail if (v,u) E then ++count; break 5.2 continue } 6. if countd-1 < fail 7.return count Main Algorithm: CRs C = 1 for d = 1..|V| C = CR(d, C) return C C parameter Complexity ©D.Moshkovits
Efficiency Lemma: The algorithm uses O(log(n)) space. Proof:
There is a constant number of variables ( d, count, u, v, countd-1). Each requires O(log(n)) space (range |V|). Complexity ©D.Moshkovits
Immerman’s Theorem Theorem[Immerman/Szelepcsenyi]: NL=coNL Proof:
(1) NON-CONN is NL-Complete (2) NON-CONNNL Hence, NL=coNL. Complexity ©D.Moshkovits
Corollary Corollary: s(n)log(n), NSPACE(s(n))=coNSPACE(s(n))
TQBF We can use the insight of Savich’s proof to show a language which is complete for PSPACE. We present TQBF, which is the quantified version of SAT. Complexity ©D.Moshkovits
TQBF Instance: a fully quantified Boolean formula
Problem: to decide if is true Example: a fully quantified Boolean formula xyz[(xyz)(xy)] Variables` range is {0,1} Complexity ©D.Moshkovits
TQBF is in PSPACE Theorem: TQBFPSPACE
Proof: We’ll describe a poly-space algorithm A for evaluating : If has no quantifiers: evaluate it If =x((x)) call A on (0) and on (1); Accept if both are true. If =x((x)) call A on (0) and on (1); Accept if either is true. in poly time Complexity ©D.Moshkovits
Algorithm for TQBF 1 1 1 1 1 xy[(xy)(xy)] y[(0y)(0y)]
(00)(00) (01)(01) (10)(10) (11)(11) 1 1 Complexity ©D.Moshkovits
Efficiency Since both recursive calls use the same space,
the total space needed is polynomial in the number of variables (the depth of the recursion) TQBF is polynomial-space decidable Complexity ©D.Moshkovits
PSAPCE Completeness Definition: A language B is PSPACE-Complete if
BPSPACE For every APSAPCE, APB. standard Karp reduction If (2) holds, then B is PSPACE-hard Complexity ©D.Moshkovits
TQBF is PSPACE-Complete
Theorem: TQBF is PSAPCE-Complete Proof: It remains to show TQBF is PSAPCE-hard: P x1x2x3…[…] “Will the poly-space M accept x?” “Is the formula true?” Complexity ©D.Moshkovits
TQBF is PSPACE-Hard Given a TM M for a language L PSPACE, and an input x, let fM,x(u, v), for any two configurations u and v, be the function evaluating to TRUE iff M on input x moves from configuration u to configuration v fM,x(u, v) is efficiently computable Complexity ©D.Moshkovits
Formulating Connectivity
The following formula, over variables u,vV and path’s length d, is TRUE iff G has a path from u to v of length ≤d (u,v,1) fM,x(u, v) u=v (u,v,d) wxy[((x=uy=w)(x=wy=v))(x,y,d/2)] w is reachable from u in d/2 steps. v is reachable from w in d/2 steps. simulates AND of (u,w,d/2) and (w,v,d/2) Complexity ©D.Moshkovits
TQBF is PSPACE-Complete
Claim: TQBF is PSPACE-Complete Proof: (s,t,|V|) is TRUE iff there is a path from s to t. is constructible in poly-time. Thus, any PSPACE language is poly-time reducible to TQBF, i.e – TQBF is PSAPCE-hard. Since TQBFPSPACE, it’s PSAPCE-Complete Complexity ©D.Moshkovits
Summary We introduced a new way to classify problems: according to the space needed for their computation. We defined several complexity classes: L, NL, PSPACE. Complexity ©D.Moshkovits
Summary Our main results were: Connectivity is NL-Complete
TQBF is PSPACE-Complete Savitch’s theorem (NLSPACE(log2)) The padding argument (extending results for space complexity) Immerman’s theorem (NL=coNL) By reducing decidability to reachability Complexity ©D.Moshkovits | 25,052 | 88,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.862419 |
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OR
1. I want to make a queen sized quilt for my sister for Christmas in the rail fence pattern. I know I cut 2 1/2" strips from each of the 3 colors, sew them together, then cut in 6 1/2" squares. How many yards of each color do I need to make queen sized?
2. Originally Posted by Raggiemom
I want to make a queen sized quilt for my sister for Christmas in the rail fence pattern. I know I cut 2 1/2" strips from each of the 3 colors, sew them together, then cut in 6 1/2" squares. How many yards of each color do I need to make queen sized?
They don't have to be 2.5".
If you want to make them wider, you can.
For instance, using three 3.5" wide strips, you't get UNFINSHED squares that were 9.5".
Different look over such a large quilt.
Both quilts are approx 80"x100".
6" blocks set 13x16 78"x96"
9" blocks set 9x11 81"x99"
3. For the 6" blocks - you'd three yards of each color.
For the 9" blocks - you'd need 3.25" yards of each color.
Not including borders, binding, etc.
4. Oh, I like the one with the 3.5" inch strips, I didn't realize how much different it would look!
5. Thanks MTS!
6. Originally Posted by Raggiemom
Oh, I like the one with the 3.5" inch strips, I didn't realize how much different it would look!
Well it's such a big area - the top of the bed is 60"x80".
You might want to make some scrap blocks and just toss them on the bed to see how it looks proportionally.
Those fabric requirements I gave you were straight out of EQ.
I'm checking the math backwards and I see you'll be fine with those amounts - again, not including the border, binding or backing.
9.5" UNFINISHED blocks:
Total blocks needed = 9x11 = 99
Four 9.5" cuts per strip (38" )
99 / 4 = 25 strips
25 x 3.5" wide cut strips = 87.5" length of fabric
87.5" / 36" = 2.5 yards
So I don't know where EQ is getting 3.25 yards but go with it - you can always use it on the block.
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SEO by vBSEO ©2011, Crawlability, Inc. | 609 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.961792 |
https://www.math.aau.at/or/doctoralschool/main.html | 1,623,670,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00119.warc.gz | 790,288,811 | 8,688 | ### Doctoral school on: Optimization over Polynomials and Semidefinite Programming
#### organized by M. Laurent (CWI, Amsterdam) and F. Rendl (University of Klagenfurt)
The doctoral school is sponsored by the research network ADONET (Algorithmic Discrete Optimization Network) funded by the European Community with 12 participating universities in 10 european countries.
### Theme of the doctoral school
The central topic of the doctoral school is the study of "sum of squares" representations of polynomials as a tool to get approximate solutions to NP-hard optimization problems. As finding such a representation amounts to solving a semidefinite program, one can approximate the global minimum of a polynomial over a set defined by polynomial inequalities and equations efficiently using semidefinite programming. Polynomial optimization includes 0/1 programming problems which can be modelled by the quadratic equations x²=x for all variables.
The purpose of the doctoral school is to introduce interested graduate students into this field and to give them a hands-on experience with software developed for minimizing polynomials. Some prior knowledge in linear algebra, numerics, algebra and optimization will be useful.
More specifically, the doctoral school offers an introduction to the following topics:
• Theoretical background from real algebraic geometry about representations of positive polynomials as sums of squares (Hilbert's 17th problem, representation theorems of Pólya, Putinar, Schmüdgen, etc.)
• The dual theory of moments (theorems about moment matrices of Curto and Fialkow, Carleman's condition, etc., used for the approximation of polynomial programs)
• Minimizing polynomials over basic closed semi-algebraic sets via the "sum of squares" approach and the "moment matrix" dual approach (convergence results, stopping criteria, extracting optimum solutions, application to combinatorial optimization)
• Exploiting symmetries to reduce the size of semidefinite programs (block-diagonalization of matrix ∗-algebras, link with representation theory, application to coding theory - improving the Delsarte bound using Terwiliger algebras, etc.)
• Algorithmic issues related to finding sum of squares representations (algorithmic alternatives to interior-point methods for large semidefinite programs, augmented Lagrangian, bundle methods, etc.)
• Barvinok's method and integer polynomial optimization (using Barvinok's rational functions for representing lattice points in polyhedra and optimizing polynomials over them) and using polynomials to model combinatorial optimization problems
### Speakers
J. Lasserre, LAAS-CNRS, Toulouse reading suggestion and slides M. Laurent, CWI, Amsterdam summary, reading suggestion and slides J. de Loera, University of California, Davis summary, reading suggestion and slides P. Parrilo, MIT, Cambridge reading suggestions F. Rendl, University of Klagenfurt summary and slides B. Reznick, University of Illinois at Urbana-Champaign summary and reading suggestion A. Schrijver, CWI, Amsterdam summary
### Tentative schedule
Here is a short file containing further details: read and print this
• Sunday
• 18.00 – 20.00: informal get together at the University in room SR i-201, mathematics department
• Monday
• 8.30 – 9.00: opening session
• 9.00 – 12.00: B. Reznick
• 16.00 – 19.00: P. Parrilo
• Tuesday
• 9.00 – 12.00: A. Schrijver
• 16.00 – 19.00: A. Schrijver, P. Parrilo (work on computer)
• Wednesday
• 9.00 – 12.00: J. Lasserre
• 12.00 – 13.00: J. Lasserre (work on computer)
• 14.00 – 21.00: hiking tour and workshop dinner
• Thursday
• 9.00 – 12.00: M. Laurent
• 16.00 – 19.00: F. Rendl
• Friday
• 9.00 – 12.00: J. de Loera
• 12.00 – 12.30: closing session
• 13.00 – 14.00: farewell lunch
### Social program
• Sunday, Sept. 11, 18.00 – 20.00: Informal get together at the Univeristy in room SR i-201, mathematics department
• Wednesday, Sept. 14, afternoon: Hiking in the Alps and Workshop Dinner at Klagenfurter Hütte
• Friday, Sept. 16, noon: Farewell Lunch at the University restaurant IQ
### Participants
• Michael Armbruster, Chemnitz University of Technology, Germany
• Ivo Bleylevens, University of Maastricht, Netherlands
• Hartwig Bosse, CWI, Amsterdam, Netherlands
• Stefan Bundfuss, Technische Universität Darmstadt, Germany
• Martin Burke, Imperial College London, UK
• Jaroslaw Byrka, CWI, Amsterdam, Netherlands
• Andrea Cassioli, Università degli Studi di Firenze, Italy
• Jakob Cimpric, University of Ljubljana, Slovenia
• Zsolt Csizmadia, Eötvös Loránd University, Budapest, Hungary
• Igor Dukanovic, University of Maribor, Slovenia
• Christian Ebenbauer, University of Stuttgart, Germany
• Gamal Elabwabi, University of Tilburg, Netherlands
• Ioannis A. Fotiou, ETH Zurich, Switzerland
• Florian Frommlet, University of Vienna, Austria
• Monia Giandomenico, University of Rome "La Sapienza", Italy
• Gaelle Giberti, Laboratoire Leibniz, Grenoble, France
• Magdalena Grüber, Humboldt-University Berlin, Germany
• Nebojsa Gvozdenovic, CWI, Amsterdam, Netherlands
• Johannes Hatzl, Graz University of Technology, Austria
• Tamas Kis, Computer and Automation Research Institute, Hungarian Academy of Sciences, Hungary
• Igor Klep, Universität Konstanz, Germany
• Etienne de Klerk, Tilburg University, Netherlands
• Dennis Michaels, Otto-von-Guericke Universitaet Magdeburg, Germany
• Marianna Nagy, Eötvös Loránd University, Budapest, Hungary
• Helfried Peyrl, ETH Zurich, Switzerland
• Gema Plaza, University of Alicante, Spain
• Janez Povh, School of Business and Management Novo Mesto, Slovenia
• Philipp Rostalski, ETH Zurich, Switzerland
• Doris Rotheigner, University of Regensburg, Germany
• Cristiano Saturni, University of Padova, Italy
• Carlo Savorgnan, Università Degli Studi Di Udine, Italy
• Markus Schweighofer, Universität Konstanz, Germany
• Marco di Summa, University of Padova, Italy
• Hakan Umit, Université Catholique de Louvain, Belgium
• Frank Vallentin, Hebrew University Jerusalem, Israel
• Dejan Veluscek, University of Ljubljana, Slovenia
• Juan Vera, Carnegie Mellon University, USA
• Hayato Waki, Tokyo Institute of Technology, Japan
• Angelika Wiegele, University of Klagenfurt, Austria
• Rico Zenklusen, ETH Zurich, Switzerland
• Stefan van Zwam, Technische Universiteit Eindhoven, Netherlands
### Accomodation and Fees
We can offer accommodation in the youth hostel in double rooms for the duration of the school (i.e. arrival on Sunday Sept. 11, 2005, departure on Saturday Sept. 17, 2005) at a rate of 22 Euro per person and night including breakfast for participants who register no later than June 26, 2005. (The youth hostel is located in walking distance to the university, the rooms are equipped with private shower and toilet.)
If you are interested in this offer, please indicate so in your registration e-mail and specify arrival and departure times. Confirmations will be sent out on June 30, 2005. All other participants will be asked to make their own accommodation arrangements.
Information about the youth hostel: www.oejhv.or.at (Address: Neckheimgasse 6, phone number +43 463 230020)
Online booking at the youth hostel: www.hihostels.com
More information on accomodation in Klagenfurt: Klagenfurt tourist information.
Travel, accommodation and living expenses must be paid by (the home institutions of) the participants. There will be no fee for participation and course material.
### Registration
To register, send an e-mail to adonet@uni-klu.ac.at containing the following information:
• Name
• Affiliation
• Current status (doctoral student / post-doc / other)
• e-mail address
• Research area of interest
• Supervisor (if applicable)
• Accomodation required (yes/no) deadline is over - please make your own arrangements
• Male/Female (if accomodation required)
### Location
The doctoral courses will take place at the University of Klagenfurt in lecture room HS B (main building, "Südtrakt") and in computer room UR z-514b.
### Travelling to Klagenfurt...
To get an idea on the location of Klagenfurt, you may have a look at this small overview map.
#### ...by car from the highway (all directions) to the University:
Coming from the West (Salzburg, Villach), you first go until "Klagenfurt West" which is the next crossing after Krumpendorf. Be careful and take at Klagenfurt-West the leftmost lane directly to Klagenfurt Center. At Klagenfurt-Wörthersee (1,5 km after Klagenfurt-West) you leave the highway. Coming from the North (St. Veit) or the East (Graz, Vienna), you follow the many tunnels in the north of Klagenfurt westwards. Immediately after the Falkenbergtunnel you take the second exit (about 300 m after the end of the Tunnel, so be alert!) directed to "Klagenfurt West". About 1,5 km after that you leave the highway at Exit "Klagenfurt-Wörthersee".
After having left the highway at Klagenfurt-Wörthersee (=Klagenfurt-See in short) you immediately come to the so called "Minimundus-Crossing" (with traffic-lights). You keep straight direction towards south. After 400 m you will see the logo of the University of Klagenfurt. Turn left into the Universitätsstraße. You have arrived at your destination.
#### ...by car from Slovenia (via Loiblpass) to the University:
Coming down the Loiblpass you go right through Unterloibl, Kirschentheuer, passing the Hollenburg and Maria Rain until you come to the first traffic lights of Klagenfurt. There you go straight ahead. After about 800 m at the next traffic lights you turn left towards the direction Wörthersee/Universität/Autobahn Villach to the so-called "Südring". You go along the Südring, passing the Wörthersee stadium. About 1 km after the stadium the road makes a big turn right. Looking ahead then you will see on your right the "Lakeside Park" and just behind it is the University of Klagenfurt.
There is a sufficient number of parking lots around the university area.
#### ...by plane or train:
Klagenfurt Airport is reachable by direct flights from Vienna and Frankfurt (Austrian Airlines), from Berlin, Cologne/Bonn, Hamburg and Hannover (Hapag-Lloyd Express) and from London Stansted and Frankfurt-Hahn (Ryanair).
NEW DESTINATIONS FROM KLAGENFURT AIRPORT: Zurich and Paris. (with a stop in Salzburg)
Styrian Spirit flies from September 6 on from Klagenfurt to Zurich and Paris (via Salzburg).
For details on travelling to Klagenfurt by train search the time-table of the Austrian Federal Railways.
#### Reaching Klagenfurt University (and the youth hostel) by bus from the main station or airport
The University of Klagenfurt is easily reachable by public transportation. The bus route system in Klagenfurt is star-shaped. The stop Heiligengeistplatz is the centre of the star. Heiligengeistplatz can be reached by bus lines from the train station (Hauptbahnhof) and from the airport (Flughafen). (For more information see the route map. The stops Hauptbahnhof (main station), Flughafen (airport) and Universität are indicated by large yellow circles.)
To reach the university from Heiligengeistplatz, lines 10, 11, 12, 20 and 21 can be used. Line 12 is the most convenient because it directly goes to the university. Get off the bus at the stop Universität or if you want to go to the hostel at the stop Jugendgästehaus, which is just one stop before Universität. Line 12 does not run, however, on Sunday.
Lines 10, 11, 20 and 21 stop at Minimundus which is close to the university. The stop Minimundus is located in Villacher Straße. If you go to the youth hostel get off at the stop Neckheimgasse, which is just one stop before Minimundus. For details see the map of the bus routes (the stop Minimundus is indicated by a red circle) and the campus map.
The travel time from Heiligengeistplatz to the university is about 10 minutes. It is recommendable to have a look at the bus schedule in order to avoid long waiting times.
The cost of a bus-ticket (valid for one hour) is 1.60 Euro and can be bought on the bus.
#### Reaching Klagenfurt University by taxi from the main station or airport
Taxis are available in front of the Arrival/Departure Hall of Klagenfurt Airport as well as in front of the train station. Typical fares and driving times are:
• Airport – University: 15-17 Euro, 20 minutes
• Main station – University: 8-10 Euro, 10 minutes
### Other useful information
Bring adequate shoes for hiking and a swimsuit. The campus is located quite close to lake "Wörthersee". You may have a look at the weather in Klagenfurt.
Information about the youth hostel: www.oejhv.or.at (Address: Neckheimgasse 6, phone number +43 463 230020)
Internet-access: Participants will have access to the University's Computer rooms. Also, wireless LAN and plug-in connections for notebooks are available at the campus area.
### Contact
For information please contact Franz Rendl (phone: +43 463 2700 3114).
### Photo download
Angelika Wiegele Last modified: Sun Oct 2 12:20:50 CEST 2005 Free counter | 3,212 | 12,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.898065 |
https://en.khanacademy.org/kmap/functions-c/x486198f6a5c3b46c:linear-equations-graphs/x486198f6a5c3b46c:intercepts/v/introduction-to-intercepts | 1,716,551,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00785.warc.gz | 201,229,405 | 77,747 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: Functions 229-236>Unit 1
Lesson 3: Intercepts
# Intro to intercepts
Learn what x- and y-intercepts are. The equations used in this video are y = 0.5x - 3 and 5x + 6y = 30.
## Want to join the conversation?
• what would happen if you were to add the Z axis to the graph and then do the math?
• good thinking dude!! you will learn this soon in 12th class in 3d geometry
• What if both x and y equal 1?
• Then...I doubt it would be counted as an intercept, as it'll have a different point of intersecting either axis. But it doesn't matter on the coordinates as you have provided only one coordinate, as a linear equation requires two.
Yes, the point is (1, 1), but then again, it's not a linear equation is it? It's a coordinate. There are infinite linear equations where x=1 and y=1, so no concrete answers on the basis of intercepts.
EDIT: Oops, didn't notice that it's a year old...
• What is the solution to this system of linear equations:
X+2y+3z=6
X+3y=0
y-z=0
• What is Z meant for? Since, Z isn't used in any linear equations that I know of.
Edit: It's confusing also with Z added to it.
• How can you find the slope of a line like how can you find the answer of a line or a slope
• To find the slope of a line, you can use the formula:
slope = (y2 - y1) / (x2 - x1),
where (x1, y1) and (x2, y2) are two points on the line. The slope represents the rate of change in the y-direction as the x-coordinate changes. A positive slope means that the line is rising, a negative slope means that the line is falling, and a slope of 0 means that the line is horizontal.
• y=-X^3 9
how do we find the x and y intercepts?
• When y=0, calculate x to find the x-intercept. When x=0, calculate y to find the y-intercept. The y-intercept is easy: 0^3*9 = 0 = y-intercept
The x-intercept is also 0: 0^3 * 9 = 0.
Thus both intercepts=0
• Does anyone know what sal looks like?
• How do you find the intercepts from a graph
• y-intercepts are when the line touches the y-axis. To find these, find the y when x = 0.
The point for a y-intercept will look like (0,y)
x-intercepts are when the line touches the x-axis. To find these, find the x when y = 0.
The point for an x-intercept will look like (x,0)
• what is the formula for this?
• Not sure what formula you are looking for, but to find x intercept let y be zero (because y is zero on the x axis), and to find y intercept let x be zero for same reasoning.
• I don't understand how to find x and y-intercepst. Like how to find say, an x-intercept of an linear equation. Or y-intercept of an equation. For example, how do you find the intercept of 4x-2=8?
• 4x-2=8 is a special type of linear equation.
Start by simplifying the equation. Solve it so your equation looks like "X = a number"
You will then have a linear equation where every point on the line will have an X-value equal to that number. For example: X=5 means X=5 for every possible value of Y that you can think of. So, you can have ordered pairs of: (5, 6); (5, 0); (5, -2); (5, -3.5); etc.
If you graph them, you will see that this type of equation creates a vertical line. It has an X-intercept, but it does not have a Y-intercept (the vertical line never crosses the Y-axis).
Hope this helps. | 960 | 3,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.949553 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/9408/2/a/dv/ | 1,600,561,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00155.warc.gz | 939,976,796 | 44,733 | # Properties
Label 9408.2.a.dv Level 9408 Weight 2 Character orbit 9408.a Self dual yes Analytic conductor 75.123 Analytic rank 0 Dimension 2 CM no Inner twists 1
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$9408 = 2^{6} \cdot 3 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 9408.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$75.1232582216$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{2})$$ Defining polynomial: $$x^{2} - 2$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 1176) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \sqrt{2}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + q^{3} + ( -2 + \beta ) q^{5} + q^{9} +O(q^{10})$$ $$q + q^{3} + ( -2 + \beta ) q^{5} + q^{9} + ( 2 + 2 \beta ) q^{11} -3 \beta q^{13} + ( -2 + \beta ) q^{15} + ( 6 - \beta ) q^{17} + ( -4 + 2 \beta ) q^{19} + ( -2 + 2 \beta ) q^{23} + ( 1 - 4 \beta ) q^{25} + q^{27} + 2 \beta q^{29} + 2 \beta q^{31} + ( 2 + 2 \beta ) q^{33} + ( -4 - 4 \beta ) q^{37} -3 \beta q^{39} + ( 6 - 3 \beta ) q^{41} -8 \beta q^{43} + ( -2 + \beta ) q^{45} + ( 4 + 6 \beta ) q^{47} + ( 6 - \beta ) q^{51} + 2 q^{53} -2 \beta q^{55} + ( -4 + 2 \beta ) q^{57} + 6 \beta q^{59} + ( -4 + 5 \beta ) q^{61} + ( -6 + 6 \beta ) q^{65} + 8 \beta q^{67} + ( -2 + 2 \beta ) q^{69} + ( -2 + 6 \beta ) q^{71} + ( 12 + 3 \beta ) q^{73} + ( 1 - 4 \beta ) q^{75} + ( 8 - 4 \beta ) q^{79} + q^{81} + 4 q^{83} + ( -14 + 8 \beta ) q^{85} + 2 \beta q^{87} + ( 10 + 3 \beta ) q^{89} + 2 \beta q^{93} + ( 12 - 8 \beta ) q^{95} + ( 4 + 3 \beta ) q^{97} + ( 2 + 2 \beta ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + 2q^{3} - 4q^{5} + 2q^{9} + O(q^{10})$$ $$2q + 2q^{3} - 4q^{5} + 2q^{9} + 4q^{11} - 4q^{15} + 12q^{17} - 8q^{19} - 4q^{23} + 2q^{25} + 2q^{27} + 4q^{33} - 8q^{37} + 12q^{41} - 4q^{45} + 8q^{47} + 12q^{51} + 4q^{53} - 8q^{57} - 8q^{61} - 12q^{65} - 4q^{69} - 4q^{71} + 24q^{73} + 2q^{75} + 16q^{79} + 2q^{81} + 8q^{83} - 28q^{85} + 20q^{89} + 24q^{95} + 8q^{97} + 4q^{99} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−1.41421 1.41421
0 1.00000 0 −3.41421 0 0 0 1.00000 0
1.2 0 1.00000 0 −0.585786 0 0 0 1.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 9408.2.a.dv 2
4.b odd 2 1 9408.2.a.dh 2
7.b odd 2 1 9408.2.a.dr 2
8.b even 2 1 1176.2.a.l 2
8.d odd 2 1 2352.2.a.bg 2
24.f even 2 1 7056.2.a.ce 2
24.h odd 2 1 3528.2.a.bc 2
28.d even 2 1 9408.2.a.ed 2
56.e even 2 1 2352.2.a.z 2
56.h odd 2 1 1176.2.a.m yes 2
56.j odd 6 2 1176.2.q.m 4
56.k odd 6 2 2352.2.q.ba 4
56.m even 6 2 2352.2.q.bg 4
56.p even 6 2 1176.2.q.n 4
168.e odd 2 1 7056.2.a.cw 2
168.i even 2 1 3528.2.a.bm 2
168.s odd 6 2 3528.2.s.bl 4
168.ba even 6 2 3528.2.s.bc 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
1176.2.a.l 2 8.b even 2 1
1176.2.a.m yes 2 56.h odd 2 1
1176.2.q.m 4 56.j odd 6 2
1176.2.q.n 4 56.p even 6 2
2352.2.a.z 2 56.e even 2 1
2352.2.a.bg 2 8.d odd 2 1
2352.2.q.ba 4 56.k odd 6 2
2352.2.q.bg 4 56.m even 6 2
3528.2.a.bc 2 24.h odd 2 1
3528.2.a.bm 2 168.i even 2 1
3528.2.s.bc 4 168.ba even 6 2
3528.2.s.bl 4 168.s odd 6 2
7056.2.a.ce 2 24.f even 2 1
7056.2.a.cw 2 168.e odd 2 1
9408.2.a.dh 2 4.b odd 2 1
9408.2.a.dr 2 7.b odd 2 1
9408.2.a.dv 2 1.a even 1 1 trivial
9408.2.a.ed 2 28.d even 2 1
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$3$$ $$-1$$
$$7$$ $$1$$
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(9408))$$:
$$T_{5}^{2} + 4 T_{5} + 2$$ $$T_{11}^{2} - 4 T_{11} - 4$$ $$T_{13}^{2} - 18$$ $$T_{17}^{2} - 12 T_{17} + 34$$ $$T_{19}^{2} + 8 T_{19} + 8$$ $$T_{31}^{2} - 8$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ 1
$3$ $$( 1 - T )^{2}$$
$5$ $$1 + 4 T + 12 T^{2} + 20 T^{3} + 25 T^{4}$$
$7$ 1
$11$ $$1 - 4 T + 18 T^{2} - 44 T^{3} + 121 T^{4}$$
$13$ $$1 + 8 T^{2} + 169 T^{4}$$
$17$ $$1 - 12 T + 68 T^{2} - 204 T^{3} + 289 T^{4}$$
$19$ $$1 + 8 T + 46 T^{2} + 152 T^{3} + 361 T^{4}$$
$23$ $$1 + 4 T + 42 T^{2} + 92 T^{3} + 529 T^{4}$$
$29$ $$1 + 50 T^{2} + 841 T^{4}$$
$31$ $$1 + 54 T^{2} + 961 T^{4}$$
$37$ $$1 + 8 T + 58 T^{2} + 296 T^{3} + 1369 T^{4}$$
$41$ $$1 - 12 T + 100 T^{2} - 492 T^{3} + 1681 T^{4}$$
$43$ $$1 - 42 T^{2} + 1849 T^{4}$$
$47$ $$1 - 8 T + 38 T^{2} - 376 T^{3} + 2209 T^{4}$$
$53$ $$( 1 - 2 T + 53 T^{2} )^{2}$$
$59$ $$1 + 46 T^{2} + 3481 T^{4}$$
$61$ $$1 + 8 T + 88 T^{2} + 488 T^{3} + 3721 T^{4}$$
$67$ $$1 + 6 T^{2} + 4489 T^{4}$$
$71$ $$1 + 4 T + 74 T^{2} + 284 T^{3} + 5041 T^{4}$$
$73$ $$1 - 24 T + 272 T^{2} - 1752 T^{3} + 5329 T^{4}$$
$79$ $$1 - 16 T + 190 T^{2} - 1264 T^{3} + 6241 T^{4}$$
$83$ $$( 1 - 4 T + 83 T^{2} )^{2}$$
$89$ $$1 - 20 T + 260 T^{2} - 1780 T^{3} + 7921 T^{4}$$
$97$ $$1 - 8 T + 192 T^{2} - 776 T^{3} + 9409 T^{4}$$ | 2,857 | 5,556 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-40 | longest | en | 0.288965 |
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## Research Methods Cont'd
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Discussion over sampling strategies and the interaction effect
Lecture number:
3
Pages:
2
Type:
Lecture Note
School:
Texas A&M University
Course:
Psyc 330 - Personality
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PSYC 330 1st Edition Lecture 3 Outline of Last Lecture I Characteristics of a theory Outline of Current Lecture II Sampling Strategies III Interaction Effect IV Reliability and Correlation a Methods of estimating reliability i Test retest ii Internal consistency split half Current Lecture Sampling Strategies Sample a Population o Want to be able to generalize conclusions so the sample should represent the population Control person and situation variables o Planned uniformity You can only generalize something to one group o Systematic variation o Randomization Interaction effect when two variable combine to produce an effect they have to be considered in combination to predict an outcome Interaction the relationship between one variable and another variable depends on a third variable means it depends A classic interaction effect will look like an X Moderator Variable another way of saying there is an interaction present its another way of describing an interaction Reliability and Correlation Reliability ability of a measure to yield consistent results o Consistency doesn t mean accurate Ex Playing darts you consistently hit the board with the dart but you fail to hit the bulls eye Thus you re consistent but not accurate Consistency is measured by correlation If the measure is reliable the correlation will be high with a low reliability the measure or correlation will be zero Reliability is measured form 0 1 you should never have a negative number This consistency typically calculating the correspondence between 2 measurements All involve some form of correlation coefficient Methods of estimating reliability Test retest Give a measure once then give it a second time in the future Calculate the correlation between the 2 Internal consistency split half Instead of giving the measure twice correlate the two halves of the measure Tells you that the different items are consistency measuring the same thing Scorer The two lists of numbers being correlated are listed proved
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## Presentation on theme: "Chapter 7 Gases."— Presentation transcript:
Chapter 7 Gases
Kinetic Theory of Gases
Particles of a gas Move rapidly in straight lines and are in constant motion. Have kinetic energy that increases with an increase in temperature. Are very far apart. Have essentially no attractive (or repulsive) forces. Have very small volumes compared to the volume of the container they occupy.
Properties of Gases Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).
Barometer A barometer measures the pressure exerted by the gases in the atmosphere. The atmospheric pressure is measured as the height in mm of the mercury column.
Learning Check A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) H2O is heavier 3) air is more dense than H2O
Solution A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense
Pressure A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area One atmosphere (1 atm) is 760 mm Hg. 1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr
Units of Pressure In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).
Learning Check A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) atm 3) x 105 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) mm Hg 2) mm Hg 3) 22,300 mm Hg
Solution A. What is 475 mm Hg expressed in atm? 2) 0.638 atm
485 mm Hg x atm = atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) mm Hg 2.00 atm x mm Hg = mm Hg 1 atm
Boyle’s Law The pressure of a gas is inversely related to its volume when T and n are constant. If volume decreases, the pressure increases.
PV Constant in Boyle’s Law
The product P x V remains constant as long as T and n do not change. P1V1 = atm x L = atm L P2V2 = atm x L = atm L P3V3 = atm x L = atm L Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)
Solving for a Gas Law Factor
The equation for Boyle’s Law can be rearranged to solve for any factor. To solve for V2, divide both sides by P2. P1V1 = P2V Boyle’s Law P2 P2 P1V1 = V2 P2 Solving for P2 P2 = P1 V1 V2
PV in Breathing Mechanics
When the lungs expand, the pressure in the lungs decreases. Inhalation occurs as air flows towards the lower pressure in the lungs.
PV in Breathing Mechanics
When the lung volume decreases, pressure within the lungs increases. Exhalation occurs as air flows from the higher pressure in the lungs to the outside.
Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T? 1. Set up a data table Conditions 1 Conditions 2 P1 = 50 mm Hg P2 = 200 mm Hg V1 = 8 L V2 = ?
Calculation with Boyle’s Law (continued)
2. When pressure increases, volume decreases. Solve Boyle’s Law for V2: P1V1 = P2V2 V = V1P1 P2 V2 = 8 L x mm Hg = L 200 mm Hg pressure ratio decreases volume
Learning Check The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? At a new pressure of 420 mm Hg, what is the new volume? 1) 60 mL 2) 120 mL 3) 240 mL
Solution The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? B) If P decreases, V increases. At a new pressure of 420 mm Hg, what is the new volume of the cylinder? 3) 240 mL
Learning Check A sample of helium gas has a volume of 6.4 L at a pressure of atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L B) 6.4 L C) 12.8 L
Solution A) 3.2 L Solve for V2: P1V1 = P2V2 V2 = V1P1 P2
V2 = L x atm = L 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant).
Learning Check A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.) 1) mm Hg 2) mm Hg 3) 1200 mm Hg
Solution 1) mm Hg Data table Conditions 1 Conditions 2 P1 = mm Hg P2 = ??? V1 = L V2 = L P2 = P1 V1 V2 600. mm Hg x L = mm Hg L
Charles’ Law The Kelvin temperature of a gas is directly related to the volume (P and n are constant). When the temperature of a gas increases, its volume increases.
