url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.mathworksheetscenter.com/mathskills/counting/countmissnumber/ | 1,685,812,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00293.warc.gz | 965,245,100 | 4,411 | # Missing Numbers Counting Worksheets
How to Spot Missing Numbers in Counting Sequences - There is an order that the numbers follow. When we identify the order or pattern based on which the numbers are increasing or decreasing, we can find out the missing numbers and the next terms. Many students find it difficult to spot missing numbers in counting sequences. Step 1: Identify Whether Numbers are in Ascending or Descending Order - The first step is to identify whether a sequence is in an incremental or decremental pattern. It is the easy and can be identified in the first look. Step 2: Evaluate Two Consecutive Terms - Take any two consecutive terms and apply a trial and error method on those to identify the pattern. You can divide these terms to find the ratio or subtract them to find the difference by which they are increasing. Step 3: Complete the Pattern - Add missing terms of the sequence and find out if the final sequence makes sense.
• ### Basic Lesson
Complete the Counting sequences. Also includes practice problems. Fill in the blanks with missing numbers.
• ### Topic Quiz
Includes 10 problems and a scoring matrix.
• ### Independent Practice 1
Complete the missing numbers problems. The answers can be found below.
• ### Independent Practice 2
Complete the counting sequences. The answers can be found below.
• ### Homework Worksheet
Features 3 practice problems with detailed examples. Complete examples are provided.
• ### Homework and Quiz Answer Key
Answers for the homework and quiz.
• ### Lesson and Practice Answer Key
Answers for both lessons and both practice sheets.
#### Why do mathematicians think Halloween and Christmas are the same?
Because 31 Oct = 25 Dec. Halloween occurs on October 31 and Christmas occurs on December 25, thus equating "oct" in October and octal, and "dec" in December and decimal. | 377 | 1,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-23 | longest | en | 0.913174 |
http://www.ck12.org/arithmetic/Decimal-Addition-Using-Front-End-Estimation/lesson/Decimal-Addition-Using-Front-End-Estimation/ | 1,438,750,511,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438043060830.93/warc/CC-MAIN-20150728002420-00325-ip-10-236-191-2.ec2.internal.warc.gz | 376,468,281 | 32,482 | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Decimal Addition Using Front-End Estimation
## Find an estimate of a decimal sum by adding only some of the given values.
0%
Progress
Practice Decimal Addition Using Front-End Estimation
Progress
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Have you ever gone to dinner with a friend?
Jessica and Daniella went to dinner. Jessica ordered a pasta dish with shrimp and lobster for $25.25 and Daniella ordered a steak dish for$18.95.
Without adding the exact values, can you estimate the total cost of both meals?
This Concept is about front-end estimation. Pay attention and you can use it to estimate the sum of these meals.
### Guidance
What is estimation?
Estimation is a method for finding an approximate solution to a problem.
For example, the sum of 22 and 51 is exactly 73. We can estimate the sum by rounding to the tens place and adding 20 + 50. Then we can say the sum of 22 and 51 is “about 70.” Before we perform an operation, an estimate gives us a rough idea of what our answer will be.
Estimation is also useful to check answers. If, after you solve a problem, you estimate to get a general idea of what an answer should be, you know if your answer is reasonable. You could say that an estimate is an “educated guess.”
By “educated guess” we don’t mean a guess that is out of nowhere. You have to have a method to figuring out the answer or estimate. Rounding numbers before adding is one way to find an estimate.
Front-end estimation is another way. Like rounding, front-end estimation uses place value; unlike rounding, it doesn’t involve any changing of values.
There are three steps to front-end estimation.
First, you add the front end, or left-most digits, to get a general estimate of the sum. Next, to refine the first estimate, you add the digits directly to the right of the front-end digits.
Finally you add the two sums together.
1. Add the front (leftmost) digits.
2. Add the digits directly to the right of the front-end.
3. Add the estimates from steps 1 and 2.
Write these steps in your notebook and then continue with the Concept.
Use front-end estimation to find the sum of 6.819 and 4.621.
We are going to use front-end estimation to add 6.819 and 4.621. Let’s begin by lining up the decimal points and taking a closer look at the numbers. Then we’ll follow the steps of front-end estimation.
6.819
4.621
1. Front-end estimation tells us to add the front or leftmost digits first, so we want to add the ones place.
$6 + 4 = 10$ .
2. Next, to refine that estimate, we’ll add the digits directly to the right of the front-end digits, or the tenths place.
$.8 + .6 = 1.4$
3. Finally, we add our two estimates together.
$10 + 1.4 = 11.4$ .
Note: The original numbers extended to the thousandth place, but we don’t need to show this by adding unnecessary zeros to our answer. Where decimals end, it is assumed that zeros extend on to infinity.
Practice using front-end estimation to find the following sums.
#### Example A
$3.5 + 2.34 = \underline{\;\;\;\;\;\;\;\;}$
Solution: $5.84$
#### Example B
$12.671 + 8.123 = \underline{\;\;\;\;\;\;\;\;}$
Solution: $20.7$
#### Example C
$15.67 + 9.345 = \underline{\;\;\;\;\;\;\;\;}$
Solution: $24.9$
Here is the original problem once again.
Jessica and Daniella went to dinner. Jessica ordered a pasta dish with shrimp and lobster for $25.25 and Daniella ordered a steak dish for$18.95.
Without adding the exact values, can you estimate the total cost of both meals?
To estimate this sum, we simply add the front-ends of each price.
$18 + 25 = 43$
The total cost of the two dinners is approximately \$43.00.
### Guided Practice
Here is one for you to try on your own.
Use front-end estimation to find the following sum.
2.93474 + 9.72155
First, we add the front-ends of the whole numbers and decimals.
$.7 + .9 = 1.6$
$2 + 9 = 11$
Now put these two sums together.
Our answer is $12.6$
### Explore More
Directions: Use front-end estimation to find the following sums.
1. $4.57 + 2.34$
2. $2.123 + 4.136$
3. $8.913 + 2.047$
4. $8.7651 + 2.345$
5. $2.436 + 4.567$
6. $8.127 + 9.431$
7. $8.214 + 7.3214$
8. $12.137 + 2.456$
9. $18.671 + 20.41$
10. $21.643 + 22.123$
11. $18.012 + 19.367$
12. $21.456 + 18.023$
13. $.48218 + .61927$
14. $6.7765 + 6.421192$
15. $.5075412 + .859931 + .373462$
### Vocabulary Language: English
Decimal
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Estimate
Estimate
To estimate is to find an approximate answer that is reasonable or makes sense given the problem.
Front-End Estimation
Front-End Estimation
Front-end estimation is a method of estimating where you only add the digits in the greatest place value.
Rounding
Rounding
Rounding is reducing the number of non-zero digits in a number while keeping the overall value of the number similar. | 1,372 | 5,163 | {"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": 28, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2015-32 | latest | en | 0.867352 |
https://devhubby.com/thread/how-to-find-the-factorial-of-100-in-c | 1,716,617,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00526.warc.gz | 187,495,870 | 25,593 | # How to find the factorial of 100 in C++?
Member
by matteo , in category: C/C++ , 8 months ago
How to find the factorial of 100 in C++?
, 8 months ago
@matteo
To find the factorial of 100 in C++, you can use a loop to multiply all the numbers from 1 to 100 together. Here is the code:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ``` ```#include using namespace std; int main() { int num = 100; unsigned long long factorial = 1; // Using unsigned long long to store large numbers for (int i = 1; i <= num; ++i) { factorial *= i; } cout << "The factorial of " << num << " is: " << factorial << endl; return 0; } ```
In this code, we start with initializing the `factorial` variable to 1. Then, using a for loop, we multiply each number from 1 to 100 with `factorial`. Finally, we print the result.
Please note that since the factorial of 100 is a large number, we are using the `unsigned long long` data type to store the result. | 280 | 937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | longest | en | 0.817481 |
http://www.readyviva.com/?s=1615477 | 1,606,331,885,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00133.warc.gz | 146,597,260 | 7,979 | # Kuta evaluate expressions one variable
### Middle School Math 2 and Accelerated Math: Timeframe ALGEBRA ...
ext: pdf date: 2020-10-26
Solve one-variable linear ... Evaluate numerical expressions ... Kuta Software. Additional activities. as described in Teacher Program Guide .
### What We're Learning Vocabulary
ext: pdf date: 2020-11-09
will learn about expressions and equations. ... evaluate To substitute the values given ... that contains one or more variables. variable A letter or a symbol that
### CHAPTER 1 Linear Equations and Inequalities - TLOK
ext: pdf date: 2020-11-06
The equation is a equation in one variable.The expressions are called of the equation. 2. The values ... We evaluate algebraic expressions to find the
### Math Worksheet Sources - Northwestern Connecticut Community ...
ext: pdf date: 2020-10-26
Linear Equations in One Variable A Linear Equations Using Inverse Matrix A Conic Sections A Mensuration A ... Evaluating expressions Number sets Adding ...
### 1.2 Writing Algebraic Expressions
ext: pdf date: 2020-11-08
1-2 Writing Algebraic Expressions ... Variable Expression Verbal Phrase ... Then evaluate the expression for 2, 3, and 4 times.
### Evaluating Variable Expressions
ext: pdf date: 2020-11-12
Evaluate each using the values given. 1) ... Evaluating Variable Expressions Date ... Create your own worksheets like this one with Infinite Pre ...
### Write each phrase as an algebraic expression.
ext: pdf date: 2020-11-19
Translating Expressions Date: _____Period:_____ Write each phrase as an algebraic expression. 1. 12 more than a ... One fourth the sum of r and 18.
### Evaluating Expressions Date Period - Create Custom Pre ...
ext: pdf date: 2020-11-05
Kuta Software - Infinite Algebra ... Evaluating Expressions Date_____ Period____ Evaluate ... Create your own worksheets like this one with Infinite Algebra 1.
### Algebra 2 revisions
ext: pdf date: 2020-11-18
9-12.A.1.14 Evaluate polynomial, rational, radical, and absolute value expressions for one or more variables. ... inequalities in one variable including
### Worksheet 2.1 Factors of Algebraic Expressions
ext: pdf date: 2020-11-17
In simplifying algebraic expressions we need to nd common factors. ... One use of the factorization of algebraic expressions or of being able to nd common algebraic
### Worksheet 2 2 Solving Equations in One Variable
ext: pdf date: 2020-11-03
in one variable. To simplify the ... Sometimes the expressions will be more complex and could involve brackets. In these cases we expand out the brackets and proceed.
### Algebra 1 revisions 2
ext: pdf date: 2020-11-12
9-12.A.1.14 Evaluate polynomial, rational, radical, and absolute value expressions for one or more variables. ... the term variable. Module 2 Kuta Software handouts
### Unit 1 Unit 2 Unit 3
ext: pdf date: 2020-11-17
Write and evaluate numerical and algebraic expressions. ... Kuta Software ... Solve literal equations for one variable.
### YO :2q
ext: pdf date: 2020-11-07
Kuta Software - Infinite Pre-Algebra One-Step Equations With Integers ... Evaluating Variable Expressions Evaluate each using the values given. 1) n2 m; ...
### Evaluating Variable Expressions - Create Custom Pre-Algebra ...
ext: pdf date: 2020-10-30
Kuta Software - Infinite Pre ... Evaluating Variable Expressions Date_____ Period____ Evaluate ... Create your own worksheets like this one with Infinite Pre-Algebra.
### Unit Course Map - Algebra I - 2011 MS
ext: pdf date: 2020-11-01
How is the Order of Operations used to evaluate algebraic expressions? ... What does the solution to an inequality in one variable mean? ... Kuta Software Box ...
### Pearson Algebra I First Nine Weeks 2013-2014
ext: pdf date: 2020-11-20
Evaluating_Variable_Expressions 3102.2.3 Describe and/or order a given set of real numbers N-RN.3. ... 2-1: Solving One-Step Equations Kuta - 1 Step EQ
### CCGPS Coordinate Algebra - Georgia Department of Education
ext: pdf date: 2020-11-03
Developing and evaluating mathematical ... interpret complicated expressions by viewing one or more of ... solve linear inequalities in one variable | 1,035 | 4,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.711379 |
https://stackoverflow.com/questions/52506679/reverse-bits-of-an-integer | 1,561,481,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999853.94/warc/CC-MAIN-20190625152739-20190625174739-00553.warc.gz | 595,471,863 | 24,182 | # Reverse bits of an integer [duplicate]
I came through an interview question. Reverse bits of a 32 bit unsigned integer. I wrote this code which is completely okay:
``````uint32_t reverseBits(uint32_t n) {
for(int i = 0, j = 31; i < j; i++, j--) {
bool iSet = (bool)(n & (1 << i));
bool jSet = (bool)(n & (1 << j));
n &= ~(1 << j);
n &= ~(1 << i);
if(iSet) n |= (1 << j);
if(jSet) n |= (1 << i);
}
return n;
}
``````
After this, there was a follow up question - If this function is called many times, how would you optimize it? I can not figure out how the solution should be optimized in that scenario.
## marked as duplicate by zch, Blastfurnace, Community♦Sep 25 '18 at 21:25
• Obligatory link to Bit Twiddling Hacks. – zch Sep 25 '18 at 21:09
• Working code that you think can be made better? This looks like a job for Coooooooode Review! I've linked to the how to ask page because it's worth a read to make sure any posts you make meet their requirements. – user4581301 Sep 25 '18 at 21:19
You can optimize the loop by using a reverse lookup table.
For more detailed information you can follow this URL from which I have taken below code.
``````// Generate a lookup table for 32bit operating system
// using macro
#define R2(n) n, n + 2*64, n + 1*64, n + 3*64
#define R4(n) R2(n), R2(n + 2*16), R2(n + 1*16), R2(n + 3*16)
#define R6(n) R4(n), R4(n + 2*4 ), R4(n + 1*4 ), R4(n + 3*4 )
// Lookup table that store the reverse of each table
unsigned int lookuptable[256] = { R6(0), R6(2), R6(1), R6(3) };
/* Function to reverse bits of num */
int reverseBits(unsigned int num)
{
int reverse_num = 0;
// Reverse and then rearrange
// first chunk of 8 bits from right
reverse_num = lookuptable[ num & 0xff ]<<24 |
// second chunk of 8 bits from right
lookuptable[ (num >> 8) & 0xff ]<<16 |
lookuptable[ (num >> 16 )& 0xff ]<< 8 |
lookuptable[ (num >>24 ) & 0xff ] ;
return reverse_num;
}
``````
• You should probably mention the source of this. – m69 Sep 25 '18 at 21:13
• Hi @m69 I have added source. – Mayur Sep 25 '18 at 21:16
• GeeksforGeeks themselves seem to have copied it from the link @zch posted, without mentioning this :-) graphics.stanford.edu/~seander/bithacks.html#BitReverseTable – m69 Sep 25 '18 at 21:19
• @m69 Thanks for mentioning the real source, I have updated source URL. – Mayur Sep 25 '18 at 21:21
• @m69 That geeks site does cite the source but only on a previous post on the same subject matter. – Blastfurnace Sep 25 '18 at 21:21 | 798 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-26 | latest | en | 0.814552 |
https://greprepclub.com/forum/please-rate-my-argument-essay-14955.html | 1,582,834,198,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00521.warc.gz | 395,880,352 | 25,703 | It is currently 27 Feb 2020, 12:09
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# Please rate my argument essay
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When Stanley Park first opened, it was the largest, most heavily used public park in town. It is still the largest park, but it is no longer heavily used. Video cameras mounted in the park's parking lots last month revealed the park's drop in popularity: the recordings showed an average of only 50 cars per day. In contrast, tiny Carlton Park in the heart of the business district is visited by more than 150 people on a typical weekday. An obvious difference is that Carlton Park, unlike Stanley Park, provides ample seating. Thus, if Stanley Park is ever to be as popular with our citizens as Carlton Park, the town will obviously need to provide more benches, thereby converting some of the unused open areas into spaces suitable for socializing.
Write a response in which you examine the stated and/or unstated assumptions of the argument. Be sure to explain how the argument depends on these assumptions and what the implications are for the argument if the assumptions prove unwarranted.
My response (including the grammatical and spelling mistakes):
The author is concerned over the drop in popularity of Stanley Park. He states that the drop is evident from the number of cars in the parking space. He also suggests that in order to increase the popularity of Stanley Park, the authorities must provide more benches in the unused space so that people can socialize. The author's argument has a lot of assumptions that may not be cogent enough to pursuade the authorities to build new benches in Stanley Park. We will look at three such assumptions and evaluate their effects on the conclusion of the argument.
The authors assumes that most of the people visiting Stanley Park use cars to get to the park. This may not be necessarily true. Most of the people who visit park are likely health conscious and thus, may prefer to walk to the park instead of using a car. Since the video cameras are placed in the parking lot, we may not be able to accurately enumerate the number of park visitors. If this is true, the author's argument that Stanley Park is losing popularity is seriously weakened. In order to make a more compelling case, the author needs to provide the actual number of park vistiors- including those who don't use a car.