Charles’ Law V and T For two conditions, Charles’ Law is written
V1 = V (P and n constant) T T2 Rearranging Charles’ Law to solve for V2 V2 = V1T2 T1
Learning Check Solve Charles’ Law expression for T2. V1 = V2 T T2
Solution V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1
Isolate T2 by dividing through by V1 V1T = V2T1 V1 V1 T2 = V2T1 V1
Calculations Using Charles’ Law
A balloon has a volume of 785 mL at 21°C. If The temperature drops to 0°C, what is the new volume of the balloon (P constant)? 1. Set up data table: Conditions 1 Conditions 2 V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K Be sure that you always use the Kelvin (K) temperature in gas calculations.
Calculations Using Charles’ Law (continued)
2. Solve Charles’ law for V2 V1 = V2 T T2 V2 = V1 T2 T1 V2 = 785 mL x K = 729 mL 294 K
Learning Check A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443°C 2) 170°C 3) – 82°C
Solution 170°C T2 = T1V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL
= 443 K – 273 K = 170°C
Gay-Lussac’s Law: P and T
The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. P1 = P2 T1 T2
Calculation with Gay-Lussac’s Law
A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) Set up a data table. Conditions 1 Conditions 2 P1 = 2.0 atm P2 = T1 = 18°C T2 = 62°C + 273 = 291 K = K ?
Calculation with Gay-Lussac’s Law (continued)
2. Solve Gay-Lussac’s Law for P2 P1 = P2 T T2 P2 = P1 T2 T1 P2 = 2.0 atm x 335 K = atm 291 K
Learning Check Use the gas laws to complete with
Increases 2) Decreases A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12.0 L to 24.0 L. D. Volume _______when T changes from 15.0 °C to 45.0°C.
Solution Use the gas laws to complete with 1) Increases 2) Decreases
A. Pressure 1) Increases, when V decreases. B. When T decreases, V 2) Decreases. C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C
Next Time We complete Chapter 7
Combined Gas Law The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P1 V1 = P2 V2 T1 T2
Combined Gas Law Calculation
A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1 Conditions 2 P1 = atm P2 = atm V1 = L (180 mL) V2 = 90.0 mL T1 = 29°C = 302 K T2 = ??
Combined Gas Law Calculation (continued)
2. Solve for T2 P1 V1 = P2 V2 T1 T2 T2 = T1 P2V2 P1V1 T2 = 302 K x atm x mL = K 0.800 atm mL T2 = 604 K – = °C
Learning Check A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?
Solution Data Table T1 = 308 K T2 = -95°C + 273 = 178K
V1 = 675 mL V2 = ??? P1 = 646 mm Hg P2 = 802 mm Hg Solve for T2 V2 = V1 P1 T2 P2T1 V2 = mL x 646 mm Hg x 178K = mL mm Hg x 308K
The volume of a gas is directly related to the number of moles of gas when T and P are constant. V1 = V2 n n2
Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) L 2) 1.8 L 3) 2.4 L
Solution 3) 2.4 L Conditions 1 Conditions 2 V1 = 1.5 L V2 = ???
n1 = mole He n2 = 1.2 moles He V2 = V1n n1 V2 = 1.5 L x moles He = L 0.75 mole He
STP The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg)
Molar Volume At STP, 1 mole of a gas occupies a volume of 22.4 L.
The volume of one mole of a gas is called the molar volume.
Molar Volume as a Conversion Factor
The molar volume at STP can be used to form conversion factors L and mole 1 mole L
Learning Check A. What is the volume at STP of 4.00 g of CH4?
1) L 2) L 3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) g 2) g 3) 1.43 g
Solution 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH mole CH4 B. 3) 1.43 g 8.00 L x 1 mole He x g He = g He 22.4 L mole He
Ideal Gas Law The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT Rearranging this expression gives the expression called the ideal gas law. PV = nRT
Universal Gas Constant, R
The universal gas constant, R, can be calculated using the molar volume of a gas at STP. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = L atm mole K Note there are four units associated with R.
Learning Check Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?
Solution What is the value of R when the STP value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L) nT (1 mole) (273K) = L mm Hg mole K
Learning Check Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N2O at 23°C, what is the pressure (mm Hg) in the tank?
Solution 1. Adjust the units of the given properties to match the units of R. V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? 2. Rearrange the ideal gas law for P. P = nRT V P = (2.8 moles)(62.4 L mm Hg)(296 K) (20.0 L) (mole K) = 2.6 x 103 mm Hg
Learning Check A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?
Solution 1. Determine the given properties.
P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mole K) = mole O2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = mole O2 x g O2 = 5.8 g O2 1 mole O2
Molar Mass of a Gas What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT ( L atm/mole K)(303K) = mole 2. Set up the molar mass relationship. Molar mass = g = g = g/mole mole mole
Gases in Equations The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. Problem: What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum? 2Al(s) Cl2 (g) AlCl3(s)
Gases in Equations (continued)
2Al(s) + 3Cl2 (g) AlCl3(s) 1.5 g ? L 1.2 atm, 300K Calculate the moles of Cl2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl2 = mole Cl2 27.0 g Al moles Al 2. Place the moles Cl2 in the ideal gas equation. V = nRT = (0.083 mole Cl2)( Latm/moleK)(300K) P atm = 1.7 L Cl2
Learning Check What volume (L) of O2 at 24°C and atm are needed to react with 28.0 g NH3? 4NH3(g) + 5O2(g) NO(g) + 6H2O(g)
Solution 1. Calculate the moles of O2 needed.
28.0 g NH3 x 1 mole NH3 x 5 mole O2 17.0 g NH mole NH3 = 2.06 mole O2 2. Place the moles O2 in the ideal gas equation. V = nRT = (2.06 moles)( L atm/moleK)(297K) P atm = L O2
Partial Pressure In a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container.
Dalton’s Law of Partial Pressures
The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. PT = P1 + P
Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles.
Total Pressure For example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L. V = 22.4 L 1.0 mole N2 0.4 mole O2 0.6 mole He 1.0 mole 0.5 mole O2 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 atm 1.0 atm 1.0 atm
Learning Check A scuba tank contains O2 with a pressure of atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank?
Solution 1. Convert the pressure in atm to mm Hg
0.450 atm x 760 mm Hg = 342 mm Hg = PO 1 atm 2. Calculate the sum of the partial pressures. Ptotal = PO + PHe 2 Ptotal = 342 mm Hg mm Hg = mm Hg
Gases We Breathe
Health Note: Scuba Diving
When a scuba diver makes a deep dive, the increased pressure causes more N2 (g) to dissolve in the blood. If a diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends." Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.
Learning Check For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium? 1) 520 mm Hg 2) mm Hg 3) mm Hg
Solution 4800 mm Hg PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm
PTotal = PO PHe 2 PHe = PTotal - PO PHe = mm Hg mm Hg = mm Hg
Similar presentations | 4,581 | 13,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-51 | latest | en | 0.893307 |
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Test 2 Time: 1.5 hours
1. Use fundamental identities and algebra to simplify
to one of the six basic trigonometric functions. Show all steps.
2. Find all exact solutions for
on the interval .
3. A triangle has angles and . The side c between these angles is 6 feet long. Find the third angle C and the lengths of the other two sides. Round the lengths to two decimal places.
4. Verify the identity . Show all steps.
5. Use the Sum and Difference Identities to find the exact values of .
6. If and , find the exact values for the following:
• .
• .
7. A triangle has sides of lengths a=5, b=7, and c=10 meters. Find its three angles and round your results to one decimal place.
8. Find all exact solutions for on the interval .
9. Let .
• Convert z to its trigonometric form.
• Use DeMoivre's Theorem to solve the equation
Write your solutions in standard form .
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# CHAPTER
Solved Problems
2.1 The latching current for a thyristor inserted between a dc source voltage of 100V and a load being
75mA. Calculate the minimum width of the gate-pulse required to turn-on the thruster when the
(i) Purely inductive having an inductance of 100 mH and (ii) Consisting of resistance and
inductance of 10 ohm and 100 mH respectively.
Sol. (i)
V =L
di
dt
i.e.,
di =
V
dt
L
i =
V
t.
L
t =
## L i 100 10-3 75 10-3
=
V
100
= 75 ms.
(ii) When load is R-L types
V =Ri+L
i =
100 10
di
dt
R
- t
V
1- e L
R
)
10
t
-3
100
=
1 - e 100 10
10
t = 100 msec
2.2
An SCR with
## dt rating of 20 Amp2 sec is used to act as a rectifier of feed a load. If a earth
fault occurs at the output of the rectifier when the input ac voltage (100 sin w t) is at its positive
peak, find the fault current and the safe time that the SCR can withstand the fault without
damage. Assume the wire resistance and thyristor resistance to be 1 ohm.
Sol. Let if be the fault current,
if =
100
input peak voltage
= 100 A.
=
1
total circuit resistance
Solution Manual 2
i2 dt = 20
Since,
tf =
20
tf
o 100
dt = 20
= 2 m sec.
1002
Thus, the safe time the fault of 100 A can be withstood without the damage of SCR is 2 m Sec.
2.3
The specification sheet for an SCR gives maximum rms on state current as 35 A. If this SCR is
used in a resistance circuit, computer average. On state current rating for half sine wave current
for conduction angles of (a) 180 (b) 90 (c) 30
## Sol. Half sine wave current waveform is shown in Fig. 3.1
i
Conduction
angle.
Im
Q1
Fig. 3.1
Iav =
Irms
I
1 p
I m sin q dq = m (1 + cos q1)
q
2p
2p 1
1
=
2p
I m2
q1
sin q dq
12
I 2 q sin 2 q p
= m 4 q1
2p 2
I2
= m
2p
12
I 2 p - q1 1 2
= m
+ sin q1
2
4
2p
12
## (a) For 180 conduction angle, q1 = 0
\
and
Iav =
Irms
Im
I
[(1 + cos 0)] = m
2p
p
I2 p 1
= m
- (0)
2p 2 4
ITAV =
12
Im
2
I rms
I
p
p
= m
=
2 Im
I av
2
I rms 35 2
= 22.282 A.
=
FF
p
## (b) For 90, conduction angle, q1 = 90.
1 cos 2 q
2 - 2
q1
12
Power Electronics
Iav =
\
and
Irms
Im
I
1 + cos q0 ] = m
[
2p
2p
I2 p 1
+ (0)
= m
2p 4 4
Form factor =
ITAV =
Im
2 2
12
Im
2 2
2p p
.
Im
2
35 2
= 15.755 A.
p
## (c) Fro 30, conduction angle, q1 = 150.
Iav =
Irms
I2 p 1
= m
+ ( -0.866)
2p 12 4
Form factor =
ITav =
2.4
Im
[1 + ( -0.866)] = 0.021 Im.
2p
12
= 0.085 Im.
0.0849035 I m
= 3.98
0.021 I m
35
= 8.79 A
3.98
## Sol. Rectangular waveform is shown in Fig. 4.1
i
360
I
t
T
nT
Fig. 4.1
Conduction angle =
\
Here
h =
Iav =
T
360
hT
360
conduction angle
I T
I
=
hT
h
I2 T
Irms =
hT
12
Solution Manual 4
Iav =
Form factor =
ITav =
I
I
.
and Irms =
2
2
I
2
=
2 I
35
Iav =
Form factor =
ITav =
360
=4
90
I
I
I
= .
and Irms =
4
2
4
I 4
=2
2 I
35
= 17.5 A.
2
360
= 12
12
Iav =
I
and Irms =
12
Form factor =
I 12
= 12
12 I
ITAV =
2.5
= 24.75 A.
## (b) For 90 conduction angle, h =
360
= 2.
180
35
12
I
12
= 10.10 A
In a class-C commutations circuit, determine the value of R, RL, and C for commutating the main
SCR when it is conducting a full current of 15 Amp. The minimum time for which this SCR is to
be reverse biased for proper commutation is 30 msec. It is given that the complementary SCR
will undergo natural commutation when its forward current falls below the holding current of
3mA. Assume supply voltage to be 100 volts.
Edc 100
= 6.66 W.
=
15
IL
Sol.
RL =
Now,
RC
Vc = Edc . 1 - e
## t = 30 ms, Edc = 200 V, Vc = 100 V.
100 = 200 1 - e RL C
Power Electronics
t
RC C
## = loge 0.5 = 0.693
t
30 10-6
=
0.693 RL 0.693 6.66
C=
C = 6.49 mF
## Select R such that
R>
Edc
3 mA
R>
100 V
3 10 - 3
2.6
\ R > 33.33 k W
Determine the values of R and C in Fig. E 2.6. The load current through the main SCR 1 is 25 A.
The minimum time for which this SCR is to be reverse biased is 30 ms. The auxiliary SCR 2
should undergo natural commutation when its forward current goes below its holding current of
2.5 mA.
+
R1
R2
C
100 V
SCR 1
SCR 2
Fig. E 2.6
Sol. In order to ensure that SCR2 will undergo natural commutation
R2
100
2.5 10-3
R2 40 K W
Where
R1 =
100
=4W
25
## The charging equation for C is
tq
R1 C
Vc = V 1 - e
tq
50 = 100 1 - e 4C
tq = 0.693 4C = 2.77 C.
C=
30 10-6
= 10.82 mF.
2.77
## The commutation capacitor C 10.82 mF
Solution Manual 6
E 2.7 For the class-C commutation, R1 = R2 = 5 W, C = 10 mF, supply voltage Edc = 100 V. Determine
the turns-off time.
Sol. Capacitor charging equation is
tc
Vc = Edc 1 - e R1 C
tc
R1 C
100 = 200 1 - e
## tc = 0.693 R1C = 0.693 5 10 106 = 34.65 msec.
E 2.8 Determine the commutating components for an auxiliary commutation method if IL = 10A, Edc
= 100 V and tq = 50 ms.
Sol. We have the equation
C=
=
Also,
Lmin =
I L max tq
Edc
10 50 10-6
= 5 mF
100
Edc 2
C
I L2
2
100
=
5 106 = 0.5 mH
10
Lmax =
0.01T 2
p2 C
T = 2.5 ms
## Lmax = 1.26 mH.
E 2.9 In class-D commutation circuit, Edc = 200 V, L = 10 mH, C = 50 mF. Determine the minimum
on time of SCR 1 and peak-value of capacitor current.
Sol. Minimum on time of SCR 1 is given by
ton(min) = p LC = 10 = 10-6 50 10-6
= 69 msec.
Peak SCR current is given by,
Ip = Edc
= 200
C
L
50 10-6 /10 10-6
= 447.21 A.
Power Electronics
E 2.10 Thyristor in Fig. 2.10 has a latching current level of 50 mA and is fired by a pulse of width 50
ms. Show that without R, SCR will fail to remain ON when the firing pulse ends. Find the
maximum value of R to ensure firing.
Neglect the SCR volt-drop and assume that the initial value of rate of rise of current remains
constant over the entire pulse width.
+
i
20 W
100 V
R.
0.5 H
Fig. E 2.10
t = L/R =
0.5
= 0.025 sec.
20
100
= 5 A.
20
Without R, SCR current i will grow exponentially as,
t
-
i = I0. 1 - e t
## Max. value of steady state current =
at
= 5 1 - e 0.025
ton = t = 50 msec,
50 10-6
## i = 5 1 - e 0.025 = 9.99 103
= 10 mA
OR
Since given that initially
\
\
di
is constant over entire time ion,
dt
ton =
I0
5
=
= 200 A/sec.
t 0.025
i
= 50 106 = 200 50 106 = 10 mA
t
\ Hence if trigger pulse with is 50 msec, SCR will fail to reach its latching current level of 50 mA.
SCR current is less by (50 10) = 40 mA. from its latching current value.
This current must be supported by additional R.
\
\
40 103 R = 100 V
R=
100
= 2.5 kW.
40 10-3 | 2,533 | 6,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.714032 |
http://shell.math.rutgers.edu/~greenfie/mill_courses/math152a/apology.html | 1,537,559,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157503.43/warc/CC-MAIN-20180921190509-20180921210909-00094.warc.gz | 218,886,828 | 1,840 | ### An apology and modification for workshop #6, Math 151:01-03, spring 2008
I asked for a writeup for problem 4 of workshop #6. I stupidly neglected to write a complete solution before assigning the problem, and therefore did not know then that an additional technique, using an integrating factor, is necessary to find a solution of the differential equation. I very much apologize for this lack of care and regret any difficulty that has occurred. I want to modify the assignment. First here's how the original assignment could be done:
Original assignment
1. Use complete English sentences to describe a mathematical model of the amount of salt, S(t), in pounds, at time t, in minutes after the start of the flow. The resulting model should be a differential equation and an initial condition: an initial value problem.
2. Solve the initial value problem using an appropriate integrating factor.
3. Use the solution of the differential equation to predict the salt concentration at the time that the tank overflows. The student will need to describe how this time is found, and what the concentration is (in appropriate units), rather than the amount of salt.
The second step of the original assignment must be modified (integrating factors are covered in Math 244, taken by engineering and physics and chemistry students, and in Math 252, taken by math majors).
Revised assignment
1. Use complete English sentences to describe a mathematical model of the amount of salt, S(t), in pounds, at time t, in minutes after the start of the flow. The resulting model should be a differential equation and an initial condition: an initial value problem. How students do this is the most important part of the writeup to me.
2. Verify that the function
``` 4t2+80t+100
S(t)= -------------
t+10```
is a solution of the initial value problem which was created in the first step. The student must show by explicit computations that this function solves the differential equation and the initial condition. Since the Existence and Uniqueness Theorem guarantees that there will be exactly one solution to the initial value problem, the student's computations show that this function must be the desired solution.
3. Use the solution of the differential equation to predict the salt concentration at the time that the tank overflows. The student will need to describe how this time is found, and what the concentration is (in appropriate units), rather than the amount of salt.
Again, I apologize for my error. Please let me know if you have any questions. I also thank Mr. Priyadashi and various students for bringing this situation to my attention. | 544 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.934886 |
https://walktocurecancer.org/submitted-p2-v-11p3-v-1-7p1-v-3p2-v/ | 1,619,061,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039560245.87/warc/CC-MAIN-20210422013104-20210422043104-00394.warc.gz | 694,176,136 | 14,860 | Submittedto:-PROF.SONALI SINGH Submitted by:AHMER HASSAN (JN170173) MANAGERIAL SCIENCE:- 2ndtrimester{2017} ASSIGNMENT ON GAME THEORYQ. Two companies are competing for the sameproduct. To improve its market share, company A decides to launch the followingstrategies. A1 = give discount coupons A2 = home delivery services A3 = free giftsThe company B decides to use mediaadvertising to promote its product. B1 = internet B2 = newspaper B3 = magazine Company B Company A B1 B2 B3 A1 6 -5 3 A2 2 -3 -7 A3 -2 7 4 Uselinear programming to determine the best strategies for both the companies. solution.
Company B Minimum Company A B1 B2 B3 A1 6 -5 3 -5 A2 2 -3 -7 -7 A3 -2 7 4 -2 Maximum 6 7 4 Minimax=-2 Maximin = 4This game has no saddle point. Sothe value of the game lies between –2 and +4. It is possible that the value ofgame may be negative or zero.
Thus, a constant k is added to all the elementsof pay-off matrix. Let k = 4, then the given pay-off matrix becomes: Company B Company A B1 B2 B3 A1 10 -1 7 A2 6 1 -3 A3 2 11 8 LetV = value of the gamep1, p2 & p3 = probabilities of selectingstrategies A1, A2 & A3 respectively.q1, q2 & q3 = probabilities of selectingstrategies B1, B2 & B3 respectively.
Company B Probability Company A B1 B2 B3 A1 10 -1 7 p1 A2 6 1 -3 p2 A3 2 11 8 p3 Probability q1 q2 q3 Company A’s objective is to maximize theexpected gains, which can be achieved by maximizing V, i.e., it might gain morethan V if company B adopts a poor strategy. Hence, the expected gain for companyA will be as follows: 10p1+6p2+2p3?V-p1+p2+11p3?V7p1-3p2+8p3?Vp1+p2+p3=1and p1, p2, p3 ? 0Dividing the above constraints by V, we get10p1/V + 6p2/V + 2p3/V ? 1-p1/V + p2/V + 11p3/V ? 17p1/V – 3p2/V + 8p3/V ? 1p1/V + p2/V + p3/V = 1/VTo simplify the problem, we put p1/V = x1, p2/V =x2, p3/V = x3In order to maximize V, company A canMinimize 1/V = x1+ x2+ x3subject to;10×1 +6×2 + 2×3 ? 1-x1 +x2 + 11×3 ? 17×1 -3×2 + 8×3 ? 1and x1, x2, x3 ? 0 Company B’sobjective is to minimize its expected losses, which can be reduced byminimizing V, i.
e., company A adopts a poor strategy. Hence, the expected lossfor company B will be as follows:10q1 – q2 + 7q3 ? V6q1 +q2 -3q3 ? V2q1 + 11q2 + 8q3 ? Vq1 + q2 + q3 = 1 and q1,q2, q3 ? 0Dividing the above constraints by V, we get10q1/V – q2/V + 7q3/V ? 16q1/V +q2/V – 3q3/V ? 12q1/V +11q2/V + 8q3/V ? 1q1/V + q2/V + q3/V = 1/V To simplify the problem, we put q1/V = y1, q2/V =y2, q3/V = y3 In order to minimize V, company B can Maximize 1/V = y1+ y2+ y3subject to10y1 – y2 + 7y3 ? 16y1 +y2 – 3y3 ? 12y1 +11y2 + 8y3 ? 1 and y1, y2, y3 ? 0
### Post Author: admin
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### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one? | 1,913 | 8,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-39 | latest | en | 0.857769 |
https://www.coursehero.com/file/6004594/EP-The-Scope-of-Logic-Wesley-C-Salmon/ | 1,516,400,669,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00483.warc.gz | 915,296,825 | 68,833 | EP The Scope of Logic - Wesley C. Salmon
EP The Scope of Logic - Wesley C. Salmon - Wesley C Salmon...
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Wesley C. Salmon, Logic (1984). THE SCOPE OF LOGIC 1 When people make statements, they may offer evidence to support them or they may not. A statement that is supported by evidence is the conclusion of an argument, and logic provides tools for the analysis of arguments. Logical analysis is concerned with the relationship between a conclusion and the evidence given to support it. When people reason, they make inferences. These inferences can be transformed into arguments, and the tools of logic can then be applied to the resulting arguments. In this way, the inferences from which they originate can be evaluated. Logic deals with arguments and inferences. One of its main purposes is to provide methods for distinguishing those which are logically correct from those which are not. 1. ARGUMENT In one of his celebrated adventures, Sherlock Holmes comes into possession of an old felt hat. Although Holmes is not acquainted with the owner of the hat, he tells Dr. Watson many things about the man -- among them, that he is highly intellectual. This assertion, as it stands, is unsupported. Holmes may have evidence for his statement, but so far he has not given it. Dr. Watson, as usual, fails to see any basis for Holmes's statement, so he asks for substantiation. "For answer Holmes clapped the hat upon his head. It came right over the forehead and settled upon the bridge of his nose. 'It is a question of cubic capacity,' said he; 'a man with so large a brain must have something in it.' " 1 Now, the statement that the owner of the hat is highly intellectual is no longer an unsupported assertion. Holmes has given the evidence, so his statement is supported. It is the conclusion of an argument. We shall regard assertions as unsupported unless evidence is actually given to support them, whether or not anyone has evidence for them. There is a straightforward reason for making the distinction in this way. Logic is concerned with arguments. An argument consists of more than just a statement; it consists of a conclusion along with supporting evidence. Until the evidence is given, we do not have an argument to examine. It does not matter who gives the evidence. If Watson had cited the size of the hat as evidence for Holmes's conclusion, we would have had an argument to examine. If we, as readers of the story, had been able to cite this evidence, again, there would have been an argument to examine. But by itself, the statement that the owner is highly intellectual is an unsupported assertion. We cannot evaluate an argument unless the evidence, which is an indispensable part of the argument, is given. To distinguish assertions for which no evidence is given from conclusions of arguments is not to condemn them. The purpose is only to make clear the circumstances in which logic is applicable and those in which it is not. If a
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Ask a homework question - tutors are online | 720 | 3,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-05 | latest | en | 0.964428 |
http://en.wikipedia.org/wiki/Color_burst | 1,397,631,594,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00091-ip-10-147-4-33.ec2.internal.warc.gz | 81,145,920 | 15,964 | # Colorburst
(Redirected from Color burst)
Horizontal sync and color burst of PAL videosignal
Colorburst is an analog video, composite video signal generated by a video-signal generator used to keep the chrominance subcarrier synchronized in a color television signal. By synchronizing an oscillator with the colorburst at the back porch (beginning) of each scan line, a television receiver is able to restore the suppressed carrier of the chrominance (color) signals, and in turn decode the color information. The most common use of colorburst is to genlock equipment together as a common reference with a vision mixer in a television studio using a multi-camera setup.
## Explanation
In NTSC, its frequency is exactly 315/88 = 3.57954[a] MHz with a phase of 180°, whereas PAL uses a frequency of exactly 4.43361875 MHz, with its phase alternating between 135° and 225° from line to line. SECAM is unique in not having a colorburst signal, since the chrominance signals are encoded using FM rather than QAM, thus the signal phase is immaterial and no reference point is needed.
Since the colorburst signal has a known amplitude, it is sometimes used as a reference level when compensating for amplitude variations in the overall signal.
## Rationale for NTSC Color burst frequency
The original black and white NTSC television standard specified a frame rate of 30 Hz and 525 lines per frame, or 15750 lines per second. The audio was encoded 4.5 MHz above the video signal. Because this was black and white, the video consisted only of luminance (brightness) information. Although all of the space in between was occupied, the line-based nature of the video information meant that the luminance data was not spread uniformly across the frequency domain; it was concentrated at multiples of the line rate. Plotting the video signal on a spectrogram gave a signature that looked like the teeth of a comb or a gear, rather than smooth and uniform.
RCA discovered that if the chrominance (color) information, which had a similar spectrum, was modulated on a carrier that was a half-integer multiple of the line rate, its signal peaks would fit neatly between the peaks of the luminance data and interference was minimized. It was not eliminated, but what remained was not readily apparent to human eyes. (Modern televisions attempt to reduce this interference further using a comb filter.)
To provide sufficient bandwidth for the chrominance signal, yet interfere only with the highest-frequency (and thus least perceptible) portions of the luminance signal, a chrominance subcarrier near 3.6 MHz was desirable. 227.5 = 455/2 times the line rate was close to the right number, and 455's small factors (5 × 7 × 13) make a divider easy to construct.
However, additional interference could come from the audio signal. To minimize interference there, it was similarly desirable for the distance between the chrominance carrier and the audio carrier to be a half-integer multiple of the line rate. The sum of these two half-integers implies that the distance between the luminance carrier and the audio carrier must be an integer multiple of the line rate. However, the original NTSC standard, with a 4.5 MHz carrier spacing and a 15750 Hz line rate, did not meet this requirement; the audio was at 285.714 times the line rate.
While existing black and white receivers could not decode a signal with a different audio carrier frequency, they could easily use the copious timing information included in the video signal to decode a slightly slower line rate. Thus, for color television, the line rate was reduced by a factor of 1.001 to 1/286 of the 4.5 MHz audio subcarrier frequency, or about 15734.2657 Hz. This reduced the frame rate to 30/1.001 ≈ 29.9700 Hz, and placed the color subcarrier at 227.5/286 = 455/572 = 35/44 of the 4.5 MHz audio subcarrier.[1]
## Crystals
An NTSC or PAL television's color decoder contains a colorburst crystal oscillator. These operate at some multiple of the colorburst frequency, such as 4 × f = 315/22 = 14.3181 MHz for NTSC.
Because so many analog color TVs were produced from the 1960s to the early 2000s, economies of scale have driven down the cost of colorburst crystals, which often find uses in various other applications, such as oscillators for microprocessors or for amateur radio (3.579 MHz is a commonQRP calling frequency in the 80-meter band).
Non-television uses of NTSC color burst frequency (generally only in the NTSC version of each device):
f = 315/88 = 3.57954[a] MHz
Component Frequency Relationship
Mattel Intellivision CPU 0.8949 MHz 1/4f
Apple II CPU (short cycles only, one in 65 cycles is longer) 1.0227 MHz 2/7f
Commodore VIC-20 CPU
Commodore 64 CPU
Atari 2600 CPU 1.1932 MHz 1/3f
Intel 8253 interval timer in IBM PC (remains in use today)
Fairchild Video Entertainment System CPU 1.7898 MHz 1/2f
Odyssey 2 CPU
Atari 8-bit family and Atari 7800 CPU
Commodore Plus/4 CPU
Nintendo Entertainment System CPU
TRS-80 Color Computer CPU 0.8949 MHz 1/4f
1.7898 MHz 1/2f
Commodore 128 CPU (FAST mode & CP/M mode) 2.0454 MHz 4/7f
Super Nintendo Entertainment System CPU 2.6847 MHz 3/4f
3.5795 MHz f
Sega Master System CPU 3.5795 MHz f
MSX CPU
Amateur radio Tx/Rx crystal for 80m band
ColecoVision CPU
CPU of IBM Personal Computer 5150 4.7727 MHz 4/3f
Commodore Amiga CPU 7.1591 MHz 2f
CPU of Tandy 1000 SX[1] (and many other IBM PC-XT clones).
NEC TurboGrafx-16 CPU | 1,383 | 5,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | longest | en | 0.930914 |
http://at.metamath.org/ilegif/frecfzennn.html | 1,721,711,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00523.warc.gz | 3,465,753 | 15,391 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > frecfzennn Unicode version
Theorem frecfzennn 9203
Description: The cardinality of a finite set of sequential integers. (See frec2uz0d 9185 for a description of the hypothesis.) (Contributed by Jim Kingdon, 18-May-2020.)