The second unstated assumption in the argument is that the author believes increasing the number of benches in the park will attract more visitors. The argument does not consider the possibility that Stanley Park may be far away from the residential areas of the town and thus, people may not visit the park even if the number of benches is raised. Contrary to that, Carlton Park is located in the heart of business district and may be more polular than the Stanley Park. The author fails to provide the major demographics of people who visit these parks. It is possible that majority of the visitors to Carlton Park are senior citizens and hence, the higher number of bences is justified. If the majority of visitors to Stanley Park are children, the increased number of benches will not affect the popularity of the park. If this is true, the argument does not hold any merit.
Finally, the author in his argument, implies that both the parks have the same facilities. It is possible that until Carlton Park was opened, people had very little choice in terms of the parks that they could visit. Maybe Carlton Park has superior facilities that attract more people across the town. In order to form a strong conclusion, the author needs to compare the positives and the negatives of each park, otherwise, the argument does not hold any water.
The argument makes a series of unstated assumptions that seriously undermine it's validity. Unless these assumptions are addressed, the author's argument to increase Stanley Park's popularity will fall apart.
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# Multiple Sequence Alignments - PowerPoint PPT Presentation
Multiple Sequence Alignments. Multiple Alignments. Generating multiple alignments Web servers Analyzing a multiple alignment what makes a ‘good’ multiple alignment? what can it tell us, why is it useful? Adjusting a multiple alignment Alignment editors and HowTo Demonstration and practice.
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Multiple Sequence Alignments
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## Multiple Sequence Alignments
### Multiple Alignments
• Generating multiple alignments
• Web servers
• Analyzing a multiple alignment
• what makes a ‘good’ multiple alignment?
• what can it tell us, why is it useful?
• Adjusting a multiple alignment
• Alignment editors and HowTo
• Demonstration and practice
### What is a Multiple Alignment?
• A comparison of sequences
• “multiple sequence alignment”
• A comparison of equivalents:
• Structurally equivalent positions
• Functionally equivalent residues
• Secondary structure elements
• Hydrophobic regions, polar residues
### Generating multiple alignments
• Pairwise sequence alignment is easy with sufficiently closely related sequences.
• Below a certain level of identity sequence alignment may become uncertain :
• twilight zone for aa sequences ~ 30%.
• In or below the twilight zone it is good to make use of additional information, eg, from evolution.
• A multiple alignment of diverse sequences is more informative than a pairwise alignment:
• residues conserved over longer period of time are under stronger evolutionary constraints.
### Multiple Sequence Alignments Algorithms
• Multiple sequence alignment uses heuristic methods only:
• With dynamic programming, computational time quickly explodes as the number of sequences increases.
• Different methods/algorithms:
• Segment-based (DiAlign, …).
• Iterative (HMMs, DiAlign, PRRP, …).
• Progressive (Clustalw, T-Coffee, MUSCLE, …).
### Progressive Alignment
• Step1: Calculate all pairwise alignments and calculate distances for all pairs of sequences.
• Step 2: Construct guide tree joining the most similar sequences using Neighbour Joining.
Step 1
Step 2
### Progressive Alignment
• Step 3: From the tree assign weights for each sequence:
• We want to down-weight nearly identical sequences and up-weight the most divergent ones.
• Step 4: Align sequences, starting at the leaves of the guide tree:
• Pairwise comparisons as well as comparison of single sequence with a group of sequences (Profile)
• Caveat: errors introduced early cannot be corrected by subsequent information
### Web servers
• ClustalW: http://www.ebi.ac.uk/Tools/clustalw2/
• T-Coffee: http://www.ebi.ac.uk/Tools/t-coffee/
• MUSCLE: http://www.ebi.ac.uk/Tools/muscle/
• DiAlign: http://dialign.gobics.de/
• ... and more at http://helix.nih.gov/apps/bioinfo/msa.html.
### Clustalw features
• Amino acid substitution matrices are varied at different alignment stages according to the divergence of the sequences to be aligned.
• Reduced gap penalties in hydrophilic regions encourage new gaps in potential loop regions rather than regular secondary structure.
Insertions and deletions are more common in loop regions than in the core of the protein!
### T-Coffee features
• More accurate than ClustalW
• Instead of amino acid substitution matrices, uses consistency in a library of pairwise alignments
Vertices represent positions in protein
sequence. Edges represent pairwise
alignments between protein sequences.
If residues I and J have many common
neighbours, their consistency is high.
j
i
### MUSCLE
• Fast implementation
• Sometimes more accurate than ClustalW or T-Coffee
### Example
• Let’s build a multiple alignment for the following sequences :
>query MKNTLLKLGVCVSLLGITPFVSTISSVQAERTVEHKVIKNETGTISISQLNKNVWVHTELGYFSGEAVPSNGLVLNTSKGLVLVDSSWDDKLTKELIEMVEKKFKKRVTDVIITHAHADRIGGMKTLKERGIKAHSTALTAELAKKNGYEEPLGDLQSVTNLKFGNMKVETFYPGKGHTEDNIVVWLPQYQILAGGCLVKSASSKDLGNVADAYVNEWSTSIENVLKRYGNINLVVPGHGEVGDRGLLLHTLDLLK>gi|2984094 MGGFLFFFLLVLFSFSSEYPKHVKETLRKITDRIYGVFGVYEQVSYENRGFISNAYFYVADDGVLVVDALSTYKLGKELIESIRSVTNKPIRFLVVTHYHTDHFYGAKAFREVGAEVIAHEWAFDYISQPSSYNFFLARKKILKEHLEGTELTPPTITLTKNLNVYLQVGKEYKRFEVLHLCRAHTNGDIVVWIPDEKVLFSGDIVFDGRLPFLGSGNSRTWLVCLDEILKMKPRILLPGHGEALIGEKKIKEAVSWTRKYIKDLRETIRKLYEEGCDVECVRERINEELIKIDPSYAQVPVFFNVNPVNAYYVYFEIENEILMGE>gi|115023|sp|P10425|MKKNTLLKVGLCVSLLGTTQFVSTISSVQASQKVEQIVIKNETGTISISQLNKNVWVHTELGYFNGEAVPSNGLVLNTSKGLVLVDSSWDNKLTKELIEMVEKKFQKRVTDVIITHAHADRIGGITALKERGIKAHSTALTAELAKKSGYEEPLGDLQTVTNLKFGNTKVETFYPGKGHTEDNIVVWLPQYQILAGGCLVKSAEAKNLGNVADAYVNEWSTSIENMLKRYRNINLVVPGHGKVGDKGLLLHTLDLLK>gi|115030|sp|P25910|MKTVFILISMLFPVAVMAQKSVKISDDISITQLSDKVYTYVSLAEIEGWGMVPSNGMIVINNHQAALLDTPINDAQTEMLVNWVTDSLHAKVTTFIPNHWHGDCIGGLGYLQRKGVQSYANQMTIDLAKEKGLPVPEHGFTDSLTVSLDGMPLQCYYLGGGHATDNIVVWLPTENILFGGCMLKDNQATSIGNISDADVTAWPKTLDKVKAKFPSARYVVPGHGDYGGTELIEHTKQIVNQYIESTSKP>gi|282554|pir||S25844 MTVEVREVAEGVYAYEQAPGGWCVSNAGIVVGGDGALVVDTLSTIPRARRLAEWVDKLAAGPGRTVVNTHFHGDHAFGNQVFAPGTRIIAHEDMRSAMVTTGLALTGLWPRVDWGEIELRPPNVTFRDRLTLHVGERQVELICVGPAHTDHDVVVWLPEERVLFAGDVVMSGVTPFALFGSVAGTLAALDRLAELEPEVVVGGHGPVAGP EVIDANRDYLRWVQRLAADAVDRRLTPLQAARRADLGAFAGLLDAERLVANLHRAHEELLGGHVRDAMEIFAELVAYNGGQLPTCLA
### ClustalW at EBI
• Many options:
• CPU mode,
• full/fast alignment,
• window length in fast mode,
• gap penalties.
### ClustalW at EBI
• Automatic display of:
• Score table
• Alignment (optional colouring)
• Tree guide
• Link to Jalview alignment editor!
### A note on the example
• It is atypical:
• It uses only three sequences.
• One should use more in order to extract reliable informations.
• It illustrates a common mistake:
• It uses too closely related sequences.
• One should use as divergent and diverse sequences as possible in order to extract relevant informations.
### A Good Multiple Alignment?
• Difficult to define…
• Good ones look pretty!
• Aligned secondary structures
• Strongly conserved residues / regions
• Comparison with known structure helps
• Bad ones look chaotic and random.
conservation
quality
consensus
?
### Multiple Alignment Features
• Barton (1993)
• “The position of insertions and deletions suggests regions where surface loops exist…
### Multiple Alignment Features
• Barton (1993)
• “The position of insertions and deletions suggests regions where surface loops exist…
• Conserved glycine or proline suggests aβ-turn...
### Multiple Alignment Features
• Barton (1993)
• “The position of insertions and deletions suggests regions where surface loops exist…
• Conserved glycine or proline suggests aβ-turn…
• Residues with hydrophobic properties conserved at i, i+2, i+4 (etc) separated by unconserved or hydrophilic residues suggests a surface β-strand…
### Multiple Alignment Features
• Barton (1993)
• “The position of insertions and deletions suggests regions where surface loops exist…
• Conserved glycine or proline suggests aβ-turn…
• Residues with hydrophobic properties conserved at i, i+2, i+4 (etc) separated by unconserved or hydrophilic residues suggests a surfaceβ-strand…
• A short run of hydrophobic amino acids (4 or 5 residues) suggests a buriedβ-strand…
### Multiple Alignment Features
• Barton (1993)
• Pairs of conserved hydrophobic amino acids separated by pairs of unconserved or hydrophilic residues suggests anα-helix with one face packed in the protein core. Similarly, an i, i+3, i+4, i+7 pattern of conserved residues.”
### Multiple Alignment Features
• Cysteine is a rare amino acid, and is often used in disulphide bonds ( pairs of conserved cysteines )
• Charged residues ( histidine, aspartate, glutamate, lysine, arginine ) and other polar residues embedded in a conserved region indicate functional importance
### Quality Assessment
• Bad residues
• Large distance from column consensus
• Bad columns
• Average distance from consensus is high – “entropy”
• Bad regions
• Profile scores
• Bad quality doesn’t always mean badly aligned!
L
I
M
I
I
L
V
E
I
V
L
A
M
P
E
R
M
K
I
D
Q
G
Q
N
M
W
D
L
V
T
W
D
Y
A
A
S
L
D
F
D
N
P
G
G
A
C
R
T
T
L
I
D
R
I
N
A
I
E
V
M
A
K
L
I
Q
### Quality Assessment
• Profiles
• A profile holds scores for each residue type (plus gaps) over every column of a multiple alignment
• Concepts:
• Consensus sequence
• Amino acid similarity
• Some multiple alignment programs use profiles to build or add to an alignment
• Any alignment, or even one sequence, can be a profile (one sequence isn’t a very good one…)
### What can we do with a multiple alignment?
• Identify subgroups (phylogeny)
• Intra-group sequence conservation
• Evolutionary relatedness (view tree)
• Identify motifs (functionality)
• Evolutionary signals
• Highly conserved residues indicate functional or structural significance!
• Widen search for related proteins
• MA better than single sequence
• Consensus sequence / profile useful
RPDDWHLHLR
GGIDTHVHFI
GFTLTHEHIC
PFVEPHIHLD
PKVELHVHLD
### What do we want to do?
• Build a homology model?
• Accuracy
• Perform phylogenetic analysis?
• Completeness
• Functional analysis of a protein family?
• Diversity
### Building the initial alignment
• Fetch related sequences and run alignment
• Clustal, Dialign, TCoffee, Muscle …
• Fetch a multiple alignment from a database and add sequences of interest
• Pfam, ProDom, ADDA …
• Start from a motif-finding procedure
• MEME, Pratt, Gibbs Sampler …
### Adjusting the alignment
• Filter alignment:
• Remove any redundancy
• Remove unrelated sequences
• Remove unwanted domains
• Recalculate alignment if necessary
• Look for conserved motifs, adjust any misalignments. Try different colour schemes and thresholds.
• One step at a time…
### Jalview Alignment Editor
Clamp, M., Cuff, J., Searle, S. M. and Barton, G. J. (2004), "The Jalview Java Alignment Editor", Bioinformatics, 20, 426-7.
HYDROPHOBIC
/ POLAR
hydrophobic
polar
BURIED INDEX
buried
surface
β-STRAND
LIKELIHOOD
probable
unlikely
HELIX LIKELIHOOD
probable
unlikely
### Colouring your alignment
• By conservation thresholds:
### Colouring your alignment
• Conservation index
Amino Acid Property Classification Schema, eg: Livingstone & Barton 1993
### Check PDB Structures
• Load MA with sequence(s) for known PDB structure
• View >> Feature Settings >> Fetch DAS Features (wait...) OR
• Right-click >> Associate Structure with Sequence >> Discover PDB ids (quicker)
• Right-click sequence name >> View PDB Entry
• Structure opens in new window – residues acquire MA colours
• Highlight residues by hovering mouse over alignment or structure
• Label residues by clicking on structure
### Compare Alignment to Structure
• Crucial way of checking alignment!
• Where are gaps / insertions /deletions ?
• In secondary structures: bad
• In surface loops: okay
• Where are our key / functional residues?
• Are they in probable active site?
• Check they are clustered
• Check they are accessible, not buried
### Demonstration and Practice
• Start Jalview (click here)
• Tools >> Preferences >>
Visual
select Maximise Window, unselect Quality, set Font Size to 8 or 9, Colour >> Clustal, uncheck Open File
Editing
check Pad Gaps When Editing
• File >> Input Alignment >> from URL (use this one)
• Get used to the controls – selecting and deselecting sequences/groups (drag mouse), dragging sequences/groups (use shift/ctrl), selecting sequence regions, hiding sequences/groups, removing columns and regions… Then explore menus and tools.
• Now load this alignment – I’ve messed up a good alignment, and now I’d like you to correct it! There are two groups of sequences and one single sequence to adjust.
### Demonstration and Practice
• View >> Feature Settings >> DAS Settings
• select Uniprot, dssp, cath, Pfam, PDBsum_ligands, PDBsum_DNAbinding, then click ‘Save as default’
• click Fetch DAS Features (then click yes at prompt) ...
• Move mouse over alignment and read information about features
• Move mouse over sequence names to check for PDB ids
• Open a PDB structure (choose any)
• View >> uncheck Show All Chains, then use up-arrow key to increase structure size.
• Hover mouse over structure (see how residues are highlighted in the sequence), then do same for sequence. Select residues in the structure by clicking them – a label will appear. Click again to remove label.
• Check position of insertions & deletions using this method. | 3,516 | 13,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.857626 |
http://perplexus.info/show.php?pid=6897&cid=44980 | 1,618,601,635,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00138.warc.gz | 74,392,980 | 4,381 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Each of A, B, C and D is a positive integer with the proviso that A ≤ B ≤ C ≤ D ≤ 20.
Determine the total number of quadruplets (A, B, C, D) such that A*B*C*D is divisible by 50.
No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 7 of 11 |
In order to be divisible by 50, two of the integers must be some combination of 5,10,15, & 20. There are 6 combinations of this [4!/(2!2!)]. Once one of these pairs is picked, there are 153 combinations of the remaining integers [18!/(2!16!)]. This yields 918 total.
Of the 6 original pairs, all satisfy being divisible by 2 except for 5,15. So this pair will require at least one even number included among the remaining two integers. To do this, the 28 odd-odd combinations [8!/(2!6!)] can be subtracted from the total. This leaves 890.
Edited on August 20, 2010, 8:06 pm
Posted by hoodat on 2010-08-20 19:58:08
Search: Search body:
Forums (0) | 331 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-17 | latest | en | 0.825407 |
https://math.stackexchange.com/questions/2756337/schwartz-kristoffel-conformal-mapping | 1,566,540,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00248.warc.gz | 544,537,280 | 31,545 | # Schwartz-Kristoffel conformal mapping
I have the following function from the Schwartz-Kristoffel conformal mapping,
\begin{align} \begin{split} \omega(\zeta) &= c\int_{}^{\zeta}\left(1-\dfrac{a_1}{t}\right)^{a_1-1}\left(1-\dfrac{a_2}{t}\right)^{a_2-1}...\left(1-\dfrac{a_k}{t}\right)^{a_k-1}dt \\ &=c\int_{}^{\zeta} \prod_{k=1}^{n-1} \left(1-\dfrac{a_k}{t}\right) ^{\alpha_k-1} dt \end{split} \end{align}
It says that since $|a_n|=1$, $|t|>1$, expanding the integrand of the above equation into series and integrating, one obtain, \begin{align} \begin{split} \omega(\zeta)=c\left\lbrace\zeta-[(a_1-1)a_1+(a_2-1)a_2+...+(a_k-1)a_k]\ln\zeta+\dfrac{e_1}{\zeta}+\dfrac{e_2}{\zeta^2}+... \right\rbrace \end{split} \end{align}
Can someone explain how do we get to that last equation above?