Hypothesis
Ref Expression
frecfzennn.1 frec
Assertion
Ref Expression
frecfzennn
Proof of Theorem frecfzennn
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 oveq2 5520 . . 3
2 fveq2 5178 . . 3
31, 2breq12d 3777 . 2
4 oveq2 5520 . . 3
5 fveq2 5178 . . 3
64, 5breq12d 3777 . 2
7 oveq2 5520 . . 3
8 fveq2 5178 . . 3
97, 8breq12d 3777 . 2
10 oveq2 5520 . . 3
11 fveq2 5178 . . 3
1210, 11breq12d 3777 . 2
13 0ex 3884 . . . 4
1413enref 6245 . . 3
15 fz10 8910 . . 3
16 0zd 8257 . . . . . . 7
17 frecfzennn.1 . . . . . . 7 frec
1816, 17frec2uzf1od 9192 . . . . . 6
1918trud 1252 . . . . 5
20 peano1 4317 . . . . 5
2119, 20pm3.2i 257 . . . 4
2216, 17frec2uz0d 9185 . . . . 5
2322trud 1252 . . . 4
24 f1ocnvfv 5419 . . . 4
2521, 23, 24mp2 16 . . 3
2614, 15, 253brtr4i 3792 . 2
27 simpr 103 . . . . 5
28 peano2nn0 8222 . . . . . . 7
29 zex 8254 . . . . . . . . . . . . . . 15
3029mptex 5387 . . . . . . . . . . . . . 14
31 vex 2560 . . . . . . . . . . . . . 14
3230, 31fvex 5195 . . . . . . . . . . . . 13
3332ax-gen 1338 . . . . . . . . . . . 12
34 0z 8256 . . . . . . . . . . . 12
35 frecfnom 5986 . . . . . . . . . . . 12 frec
3633, 34, 35mp2an 402 . . . . . . . . . . 11 frec
3717fneq1i 4993 . . . . . . . . . . 11 frec
3836, 37mpbir 134 . . . . . . . . . 10
39 omex 4316 . . . . . . . . . 10
40 fnex 5383 . . . . . . . . . 10
4138, 39, 40mp2an 402 . . . . . . . . 9
4241cnvex 4856 . . . . . . . 8
43 vex 2560 . . . . . . . 8
4442, 43fvex 5195 . . . . . . 7
45 en2sn 6290 . . . . . . 7
4628, 44, 45sylancl 392 . . . . . 6
4746adantr 261 . . . . 5
48 fzp1disj 8942 . . . . . 6
4948a1i 9 . . . . 5
50 f1ocnvdm 5421 . . . . . . . . . 10
5119, 50mpan 400 . . . . . . . . 9
52 nn0uz 8507 . . . . . . . . 9
5351, 52eleq2s 2132 . . . . . . . 8
54 nnord 4334 . . . . . . . 8
55 ordirr 4267 . . . . . . . 8
5653, 54, 553syl 17 . . . . . . 7
5756adantr 261 . . . . . 6
58 disjsn 3432 . . . . . 6
5957, 58sylibr 137 . . . . 5
60 unen 6293 . . . . 5
6127, 47, 49, 59, 60syl22anc 1136 . . . 4
62 1z 8271 . . . . . 6
63 1m1e0 7984 . . . . . . . . . 10
6463fveq2i 5181 . . . . . . . . 9
6552, 64eqtr4i 2063 . . . . . . . 8
6665eleq2i 2104 . . . . . . 7
6766biimpi 113 . . . . . 6
68 fzsuc2 8941 . . . . . 6
6962, 67, 68sylancr 393 . . . . 5
7069adantr 261 . . . 4
71 peano2 4318 . . . . . . . . 9
7253, 71syl 14 . . . . . . . 8
7372, 19jctil 295 . . . . . . 7
74 0zd 8257 . . . . . . . . . 10
75 id 19 . . . . . . . . . 10
7674, 17, 75frec2uzsucd 9187 . . . . . . . . 9
7753, 76syl 14 . . . . . . . 8
7852eleq2i 2104 . . . . . . . . . . 11
7978biimpi 113 . . . . . . . . . 10
80 f1ocnvfv2 5418 . . . . . . . . . 10
8119, 79, 80sylancr 393 . . . . . . . . 9
8281oveq1d 5527 . . . . . . . 8
8377, 82eqtrd 2072 . . . . . . 7
84 f1ocnvfv 5419 . . . . . . 7
8573, 83, 84sylc 56 . . . . . 6
8685adantr 261 . . . . 5
87 df-suc 4108 . . . . 5
8886, 87syl6eq 2088 . . . 4
8961, 70, 883brtr4d 3794 . . 3
9089ex 108 . 2
913, 6, 9, 12, 26, 90nn0ind 8352 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wa 97 wal 1241 wceq 1243 wtru 1244 wcel 1393 cvv 2557 cun 2915 cin 2916 c0 3224 csn 3375 class class class wbr 3764 cmpt 3818 word 4099 csuc 4102 com 4313 ccnv 4344 wfn 4897 wf1o 4901 cfv 4902 (class class class)co 5512 freccfrec 5977 cen 6219 cc0 6889 c1 6890 caddc 6892 cmin 7182 cn0 8181 cz 8245 cuz 8473 cfz 8874 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-13 1404 ax-14 1405 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 ax-coll 3872 ax-sep 3875 ax-nul 3883 ax-pow 3927 ax-pr 3944 ax-un 4170 ax-setind 4262 ax-iinf 4311 ax-cnex 6975 ax-resscn 6976 ax-1cn 6977 ax-1re 6978 ax-icn 6979 ax-addcl 6980 ax-addrcl 6981 ax-mulcl 6982 ax-addcom 6984 ax-addass 6986 ax-distr 6988 ax-i2m1 6989 ax-0id 6992 ax-rnegex 6993 ax-cnre 6995 ax-pre-ltirr 6996 ax-pre-ltwlin 6997 ax-pre-lttrn 6998 ax-pre-apti 6999 ax-pre-ltadd 7000 This theorem depends on definitions: df-bi 110 df-dc 743 df-3or 886 df-3an 887 df-tru 1246 df-fal 1249 df-nf 1350 df-sb 1646 df-eu 1903 df-mo 1904 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-ne 2206 df-nel 2207 df-ral 2311 df-rex 2312 df-reu 2313 df-rab 2315 df-v 2559 df-sbc 2765 df-csb 2853 df-dif 2920 df-un 2922 df-in 2924 df-ss 2931 df-nul 3225 df-pw 3361 df-sn 3381 df-pr 3382 df-op 3384 df-uni 3581 df-int 3616 df-iun 3659 df-br 3765 df-opab 3819 df-mpt 3820 df-tr 3855 df-eprel 4026 df-id 4030 df-po 4033 df-iso 4034 df-iord 4103 df-on 4105 df-suc 4108 df-iom 4314 df-xp 4351 df-rel 4352 df-cnv 4353 df-co 4354 df-dm 4355 df-rn 4356 df-res 4357 df-ima 4358 df-iota 4867 df-fun 4904 df-fn 4905 df-f 4906 df-f1 4907 df-fo 4908 df-f1o 4909 df-fv 4910 df-riota 5468 df-ov 5515 df-oprab 5516 df-mpt2 5517 df-1st 5767 df-2nd 5768 df-recs 5920 df-irdg 5957 df-frec 5978 df-1o 6001 df-2o 6002 df-oadd 6005 df-omul 6006 df-er 6106 df-ec 6108 df-qs 6112 df-en 6222 df-ni 6402 df-pli 6403 df-mi 6404 df-lti 6405 df-plpq 6442 df-mpq 6443 df-enq 6445 df-nqqs 6446 df-plqqs 6447 df-mqqs 6448 df-1nqqs 6449 df-rq 6450 df-ltnqqs 6451 df-enq0 6522 df-nq0 6523 df-0nq0 6524 df-plq0 6525 df-mq0 6526 df-inp 6564 df-i1p 6565 df-iplp 6566 df-iltp 6568 df-enr 6811 df-nr 6812 df-ltr 6815 df-0r 6816 df-1r 6817 df-0 6896 df-1 6897 df-r 6899 df-lt 6902 df-pnf 7062 df-mnf 7063 df-xr 7064 df-ltxr 7065 df-le 7066 df-sub 7184 df-neg 7185 df-inn 7915 df-n0 8182 df-z 8246 df-uz 8474 df-fz 8875 This theorem is referenced by: frecfzen2 9204
Copyright terms: Public domain W3C validator | 3,521 | 6,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.091206 |
https://math.answers.com/basic-math/32_is_what_percent_of_64 | 1,716,151,697,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00673.warc.gz | 340,713,307 | 49,159 | 0
# 32 is what percent of 64?
Updated: 4/28/2022
Shakieramath
Lvl 1
14y ago
32 out of 64 is arithmetically/fractionally written as 32/64.
When simplified by cancelling down, this is the equivalent of 1/2
Next: the term percentage actually means per [or "out of"] cent [100 = cent is a derivative word meaning 100]. So percent means 'per 100' ... or 'for every 100'. Essentially it means, 'divided by 100'.
This means that in giving a fraction or ratio (such as 1/2 - above) in percentage terms, we have effectively divided the answer by 100, simply in using the term 'percent'. Therefore, to keep the integrity of the fraction, we must multiply by 100, as what we do to the bottom line, we must also do to the top line - which must be multiplied by 100.
Arithmetically, the expression, then, is written as:
1/2 X 100/1 ... [because in stating that the answer is a percentage, we are "automatically" diving by 100.]
=> 100/2 = 50, effectively calculates the top line figure, even though the calculation itself is in both - top and bottom line form.
So, ultimately, in percentage terms, the answer is 50/100, or, as we say it,
50 percent.
Wiki User
14y ago
Wiki User
6y ago
32 is exactly 50%, or half of 64.
Take 64 and divide it by two, and you get the answer.
Or you can cross multiply:
(64/100)*(32/X)
X = (32*100)/64
X = 50% | 383 | 1,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.945737 |
https://socratic.org/questions/if-y-varies-directly-as-z-and-inversely-as-x-and-y-27-and-z-3-when-z-2-how-do-yo | 1,575,916,586,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540521378.25/warc/CC-MAIN-20191209173528-20191209201528-00132.warc.gz | 546,737,980 | 5,928 | # If y varies directly as z and inversely as x and y = 27 and z = -3 when z = 2, how do you find x when y = 9 and z = -5?
There is some mistake in the question. $z = - 3 , z = 2$ are the confusing . One of them should be $x$. | 77 | 226 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 3.375 | 3 | CC-MAIN-2019-51 | latest | en | 0.949371 |
https://www.sarthaks.com/1226579/calculate-cot-2-30-2cos-2-60-3-4-sec-2-45-4sin-2-30 | 1,675,267,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00327.warc.gz | 999,208,102 | 14,899 | # Calculate : cot^(2)30^(@)-2cos^(2)60^(@)-(3)/(4)sec^(2)45^(@)-4sin^(2)30^(@).
+1 vote
56 views
closed
Calculate : cot^(2)30^(@)-2cos^(2)60^(@)-(3)/(4)sec^(2)45^(@)-4sin^(2)30^(@).
+1 vote
by (72.0k points)
selected
cot^(2)30^(@)-2 cos^(2)60^(@)-(3)/(4) sec^(2)45^(@)-4 sin^(2)30^(@)
=(sqrt(3))^(2)-2xx((1)/(2))^(2)-(3)/(4) xx (sqrt(2))^(2)-4xx((1)/(2))^(2)
=3-2xx(1)/(4)-(3)/(4)xx2-4xx(1)/(4)
=3-(1)/(2)-(3)/(2)-1=(6-1-3-2)/(2)=(6-6)/(2)=(0)/(2)=0. | 243 | 454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-06 | latest | en | 0.446267 |
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# Vedic Math Tricks
May 11, 2022 | Rockville
Vedic math is a system of mathematics that was designed and published by Indian Hindu monk and mathematician, Jagadguru Shri Bharathi Krishna Tirthaji between A.D. 1911 and 1918.
Vedic math, more commonly known as mental mathematics, is a collection of Methods, (or Sutras) to solve numerical computations quickly and faster. It consists of 16 Sutras (Formulae) and 13 sub-sutras (Sub Formulae). Sutras are short formulas used to carry out difficult mathematical calculations mentally in an easy and simple manner. You can solve many difficult and time-consuming math quickly, using these Vedic math tricks.
Let's learn a few tricks!
1. Squaring Of A Number Whose Unit Digit Is 5:
You can quickly find the square of a two-digit number ending with 5 using this Vedic Math trick. Take all the numbers but the last one (call that last number N), then multiply N(N+1). Finally, tack 25 onto the end!
For example Find (55) ² =?
Step 1. 55 x 55 = . . 25 (end terms)
Step 2. 5 x (5 + 1) = 30
Step 3. Place 25 on the end of the end of the product.
So our answer will be 3025.
Test it out! Try to find the squares 85 & 95.
2. Multiplying a Large Number By 5:
Memorizing your times tables is a great way to reinforce your mental math skills, but what happens when a number is too big to easily multiply by 5? Take any number, and depending on its even or odd nature, divide the number by 2 (get half of the number), then add a 0 or a 5 (depending on if it is even or odd) to the end of your answer.
Even Number:
2464 x 5 =?
Step 1. 2464 / 2 = 1232
Step 2. Place a 0 onto the end of the quotient
So, 2464 x 5 = 12320
Odd Number:
3775 x 5
Step 1. Subtract one
( 3775 - 1)
Step 2. Divide the difference by 2
3774 / 2 = 1887
Step 2. Place a 5 on the end of the quotient
3775 x 5 = 18875
Now try —- 1234 x 5, 123 x 5
3. Subtraction From 1000, 10000, 100000:
Sometimes, children find it difficult to subtract numbers from 1000, 10000, or 100000, because of extensie borrowing. Here is an easy method using Vedic that streamlines the borrowing process.
For example:
1000 – 573 = ?
Subtract the first two digits in from 9 and then subtract the last digit from 10.
Step 1. 9 – 5 = 4
Step 2. 9 – 7 = 2
Step 3. 10 – 3 = 7
So, the answer is: (1000 – 573) = 427
The process is the same for 10,000 and 100000 regardless of the amount of zeros.
Try to solve these sums using the mentioned Vedic math tricks.
1000 – 757, 10,000 – 1029, 10,000 – 1264, 1000 – 337.
4. Dividing A Large Number By 5:
This trick will get you the result quickly by dividing a large digit number by 5. All you need to follow only two steps, in the first step multiply the number by 2 while in the second step move the decimal point.
So let's learn the steps
1st step. Multiply the number by 2
2nd step: Move the decimal point to the left.
For example: 245 / 5 = ?
Step 1. 245 * 2 = 490
Step 2. Move the decimal: 49.0 or just 49
Let’s try another: 2129 / 5
Step 1: 2129 * 2 = 4258
Step2: Move the decimal: 425.8 or just 425
Now you try to solve 16951/5, 2112/5, 4731/5
5. Multiply Any Two-digit Number By 11
This Vedic trick can help you tackle tricky numbers like multiples of 11!.
Just add the first and last digits together. Then put the sum in between the first and last digit!
For example:
Solve 32 x 11
32 x 11 = 3 “(3+2)” 2 = 352
So, the answer is: 32 x 11 =352
Another Example:
52 x 11 = 5 “(5+2)” 2 = 572
Now try 35 x 11, 19 x 11, 18 x 11.
Vedic math showcases a different perspective on how to do the math. Tricks like these can do wonders if properly executed. At first, you might find them a bit complex, but if you practice, them, they can be a wonderful help. Leave a comment and let us know how testing these out went! | 1,177 | 3,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-33 | latest | en | 0.860797 |
https://pwntestprep.com/2016/04/it-is-given-that-sin-a-k-where-a-is-an-angle-measured-in-radians-and-pi-a-3pi2/ | 1,719,236,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00791.warc.gz | 411,950,207 | 17,808 | It is given that sin A = K , where A is an angle measured in radians and (pi)< A < 3(pi)/2 . If sin B = K which of the following could be the value of B ?
A) A – (pi)
B) (pi) + A
C) 2(pi) – A
D) 3(pi) – A
If A is between and , then it’s in the 3rd quadrant, which means its sine is negative. So you know is that K is negative.
Knowing that sines are negative in the 3rd and 4th quadrants, which of the answer choices will land you in one of those quadrants, knowing that A is in the 3rd quadrant?
A – π will put you in the 1st quadrant.
π + A will also put you in the 1st quadrant.
2π – A will put you in the 2nd quadrant.
3π – A will put you in the 4th quadrant. That works—the sine is negative there!
If you’re having trouble with these calculations, you might find it useful to plug in a value that works for A, like A = 1.25π. then take the sine of each choice with your calculator to see which one ends up negative. | 278 | 927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.960259 |
http://mathematica.stackexchange.com/questions/19965/multiply-a-list-of-matrices-by-a-list-of-vectors | 1,455,094,221,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701159031.19/warc/CC-MAIN-20160205193919-00294-ip-10-236-182-209.ec2.internal.warc.gz | 152,758,488 | 17,090 | Multiply a list of matrices by a list of vectors
How does one multiply a list of matrices by a list of vectors, elementwise? For example, multiplying
A = {IdentityMatrix[2], 2*IdentityMatrix[2]}
x = {{1, 1}, {2, -2}}
should return
{{1, 1}, {4, -4}}
Neither Dot nor Times accomplishes this; both have the wrong dimensions. A cumbersome way would be
result = {{0, 0}, {0, 0}}
Do[result[[i]] = A[[i]].x[[i]], {i, 2}]
but surely there is a cleaner way.
- | 143 | 458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-07 | longest | en | 0.803006 |
https://converter.ninja/volume/us-customary-cups-to-centiliters/961-uscup-to-cl/ | 1,623,607,603,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610196.46/warc/CC-MAIN-20210613161945-20210613191945-00614.warc.gz | 181,603,546 | 5,327 | # 961 US customary cups in centiliters
## Conversion
961 US customary cups is equivalent to 22736.12952765 centiliters.[1]
## Conversion formula How to convert 961 US customary cups to centiliters?
We know (by definition) that: $1\mathrm{uscup}\approx 23.65882365\mathrm{centiliter}$
We can set up a proportion to solve for the number of centiliters.
$1 uscup 961 uscup ≈ 23.65882365 centiliter x centiliter$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{centiliter}\approx \frac{961\mathrm{uscup}}{1\mathrm{uscup}}*23.65882365\mathrm{centiliter}\to x\mathrm{centiliter}\approx 22736.12952765\mathrm{centiliter}$
Conclusion: $961 uscup ≈ 22736.12952765 centiliter$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 centiliter is equal to 4.39828599139477e-05 times 961 US customary cups.
It can also be expressed as: 961 US customary cups is equal to $\frac{1}{\mathrm{4.39828599139477e-05}}$ centiliters.
## Approximation
An approximate numerical result would be: nine hundred and sixty-one US customary cups is about twenty-two thousand, seven hundred and thirty-six point one three centiliters, or alternatively, a centiliter is about zero times nine hundred and sixty-one US customary cups.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 407 | 1,464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-25 | latest | en | 0.713665 |
https://www.jiskha.com/display.cgi?id=1327890631 | 1,503,209,793,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00083.warc.gz | 913,414,579 | 4,255 | # Chemistry
posted by .
1) Determine the value of delta H net for the following equation:
SnBr2(s) + TiCl4(l) -> TiBr2(s) + SnCl4(l)
I am not sure how to solve this.
• Chemistry -
delta Hf = dHf
dHrxn = (n*dHf products) - (n*dHf reactants)
• Chemistry -
for TiCl4 i got -804.2 and for SnCl4 I got -511.3 and then i subtracted them to get -292.9 but I don't think this is right.
• Chemistry -
You didn't use ALL of the products nor ALL of the reactants.
• Chemistry -
ok
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https://scitechinstitute.org/listing/real-world-math-project-based-learning-spreadsheet-app/ | 1,652,688,462,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00743.warc.gz | 590,389,588 | 30,764 | ### Submit a Resource
Real World Math: Project Based Learning - Spreadsheet App
This lesson involves students formatting a spreadsheet so that it is a workable math app. Students are more than aware of what apps look like and what they can do. By applying math formulas in spreadsheets, students can get an introduction to design and coding, and get a chance to apply their math knowledge to something someone else might actually use.
This is an easy project I thought of when my class had some time to fill. Of course, it helps if they are already familiar with working with spreadsheets, but if not – it doesn’t take too much time to get up to speed. The assignment was to create a spreadsheet page that would calculate temperatures in degrees Fahrenheit to Celsius, and vice versa. They needed to make sure that their formula would follow the order of operations correctly. In this case, this was accomplished simply enough by designating a cell for the user to enter a temperature, and then targeting it in another cell with a formula, such as “=(C6-32)*5/9 “. Students get really engaged when they start exploring the design elements of the project. What colors should they use? What font size is appropriate? Are the dimensions appropriate for a cell phone’s or tablet’s screen? What should it look like if the device is rotated? What would make a good title? Are there functional buttons that should be added? The design aspect of the activity is the main reason why I’ve included it with Project Based Lessons. From the commercial side of app markets, to refining their product, there are many ways you can approach this and extend it further.
Most math formulas or functions should be appropriate for the task. The area of geometric shapes would rely on values entered in several cells and an approximate value for pi. Use the activity to investigate how math formulae can be devised; for instance, the relationship between the area formulas of a parallelogram, triangle, and trapezoid. Use proportions to develop the conversion formulas for temperature. On a deeper level, have a class discussion on the usefulness of formula apps and whether it’s important for someone to remember a formula or not.
Spreadsheet applications are very versatile tools that can be used in many ways. If you are unfamiliar with them, the best way to learn is to create one and explore what you can do. Each time you’ll learn something new; if there’s something particular you want to accomplish, you can usually find the answer online or from other users.
A spreadsheet page is filled with boxes or “cells” that are arranged by columns in letters and rows by number. A cell is identified by the row and column it lies in; for instance, “B10”. The size of rows and columns can be adjusted by grabbing the border between them and pulling in the desired direction. You’ll find many familiar word processing functionality such as color, font size, and so on. If you want to perform a function, then you’ll type an “=” sign in a cell followed by the function. For example, type “6” in B10, “8” in B11, and “=B10+B11” in B12. Hit enter and the result “14” will appear in cell B12. If you change any of the values in B10 or B11, you’ll find the sum is automatically adjusted. There are some differences in the math symbols you’ll use. The asterisk * is used for multiplication, carrot ^ for exponents, and back slash / for division.
When students are finalizing their “apps”, have them go to the View menu and select to hide gridlines for a cleaner appearance.
Objectives
• Apply mathematical formulas in spreadsheets
• Create a workable calculation app using the order of operations
• Apply design and functional elements
SheetsApp example
Technology Topics
Online & Software Tools
Math Topics
Algebra & Pre-Algebra, Geometry
K-6, Middle School, High School, Educator
Real World Math
Lesson Plan | 892 | 3,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.921301 |
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Tutorial TUTORIAL TO DIFFERENTIATE THE DAYS OF NIGHT
Notices
Tools, Tutorials & Resources Various tools to help you develop your hacks can be found here.New threads in this forum are to be approved by a moderator before they are displayed.
#1
August 16th, 2011 (10:29 AM). Edited August 17th, 2011 by Wesley FG.
Wesley FG Join Date: Oct 2008 Location: Brazil Gender: Nature: Adamant Posts: 235
TUTORIAL TO DIFFERENTIATE THE DAYS OF NIGHT
==================================================
Things Necessary:
==================================================
Rom Pokémon (Fr/Lg/R/S/E)
DNS
Scripter Editor ( I use XSE)
==================================================
Tutorial
==================================================
This tutorial is very simple, it will read just give a script ready to modify as you please, and do unimaginable things with your good imagination and celebrate.
The first thing to do is enter the routine of DNS in your rom and observe the following information:
In my case I will create the windows and street lights at night, note the byte 0x0300553C, this is where the byte RTC was installed in ROM and the information is in the following order as shown: year (2bytes), 00 month (1byte), day (1 byte), 00 hour (1byte), minute (1byte), second (1byte). With this information we can do unimaginable things, such as scripts that occur only on certain days of the week, among other things.
Now open your emulator to verify this information, and go to the Memory Viewr and opened this tab:
With this we can see that the information in the order described in 300553C start, but for use in a script and know whether it is day or night you must use the information of time, which is in the blue square, and will be 6 bytes in front of 300553C, using the scientific calculator (scientific mode) windows and adding 300553C +6 = 3005542, ie the direction that we will use to compare the hours will be 3005542!
Ready now define the hours that will be in my case at night from 19:00 PM the night (hex 13) to 4:00 am (4 hexadecimal) now that we have all this information we will make the script!
And looks like this: (XSE)
Quote:
#dynamic 0x900000 copybyte 0x202E8DC 0x3005542 compare LASTRESULT 0x13 if 0x4 goto @night compare LASTRESULT 0x4 if 0x3 goto @night goto @day end #org @night Setmaptile …… End #org @day Nop End
Now we analyze, with the direction used copybyte found 0x3005542, which is where the information of the hour, and then made whether the hours are in the range between 19 and 4 hours, so the script that will occur and the @ night if the time is not in this range, it drop to @ day.
Now use imagination to create a tile with a lamp pole with lights, or lights in the window, make a script to change the tiles setmaptile overnight and ready.
Also you can compare and know which day of week, etc..
In my case I used a rom Ruby, FireRed also in use in the same way will only offset the seedlings in the RTC memory viewer just look in DNS, the same way I did mine in Ruby.
I use the translator, bye.
=============================================================================
CREDITS
==============================================================================
Dark Rayquaza, Derlo, Dante, ZodiacDagreath (for help-me with dudes)
PrimerDialkga (DNS Creator)
Kalos Demaker. Soon!!
My Devianart:http://wesleyfg.deviantart.com/
#2
August 19th, 2011 (3:52 AM).
Jambo51 Glory To Arstotzka Join Date: Jun 2009 Gender: Male Nature: Quiet Posts: 731
This is a good tutorial, and is quite easily read considering that English isn't your natural language, and you translated it.
However, the offset you showed for LASTRESULT is only applicable for either Ruby or for your language of Ruby (one or the other, I forget exactly which).
For BPRE, it's at 0x020370D0, for example.
It's better to use comparefarbytetobyte 0x3005542 [Time] as this doesn't require a copy to a RAM address which will be different between versions.
Hope this helps.
Hey guys, please check out my recreations of the gen 1 and 2 music on my custom engine at my SoundCloud! - Here!
#3
April 22nd, 2012 (6:12 AM).
Smoley' Join Date: Apr 2012 Gender: Male Posts: 13
Thanks You ! It'll be veryuseful for my hack,
#4
April 22nd, 2012 (6:45 PM).
DarkFlameSquirrel Music Freak Join Date: Apr 2012 Location: Texas, USA Gender: Male Nature: Gentle Posts: 7
When I load Pokemon Emerald with DNS, I get an option to insert RTC. Doesn't Emerald Version already have Real Time Clock or some kind of time feature?
And I seriously hope this is a DNS tutorial and not a wrong thread D:
"I will not live in misery." -Bleeding Through | 1,152 | 4,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-48 | latest | en | 0.854077 |
https://www.onlinemathlearning.com/igcse-2020-0580-11-may.html | 1,718,930,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00296.warc.gz | 809,395,797 | 10,366 | # IGCSE Maths 2020 0580/11 May/June
Cambridge CIE IGCSE Past Papers and solutions.
Questions and Worked Solutions for IGCSE 2020 0580/11 May/June Paper 1.
Related Pages
More IGCSE Past Papers
IGCSE 2020 0580/11 May/June (pdf)
1. Write down the value of the 7 in the number 570296.
2. The table shows the temperature, in °C, at midday on the first day of each month during one year in a city.
Calculate the mean of these temperatures.
3. Write these numbers in order, starting with the smallest.
4. (a) On each shape draw all the lines of symmetry.
(b) Write down the order of rotational symmetry of this shape.
5. In the triangle ABC, AB = AC and angle BAC = 38°.
BCD is a straight line.
Work out angle ACD.
6. (a) Diego flies from Madrid to Buenos Aires.
His flight leaves at 20 55 and arrives at 03 50 local time.
The local time in Buenos Aires is 5 hours behind the local time in Madrid.
Work out, in hours and minutes, the time the flight takes.
(b) Diego changes 200 euros into Argentine Peso.
The exchange rate is 1 euro = 24.8 pesos.
Work out how many pesos he receives.
(c) The distance between Madrid and Buenos Aires is 10050km.
Diego’s return flight takes 12 hours 30 minutes.
Calculate the average speed, in km/h, for the return flight.
1. Rectangle A measures 3 cm by 8 cm.
Five rectangles congruent to A are joined to make a shape.
Work out the perimeter of this shape.
2. Find the highest odd number that is a factor of 60 and a factor of 90.
3. (a) Write PQ as a column vector.
(b) Write 3PQ as a single vector.
4. Work out the size of one interior angle of a regular 9-sided polygon.
5. A cone has radius 4.5 cm and height 10.4cm.
Calculate, in terms of r, the volume of the cone.
(The volume, V, of a cone with radius r and height h is V = 1/3 πr2h)
6. (a) The nth term of a sequence is 60-8n.
Find the largest number in this sequence.
(b) Here are the first five terms of a different sequence.
12 19 26 33 40
Find an expression for the nth term of this sequence.
7. Factorise completely.
21a2 + 28ab
8. The diagram shows a trapezium.
Work out the value of x.
9. Simplify.
4p5q3 × p2q-4
10. (a) Write the number 0.0605 in standard form.
(b) Calculate (1.63 × 1012) × (2.47 × 10-1).
11. Expand and simplify.
(x - 5)(x - 7)
12. Mrs Salaman gives her class two mathematics tests.
The scatter diagram shows information about the marks each student scored.
(a) Write down the highest mark scored on test 1.
(b) Write down the type of correlation shown in the scatter diagram.
(c) Draw a line of best fit on the scatter diagram.
(d) Hamish scored a mark of 40 on test 1.
He was absent for test 2.
Use your line of best fit to find an estimate for his mark on test 2.
13. The length, l cm, of a sheet of paper is 29.7 cm, correct to the nearest millimetre.
Complete this statement about the value of l.
14. Without using a calculator, work out
You must show all your working and give your answer as a fraction in its simplest form.
15. Lucia invests \$5000 at a rate of 4.5% per year compound interest.
Calculate the value of her investment at the end of 7 years.
16. (a) Find the equation of line L in the form y = mx + c.
(b) On the grid, draw a line that is perpendicular to line L.
17. Explain why triangle ABC is similar to triangle PQR.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 958 | 3,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.884914 |
http://en.wikipedia.org/wiki/QDGC | 1,409,149,757,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829393.78/warc/CC-MAIN-20140820021349-00374-ip-10-180-136-8.ec2.internal.warc.gz | 69,154,077 | 11,807 | # QDGC
Jump to: navigation, search
QDGC - Quarter Degree Grid Cells (or QDS - Quarter degree Squares) are a way of dividing the longitude latitude degree square cells into smaller squares, forming in effect a system of geocodes. Historically QDGC has been used in a lot of African atlases. Several African biodiversity projects uses QDGC, among which The atlas of Southern African Birds[1] is the most prominent one. In 2009 a paper by Larsen et al. [2] describes the QDGC standard in detail.
## Mechanics
QDGC represents a way of making (almost) equal area squares covering a specific area to represent specific qualities of the area covered. The squares themselves are based on the degree squares covering earth.
Around the equator we have 360 longitudinal lines, and from the north to the south pole we have 180 latitudinal lines. Together this gives us 64800 segments or tiles covering earth. The form of the squares becomes more rectangular the longer north we come. At the poles they are not square or even rectangular at all, but end up in elongated triangles.
Each degree square is designated by a full reference to the main degree square. S01E010 is a reference to a square in Tanzania. S means the square is south of equator, and E means it is East of the zero meridian. The numbers refer to longitudinal and latitudinal degree.
A square with no sublevel reference is also called QDGC level 0. This is square based on a full degree longitude by a full degree latitude. The QDGC level 0 squares are themselves divided into four.
A B C D
To get smaller squares the above squares are again divided in four - giving us a total of 16 squares within a degree square. The names for the new level of squares are named the same way. The full reference of a square could then be:
• S01E010AD
The number of squares for each QDGC level can be calculated with this formula:
number of squares = (2d)2
(where d is QDGC level)
Table showing level, number of squares and an example reference:
Level Squares Example 0 1 S01E010 1 4 S01E010A 2 16 S01E010AD 3 64 S01E010ADC 4 256 S01E010ADCB 5 1024 S01E010ADCBD 6 4096 S01E010ADCDBA
To decide which name a specific longitude latitude value belongs to it is possible to use this code:
Download shapefiles datasets for Africa, South America, India and China here:
An online service to upload and convert .dbf-files files is described on this webpage:
## References
1. ^ HARRISON, J.A., ALLAN, D.G., UNDERHILL, L.G., HERREMANS, M., TREE, A.J., PARKER, V. & BROWN, C.J. (1997) The atlas of southern African birds. Vols 1 and 2, BirdLife South Africa, Johannesburg, South Africa.
2. ^ R. Larsen, T. Holmern, S. D. Prager, H. Maliti and E. Røskaft (2009) Using the extended quarter degree grid cell system to unify mapping and sharing of biodiversity data, African Journal of Ecology. Volume 47, Issue 3 , Pages382 - 392 | 730 | 2,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.908235 |
https://www.mindedge.com/page/basic-math-statistics-for-healthcare | 1,560,753,132,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998440.47/warc/CC-MAIN-20190617063049-20190617085049-00319.warc.gz | 813,026,482 | 7,684 | # MindEdge's Basic Math and Statistics for Healthcare
MindEdge's Basic Math and Statistics for Healthcare learning resource provides a comprehensive course in the mathematical and statistical knowledge needed for the healthcare professions. With real-world examples from the field of healthcare, students practice skills ranging from solving basic arithmetic equations to interpreting the kind of complex statistical data that they will encounter in published healthcare research. The learning resource draws on video, graphics, slideshows, and case studies to convey mathematical concepts and techniques, while students achieve mastery by engaging with exercises, puzzles, games, review tests, and problem sets. The adaptive learning included throughout the learning resource allows students to pinpoint weaknesses in their skills to more efficiently master the material.
The learning resource is designed to address the needs of students with a range of mathematical competencies. The early modules lay a solid foundation in arithmetic and algebraic expressions and equations, including those involving integers, fractions, and decimals. Later modules teach the fundamentals of statistics and data interpretation. Learners study quantitative decision-making tools and techniques, which tie into real-world, healthcare case studies.
This modular course can be tailored to your school with webtexts, ebooks, and optional trade paperbacks available. It seamlessly integrates into all learning management systems.
MindEdge's Basic Math and Statistics for Healthcare content builds skills in:
• Simplifying expressions involving positive and negative integers, fractions, and decimals
• Solving algebraic equations by performing addition, subtraction, multiplication, and division
• Understanding and working with percentages
• Graphing points and lines on coordinate axes
• Working with inequalities
• Calculating the mean, median, mode, quartiles, range and interquartile range on a set of quantitative data
• Evaluating data and identifying the best form for displaying statistical data for one variable from several different kinds of graphs: bar chart, pie chart, histogram, stem plot and box plot.
• Identifying the appropriate numerical measures of and graphical display for a given classification in different contexts of statistics in two variables
• Working with conditional percentages in a two-way table
• Understanding regression analysis
• Understanding, interpreting, and estimating correlation between two variables
• Calculating probabilities and conditional probabilities
• Understanding and recognizing dependent, independent, and disjoint events
• Calculating probabilistic unions and intersections
• Constructing frequency tables and probability trees
The seven modules of this learning resource are as follows:
1. Basic Numeracy and Calculation Skills
2. Fractions, Decimals, and Percentages
3. Basic Algebra
4. Descriptive Statistics for a Single Variable
5. Descriptive Statistics for Two Variables
6. Correlation and Regression
7. Probability | 536 | 3,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.902397 |
https://www.jaapsch.net/puzzles/eightbll.htm | 1,722,814,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00133.warc.gz | 656,972,726 | 2,961 | # Diamond 8-Ball Puzzle
This puzzle consists of 8 numbered balls which can roll around a 3x3 square. There is a restriction that differentiates it from a simple 3x3 version of the fifteen puzzle. There are walls on the left and right side of the centre position, so that a ball can enter or leave the centre in a vertical direction only. In the solved state the centre spot is empty, and balls are in clockwise numerical order starting from 1 at the top left corner.
This puzzle was invented by Joshua Frankel, and manufactured by Binary Arts, now called ThinkFun.
If your browser supports JavaScript, then you can play Diamond Eightball by clicking the link below:
## The number of positions:
If we count only those positions with the gap in the centre, then there are 8 pieces which can be any order, giving a maximum of 8! = 40,320 positions. As with the fifteen puzzle, only even permutations are achievable, so there are actually 8!/2 = 20,160 positions with the centre empty, or 9!/2 = 181,440 positions with the empty spot anywhere.
Note however that there are really two solutions - you could solve it with the 1 ball at the bottom right, and then turn the puzzle upside down.
I performed a computer search for this puzzle. The following table shows how many positions there are (with the centre empty) for each number of moves from the solved position. A move consists of moving the space around the left or right side in either direction, shifting five pieces in the process. The first column shows the number of positions at each distance from only one of the two solutions, the second if either solution is allowed.
Moves One solution Two solutions
012
148
21224
33264
482156
5204360
6496792
71,1531640
82,4313202
94,3255132
105,8295556
114,4082916
121,133306
13502
Total 20,160 20,160
Avg Depth9.51969.0299
## Notation:
There are 2 loops in the puzzle; the left half and the right half. Denote these by L and R. A clockwise shift of all the five balls in a loop is indicated by the relevant letter (L or R), and an anti-clockwise shift by the letter followed by an apostrophe (i.e. by L' or R').
## Solution:
Phase 1: Solve balls 3,4,5, and 6.