Let us make some preparations at first.
Preparation 1. Thanks to the fact that $\left|a_k\right|=1$ and that $\left|t\right|>1$, we have $\left|a_k/t\right|<1$, for which the logarithmic function yields the following expansion $$\log\left(1-\frac{a_k}{t}\right)=-\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m.$$
Preparation 2. Note that the exponential function is holomorphic on the whole complex plane, i.e., $\forall\,z\in\mathbb{C}$, $$\exp z=\sum_{p=0}^{\infty}\frac{1}{p!}z^p.$$
With the above two preparations, we may rewrite the original integrand as follows. \begin{align} I&=\prod_{k=1}^{n-1}\left(1-\frac{a_k}{t}\right)^{a_k-1}\\ &=\prod_{k=1}^{n-1}\exp\left[\left(a_k-1\right)\log\left(1-\frac{a_k}{t}\right)\right]&\text{(using $a^b=e^{b\log a}$)}\\ &=\exp\left[\sum_{k=1}^{n-1}\left(a_k-1\right)\log\left(1-\frac{a_k}{t}\right)\right]&\text{(using $e^{a_1}e^{a_2}\cdots=e^{a_1+a_2+\cdots}$)}\\ &=\exp\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]&\text{(using preparation 1)}\\ &=\sum_{p=0}^{\infty}\frac{1}{p!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^p.&\text{(using preparation 2)} \end{align} This last line implies that the integrand is no more than a power series with respect to $1/t$. That is, as one expands all brackets and parentheses therein, the result reads $$I=c_0+c_1\left(\frac{1}{t}\right)+c_2\left(\frac{1}{t}\right)^2+\cdots=c_0+\frac{c_1}{t}+\frac{c_2}{t^2}+\cdots,$$ where each $c_j$ is a constant. These constants can be determined from the expression of $I$ from above. For example, $c_0$ is the constant coefficient, which corresponds to the $p=0$ term in the expression (because any $p>0$ term would include $1/t$). When $p=0$, we have $$\frac{1}{p!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^p=\frac{1}{0!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^0=1.$$ Thus $$c_0=1.$$ Likewise, $c_1$ corresponds to the $p=1,m=1$ term, because the least order of $t$ in any $p>1$ term is $\left(1/t\right)^p$, while the least order of $t$ in any $p=1,m>1$ term is $\left(1/t\right)^m$. Taking $p=m=1$, and we eventually have $$c_1=-\sum_{k=1}^{n-1}\left(1-a_k\right)a_k.$$
Now, the integration of the integrand is of the form $$\int^{\zeta}I{\rm d}t=\int^{\zeta}\left(c_0+\frac{c_1}{t}+\frac{c_2}{t^2}+\frac{c_3}{t^3}+\cdots\right){\rm d}t=c_0\zeta+c_1\log\zeta-\frac{c_2}{\zeta}-\frac{c_3}{2\zeta^2}+\cdots,$$ or, by using the above values for $c_0$ and $c_1$, $$\int^{\zeta}I{\rm d}t=\zeta-\left(\sum_{k=1}^{n-1}\left(1-a_k\right)a_k\right)\log\zeta+\frac{e_1}{\zeta}+\frac{e_2}{\zeta^2}+\cdots,$$ where $e_j$'s are constants made up of $c_j$'s, e.g., $e_1=-c_2$ and $e_2=-c_3/2$.
Finally, there is a common factor $c$ in front of all terms. Taking this term into consideration, and one obtains $$\omega(\zeta)=c\int^{\zeta}I{\rm d}t=c\left[\zeta-\left(\sum_{k=1}^{n-1}\left(1-a_k\right)a_k\right)\log\zeta+\frac{e_1}{\zeta}+\frac{e_2}{\zeta^2}+\cdots\right],$$ as is expected.
• Hi hypernova. Thanks for your answer! Do you mind to have a live chat discussion? I still cannot understand some of the steps above. – BeeTiau Apr 27 '18 at 16:12
• Do you mind to expand the derivation from step #2 to #3 and from #3 to #4 above? – BeeTiau Apr 27 '18 at 16:14
• Or perhaps, explain what was actually used to derive step #2 to #3 and from #3 to #4 above? – BeeTiau Apr 27 '18 at 16:20
• @BeeTiau: Sure! Apology for this delayed response. Can we start a live chat now? Plus, were you asking how the integrand is expanded to series or how the $c_0$ and $c_1$ are determined? – hypernova Apr 27 '18 at 16:23
• Hi hypernova, I think if you could help me to expand the derivation, that would great. I think that's better for the benefit of everyone than just to have a live chat that I previously asked for. – BeeTiau Apr 27 '18 at 16:29 | 1,916 | 4,820 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-35 | latest | en | 0.480451 |
https://jp.mathworks.com/matlabcentral/cody/problems/730-how-many-trades-represent-all-the-profit/solutions/331034 | 1,579,854,735,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250616186.38/warc/CC-MAIN-20200124070934-20200124095934-00207.warc.gz | 513,419,637 | 15,619 | Cody
# Problem 730. How many trades represent all the profit?
Solution 331034
Submitted on 8 Oct 2013 by lkjslkjdlk
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% trades = [1 3 -4 2 -1 2 3] y_correct = 2; assert(isequal(trade_profit(trades),y_correct))
trades = 1 3 -4 2 -1 2 3
2 Pass
%% trades = [1 2 3 -5] y_correct = 1; assert(isequal(trade_profit(trades),y_correct))
trades = 1 2 3 -5
3 Pass
%% trades = [1 2 3 4 5 6] y_correct = 6; assert(isequal(trade_profit(trades),y_correct))
trades = 1 2 3 4 5 6
4 Pass
%% trades = [-2 3 -4 5 -6 1 2 3 4 5] y_correct = 3; assert(isequal(trade_profit(trades),y_correct))
trades = -2 3 -4 5 -6 1 2 3 4 5 | 303 | 784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-05 | latest | en | 0.554913 |
http://blog.myrank.co.in/2017/01/12/ | 1,566,581,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00056.warc.gz | 29,135,810 | 21,038 | # Logic Puzzles 2
Puzzle 1: What number or numbers comes next in this series?Puzzle 2: Suppose today is Thursday. What day of the week will it be 19 days from today? Puzzle 3: What is the odd one out?Puzzle 4: What is the factorial of 5? A. 1 B. 60 C. 120 D. 180 Puzzle 5: The average of three numbers is 29. Read more about Logic Puzzles 2[…]
# Continuous functions
A function f(x) is said to be continuous, if it is continuous at each point of its domain.Everywhere continuous function: A function f(x) is said to be everywhere continuous if it is continuous on the entire real line (-∞, ∞) i.e. On R. Some fundamental results on continuous functions: Here, we list some fundamentals result Read more about Continuous functions[…] | 202 | 735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-35 | longest | en | 0.888344 |
https://www.enotes.com/homework-help/what-y-dy-dx-sin-lnx-x-255786 | 1,521,616,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00312.warc.gz | 771,587,132 | 9,859 | # What is y if dy/dx=sin(lnx)/x?
hala718 | Certified Educator
Given dy/dx = sin(lnx)/x
We need to find y.
Then we will find the integral of dy/dx.
Let u = ln x ==> du = 1/x dx ==> xdu = dx
==> dy/dx = sinu / x
==> dy = sin(u) /x * dx
==> Int dy = Int sin(u)/x * x du
==> y = Int sinu du
==> y= -cos u + C
Now we will substitute with u= ln x
==> y= -cos ( lnx ) + C
giorgiana1976 | Student
To determine the primitive y, we'll have to calculate the indefinite integral of dy.
We'll solve by replacing ln x by t:
ln x = t
We'll differentiate both sides:
dx/x = dt
We'll re-write the integral in the new variable t:
Int dy = Int sin(lnx) dx/x
Int sin(lnx) dx/x = Int sin t dt
Int sin t dt = - cos t + C
We'll replace t by ln x:
Int sin(lnx) dx/x = - cos (ln x) + C
The primitive function is: y = - cos (ln x) + C. | 285 | 835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-13 | latest | en | 0.751441 |
https://notebooks-prod.herokuapp.com/previews?key=570e873588217b6f69000557%2F0dd6c532-0be2-4788-9d8c-3a2643d2d613%2Fpreview.html&vf=false | 1,568,556,002,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571506.61/warc/CC-MAIN-20190915134729-20190915160729-00294.warc.gz | 594,928,178 | 469,670 | Notebook
# Generalized Method of Moments with ARCH and GARCH Models¶
By Delaney Granizo-Mackenzie and Andrei Kirilenko developed as part of the Masters of Finance curriculum at MIT Sloan.
Part of the Quantopian Lecture Series:
AutoRegressive Conditionally Heteroskedastic (ARCH) occurs when the volatility of a time series is also autoregressive.
In [1]:
import cvxopt
from functools import partial
import math
import numpy as np
import scipy
from scipy import stats
import statsmodels.api as sm
from statsmodels.stats.stattools import jarque_bera
import matplotlib.pyplot as plt
## Simulating a GARCH(1, 1) Case¶
We'll start by using Monte Carlo sampling to simulate a GARCH(1, 1) process. Our dynamics will be
$$\sigma*1 = \sqrt{\frac{a_0}{1-a_1-b_1}}$$$$\sigma_t^2 = a_0 + a_1 x*{t-1}^2+b*1 \sigma*{t-1}^2$$$$x_t = \sigma_t \epsilon_t$$$$\epsilon \sim \mathcal{N}(0, 1)$$
Our parameters will be $a_0 = 1$, $a_1=0.1$, and $b_1=0.8$. We will drop the first 10% (burn-in) of our simulated values.
In [2]:
# Define parameters
a0 = 1.0
a1 = 0.1
b1 = 0.8
sigma1 = math.sqrt(a0 / (1 - a1 - b1))
In [3]:
def simulate_GARCH(T, a0, a1, b1, sigma1):
# Initialize our values
X = np.ndarray(T)
sigma = np.ndarray(T)
sigma[0] = sigma1
for t in range(1, T):
# Draw the next x_t
X[t - 1] = sigma[t - 1] * np.random.normal(0, 1)
# Draw the next sigma_t
sigma[t] = math.sqrt(a0 + b1 * sigma[t - 1]**2 + a1 * X[t - 1]**2)
X[T - 1] = sigma[T - 1] * np.random.normal(0, 1)
return X, sigma
Now we'll compare the tails of the GARCH(1, 1) process with normally distributed values. We expect to see fatter tails, as the GARCH(1, 1) process will experience extreme values more often.
In [4]:
X, _ = simulate_GARCH(10000, a0, a1, b1, sigma1)
X = X[1000:] # Drop burn in
X = X / np.std(X) # Normalize X
def compare_tails_to_normal(X):
# Define matrix to store comparisons
A = np.zeros((2,4))
for k in range(4):
A[0, k] = len(X[X > (k + 1)]) / float(len(X)) # Estimate tails of X
A[1, k] = 1 - stats.norm.cdf(k + 1) # Compare to Gaussian distribution
return A
compare_tails_to_normal(X)
Out[4]:
array([[ 1.54888889e-01, 2.41111111e-02, 1.11111111e-03,
0.00000000e+00],
[ 1.58655254e-01, 2.27501319e-02, 1.34989803e-03,
3.16712418e-05]])
Sure enough, the tails of the GARCH(1, 1) process are fatter. We can also look at this graphically, although it's a little tricky to see.
In [5]:
plt.hist(X, bins=50)
plt.xlabel('sigma')
plt.ylabel('observations');
In [6]:
# Sample values from a normal distribution
X2 = np.random.normal(0, 1, 9000)
both = np.matrix([X, X2])
In [7]:
# Plot both the GARCH and normal values
plt.plot(both.T, alpha=.7);
plt.axhline(X2.std(), color='yellow', linestyle='--')
plt.axhline(-X2.std(), color='yellow', linestyle='--')
plt.axhline(3*X2.std(), color='red', linestyle='--')
plt.axhline(-3*X2.std(), color='red', linestyle='--')
plt.xlabel('time')
plt.ylabel('sigma');
What we're looking at here is the GARCH process in blue and the normal process in green. The 1 and 3 std bars are drawn on the plot. We can see that the blue GARCH process tends to cross the 3 std bar much more often than the green normal one.
## Testing for ARCH Behavior¶
The first step is to test for ARCH conditions. To do this we run a regression on $x_t$ fitting the following model.
$$x*t^2 = a_0 + a_1 x*{t-1}^2 + \dots + a*p x*{t-p}^2$$
We use OLS to estimate $\hat\theta = (\hat a_0, \hat a_1, \dots, \hat a_p)$ and the covariance matrix $\hat\Omega$. We can then compute the test statistic
$$F = \hat\theta \hat\Omega^{-1} \hat\theta'$$
We will reject if $F$ is greater than the 95% confidence bars in the $\mathcal(X)^2(p)$ distribution.
To test, we'll set $p=20$ and see what we get.
In [8]:
X, _ = simulate_GARCH(1100, a0, a1, b1, sigma1)
X = X[100:] # Drop burn in
p = 20
# Drop the first 20 so we have a lag of p's
Y2 = (X**2)[p:]
X2 = np.ndarray((980, p))
for i in range(p, 1000):
X2[i - p, :] = np.asarray((X**2)[i-p:i])[::-1]
model = sm.OLS(Y2, X2)
model = model.fit()
theta = np.matrix(model.params)
omega = np.matrix(model.cov_HC0)
F = np.asscalar(theta * np.linalg.inv(omega) * theta.T)
print np.asarray(theta.T).shape
plt.plot(range(20), np.asarray(theta.T))
plt.xlabel('Lag Amount')
plt.ylabel('Estimated Coefficient for Lagged Datapoint')
print 'F = ' + str(F)
chi2dist = scipy.stats.chi2(p)
pvalue = 1-chi2dist.cdf(F)
print 'p-value = ' + str(pvalue)
# Finally let's look at the significance of each a_p as measured by the standard deviations away from 0
print theta/np.diag(omega)
(20, 1)
F = 341.736612805
p-value = 0.0
[[ 82.59140112 71.46296619 42.80265659 5.9693111 5.61141539
30.89781715 36.54379448 21.58236845 30.66660757 -10.92922828
13.41504385 48.73313057 10.17361295 -70.71538298 23.39274787
40.53162163 56.12994402 -28.95568019 42.05335583 35.05024461]]
## Fitting GARCH(1, 1) with MLE¶
Once we've decided that the data might have an underlying GARCH(1, 1) model, we would like to fit GARCH(1, 1) to the data by estimating parameters.
To do this we need the log-likelihood function
$$\mathcal{L}(\theta) = \sum_{t=1}^T - \ln \sqrt{2\pi} - \frac{x_t^2}{2\sigma_t^2} - \frac{1}{2}\ln(\sigma_t^2)$$
To evaluate this function we need $x_t$ and $\sigma_t$ for $1 \leq t \leq T$. We have $x_t$, but we need to compute $\sigma_t$. To do this we need to make a guess for $\sigma_1$. Our guess will be $\sigma_1^2 = \hat E[x_t^2]$. Once we have our initial guess we compute the rest of the $\sigma$'s using the equation
$$\sigma*t^2 = a_0 + a_1 x*{t-1}^2 + b*1\sigma*{t-1}^2$$
In [9]:
X, _ = simulate_GARCH(10000, a0, a1, b1, sigma1)
X = X[1000:] # Drop burn in
In [10]:
# Here's our function to compute the sigmas given the initial guess
def compute_squared_sigmas(X, initial_sigma, theta):
a0 = theta[0]
a1 = theta[1]
b1 = theta[2]
T = len(X)
sigma2 = np.ndarray(T)
sigma2[0] = initial_sigma ** 2
for t in range(1, T):
# Here's where we apply the equation
sigma2[t] = a0 + a1 * X[t-1]**2 + b1 * sigma2[t-1]
return sigma2
Let's look at the sigmas we just generated.
In [11]:
plt.plot(range(len(X)), compute_squared_sigmas(X, np.sqrt(np.mean(X**2)), (1, 0.5, 0.5)))
plt.xlabel('Time')
plt.ylabel('Sigma');
Now that we can compute the $\sigma_t$'s, we'll define the actual log likelihood function. This function will take as input our observations $x$ and $\theta$ and return $-\mathcal{L}(\theta)$. It is important to note that we return the negative log likelihood, as this way our numerical optimizer can minimize the function while maximizing the log likelihood.
Note that we are constantly re-computing the $\sigma_t$'s in this function.