1. Do any moves to bring ball 5 to the top right corner, i.e. the position where 3 will be when solved.
2. If ball 4 lies on the right (directly below the 5, or in the bottom right corner), then do RR LL R'R' so that 4 lies in the left loop and 5 is again top right.
3. Shift the left loop to bring ball 4 to the top position, and then do R. You now should have ball 4 at the top right, and 5 directly below it.
4. If ball 3 lies at the bottom right corner, then do R L R'.
5. Shift the left loop to bring ball 3 to the top position, and then do R. You now should have balls 3, 4 and 5 in their correct positions.
6. Shift the left loop to bring ball 6 to the bottom position, where it belongs.
Phase 2: Solve the rest.
1. Find ball 7. Depending on its position, do one of the following sequences:
top-left: L'RL'R' LLRL'L'R'
top: RLL R'LR LLR'
left: R'LRLL R'L'L'RL
2. Find ball 8. Depending on its position, do one of the following sequences:
top-left: LR'L'RL'R'LLRL
top: L'R'L'RL'R'LLRL'
## Nice sequences:
LRLRLRL: Constructs the second solution, with the 1 at the bottom right corner. | 866 | 3,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.912889 |
https://www.electro-tech-online.com/threads/either-a-good-idea-or-i-dont-know-what-im-talking-about.5613/ | 1,701,769,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00298.warc.gz | 838,555,045 | 25,084 | Continue to Site
# Either a good idea, or I don't know what I'm talking about.
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Not open for further replies.
#### Electric Rain
##### New Member
Like you can tell from my subject, I don't know if this is either a good idea, or if I don't know what I'm talking about. :lol: It's an idea for "endless" power. Say I have two motors, both linked together by their terminals and with a belt on their shafts. Could I jump start one of them, which would turn the other one, making it generate power through it's terminals, going to the other motor to make it keep spinning after the jump start and therefore just keeping them both spinning with no power, then take power from them to power something else? I can explain further if anyone doesn't get it, I'll even make a diagram. But if you do get it, could someone tell me what I didn't think through? It seems like it would work, but SOMEthing must be wrong. :cry: And um... go easy on me if that was just a complete waste of space in these forums please... So would it work even if it were a very small amount of power I could draw? Thanks a lot.
Rain
Re: Either a good idea, or I don't know what I'm talking abo
Electric Rain said:
Like you can tell from my subject, I don't know if this is either a good idea, or if I don't know what I'm talking about. :lol: It's an idea for "endless" power. Say I have two motors, both linked together by their terminals and with a belt on their shafts. Could I jump start one of them, which would turn the other one, making it generate power through it's terminals, going to the other motor to make it keep spinning after the jump start and therefore just keeping them both spinning with no power, then take power from them to power something else? I can explain further if anyone doesn't get it, I'll even make a diagram. But if you do get it, could someone tell me what I didn't think through? It seems like it would work, but SOMEthing must be wrong. :cry: And um... go easy on me if that was just a complete waste of space in these forums please... So would it work even if it were a very small amount of power I could draw? Thanks a lot.
Rain
Sorry, but it's a stupid idea :lol:
It's called 'perpetual motion', and is impossible.
In your particular case (which has been suggested ever since electric motors were invented), there are a number of flaws - basically due to efficiency and losses. A motor puts out less mechanical power than it takes in electrical power, and a generator puts out less electrical power than the mechanical power it uses.
For your scheme to work, it would require 100% efficiency from both motor and generator - which is impossible!. In order to actually take power from the scheme it would require efficiency higher than 100%.
There have been many 'pertetual motion' machines designed over the years, all have the same basic need for greater than 100% efficiency.
your idea is simillar to the perpetual motion machine. start the thing and it will work forever.
remember one thing that the total energy of a system remains constant. so u cant generate energy from nowhere. and power is the rate of expending energy. i hope u r understanding what i am saying (u would if u have studied physics).
lets assume it works out for a short time. the motors are running. then u operate something from the motors. that something is the load of the motors. it will take energy from the motors and eventually the motors will be exhausted of all the energy they had. to continuously operate that load the motors need a continuous supply of energy which it doesnt have.
and think of all the mechanical losses. friction :?: :?:
i hope i went easy on u
Alright then. I'm just starting out, so I don't know any of this. But thanks for telling me all of that guys, it's all good stuff to know. :| Anyway, if you will since I now know it wouldn't work, forget I asked. :lol: Thanks.
Rain
Perpetual motion machines are theoretically impossible and would violate the second law of thermodynamics.But an interesting topic none the less, check out maxwells demon.
The question is good in that the discussion helps to convey an understanding of "what goes in must come out" and the numerous variations of that statement. Now and then we'll see where someone wants to do something - and hasn't thought that part thru. A popular one seems to be the budding audiophile who wants to design a 4 kw amplifier to run from his car cigarette lighter. The first thought on an experienced person's mind is "where's he goning to get the power..." No need to apologize. We all started at the beginning too.
There is actually one design for a perpetual motion machine that works, and provides constant output energy for a fixed setup cost. It's rather complex, and I'm not sure of the symbols for the parts, so let me explain it and hopefully I will get across what I mean. Take some Triticum aestivum that has been heat set into a rectangular section, approximately 1/2" high, 4" long, and 3" wide. Coat one side with pasturized bovine colloidal gel. Strap this onto the dorsal side of a Felis domesticus, and drop from approximately waist height.
The device will begin rotating as counter acting laws fight for dominance. It will make a lot of noise, and be pretty messy, but should rotate just above the ground indefinitely.
Enjoy!
[groan] I can't believe somebody actually posted the old buttered-toast/housecat line here...
ChrisP said:
[groan] I can't believe somebody actually posted the old buttered-toast/housecat line here...
I don't see the words butter, toast, or house cat anywhere in my post! :wink:
Gandledorf said:
ChrisP said:
[groan] I can't believe somebody actually posted the old buttered-toast/housecat line here...
I don't see the words butter, toast, or house cat anywhere in my post! :wink:
True, but you've probably launched several of our fellow trolls in pursuit of Branta canadensis.
Gandledorf said:
The device will begin rotating as counter acting laws fight for dominance. It will make a lot of noise, and be pretty messy, but should rotate just above the ground indefinitely.
May be so, but it's not pertetual motion, it takes it's input power from gravity and small furry rodents
Ron H said:
Gandledorf said:
ChrisP said:
[groan] I can't believe somebody actually posted the old buttered-toast/housecat line here...
I don't see the words butter, toast, or house cat anywhere in my post! :wink:
True, but you've probably launched several of our fellow trolls in pursuit of Branta canadensis.
And here I was hoping for a good hunt for some Gallinago gallinago.
If any of you are gullible and want to waste several years of your life, take a look at **broken link removed**.
Nigel Goodwin said:
Gandledorf said:
The device will begin rotating as counter acting laws fight for dominance. It will make a lot of noise, and be pretty messy, but should rotate just above the ground indefinitely.
May be so, but it's not pertetual motion, it takes it's input power from gravity and small furry rodents
I guess the true question is whether the device rotates fast enough to pay off the energy cost of the rats, and whether said device will continue to operate when the mammalian component dies.
Has the cat morphed into a rat, or did I miss something?
Ron H said:
Has the cat morphed into a rat, or did miss something?
No Nigel just mentioned that technically the device required the energy supplied by gravity and small furry rodents.
BTW, we are all dorks. This is concrete proof.
Ron H said:
Has the cat morphed into a rat, or did I miss something?
Nigel Goodwin said:
Ron H said:
Has the cat morphed into a rat, or did I miss something? | 1,752 | 7,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.983457 |
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``````# solves *bounded* LPs of the form:
# max cx
# sub to: Ax <= b
from sympy import *
from itertools import combinations
# enumerates all the vertices of {x | Ax <= b}
def enumeratevertices(A, b):
m, n = A.rows, A.cols
for rowlist in combinations(range(m), n):
Ap = A.extract(rowlist, range(n))
bp = b.extract(rowlist, [0])
if Ap.det() != 0:
xp = Ap.LUsolve(bp)
d = A * xp - b
feasible = True
for i in range(m):
if d[i] > 0:
feasible = False
if feasible:
yield xp
# finds the optimum using vertex enumeration
def findoptimum(A, b, c):
m, n = A.rows, A.cols
bestvalue, bestvertex = None, None
for vertex in enumeratevertices(A, b):
if bestvalue is None or (vertex.T*c)[0] > bestvalue:
bestvalue = (vertex.T * c)[0]
bestvertex = vertex
return bestvertex
def solve(A, b, c):
x = findoptimum(A, b, c)
if not x:
print 'LP is infeasible'
else:
print 'Vertex', x.T, 'is optimal'
print 'Optimal value is', c.T*x
if __name__ == '__main__':
A = Matrix([[-10, -6, -9, -10],
[ 8, -6, -5, -5],
[ -7, -1, -9, 3],
[ -1, -4, 5, 10],
[ 1, 2, 0, 10],
[ 2, -9, 3, -8],
[ -8, -1, -8, 1],
[ 7, -10, 4, -4],
[-10, 2, 5, 8],
[ -7, 9, 4, -4],
[ -1, 0, 0, 0],
[ 0, -1, 0, 0],
[ 0, 0, -1, 0],
[ 0, 0, 0, -1]])
b = Matrix([9, 7, 3, 4, 8, 0, 3, 2, 4, 8, 0, 0, 0, 0])
c = Matrix([2, -2, -3, 8])
solve(A, b, c)``````
× | 645 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-13 | longest | en | 0.47335 |
https://mypages.iit.edu/~smile/weekly/bc091200.htm | 1,660,468,898,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00064.warc.gz | 400,460,225 | 1,930 | High School Biology Chemistry Smile Meeting
12 September 2000
Notes prepared by Earl Zwicker
Glenda Ellis (Williams School)
marked a 30 meter course around the perimeter of the room, gave us stop watches, and we measured the time it took for each of us to walk the course. From this, the walking speeds for each of us were calculated and compared. Speeds ranged from 1.20 to 1.67 meters/second. Ken Schug recalculated the speeds in miles/hour! And we then had an interesting discussion about how our speeds compare with walking and running speeds of various animals:
• falcon in a dive
• elephant
• cheetah
What a great way to get student's ego involved in an experiment, at the same time learning about the concept of speed and how it is calculated!
Erma Lee (Williams School)
turned us into bean counters! She gave us dried, red beans and asked us to estimate how many beans it takes to go around a person's shoe. Then we had to measure it! Erma asked us to trace an outline of our shoe on an 8.5 in x 11 in paper, and then we did the measurement on the tracings, laying the beans end-to-end. The distance around was measured in "bean lengths." She asked us to measure the surface area within the tracing, a sort of surface area of a shoe. To do this we placed as many beans as possible within the outline and in a single layer, and counted them. We repeated these measurements for our hands using the same techniques. We found that the surface area of our hands was about 75% that for feet, and the perimeter of a hand (fingers closed) was about 80% that for feet. What a rich learning experience! - involving measurement, observation, comparison and conclusions. Thanks, Erma!
Karlene Joseph (Lane Tech HS)
had us write hidden messages on 8.5 in x 11 in paper by using phenolphthalein as an invisible ink. When we sprayed the paper with an ammonia solution (a base), the phenolphthalein turned red, and our messages appeared. Spraying with vinegar (acetic acid) caused the red to turn clear so the messages would disappear!
Now that she had us "hooked," Karlene did a titration of 15 ml of vinegar with ammonia using phenolphthalein as an indicator, and we found that about 12 ml of ammonia was needed to turn the vinegar to deep pink. Thus, the molarity of the ammonia was 15/12 = 1.25 times that of the vinegar. A wonderful phenomenological approach, Karlene!
Great ideas! Don't miss the next one!
SEE YOU THERE! | 570 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-33 | latest | en | 0.967264 |
https://www.convertunits.com/from/mic/to/microgram | 1,670,011,793,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00555.warc.gz | 772,240,444 | 23,851 | ## ››Convert mic to microgram
mic microgram
How many mic in 1 microgram? The answer is 1.
We assume you are converting between mic and microgram.
You can view more details on each measurement unit:
mic or microgram
The SI base unit for mass is the kilogram.
1 kilogram is equal to 1000000000 mic, or 1000000000 microgram.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between mic and micrograms.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of mic to microgram
1 mic to microgram = 1 microgram
5 mic to microgram = 5 microgram
10 mic to microgram = 10 microgram
20 mic to microgram = 20 microgram
30 mic to microgram = 30 microgram
40 mic to microgram = 40 microgram
50 mic to microgram = 50 microgram
75 mic to microgram = 75 microgram
100 mic to microgram = 100 microgram
## ››Want other units?
You can do the reverse unit conversion from microgram to mic, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Microgram
The SI prefix "micro" represents a factor of 10-6, or in exponential notation, 1E-6.
So 1 microgram = 10-6 grams-force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 439 | 1,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.825732 |
https://student-baba.com/is-drift-velocity-independent-of-time/ | 1,695,337,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00868.warc.gz | 601,939,113 | 20,229 | # Is Drift Velocity Independent of time?
Electricity is an important part of our day to day life and we all know about the current and electricity. Current is created due to moving electron and electron moves due to drift velocity but the thing which makes me wonder is, is drift velocity independent of time ? and therefore I did some research on this.
Yes, drift velocity is independent of time as shown in the equation of drift velocity.
Let’s see by deriving the equation of drift velocity and find how exactly is drift velocity independent of time ?
## Drift velocity independent of time:
So before learning about how drift velocity is independent of time we need to first define what drift velocity is:
### What is drift velocity ?
In a conductor, when an electric field is applied across a conductor, free-electron experiences a force in the direction of higher potential from a lower potential and while it keeps moving with the force due to thermal velocity it drifts in the direction of force and this is called as drift velocity.
This drift velocity is the reason due to which electron moves inside a conductor and current is produced.
Current flowing inside a conductor due to drift velocity
Yes, drift velocity is independent of time. Drift velocity is the reason due to which current flows inside a conductor. In the derivation of the equation of drift velocity, we can see that acceleration “a” is constant and relaxation time “乁” is kept constant.
So current and drift velocity does not depend on time.
### Derivation of drift velocity equation:
Electrons that are present inside the conductors, moves in random different directions, which causes a collision between them.
Electric current is caused when all the electrons present inside a conductor moves in one particular direction but in the above scenarios, it’s totally opposite of it which causes random collisions everwhere.
Now, there’s no fixed time when these collisions happen inside a conductor, and therefore let’s suppose the time between the collision of any two pair of the electron is T1, In the same way, let’s take T2 for the second pair of electrons and this continues for all the pair of electron present inside a conductor.
Let’s, for example, suppose the total number of pairs be ‘n’ so we will calculate an average from T1 to Tn and this average is named as Relaxation time which can be supposed as the time period between collision between two electrons.
Relaxation time: “乁” is the time period for electron elapsed between two consecutive collisions.
Relaxation time formula
We know that: V= U + at
Where, V = final velocity, U = initial velocity, a = acceleration and t = time.
If initial velocity ‘U’ = 0 then, V = at
Now we know that acceleration:
We know that.
This average velocity is due to the application of E in one direction. Hence this is drift velocity.
In the equation of drift velocity ‘Vd‘ depends upon the electric field ‘E ‘and Relaxation time ‘乁’.
Conclusion:
In the equation of drift velocity ‘Vd‘ depends upon electric field and relaxation time. If temperature increases the relaxation time will decrease resulting in the decrease of drift velocity.
For a constant temperature relaxation time is constant so we can say that yes, drift velocity is independent of the time at a constant temperature.
#### F.A.Q:
Q.1 How drift velocity is responsible for the current ?
This drift velocity is responsible for current because this is not cancelled out. This drift velocity is very small but still, it is effective to make current because number of electron carrying in the same direction is extremely large.
Q.2 Why is acceleration constant here ?
The acceleration only depends on the applied electric field on a conductor (due to potential difference created by a battery).
Q.3 Why is the relaxation time also constant ?
The relaxation time is the average time. A collision of an electron is related to kinetic energy, therefore, it depends upon the temperature. So for a particular temperature relaxation time is kept constant.
Q.4 How does acceleration of an object and its dependency of time-related ?
The average value is used in the above equations. Initial velocity ‘U’ is zero because the motion of an electron in the absence of an electric field will be distributed in all the direction of conductor resulting in the net velocity to be zero. The equation used here is V= U+at | 894 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-40 | longest | en | 0.941726 |
https://rememberingsomer.com/number-between-1-and-20-with-5-factors/ | 1,653,294,129,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00467.warc.gz | 549,691,953 | 5,607 | Problem 1: The area the a rectangle-shaped garden is 7 square yards. Perform all feasible whole-number size the garden can have.
You are watching: Number between 1 and 20 with 5 factors
Solution: 1 yd x 7 yd
The whole-number dimensions, 1 and 7, the the rectangle-shaped garden above, room the factors of the number 7.
Problem 2: The area of a rectangular garden is 8 square yards. Perform all possible whole-number size the garden can have.
Solution: 1 yd x 8 yd, 2 yd x 4 yd
The whole-number dimensions, 1, 2, 4 and 8, of the rectangle-shaped gardens in trouble 2, are the factors of the number 8. In trouble 1, the number 7 has only 2 factors. The number 7 is prime. In problem 2 above, the number 8 has 4 factors. The number 8 is composite.
Definitionsprime number has just two factors: 1 and also itself.A composite number has more than 2 factors.The number 1 is neither prime no one composite.
When the area of a rectangle is a element number, there is only one collection of feasible dimensions for that rectangle. When the area the a rectangle is a composite number, there space two or an ext sets of possible dimensions for the rectangle. Each set of dimensions is a pair that factors.
To recognize if a number is element or composite, follow this steps:
Find all components of the number.If the number has only 2 factors, 1 and itself, then it is prime.If the number has more than 2 factors, climate it is composite.
Example 1: Is the number 2 prime or composite?
Solution: The components of 2 are 1 x 2. 2 is prime.
Example 2: Is the number 9 element or composite?
Solution: The determinants of 9 are 1 x 9, 3 x 3. 9 is composite.
We have identified if a solitary number is element or composite. Let"s look at a range of number to check out if they room prime or composite. Please keep in mind that each range of numbers offered in instances 3, 4 and 5 listed below are inclusive.
Example 3: Find all prime numbers in between 2 and also 9.
factors the 2: 1 x 2 2 is prime factors that 3: 1 x 3 3 is prime factors that 4: 1 x 4, 2 x 2 4 is composite factors that 5: 1 x 5 5 is prime factors that 6: 1 x 6, 2 x 3 6 is composite factors of 7: 1 x 7 7 is prime factors of 8: 1 x 8, 2 x 4 8 is composite factors that 9: 1 x 9, 3 x 3 9 is composite
Solution: The prime numbers between 2 and also 9 are 2, 3, 5 and also 7.
Example 4: Find every prime numbers in between 10 and also 19.
factors the 10: 1 x 10, 2 x 5 10 is composite factors of 11: 1 x 11 11 is prime factors that 12: 1 x 12, 2 x 6, 3 x 4 12 is composite factors of 13: 1 x 13 13 is prime factors that 14: 1 x 14, 2 x 7 14 is composite factors that 15: 1 x 15, 3 x 5 15 is composite factors the 16 1 x 16, 4 x 4 16 is composite factors that 17: 1 x 17 17 is prime factors that 18: 1 x 18, 3 x 6 18 is composite factors the 19: 1 x 19 19 is prime
Solution: The element numbers between 10 and 19 room 11, 13, 17 and also 19.
Example 5: Find every prime numbers between 20 and also 29.
factors of 20: 1 x 20, 2 x 10, 4 x 5 20 is composite factors that 21: 1 x 21, 3 x 7 21 is composite factors the 22: 1 x 22, 2 x 11 22 is composite factors the 23: 1 x 23 23 is prime factors the 24: 1 x 24, 2 x 12, 3 x 8, 4 x 6 24 is composite factors that 25: 1 x 25, 5 x 5 25 is composite factors of 26: 1 x 26, 2 x 13 26 is composite factors of 27: 1 x 27, 3 x 9 27 is composite factors the 28: 1 x 28, 2 x 14, 4 x 7 28 is composite factors that 29: 1 x 29 29 is prime
Solution: The prime numbers in between 20 and 29 space 23 and 29.
Example 6: Is the number 31 prime or composite? explain your answer using complete sentences.
Solution 1: The number 31 is prime due to the fact that its only factors are one and also itself.
Solution 2: Thirty-one is a element number. This is since the number 31 has only 2 factors: 1 and also 31.
Solution 3: I separated the number 31 by all numbers in between 1 and 31 and found no factors other 보다 one and also thirty-one. Therefore, 31 is prime.
There are many feasible ways to explain the systems to this problem. These are simply three feasible explanations.
Summary: A prime number has only two factors: 1 and also itself. A composite number has an ext than 2 factors. The number 1 is no prime no one composite.
The element numbers in between 2 and also 31 space 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 because each of this numbers has actually only two factors, itself and also 1.
### Exercises
Directions: read each question below. Pick your answer by clicking its button. Feedback to her answer is noted in the results BOX. If you make a mistake, pick a different button.
1 Each that the adhering to numbers is composite EXCEPT:30, 31, 32, 33, 34, 35, 36, 37, 38, 3932 and 3930 and also 3431 and 37All compositeRESULTS BOX:
2 The element numbers between 40 and also 49 are:42, 43 and also 4741, 43 and 47 43, 45 and also 47None the the above.RESULTS BOX:
3 The element numbers between 50 and also 59 are:53 and 5951 and 59 53 and also 57None of the above.RESULTS BOX:
4 The prime numbers in between 60 and 69 are:63 and also 6961 and also 67 60 and also 65None of the above.RESULTS BOX:
5.See more: Can I Dissolve Tylenol In Water, Robot Or Human The element numbers in between 20 and 69 are:21, 23, 29, 31, 37, 41, 43, 47, 53, 63 and 6923, 29, 31, 33, 37, 41, 43, 47, 59, 61 and also 6723, 29, 31, 37, 41, 43, 47, 53, 59, 61 and also 67None the the above.RESULTS BOX: .tags a { color: #fff; background: #909295; padding: 3px 10px; border-radius: 10px; font-size: 13px; line-height: 30px; white-space: nowrap; } .tags a:hover { background: #818182; } Home Contact - Advertising Copyright © 2022 rememberingsomer.com #footer {font-size: 14px;background: #ffffff;padding: 10px;text-align: center;} #footer a {color: #2c2b2b;margin-right: 10px;} | 1,869 | 5,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-21 | latest | en | 0.852398 |
https://www.coursehero.com/file/1385663/420Hw06/ | 1,487,858,707,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00126-ip-10-171-10-108.ec2.internal.warc.gz | 828,380,301 | 46,415 | # 420Hw06 - STAT 420 (10 points) (due Friday, March 7, by...
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STAT 420 Spring 2008 Homework #6 (10 points) (due Friday, March 7, by 4:00 p.m.) 1. Can a corporation’s annual profit be predicted from information about the company’s chief executive officer (CEO)? Forbes (May, 1999) presented data on company profit ( y ), (in \$ millions), CEO’s annual income ( x 1 ) (in \$ thousands), and percentage of the company’s stock owned by the CEO ( x 2 ). Company Profit, y CEO Income, x 1 Stock, x 2 Gap 824.5 . Drexler 3,743 . 1.71% Intel 6,068.0 . Grove 52,598 . .13 % Gateway 2000 346.4 . Waitt 855 . 43.93 % HJ Heinz 746.9 . O’Reilly 2,916 . 1.63 % Conseco 630.7 . Hilbert 124,579 . 3.64 % Citicorp 5,807.0 . Reed 6,200 . .22 % Cisco Systems 1,362.3 . Chambers 560 . .06 % General Electric 9,296.0 . Welch 40,626 . .03 % America Online 254.0 . Case 26,917 . .54 % Computer Associates 570.0 . Wang 10,614 . 3.79 % Lockheed Martin 1,001.0 . Augustine 2,533 . .01 % Bear Stearns 538.6 . Cayne 23,215 . 3.44 % Source : “Compensation Fit for a King,”
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## This note was uploaded on 02/10/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
### Page1 / 3
420Hw06 - STAT 420 (10 points) (due Friday, March 7, by...
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Ask a homework question - tutors are online | 534 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | longest | en | 0.799996 |
http://inst.eecs.berkeley.edu/~cs281a/fa08/announcements.html | 1,516,544,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00165.warc.gz | 170,332,604 | 2,288 | ### Announcements
• Wednesday, November 26
• In addition to his regularly scheduled office hours (Tues, 11--12, Thurs 3:30--4:30), Prof. Wainwright will be holding an additional hour of office hours today (Monday) from 3--4 pm in 421 Evans Hall.
• Homework solutions posted for home2sol.pdf and home3sol.pdf
• New homework posted: HW5 and here is the HW5_Data.zip.
• Paper on Variational Methods. (Wainwright, Jordan).
• Reminder that there is no class tomorrow (Tuesday, Nov. 11th)for the holiday. Enjoy!
• For anybody that didn't collect their homework on Friday, Homework 2 will be returned in class on Tuesday. (They will be alphabetized this time!)
• Tuesday, October 28
• Homework 4 has now been posted. It's due November 4th, 2008.
• Wednesday, October 22
• Minor clarifications on the homework. For problem 3.2, assume that $\theta \in \theta^d$ where $d=1$. Hence, the Gaussian random variables are univariate. For 3.1 (c), \sqrt{ \det(\Gamma) } should be in the numerator, not the denominator. The \sqrt{2 pi} term should read (2 pi)^{d/2}, there is a missing $d$ in the problem set.
• THIS WEEK ONLY (Tuesday Oct 21). Prof. Wainwright will not have his usual OH from 11--12. Instead, Sahand will have an extra office hour at the same time; location: 611 Soda (same place as Sahand's usual office hours)
• A clarification regarding Prof. Wainwright's Thursday office hours (every week), 3:30 -- 4:30. Prof. Wainwright typically starts talking to students after class at 3:30. If there is a group of students, everybody goes over to Evans Hall 330. If you want to join later, try Evans Hall 330, or Prof. Wainwright's office, Evan Hall 421.
• Solutions to Homework 1 have been posted. Solutions
• Homework 3 has been posted. homework 3
• Sunday, September 28
• The due date for homework 2 has been postponed to Tuesday, October 7th.
• Please note The Disabled Students' Program (DSP) is looking for a note-taker! If you take clear, well-organized notes, this is a good opportunity for you to assist a fellow student and receive pay. Please come to the Disabled Students' Program Office and fill out an application. We are located in 260 Cesar Chavez right beside The Golden Bear Cafe. The application is also available online at dsp.berkeley.edu on the 'Note Takers Needed This Semester' link. If you have any questions, please send an email to dspnotes@berkeley.edu.
• Homework 2 has been posted. homework 2
• Sunday, September 21
• Typo in homework due date. It is due Thurs. September 18th.
• Monday, September 8
• Office hours and discussion section times and locations have been posted.
• Sahand will be holding a temporary office hours from 11-12 tomorrow at Brewed Awakening until a final office hours can be set.
• Chapter 2 can be found here.
• Thursday, August 29
• Welcome to CS281a/Stat241a | 732 | 2,813 | {"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} | 3.046875 | 3 | CC-MAIN-2018-05 | latest | en | 0.934798 |
https://usethinkscript.com/threads/reverse-relative-strength-using-donchian-channels-for-thinkorswim.9476/ | 1,720,867,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00499.warc.gz | 475,621,581 | 21,657 | # Reverse Relative Strength using Donchian Channels For ThinkOrSwim
#### Pensar
##### Expert
VIP
@Pensar- I think this is pretty close, but I have a couple more questions/requests if I may. I watched another playing of the webinar and caught this slide that shows the setting he is using for the Swing Energy indicator. You can See chart here
The settings show ("b", 10, 50, 75, Green, Yellow, Red). He said there are 2 versions of this indicator- a "Buy" and a "Sell" version. I'm sure that is what the "b" is in the setting. I think the 10 is the Doncian channel length, and 50, 75 are the plotting limits to change colors.
According to the author, You should have a high fully charged value, meaning high green bars for a fully charged buy. This would require running only the "buy" version. You would also have a high fully charged value, meaning high green bars for a fully charged sell also. Is this something you can help me with? Thanks in advance.
@hockeycoachdoug Now that's a great clue. Of course, this code will not be an exact match, given that they very likely use a modified version of the donchian channel.
Code:
``````declare lower;
input direction = {default "buy", "sell"};
input short_donchian = 10;
input long_donchian = 50;
input high_value = 75;
input low_value = 25;
input averagelength = 2;
input averagetype = averagetype.SIMPLE;
def hh = highest(high,long_donchian);
def ll = lowest(low,short_donchian);
def strength;
switch (direction) {
case sell: strength = movingaverage(averagetype,100*(close-ll)/(hh-ll),averagelength);
default: strength = movingaverage(averagetype,100*(hh-close)/(hh-ll),averagelength); }
plot RS = strength;
RS.setpaintingstrategy(paintingstrategy.histogram);
RS.assignvaluecolor(if rs > 75 then color.uptick else if rs > 25 then color.yellow else color.downtick);
RS.setlineweight(3);
plot high_strength = high_value;
high_strength.setdefaultcolor(color.gray);
high_strength.setpaintingstrategy(paintingstrategy.horizontal);
high_strength.setstyle(curve.short_dash);
plot low_strength = low_value;
low_strength.setdefaultcolor(color.gray);
low_strength.setpaintingstrategy(paintingstrategy.horizontal);
low_strength.setstyle(curve.short_dash);``````
I think this is the the closest I can reverse-engineer it without actually seeing the other indicator's code. In all honesty, though, if their little code differences are the only key to profitable v.s. unprofitable, there's bigger problems to worry about. A quote I read once, "Those who can trade do and those who can't trade sell indicators", or something like that. Really makes one think about another saying, "there's a market for everything".
Eye-Candy of both directions -
Merry Christmas to you!
Last edited by a moderator:
curious question, what does the red/yellow/green indicate, could not interpret it right, also is this mainly for daily chart and above timeframe?
@Pensar- I don't think we have this quite right yet. I watched the webinar again to ask question about the settings shown in the above picture. This was my question-
### Can you discuss what the settings are for your TOP Swing Energy indicator. Is it a function of distance from your TOP Breakout Levels (Similar to Donchian Channel). I am trying to recreate a similar indicator. Thanks.
Here was their response-
Mark tend to use 10 bars lookback for TOP Swing Energy. The indicator focuses on movement over the lookback period but not like Donchian Channel. More from recent extremes.
We are currently working on identifying indicators to release in 2022, and it has not been decided if the TOP Swing Energy will be one of them
They don't use the actual Donchian - they retitled a similar indicator and called it TOP Trend Breakout Levels - They typically use a 25 bar high and 25 bar low and it is basically a 25 bar Donchian channel.
I assume the settings shown are "buy"/"sell", look back period 10 days, 25 is the TOP Trend Breakout length, and the 25 and 75 I think are the plot boundaries to be either red or bars. They said there is a separate Buy and Sell versions. I now think they are tracking how close the last 10 bars are to the upper or lower channel. Any thoughts on how to code something like that?
Last edited:
Does anyone here have the buy and sell signals for the upper chart?
The Buy/sell signals are from their proprietary indicator known as HedgeFundTrender or also called HedgeFundTrader or HFT. It is a proprietary indicator, but I'm sure you will find many versions of it available on the web
Do their signals look like Trend reversals signals? I suspected it as when he shows his active signals that those not repainted yet?
I don't think I understand your question. The signals they are trading come from TOP Hedge Fund Trender. They made a scan to scan for the buy/sell HFT signals of which there often are many. They then use their TOP Swing Energy indicator as a filter to hopefully filter out the weak, unenergized trades. I don't think their actual indicators repaint if that is your question.
@Pensar not sure if you are still watching this thread after it was moved. Attached is a screen shot showing the variables inputs. They said they use both a buy and a sell version so that is the "b" for buy. According to TOPTradeTools, 10 is the look back period, I think the 50 is the length of their doncian channel. They actually use something called TOP Trend Breakout Levels which is a Doncian channel. I think the 75 and 25 are the limits on the histogram for green and red. Obviously the toggle need to be part of the buy/sell. I assume it has to do with ATR proximity to the Doncian Channel. Any additional thoughts or ideas are appreciated. This could be a good filter for any number of methodologies that generate actual buy/sell signals.
@hockeycoachdoug I just tried creating an version using ATR and the results are nowhere near to the sample pics . . . the best match seems to be what I've posted earlier. I'm going to try some other things yet, but currently you have the closest match I can get.
Many Thanks @Pensar. I appreciate your help. I will keep an eye on this thread. Hopefully someone else might chime in with some ideas also.
Last edited:
@hockeycoachdoug Now that's a great clue. Of course, this code will not be an exact match, given that they very likely use a modified version of the donchian channel.
I think this is the the closest I can reverse-engineer it without actually seeing the other indicator's code. In all honesty, though, if their little code differences are the only key to profitable v.s. unprofitable, there's bigger problems to worry about. A quote I read once, "Those who can trade do and those who can't trade sell indicators", or something like that. Really makes one think about another saying, "there's a market for everything".
Eye-Candy of both directions -
Merry Christmas to you!
Can you post the code for the one on the bottom? Thanks,
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What are the benefits of VIP Membership? | 1,844 | 8,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.829124 |
http://en.wikibooks.org/wiki/Java_Programming/Basic_arithmetic | 1,394,535,529,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011176878/warc/CC-MAIN-20140305091936-00086-ip-10-183-142-35.ec2.internal.warc.gz | 57,277,441 | 14,221 | # Arithmetic expressions
Primitive Types Java Programming Arithmetic expressions Literals
Navigate Language Fundamentals topic: ( v • d • e )
In order to do arithmetic in Java, one must first declare at least one variable. Typically one declares a variable and assigns it a value before any arithmetic is done. Here's an example of declaring an integer variable:
Code section 3.59: Variable assignation. ```int x = 5; ```
After creating a variable, one can manipulate its value by using Java's operators: `+` (addition), `-` (subtraction), `*` (multiplication), `/` (integer division), `%` (modulo or remainder), `++` (pre- & postincrement by one), `--` (pre- & postdecrement by one).