In [12]:
def negative_log_likelihood(X, theta):
T = len(X)
# Estimate initial sigma squared
initial_sigma = np.sqrt(np.mean(X ** 2))
# Generate the squared sigma values
sigma2 = compute_squared_sigmas(X, initial_sigma, theta)
# Now actually compute
return -sum(
[-np.log(np.sqrt(2.0 * np.pi)) -
(X[t] ** 2) / (2.0 * sigma2[t]) -
0.5 * np.log(sigma2[t]) for
t in range(T)]
)
Now we perform numerical optimization to find our estimate for $$\hat\theta = \arg \max*{(a_0, a_1, b_1)}\mathcal{L}(\theta) = \arg \min*{(a_0, a_1, b_1)}-\mathcal{L}(\theta)$$
We have some constraints on this
$$a_1 \geq 0, b_1 \geq 0, a_1+b_1 < 1$$
In [13]:
# Make our objective function by plugging X into our log likelihood function
objective = partial(negative_log_likelihood, X)
# Define the constraints for our minimizer
def constraint1(theta):
return np.array([1 - (theta[1] + theta[2])])
def constraint2(theta):
return np.array([theta[1]])
def constraint3(theta):
return np.array([theta[2]])
cons = ({'type': 'ineq', 'fun': constraint1},
{'type': 'ineq', 'fun': constraint2},
{'type': 'ineq', 'fun': constraint3})
# Actually do the minimization
result = scipy.optimize.minimize(objective, (1, 0.5, 0.5),
method='SLSQP',
constraints = cons)
theta_mle = result.x
print 'theta MLE: ' + str(theta_mle)
theta MLE: [ 0.89882685 0.10272342 0.81126578]
Now we would like a way to check our estimate. We'll look at two things:
1. How fat are the tails of the residuals.
2. How normal are the residuals under the Jarque-Bera normality test.
We'll do both in our check_theta_estimate function.
In [14]:
def check_theta_estimate(X, theta_estimate):
initial_sigma = np.sqrt(np.mean(X ** 2))
sigma = np.sqrt(compute_squared_sigmas(X, initial_sigma, theta_estimate))
epsilon = X / sigma
print 'Tails table'
print compare_tails_to_normal(epsilon / np.std(epsilon))
print ''
_, pvalue, _, _ = jarque_bera(epsilon)
print 'Jarque-Bera probability normal: ' + str(pvalue)
check_theta_estimate(X, theta_mle)
Tails table
[[ 1.58888889e-01 2.31111111e-02 1.22222222e-03 0.00000000e+00]
[ 1.58655254e-01 2.27501319e-02 1.34989803e-03 3.16712418e-05]]
Jarque-Bera probability normal: 0.547561671115
## GMM for Estimating GARCH(1, 1) Parameters¶
We've just computed an estimate using MLE, but we can also use Generalized Method of Moments (GMM) to estimate the GARCH(1, 1) parameters.
To do this we need to define our moments. We'll use 4.
1. The residual $\hat\epsilon_t = x_t / \hat\sigma_t$
2. The variance of the residual $\hat\epsilon_t^2$
3. The skew moment $\mu_3/\hat\sigma_t^3 = (\hat\epsilon_t - E[\hat\epsilon_t])^3 / \hat\sigma_t^3$
4. The kurtosis moment $\mu_4/\hat\sigma_t^4 = (\hat\epsilon_t - E[\hat\epsilon_t])^4 / \hat\sigma_t^4$
In [15]:
# The n-th standardized moment
# skewness is 3, kurtosis is 4
def standardized_moment(x, mu, sigma, n):
return ((x - mu) ** n) / (sigma ** n)
GMM now has three steps.
Start with $W$ as the identity matrix.
1. Estimate $\hat\theta*1$ by using numerical optimization to minimize $$\min*{\theta \in \Theta} \left(\frac{1}{T} \sum*{t=1}^T g(x_t, \hat\theta)\right)' W \left(\frac{1}{T}\sum*{t=1}^T g(x_t, \hat\theta)\right)$$
2. Recompute $W$ based on the covariances of the estimated $\theta$. (Focus more on parameters with explanatory power) $$\hat W*{i+1} = \left(\frac{1}{T}\sum*{t=1}^T g(x_t, \hat\theta_i)g(x_t, \hat\theta_i)'\right)^{-1}$$
3. Repeat until $|\hat\theta_{i+1} - \hat\theta_i| < \epsilon$ or we reach an iteration threshold.
Initialize $W$ and $T$ and define the objective function we need to minimize.
In [16]:
def gmm_objective(X, W, theta):
# Compute the residuals for X and theta
initial_sigma = np.sqrt(np.mean(X ** 2))
sigma = np.sqrt(compute_squared_sigmas(X, initial_sigma, theta))
e = X / sigma
# Compute the mean moments
m1 = np.mean(e)
m2 = np.mean(e ** 2) - 1
m3 = np.mean(standardized_moment(e, np.mean(e), np.std(e), 3))
m4 = np.mean(standardized_moment(e, np.mean(e), np.std(e), 4) - 3)
G = np.matrix([m1, m2, m3, m4]).T
return np.asscalar(G.T * W * G)
def gmm_variance(X, theta):
# Compute the residuals for X and theta
initial_sigma = np.sqrt(np.mean(X ** 2))
sigma = np.sqrt(compute_squared_sigmas(X, initial_sigma, theta))
e = X / sigma
# Compute the squared moments
m1 = e ** 2
m2 = (e ** 2 - 1) ** 2
m3 = standardized_moment(e, np.mean(e), np.std(e), 3) ** 2
m4 = (standardized_moment(e, np.mean(e), np.std(e), 4) - 3) ** 2
# Compute the covariance matrix g * g'
T = len(X)
s = np.ndarray((4, 1))
for t in range(T):
G = np.matrix([m1[t], m2[t], m3[t], m4[t]]).T
s = s + G * G.T
return s / T
Now we're ready to the do the iterated minimization step.
In [17]:
# Initialize GMM parameters
W = np.identity(4)
gmm_iterations = 10
# First guess
theta_gmm_estimate = theta_mle
# Perform iterated GMM
for i in range(gmm_iterations):
# Estimate new theta
objective = partial(gmm_objective, X, W)
result = scipy.optimize.minimize(objective, theta_gmm_estimate, constraints=cons)
theta_gmm_estimate = result.x
print 'Iteration ' + str(i) + ' theta: ' + str(theta_gmm_estimate)
# Recompute W
W = np.linalg.inv(gmm_variance(X, theta_gmm_estimate))
check_theta_estimate(X, theta_gmm_estimate)
Iteration 0 theta: [ 0.898825 0.10283861 0.81129305]
Iteration 1 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 2 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 3 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 4 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 5 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 6 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 7 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 8 theta: [ 0.89881434 0.1027472 0.81118925]
Iteration 9 theta: [ 0.89881434 0.1027472 0.81118925]
Tails table
[[ 1.58888889e-01 2.32222222e-02 1.22222222e-03 0.00000000e+00]
[ 1.58655254e-01 2.27501319e-02 1.34989803e-03 3.16712418e-05]]
Jarque-Bera probability normal: 0.547586182431
## Predicting the Future: How to actually use what we've done¶
Now that we've fitted a model to our observations, we'd like to be able to predict what the future volatility will look like. To do this, we can just simulate more values using our original GARCH dynamics and the estimated parameters.
The first thing we'll do is compute an initial $\sigma_t$. We'll compute our squared sigmas and take the last one.
In [18]:
sigma_hats = np.sqrt(compute_squared_sigmas(X, np.sqrt(np.mean(X**2)), theta_mle))
initial_sigma = sigma_hats[-1]
initial_sigma
Out[18]:
2.9977470389171796
Now we'll just sample values walking forward.
In [19]:
a0_estimate = theta_gmm_estimate[0]
a1_estimate = theta_gmm_estimate[1]
b1_estimate = theta_gmm_estimate[2]
X_forecast, sigma_forecast = simulate_GARCH(100, a0_estimate, a1_estimate, b1_estimate, initial_sigma)
In [20]:
plt.plot(range(-100, 0), X[-100:], 'b-')
plt.plot(range(-100, 0), sigma_hats[-100:], 'r-')
plt.plot(range(0, 100), X_forecast, 'b--')
plt.plot(range(0, 100), sigma_forecast, 'r--')
plt.xlabel('Time')
plt.legend(['X', 'sigma']);
One should note that because we are moving foward using a random walk, this analysis is supposed to give us a sense of the magnitude of sigma and therefore the risk we could face. It is not supposed to accurately model future values of X. In practice you would probably want to use Monte Carlo sampling to generate thousands of future scenarios, and then look at the potential range of outputs. We'll try that now. Keep in mind that this is a fairly simplistic way of doing this analysis, and that better techniques, such as Bayesian cones, exist.
In [21]:
plt.plot(range(-100, 0), X[-100:], 'b-')
plt.plot(range(-100, 0), sigma_hats[-100:], 'r-')
plt.xlabel('Time')
plt.legend(['X', 'sigma'])
max_X = [-np.inf]
min_X = [np.inf]
for i in range(100):
X_forecast, sigma_forecast = simulate_GARCH(100, a0_estimate, a1_estimate, b1_estimate, initial_sigma)
if max(X_forecast) > max(max_X):
max_X = X_forecast
elif min(X_forecast) < min(max_X):
min_X = X_forecast
plt.plot(range(0, 100), X_forecast, 'b--', alpha=0.05)
plt.plot(range(0, 100), sigma_forecast, 'r--', alpha=0.05)
# Draw the most extreme X values specially
plt.plot(range(0, 100), max_X, 'g--', alpha=1.0)
plt.plot(range(0, 100), min_X, 'g--', alpha=1.0);
This presentation is for informational purposes only and does not constitute an offer to sell, a solicitation to buy, or a recommendation for any security; nor does it constitute an offer to provide investment advisory or other services by Quantopian, Inc. ("Quantopian"). Nothing contained herein constitutes investment advice or offers any opinion with respect to the suitability of any security, and any views expressed herein should not be taken as advice to buy, sell, or hold any security or as an endorsement of any security or company. In preparing the information contained herein, Quantopian, Inc. has not taken into account the investment needs, objectives, and financial circumstances of any particular investor. Any views expressed and data illustrated herein were prepared based upon information, believed to be reliable, available to Quantopian, Inc. at the time of publication. Quantopian makes no guarantees as to their accuracy or completeness. All information is subject to change and may quickly become unreliable for various reasons, including changes in market conditions or economic circumstances. | 5,228 | 16,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-39 | latest | en | 0.570444 |
http://rockapproll.com/app/Everyday-Mathematics--Monster-Squeeze- | 1,503,551,954,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133032.51/warc/CC-MAIN-20170824043524-20170824063524-00346.warc.gz | 354,539,311 | 17,484 | # Everyday Mathematics® Monster Squeeze™
McGraw-Hill’s Monster Squeeze game reinforces number recognition and offers a quick and easy way to practice number line concepts and number comparisons. ...
### Description
McGraw-Hill’s Monster Squeeze game reinforces number recognition and offers a quick and easy way to practice number line concepts and number comparisons. This two-player game runs on the iPad, iPhone, and iPod Touch. Players take turns finding the secret number on a number line. In each round, the monster designates a secret number on a number line. Players try to guess this number by clicking one of the numbers. If the number chosen is larger than the secret number, that number and all larger numbers on the number line are covered by the monster. If the number chosen is smaller than the secret number, that number and all smaller numbers on the number line are covered by the monster. Players take turns clicking numbers until one of them identifies the secret number and wins the round. There are a total of 5 rounds in a game. Features of Monster Squeeze: - 2 players (for extra practice, a single player can play as Player 1 and Player 2) - 5 competitive rounds of play - Practices number line concepts and comparisons of numbers - Customizable number lines: available number lines from 1 to 10, 5 to 15, 10 to 20, 15 to 25, and 20 to 30 - Full tutorial - Compatible with iPad, iPhone, and iPod Touch Look for these other Everyday Mathematics® Games Apps: - Addition Top-It™ - Subtraction Top-It™ - Beat the Computer™ Multiplication - Name That Number™ - Equivalent Fractions™ - Tric-Trac™ - Baseball Multiplication™ 1–6 Facts - Baseball Multiplication™ 1–12 Facts - Divisibility Dash™ Everyday Mathematics® web site: everydaymath.com
### Technical specifications
Version: 1.0
Size: 14.9 MB
System:
Price: 1,81 €
Developed by McGraw-Hill School Education Group
Day of release: 2011-01-18
Recommended age: 4+ | 453 | 1,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.82897 |
https://foreach.id/EN/common/power/kilocalorieIT%7Cminute-to-kilocalorieIT%7Chour.html | 1,618,579,411,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00192.warc.gz | 366,069,807 | 10,797 | # Convert kilocalorie (IT)/minute to kilocalorie (IT)/hour (kcal/m to kcal/h)
Batch Convert
• kilocalorie (IT)/hour [kcal/h]
• kilocalorie (IT)/minute [kcal/m]
Copy
_
Copy
• kilocalorie (IT)/hour [kcal/h]
• kilocalorie (IT)/minute [kcal/m]
## Kilocalorie (IT)/minute to Kilocalorie (IT)/hour (kcal/m to kcal/h)
### Kilocalorie (IT)/minute (Symbol or Abbreviation: kcal/m)
Kilocalorie (IT)/minute is one of power units. Kilocalorie (IT)/minute abbreviated or symbolized by kcal/m. The value of 1 kilocalorie (IT)/minute is equal to 69.78 watt. In its relation with kilocalorie (IT)/hour, 1 kilocalorie (IT)/minute is equal to 60 kilocalorie (IT)/hour.
#### Relation with other units
1 kilocalorie (IT)/minute equals to 69.78 watt
1 kilocalorie (IT)/minute equals to 0.06978 kilowatt
1 kilocalorie (IT)/minute equals to 0.00006978 megawatt
1 kilocalorie (IT)/minute equals to 69,780,000,000,000 picowatt
1 kilocalorie (IT)/minute equals to 69,780,000,000 nanowatt
1 kilocalorie (IT)/minute equals to 69,780,000 microwatt
1 kilocalorie (IT)/minute equals to 69,780 milliwatt
1 kilocalorie (IT)/minute equals to 6,978 centiwatt
1 kilocalorie (IT)/minute equals to 697.8 deciwatt
1 kilocalorie (IT)/minute equals to 6.978 dekawatt
1 kilocalorie (IT)/minute equals to 0.6978 hectowatt
1 kilocalorie (IT)/minute equals to 6.978e-8 gigawatt
1 kilocalorie (IT)/minute equals to 0.093577 horsepower
1 kilocalorie (IT)/minute equals to 0.094874 horsepower (metric)
1 kilocalorie (IT)/minute equals to 0.0071135 horsepower (boiler)
1 kilocalorie (IT)/minute equals to 0.093539 horsepower (electric)
1 kilocalorie (IT)/minute equals to 0.093533 horsepower (water)
1 kilocalorie (IT)/minute equals to 0.094874 pferdestarke
1 kilocalorie (IT)/minute equals to 238.1 Btu (IT)/hour
1 kilocalorie (IT)/minute equals to 3.9683 Btu (IT)/minute
1 kilocalorie (IT)/minute equals to 0.066139 Btu (IT)/second
1 kilocalorie (IT)/minute equals to 238.26 Btu (th)/hour
1 kilocalorie (IT)/minute equals to 3.971 Btu (th)/minute
1 kilocalorie (IT)/minute equals to 0.066183 Btu (th)/second
1 kilocalorie (IT)/minute equals to 0.0002381 MBtu (IT)/hour
1 kilocalorie (IT)/minute equals to 0.2381 MBH
1 kilocalorie (IT)/minute equals to 0.019842 ton (refrigeration)
1 kilocalorie (IT)/minute equals to 60 kilocalorie (IT)/hour
1 kilocalorie (IT)/minute equals to 0.016667 kilocalorie (IT)/second
1 kilocalorie (IT)/minute equals to 60.04 kilocalorie (th)/hour
1 kilocalorie (IT)/minute equals to 1.0007 kilocalorie (th)/minute
1 kilocalorie (IT)/minute equals to 0.016678 kilocalorie (th)/second
1 kilocalorie (IT)/minute equals to 60,000 calorie (IT)/hour
1 kilocalorie (IT)/minute equals to 1,000 calorie (IT)/minute
1 kilocalorie (IT)/minute equals to 16.667 calorie (IT)/second
1 kilocalorie (IT)/minute equals to 60,040 calorie (th)/hour
1 kilocalorie (IT)/minute equals to 1,000.7 calorie (th)/minute
1 kilocalorie (IT)/minute equals to 16.678 calorie (th)/second
1 kilocalorie (IT)/minute equals to 185,280 foot pound-force/hour
1 kilocalorie (IT)/minute equals to 3,088 foot pound-force/minute
1 kilocalorie (IT)/minute equals to 51.467 foot pound-force/second
1 kilocalorie (IT)/minute equals to 697,800,000 erg/second
1 kilocalorie (IT)/minute equals to 0.06978 kilovolt ampere
1 kilocalorie (IT)/minute equals to 69.78 volt ampere
1 kilocalorie (IT)/minute equals to 69.78 newton meter/second
1 kilocalorie (IT)/minute equals to 69.78 joule/second
1 kilocalorie (IT)/minute equals to 6.978e-8 gigajoule/second
1 kilocalorie (IT)/minute equals to 0.00006978 megajoule/second
1 kilocalorie (IT)/minute equals to 0.06978 kilojoule/second
1 kilocalorie (IT)/minute equals to 0.6978 hectojoule/second
1 kilocalorie (IT)/minute equals to 6.978 dekajoule/second
1 kilocalorie (IT)/minute equals to 697.8 decijoule/second
1 kilocalorie (IT)/minute equals to 6,978 centijoule/second
1 kilocalorie (IT)/minute equals to 69,780 millijoule/second
1 kilocalorie (IT)/minute equals to 69,780,000 microjoule/second
1 kilocalorie (IT)/minute equals to 69,780,000,000 nanojoule/second
1 kilocalorie (IT)/minute equals to 69,780,000,000,000 picojoule/second
1 kilocalorie (IT)/minute equals to 251,210 joule/hour
1 kilocalorie (IT)/minute equals to 4,186.8 joule/minute
1 kilocalorie (IT)/minute equals to 251.21 kilojoule/hour
1 kilocalorie (IT)/minute equals to 4.1868 kilojoule/minute
### Kilocalorie (IT)/hour (Symbol or Abbreviation: kcal/h)
Kilocalorie (IT)/hour is one of power units. Kilocalorie (IT)/hour abbreviated or symbolized by kcal/h. The value of 1 kilocalorie (IT)/hour is equal to 1.163 watt. In its relation with kilocalorie (IT)/minute, 1 kilocalorie (IT)/hour is equal to 0.016667 kilocalorie (IT)/minute.
#### Relation with other units
1 kilocalorie (IT)/hour equals to 1.163 watt
1 kilocalorie (IT)/hour equals to 0.001163 kilowatt
1 kilocalorie (IT)/hour equals to 0.000001163 megawatt
1 kilocalorie (IT)/hour equals to 1,163,000,000,000 picowatt
1 kilocalorie (IT)/hour equals to 1,163,000,000 nanowatt
1 kilocalorie (IT)/hour equals to 1,163,000 microwatt
1 kilocalorie (IT)/hour equals to 1,163 milliwatt
1 kilocalorie (IT)/hour equals to 116.3 centiwatt
1 kilocalorie (IT)/hour equals to 11.63 deciwatt
1 kilocalorie (IT)/hour equals to 0.1163 dekawatt
1 kilocalorie (IT)/hour equals to 0.01163 hectowatt
1 kilocalorie (IT)/hour equals to 1.163e-9 gigawatt
1 kilocalorie (IT)/hour equals to 0.0015596 horsepower
1 kilocalorie (IT)/hour equals to 0.0015812 horsepower (metric)
1 kilocalorie (IT)/hour equals to 0.00011856 horsepower (boiler)
1 kilocalorie (IT)/hour equals to 0.001559 horsepower (electric)
1 kilocalorie (IT)/hour equals to 0.0015589 horsepower (water)
1 kilocalorie (IT)/hour equals to 0.0015812 pferdestarke
1 kilocalorie (IT)/hour equals to 3.9683 Btu (IT)/hour
1 kilocalorie (IT)/hour equals to 0.066139 Btu (IT)/minute
1 kilocalorie (IT)/hour equals to 0.0011023 Btu (IT)/second
1 kilocalorie (IT)/hour equals to 3.971 Btu (th)/hour
1 kilocalorie (IT)/hour equals to 0.066183 Btu (th)/minute
1 kilocalorie (IT)/hour equals to 0.001103 Btu (th)/second
1 kilocalorie (IT)/hour equals to 0.0000039683 MBtu (IT)/hour
1 kilocalorie (IT)/hour equals to 0.0039683 MBH
1 kilocalorie (IT)/hour equals to 0.00033069 ton (refrigeration)
1 kilocalorie (IT)/hour equals to 0.016667 kilocalorie (IT)/minute
1 kilocalorie (IT)/hour equals to 0.00027778 kilocalorie (IT)/second
1 kilocalorie (IT)/hour equals to 1.0007 kilocalorie (th)/hour
1 kilocalorie (IT)/hour equals to 0.016678 kilocalorie (th)/minute
1 kilocalorie (IT)/hour equals to 0.00027796 kilocalorie (th)/second
1 kilocalorie (IT)/hour equals to 1,000 calorie (IT)/hour
1 kilocalorie (IT)/hour equals to 16.667 calorie (IT)/minute
1 kilocalorie (IT)/hour equals to 0.27778 calorie (IT)/second
1 kilocalorie (IT)/hour equals to 1,000.7 calorie (th)/hour
1 kilocalorie (IT)/hour equals to 16.678 calorie (th)/minute
1 kilocalorie (IT)/hour equals to 0.27796 calorie (th)/second
1 kilocalorie (IT)/hour equals to 3,088 foot pound-force/hour
1 kilocalorie (IT)/hour equals to 51.467 foot pound-force/minute
1 kilocalorie (IT)/hour equals to 0.85778 foot pound-force/second
1 kilocalorie (IT)/hour equals to 11,630,000 erg/second
1 kilocalorie (IT)/hour equals to 0.001163 kilovolt ampere
1 kilocalorie (IT)/hour equals to 1.163 volt ampere
1 kilocalorie (IT)/hour equals to 1.163 newton meter/second
1 kilocalorie (IT)/hour equals to 1.163 joule/second
1 kilocalorie (IT)/hour equals to 1.163e-9 gigajoule/second
1 kilocalorie (IT)/hour equals to 0.000001163 megajoule/second
1 kilocalorie (IT)/hour equals to 0.001163 kilojoule/second
1 kilocalorie (IT)/hour equals to 0.01163 hectojoule/second
1 kilocalorie (IT)/hour equals to 0.1163 dekajoule/second
1 kilocalorie (IT)/hour equals to 11.63 decijoule/second
1 kilocalorie (IT)/hour equals to 116.3 centijoule/second
1 kilocalorie (IT)/hour equals to 1,163 millijoule/second
1 kilocalorie (IT)/hour equals to 1,163,000 microjoule/second
1 kilocalorie (IT)/hour equals to 1,163,000,000 nanojoule/second
1 kilocalorie (IT)/hour equals to 1,163,000,000,000 picojoule/second
1 kilocalorie (IT)/hour equals to 4,186.8 joule/hour
1 kilocalorie (IT)/hour equals to 69.78 joule/minute
1 kilocalorie (IT)/hour equals to 4.1868 kilojoule/hour
1 kilocalorie (IT)/hour equals to 0.06978 kilojoule/minute
### How to convert Kilocalorie (IT)/minute to Kilocalorie (IT)/hour (kcal/m to kcal/h):
#### Conversion Table for Kilocalorie (IT)/minute to Kilocalorie (IT)/hour (kcal/m to kcal/h)
kilocalorie (IT)/minute (kcal/m) kilocalorie (IT)/hour (kcal/h)
0.01 kcal/m 0.6 kcal/h
0.1 kcal/m 6 kcal/h
1 kcal/m 60 kcal/h
2 kcal/m 120 kcal/h
3 kcal/m 180 kcal/h
4 kcal/m 240 kcal/h
5 kcal/m 300 kcal/h
6 kcal/m 360 kcal/h
7 kcal/m 420 kcal/h
8 kcal/m 480 kcal/h
9 kcal/m 540 kcal/h
10 kcal/m 600 kcal/h
20 kcal/m 1,200 kcal/h
25 kcal/m 1,500 kcal/h
50 kcal/m 3,000 kcal/h
75 kcal/m 4,500 kcal/h
100 kcal/m 6,000 kcal/h
250 kcal/m 15,000 kcal/h
500 kcal/m 30,000 kcal/h
750 kcal/m 45,000 kcal/h
1,000 kcal/m 60,000 kcal/h
100,000 kcal/m 6,000,000 kcal/h
1,000,000,000 kcal/m 60,000,000,000 kcal/h
1,000,000,000,000 kcal/m 60,000,000,000,000 kcal/h
#### Conversion Table for Kilocalorie (IT)/hour to Kilocalorie (IT)/minute (kcal/h to kcal/m)
kilocalorie (IT)/hour (kcal/h) kilocalorie (IT)/minute (kcal/m)
0.01 kcal/h 0.00016667 kcal/m
0.1 kcal/h 0.0016667 kcal/m
1 kcal/h 0.016667 kcal/m
2 kcal/h 0.033333 kcal/m
3 kcal/h 0.05 kcal/m
4 kcal/h 0.066667 kcal/m
5 kcal/h 0.083333 kcal/m
6 kcal/h 0.1 kcal/m
7 kcal/h 0.11667 kcal/m
8 kcal/h 0.13333 kcal/m
9 kcal/h 0.15 kcal/m
10 kcal/h 0.16667 kcal/m
20 kcal/h 0.33333 kcal/m
25 kcal/h 0.41667 kcal/m
50 kcal/h 0.83333 kcal/m
75 kcal/h 1.25 kcal/m
100 kcal/h 1.6667 kcal/m
250 kcal/h 4.1667 kcal/m
500 kcal/h 8.3333 kcal/m
750 kcal/h 12.5 kcal/m
1,000 kcal/h 16.667 kcal/m
100,000 kcal/h 1,666.7 kcal/m
1,000,000,000 kcal/h 16,667,000 kcal/m
1,000,000,000,000 kcal/h 16,667,000,000 kcal/m
#### Steps to Convert Kilocalorie (IT)/minute to Kilocalorie (IT)/hour (kcal/m to kcal/h)
1. Example: Convert 37 kilocalorie (IT)/minute to kilocalorie (IT)/hour (37 kcal/m to kcal/h).
2. 1 kilocalorie (IT)/minute is equivalent to 60 kilocalorie (IT)/hour (1 kcal/m is equivalent to 60 kcal/h).
3. 37 kilocalorie (IT)/minute (kcal/m) is equivalent to 37 times 60 kilocalorie (IT)/hour (kcal/h).
4. Retrieved 37 kilocalorie (IT)/minute is equivalent to 2220 kilocalorie (IT)/hour (37 kcal/m is equivalent to 2220 kcal/h).
▸▸
▸▸ | 3,724 | 10,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.616358 |
https://www.wyzant.com/resources/answers/topics/density?page=2 | 1,623,761,935,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00348.warc.gz | 1,007,888,486 | 14,778 | 87 Answered Questions for the topic Density
03/18/19
#### Let f(x) = k(9x − x^2) if 0 ≤ x ≤ 9
Let f(x) = k(9x − x2) if 0 ≤ x ≤ 9 and f(x) = 0 if x < 0 or x > 9.(a) For what value of k is f a probability density function? k= ?(b) For that value of k, find P(X > 1). P(X > 1) =... more
03/18/19
#### A milk truck carries milk with density 64.6
A milk truck carries milk with density 64.6 lb/ft^(3) lb/ft3 in a horizontal cylindrical tank with diameter 6 ft.(a) Find the force exerted by the milk on one end of the tank when the tank is full.... more
Density
10/07/18
#### Working out density of liquid c
Liquid A has a density of 0.7g/cm^3.Liquid B has a density of 1.6g/cm^3. 140 g of liquid A and 128g of liquid B are mixed to make liquid C. What is the density of liquid C?
Density
10/01/18
#### A certain liquid has a density of 0.70g/m. Find the mass of 6.4l of this liquid
What is the mass of 6.4 of this liquid
Density
09/16/18
#### if ethanol has a density of 0.785g/cm3, calculate the volume of 82.5g of ethanol
if ethanol has a density of 0.785g/cm3, calculate the volume of 82.5g of ethanol
Density Volume Length Height
08/27/18
#### How do I find the length with the given density, mass, width and a height?
Okay so the problem is: Platinum has a density of 21.4 grams per cubic centimeter. A bar of platinum has a mass of about 1000 grams. It has a width of 51 millimeters and a height of 9.7... more
Density
06/20/18
#### calculate weight
A stone of mass 3kg and density 2gcm-3 is suspended by a spring from a spring balance. the stone is completely submerged in a liquid of density 1.3gcm-3. calculate the weight that the balance reads... more
Density
06/19/18
#### Density g/cm cubed
An object weighing 9.6 g is placed in a graduated cylinder displacing the volume from 10.0 mL to 13.2 mL. What is it’s density in g/ cm cubed?
Density
06/18/18
#### Determine the mass of instrument
A balloon used to carry instrument for meteorogical department up into the atmosphere has a capacity of 30m3 and is filledd with hydrogen. the total mass of balloon and hydrogen is 3kg. Determine... more
Density
06/18/18
#### calculate the density of the wood
A bar of brass of mass 180g and density 8.5gm-3 is attached to a piece of wood of mass 70g and both are totally immersed in water. the apparent mass is 120g. Calculate the density of the wood
Density
06/12/18
#### Calculate the height of the mountain
The pressure at the base of a mountain is 76cm of mercury and at the top it is 65cm of mercury. Calculate the height of the mountain given that the density of air is 1.25kgm-3 and that of mercury... more
Density
06/12/18
#### find the mass of gold contained in the crown
find the mass of gold contained in the crown, if the crown made of an alloy of gold and silver has a volume 60 centimeter cubic and mass of 1.05kg, and the density of gold is 19.3g/centimeter cubic... more
03/26/18
#### Maximum yield
An apple orchard has an average yield of 32 bushels of apples per tree if tree density is 22 trees per acre. For each unit increase in tree density, the yield decreases by 2 bushels per tree. How... more
01/30/18
#### Calculate the Partial pressure of each gas at STP, and calculate the density at STP.
At STP, air can be considered to be a mixture of ideal gases: 21.1% O<sub>2</sub>, 78.0% N<sub>2</sub> and 1.00% Ar by volume. a) Calculate the partial pressure of each... more
Density
01/19/18
#### What is the density of an object?
An object with a mass of 10 g and a volume of 5 cm3 was placed into a container of water.
Density
01/12/18
#### a light alloy consists of 70 percent aluminium and 30 percent magnesium by mass calculate density of the alloy given thet density of aluminium is 2700 kilograms
per metre cubed and density of magnesium is 1740 kilograms per metre cubed
12/10/17
#### Can you help me find a counterexample to the following;
For a subset x⊆R: 'X is dense in R implies that x∩Q≠Ø
Density
11/29/17
#### A chef fills a 50ml container with 43.5g of cooking oil. What is the density of the oil
What is the density of the oil and how do you find it.
Density
11/13/17
#### How to calculate the density of the ring? ( given in g/cm 3 )
A ring has a mass of 38.6 and volume of 2 3 cm.
11/09/17
#### PLEASE TELL ME HOW TO WORK IT OUT, DON'T TELL ME THE ANSWER,THANK YOU!
If you dissolve 300 g of salt in 10 litres of water, what is the density of that solution of salty water? In this particular example, assume that the total volume doesn’t change when salt is added... more
Density
10/10/17
#### Finding density
A block of lead has dimensions of 4.5 cm by 5.2 cm by 6.0 cm. The block has a mass of 1587 g. Calculate the density of lead.
Density Mass
09/13/17
#### What is the mass of 15.0 cm3 of iron?
The density of iron is 7.86 g/cm3. What is the mass of 15.0 cm3 of iron?
05/05/17
#### Alloy Question
An alloy is made from a mixture of gold and silver. The density of gold is 19.3 g/cm^3. The density of silver is 10.5g/cm^3. A ring is made from the alloy. The mass of the ring is 10g and its... more
Density
04/25/17
#### A measure of how much substance is packed into a space
What is density ? What is mass divided by vol? What is water ? What is element?
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 1,623 | 5,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-25 | latest | en | 0.831241 |
https://www.smartickmethod.com/blog/math/operations-and-algebraic-thinking/division/practice-dividing/ | 1,576,183,428,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540545146.75/warc/CC-MAIN-20191212181310-20191212205310-00070.warc.gz | 859,780,423 | 13,602 | # Smartick - Math, one click away
Sep20
## Practice Dividing with and without Remainders
Dividing is equitable distribution, i.e. splitting between equal parts or groups.
##### What is division?
• Dividend: The number that you have to divide.
• Divisor: The number that the dividend is being divided by
• Quotient: The result of the division.
• Remainder: What is left over after the division
##### Dividing with No Remainders
We want to distribute these 12 balls into 3 boxes in equal parts, therefore there must be the same number of balls in each of the boxes.
Let’s start distributing the balls in each box until finally there are 4 balls in each box.
So the division we have done is 12 balls / 3 boxes = 4 balls in each box.