Code listing 3.10: Operators.java ```public class Operators { ``` ``` public static void main(String[] args) { ``` ``` int x = 5; ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Addition ---"); ``` ``` x = 5; ``` ``` System.out.println("x + 2 = " + (x + 2)); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Subtraction ---"); ``` ``` x = 5; ``` ``` System.out.println("x - 4 = " + (x - 4)); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Multiplication ---"); ``` ``` x = 5; ``` ``` System.out.println("x * 3 = " + (x * 3)); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- (Integer) Division ---"); ``` ``` x = 5; ``` ``` System.out.println("x / 2 = " + (x / 2)); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Modulo (Remainder) ---"); ``` ``` x = 5; ``` ``` System.out.println("x % 2 = " + (x % 2)); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Preincrement by one ---"); ``` ``` x = 5; ``` ``` System.out.println("++x = " + (++x )); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Predecrement by one ---"); ``` ``` x = 5; ``` ``` System.out.println("--x = " + (--x )); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Postincrement by one ---"); ``` ``` x = 5; ``` ``` System.out.println("x++ = " + (x++ )); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` ``` ``` System.out.println("--- Postdecrement by one ---"); ``` ``` x = 5; ``` ``` System.out.println("x-- = " + (x-- )); ``` ``` System.out.println("x = " + x); ``` ``` System.out.println(); ``` ``` } ``` ```} ```
Console for Code listing 3.10 ```x = 5 --- Addition --- x + 2 = 7 x = 5 --- Subtraction --- x - 4 = 1 x = 5 --- Multiplication --- x * 3 = 15 x = 5 --- (Integer) Division --- x / 2 = 2 x = 5 --- Modulo (Remainder) --- x % 2 = 1 x = 5 --- Preincrement by one --- ++x = 6 x = 6 --- Predecrement by one --- --x = 4 x = 4 --- Postincrement by one --- x++ = 5 x = 6 --- Postdecrement by one --- x-- = 5 x = 4 ```
The division operator rounds towards zero: `5/2` is 2, and `-5/2` is -2. The remainder operator has the same sign as the left operand; it is defined such that `((a/b)*b) + (a%b)` is always equal to a. The preincrement, predecrement, postincrement, and postdecrement operators are special: they also change the value of the variable, by adding or subtracting one. The only difference is that preincrement/decrement returns the new value of the variable; postincrement returns the original value of the variable.
Question 3.8: Consider the following code:
Question 3.8: Question8.java ```public class Question8 { ``` ``` public static void main(String[] args) { ``` ``` int x = 10; ``` ``` x = x + 10; ``` ``` x = 2 * x; ``` ``` x = x - 19; ``` ``` x = x / 3; ``` ``` System.out.println(x); ``` ``` } ``` ```} ```
What will be printed in the standard output?
Output for Question 3.8 ```7 ```
`int x = 10;` => 10
`x = x + 10;` => 20
`x = 2 * x;` => 40
`x = x - 19;` => 21
`x = x / 3;` => 7
When using several operators in the same expression, one must consider Java's order of precedence. Java uses the standard PEMDAS (Parenthesis, Exponents, Multiplication and Division, Addition and Subtraction) order. When there are multiple instances of the same precedence, Java reads from left to right. Consider what the output of the following code would be:
Code section 3.60: Several operators. ```System.out.println(10*5 + 100/10 - 5 + 7%2); ```
Console for Code section 3.60 ```56 ```
The following chart shows how Java would compute this expression:
Figure 3.1: Computation of an arithmetic expression in the Java programming language
Besides performing mathematical functions, there are also operators to assign numbers to variables (each example again uses the variable initialized as `x = 5`):
Code listing 3.11: Assignments.java ```public class Assignments { ``` ``` public static void main(String[] args) { ``` ``` int x = 5; ``` ``` x = 3; ``` ``` System.out.println("Assignment (x = 3) : " + x); ``` ``` ``` ``` x = 5; ``` ``` x += 5; ``` ``` System.out.println("Assign x plus another integer to itself (x += 5): " + x); ``` ``` ``` ``` x = 5; ``` ``` x -= 4; ``` ``` System.out.println("Assign x minus another integer to itself (x -= 4): " + x); ``` ``` ``` ``` x = 5; ``` ``` x *= 6; ``` ``` System.out.println("Assign x multiplied by another integer to itself (x *= 6): " + x); ``` ``` ``` ``` x = 5; ``` ``` x /= 5; ``` ``` System.out.println("Assign x divided by another integer to itself (x /= 5): " + x); ``` ``` } ``` ```} ```
Console for Code listing 3.11 ```Assignment (x = 3) : 3 Assign x plus another integer to itself (x += 5): 10 Assign x minus another integer to itself (x -= 4): 1 Assign x multiplied by another integer to itself (x *= 6): 30 Assign x divided by another integer to itself (x /= 5): 1 ```
## Using bitwise operators within Java
Java has besides arithmetic operators a set of bit operators to manipulate the bits in a number, and a set of logical operators. The bitwise logical operators are
Operator Function Value of
x before
Example
input
Example
output
Value of
x after
`&` Bitwise AND 7 `x&27` 3 7
`|` Bitwise OR 7 `x|27` 31 7
`^` Bitwise XOR 7 `x^27` 28 7
`~` Bitwise inversion 7 `~x` -8 7
Besides these logical bitwise functions, there are also operators to assign numbers to variables (`x = -5`):
Operator Function Example
input
Example output
`&=` Assign `x` bitwisely ANDed with another value to itself `x &= 3` 3
`|=` Assign `x` bitwisely ORed with another value to itself `x` |`= 3` -5
`^=` Assign `x` bitwisely XORed with another value to itself `x ^= 3` -8
`<<=` Assign `x` divided by another integer to itself `x <<= 1` -10
`>>=` Assign `x` bitwisely negated with another value to itself `x >>= 1` -3
`>>>=` Assign `x` bitwisely negated with another value to itself `x >>>= 1` 2,305,843,009,213,693,949 (64 bit)
The shift operators are used to shift the bits to the left or right, which is also a quick way to multiply/divide by two:
Operator Function Value of
x before
Example
input
Example output Value of
x after
`<<` Logical shift left -15 `x << 2` -60 -15
`>>` Arithmetic shift right -15 `x >> 3` -2 -15
`>>>` Logical shift right -15 `x >>> 3` 2,305,843,009,213,693,937 (64 bit) -15
Primitive Types Java Programming Arithmetic expressions Literals | 2,135 | 7,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-10 | longest | en | 0.629297 |
http://hvac-talk.com/vbb/showthread.php?923172-to-amp-or-not-to-amp | 1,369,168,718,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700563008/warc/CC-MAIN-20130516103603-00013-ip-10-60-113-184.ec2.internal.warc.gz | 130,776,421 | 14,084 | # Thread: to amp or not to amp
1. Regular Guest
Join Date
Jun 2006
Posts
308
## to amp or not to amp
on heating elements or just loads in general,,, whats the theory is it more resistance more amp draw or the other way around ? im thinking the more current /electrons traveling through the wire/cunductor/element, the more amp draw.
2. Just think about it in terms of Ohm's Law:
E = I x R
Where:
E = Voltage
I = Current
R = Resistance
If you rearrange the equation to solve for Current, it would look like this:
I = E / R
So you can see from this that if the voltage is constant, a higher resistance will result in a lower current.
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• | 199 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-20 | latest | en | 0.882504 |
https://www.aqua-calc.com/convert/pressure/pound-force-per-square-millimeter-to-kilopascal | 1,606,575,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00586.warc.gz | 565,989,830 | 11,118 | # Convert pounds-force per (square millimeter) to kilopascals
## [lbf/mm² to kPa] (lbf:pound-force, mm:millimeter, kPa:kilopascal)
### pounds-force per square millimeter to kilopascals conversion cards
• 1
through
20
pounds-force per square millimeter
• 1 lbf/mm² to kPa = 4 448.22161 kPa
• 2 lbf/mm² to kPa = 8 896.44322 kPa
• 3 lbf/mm² to kPa = 13 344.66483 kPa
• 4 lbf/mm² to kPa = 17 792.88644 kPa
• 5 lbf/mm² to kPa = 22 241.10805 kPa
• 6 lbf/mm² to kPa = 26 689.32966 kPa
• 7 lbf/mm² to kPa = 31 137.55127 kPa
• 8 lbf/mm² to kPa = 35 585.77288 kPa
• 9 lbf/mm² to kPa = 40 033.99449 kPa
• 10 lbf/mm² to kPa = 44 482.2161 kPa
• 11 lbf/mm² to kPa = 48 930.43771 kPa
• 12 lbf/mm² to kPa = 53 378.65932 kPa
• 13 lbf/mm² to kPa = 57 826.88093 kPa
• 14 lbf/mm² to kPa = 62 275.10254 kPa
• 15 lbf/mm² to kPa = 66 723.32415 kPa
• 16 lbf/mm² to kPa = 71 171.54576 kPa
• 17 lbf/mm² to kPa = 75 619.76737 kPa
• 18 lbf/mm² to kPa = 80 067.98898 kPa
• 19 lbf/mm² to kPa = 84 516.21059 kPa
• 20 lbf/mm² to kPa = 88 964.4322 kPa
• 21
through
40
pounds-force per square millimeter
• 21 lbf/mm² to kPa = 93 412.65381 kPa
• 22 lbf/mm² to kPa = 97 860.87542 kPa
• 23 lbf/mm² to kPa = 102 309.09703 kPa
• 24 lbf/mm² to kPa = 106 757.31864 kPa
• 25 lbf/mm² to kPa = 111 205.54025 kPa
• 26 lbf/mm² to kPa = 115 653.76186 kPa
• 27 lbf/mm² to kPa = 120 101.98347 kPa
• 28 lbf/mm² to kPa = 124 550.20508 kPa
• 29 lbf/mm² to kPa = 128 998.42669 kPa
• 30 lbf/mm² to kPa = 133 446.6483 kPa
• 31 lbf/mm² to kPa = 137 894.86991 kPa
• 32 lbf/mm² to kPa = 142 343.09152 kPa
• 33 lbf/mm² to kPa = 146 791.31313 kPa
• 34 lbf/mm² to kPa = 151 239.53474 kPa
• 35 lbf/mm² to kPa = 155 687.75635 kPa
• 36 lbf/mm² to kPa = 160 135.97796 kPa
• 37 lbf/mm² to kPa = 164 584.19957 kPa
• 38 lbf/mm² to kPa = 169 032.42118 kPa
• 39 lbf/mm² to kPa = 173 480.64279 kPa
• 40 lbf/mm² to kPa = 177 928.8644 kPa
• 41
through
60
pounds-force per square millimeter
• 41 lbf/mm² to kPa = 182 377.08601 kPa
• 42 lbf/mm² to kPa = 186 825.30762 kPa
• 43 lbf/mm² to kPa = 191 273.52923 kPa
• 44 lbf/mm² to kPa = 195 721.75084 kPa
• 45 lbf/mm² to kPa = 200 169.97245 kPa
• 46 lbf/mm² to kPa = 204 618.19406 kPa
• 47 lbf/mm² to kPa = 209 066.41567 kPa
• 48 lbf/mm² to kPa = 213 514.63728 kPa
• 49 lbf/mm² to kPa = 217 962.85889 kPa
• 50 lbf/mm² to kPa = 222 411.0805 kPa
• 51 lbf/mm² to kPa = 226 859.30211 kPa
• 52 lbf/mm² to kPa = 231 307.52372 kPa
• 53 lbf/mm² to kPa = 235 755.74533 kPa
• 54 lbf/mm² to kPa = 240 203.96694 kPa
• 55 lbf/mm² to kPa = 244 652.18855 kPa
• 56 lbf/mm² to kPa = 249 100.41016 kPa
• 57 lbf/mm² to kPa = 253 548.63177 kPa
• 58 lbf/mm² to kPa = 257 996.85338 kPa
• 59 lbf/mm² to kPa = 262 445.07499 kPa
• 60 lbf/mm² to kPa = 266 893.2966 kPa
• 61
through
80
pounds-force per square millimeter
• 61 lbf/mm² to kPa = 271 341.51821 kPa
• 62 lbf/mm² to kPa = 275 789.73982 kPa
• 63 lbf/mm² to kPa = 280 237.96143 kPa
• 64 lbf/mm² to kPa = 284 686.18304 kPa
• 65 lbf/mm² to kPa = 289 134.40465 kPa
• 66 lbf/mm² to kPa = 293 582.62626 kPa
• 67 lbf/mm² to kPa = 298 030.84787 kPa
• 68 lbf/mm² to kPa = 302 479.06948 kPa
• 69 lbf/mm² to kPa = 306 927.29109 kPa
• 70 lbf/mm² to kPa = 311 375.5127 kPa
• 71 lbf/mm² to kPa = 315 823.73431 kPa
• 72 lbf/mm² to kPa = 320 271.95592 kPa
• 73 lbf/mm² to kPa = 324 720.17753 kPa
• 74 lbf/mm² to kPa = 329 168.39914 kPa
• 75 lbf/mm² to kPa = 333 616.62075 kPa
• 76 lbf/mm² to kPa = 338 064.84236 kPa
• 77 lbf/mm² to kPa = 342 513.06397 kPa
• 78 lbf/mm² to kPa = 346 961.28558 kPa
• 79 lbf/mm² to kPa = 351 409.50719 kPa
• 80 lbf/mm² to kPa = 355 857.7288 kPa
• 81
through
100
pounds-force per square millimeter
• 81 lbf/mm² to kPa = 360 305.95041 kPa
• 82 lbf/mm² to kPa = 364 754.17202 kPa
• 83 lbf/mm² to kPa = 369 202.39363 kPa
• 84 lbf/mm² to kPa = 373 650.61524 kPa
• 85 lbf/mm² to kPa = 378 098.83685 kPa
• 86 lbf/mm² to kPa = 382 547.05846 kPa
• 87 lbf/mm² to kPa = 386 995.28007 kPa
• 88 lbf/mm² to kPa = 391 443.50168 kPa
• 89 lbf/mm² to kPa = 395 891.72329 kPa
• 90 lbf/mm² to kPa = 400 339.9449 kPa
• 91 lbf/mm² to kPa = 404 788.16651 kPa
• 92 lbf/mm² to kPa = 409 236.38812 kPa
• 93 lbf/mm² to kPa = 413 684.60973 kPa
• 94 lbf/mm² to kPa = 418 132.83134 kPa
• 95 lbf/mm² to kPa = 422 581.05295 kPa
• 96 lbf/mm² to kPa = 427 029.27456 kPa
• 97 lbf/mm² to kPa = 431 477.49617 kPa
• 98 lbf/mm² to kPa = 435 925.71778 kPa
• 99 lbf/mm² to kPa = 440 373.93939 kPa
• 100 lbf/mm² to kPa = 444 822.161 kPa
#### Foods, Nutrients and Calories
CLASSIC CHEF SALAD WITH RANCH DRESSING, UPC: 5051379085063 contain(s) 86 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Triiron tetraoxide [Fe3O4] weighs 5 170 kg/m³ (322.75256 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-124, liquid (R124) with temperature in the range of -40°C (-40°F) to 82.23°C (180.014°F)
#### Weights and Measurements
ounce per square millimeter (oz/mm²) is a non-metric measurement unit of surface or areal density
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
kcal to J conversion table, kcal to J unit converter or convert between all units of energy measurement.
#### Calculators
Price conversions and cost calculator for materials and substances | 2,775 | 5,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.06953 |
http://openstudy.com/updates/558648e9e4b0b31a8a423e3a | 1,495,935,960,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00494.warc.gz | 355,100,340 | 8,738 | • anonymous
Darrell tried to solve the equation x2 + 10x = -9 by completing the square. His work and some of his answers are shown below. Each red blank, marked with a letter A, should be filled in with the same number. Each blue blank, marked with the letter B, should be filled in with the same expression. Each green blank, marked with the letter C, should be filled in with the same number. Enter the number, expression, and number that fill in blanks A, B, and C correctly. Separate your answers with commas, like this: 42, 3x-4, 53
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Not the answer you are looking for? Search for more explanations. | 377 | 1,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.534715 |
https://caminorealaf.es/how-do-sports-odds-work-with-pictures/ | 1,709,549,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00028.warc.gz | 148,670,348 | 10,877 | How Do Sports Odds Work? With Pictures
# How Do Sports Odds Work? With Pictures
Each different format for betting odds will have a different formula for calculating your winnings. All of the previous wager types are bets placed on single events. Parlays involve taking multiple bets and bundling them together. This increases the potential payout for each wager, but also comes with an added risk. If even one of your bets is incorrect, you lose the entire parlay wager. The over/under is the number of points oddsmakers expect will be the total score for the contest .
If you wagered on Team A to win the Super Bowl and they end up winning, you are going to need to risk more than \$100 to win \$100 as indicated by the «-» symbol preceding the payouts. In this particular example, a bettor would need to wager \$150 to win \$100. With American odds, the number represented is how much you have to bet to win \$100. If the number is preceded with a «+» sign, then you would risk less than \$100 to win a wager worth \$100. And, the opposite is true when the number is preceded with a «-» where the bettor would have to risk more than \$100 just to win \$100 of the wager.
## Fractional And American
This AFC West divisional matchup has the potential of being one of the more fun games of this early season. Both Free Sports Picks And Betting Predictions teams bring electric offenses into the game, featuring two quarterbacks that can really chuck it. The Raiders are a very good team, but I think they finally end up on the wrong side of a close game in Week 4. As with any odds format, the potential return and profit from a bet is relevant to the stake amount. Using a \$100 bet example is the easiest way to calculate profits with American odds, but below is a simple formula to work out potential returns for any stake.
## Online Blackjack
The little amount of money you spend for books and practice software will be the best investment you will make if you are serious. Practice in the casino is limited to just backcounting, you can stand all day long if you can and practice counting for free, this tactic helped me alot when I was learning. So, you might want to get Wong’s Professional Blackjack which details the count and is more suited to the modern game.
If you want to place a baseball-related bet on the Vegas odds, for example, watch at least 3 or 4 games and check the score histories before making your bet. Check the «over-under» score to bet how many points will be won total. The over-under score is a median guess at how many points total will be won by each team. Ask the ticket master what the over-under score is for a certain team, then decide whether you believe more («over») or less («under») points will be scored. Write down the rotation numbers of the teams you want to bet on.
## Should I Bet With Point Spreads?
MyBookie is North America Trusted Sportsbook & Bookmaker, Offering top sporting action in the USA & abroad. The odds for the solid number will usually be less, as there are now 2 outcomes that can favour the bettor, not 1. Geoff Clark does a quick breakdown from a betting perspective of four games on the Bet Slippin’ Podcast for the NBA’s November 18 slate. I wrote up a guide to golf betting that explains these unique features so you can feel comfortable wagering on PGA events too. Use the rotation number/numbers and the name of the team/teams you want to bet on. The rotation number appears to the left of each team and is used as a unique identifier so that there is no confusion about what you want.
He will be a big part of getting James Wiseman’s development on the right track and could help in some international scouting. The NBA Draft Lottery is next Tuesday night and the Warriors could have 2 selections if the ping pong balls drop their way. The guys talk about all the different scenarios, who they would take if the Warriors get Minnesota’s pick and if they would package the 2 picks to move up or get a big star. Anthony, Marcus, Tim & Ethan are ready to break down all the Warriors options at #7 & #14 in the upcoming NBA Draft. The guys talk about what the W’s should do if Jonathan Kuminga is available at #7. Some other interesting options for the W’s at #7 are Barnes, Bouknight and Davion Mitchell.
To convert fractional odds to decimal odds, you need to turn the fraction into decimal form and then add 1. To convert a fraction to a decimal, you simply divide the numerator by the denominator . The decimal odds reflect your total returns, meaning both your profit and original stake. So if you win a £10 bet with 2.5 odds, you’ll get £25 back – your £10 stake plus £15 profit. Most sports betting sites have functions to calculate your potential winnings in real time.
The new bet odds calculator is an essential tool if you are placing wagers. Now, with the betting odds calculator, you can easily figure out the risk versus reward of your wagers. There are many ways to bet on sports but one of the most popular ways for football and basketball is called the point spread. The point spread is a margin of victory and as a bettor, you determine which side will beat the margin. At WSN, we have created a simple-to-use and free betting odds calculator and converter that lets you convert American, decimal, fractional, and implied odds. Understanding odds conversion to the percentage and implied probability behind the odds on offer is key to assessing the potential value in a particular betting market. | 1,180 | 5,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.945982 |
https://www.toccochicago.com/2022/07/28/how-do-you-scale-an-object-in-one-direction-in-autocad/ | 1,713,485,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00312.warc.gz | 923,139,976 | 11,127 | ## How do you scale an object in one direction in AutoCAD?
To scale in one direction: 1. Select the object, right click-scale, snap to point (left click on ) from wich you want the object to be scaled, enter the scale value, enter, done.
How do you squish something in AutoCAD?
You can copy (Control C) or cut (Control X) the original object and paste it back in as a block (control Shift V) and then if you go to Modify / Properties you will have the option to change the scaling of the X, Y or Z independently. You can then explode it and work with it again.
How do you scale down drawings?
How to Scale Things Down for a Drawing
1. Measure the objects that you want to draw.
3. Scale the items by use of ratios.
4. Divide the 10 by the measured length of the wall, 120, which looks like this as a fraction: 10/120.
5. Set up a ration for the height of the wall, too.
### How do I set up nanoCAD?
2. Run LicServSetup.exe.
3. Request a license, after which you will be prompted to install a License Server.
4. Install nanoCAD on workstation computers.
How do you scale a single axis in AutoCAD?
Insert block. In the dialogue box uncheck the Uniform Scale option then enter the X, Y and Z scale.
What is annotative scale in AutoCAD?
Annotative scaling is the process in which you select a scale for a drawing and all the annotative text, dimensions, blocks and hatches change to reflect the scale. This can also be set independently for each viewport so multiple scales can show on one sheet drawing.
#### How do I scale a drawing in AutoCAD?
How to scale up in AutoCAD – Window select the object(s), type SCALE, and then specify a number larger than 1. Hit Enter. The size of the object(s) will SCALE UP by that scale factor.
How do you scale things?
To scale an object to a smaller size, you simply divide each dimension by the required scale factor. For example, if you would like to apply a scale factor of 1:6 and the length of the item is 60 cm, you simply divide 60 / 6 = 10 cm to get the new dimension. | 488 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.867544 |
http://math.stackexchange.com/questions/127253/positive-definite-bilinear-form-question?answertab=oldest | 1,462,218,628,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117405.91/warc/CC-MAIN-20160428161517-00018-ip-10-239-7-51.ec2.internal.warc.gz | 175,369,714 | 16,060 | # Positive definite bilinear form question
I'm given that a quadratic form $q$ is positive definite if $q(v) > 0 \quad \forall 0 \neq v \in V$ and equivalently for bilinear form $\tau$. Does this mean that in a positive definite bilinear form there can exists $v,w \in V$ such that $\tau (v,w) \leq 0$ ?
-
If $V=\mathbb{R},$ $\tau:(x,y)\mapsto xy$ is positive definite but $\tau(1,-1)=-1.$ | 125 | 392 | {"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} | 2.859375 | 3 | CC-MAIN-2016-18 | latest | en | 0.719737 |
https://www.scribd.com/document/397628/MAE-360-hwk3 | 1,550,495,416,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00623.warc.gz | 977,860,419 | 42,513 | You are on page 1of 3
# J.
Murray
Hwk 3
MAE 360
Question 1:
Find the dimensionless velocities at at least 50 points along the airfoil surface.
To find these velocities the equation for the u values is (1) for this equation t is
the panel distribution according to the cosine distribution. X is the distance from
the origin along the cord from the leading edge to the trailing edge. Z is the
location of the point on the upper surface of the airfoil at a distance x along the
airfoil.
t (i +1) − x j ti − x j −1
1
u = 1+ − A b ( )
(t ( i +1) − x j ) + Z 2j (t i − x j ) + Z 2j
(1)
2π
2 2
t ( i +1) − x j t − xj
tan −1 − tan −1 i
Z Z
A= j j (this is a matrix) (2)
Zj
−1
1
w = −
Zj
Zj
(
A b )
(t ( i +1) − x j ) + Z j (t i − x j ) + Z 2j
(3)
2π
2 2 2
Using these equations it is possible to find the velocities of u and w of a point
along the upper surface of the airfoil.
Question 2:
Find and plot the pressure coefficients as a function of the dimensionless x at these
points.
Figure (1) is the plot that these points provided.
Cp Distribution on a NACA 0012 Airfoil at 0 Degree Angle of Attack
-0.5
0
Cp
0.5
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Chord (Unitless)
Figure 1: MATLab Figure of the airfoil Cp
Question 3:
Compare your results with the following. Comment on how well or how poorly the
pressures compare.
The two plots compare very well. The plots are almost identical expect
where the x value approaches the leading and trailing edges. As the x value
approaches these edges the equation for the plot causes a discontinuity. This
discontinuity can not be plotted accurately using discrete methods. The plots
created using this method are very accurate until very close to the stagnation
points. As figure (2) and (3) show when the gridlines and ticks are the same, the
plots are the same until very close to 0 and 1 on the chord axis.
Cp Distribution on a NACA 0012 Airfoil at 0 Degree Angle of Attack
-0.5
0
Cp
0.5
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Chord (Unitless)
Figure 2 Plot of the Pressure Coefficient
Figure 3 Homework Assignment 3 Provided Figure
The code:
function aero=hmk3
clc
warning off;
global Npanel mu t
N =51; Npanel = N-1; % there are N endpoints and Npanel panels
% locations of first and last endpoints - varies depending on the
number of
% panels. Determine using trial and error.xs =.0018; xf =.9985;
xs=.009;
xf=.9985;
c=linspace(xs,xf);
% distribute panels according to a cosine distribution
ths = pi-acos((.5-xs)/.5);
thf = acos((xf-.5)/.5);
dth = (thf-ths)/Npanel;
th = [ths:dth:thf];
t= 1/2 + 1/2*cos(th);
%assign field point locations at the center of the panels
for i =1 : Npanel
x(i) =(t(i) + t(i+1))/2;
end
% this is the upper surface of a NACA 0012 airfoil
Z=.6.*(.2969.*sqrt(x)-.126.*x-.3516.*x.^2+.2843.*x.^3-
.1015.*x.^4);
ma=max(Z);
Tau=ma;
rho=.5*Tau^2*1;
E=rho/2;
% creat the A matrix and the rhs b vector
for j = 1:Npanel
b(j)=2*pi;
for i = 1:Npanel
A(j,i)=(atan2(t(i+1)-x(j),Z(j))-atan2(t(i)-x(j),Z(j)))/Z(j);
end
end
%solve for the doublet strengths
mu = inv(A)*b';
for i=1:Npanel
Psi(i)=Z(i)-sum(mu(i)*atan((t(i)-x(i))/Z(i)));
for j=1:Npanel
U(j,i)=[(t(i+1)-x(j))/((t(i+1)-x(j))^2+(Z(j))^2)-((t(i)-
x(j))/((t(i)-x(j))^2+(Z(j))^2))];
W(j,i)=[(Z(j)/((t(i+1)-x(j))^2+(Z(j))^2))-(Z(j)/((t(i)-
x(j))^2+(Z(j))^2))];
end
end
u=1+(1/(2*pi)).*U*mu;
w=(-1/(2*pi)).*W*mu;
for i=1:Npanel
Cp(i)=1-(u(i))^2-(w(i))^2;
end
figure(2)
plot(x,Cp);
set(gca,'YDir','reverse'); | 1,391 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.851565 |
https://www.reddit.com/r/HomeworkHelp/comments/1ejizp/high_school_igcse_physics_simple_question_about/ | 1,500,873,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424721.74/warc/CC-MAIN-20170724042246-20170724062246-00513.warc.gz | 842,487,542 | 26,508 | This is an archived post. You won't be able to vote or comment.
[–] 0 points1 point (20 children)
Holy crap IGCSE exams have changed since I did them.
About question one, why would the current pass through the two bulbs on the left? To me it seems like a short circuit and no current would pass through them.
But let's assume you are correct and the current does pass through the lamps on the left, I fail to see how you concluded that the question would be invalid. Yes, not all the lamps would have the same voltage running through them but that's because not the same current is running through them. The two lamps on the right each get 12V and the same current (lets call it X), and the two lamps on the left each get 6V and the same current (Y). So two lamps on the right each get X Amperes, two lamps on the left each get Y Amperes. Or did I misunderstood your point?
[–][S] 0 points1 point (15 children)
Why would the current not pass? The circuit isn't open as the lamps are still connected to each other. The resistance of the of the open junction doesn't affect the circuit on the left as long as the lamps are connected(right?).
Anywho, I wrote it incorrectly. The question was asking what Voltage flows through each lamp. It didn't specify which lamp they are talking about, so the voltage does indeed vary.
[–] 0 points1 point (13 children)
If by open junction you mean the line that is right next to the number 3 then you are 100% correct. However, I am talking about the closed junction painted (poorly, I apologize :p) in red:
http://imgur.com/SF647V4
If that is indeed a wire, then current always takes the path of least resistance, that is the short circuit.
[–][S] 0 points1 point (12 children)
....... this was only a sign of my poor drawing skills, not a wire.
[–] 0 points1 point (9 children)
Well then good man, if it isn't a wire, your answer are correct.
[–][S] 0 points1 point (8 children)
Then why would they ask me about a voltage that flows through each lamp? Doesn't that insinuate uniformity of the voltage, when in fact it isn't (considered it is connected to junction 2)
[–] 0 points1 point (7 children)
Doesn't that insinuate uniformity of the voltage
The resistance are the same but the current passing through the lamps isn't. Why would that insinuate uniformity of the voltage?
[–][S] 0 points1 point (6 children)
I am saying that the question insinuates uniformity of the voltage, not that I think of it in that matter. To ask what Voltage each current receives, means that the man made this question intended for all the lamps in the whole circuit to have the same voltage, when this isn't true always.
[–] 0 points1 point (5 children)
Not really. If each lamp receives a different voltage, or the same voltage, the question would be phrased the same way.
You are doing the same mistake I did when I took my IGCSEs, don't try to play the examiner and assume answers based on the way the question was asked or whether the question was asked at all. Study the material and think about each question carefully and good things happen.
[–][S] 0 points1 point (4 children)
So this takes us all the way back. Both of us are wrong? Trust me, out of all the things, the circuit is the least explained matter in IGCSE. The things we take do not usually prepare us for circuits with functions unusual to us. It doesn't really depend on studying and becomes a mixture of luck, with self-logic.
[–]Physics, math, programming, whatever 0 points1 point (1 child)
The horizontal part is a wire though, right?
Is it Option A
or Option B
[–][S] 1 point2 points (0 children)
Option A. Does that cause any difference? The wire poses no difference as far as I know.
[–]Physics, math, programming, whatever 0 points1 point (0 children)
[–][S] 0 points1 point (3 children)
Correct me if I am wrong, but isn't a short circuit when you connect a parallel wiring circuit to a resistor to increase the voltage across the intended resistor?
[–]Physics, math, programming, whatever 0 points1 point (0 children)
The two bulbs on the left can be simplified to one resistor, since they're in parallel. They end up just looking like a dangling resistor on the circuit with only one end connected, therefore they're doing nothing.
[–] 0 points1 point (1 child)
Not really. Short circuit is a electrical circuit that allows a current to travel along an unintended path, often where essentially no (or a very low) resistance is encountered. (From wikipedia)
It's basically saying, if, in an electrical circuit the passing current has to go to through wire A or wire B which are in series, more current will go to the wire with lower resistance. Short circuit is when one of the wires has no or very low resistance such that little current passes through the other wire and the wire with little or no resistance gets almost all the current. That seems to be the case when the current faces the junction on the left. It either has to pass through two lamps with high resistance, or through the wire (that I showed you in red in the comment above) with no resistance. Thus, short circuit.
Edit: Drawing circuits correctly is important to avoid confusion. If it's not a wire, then no short circuit, your answer is correct.
[–][S] 0 points1 point (0 children)
I am only taking O.L so unfortunately my idea of a short circuit is connecting 2 ends of wires containing no resistance in parallel to a resistor to increase the voltage entering the intended resistor. The more you know, I guess.
[–]Physics, math, programming, whatever 0 points1 point (10 children)
Wait, what's going on here? Voltage is not current... Current is what goes through (measured in amps), voltage is what goes across (measured in volts). Question one asks what happens if the switch is in position 2. Well, current comes out of the battery, flows through the switch, through the bulbs on the right, and back to the battery. There is no path for current to flow through the bulbs on the left. The only bulbs on will be the two on the right.
As for the second question, i'm confused about if it's asking for voltage or current. In any case, both bulbs will have a voltage of 12V across them because they're in parallel. If they were in series, we could say that the voltage across each one would be 6V because they have equal resistance. However, since they're in parallel, they do split the current in half. We can't determine the current though if we don't know the resistance of the bulbs.
[–][S] 0 points1 point (9 children)
Why wouldn't a current go through or a voltage go across the circuit on the left? It is a circuit parallel to the wire carrying the current, therefore a current can pass through it as if it were a circuit connected parallel to the wire. Correct?
[–]Physics, math, programming, whatever 1 point2 points (8 children)
It's physically parallel to the wire carrying current, yes. But it's not actually connected to anything. The two lamps on the left are shorted by the wire running horizontally on the bottom.
In this image, the green circled is all one node: http://i.imgur.com/CSzZzL3.png
It can therefore be simplified to this: http://i.imgur.com/7oN0R65.png
(sorry for the horrible drawing, but you get the idea)
The two bulbs on the left do not complete a circuit. Current can't flow through them because there's nowhere for it to come out!
[–] 0 points1 point (0 children)
That was my point exactly earlier. But he said that the wire between the two lamps on the left (i colored it red in a picture in one of my comments above) was just a drawing mistake and not a wire at all. So that makes the the two lamps on the left in series with the wire carrying current.
[–][S] 0 points1 point (6 children)
http://i.imgur.com/v5BzTBd.png
Can the current not flow in the direction of the arrows? I mean, shouldn't the little junction at the bottom be disregarded since the lamps can be connected normally without its existence, therefore, allowing the current to pass normally, since it has no effect?
[–]Physics, math, programming, whatever 0 points1 point (5 children)
Nope. The junction at the bottom is providing a path for current to flow with zero resistance, so no current goes up through the resistors.
If you have java, check out this circuit
If not, here's a screenshot: http://i.imgur.com/pkhK9zW.png The info at the bottom is showing the current and voltage through the blue resistor. The yellow dots represent current
[–]Physics, math, programming, whatever 0 points1 point (0 children)
Here it is without the bottom wire: http://imgur.com/hjhejcB
[–][S] 0 points1 point (0 children)
I am utterly perplexed here. I understand that a current flows through the wire at the bottom, but since the circuit is connected in parallel to the wire, doesn't that mean the current is divided and can flow through the circuit on the left? Anywho, doesn't voltage travel regardless of current(or at least I was taught), therefore it cab receive a voltage, right?
[–][S] 0 points1 point (2 children)
I understand now. I greatly apologise for lugging you for so long. Thank you a great deal
[–]Physics, math, programming, whatever 0 points1 point (1 child)
No problem. I understand that this stuff is hard to grasp at first, but once you get more experience under your belt, it becomes second nature.
[–][S] 0 points1 point (0 children)
Roger that. I already know the rules by heart, but it seems that, like mathematics, electricity is best learnt by practice (in an exam environment, with only 2 minutes left on the clock, it made sense to me to disregard the wire and consider it a normally connected circuit and not a short circuit, especially that we barely even mentioned what a short-circuit is in the syllabus). Anywho, I am aspiring to take on an Electric Engineering degree after high school. Can you provide any source where I can practice high-school level Electricity word problems? | 2,383 | 9,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-30 | latest | en | 0.971656 |
http://mathoverflow.net/questions/40165/zariski-dense-subgroup-of-algebraic-group/40182 | 1,469,351,115,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823989.0/warc/CC-MAIN-20160723071023-00199-ip-10-185-27-174.ec2.internal.warc.gz | 166,421,657 | 13,488 | # Zariski dense subgroup of algebraic group [closed]
Let $G$ be a Zariski dense subgroup of algebraic group $H.$ Then is $H$ the Zariski closure of $G$?
Conversely, if $G$ is a subgroup of an algebraic group $H$ and if $H$ is the Zariski closure of $G,$ then is $G$ Zariski dense in $H$?
What are the equivalent conditions of being a Zariski dense subgroup of an algebraic group?
-
## closed as off-topic by Venkataramana, Felipe Voloch, Peter Humphries, Daniel Loughran, Tom De MedtsJul 14 at 7:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Venkataramana, Felipe Voloch, Peter Humphries, Daniel Loughran, Tom De Medts
If this question can be reworded to fit the rules in the help center, please edit the question.
Either I don't understand the question, or this is just a trivial consequence of the definition of "dense". – Martin Brandenburg Sep 27 '10 at 15:45
It seems like an elementary exercise in general topology, and it has nothing to do with Zariski topology or algebraic groups. – Qfwfq Sep 27 '10 at 15:49
I am also not sure, whether I understand the question properly; but it seems to me that one should distinguish carefully between the group scheme and its geometric points here.
To fix ideas, let $K$ be an algebraically closed field and $G/K$ an algebraic group. An algebraic subgroup of $G$ is by definition a closed subgroup scheme of $G$. Furthermore there is the abstract group $G(K)$, and we have the Zariski topology on $G(K)$. There is a bijection $H\mapsto H(K)$ from the set of smooth algebraic subgroups of $G$ to the set of closed subgroups of $G(K)$. (Cf. Milne's lecture notes on algebraic groups for example.)
If $\Gamma$ is a subgroup of $G(K)$, then there is a unique smooth algebraic subgroup $\Gamma^{zar}$ of $G$, such that $\Gamma^{zar}(K)=\overline{\Gamma}$. Some people call the algebraic group $\Gamma^{zar}$ the Zariski closure of $\Gamma$.
Assume now that $G$ is smooth. With the above terminology we see:
a) If $\Gamma$ is Zariski dense in $G(K)$, then $\Gamma^{zar}=G$.
b) If $H$ is a smooth algebraic subgroup of $G$ and $H(K)$ is Zariski dense in $G(K)$, then $H(K)=G(K)$ and consequently $H=G$.
And one cannot drop the smoothness assumption on $G$ here.
- | 649 | 2,390 | {"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} | 2.71875 | 3 | CC-MAIN-2016-30 | latest | en | 0.915048 |
https://studylib.net/doc/25182901/8th-and-9th-forces-and-motion- | 1,545,222,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832259.90/warc/CC-MAIN-20181219110427-20181219132427-00229.warc.gz | 741,161,970 | 32,572 | # 8th & 9th Forces and Motion
advertisement
```
I can conceptually and mathematically
describe forces and motion.
Distance is the length of a path between two
points.
Displacement is the direction and distance
between a beginning point and an ending
point.
Sometimes it is necessary to combine
displacements mathematically.
We can do this by using vectors.
A vector is a quantity that has a magnitude
(how much) and direction.
A scalar is a quantity that has a magnitude
without a direction.
Speed is how fast something moves.
There are 2 ways to describe the speed of an
object:
◦ a) average speed
◦ b) instantaneous speed
Average speed is the average value of speed
for an object over the entire duration of the
trip.
The mathematical equation for
average speed is:
Average Speed = Total Distance
Total Time
v=d
cccct
V speed
d distance
t time
Instantaneous Speed is the speed measured
at a particular instant in time.
Distance and time can be displayed on a
distance-time graph.
Speed is the slope of a line on a distancetime graph.
Velocity is speed with a direction.
For example:
◦ the speed of a Camaro may be 75 miles per hour,
◦ the velocity may be 75 miles per hour EAST.
Acceleration is the rate at which velocity
changes.
Acceleration is a vector.
Acceleration can be described as a change in:
◦ Speed
◦ Direction
◦ or both
Free fall is when an object falls towards Earth
due to earth’s gravity.
Objects falling near Earth’s surface accelerate
downward at a rate of 9.8 m/s2
You can calculate the acceleration of an
object by dividing the change in velocity by
the total time:
Acceleration = change in velocity
total time
= (v2 – v1 )
t
Acceleration is the slope of a speed-time
graph.
Average acceleration is a measure of the
average value for the acceleration of an
object over a given amount of time.
Instantaneous acceleration is a measure of
the acceleration of an object at any given
instant in time.
I can conceptually and mathematically
describe forces and motion.
``` | 585 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-51 | latest | en | 0.915168 |
www.khalidelassaad.com | 1,725,729,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00284.warc.gz | 830,896,536 | 3,168 | # Ulam Spiral
#### YouTube: Prime Spirals - Numberphile
Do ya like prime numbers?
This is a simple pair of scripts written that produce an Ulam spiral and a random spiral (with the same percentage of white squares as the Ulam spiral) in `tkinter`. The program does not use a sieve to determine if a number is prime. Instead, it uses the slower method of calculating modulo every number up to the input’s square root.
I watched the YouTube video linked above and thought it was interesting, so I decided to recreate it.
An Ulam spiral, starting from the red square in the middle. Prime numbers are white, non-prime are black. Note the diagonals that emerge in the distribution of the prime numbers.
Here is a random spiral, where each number has the same probability of showing up white as any other number. That probability is the number of white cells over number of total cells in the above image.
No diagonals emerge! Prime numbers, seemingly patternless, actually exhibit some predictability and order compared to (not so) true randomness.
Neat!
Also included in `spiral.py` is a function that takes an integer and returns a hexcode. Passing incrementing numbers to this function sweeps through a rainbow of colors. I cannot for the life of me remember why I made this. | 273 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.899906 |
https://cracku.in/blog/cat-mixture-and-alligations-questions-pdf/ | 1,721,825,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00781.warc.gz | 147,353,345 | 40,046 | # CAT Mixture and Alligations Questions PDF [Most Important]
0
395
Mixture and Alligations is one of the important topics in the CAT Quants (Arithmetic) section. These questions are not very tough, so ensure that you do not miss out on these questions. Know all the Important formulas from Mixtures & Alligations. Solve & practice more questions from CAT Mixture and Alligations so that you get a hang of these questions. You can check out these CAT Mixture and Alligation questions from the CAT Previous year papers. In this post, we will look into some important Mixture and Alligation Questions for CAT. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download these Important Mixture and Alligation Questions for CAT (with detailed answers) PDF along with the video solutions below, which is completely Free.
Question 1:Â A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]
a)Â 2 : 3
b)Â 1 : 2
c)Â 1 : 3
d)Â 3 : 4
Solution:
After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.
Instructions
DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.
Question 2:Â What is the maximum amount of sucrose (to the nearest gram) that can be added to one-gram of saccharin such that the final mixture obtained is atleast 100 times as sweet as glucose?
a)Â 7
b)Â 8
c)Â 9
d)Â 100
Solution:
For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.
1 gm of saccharin = 675
Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.
So, (675 + N) / (N+1) = 74
=> 675 + N = 74N + 74
=> 601 = 73N => N = 8.23
When N=9, sweetness will be S = (675+9)/10 = 684/10 = 68.4
When N=8, sweetness will be S = (675+8)/9 = 683/9 = 75.8
So, option b) is the correct answer.
Question 3:Â Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?
a)Â 1.3
b)Â 1
c)Â 0.6
d)Â 2.3
Solution:
The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30
Option a) is the correct answer.
Question 4:Â Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?
a)Â 27:14
b)Â 27:13
c)Â 27:16
d)Â 27:18
Solution:
The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4
Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$
Let the ratio in which they should be mixed be equal to X:1.
Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$
The total volume of water is $\frac{2X}{9}+\frac{4}{13}$
They are in the ratio 3:1
Hence, $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$
Therefore, $91X+81=78X+108$
Therefore $X = \frac{27}{13}$
Question 5: Consider three mixtures — the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has
a)Â The same amount of water and liquid B
b)Â The same amount of liquids B and C
c)Â More water than liquid B
d)Â More water than liquid A
Solution:
The proportion of water in the first mixture is $\frac{1}{3}$
The proportion of Liquid A in the first mixture is $\frac{2}{3}$
The proportion of water in the second mixture is $\frac{1}{4}$
The proportion of Liquid B in the second mixture is $\frac{3}{4}$
The proportion of water in the third mixture is $\frac{1}{5}$
The proportion of Liquid C in the third mixture is $\frac{4}{5}$
As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$
The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$
The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$
The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$
Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$
From the given choices, only option C Â is correct.
Question 6:Â Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio
a)Â 17 : 25
b)Â 18 : 25
c)Â 19 : 24
d)Â 21 : 25
Solution:
The selling price of the mixture is Rs.40/kg.
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x.
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 =Â $\frac{3a+2b}{5}Â$
$\frac{3a+2b}{5} = \frac{40}{1.1}$
$3.3a+2.2b=200$ ——–(1)
The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 =Â $\frac{2a+3b}{5}$
$\frac{2a+3b}{5} = \frac{40}{1.05}$
$2.1a+3.15b=200$Â ——(2)
Equating (1) and (2), we get,
$3.3a+2.2b = 2.1a+3.15b$
$1.2a=0.95b$
$\frac{a}{b} = \frac{0.95}{1.2}$
$\frac{a}{b} = \frac{19}{24}$
Therefore, option C is the right answer.
Question 7:Â A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is
a)Â 16
b)Â 26
c)Â 20
d)Â 22
Solution:
Let the price of paint B be x.
Price of paint A = x+8
We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture.
The seller sells 10 litres of paint at Rs.264 earning a profit of 10%.
=> The cost price of 10 litres of the paint mixture = Rs. 240
Therefore, the cost of 1 litre of the mixture = Rs.24
We have to find the highest possible cost of paint B.
When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal.
=> (x+x+8)/2 = 24
2x = 40
x = Rs. 20
Therefore, option C is the right answer.
Question 8:Â A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
a)Â 30%
b)Â 40%
c)Â 50%
d)Â 60%
Solution:
Let the volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $\frac{1}{2}$ = 50%
Hence, 50 is the correct answer.
Question 9:Â A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
a)Â 30.3
b)Â 35.2
c)Â 25.4
d)Â 20.5
Solution:
Final quantity of alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml
Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml
Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %
Question 10:Â There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio
a)Â 251 : 163
b)Â 239 : 161
c)Â 220 : 149
d)Â 229 : 141
Solution:
It is given that in drum 1, A and B are in the ratio 18 : 7.
Let us assume that in drum 2, A and B are in the ratio x : 1.
It is given that drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.
By equating concentration of A
$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$
$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$
$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$
$\Rightarrow$ $x = \dfrac{239}{161}$
Therefore, we can say that in drum 2, A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.
Question 11:Â A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
a)Â 30%
b)Â 40%
c)Â 50%
d)Â 60%
Solution:
Let the volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $\frac{1}{2}$ = 50%
Hence, 50 is the correct answer.
Question 12:Â The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is
a)Â 3 : 10
b)Â 1 : 3
c)Â 1 : 4
d)Â 2 : 5
Solution:
Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.
It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.
$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$
$\Rightarrow$ $a+2b+3c = 120$ … (1)
If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.
$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$
$\Rightarrow$Â $3a+2b+c = 180$ … (2)
From equation (1) and (2), we can say that
$\Rightarrow$ $b+2c = 45$
$\Rightarrow$ $b = 45 – 2c$
Also, on subtracting (1) from (2), we get
$a – c = 30$
$\Rightarrow$ $a = 30 + c$
In solution D, B and C are mixed in the ratio 2 : 7
So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$
Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$
Hence, option B is the correct answer.
Question 13:Â The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is
a)Â 15
b)Â 13
c)Â 12
d)Â 14
Solution:
Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.
The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.
The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.
In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water
Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A
i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.
after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.
After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.
Each transfer can be captured through the following table.
Percentage of salt in vessel A =$\ \frac{\ 70}{500}\times\ 100$
=14%
Question 14:Â A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?
Solution:
Initially the amount of Dye and Water are 16,24 respectively.
To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.
Now, the Dye and Water arr 16,40 respectively.
After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.
To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.
Hence , 8 is correct answer.
Question 15: Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is
a)Â 90%
b)Â 94%
c)Â 92%
d)Â 89%
Solution:
Initially let’s consider A and B as one component
The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.
Let the percentage of alcohol in component 1 be ‘x’.
Using allegations , $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84
Percentage of alcohol in A = 60% => Let’s percentage of alcohol in B = x%
The resultant mixture has 84% alcohol. ratio = 1:3
Using allegations , $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$
=> x= 92%
Question 16:Â The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is
The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% :Â $\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$ | 5,039 | 15,802 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.933403 |
https://www.instructables.com/id/Prerequisites/ | 1,534,392,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210413.14/warc/CC-MAIN-20180816034902-20180816054902-00114.warc.gz | 933,733,708 | 11,957 | # Prerequisites
250
## Introduction: Prerequisites
The following information is a single lesson in a larger project. Find more great projects here.
Lesson Overview:
Now we'll learn what we'll need to do before jumping into the full kit!
## Step 1: Introduction
Before starting the Arduino Basic Kit projects, it is helpful to have some foundational knowledge of electronics!
The following topics are not required for completing projects, but they are highly recommended. We'll go over some of the basics in this lesson and also provide links for more information.
1. Continue to the next step.
## Step 2: Current, Voltage and Resistance
In order to talk about electrical circuits, you need to be familiar with the terms voltage, current, and resistance. These terms are defined below.
Refer to the Arduino Get to Know Your Tools project.
1. Current (measured in amperes, or amps; with the A symbol) is the amount of electrical charge flowing through a specific point in your circuit.
2. Voltage (measured in volts; with the V symbol) is the difference in energy across one point in a circuit and another.
3. Resistance (measured in ohms; with the Ω symbol) is how much a component impedes the flow of electrical energy.
4. Continue to the next step.
## Step 3: Ohm's Law
Current, voltage, and resistance are all related to each other in an electrical circuit. When you change one of these in a circuit, it affects the others. The relationship between them is known as Ohm's Law, named for Georg Simon Ohm, who discovered it.
Ohm's Law applies to the voltage of, current through, and resistance of a single component - such as a resistor.
VOLTAGE (V) = CURRENT (I) x RESISTANCE (R)
Refer to the Arduino Get to Know Your Tools project.
1. Continue to the next step.
## Step 4: Circuit Diagrams
In this project, you have seen circuits laid out on a breadboard or real components connected to each other with wires. In future projects we will also be using circuit diagrams, which use symbols to represent components.
A circuit diagram is like a flowchart for your circuit, giving you a map of which components are connected to each other.
The component symbols that we use in the Arduino Basic Kit projects are shown in the chart below.
1. Many of the circuit diagrams symbols look like the components they represent. For example, the symbol for a piezo speaker looks like a loudspeaker cone!
2. Continue to the next step.
## Step 5: Voltage Dividers (1)
It is useful to know a few things about voltage dividers before embarking on projects that use analog sensors. The voltage divider that we use is simply two resistors in series between high and low voltage points. The resistors share the total voltage and you can create a custom voltage level between the two resistors by changing their values.
To figure out what the voltage is over each resistor, you use the following system of equations, considering the fact that the current flowing through each resistor is the same!
I = V1/R1 = V2/R2
V1 + V2 = total Voltage = 5 V
The voltage that you usually care about is the output (V-out), which is equal to V2.
V-out = V2 = (5 x R2) / (R1 + R2)
For more information, and the full derivation of the voltage divider equation, visit:
https://en.wikipedia.org/wiki/Voltage_divider
1. As a general rule, the larger resistor in a voltage divider will have the larger voltage across it!
2. Continue to the next step to see an application of voltage dividers.
## Step 6: Voltage Dividers (2)
Voltage dividers are used in many projects with analog sensors. The light sensor, for example, changes its resistance when light is shining on it. Under low light conditions, the sensor is 180 k-ohms. Under bright lights, it is 506 ohms.
Instead of detecting resistance directly, the Arduino needs to sense a voltage. We use the voltage divider shown below to convert the resistance into an output voltage signal.
1. In the light sensor example, if the sensor is hit by medium-level light and has a resistance of 1000 ohms, the output voltage that reaches the Arduino board will be 4.5 volts.
2. As a general rule when more light hits the sensor, its resistance decreases and the output voltage in this circuit increases.
3. Continue to the next step.
## Step 7: Multimeter (recommended)
Finally, a multimeter is a useful tool for that can verify the amount of resistance of a component, the voltage across a component, or the amount of current running through your circuit. There is a multimeter available in the circuit simulator.
1. In the light sensor example, if the sensor is hit by medium-level light and has a resistance of 1000 ohms, the output voltage that reaches the Arduino board will be 4.5 volts.
2. As a general rule when more light hits the sensor, its resistance decreases and the output voltage in this circuit increases.
3. Continue to the next step.
## Step 8: Review
Once you have reinforced your general knowledge of the prerequisites, you can continue on to create cool projects with your Arduino!
The instructions for each project will provide you with enough information to complete it, even if your electronics foundation is a little rusty. However, be sure to look out for the following terms, and review the concepts when needed:
Current
Voltage
Resistance
Ohm's Law
Circuit diagram
Voltage divider
Multimeter
Let's get started with the next project: Getting to know your tools!
Congratulations, you have completed this project!
Check out other great projects here.
## Recommendations
• ### Arduino Class
71,681 Enrolled | 1,220 | 5,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.931016 |
http://slideplayer.com/slide/3281558/ | 1,524,358,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.57/warc/CC-MAIN-20180422002521-20180422022521-00360.warc.gz | 287,497,422 | 22,705 | BIOE 220/RAD 220 REVIEW SESSION 3 February 6, 2011.
Presentation on theme: "BIOE 220/RAD 220 REVIEW SESSION 3 February 6, 2011."— Presentation transcript:
BIOE 220/RAD 220 REVIEW SESSION 3 February 6, 2011
What We’ll Cover Today Review some issues from HW 3 Review important points from each modality What is bright in what modality? Brain development review
Common issues in HW 3 Make sure to use sufficient significant figures when solving math problems. Premature rounding can lead to errors in your final answer When calculating the intensity of reflected ultrasound, account for forward attenuation, reflection amount, then return attenuation When working with dB, add the values. When working with ratios, multiply the values We loosely refer to dB in class with or without the minus sign, but the energy is only decreasing after being transmitted from the transducer, so each effect will be –dB (ratio <1) Bright areas in fetal brain ultrasound: Bone, Choroid Plexus, Sulci Dark areas: Ventricles (CSF), Sinuses (blood), areas far from transducer (attenuation), areas beyond bond (no energy)
MRI – Important topics (page 1) Proton density, M 0, B 0, M z, M xy B 1 field (RF pulse): = M 0 x B 1 Larmor frequency: f 0 = gamma_bar * B 0 = 42.58 MHz/T*B 0 How signal is measured Image contrast: T1, T2, T2* (1/T2* = 1/T2 + 1/T2’) Off-resonance Pulse sequences: TE, TR, TI, what parameters for given contrast, spin echoes vs gradient echoes
MRI – Important topics (page 2) M xy = M 0 (1-exp(-TR/T1))exp(-TE/T2) Gadolinium, and how it affects measurement Magnetic gradients Diffusion-Weighted MRI How we collect multiple lines of k-space, how many slices can we collect at once? (T scan = N y * TR) Noise dependencies: SNR ~ f(TE,TR,T1,T2,M 0 )B 0 (voxel volume)*sqrt(T sampling) Know that bandwidth and sampling time are inversely related
X-Rays: Radiography X-ray energy: E(keV) = 1.24/λ(in nm), c = λ*f (f=3x10 8 m/s) X-ray interactions: Compton Scattering, Photoelectric Effect, (absorption) Linear Attenuation Coefficient (units are cm -1 ) Mass attenuation coefficient * density = LAC Proportional to Z, density, energy Film convention: Dark = high x-ray exposure X-ray resolution (5-10 line pairs/mm for non mammography, better for mammography) X-ray tubes: kVp, mAs, effect of each on the x-ray spectrum and image quality Dangers of x-ray exposure, how much comes from natural vs man-made sources, beam hardening Image formation, magnification Scatter, effect of SPR, ways to improve it Angiography: Iodine contrast, DSA
X-Rays: CT 360 degrees of radiographs are taken and backprojected to form 2d image These are stacked to form a full 3d volume (modern collection actually uses a helix pattern) Filtered backprojection corrects for blurriness of simple recon Units are normalized to water LAC: HU = (mu – mu water )/mu water * 1000 LACs depend on x-ray energy Be familiar with general ranges of HU for various tissues Windows and levels for CT give range of visible contrast Bone window: W = 2500, L = 1000 Soft tissue window: W = 600, L = -100 Contrast is already normalized: C = A - B Typical resolution: 1-2 mm x 1-2 mm in plane X-Ray dose is much higher than for radiograph (why?)
Angiography Iodine contrast for CT, Gadolinium contrast for MRI MRI can also use time of flight non-contrast Understand presaturation Understand flow voids
Ultrasound Transducer transmits oscillating pressure wave (1-10MHz) into body Duration of sine wave is limited to control axial resolution Soft tissue speed of sound is 1540 m/s Air much lower, bone twice as high Speed of sound is independent of applied frequency To image: send pulse, listen to echos, draw that “line”, move to next line and send new pulse At boundaries between different acoustic impedances, energy is reflected: % reflected = (Z 2 -Z 1 ) 2 /(Z 2 +Z 1 ) 2 Soft tissue impedance = 1.62, air =.0004, bone = 7.8 (these numbers should be given to you) Tissue is made up of many small reflectors, these reflect signal to transducer “speckle” results from distribution of micro scatterers Attenuation occurs in both directions, depends on tissue type and depth of penetration Soft tissue is 1 dB/cm/MHz, water is very low, bone is 20 times higher We measure relative sound intensity, generally compared to transmitted energy Relative sound intensity (dB) = 10 * log(I / I 0 ) Time gain control amplifies signal to offset expected attenuation Axial Resolution = Spatial Pulse Length / 2 = (n * lambda) / 2 Lateral resolution = beam width, smaller with increasing frequency for fixed transducer, but higher frequency has less penetration (higher attenuation)
What’s bright? Radiographs: Bright means less film exposure More attenuation has occurred, either from going through more tissue, or higher Z material (bone, metal, etc) CT: Bright means high LAC Cortical bone will always be bright, Iodine is bright because of high Z Ultrasound: Bright means high scattering or reflection Sulci, Bone, Choroid plexus all have high scattering Interface between tissue and air or tissue and bone will have very high % reflection
What’s dark? Radiographs: Dark means more film exposure Less attenuation has occurred, from going through mostly air CT: Dark means low LAC Low density (low Z) materials will be darker, tissues are mostly similar to water, fat is slightly darker Ultrasound: Dark means low scattering or reflection Fluid (CSF, blood) have fewer scatterers than soft tissue Anything beyond a highly reflecting interface will look dark, since little energy reaches it Anything far from the transducer will appear dark, because most of the energy has been attenuated
MRI: What’s bright/dark? MRI: Brightness depends on contrast of scan More “rigid” materials generally have shorter T1 and T2 In proton density map, brightness is similar across many tissues (WM is ~20% darker) In T1 weighting, anything with short T1 will recover signal more quickly and will be brighter -> more “rigid” things will be brighter -> white matter brighter than gray matter In T2 weighting, anything with short T2 will decay more quickly and be darker -> more “fluid” things will be brighter -> CSF is brightest, gray matter is brighter than white matter In a FLAIR sequence, the CSF signal is nulled, and appears dark. Gray matter is brighter than white matter between we still use T2 weighting Fat hasn’t been discussed much yet in lecture, but it has a short T1 and T2, with a higher PD than other tissues Cortical bone has T2 that’s so short it always decays completely before we can make a measurement -> always dark
BRAIN DEVELOPEMENT Review of Brain Lecture 7
Possible test material Neural plate Neural tube Rhombencephalon Mesencephalon Diencephalon Telencephalon Cavum septum pellucidum Anterior fontinelle Posterior fontinelle Caudothalamic groove Germinal matrix hemorrhage Hydrocephalus Anencephaly Holoprosencephaly Agenesis of the corpus callosum What adult structures derive from pre-natal structures? Order of sulcal development Order of myelination of white matter? Where are the fontinelles, what surrounds them, when do they close?
VISUAL AND AUDITORY SYSTEMS Review of Brain Lecture 8
Visual System Retina Optic nerves chiasm optic tracts LGN Visual cortex
Visual System
How do we image the auditory system? Sensorineural: MRI “Retrocochlear lesions” Presence of CN 8 (Auditory nerve) Masses Infarcts Conductive: CT Temporal bone Very high resoluBon, axial reconstruct into coronal plane Visualize Tympanic membrane Ossicles OBc capsule
Possible test material Retina Optic nerve Optic chiasm Optic tracts Lateral geniculate nucleus Optic radiations External auditory canal Tympanic membrane Middle ear ossicles Oval window Cochlea Semicircular canals Internal auditory canal | 1,897 | 7,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-17 | latest | en | 0.808288 |
https://discuss.dizzycoding.com/python-functions-with-multiple-parameter-brackets/ | 1,701,306,700,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.15/warc/CC-MAIN-20231130000127-20231130030127-00515.warc.gz | 256,547,343 | 12,911 | # Python functions with multiple parameter brackets
Posted on
### Question :
Python functions with multiple parameter brackets
I’ve been having trouble understanding what `h(a)(b)` means. I’d never seen one of those before yesterday, and I couldn’t declare a function this way:
``````def f (a)(b):
return a(b)
``````
When I tried to do `def f (a, b):`, it didn’t work either. What do these functions do? How can I declare them? And, finally, what’s the difference between `f(a, b)`and `f(a)(b)`?
Functions with multiple parameter brackets don’t exist, as you saw when you tried to define one. There are, however, functions which return (other) functions:
``````def func(a):
def func2(b):
return a + b
return func2
``````
Now when you call `func()` it returns the inner `func2` function:
``````>>> func2 = func(1) # You don't have to call it func2 here
>>> func2(2)
3
``````
But if you don’t need the inner function later on, then there’s no need to save it into a variable and you can just call them one after the other:
``````>>> func(1)(2) # func(1) returns func2 which is then called with (2)
3
``````
This is a very common idiom when defining decorators that take arguments.
Notice that calling `func()` always creates a new inner function, even though they’re all named `func2` inside of the definition of our `func`:
``````>>> f1 = func(1)
>>> f2 = func(1)
>>> f1(1), f2(1)
(2, 2)
>>> f1 is f2
False
``````
And, finally, what’s the difference between `f(a, b)`and `f(a)(b)`?
It should be clear now that you know what `f(a)(b)` does, but to summarize:
• `f(a, b)` calls `f` with two parameters `a` and `b`
• `f(a)(b)` calls `f` with one parameter `a`, which then returns another function, which is then called with one parameter `b`
`f(a)(b)` just means that the expression `f(a)` returns a value that is itself callable. It’s a short form of
``````g = f(a)
g(b)
``````
You might be more comfortable adding a pair of redundant parentheses to emphasize that this is not a single syntactic construct.
``````(f(a))(b) # f(a) is evaluated first, then the result is applied to b
``````
It is exactly analogous to the same doubling of square brackets for indexing nested dictionaries.
``````d1[x][y]
``````
is equivalent to
``````d2 = d1[x]
d2[y]
``````
Lets say we have an expression like
``````f(a)(b)
``````
then, `f(a)` returns a function itself which gets invoked with argument `b`. Consider the following example
``````def f(a):
def g(b):
return a * b
return g
``````
Then `f(5)(4)` evaluates to `5 * 4`, since `f(5)` returns a function which is basically
``````def g(b):
return 5 * b
``````
One could now do stuff like this
``````mult_by_5 = f(5)
[mult_by_5(x) for x in range(10)]
``````
Let’s be fancy, what about more nested functions?:
``````def f(a):
def g(b):
def h(c):
return a * b *c
return h
return g
f(2)(3)(4) # 24
`````` | 855 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | latest | en | 0.83572 |
https://vitalik.ca/general/2017/03/11/a_note_on_charity.html?ref=hackernoon.com | 1,660,642,901,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00370.warc.gz | 545,592,200 | 5,304 | # A Note On Charity Through Marginal Price Discrimination
2017 Mar 11 See all posts
A Note On Charity Through Marginal Price Discrimination
Updated 2018-07-28. See end note.
The following is an interesting idea that I had two years ago that I personally believe has promise and could be easily implemented in the context of a blockchain ecosystem, though if desired it could certainly also be implemented with more traditional technologies (blockchains would help get the scheme network effects by putting the core logic on a more neutral platform).
Suppose that you are a restaurant selling sandwiches, and you ordinarily sell sandwiches for $7.50. Why did you choose to sell them for$7.50, and not $7.75 or$7.25? It clearly can't be the case that the cost of production is exactly $7.49999, as in that case you would be making no profit, and would not be able to cover fixed costs; hence, in most normal situations you would still be able to make some profit if you sold at$7.25 or $7.75, though less. Why less at$7.25? Because the price is lower. Why less at $7.75? Because you get fewer customers. It just so happens that$7.50 is the point at which the balance between those two factors is optimal for you.
Notice one consequence of this: if you make a slight distortion to the optimal price, then even compared to the magnitude of the distortion the losses that you face are minimal. If you raise prices by 1%, from $7.50 to$7.575, then your profit declines from $6750 to$6733.12 - a tiny 0.25% reduction. And that's profit - if you had instead donated 1% of the price of each sandwich, it would have reduced your profit by 5%. The smaller the distortion the more favorable the ratio: raising prices by 0.2% only cuts your profits down by 0.01%.
Now, you could argue that stores are not perfectly rational, and not perfectly informed, and so they may not actually be charging at optimal prices, all factors considered. However, if you don't know what direction the deviation is in for any given store, then even still, in expectation, the scheme works the same way - except instead of losing $17 it's more like flipping a coin where half the time you gain$50 and half the time you lose $84. Furthermore, in the more complex scheme that we will describe later, we'll be adjusting prices in both directions simultaneously, and so there will not even be any extra risk - no matter how correct or incorrect the original price was, the scheme will give you a predictable small net loss. Also, the above example was one where marginal costs are high, and customers are picky about prices - in the above model, charging$9 would have netted you no customers at all. In a situation where marginal costs are much lower, and customers are less price-sensitive, the losses from raising or lowering prices would be even lower.