Division is an inverse operation to multiplication and can be considered as repeated subtraction.
For example if we divide 12 by 3:
12 – 3 = 9
9 – 3 = 6
6 – 3 = 3
3 – 3 = 0
The number of times we can subtract 3 from 12 is 4, so 12/3 = 4.
##### Dividing with Remainders or Inexact Divisions
The remainder is the amount left over after dividing one number by another.
For example: if we divide 5 balls between 2 boxes, we have two balls in each box and we are left with one, so 5/ 2 = 2 and the remainder is 1.
In Smartick you can find many division problems with or without remainders. Join in and practice elementary math in a fun way! | 355 | 1,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.916581 |
https://www.thewinningbets.com/predictions-portugal/ligapro/prediction-of-academico-de-viseu-gd-chaves-19-january-2020-28696 | 1,579,867,422,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00272.warc.gz | 1,125,684,785 | 19,715 | Match Day 17 of the League Portugal - LigaPro
• D
• D
• W
• W
• L
GD Chaves
• W
• D
• L
• L
Do you want to increase your chances of winning? Have a look at our prediction on Académico de Viseu and GD Chaves. For the day 17 of the championship of Portugal - LigaPro we have calculated the percentages of the two teams to increase your earning potential If you want to bet using a mathematical method, based on the analysis of league trends and other variables, subscribe to the group of The Winning Bets.
The gap between the two teams makes our prediction quite safe, even if the surprise is behind the angalo. When the prices of the two teams are so different, we must carefully consider the decision to be made Since GD Chaves has well 5 points less compared to Académico de Viseu, we can deduce that on a technical level away team is superior to its opponent. In 2 home victories on 7 matches, Académico de Viseu scored 6 and immediately 8 goals. GD Chaves scored 9 goals and conceded 10, totaling 3 victor out of home on 8 meetings.
## Analysis of the State of Form
Studying the performance in the league, we deduce the state of form of the formations that will take the field. Let's see how they behaved in the last 5 games. The table shows the date of the match and the name of the two teams in the order. After this basic information, you can find the goals scored by the two teams in the entire match or only in the first half, in addition to the ranking. before the start of the meeting.
## Analysis of the State of Form: Académico de Viseu
As for Académico de Viseu , it is a team that has had good results in the last five games, drawing and winning most of the time for a total of 8 points Will you be able to keep this trend?.
### Analysis of the State of Form: GD Chaves
The team GD Chaves shows up to the clash with performances under the average, has lost and tied more times than it has won. his score, or will he fall even further?.
## Stats
For you the average goal of Académico de Viseu and GD Chaves and their percentage of Under and Over relative to the championship Portugal - LigaPro .These data will allow you an in-depth analysis of the two teams and of the match they will play .
If we observe the data, it is immediately obvious that the average of the total Goals (facts and suffered) is Goals for the Académico de Viseu and of Goals for the GD Chaves.
Let's merge these two values into a single one, and compare it with the championship average.We will note that the merged average, of Gol, is with the average total goals on all championship, of 2.89 Goals. The average goals of the two teams of the first half generate a unit value of . We compare this value with 1.11, the average goal of the first half of the championship, to discover that is of the statistics regarding the entire tournament. The average goals relative to the second times of the two teams is of the whole tournament. is the value calculated for the two challengers and 1.78 is given of the championship.
We arrive at the summary table, in which you will find all the values that we have explained the calculation in. You can understand by yourself the rhythms of goals maintained by the two teams Académico de Viseu and GD Chaves, and how much is It is probable that goals will be scored based on the past and we have the same data for the first and second halves as well.
Average of the goals / game
Under 0.5 Goals
Over 0.5 Goals
Under 1.5 Goals
Over 1.5 Goals
Under 2.5 Goals
Over 2.5 Goals
Under 3.5 Goals
Over 3.5 Goals
Under 4.5 Goals
Over 4.5 Goals
Goal
No Goal
## Full Match
Below you will find the probability tables we recorded for the game Académico de Viseu vs GD Chaves . We have very complete statistics, ranging from the result of the game, to the analysis of the goal and who will score during the match, to the analysis of the latter in the two times taken individually.
The next tables will show the percentages as regards the exact result and the final result
### Winner of the Match
The data shown on these tables will show how GD Chaves is the favorite team for this match.
Percentage
GD Chaves (Away Win)
Draw
Look at the table below for all the percentages calculated by our algorithm.
Analyzing the previous table we can also deduce that it is low the probability that both teams score (37.57%).
The possibility that one or both teams score is low, our system has established these possible outcomes for the game in question:
H - A 0 1 2 3 4
0
1
2
3
4
### Académico de Viseu vs GD Chaves: First Half
Below we show the analysis of the match evaluating only the performances of the two teams in the first half.
The possibility that one or both teams score in the first half is low.
The algorithm indicates the presence low of the probabilistic conditions for which both teams could score a goal in the first half. The percentage is 3.05%.
### Académico de Viseu vs GD Chaves: Second Half
The possibility that one or both teams score in the second half is low.
There is a probability of 31.29% that both teams score a goal during the second half.
### Summary of Predictions
In short, the summary of the prediction of this football match
Discover
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##### 3 Most Probable Results 2T
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149314 | 1,308 | 5,498 | {"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} | 3.640625 | 4 | CC-MAIN-2020-05 | latest | en | 0.943018 |
http://www.math.ucsd.edu/~helton/WAVRIK.html | 1,516,124,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00794.warc.gz | 507,456,426 | 2,058 | Posted Dec 1, 1995
## Rules for the Computer Simplification of the Formulas in Operator Theory and Linear Systems
### By J. W. Helton and J. J. Wavrik
This article formulates and treats questions in operator theory arising from computer simplification of formulas commonly found in the study of operator models. Operator model theory originated with Moshe Livsic and subsequently became one of the main branches of operator theory. In studying a particular operator model polynomials in certain expressions occur repeatedly. This makes it a natural area for exploring computer algebra simplification.
The purpose of a simplification theory is to provide a means for replacing complex expressions by expressions which are ``simpler'' in some sense. The main task is to obtain a list of rules each of which replaces a ``complicated'' monomial which occurs in an expression by a sum of ``simpler'' monomials. To simplify such an expression one applies the rules to the expression until no further reduction is possible. The result is called an N-Form (normal form) for the original expression. The reduction of an expression to an N-Form can be easily implemented on a computer. This article provides a collection of reduction rules for expressions which arise in the Nagy-Foias operator model. Our simplifying rules were obtained by applying an algorithm for computing a Gröbner basis for an ideal in a polynomial ring. It is applied to the ideal generated by a set of fundamental relations which obviously hold for NF calculations. We conjecture that all appropriate relations are in this ideal. The algorithm produces an infinite set of rules in the NF case. The traditional operator theorist's functional calculus is used to produce a nice formulation of the rules as a finite set.
If a set of generators is a Gröbner Basis, then the reduction to an N-form has very nice properties; the N-form is independent of the order in which rules are applied and equality of N-forms can be used as a test of equivalence of expressions mod the ideal. We have established that our rules form a Gröbner Basis for several situations. A proof outline and computer tests provide strong evidence that this is true for the NF case as well.
The results here have potential applications to engineering systems theory since this algebraic structure occurs in formulas arising in H control. Also since the entries of a unitary 2x2 block matrix have algebraic structure similar to the NF model, this applies to `all pass' functions from engineering. | 510 | 2,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.944798 |
http://swag.outpostbbs.net/TURTLE/REKPIC15.PAS.html | 1,582,699,257,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00020.warc.gz | 139,314,580 | 2,649 | ``````(* ÚÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ¿
³ Programated by Vladimir Zahoransky ³
³ Contact : zahoran@cezap.ii.fmph.uniba.sk ³
³ Program tema : Gen. v. for n-angle in circle no rekusion ³
ÀÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ *)
{ This program is general no rekusion version of this effekt.
Program draw the n-angle in circle ... .
Part one : Translations
This is nice effekt, but not so easy to undestand !
Ú s s ¿
Pt = ³- Ä , - Ä * cotg(Pi/L) ³
À 2 2 Ù
(See general pictute in rekpic11.pas)
Here is rectangular triangle PsPtS.
In general is angle = 380/(2*L) = 180/L .
Suma angles inside = (n-2)*180 = n*180 - centre circle (2*180)
The angle in central triangle is 180/L.
Well, if you use rectangular triangle, we can define trigoniometry
functions. (sinus, cosinus ...)
We muth use here ZmenXY for absolute coordinates.
We muth use here ZmenSmer in absolute version for nagle :
L-2 90L - 180L - 360 90*(4-L)
90 - ---- *180 = ---------------- = --------
L L L
Part two : Draw n-angle. It is just poly. Here is other mathematical
problem. How s ? (lenght of triangle and r for circle)
IF you see general picture in rekpic11.pas then it is just other
work with rectangular triangle PsPtS.
S is just prepona. (maximal lenght in rectangular triangle)
Part tree - Draw circle.
This is diffycult problem. In muth is approximate relation :
sin(alfa)<alfo<tg(alfa) (alfa -> 0)
Well for us is interesting sin(alfa)=tg(alfa) (alfa -> 0)
(Just for eliminate anomals)
Here is interesting to use sinus sentens.
d s In program alfa = 60.
--------- = --------
sin(Pi/alfa) tg(pi/L)
d - lenght of dopredu in
d = s*sin(Pi/alfa)/tg(pi/L) circumference.
Before we use it, you muth rotate the turtle left :
180/L - 180/n (n is there alfa).
It is a angle of PtAB - alfa. (This is angle for
eliminate anomals with tg(alfa) It is not so easy
to undestand, but make it correkt. The effekt with
circle is good for alfa >40.
Then decrement level of drawing and work part 1, wihle
n>0.
Well, this is full documentation with math relations for this
effekt.
}
Uses okor;
const L=6;
Alfa=80;
Type Mykor=object(kor)
Procedure Config(s:real);
Procedure poly(n:integer;s,u:real);
Procedure triangle(s:real);
Procedure circle(n:integer;s:real);
Procedure draw(n:integer;s:real);
End;
var K:Mykor;
poc:integer;
Procedure Mykor.Config(s:real);
Begin
ph;
Zmenxy(-s/2,-s/2* (cos(pi/L) / sin(pi/L)) );
ZmenSmer(90*(4-L)/L);
pd;
End;
Procedure Mykor.poly(n:integer;s,u:real);
Begin
While n>0 do Begin
Dopredu(s);
Vpravo(u);
Dec(n);
End;
End;
Procedure Mykor.triangle(s:real);
Begin
poly(L,s,360/L);
End;
Procedure Mykor.circle(n:integer;s:real);
Begin
Vlavo(180/L-180/n);
Poly(n,s,360/n);
End;
Procedure Mykor.draw(n:integer;s:real);
Begin
While n>0 do Begin
Inc(poc);
Zmenfp(poc);
Config(s);
Triangle(s);
s:=s/cos(pi/L);
circle(Alfa,s*sin(pi/Alfa)/(sin(pi/L)/cos(pi/L)));
dec(n);
End;
End;
Begin
Poc:=0;
With k do Begin
Init(0,0,0);
Draw(8,80);
CakajKlaves;
Koniec;
End;
End.
`````` | 1,074 | 3,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-10 | latest | en | 0.668656 |
http://www.cartographersguild.com/showthread.php?t=5372&mode=threaded | 1,462,300,699,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121737.31/warc/CC-MAIN-20160428161521-00173-ip-10-239-7-51.ec2.internal.warc.gz | 414,696,034 | 27,847 | # Thread: How to progressively map a plane to a semisphere?
1. ## How to progressively map a plane to a semisphere?
this is somehow related to this question
http://www.gimptalk.com/forum/%22ben...re-t39911.html
Anyway my qustion may be a bit different:
suppose a plane, as a rectangular and a bit elastic grid,as a net
Now from below the grid ,in its very center a sphere, no larger then the plane, start to pop out,springing upward , pushing up the net
Beyond a certain point of push the elasticity of the net is no sufficient, starting where tension is higher )the net anchors to the plane broke
More the net is free more it wrap around the sphere
But here the difference with the original question, i would like this applied to the Semisphere not to the sphere
meaning that when half of the sphere pop out, the net should be totally wrapped around the surface of the semisphere:
all anchor points previously on the periphery of the rectangle transferred on the Equator, on the maximum circumference of the Sphere, where intersect the plane.
Now you may replace the word "grid" with "original image"
to understand better the point
well i would really like have a similar mapping script and what i found difficult is not map the plane to the semi-sphere but all the "in betweens", the progression
I believe a script for Mathmap will be the easier and more handy solution (as name said is a gimp plugin created for mapping, and ,bigger point support animation and customized script )
But anyway if that may solved by a gimp script or plugin would be the same till the output may be a animation.
I just think that MathMap, and soon GMIC offer a easier and more flexible solution,and what relevant a good preview of the results, but even as a gimp script will be handy
But i'm not good to write scripts, even less in this case because i can't figure out a formula for a similar mapping
I just hope cartographers may ,if not actually code (will be the best ) found the formula to be applied
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• | 482 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-18 | latest | en | 0.919093 |
https://thedesigningfairy.com/for-a-to-have-0-as-an-eigenvalue-k-must-be/ | 1,656,476,950,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00483.warc.gz | 612,450,966 | 5,200 | Does my proof hold up to prove that \$0\$ is the only eigenvalue of \$A\$ if \$A^2 = 0\$?
Let \$A\$ be an \$n imes n\$ matrix.\$A^2 = A imes A\$ because of matrix multiplication.If \$A = k\$, where \$k eq 0\$, then \$A^2 eq 0\$. Thus if \$A^2 = 0\$, \$A\$ is the zero matrix.
You are watching: For a to have 0 as an eigenvalue, k must be
If \$A\$ is the zero matrix, det(\$A^2) = 0\$.\$lambda eq 0\$ if \$A\$ is non-singular. \$A\$ is a non-singular matrix IFF \$det(A)\$ \$ eq 0\$. The charcteristic equation is false when \$lambda = 0\$ because that a non-singular matrix so it can only it is in true when the matrix is singular, make \$det(A) = 0\$.\$lambda = 0\$,\$ ightarrow\$\$eginvmatrix0I - 0endvmatrix= 0\$
It is wrong. Friend seem to think that if \$A eq0\$, climate \$A^2 eq0\$, i beg your pardon is false. Take it \$A=left(eginsmallmatrix0&1\0&0endsmallmatrix ight)\$, for instance.
If \$A\$ had an eigenvalue \$lambda eq0\$, climate there would be a vector \$v eq0\$ such that \$A.v=lambda v\$. So,eginalignA^2.v&=A.(A.v)\&=A(lambda v)\&=lambda(A.v)\&=lambda^2v\& eq0endalignand therefore \$A^2 eq0\$.
What you wrote is a mess.
If \$A = k\$, wherein \$k eq 0\$, then \$A^2 eq 0\$. Therefore if \$A^2 = 0\$, \$A\$ is the zero matrix.
This is false: think about \$A=eginpmatrix0&1\0&0endpmatrix\$.
The rest I cannot also comment on.
What you desire to perform is the following. Expect there is one eigenvalue \$lambda eq 0\$. Then there is an eigenvector \$vin hedesigningfairy.combb C^nsetminus\$ such that \$Av=lambda v\$. Yet then \$A^2v=lambda^2v eq 0\$, therefore the matrix \$A^2\$ cannot be \$0\$.
There are non-zero matrices \$A\$ whereby \$A^2 = 0\$ so your proof go not hold true. If \$A^2 = 0\$, we currently know that \$A\$ is non-invertible since if \$A\$ to be invertible, we might multiply both sides through \$A^-1\$ and also get \$A = 0\$, yet \$0\$ is not invertible. Us do know \$0\$ is an eigenvalue the \$A\$ though due to the fact that we deserve to multiply \$A\$ with any column that \$A\$ to attain \$0\$. We now have actually to present \$0\$ is the just eigenvalue. Mean there is an eigenvalue \$lambda eq 0\$. Then, \$Ax = lambda x\$ and also \$0 = A^2x = AAx = Alambda x = lambda Ax\$. Due to the fact that \$lambda eq 0\$, we acquire \$Ax = 0 = 0x\$ for part eigenvector \$x\$, yet this method \$0\$ is one eigenvalue for \$x\$, i beg your pardon contradicts \$lambda eq 0\$. Thus, \$lambda = 0\$ is the just eigenvalue because that \$A\$.
See more: The Primary Objective Of Financial Accounting Information Is To Provide Useful Information To:
Here is an alternative proof: \$x^2\$ is an annihilating polynomial for the matrix \$A\$, for this reason the minimal polynomial because that \$A, m_A(x)\$ divides \$x^2implies m_A(x)=x, x^2\$. \$0\$ is the just root that the minimal polynomial, so \$A\$ just has actually one distinct eigenvalue: \$0\$.