So what is the point of all this? Well, suppose that our sandwich shop changes its policy: it sells sandwiches for $7.55 to the general public, but lowers the prices to$7.35 for people who volunteered in some charity that maintains some local park (say, this is 25% of the population). The store's new profit is $$\6682.5 \cdot 0.25+\6742.5 \cdot 0.75=\6727.5$$ (that's a $22.5 loss), but the result is that you are now paying all 4500 of your customers 20 cents each to volunteer at that charity - an incentive size of$900 (if you just count the customers who actually do volunteer, $225). So the store loses a bit, but gets a huge amount of leverage, de-facto contributing at least$225 depending on how you measure it for a cost of only $22.5. Now, what we can start to do is build up an ecosystem of "stickers", which are non-transferable digital "tokens" that organizations hand out to people who they think are contributing to worthy causes. Tokens could be organized by category (eg. poverty relief, science research, environmental, local community projects, open source software development, writing good blogs), and merchants would be free to charge marginally lower prices to holders of the tokens that represent whatever causes they personally approve of. The next stage is to make the scheme recursive - being or working for a merchant that offers lower prices to holders of green stickers is itself enough to merit you a green sticker, albeit one that is of lower potency and gives you a lower discount. This way, if an entire community approves of a particular cause, it may actually be profit-maximizing to start offering discounts for the associated sticker, and so economic and social pressure will maintain a certain level of spending and participation toward the cause in a stable equilibrium. As far as implementation goes, this requires: • A standard for stickers, including wallets where people can hold stickers • Payment systems that have support for charging lower prices to sticker holders included • At least a few sticker-issuing organizations (the lowest overhead is likely to be issuing stickers for charity donations, and for easily verifiable online content, eg. open source software and blogs) So this is something that can certainly be bootstrapped within a small community and user base and then let to grow over time. Update 2017.03.14: here is an economic model/simulation showing the above implemented as a Python script. Update 2018.07.28: after discussions with others (Glen Weyl and several Reddit commenters), I realized a few extra things about this mechanism, some encouraging and some worrying: • The above mechanism could be used not just by charities, but also by centralized corporate actors. For example, a large corporation could offer a bribe of$40 to any store that offers the 20-cent discount to customers of its products, gaining additional revenue much higher than \$40. So it's empowering but potentially dangerous in the wrong hands... (I have not researched it but I'm sure this kind of technique is used in various kinds of loyalty programs already)
• The above mechanism has the property that a merchant can "donate" $$\x$$ to charity at a cost of $$\x^{2}$$ (note: $$x^{2}<x$$ at the scales we're talking about here). This gives it a structure that's economically optimal in certain ways (see quadratic voting), as a merchant that feels twice as strongly about some public good will be inclined to offer twice as large a subsidy, whereas most other social choice mechanisms tend to either undervalue (as in traditional voting) or overvalue (as in buying policies via lobbying) stronger vs weaker preferences. | 1,428 | 6,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.954426 |
https://www.etcnbusiness.com/ebooks/3-k-factor-critical-graphs-and-toughness | 1,555,841,906,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530527.11/warc/CC-MAIN-20190421100217-20190421122217-00287.warc.gz | 673,829,084 | 9,614 | By Shi M., Yuan X., Cai M.
Read Online or Download (3,k)-Factor-Critical Graphs and Toughness PDF
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Additional resources for (3,k)-Factor-Critical Graphs and Toughness
Example text
The synopsis construction is typically defined through either node or edge contractions. The key is to define a synopsis which retains the relevant structural property of the underlying graph. In [7], the algorithm in [177] is used in order to collapse the dense regions of the graph, and represent the summarized graph in terms of sparse regions. The resulting contracted graph still retains important structural properties such as the connectivity of the graph. In [46], a randomized summarization technique is used in order to determine frequent patterns in the underlying graph.
This summarization is then leveraged in order to determine the frequent subgraph patterns from the data. Bounds are derived in [46] on the false positives and false negatives with the use of such an approach. Another challenging variation is when the frequent patterns are overlaid on a very large graph, as a result of which patterns may themselves be very large subgraphs. An algorithm called TSMiner was proposed in [110] to determine frequent structures in very large scale graphs. Graph pattern mining has numerous applications for a variety of applications.
Matsumoto. An Application of Boosting to Graph Classification, NIPS Conf. 2004. [19] J. Leskovec, J. Kleinberg, C. Faloutsos. Graph Evolution: Densification and Shrinking Diameters. ACM Transactions on Knowledge Discovery from Data (ACM TKDD), 1(1), 2007. [20] K. Liu and E. Terzi. Towards identity anonymization on graphs. ACM SIGMOD Conference 2008. [21] R. Kumar, P Raghavan, S. Rajagopalan, D. Sivakumar, A. Tomkins, E. Upfal. The Web as a Graph. ACM PODS Conference, 2000. An Introduction to Graph Data 11 [22] S.
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### (3,k)-Factor-Critical Graphs and Toughness by Shi M., Yuan X., Cai M.
by Paul
4.5
Rated 4.76 of 5 – based on 49 votes | 794 | 3,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-18 | latest | en | 0.908124 |
https://www.w3resource.com/csharp-exercises/basic-algo/csharp-basic-algorithm-exercises-2.php | 1,723,468,904,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00509.warc.gz | 817,688,081 | 27,649 | C# - Get the absolute difference between n and 51
# C#: Get the absolute difference between n and 51
## C# Sharp Basic Algorithm: Exercise-2 with Solution
Write a C# Sharp program to get the absolute difference between n and 51. If n is broader than 51 return triple the absolute difference.
Visual Presentation:
Sample Solution:
C# Sharp Code:
``````using System;
// Namespace declaration
namespace exercises
{
// Class declaration
class Program
{
// Main method - entry point of the program
static void Main(string[] args)
{
// Calling the 'test' method and displaying the returned values
Console.WriteLine(test(53)); // Output: 6
Console.WriteLine(test(30)); // Output: 21
Console.WriteLine(test(51)); // Output: 0
Console.ReadLine(); // Keeping the console window open
}
// Method to perform a calculation based on a given integer 'n'
public static int test(int n)
{
const int x = 51; // Constant value x assigned as 51
// Checking if 'n' is greater than x
if (n > x)
{
// Returning the result of (n - x) multiplied by 3
return (n - x) * 3;
}
// Returning the absolute difference between x and n
return x - n;
}
}
}
```
```
Sample Output:
```6
21
0```
Flowchart:
C# Sharp Code Editor:
Improve this sample solution and post your code through Disqus
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
| 344 | 1,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.657768 |
https://alumniagri.in/task/find-the-diagonal-of-the-rectangle-whose-length-is-35cm-and-9929517 | 1,701,814,864,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00543.warc.gz | 124,052,548 | 4,471 | Math, asked by vaishalichaudhari, 13 days ago
Find the diagonal of the rectangle whose length is 35cm and breadth is 12cm
5
420
by pythagoras theoram
we know a^2 + b^2 =c^2
144 + 35^2=420^2
12
the length of a diagonal is 37 cm
Explanation
12 ^2 + 35^2 = diagonal^2
144 + 1225 =1369= 37^2
Diagonal= 37cm
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Social Sciences, 2 months ago | 174 | 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} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.76597 |
http://www.instructables.com/id/Fused-Pyramid/ | 1,527,412,178,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00568.warc.gz | 408,264,296 | 12,000 | # Fused Pyramid
621
3
2
Published
## Introduction: Fused Pyramid
The Cheops pyramid in Egypt is a stack of 201 layers of stones. What could be more natural than to use Fused Filament Fabrication, since the inherent drawback of visible layers, in this case becomes an asset, as this makes the print more realistic. My work consists of studying how to set the slicing parameter (layer thickness, scale), taking into account the printer constraints, (bed size) in order to print the pyramid in a way that corresponds with the actual thickness of the layers of the real Cheops pyramid.
## Step 1: Model Construction
In order to modelize the thickness of the stone layers, I used the study from the French archeologist Georges Goyon entitled "Les rangs d'assises de la Grande Pyramide" BIFAO 78 (1978), p. 405-413.
There are 201 physical layers ranging between 49.5cm for the thinnest layer no. 173 to 150cm to the thicker layer no.1. I therefore introduced 201 boxes into Blender, each of them having the measure given by the report at a scale of 1/10000. Then I stacked the boxes along the z-axis. The X-Y dimensions of each box were set such to follow the slope of the apothem. (14/11/) and the first box (base of the pyramid) measures 230.35m. Thus, at the end, I obtained a model which has the same number of boxes as the physical layers of the pyramid. To finalize the model I merged all the boxes (modifier "union" in Blender) together to obtain a single watertight shell that passed the 3D toolbox of Blender.
## Step 2: Slicing the Model
Here comes the main target of this work which attempts to reproduce the layers of the pyramid using the printed layer height set into the slicer.
Let's find now the scale and the printing layer thickness that make a print which has the same number (n) of layers as the pyramid (201) and that fits inside the printing bed. The printer I use is a PrusaI3 with a heated bed which has a print area of 18cm by 18cm.
First of all I took the average stone thickness (l) of 69cm given by the study of G.Goyon and made it correspond with the printing layer height (l'). The printed layer height l' and the total print height z' of the pyramid are linked by the relationship z'=n x l' . For instance with a layer height of 300u, we will get a print which is 201x0.3mm= 60.3mm which is a good dimension to print.
Then, since 0.3mm (300u) corresponds with 690mm (69cm), the scale is 0.3/690=4.35/10000. Therefore we can calculate the printed base lenght as the real length 230.35m multiplied by the scale 230.35 x 4.35/10000=0.1102m or 10.02cm. Again a good dimension for most FFF printers.
Finally I used Cura to slice the model, with walls of 1.2mm (3x noozle size of 0.4mm)) and an infill of 10%. I added a brim of 30 loops around in order to prevent warping and obtain a nice flat base. One can see on the screenshots of Cura that the scale of 4.35 and a layer height of 300u ends up with exactly 201 printed layers with the dimensions calculated above.
## Step 3: Printing the Model
As already mentionned , I have used a PrusaI3 for that print. This open source printer is quite accurate and thanks to the heating bed allows a good adhesion of the print. I took a PLA filament from ColorFabb which was the nearest color to the pyramid that I found on my shelf and set the printing temperature to 220°C for the filament and 60°C for the bed. The print time is 3hours15 and the print weight is 60grams.
## Step 4: Print Results
Here we are.
The overall impression of the pyramid geometry is well reproduced by the print. The print layer number being the same as the real stone layer number brings an additional touch of reality, which in my opinion is more authentic than a pyramid made of flat surfaces. Thanks to this feature, the print can be used as a teaching support to illustrate how the external shell of the pyramid was made.
On the print quality side, one can see that the pyramid has a little warp on one corner, because the printed lines of the brim were not touching closely to each other. As a consequence, the print started to lift up because the brim was not sticking well enough. A bed levelling to bring the nozzle closer to the bed allowed to me to resolve this issue, and there was no more warping.
## Step 5: Conclusion
This work is an attempt to reproduce the pyramid construction using the inherent drawback of the FFF technology where the printed layer thickness is in relationship with the average thickness of the real stone layer of the pyramid. A further study would be to exactly emulate each of the stone layers by a dedicated printed layer height. This may be possible by editing the GCODE such that the layer thickness changes during the print. I feel that the project succeeded in using the FFF drawbacks to give an authentic and textured surface that corresponds with the actual appearance of the Pyramid of Cheops.
## Recommendations
• ### 3D Printing With Circuits Class
4,249 Enrolled
• ### Make it Move Contest
We have a be nice policy. | 1,207 | 5,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.910746 |
https://math.stackexchange.com/questions/3045520/definition-of-renaming-in-lambda-calculus | 1,558,538,392,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256858.44/warc/CC-MAIN-20190522143218-20190522165218-00315.warc.gz | 534,632,232 | 30,974 | Definition of renaming in lambda calculus
From "Type Theory and Formal Proof, An Introduction" by Rob Nederpelt and Herman Geuvers:
Definition 1.5.1 (Renaming; $$M^{x \to y}$$; $$=_{\alpha}$$) Let $$M^{x \to y}$$ denote the result of replacing every free occurrence of $$x$$ in $$M$$ by $$y$$. The relation 'renaming', expressed with symbol $$=_{\alpha}$$, is defined as follows: $$\lambda x . M =_{\alpha} \lambda y . M^{x \to y}$$, provided that $$y \notin FV(M)$$ and $$y$$ is not a binding variable in $$M$$.
$$FV$$ denotes the set of free variables of a $$\lambda$$-term.
Are the conditions $$y \notin FV(M)$$ and $$y$$ is not a binding variable in $$M$$, the same as requiring that $$y$$ does occur in $$M$$? My understanding is that an occurrence of a variable can be either free, bound, or binding. If an occurrence of $$y$$ is bound in $$M$$, does this not also require that $$y$$ is a binding variable in $$M$$? | 273 | 925 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-22 | latest | en | 0.858454 |
https://physics.stackexchange.com/questions/374590/owen-biesels-paper-on-the-precession-of-mercury-s-perihelion?noredirect=1 | 1,585,866,425,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00443.warc.gz | 633,894,590 | 31,690 | # Owen Biesel's paper on 'The Precession of Mercury’s Perihelion'
I have seen this paper was referred to this question How do you calculate the anomalous precession of Mercury? . I am also writing a project paper on Mercury's Perihelion Precession. I will use the calculation for the precession used in this paper. But there are some steps in the paper that i did not understand. I do not have enough reputation to comment on that question.So i am writing down the parts that i did not get. Any help will be appreciated.
(1)(page 9 in the paper)
As you can see there are two terms in the final equation. But when they plugged in the observational values. they only took $2.515*10^7$ but did not use $4.88*10^{-15}$ . WHY?
(2)(Page 4) I tried to calculate this integration on Mathematica.But gives different result.
(3)(page 8)I did not get the how the Taylor series expansion led to the equation which is marked in red arrow (1st arrow). also the integration which is marked in red arrow.(2nd arrow).
## 1 Answer
As far as I can tell from the text, $4.88 \times 10^{-15}$ is the upper bound on the second term in the equation for $\phi_+ - \phi_-$. This is a very small number, meaning the first term alone, $2.515 \times 10^{-7}$, is a good approximation for the equation, and hence the value of $\phi_+ - \phi_-$ as a whole.
In the case of the first arrow, simply replace $1- \left( \frac{\epsilon}{r} \right)^{-1/2}$ with $1+\epsilon/2r$ in the integration.
In the case of the second arrow, integrate the second term between $R_+$ and $R_-$ with respect to $r$.
• One of my friend told me the exact same thing yesterday. Thanks ,man. – user1157 Dec 17 '17 at 17:26 | 448 | 1,677 | {"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} | 3.109375 | 3 | CC-MAIN-2020-16 | latest | en | 0.933741 |
https://stage.geogebra.org/m/xBg4a4YQ | 1,716,975,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00311.warc.gz | 470,658,509 | 27,902 | Google Classroom
GeoGebra Classroom
# Ellipse (Graph & Equation Anatomy)
Move the various sliders in this applet around to investigate what happens to the graph of an ellipse as you change various parameters within the standard form of its equation. Key questions: Is it ever possible for the graph of an ellipse to become a circle? If so, under what condition(s) will this happen? Is it possible to tell the location of an ellipse's center just by looking at its equation? If so, how? How can the length of an ellipse's semimajor & semiminor axes be determined just by looking at its equation? What must happen to ensure that the graph's major axis is horizontal? Vertical? What must happen to ensure that the graph's minor axis is horizontal? Vertical?
Questions are located above the applet. | 169 | 796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.950105 |
https://ww2.mathworks.cn/matlabcentral/profile/authors/2608514 | 1,675,075,994,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00532.warc.gz | 642,876,698 | 26,214 | Community Profile
# Pink_panther
Engineering
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totofufu.github.io | 1,556,186,185,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578716619.97/warc/CC-MAIN-20190425094105-20190425120105-00013.warc.gz | 175,120,205 | 5,558 | ## Summary
We implemented a parallel SAT solver on the Andrew Unix machines capable of accepting any number of variables, clauses, and clause length. Our parallel implementation proved to be faster than our serial version, and we also created a random SAT expression generator to test large inputs quickly. We ultimately achieved a 4-5x speedup on our largest graph containing 100,000 variables and 100,000 clauses with a maximum clause length of 10,000.
## Background
The k-SAT problem involves taking an expression in conjunctive normal form (CNF) and determining if it is possible to assign boolean values to every variable such that the CNF evaluates to true. For example, the expression ((x1 or ~x2) and (x3 or x4)) is satisfiable because we can assume all 4 literals to true, and the CNF will evaluate to true, but the expression ((x1 or ~x2) and (x2) and (~x1)) is not satisfiable because there is no assignment of boolean values that will make this CNF evaluate to true.
Solving the k-SAT problem is a common NP-complete problem, and brute force algorithms normally run in O(2^n) time, although simplifications can be made. The entire problem space is modeled as a decision tree, where each node splits into two more children by considering both true or false values for a specific variable.
We created a brute-force algorithm that attempts every single possible solution to an input SAT expression. We specify that an input SAT expression must consist of variables enumerated from x1, x2, x3, etc., and the negations of variables to create other literals are also valid. Our output will specify whether the input expression is satisfiable (in which case the satisfiable solution/s will be outputted) or whether the input expression has no solution (and isn’t satisfiable). We believe that in the sequential version, the computationally expensive operations lie in the sheer number of possible solutions that must be attempted, which is upper-bounded by 2^num_vars each time, where num_vars is the number of variables in the input expression.
After this initial brute-force attempt, we decided to look through some algorithms specifically targeted towards solving SAT expressions. We landed on the Davis-Putnam-Logemann-Loveland (DPLL) algorithm. The DPLL algorithm describes a recursive strategy to brute force every possible solution but also eliminate any definitive clauses and simplify the expression further so that some paths can be ruled out. Specifically, the algorithm assigns the first variable (which we’ll call x1) a value of true and transforms the SAT expression by getting rid of any clauses with x1 (since that entire clause must evaluate to true) and eliminating ~x1 (negation of x1) within any other clauses. It recurses on the remaining variables by doing the same thing for x2, etc. At the very end, if we have all variables that are of the same polarity (eg. only x1, x2, x3, etc. and no negations), then we know that this expression must be satisfiable because we can assign a truth value of true to every single remaining variable without conflict with possible literals that might be negations. If we’re left with only empty clauses however, then we know that all the variables were eliminated because their value was false, so this SAT expression is not satisfiable.
The DPLL approach also specifies Unit Propagation (UP) and Pure Literal Elimination (PLE). Unit Propagation states that if a clause has length 1, then the literal inside that clause can only have a truth value of true, so there’s no need to try assigning false to these literals. This eliminates a good portion of the decision tree. Pure Literal Elimination states that a variable that only occurs with one polarity is pure and can always be assigned a truth value of true. This also eliminates a good portion of possible solutions and can significantly reduce the size of our decision tree. Note that both UP and PLE are optimizations that can be performed in each recursive call because the SAT expression is constantly being simplified, and so clauses can easily reduce to size 1, which UP would take advantage of, and variables can also become pure after simplification, which PLE would take advantage of.
In terms of the parallelism that we could take advantage of, we felt that it was the sheer number of possible solutions that the solver had to attempt that made it run extremely slowly. The dependencies in the solver primarily focused on creating the decision tree top-down. Specifically, none of the child nodes could be created until the parent node was created because the later representations built on the earlier ones. In addition, we could cache the representations of the possible solutions because many of the expressions could simplify to similar representations. For example, the SAT expression ((x1 or x2) and (x3)) has the same representation when x1 = T, x2 = T, x3 = F and when x1 = T, x2 = F, x3 = F because the first clause is entirely gone from both representations. Through caching, we felt that we would be able to memoize the results and speed up computation.
## Approach
For our initial brute-force algorithm, we brute force every single possible solution by using pthreads to speed up the computations: specifically, we keep some shared variables between the pthreads and constantly check to see if any of the attempted solutions have been satisfiable and terminate all threads if we find a satisfiable solution. Otherwise, we run through all possible solutions and report that there is no solution found.
Our sequential DPLL solution involves representing each subproblem as a clause map. This representation maps the index of each clause to a set of its current literals (which changes throughout the decision tree). In addition, we also keep a vars map that maps each variable to which indexed clauses the variable (and its negation) appears in. This map is always constant and helps us pinpoint variable locations in constant time instead of having to search through the entire clauses map each time.
Our parallel approach primarily involved figuring out a clever way to assign threads to the possible solutions in the decision tree. We initially started off with a single work queue from which threads would dequeue a representation of a possible solution (essentially a node from our decision tree) and generate the children of this node. We represent every node as a "clause map", which is essentially just a modified version of the original SAT expression that we are testing for satisfiability. From this, each thread can then generate child nodes and enqueue more modified clause maps onto the work queue. This essentially created a BFS-style search algorithm. We decided to change it to a DFS-style search algorithm because with a BFS search, the solver would only ever terminate once we hit the very last level of the decision tree. With a DFS-style approach however, we would hit the leaf nodes faster and thus have a likelier chance of terminating sooner. Even in the worst case with DFS, where the termination condition would be at the later leaf nodes that the solver checked, BFS would have an approximately equal solution time.
We also implemented caching by memoizing the representations of the clauses at each node and saving the boolean value of the clause map in case later solutions also reached the same clause map, as explained in the “Background” portion above. This essentially helped remove large subtrees by eliminating all possible children nodes from a cached node.
Another optimization we implemented was targeting maximal simplification of a SAT representation. Whenever we assign a truth value to a variable, we eliminate all clauses containing this variable and simplify the starting SAT expression significantly. We initially assigned truth values to variables in numerical order, but we decided to instead sort the variables by decreasing frequency and start off by assigning truth values to the variables with the highest frequency of appearances in the SAT expression. This reduces the SAT expression by quite a large amount and searches a smaller number of subtrees first to more quickly reach the leaf nodes in case the satisfiable solution can be reached there. We essentially optimize for as simplified a solution as we can possibly get so that UP and PLE can reduce the expression even further.
Lastly, we used C++ for our programs and targeted the Andrew Linux servers.
## Results
We display the results of our DPLL sequential, baseline implementation on our largest graphs, comparing them against the time it took against our DPLL multi-threaded implementations. We decided to omit the brute force baseline implementation because it would've taken much, much longer (on the scale of hours) to finish running against our largest graphs.
We created a random SAT expression generator that takes in a variable number of variables, clauses, and max clause length. The first graph below is run on a randomly generated file containing 10,000 variables and 10,000 clauses with a max clause length of 1,000. The second graph below is run on a randomly generated file containing 100,000 variables and 100,000 clauses with a max clause length of 10,000. The second file is clearly much larger than the first file. The "Base" times below are for our sequential DPLL algorithm, the "Multi-Caching" uses multi-threading and caching, and the "Multi-Sorting" uses multi-threading with sorting frequencies.
As you can see below, the multi-threaded implementations are only a few seconds faster than the DPLL implementation for the first graph, but on the second graph, the multi-threaded implementations achieve a 4-5x speedup. Because the second graph describes a much larger file (112 MB!), we believe that the effects of our multithreaded implementation scale better over a larger problem set and result in a larger speedup.
## References
DPLL Algorithm:
https://en.wikipedia.org/wiki/DPLL_algorithm
DFS:
Parallel SAT Solver paper (2012):
Parallel Implementations of Backtracking Algorithms:
http://rsim.cs.illinois.edu/arch/qual_papers/systems/7.pdf
## List of Work
Ashley’s work:
Random expression generator
DPLL sequential algorithm
DPLL caching
Eileen’s work:
Brute force SAT solver
DPLL parallel trees
Final writeup / website management | 2,078 | 10,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.871223 |
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# 2 Digit by 2 Digit Multiplication using Partial Products {Fall Theme}
Resource Type
Common Core Standards
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PDF (Acrobat) Document File
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Product Description
This resource includes 2 digit by 2 digit multiplication practice pages in a beautiful fall theme! Two different versions of each practice page are included, so that students can solve using flexible strategies or the traditional algorithm. Color and black and white pages are included.
★Page 4: Multiplying 2-digit by 2-digit anchor chart (color clip art)
★Pages 5-22: 2-digit by 2-digit practice pages and answer keys (includes a partial product box and open box format for each page)
★Pages 23-26 Fall and non-seasonal multiplication dice roll pages- can be used at math stations or centers (students write their own equations).
******************
★Pages 27 - 49: Black and white clip art versions of all pages
Please check the preview to see everything that is included!
Please note: The multiplication worksheets are the same for both the color and black and white clip art. I included both so you could choose. There are a total of 6 unique multiplication practice pages in two formats (partial product and open boxes), the corresponding answer keys, and 1 anchor chart in this set.
This set also includes two dice roll pages that can be used at math stations or centers. Students roll dice and create their own equations. These pages may be used over and over and are great for homework or extra practice.
The number sentences on each page are written horizontally with work space beneath for students to show work and solve. I teach my students to multiply using partial product boxes (also called area method) before moving on to the traditional algorithm. This method is a natural transition from using manipulatives or grid paper and has the added advantage of continued practice with decomposing numbers and using the distributive property. I have included a mini anchor chart demonstrating the method for multiplying using partial products.
You may have several students using the partial product boxes and others using the traditional algorithm. Each worksheet comes in two formats- with partial product boxes and with open boxes (without lines). Use the open box for students using the traditional algorithm as it has more vertical space for students to work. Students will rewrite the problem vertically (practicing correct place value alignment) before solving.
Please let me know if you have any questions!
Thank you!
-The Knitted Apple
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You may also be interested in:
2 Digit by 2 Digit Multiplication {Christmas Theme}
2 Digit by 2 Digit Multiplication {Winter Theme}
2 Digit by 2 Digit Multiplication {Spring Theme}
2 Digit by 2 Digit Multiplication {Summer Theme}
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\$3.00 | 721 | 3,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.897417 |
https://www.tutorpace.com/algebra/variables-in-math-online-tutoring | 1,550,876,405,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249406966.99/warc/CC-MAIN-20190222220601-20190223002601-00498.warc.gz | 1,013,904,185 | 10,903 | # Variables in Math
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Variables are the letters or alphabets like x, a b, c, y, p, t etc. and they are predominantly used in math almost everywhere. These variables are especially useful in algebraic expressions and in solving algebraic equations. The significance of variables is due to the reason that variables can be assigned any number according to the given question and this helps us evaluate the values of the algebraic expressions in different cases. Therefore variables in math are very helpful and play a significant role in math calculations.
Example 1: Solve the given equation to find the value of the variable, 7a + 4 = -10
Given equation: 7a + 4 = -10
In the above equation, the variable is the alphabet ‘a’ since variables are always represented by letters.
Solving the equation means to find the value of ‘a’ in the given equation.
7a + 4 = -10 -> 7a = -10 – 4 -> 7a = -14
This gives: a = -14/7 ->a = -2.
Hence the value of the variable, a = -2.
Example 2: Solve the given equation to find the value of the variable, 6x – 5 =19
Given equation: 6x – 5 = 19
In the above equation, the variable is the alphabet ‘x’ since variables are always represented by letters.
Solving the equation means to find the value of ‘x’ in the given equation.
6x – 5 = 19 -> 6x = 19 + 5-> 6x = 24.
This gives: x = 24/6 -> x = 4
Hence the value of the variable, x = 4. | 399 | 1,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.889442 |
https://www.electricalvolt.com/low-reactance-grounding/ | 1,716,782,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00512.warc.gz | 637,699,920 | 23,629 | Low Reactance Grounding
Adding inductive reactance from a neutral point to the ground is called reactance grounding. The inductive reactance is called the reactor.
Using reactance it is possible to limit the three-phase fault current of in order of kA to a relatively low value in order of 200-800 ampere. Even after adding low reactance to neutral, it is not ensured that whether it is reactance grounding or not. The following criteria must be fulfilled to ensure it is reactance grounding.
• At the time of the fault, the ratio of zero sequence impedance to positive sequence impedance is greater than 3.
Xₒ/X1 > 3
• After the fault, the reactance of the system and the reactance added to neutral may have resonance during the fault. To avoid this, the ratio of zero sequence impedance to positive sequence impedance must be equal to or less than 3.
Xₒ/X1 ≤ 3
• During fault clearing, the ratio of zero sequence impedance to positive sequence impedance is greater than 3.
Xₒ/X1 > 3
In a synchronous generator, the ground fault currents are higher than the three-phase current at the terminals. The high fault current cause excessive heating in the generator winding. To limit the fault current low reactance in the neutral circuit is added,
In transmission and distribution systems, without directly connected rotating machines, the neutrals of the transformers are usually effectively grounded.
The phase-to-ground fault current should be in the range of 25% to 100% of the three-phase fault current. ground fault current of less than 25% of the three-phase current may cause damage due to transient overvoltages. Choose the value of reactance needed to limit the ground-fault current to the preferred amount.
The ratio is
• Xₒ/X1 = 10 when 25%
• Xₒ/X1 = 1 when 100%
• Xₒ/X1 = 3 when 60%
Limiting the ground-fault current to 60% of the three-phase fault current is the borderline between effective grounding and reactance grounding.
When installing a 100% reactor in a generator neutral, the system is not reactance grounded but effective grounded, by definition, and the maximum fault-current contribution of this generator to a line-to-ground fault anywhere in the system outside of the generator will be its three-phase fault current.
Cons of Reactive grounding
The inductance connected between the neutral and ground may resonate with the shunt capacitance to the ground under arcing ground fault conditions. Under this condition, it creates very high transient overvoltage in the electrical power system. This is similar to the phenomenon that happens with ungrounded systems. To avoid this, at least 60 % of the three-phase current must flow through the reactor. If the three-phase current is 20,000A, 6000 amperes current must flow through the reactor. Thus, very high current flow through the reactor to avoid the condition of transient overvoltage, and therefore reactance grounding is rarely used. | 632 | 2,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.909154 |
https://discourse.openredstone.org/t/shinrods-builder-application/3123 | 1,708,658,244,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00538.warc.gz | 219,187,348 | 6,242 | # Shinrod's Builder Application
Shinrod
### What do you like the most about redstone?:
Be able to build whatever you want
### What does the thing do?:
It adds up to 16 signals in hex and outputs two lines X and Y such as SUM(inputs) = 16*X + Y.
For two inputs A and B, it works as follows :
``````if A + B < 16:
X = 0
Y = A + B
# implemented as Y = 15 - (15 - A - B)
else:
# A + B > 15
X = 1
Y = A + B - 16
# implemented as Y = B - (15 - A) - 1
``````
### Image(s) and/or video(s) of the device:
1 slice of the adder :
The inputs are the barrels with green numbers (the two barrel on the middle left).
The top part (orange) computes Y when A + B < 16.
The bottom part (blue) computes Y when A + B > 15.
The yellow line powers on if A + B > 15. It turns off the Top Y and put X to 1.
The green line inject Y back in the system for the next layers. (I didn’t include the comparator line here for clarity. It is represented by the green glass).
Here, A = 13 and B = 8. 13+8 = 21 = 16*1 + 5. So X = 1 and Y = 5.
1. Here is an example with A = 3 and B = 8 (which gives X = 0 and Y = 11):
1. And here is an edge case with A = 13 and B = 3, so that A+B = 16 (giving X = 1 and Y = 0).
1. And here is the full adder, with slices combined:
The inputs are the barrels of the middle column.
1. And the results for the sum of the range [1, 15] (sum= 120 = 16*7 + 8, giving X = 7 and Y = 8) :
I reproduced the first slice on my plot, you can visit if you want to see it: `/p visit Shinrod`
### What do you plan on making for your build trial?:
I have heard there is a lot of ALU, so let’s settle for that. I have a design with SUM, AND, OR, XOR, left/right shift and negations of the output (NAND, NOR, NXOR…).
### Do you agree with the rules?:
Yes
This application has been accepted! Whenever both you and a staff member are free, feel free to ask them for a trial. You are able to try again after failing and waiting 24 hours. It is always recommended to do a practice trial with another member before starting your real one and to practice the questions found at ORE Binary Quiz to help prepare for some of the trial questions.
1 Like | 642 | 2,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.931745 |
https://discourse.processing.org/t/the-part-that-threw-me-off-so-that-as-it-continues-to-increase/10048 | 1,656,411,304,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00091.warc.gz | 252,591,528 | 6,447 | # The part that threw me off, "..., so that as it continues to increase,..."
Not sure if I’m not understanding the subject or if this sentence was not the best way to communicate behavior?
Page 214 of “Getting Started with p5.js”
The author communicates the arc behavior as such:
“If it has, we set the value of x to a negative value, so that as it continues to increase, it will enter the screen from the left.”
The part that threw me off, “…, so that as it continues to increase,…”
I don’t see x continuing to increase? It seems to me that x is reset to -40 after “… x has increased beyond the width of the screen (plus the radius of the shape).”
If I’m correct, maybe a better way of communicating this behavior:
“If it has, we reset the value back to -RADIUS so that arc will again enter the screen from the left.”
Kind regards,
Can you post more context here? Not everybody has the book that you’re talking about.
1 Like
Kevin,
Maybe I posted in the wrong category? This post was meant for authors of book and those reading the book.
But I’m happy to give more details: here’s a screen capture of pages with example: https://www.screencast.com/t/y4ASOsVBJdL
Kind regards,
Not an author, but I’ll mention that the code and passage in question is:
``````var radius = 40;
var speed = 0.5;
function setup() {
createCanvas(240, 120);
}
function draw() {
background(0);
x += speed; // Increase the value of x
if (x > width+radius) { // If the shape is off screen
x = -radius; // move to the left edge
}
}
``````
On each trip through draw(), the code tests to see if the value of x has increased beyond the width of the screen (plus the radius of the shape). If it has, we set the value of x to a negative value, so that as it continues to increase, it will enter the screen from the left. See Figure 8-1 for a diagram of how it works.
No matter what, x will always continue to increase, because of the first line in draw:
``````x += speed; // Increase the value of x
``````
Literally, every time draw `continue`s for another frame, x increases, regardless of what value it is set to. “Continue” is what draw does as it loops (and is also a keyword in JavaScript that means “go on to the next iteration of the loop”) https://www.w3schools.com/js/js_break.asp. I wonder if the confusion might be that you were thinking of “as it continuously increases” – that is, in an unbroken and uninterrupted way, which x does not do (it jumps around, but keeps on marching up from where-ever you set it).
Not sure if we’re getting into semantics here, or I’m still not clear. Either way, I’m happy to discuss.
Okay, got some pad and paper out and started to follow the code step-by-step.
I did some multiplication and see that on iteration 214, x = 321, which is greater than width+radius. Therefore x is assigned the value -40. On iteration 215, x = -38.5, and so on…
So, how is x “increasing”? The value of x went from 321 to -40 on iteration 214. That doesn’t look like an increase to me
Thanks for help.
Yes, this sounds like it might be semantic.
Still, in case it is conceptual: A balloon is full of helium. The helium causes it to rise. When it touches the ceiling we bump the balloon down below the table. The balloon was rising before we bumped it down. It will continue to rise after we bumped it, because it is still full of helium. As it continues to rise, it will rise above the edge of the table. In fact no matter how many times we bump it down, it will continue to rise – we have been bumping it down all afternoon and it has been continuing to rise all afternoon. The important point to understand about a helium balloon is not it’s position, it is that it imparts a continuous rising action on whatever it happens to be doing.