## Not the prize you're feather for? Browse other questions tagged linear-algebra matrices proof-verification eigenvalues-eigenvectors or asking your own question.
Queries in the proof of a square procession \$A\$ is invertible if and also only if \$lambda = 0\$ is not an eigenvalue of \$A\$
carry out a proof or counterexample. If \$M\$ is an \$n imes n\$ procession that is diagonal, then every non-zero worth in \$M\$ is one eigenvalue of \$M\$.
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See the recommended documentation of this function
Manual Scilab >> Biblioteca de Gráficos > Datatips > orthProj
orthProj
Computes the orthogonal projection of a point to a polyline in the plane.
Calling Sequence
`[d,ptp,ind,c] = orthProj(data,pt)`
Arguments
data
a n by 2 array. The first column contains the X coordinates of the poyline, while the second one contains the Y coordinates.
pt
an array with 2 entries: the coordinates of the point
d
The minimum distance between the given point and a polyline data point.
ptp
an array with 2 entries: the coordinates of the point that realizes the minimum distance
ind
The polyline closest point lies on the segment `[ind ind+1]`.
c
The interpolation coefficient of the orthonal projection.
Description
Computes the orthogonal projection of a point to a polyline in the plane.
Examples
```x=linspace(0,1,30)';
y=sin(4*x.^3)/2;
clf();a=gca();a.isoview='on';
plot(x,y)
pt=[0.7 0.3];
plot(pt(1),pt(2),'xb')
[d,ptp,ind,c]=orthProj([x y],pt);
plot(x(ind:ind+1),y(ind:ind+1),'+r')
xpoly([pt(1);ptp(1)],[pt(2);ptp(2)])
e=gce();e.polyline_style=4;e.arrow_size_factor = 1.5;
pt=[0.75 -0.3];
plot(pt(1),pt(2),'xb')
[d,ptp,ind,c]=orthProj([x y],pt);
plot(x(ind:ind+1),y(ind:ind+1),'+r')
xpoly([pt(1);ptp(1)],[pt(2);ptp(2)])
e=gce();e.polyline_style=4;e.arrow_size_factor = 1.5;```
Authors
• Serge Steer, INRIA
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# FuelCellSolns - Design of Alternative Energy Systems Fuel...
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1 ME 417 Design of Alternative Energy Systems Fuel Cell Problems Solutions 1. Determine the ideal voltage, current, and required mass flow rate of hydrogen for an ideal air/hydrogen fuel cell operating at 101 kPa and 350 K that is producing 25 kW. Solution: Our general energy equation is s s T - h h w products j j reactants i i FC products j j reactants i i elec ± ² ³ ´ µ ν - ν ν - ν = For a hydrogen/air fuel cell the balanced chemical reaction equation is given by H 2 + ½O 2 + ½(3.76)N 2 H 2 O + ½(3.76)N 2 which allows use to write our energy equation as { } s ) 88 . 1 ( s s ) 88 . 1 ( s ) 5 . 0 ( s T - h ) 88 . 1 ( h h ) 88 . 1 ( h ) 5 . 0 ( h w p , N O H r , N O H FC p , N O H r , N O H elec 2 2 2 2 2 2 2 2 2 2 - - + + - - + + = We assume that we have an isothermal fuel cell at 350 K, so that T FC = T p = T r = 350 K So we can now evaluate our enthalpies kmole / kJ 1503 ) 8468 9971 ( 0 h h h 2 2 2 H H , f H = - + = Δ + = kmole / kJ 1531 ) 8682 213 , 10 ( 0 h h h 2 2 2 O O , f O = - + = Δ + = kmole / kJ 070 , 240 ) 9904 652 , 11 ( 820 , 241 h h h O H O H , f O H 2 2 2 - = - + - = Δ + = We have not included the enthalpies for the N 2 , since we know that they will cancel out. To evaluate the entropies we need to have our partial pressures. Calculating them we have 0.30 88 . 1 5 . 0 1 1 y 2 H = + + = 0.15 88 . 1 5 . 0 1 5 . 0 y 2 O = + + = 0.55 88 . 1 5 . 0 1 88 . 1 y r , N 2 = + + = 0.35 88 . 1 1 1 y O H 2 = + =
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ME 417 Design of Alternative Energy Systems 2 0.65 88 . 1 1 88 . 1 y p , N 2 = + = Going to the ideal gas tables at 350 K we find ) K kmole /( kJ 21 . 135 s o H 2 = ) K kmole /( kJ 765 . 209 s o O 2 = ) K kmole /( kJ 173 . 196 s o r , N 2 = ) K kmole /( kJ 125 . 194 s o O H 2 = ) K kmole /( kJ 173 . 199 s o p , N 2 = Now substituting }) 65 . 0 ln{ ) 314 . 8 ( (196.173 ) 88 . 1 ( }) 35 . 0 ln{ ) 314 . 8 ( 94.125 1 ( }) 55 . 0 ln{ ) 314 . 8 ( (196.173 ) 88 . 1 ( }) 15 . 0 ln{ ) 314 . 8 ( 09.768 2 ( ) 5 . 0 ( }) 3 . 0 ln{ ) 314 . 8 ( 21 . 135 ( (298) - 240,070) ( ) 531 1 ( ) 5 . 0 ( 503 1 w elec ± ² ³ ´ µ - - - - - + - + - - - + = kJ/kmole 230,360 2) (350)(34.2 - 242,340
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## This note was uploaded on 07/25/2008 for the course ME 417 taught by Professor Somerton during the Spring '07 term at Michigan State University.
### Page1 / 7
FuelCellSolns - Design of Alternative Energy Systems Fuel...
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Ask a homework question - tutors are online | 1,087 | 2,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.737171 |
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# irfan
2016 以来アクティブ
バッジを表示
#### Content Feed
false negative and false positive ?
how to write false negative and false positive code for random signal
6年弱 前 | 0 件の回答 | 0
### 0
How to choose Threshold when Add AWGN to signal
Hello, sir, I am adding AWGN noise to my Random signal but when I add it I also need to change my threshold value for the signa...
6年弱 前 | 1 件の回答 | 0
### 1
how can we calculate false positive and false negative for signal ? can any one provide me matlab code ?
if signal recovered its false negative and if signal is not recovered its false positive what i know that for ii=1:length(X_...
6年弱 前 | 0 件の回答 | 0
### 0
hello sir i am writing one equation in matalb but i am facing problem in it when i write that equation it give me wrong answer can anyone help me to write that equations for me
i am solving nonlinear equation but i am facing problem in these equation can anyone help me to write that equations for me i ha...
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### 0
hello sir i want to plot digram for SNR Vs false negative and positive ..
i am confused how to write matlab code for false negative and false positive can anyone provide me the matlab code for false neg...
6年弱 前 | 0 件の回答 | 0
### 0
subscript indices must either be real positive integers or logicals( X_re(k_r)=XD(1,kd); can anyone help me why i am getting this error
clc clear all N=1024; tic; k=50; d=28; aa=6; b=10; a=3; X=zeros(1,N); % xp_unsolved2=[]; % X_p=randperm(N,k); % X_v...
6年弱 前 | 1 件の回答 | 0
### 1
Undefined operator '^' for input arguments of type 'cell'.
c0=1; c1=2; c3=4; A={((c0/2)-(c1*c2/6)+(c2^3/27))-[(c0/2)-(c1*c2/6)+((c2^3/27)^2)+(c1/3-c2^2)^3]^1/2}^1/3; i am ge...
6年弱 前 | 1 件の回答 | 0
### 1
hello sir can any one give me a matlab code of moment preserving problem of tsai .
i want to use it for my assignment
6年弱 前 | 0 件の回答 | 0
### 0
hello sir i have three different graph from three different code i want to get one performance graph ho i will do that ?
i have try save(filename) this syntax of Matlab but its not working in my case can you tell me which is the better way to plot t...
6年弱 前 | 1 件の回答 | 0
### 1
Subscript indices must either be real positive integers or logicals.
k_est=k_p(bb,:); XF_est=xf(bb).XF; num_2=num_2+1; kkk2(num_2,:)=k_est; for iii=1:length(k_es...
6年弱 前 | 1 件の回答 | 0
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hello sir i want real values but when i write my function in matlab s(ii)==real (s(ii)); it give me complex value how to solve it
ii=1:2; s(ii)=real(s(ii)); s(ii)=round(log(z(ii))*N/j/2/pi)+1; this is my short function please guide me where i a...
6年弱 前 | 1 件の回答 | 0
### 1
The signal-to-noise ratio must be a real scalar?
i get this answer when my SNR is Vector.i want to plot SNR for [5,10,15,20,25,30,35,40,35,50] but when i write in vector form i ...
6年弱 前 | 1 件の回答 | 0
### 1
hello sir i want to find a nonzero number in my total error signal ?
my original signal is X and my Recovered signal is X_re and the total Error which i count between them is Error=X-X_re; now i h...
6年弱 前 | 1 件の回答 | 0
### 1
snr vs recover signal
for snr_num=1:9; snr=snr_num*5+5; for exp_num=1:100; Sir i want to plot snr for my recover signal and i want ...
### 0
hello sir i want to ask question about threshold value how we will choose a threshold value for a random signal? for example i have random signal N=1024 and K=20 for my sparse signal i want to calculate the threshold values ?
i am downsampling my signal and i take fft of my downsampling signal after that i want to select the appropriate threshold value...
### 0
hello sir i am calculating mean square for my two signals but its giving me constant mean square error..?
clc clear all N=1024; k=8; d=16; a=1; b=5; X=zeros(1,N); X_p=randperm(N,k); X_v1=rand(1,k); X_v...
### 0
how to genrate random signal
N=1024;
### 1
hello sir i want to calculate mean square error for my all possible value for the system how to calculate it ?
k0=1; ss=[0,1,2,3]; kk=0; poss=[]; for ii=1:d poss=[poss,k0+(ii-1)*(N/d)]; end for ii=1:length(poss) k1=...
### 1
i am solving over-determined system but when i run my code its give me result inner dimension mismatch how to solve it
l=0:28 N=28 for k=1:7 k1=k+(n*7); k2=k+(n*7); M=[1 1 ; e^(j*2*3.14)*(k1-1)*(l/N) e^(j*2*3.14)*(k2-1)*(l/N) ...
### 1
how we will do bucktization in matlab anyone have a code of bucketization process in matlab so please send me ?
i want to do bucketization after downsampling and the solve that system.i am looking forward for your help | 1,426 | 4,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.690414 |
https://www.scirra.com/forum/multipath_t86232 | 1,502,957,789,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102993.24/warc/CC-MAIN-20170817073135-20170817093135-00523.warc.gz | 955,552,094 | 10,390 | # Multipath
Get help using Construct 2
### » Mon Jul 15, 2013 12:06 pm
Hello,
I'm working on a tower defense game in which i planed multi paths for enemies to reach the tower.But enemies selecting only shortest path.Can any one help how to select random path.
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### » Mon Jul 15, 2013 12:11 pm
If you use the Pathfinding, the A* algorithm behind it will only give you the shortest path. You'll need to do the paths yourself if you want it in another way.
As I did the same as you, here is how I managed multipath : waitpoints. Just put some waitpoints on the paths, and associate a "waitpoint route" for every spawner or every mob (depending on what you want). Then, instead of looking for a path to the exit, look for a path to next waiting point, and again until you are at the objective.
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### » Mon Jul 15, 2013 1:25 pm
You could also blockade one of the ways with an invisible object
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### » Mon Jul 15, 2013 7:30 pm
> You could also blockade one of the ways with an invisible object
What happens to the next minion on the same path.. if it needs to take a different route?
How about if we could have multi target.. and assign certain sequence..
say minion1 = targets (1,3)
minion 2 = targets (2,3)
so that way minion1 would take a different route collecting target1 and then target3 .. than minion2 that'd take 2 first and then 3.
Eventually both reach 3 in different ways. (ofcourse the way you place targets 1 and 2 matter)
@Ashley ?
Krish
Hey! Did you know that you can hire me to make your games?
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### » Mon Jul 15, 2013 7:41 pm
Here is one way to do something like that:
Path demo
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Reputation: 7,292 | 535 | 1,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.909992 |
http://www.it2051229.com/javacredvalidation.html | 1,516,367,327,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00032.warc.gz | 481,711,573 | 5,924 | Credit card companies use built-in "checksums" as added security measures when creating the account numbers on credit cards. This means that there are only certain valid credit card numbers, and validity can instantly be detected by using an algorithm that may involve adding up parts of the numbers or performing other checks. In this problem, you will implement a security algorithm on a credit card number, and your program will indicate whether or not the card number is valid, according to the security check. Note that this algorithm is purely made-up; don't try to use it to detect fake credit card numbers!
To check that the credit card number is valid, verify that it conforms to the following rules:
• The first digit must be a 4.
• The fourth digit must be one greater than the fifth digit
• The product of the first, fifth, and ninth digits must be 24
• The sum of all digits must be evenly divisible by 4
• The sum of the first four digits must be one less than the sum of the last four digits
• If you treat the first two digits as a two-digit number, and the seventh and eight digits as a two digit number, their sum must be 100.
Hint: valid card numbers according to this set of rules include 480760521766 and 409434602754. You should use these for testing out your program.
Your job is to create a CreditCard class that represents a card number as a String. Hence you will have an instance variable of type String to store this. In addition the CreditCard class must have a mutator method called check( ) that determines if the credit card is valid or not. This information (whether the card is valid or not) should be stored in an instance variable of type boolean. Your class should also have an accessor method called isValid( ) that returns the value of this boolean. Your class CreditCard should also have an instance variable of type int called errorCode. This variable should be initialized to 0 and changed to an integer between 1 and 6 if the CreditCard object fails one of the tests above. That is, the check( ) method should change errorCode to the number corresponding to the first test the number failed. If the number does not fail any tests then you can leave the errorCode variable at 0. Finally, you should have an accessor method called getErrorCode( ) that returns the value of the variable errorCode. You might find this method useful when you are verifying that your check( ) method works correctly.
Attached you will find a test class called CreditCardTester. Your CreditCard class must work with the test class provided without modification. You must NOT alter the test class.
To complete this problem, you will need to use various methods of the String class, described in Horstmann and in the Java online documentation for String. In particular, you should use methods for getting the character at a certain index and for converting characters to Strings. You will also need to use the Integer.parseInt method to convert a String to an int. | 615 | 2,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-05 | latest | en | 0.915559 |
https://root.cern.ch/doc/master/MnParameterScan_8cxx_source.html | 1,675,147,140,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00756.warc.gz | 530,707,172 | 6,601 | ROOT Reference Guide
MnParameterScan.cxx
Go to the documentation of this file.
1// @(#)root/minuit2:$Id$
2// Authors: M. Winkler, F. James, L. Moneta, A. Zsenei 2003-2005
3
4/**********************************************************************
5 * *
6 * Copyright (c) 2005 LCG ROOT Math team, CERN/PH-SFT *
7 * *
8 **********************************************************************/
9
11#include "Minuit2/FCNBase.h"
12
13namespace ROOT {
14
15namespace Minuit2 {
16
18 : fFCN(fcn), fParameters(par), fAmin(fcn(par.Params()))
19{
20}
21
22MnParameterScan::MnParameterScan(const FCNBase &fcn, const MnUserParameters &par, double fval)
23 : fFCN(fcn), fParameters(par), fAmin(fval)
24{
25}
26
27std::vector<std::pair<double, double>> MnParameterScan::
28operator()(unsigned int par, unsigned int maxsteps, double low, double high)
29{
30 // do the scan for parameter par between low and high values
31
32 // if(maxsteps > 101) maxsteps = 101;
33 std::vector<std::pair<double, double>> result;
34 result.reserve(maxsteps + 1);
35 std::vector<double> params = fParameters.Params();
36 result.push_back(std::pair<double, double>(params[par], fAmin));
37
38 if (low > high)
39 return result;
40 if (maxsteps < 2)
41 return result;
42
43 if (low == 0. && high == 0.) {
44 low = params[par] - 2. * fParameters.Error(par);
45 high = params[par] + 2. * fParameters.Error(par);
46 }
47
48 if (low == 0. && high == 0. && fParameters.Parameter(par).HasLimits()) {
50 low = fParameters.Parameter(par).LowerLimit();
52 high = fParameters.Parameter(par).UpperLimit();
53 }
54
55 if (fParameters.Parameter(par).HasLimits()) {
57 low = std::max(low, fParameters.Parameter(par).LowerLimit());
59 high = std::min(high, fParameters.Parameter(par).UpperLimit());
60 }
61
62 double x0 = low;
63 double stp = (high - low) / double(maxsteps - 1);
64 for (unsigned int i = 0; i < maxsteps; i++) {
65 params[par] = x0 + double(i) * stp;
66 double fval = fFCN(params);
67 if (fval < fAmin) {
68 fParameters.SetValue(par, params[par]);
69 fAmin = fval;
70 }
71 result.push_back(std::pair<double, double>(params[par], fval));
72 }
73
74 return result;
75}
76
77} // namespace Minuit2
78
79} // namespace ROOT
Option_t Option_t TPoint TPoint const char GetTextMagnitude GetFillStyle GetLineColor GetLineWidth GetMarkerStyle GetTextAlign GetTextColor GetTextSize void char Point_t Rectangle_t WindowAttributes_t Float_t Float_t Float_t Int_t Int_t UInt_t UInt_t Rectangle_t result
Interface (abstract class) defining the function to be minimized, which has to be implemented by the ...