Put another way, there is a force (x+=) applied, which we understand acts to increase the value – independent of other forces which are also applied. Put another way, in the code we are talking about velocity, not position. We can make an instantaneous change to the position, but the velocity remains positive.
Yes, I see that += is always increasing. I’m still not sure I would have articulated the behavior as “increasing”, but I totally see your point.
More importantly I learned something valuable, due to poignant analogies and metaphors. Very good teaching method.
Muchas gracias!
P.S. I’d also like to add that this p5.js community is great. The people are helpful, talented, generous with their time, and kind.
1 Like | 1,066 | 4,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-27 | latest | en | 0.911486 |
https://www.physicsforums.com/threads/partial-derivatives.304766/ | 1,508,322,655,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00809.warc.gz | 1,028,192,928 | 15,226 | # Partial derivatives
1. Apr 3, 2009
### Pietair
1. The problem statement, all variables and given/known data
If z = 1 / (x^2+y^2-1)
show that x(dz/dx)+y(dz/dy)=2z(1+z)
2. The attempt at a solution
z = (x^2+y^2-1)^-1
dz/dx = -2x(x^2+y^2-1)^-2 = -2x * z^2
dz/dy = -2y(x^2+y^2-1)^-2 = -2y * z^2
(-2x^2 * z^2) - (2y^2 * z^2) = 2z(1+z)
I can express x and y in something like z and x/y:
x = z^-1 - (z^-1*y^2)
y = z^-1 - (z^-1*x^2)
Though substituting this values in the obtained equation doesn't get me near the answer...
2. Apr 3, 2009
### HallsofIvy
Staff Emeritus
Yes, exactly right. Why would you want to do the following?
3. Apr 3, 2009
### Pietair
Because I have no idea how to get rid of the x and y...
4. Apr 3, 2009
### n!kofeyn
There's a negative sign missing in the formula you need to show. It should be
$$x \frac{dz}{dx} + y \frac{dz}{dy} = -2z(1+z)$$
5. Apr 4, 2009
### n!kofeyn
It sounds like you're trying to get x and y in terms of z once you have found the partial derivatives. Leave the partials in terms of x and y, and then expand out the -2z(1+z) in terms of what z equals. This will help you verify the formula (please see my previous post for the correction to the problem statement). | 452 | 1,223 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.840862 |
https://www.studentlance.com/solution/strayer-acc560-acc560-week-8-quiz-ch-11-latest-2015 | 1,521,554,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00449.warc.gz | 881,343,075 | 7,218 | # Strayer Acc560/Acc560 Week 8 Quiz Ch 11 (Latest 2015) - 93870
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Question Multiple Choice Question 129 Dillon has a standard of 1.5 pounds of materials per unit, at \$6 per pound. In producing 2,000 units, Dillon used 3,100 pounds of materials at a total cost of \$18,135. Dillon's materials price variance is \$1,050 F. \$135 U. \$465 F. \$600 F. Multiple Choice Question 81 Scorpion Production Company planned to use 1 yard of plastic per unit budgeted at \$81 a yard. However, the plastic actually cost \$80 per yard. The company actually made 3,900 units, although it had planned to make only 3,300 units. Total yards used for production were 3,960. How much is the total materials variance? \$3,960 F \$48,600 U \$4,860 U \$900 U Multiple Choice Question 51 Standard costs may show past cost experience. help establish expected future costs. all of these. are the budgeted cost per unit in the present. Multiple Choice Question 68 Allowance for spoilage is part of the direct labor price standard. labor quantity standard. materials price standard. materials quantity standard. Multiple Choice Question 99 A company uses 8,400 pounds of materials and exceeds the standard by 300 pounds. The quantity variance is \$1,800 unfavorable. What is the standard price? \$2 \$4 Cannot be determined from the data provided. \$6 Multiple Choice Question 152 The balanced scorecard approach evaluates performance using about 10 different perspectives in order to effectively incorporate all areas of the organization. uses rather vague, open statements when setting objectives in order to allow managers and employees flexibility. normally sets the financial objectives first, and then sets the objectives in the other perspectives to accomplish the financial objectives. uses only financial measures to evaluate performance. Multiple Choice Question 78 Hofburg’s standard quantities for 1 unit of product include 2 pounds of materials and 1.5 labor hours. The standard rates are \$2 per pound and \$7 per hour. The standard overhead rate is \$8 per direct labor hour. The total standard cost of Hofburg’s product is \$14.50. \$26.50. \$22.50. \$17.00. Multiple Choice Question 54 If a company is concerned with the potential negative effects of establishing standards, it should set loose standards that are easy to fulfill. not employ any standards. set tight standards in order to motivate people. offer wage incentives to those meeting standards. Multiple Choice Question 56 Ideal standards are rigorous but attainable. reflect optimal performance under perfect operating conditions. will always motivate employees to achieve the maximum output. are the standards generally used in a master budget. Multiple Choice Question 122 Shipp, Inc. manufactures a product requiring two pounds of direct material. During 2013, Shipp purchases 24,000 pounds of material for \$99,200 when the standard price per pound is \$4. During 2013, Shipp uses 22,000 pounds to make 12,000 products. The standard direct material cost per unit of finished product is \$8.00. \$8.53. \$8.27. \$9.01. Multiple Choice Question 113 Monster Company produces a product requiring 3 direct labor hours at \$16.00 per hour. During January, 2,000 products are produced using 6,300 direct labor hours. Monster’s actual payroll during January was \$98,280. What is the labor quantity variance? \$2,520 F \$4,800 U \$4,800 F \$2,280 U Multiple Choice Question 103 Edgar, Inc. has a materials price standard of \$2.00 per pound. Six thousand pounds of materials were purchased at \$2.20 a pound. The actual quantity of materials used was 6,000 pounds, although the standard quantity allowed for the output was 5,400 pounds. Edgar, Inc.'s materials price variance is \$120 U. \$1,200 U. \$1,200 F. \$1,080 U. Multiple Choice Question 105 Edgar, Inc. has a materials price standard of \$2.00 per pound. Six thousand pounds of materials were purchased at \$2.20 a pound. The actual quantity of materials used was 6,000 pounds, although the standard quantity allowed for the output was 5,400 pounds. Edgar, Inc.'s total materials variance is \$2,400 F. \$2,520 U. \$2,520 F. \$2,400 U. Multiple Choice Question 73 Oxnard Industries produces a product that requires 2.6 pounds of materials per unit. The allowance for waste and spoilage per unit is .3 pounds and .1 pounds, respectively. The purchase price is \$2 per pound, but a 2% discount is usually taken. Freight costs are \$.10 per pound, and receiving and handling costs are \$.07 per pound. The hourly wage rate is \$12.00 per hour, but a raise which will average \$.30 will go into effect soon. Payroll taxes are \$1.20 per hour, and fringe benefits average \$2.40 per hour. Standard production time is 1 hour per unit, and the allowance for rest periods and setup is .2 hours and .1 hours, respectively. The standard direct labor rate per hour is \$15.90. \$15.60. \$12.30. \$12.00. Multiple Choice Question 41 The difference between a budget and a standard is that a budget expresses a total amount, while a standard expresses a unit amount. a budget expresses what costs were, while a standard expresses what costs should be. a budget expresses management's plans, while a standard reflects what actually happened. standards are excluded from the cost accounting system, whereas budgets are generally incorporated into the cost accounting system.
Solution Description
Question
Multiple Choice Question 129
Dillon has a standard of 1.5 pounds of materials per unit, at \$6 per pound. In producing 2,000 units, Dillon used 3,100 pounds of materials at a total cost of \$18,135. Dillon's materials price variance is
\$1,050 F.
\$135 U.
\$465 F.
\$600 F.
Multiple Choice Question 81
Scorpion Production Company planned to use 1 yard of plastic per unit budgeted at \$81 a yard. However, the plastic actually cost \$80 per yard. The company actually made 3,900 units, although it had planned to make only 3,300 units. Total yards used for production were 3,960. How much is the total materials variance?
\$3,960 F
\$48,600 U
\$4,860 U
\$900 U
Multiple Choice Question 51
Standard costs
may show past cost experience.
help establish expected future costs.
all of these.
are the budgeted cost per unit in the present.
Multiple Choice Question 68
Allowance for spoilage is part of the direct
labor price standard.
labor quantity standard.
mat
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# E let x represent the average number of putts and y
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(e) Let x represent the average number of putts and y represent the average number of strokes. Use the definition of the correlation coefficient above to determine the measurement units of the correlation coefficient r in terms of putts and strokes. x y I II III I II IV III IV
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Chance/Rossman, 2015 ISCAM III Investigation 5.7 360 (f) If the correlation coefficient between two variables equals zero, what do you think the scatterplot will look like? (g) Suppose we find the correlation coefficient of a variable with itself. Substitute x i in for y i (and so x for y and s x for s y ) in the above equation. Simplify. What is the correlation coefficient equal to? (h) Do you think the correlation coefficient will be a resistant measure of association? Explain. (i) The following scatterplots display 7 pairs of variables for these golfers. Rank these graphs in order from strongest negative correlation to strongest positive correlation. A: B: C: D:
Chance/Rossman, 2015 ISCAM III Investigation 5.7 361 E: F: G: (j) Use technology to determine the correlation coefficient for each of the above scatterplots x In R: > cor(x, y) x In Minitab: Choose Stat > Basic Statistics > Correlation and enter the pair of variables. You can unselect the “Show p - values” box for now. Alternatively, type MTB> corr cx cy. Record the values of these correlation coefficients below: Strongest negative birdie conversion and average putts Medium negative money and average score Weak negative money and average putts No association driving distance and average putts Weak positive money and driving distance Medium positive money and birdie average Strongest positive birdie average and birdie conversion (k) Based on these correlation coefficient values and/or the definition/formula, what do you think is the largest value that r can assume? What is the smallest value? [ Hint : It’s not zero.] Largest: Smallest: (l) If the association is negative, what values will r have? What if the association is positive? Strongest negative: Medium negative: Weak negative: No association: Weak positive Medium positive: Strongest positive:
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Chance/Rossman, 2015 ISCAM III Investigation 5.7 362 (m) What does a correlation coefficient equal to zero signify? (n) What does a correlation coefficient close to 1 or ± 1 signify? (o) Which has a stronger correlation coefficient with scoring average: driving distance or average putts? Does this support the cliché? Explain. Study Conclusions The correlation coefficient for scoring average and average putts indicates a moderately strong positive linear association ( r = 0.444) whereas the correlation coefficient for scoring average and driving indicates a weaker negative association ( r = ˗0 .265). This appears to support that putting performance is more strongly related to a PGA golfer’s overall scoring average than the golfer’s driving distance, as the cliché would suggest. We must keep in mind that these data are only for only the first 2.5 months of the season (when most golfers have played only around 6 8 events) and may not be representative of the scores and money earnings later in the year.
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https://davida.davivienda.com/viewer/printable-third-grade-math-worksheets.html | 1,718,274,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00204.warc.gz | 173,400,757 | 8,284 | These pages are part of a comprehensive unit aligned to common core math standards. These are printable downloads with questions and answer keys provided. Web result this compilation of 3rd grade printable math worksheets that will have your learners engaged and practicing valuable skills. Web result third grade worksheets and printables. Exercises also include multiplying by whole tens and whole hundreds and some column. Round a number to the nearest 10, 100 or 1000; 3 digit easter subtraction math worksheet.
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View Full Version : Bloodability with EGO
teloft
01-30-2004, 02:48 PM
This topic could be used for housrules
Currently Im not so satisfied by the way blood abilityes are handled.
I dont like the idee of ruling blood type thet only alows acsess to a list of abs.
I think its to formalised.
What I came up with last nigth when I was going to bed was somthing I thougt of as briliant. A bit complex. but I still like it now the morning after.
What if the blood abs had alignment of there own. And ego like powerfull items.
Now when you create your character, you have his bloodline, or history.
The history gives you the list of abilitys your charater can have acsess to. So you dont have acsess to any of the abilityes thet no one of your line/history has had.
If the abilitys an alignment difrent to yours you save agenst the abilitys ego
the save is based on what the ability is. chancing form is perhaps a fort save while gaining a spelllike ability is perhaps a will save.
The total ego points of your abilitys combined can not be graiter then your blood score.
the graitness of the ability is now mesured by its ego score.
Now an example.
I have here a LG paladin
with Bloodscore of 15
and blood abilitys
"Higthen ability cha" LG of EGO 10
"Bloodmark" LN of EGO 5
Now he slays a CE creature
with Bloodscore of 35
and blood abilitys
"aura of doom" CE of Ego 25
"withering tuch" NE of Ego 10
By a critical blow trow the hart, he instantly gains higer bloodscore. but in the process he must save versus the abilityes of his foe.
"aura of doom" faild save
"withering tuch" saved
if we use the slayn blood*10 for RP we get 350RP
15*4 = 60
+ 16*4 = 124
+ 17*4 = 192
+ 18*4 = 264
+ 19*4 = 340
So the New blood score is 19
But how sould we calculate the new ego points of the abilityes
he sould have
"aura of doom" CE of Ego 25
"Higthen ability cha" LG of EGO 10
"Bloodmark" LN of EGO 5
But as this is not possible with only a score of 19 the older abilityes must save agenst the new manifested ability. just like items on a character when a fireball hits him.
"Bloodmark" gets lucky and saves
"Higthen ability +1 cha" fails his save
So the newly gaind ability has ego of 14 and is downgraided in power.
0 no manifestasion / minor visible marks of no power
1-10 Minor ability
11-20 Major ability
21+ Great ability
with Bloodscore of 19
and blood abilitys
"aura of doom" CE of Ego 14
"Bloodmark" LG of EGO 5
"Higthen ability cha" LN of EGO 0
this paladin has a conflict with his blood. tread his conflicting blood as a magic item with strong EGO
Edit:
I was thinking of having each type of blood as its own ability.
So in order to have Azary Dominant in your blood. he would have to have some EGO. This represeant not only the fact thet you have been in contact with Azary's blood. But his character is now in your blood. Trying to manipulate you.
Note, if PC almost looses his controle of the character by means of his blood, the character becomes a NPC, like in ravenloft.
I have come up with a investitour action of inner conflict witch can be used with this system of blood.
When the regent finds out thet he has unwanted efects about him. he can contest this efect, efectively returning to a better way. Now in order to do this. the regent must spend RP in order to boost one of his Bloods abilityes EGO's (one RP spent gives +1 EGO for one action) Then he spends his action in inner conflict where the aided blood contests the unwanted blood. But this action can backfire. For the contested blood is alowed a save. if it passes its save, it can now atempt to contest the rival blood.
What hapens when I have a blood increas. or if I have higer blood score then I have blood abilityes total EGO.
well, one of the blood abilityes will increas its ego to fully empower the total bloodscore. The Blood ability thet recives the score is determand randomly by the value of its EGO. RP can be spent in order to supress one of the abilityes ego or to increas it. This will alow the strongest ability the most chence to gain power, and it will alow the player to modify the outcome by using RP. This will slow regents down thet have unfavorable mix of abilityes.
___
what do you, good folks of this side, think of this?
teloft
02-01-2004, 08:02 PM
> did you read my "ego" post
It`s gonna take me awhile to further develop my opinions about it. On
the one hand, it sounds really cool but on the other hand it doesn`t
sound all that different from requiring a skill check or something. By
ability isn`t gonna go off on its own or show signs of being manic
depressive or anything...
We`ll see. I`m not sure what to make of it yet. But yeah I saw it.
Thanks for posting it.
--Lord Rahvin
the feel I was trying to get here is the inner conflict of a regent thet is batteling with unwanted blood within his vains.
the feel of a good and kind ruler, thet had apperad to be a good ruler before. and is now corruped by foul blood. Bu not turning the regent rigth away one way or a noter. the regent can keep his power. slowly corupting his realm along with his own coruption. Without anyone there taking notice rigt away, until its to late.
a nother effort is to express the blend of bloodlines. when breeding the blooded in order to gain strong bloodlines. you will have more complex issues to deal with. wether one ability will have the overhand over the other.
I hav had more then one idee how the personalityes of the old gods sould afect the regent.
1. every ability has a personality.
2. the personality is a seperaid ability. where tha goals of the old god will still be alive. afecting the regent. this can alow a PC to play out a inner conflict betvine himselfe and the blood.
3. the personality of the blood is tied to the shape ability. where the personality of the old god in will always manifest itselfe by chansing the regents form. Bloodform. thent it will only be based on the personality of the divine whether the regents starts chansing his form or not.
I have also been thinking about the ability of the regent to manifest his own personality as a blood ability. So his goals can be carried on to the next carrier of the blood. Even so thet the regent dosent realy die if he has this ability. but lives on in the blood of his new hoast, lives on as blood ability with its own ego. and can manipulate in a difrent manner then ever before.
its the Conflict them Im after. when the regents has a inner conflict it can be fun to play. Now how this shall manifest it selfe in the hous-rules of the DMs is up to them. | 1,702 | 6,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-24 | latest | en | 0.932735 |
https://www.exceldemy.com/excel-count-duplicate-values-in-multiple-columns/ | 1,709,388,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00729.warc.gz | 751,044,217 | 77,263 | # How to Count Duplicate Values in Multiple Columns in Excel (6 Ways)
Get FREE Advanced Excel Exercises with Solutions!
If you are looking for some of the easiest ways to count duplicate values in multiple columns in Excel, then you will find this article useful. Counting duplicates in multiple columns is quite different from counting duplicates in a single column. So, let’s explore the methods to count the duplicates easily.
## How to Count Duplicate Values in Multiple Columns in Excel: 6 Ways
Here, we have the sales records of a company containing the records for January and February and we will try to count the duplicate sales values between the January and February columns.
We have used Microsoft Excel 365 version here, you can use any other versions according to your convenience.
### Method-1: Using COUNTIF Function to Count Duplicate Values in Multiple Columns
Here, we will use the COUNTIF function to count the duplicate sales values between the January and February columns and so that we have added an extra column Helper.
Steps:
➤ Type the following formula in cell E4
`=COUNTIF(\$C\$4:\$D\$11,C4)`
Here, \$C\$4:\$D\$11 is the range, C4 is the criteria corresponding to the cell with the formula.
➤ Press ENTER and drag down the Fill Handle tool.
Then, you will get 2 for the duplicate values in the two columns as you can see the duplicate values have been indicated with red, blue, and green color boxes.
Now, we will count the number of duplicate values by using the following formula
`=COUNTIF(E4:E11,2)`
Here, E4:E11 is the range and 2 is the criteria for indicating duplicates.
After pressing ENTER, you will get 3 as the number of the duplicate values between the January and February columns.
### Method-2: Combination of SUM, IF, ISNA, and MATCH Functions to Count Duplicate Values
In this section, we will be using the SUM function, IF function, ISNA function, and MATCH function to count the duplicate values in the January and February columns.
Steps:
➤ Type the following formula in cell D12
`=SUM(IF(ISNA(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)),0,1))`
Here, \$C\$4:\$C\$11 is the range of the January column and \$D\$4:\$D\$11 is the range of the February column.
• `MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)` → returns the relative position of the duplicate values in the range \$D\$4:\$D\$11 corresponding to the range \$C\$4:\$C\$11.
Output → `{#N/A;3;#N/A;#N/A;8;#N/A;#N/A;4}`
• `ISNA(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0))` → ISNA returns TRUE for #N/A error otherwise FALSE
Output → `{TRUE;FALSE;TRUE;TRUE;FALSE;TRUE;TRUE;FALSE}`
• `IF(ISNA(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)),0,1)` → IF return 0 for TRUE and 1 for FALSE
Output → `{0;1;0;0;1;0;0;1}`
• `SUM(IF(ISNA(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)),0,1)) `becomes
`SUM({0;1;0;0;1;0;0;1})`
Output → 3
After pressing ENTER, you will get the total number of duplicate sales values as 3 between the January and February columns.
Related Content: How to Count Duplicates in Two Columns in Excel
### Method-3: Using AND Function, COUNTIF Function to Count Duplicate Values in Multiple Columns
Here, we will use the AND function, COUNTIF function to count the duplicate sales values between the January and February columns.
Steps:
➤ Type the following formula in cell E4
`=AND(COUNTIF(\$C\$4:\$C\$11,C4),COUNTIF(\$D\$4:\$D\$11,C4))`
Here, \$C\$4:\$C\$11 is the range of the January column and \$D\$4:\$D\$11 is the range of the February column.
• `COUNTIF(\$C\$4:\$C\$11, C4)` → returns the number of the value in cell C4 in the range \$C\$4:\$C\$11
Output → 1
• `COUNTIF(\$D\$4:\$D\$11, C4)` → returns the number of the value in cell C4 in the range \$D\$4:\$D\$11
Output → 0
• `AND(COUNTIF(\$C\$4:\$C\$11,C4),COUNTIF(\$D\$4:\$D\$11,C4)) `becomes
`AND(1,0)`
Output → FALSE
➤ Press ENTER and drag down the Fill Handle tool.
In this way, you will get TRUE for the duplicate values and FALSE for the unique values and now we will count the number of TRUE in the Helper column to get the number of duplicate values.
➤ Type the following formula in cell E12
`=COUNTIF(E4:E11,TRUE)`
Here, E4:E11 is the range, and TRUE is the criteria.
➤ Press ENTER.
Finally, you will be able to count the duplicate values in the January and February columns which are 3.
Read More: How to Count Duplicate Rows in Excel
### Method-4: Using Excel SUMPRODUCT, COUNTIF Functions to Count Duplicate Values in Multiple Columns
You can use the SUMPRODUCT function, COUNTIF function to count the duplicate values in the columns; January and February.
Steps:
➤ Type the following formula in cell D12
`=SUMPRODUCT(COUNTIF(\$C\$4:\$C\$11,\$D\$4:\$D\$11))`
Here, \$C\$4:\$C\$11 is the range of the January column and \$D\$4:\$D\$11 is the range of the February column.
• `COUNTIF(\$C\$4:\$C\$11,\$D\$4:\$D\$11) `gives the matched number of cells between the ranges \$C\$4:\$C\$11 and \$D\$4:\$D\$11
Output → `{0;0;1;1;0;0;0;1}`
• `SUMPRODUCT(COUNTIF(\$C\$4:\$C\$11,\$D\$4:\$D\$11)) `becomes
`SUMPRODUCT({0;0;1;1;0;0;0;1})`
Output → 3
➤ Press ENTER.
Afterward, you will get the number of duplicate values in the January and February columns which is 3.
### Method-5: Combination of SUMPRODUCT, ISNUMBER, MATCH Functions to Count Duplicate Values
Here, we will use the SUMPRODUCT function, ISNUMBER function, MATCH function to determine the duplicates between the January and February columns.
Steps:
➤ Type the following formula in cell D12
`=SUMPRODUCT(--(ISNUMBER(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0))))`
Here, \$C\$4:\$C\$11 is the range of the January column and \$D\$4:\$D\$11 is the range of the February column.
• `MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)` → returns the relative position of the duplicate values in the range \$D\$4:\$D\$11 corresponding to the range \$C\$4:\$C\$11.
Output → `{#N/A;3;#N/A;#N/A;8;#N/A;#N/A;4}`
• `ISNUMBER(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0))` becomes
`ISNUMBER({#N/A;3;#N/A;#N/A;8;#N/A;#N/A;4})`
Output → `{FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE;TRUE}`
• `--(ISNUMBER(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0))) `becomes
`--({FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE;TRUE})` → Double negation (–) will convert TRUE into 1 and FALSE into 0
Output → `({0;1;0;0;1;0;0;1})`
• `SUMPRODUCT(--(ISNUMBER(MATCH(\$C\$4:\$C\$11,\$D\$4:\$D\$11,0)))) `becomes
`SUMPRODUCT({0;1;0;0;1;0;0;1})`
Output → 3
After pressing ENTER, you will get the total number of duplicate sales values between the January and February columns.
### Method-6: Using Conditional Formatting to Count Duplicate Values in Multiple Columns in Excel
Here, we will use Conditional Formatting to count the duplicate values in the January and February columns.
Steps:
➤ Select the January and February columns.
➤ Go to Home Tab >> Styles Group >> Conditional Formatting Dropdown >> Highlight Cells Rules Option >> Duplicate Values Option.
Then, the Duplicate Values wizard will appear.
➤ Select the Custom Format option in the second box.
Now, the Format Cells dialog box will open up.
➤ Select Fill Option, choose the Yellow Color, and press OK.
Again, press OK in the Duplicate Values dialog box.
In this way, the cells with duplicate values will be highlighted.
➤ Select any of the columns between January and February.
➤ Go to Home Tab >> Editing Group >> Find & Select Dropdown >> Find Option.
After that, the Find and Replace Dialog Box will pop up.
➤Select the Format Option.
Then, the Find Format Dialog Box will appear.
➤ Select Fill Option, choose the Yellow Color, and press OK.
Afterward, the following Preview section will appear.
➤ Click the Find All option.
Eventually, you will see the number of yellow color cells which is 3 on the left bottom corner of the dialog box.
## Practice Section
For doing practice by yourself we have provided a Practice section like below in a sheet named Practice. Please do it by yourself.
## Conclusion
In this article, we tried to cover the ways to count duplicate values in multiple columns in Excel. Hope you will find it useful. If you have any suggestions or questions, feel free to share them in the comment section.
## Related Articles
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Advanced Excel Exercises with Solutions PDF | 2,592 | 9,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | latest | en | 0.893659 |
https://isabelle.in.tum.de/dist/library/HOL/HOL-ex/ThreeDivides.html | 1,679,396,455,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00186.warc.gz | 399,081,533 | 7,240 | # Theory ThreeDivides
(* Title: HOL/ex/ThreeDivides.thy
Author: Benjamin Porter, 2005
*)
section ‹Three Divides Theorem›
theory ThreeDivides
imports Main "HOL-Library.LaTeXsugar"
begin
subsection ‹Abstract›
text ‹
The following document presents a proof of the Three Divides N theorem
formalised in the Isabelle/Isar theorem proving system.
{\em Theorem}: $3$ divides $n$ if and only if $3$ divides the sum of all
digits in $n$.
{\em Informal Proof}:
Take $n = \sum{n_j * 10^j}$ where $n_j$ is the $j$'th least
significant digit of the decimal denotation of the number n and the
sum ranges over all digits. Then $$(n - \sum{n_j}) = \sum{n_j * (10^j - 1)}$$ We know $\forall j\; 3|(10^j - 1)$ and hence $3|LHS$,
therefore $$\forall n\; 3|n \Longleftrightarrow 3|\sum{n_j}$$
‹□›
›
subsection ‹Formal proof›
subsubsection ‹Miscellaneous summation lemmas›
text ‹If $a$ divides ‹A x› for all x then $a$ divides any
sum over terms of the form ‹(A x)*(P x)› for arbitrary $P$.›
lemma div_sum:
fixes a::nat and n::nat
shows "∀x. a dvd A x ⟹ a dvd (∑x<n. A x * D x)"
proof (induct n)
case 0 show ?case by simp
next
case (Suc n)
from Suc
have "a dvd (A n * D n)" by (simp add: dvd_mult2)
with Suc
have "a dvd ((∑x<n. A x * D x) + (A n * D n))" by (simp add: dvd_add)
thus ?case by simp
qed
subsubsection ‹Generalised Three Divides›
text ‹This section solves a generalised form of the three divides
problem. Here we show that for any sequence of numbers the theorem
holds. In the next section we specialise this theorem to apply
directly to the decimal expansion of the natural numbers.›
text ‹Here we show that the first statement in the informal proof is
true for all natural numbers. Note we are using \<^term>‹D i› to
denote the $i$'th element in a sequence of numbers.›
lemma digit_diff_split:
fixes n::nat and nd::nat and x::nat
shows "n = (∑x∈{..<nd}. (D x)*((10::nat)^x)) ⟹
(n - (∑x<nd. (D x))) = (∑x<nd. (D x)*(10^x - 1))"
text ‹Now we prove that 3 always divides numbers of the form $10^x - 1$.›
lemma three_divs_0:
shows "(3::nat) dvd (10^x - 1)"
proof (induct x)
case 0 show ?case by simp
next
case (Suc n)
let ?thr = "(3::nat)"
have "?thr dvd 9" by simp
moreover
have "?thr dvd (10*(10^n - 1))" by (rule dvd_mult) (rule Suc)
hence "?thr dvd (10^(n+1) - 10)" by (simp add: nat_distrib)
ultimately
have"?thr dvd ((10^(n+1) - 10) + 9)"
by (simp only: ac_simps) (rule dvd_add)
thus ?case by simp
qed
text ‹Expanding on the previous lemma and lemma ‹div_sum›.›
lemma three_divs_1:
fixes D :: "nat ⇒ nat"
shows "3 dvd (∑x<nd. D x * (10^x - 1))"
by (subst mult.commute, rule div_sum) (simp add: three_divs_0 [simplified])
text ‹Using lemmas ‹digit_diff_split› and
‹three_divs_1› we now prove the following lemma.
›
lemma three_divs_2:
fixes nd::nat and D::"nat⇒nat"
shows "3 dvd ((∑x<nd. (D x)*(10^x)) - (∑x<nd. (D x)))"
proof -
from three_divs_1 have "3 dvd (∑x<nd. D x * (10 ^ x - 1))" .
thus ?thesis by (simp only: digit_diff_split)
qed
text ‹
We now present the final theorem of this section. For any
sequence of numbers (defined by a function \<^term>‹D :: (nat⇒nat)›),
we show that 3 divides the expansive sum $\sum{(D\;x)*10^x}$ over $x$
if and only if 3 divides the sum of the individual numbers
$\sum{D\;x}$.
›
lemma three_div_general:
fixes D :: "nat ⇒ nat"
shows "(3 dvd (∑x<nd. D x * 10^x)) = (3 dvd (∑x<nd. D x))"
proof
have mono: "(∑x<nd. D x) ≤ (∑x<nd. D x * 10^x)"
by (rule sum_mono) simp
txt ‹This lets us form the term
\<^term>‹(∑x<nd. D x * 10^x) - (∑x<nd. D x)››
{
assume "3 dvd (∑x<nd. D x)"
with three_divs_2 mono
show "3 dvd (∑x<nd. D x * 10^x)"
by (blast intro: dvd_diffD)
}
{
assume "3 dvd (∑x<nd. D x * 10^x)"
with three_divs_2 mono
show "3 dvd (∑x<nd. D x)"
by (blast intro: dvd_diffD1)
}
qed
subsubsection ‹Three Divides Natural›
text ‹This section shows that for all natural numbers we can
generate a sequence of digits less than ten that represent the decimal
expansion of the number. We then use the lemma ‹three_div_general› to prove our final theorem.›
text ‹\medskip Definitions of length and digit sum.›
text ‹This section introduces some functions to calculate the
required properties of natural numbers. We then proceed to prove some
properties of these functions.
The function ‹nlen› returns the number of digits in a natural
number n.›
fun nlen :: "nat ⇒ nat"
where
"nlen 0 = 0"
| "nlen x = 1 + nlen (x div 10)"
text ‹The function ‹sumdig› returns the sum of all digits in
some number n.›
definition
sumdig :: "nat ⇒ nat" where
"sumdig n = (∑x < nlen n. n div 10^x mod 10)"
text ‹Some properties of these functions follow.›
lemma nlen_zero:
"0 = nlen x ⟹ x = 0"
by (induct x rule: nlen.induct) auto
lemma nlen_suc:
"Suc m = nlen n ⟹ m = nlen (n div 10)"
by (induct n rule: nlen.induct) simp_all
text ‹The following lemma is the principle lemma required to prove
our theorem. It states that an expansion of some natural number $n$
into a sequence of its individual digits is always possible.›
lemma exp_exists:
"m = (∑x<nlen m. (m div (10::nat)^x mod 10) * 10^x)"
proof (induct "nlen m" arbitrary: m)
case 0 thus ?case by (simp add: nlen_zero)
next
case (Suc nd)
obtain c where mexp: "m = 10*(m div 10) + c ∧ c < 10"
and cdef: "c = m mod 10" by simp
show "m = (∑x<nlen m. m div 10^x mod 10 * 10^x)"
proof -
from ‹Suc nd = nlen m›
have "nd = nlen (m div 10)" by (rule nlen_suc)
with Suc have
"m div 10 = (∑x<nd. m div 10 div 10^x mod 10 * 10^x)" by simp
with mexp have
"m = 10*(∑x<nd. m div 10 div 10^x mod 10 * 10^x) + c" by simp
also have
"… = (∑x<nd. m div 10 div 10^x mod 10 * 10^(x+1)) + c"
by (subst sum_distrib_left) (simp add: ac_simps)
also have
"… = (∑x<nd. m div 10^(Suc x) mod 10 * 10^(Suc x)) + c"
also have
"… = (∑x∈{Suc 0..<Suc nd}. m div 10^x mod 10 * 10^x) + c"
by (simp only: sum.shift_bounds_Suc_ivl)
also have
"… = (∑x<Suc nd. m div 10^x mod 10 * 10^x)"
by (simp add: atLeast0LessThan[symmetric] sum.atLeast_Suc_lessThan cdef)
also note ‹Suc nd = nlen m›
finally
show "m = (∑x<nlen m. m div 10^x mod 10 * 10^x)" .
qed
qed
text ‹\medskip Final theorem.›
text ‹We now combine the general theorem ‹three_div_general›
and existence result of ‹exp_exists› to prove our final
theorem.›
theorem three_divides_nat:
shows "(3 dvd n) = (3 dvd sumdig n)"
proof (unfold sumdig_def)
have "n = (∑x<nlen n. (n div (10::nat)^x mod 10) * 10^x)"
by (rule exp_exists)
moreover
have "3 dvd (∑x<nlen n. (n div (10::nat)^x mod 10) * 10^x) =
(3 dvd (∑x<nlen n. n div 10^x mod 10))"
by (rule three_div_general)
ultimately
show "3 dvd n = (3 dvd (∑x<nlen n. n div 10^x mod 10))" by simp
qed
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