Definition: FCNBase.h:45
std::vector< std::pair< double, double > > operator()(unsigned int par, unsigned int maxsteps=41, double low=0., double high=0.)
MnParameterScan(const FCNBase &, const MnUserParameters &)
API class for the user interaction with the parameters; serves as input to the minimizer as well as o...
double Error(unsigned int) const
std::vector< double > Params() const | 828 | 2,948 | {"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} | 2.59375 | 3 | CC-MAIN-2023-06 | longest | en | 0.141724 |
http://www.hpmuseum.org/forum/showthread.php?tid=7955&pid=70844&mode=threaded | 1,550,941,761,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249508792.98/warc/CC-MAIN-20190223162938-20190223184938-00349.warc.gz | 355,414,747 | 11,168 | Little explorations with HP calculators (no Prime)
03-31-2017, 07:35 AM (This post was last modified: 03-31-2017 07:40 AM by pier4r.)
Post: #87
pier4r Senior Member Posts: 1,921 Joined: Nov 2014
RE: Little explorations with the HP calculators
Since I see that there are like doubts, I share one solution from brilliant (note sure if it is correct but it seems so). I had to give up the chance to solve the problem but it is al right since I can just force myself to come up with my solution (maybe considering concave pentagons as wrote before).
So to not spoiler it, I put the written solution and the image created by one user as links, so they do not immediately show up.
http://i.imgur.com/hyiTDIg.png - written solution.
http://i.imgur.com/METbYu2.png - construction.
I have to analyze it a bit better though doing my construction, otherwise I cannot be sure I followed properly.
side note: I thought I was getting random problems from the brilliant collection about the Geometry topic. In fact, I was getting random problem from the brilliant collection on _one_ subtopic of the Geometry topic, which has 10+ topics. So there are even more problems and variety that I expected.
Wikis are great, Contribute :)
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Messages In This Thread Little explorations with HP calculators (no Prime) - pier4r - 03-15-2017, 11:11 PM RE: Little explorations with the HP calculators - Thomas Okken - 03-15-2017, 11:56 PM RE: Little explorations with the HP calculators - pier4r - 03-16-2017, 12:01 AM RE: Little explorations with the HP calculators - Logan - 03-16-2017, 11:06 AM RE: Little explorations with the HP calculators - ekubaskie - 03-16-2017, 02:07 AM RE: Little explorations with the HP calculators - Csaba Tizedes - 03-16-2017, 08:09 AM RE: Little explorations with the HP calculators - TASP - 03-16-2017, 06:29 PM RE: Little explorations with the HP calculators - pier4r - 03-21-2017, 10:40 PM RE: Little explorations with the HP calculators - Joe Horn - 03-21-2017, 11:41 PM RE: Little explorations with the HP calculators - Dieter - 03-28-2017, 05:46 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 06:17 PM RE: Little explorations with the HP calculators - pier4r - 03-22-2017, 08:37 AM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 12:14 PM RE: Little explorations with the HP calculators - Joe Horn - 03-22-2017, 03:27 PM RE: Little explorations with the HP calculators - pier4r - 03-22-2017, 04:45 PM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 11:11 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 08:07 AM RE: Little explorations with the HP calculators - Simone Cerica - 03-23-2017, 08:58 AM RE: Little explorations with the HP calculators - John Keith - 03-22-2017, 11:25 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 08:28 AM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 10:00 AM RE: Little explorations with the HP calculators - Joe Horn - 03-23-2017, 12:19 PM RE: Little explorations with the HP calculators - pier4r - 03-23-2017, 01:23 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 01:45 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 03:23 PM RE: Little explorations with the HP calculators - pier4r - 03-24-2017, 10:04 PM RE: Little explorations with the HP calculators - Dieter - 03-24-2017, 10:54 PM RE: Little explorations with the HP calculators - pier4r - 03-25-2017, 08:17 AM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 12:14 PM RE: Little explorations with the HP calculators - Dieter - 03-27-2017, 06:01 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 06:14 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 06:30 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 06:57 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 07:07 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:22 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 07:39 PM RE: Little explorations with the HP calculators - Dieter - 03-27-2017, 08:55 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 10:35 PM RE: Little explorations with HP calculators (no Prime) - RMollov - 05-06-2018, 06:13 AM RE: Little explorations with HP calculators (no Prime) - Gerson W. Barbosa - 05-22-2018, 03:03 AM RE: Little explorations with HP calculators (no Prime) - RMollov - 05-23-2018, 11:04 AM RE: Little explorations with the HP calculators - Thomas Okken - 03-27-2017, 12:54 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 02:27 PM RE: Little explorations with the HP calculators - Thomas Okken - 03-27-2017, 03:44 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 03:55 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 05:29 PM RE: Little explorations with the HP calculators - Joe Horn - 03-27-2017, 07:24 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:35 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:01 PM RE: Little explorations with the HP calculators - Joe Horn - 03-27-2017, 08:08 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 08:14 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:36 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 08:44 PM RE: Little explorations with the HP calculators - Han - 03-27-2017, 08:07 PM RE: Little explorations with the HP calculators - pier4r - 03-27-2017, 07:58 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-27-2017, 10:12 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 10:44 AM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 11:07 AM RE: Little explorations with the HP calculators - Dieter - 03-28-2017, 05:33 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 06:13 PM RE: Little explorations with the HP calculators - Han - 03-28-2017, 07:32 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 08:15 PM RE: Little explorations with the HP calculators - Dieter - 03-29-2017, 07:54 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 08:48 AM RE: Little explorations with the HP calculators - Dieter - 03-29-2017, 01:07 PM RE: Little explorations with the HP calculators - Han - 03-28-2017, 07:36 PM RE: Little explorations with the HP calculators - pier4r - 03-28-2017, 09:32 PM RE: Little explorations with the HP calculators - Han - 03-29-2017, 01:58 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 06:02 AM RE: Little explorations with the HP calculators - Simone Cerica - 03-29-2017, 09:51 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 11:04 AM RE: Little explorations with the HP calculators - Han - 03-29-2017, 11:31 AM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 11:49 AM RE: Little explorations with the HP calculators - Han - 03-29-2017, 06:15 PM RE: Little explorations with the HP calculators - Dieter - 03-30-2017, 12:26 PM RE: Little explorations with the HP calculators - Han - 03-30-2017, 04:14 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-30-2017, 05:47 PM RE: Little explorations with the HP calculators - Han - 03-30-2017, 06:36 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-30-2017, 07:33 PM RE: Little explorations with the HP calculators - pier4r - 03-29-2017, 06:35 PM RE: Little explorations with the HP calculators - Isaac S. Friedman - 03-29-2017, 06:58 PM RE: Little explorations with the HP calculators - pier4r - 03-30-2017, 08:30 AM RE: Little explorations with the HP calculators - Dieter - 03-30-2017, 12:13 PM RE: Little explorations with the HP calculators - pier4r - 03-30-2017, 06:02 PM RE: Little explorations with the HP calculators - Han - 03-30-2017, 08:19 PM RE: Little explorations with the HP calculators - Dieter - 03-30-2017, 09:22 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-30-2017, 10:38 PM RE: Little explorations with the HP calculators - pier4r - 03-30-2017, 08:41 PM RE: Little explorations with the HP calculators - Han - 03-30-2017, 10:41 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017 07:35 AM RE: Little explorations with the HP calculators - SlideRule - 04-06-2017, 12:47 AM RE: Little explorations with the HP calculators - pier4r - 04-07-2017, 10:30 AM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 10:01 AM RE: Little explorations with the HP calculators - DavidM - 03-31-2017, 03:12 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 05:27 PM RE: Little explorations with the HP calculators - DavidM - 03-31-2017, 06:34 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 07:34 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 08:33 PM RE: Little explorations with the HP calculators - DavidM - 03-31-2017, 09:41 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 09:51 PM RE: Little explorations with the HP calculators - DavidM - 04-01-2017, 02:27 PM RE: Little explorations with the HP calculators - Csaba Tizedes - 04-01-2017, 11:45 PM RE: Little explorations with the HP calculators - Han - 03-31-2017, 09:19 PM RE: Little explorations with the HP calculators - pier4r - 03-31-2017, 09:42 PM RE: Little explorations with the HP calculators - pier4r - 04-01-2017, 07:58 PM RE: Little explorations with the HP calculators - DavidM - 04-01-2017, 09:45 PM RE: Little explorations with the HP calculators - pier4r - 04-02-2017, 08:34 AM RE: Little explorations with the HP calculators - DavidM - 04-02-2017, 04:01 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 06:24 PM RE: Little explorations with the HP calculators - DavidM - 04-02-2017, 06:19 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 08:52 AM RE: Little explorations with the HP calculators - DavidM - 04-03-2017, 03:03 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 06:47 PM RE: Little explorations with the HP calculators - DavidM - 04-03-2017, 09:59 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 10:12 PM RE: Little explorations with the HP calculators - pier4r - 04-02-2017, 06:00 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 07:17 AM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 05:05 PM RE: Little explorations with the HP calculators - DavidM - 04-03-2017, 06:50 PM RE: Little explorations with the HP calculators - pier4r - 04-03-2017, 06:57 PM RE: Little explorations with the HP calculators - pier4r - 04-04-2017, 08:25 AM RE: Little explorations with the HP calculators - pier4r - 04-04-2017, 12:22 PM RE: Little explorations with the HP calculators - DavidM - 04-04-2017, 06:26 PM RE: Little explorations with the HP calculators - pier4r - 04-04-2017, 06:36 PM RE: Little explorations with the HP calculators - ttw - 04-14-2017, 08:53 AM RE: Little explorations with the HP calculators - John Keith - 04-14-2017, 12:26 PM RE: Little explorations with the HP calculators - Joe Horn - 04-15-2017, 05:34 AM RE: Little explorations with the HP calculators - pier4r - 04-05-2017, 10:49 AM RE: Little explorations with the HP calculators - pier4r - 04-06-2017, 10:43 PM RE: Little explorations with the HP calculators - SlideRule - 04-06-2017, 11:24 PM RE: Little explorations with the HP calculators - rprosperi - 04-06-2017, 11:37 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 05:12 PM RE: Little explorations with the HP calculators - pier4r - 04-07-2017, 07:54 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 02:29 AM RE: Little explorations with the HP calculators - pier4r - 04-07-2017, 09:46 AM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 02:37 PM RE: Little explorations with the HP calculators - SlideRule - 04-07-2017, 12:57 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 02:52 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 10:09 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 05:19 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 05:41 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 06:00 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 08:52 PM RE: Little explorations with the HP calculators - pier4r - 04-07-2017, 09:26 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 10:04 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 10:19 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 11:39 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 11:52 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-08-2017, 12:40 AM RE: Little explorations with the HP calculators - Han - 04-08-2017, 01:17 AM RE: Little explorations with the HP calculators - pier4r - 04-08-2017, 09:45 AM RE: Little explorations with the HP calculators - Han - 04-07-2017, 10:00 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 02:38 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 02:57 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 04:11 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017, 05:01 PM RE: Little explorations with the HP calculators - Dieter - 04-07-2017, 05:40 PM RE: Little explorations with the HP calculators - pier4r - 04-07-2017, 07:59 PM RE: Little explorations with the HP calculators - SlideRule - 04-07-2017, 06:58 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 07:19 PM RE: Little explorations with the HP calculators - Han - 04-07-2017, 10:23 PM RE: Little explorations with the HP calculators - Han - 04-08-2017, 02:44 AM RE: Little explorations with the HP calculators - Han - 04-08-2017, 03:37 AM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-08-2017, 03:01 PM RE: Little explorations with the HP calculators - Han - 04-08-2017, 05:19 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-08-2017, 09:17 PM RE: Little explorations with the HP calculators - Han - 04-08-2017, 10:16 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-08-2017, 10:53 PM RE: Little explorations with the HP calculators - pier4r - 04-09-2017, 07:55 AM RE: Little explorations with the HP calculators - SlideRule - 04-09-2017, 02:18 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-09-2017, 03:40 PM RE: Little explorations with the HP calculators - pier4r - 04-08-2017, 06:23 PM RE: Little explorations with the HP calculators - pier4r - 04-09-2017, 05:44 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-09-2017, 07:17 PM RE: Little explorations with the HP calculators - brickviking - 04-10-2017, 02:22 AM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-10-2017, 03:39 AM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-10-2017, 12:37 AM RE: Little explorations with the HP calculators - pier4r - 04-10-2017, 08:00 AM RE: Little explorations with the HP calculators - Dieter - 04-10-2017, 09:30 AM RE: Little explorations with the HP calculators - Dieter - 04-10-2017, 10:53 AM RE: Little explorations with the HP calculators - pier4r - 04-10-2017, 12:11 PM RE: Little explorations with the HP calculators - SlideRule - 04-10-2017, 01:01 PM RE: Little explorations with the HP calculators - pier4r - 04-10-2017, 02:30 PM RE: Little explorations with the HP calculators - pier4r - 04-10-2017, 09:51 AM RE: Little explorations with the HP calculators - Dieter - 04-10-2017, 10:05 AM RE: Little explorations with the HP calculators - telemachos - 04-10-2017, 03:22 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-10-2017, 04:49 PM RE: Little explorations with the HP calculators - pier4r - 04-10-2017, 09:20 PM RE: Little explorations with the HP calculators - pier4r - 04-14-2017, 08:30 PM RE: Little explorations with the HP calculators - pier4r - 04-15-2017, 06:59 AM RE: Little explorations with the HP calculators - pier4r - 04-18-2017, 04:15 PM RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-18-2017, 05:52 PM RE: Little explorations with the HP calculators - pier4r - 04-18-2017, 06:10 PM RE: Little explorations with the HP calculators - pier4r - 04-19-2017, 12:22 PM RE: Little explorations with the HP calculators - DavidM - 04-19-2017, 02:49 PM RE: Little explorations with the HP calculators - pier4r - 04-19-2017, 03:56 PM RE: Little explorations with the HP calculators - DavidM - 04-19-2017, 06:40 PM RE: Little explorations with the HP calculators - John Keith - 04-28-2017, 01:24 PM RE: Little explorations with the HP calculators - pier4r - 04-19-2017, 07:23 PM RE: Little explorations with the HP calculators - ttw - 04-19-2017, 09:58 PM RE: Little explorations with the HP calculators - pier4r - 04-23-2017, 07:24 PM RE: Little explorations with the HP calculators - ttw - 04-24-2017, 05:01 AM RE: Little explorations with the HP calculators - peacecalc - 04-24-2017, 07:23 PM RE: Little explorations with the HP calculators - pier4r - 04-24-2017, 08:29 PM RE: Little explorations with the HP calculators - pier4r - 04-28-2017, 03:01 PM RE: Little explorations with the HP calculators - Gene - 04-28-2017, 05:22 PM RE: Little explorations with the HP calculators - pier4r - 04-30-2017, 02:01 PM RE: Little explorations with the HP calculators - pier4r - 05-01-2017, 05:36 PM RE: Little explorations with the HP calculators - DavidM - 05-02-2017, 03:45 PM RE: Little explorations with the HP calculators - pier4r - 05-02-2017, 03:51 PM RE: Little explorations with the HP calculators - DavidM - 05-02-2017, 04:17 PM RE: Little explorations with the HP calculators - pier4r - 05-02-2017, 09:51 AM RE: Little explorations with the HP calculators - pier4r - 05-28-2017, 10:50 PM RE: Little explorations with the HP calculators - rprosperi - 05-29-2017, 12:04 AM RE: Little explorations with the HP calculators - pier4r - 05-29-2017, 02:54 PM RE: Little explorations with the HP calculators - rprosperi - 05-29-2017, 04:32 PM RE: Little explorations with the HP calculators - pier4r - 05-06-2017, 03:27 PM RE: Little explorations with the HP calculators - DavidM - 05-06-2017, 04:51 PM RE: Little explorations with the HP calculators - pier4r - 05-06-2017, 05:14 PM RE: Little explorations with the HP calculators - pier4r - 05-14-2017, 04:12 PM RE: Little explorations with the HP calculators - grsbanks - 05-29-2017, 08:26 PM RE: Little explorations with the HP calculators - pier4r - 05-29-2017, 10:35 PM RE: Little explorations with the HP calculators - pier4r - 06-01-2017, 07:00 PM RE: Little explorations with the HP calculators - Vtile - 06-01-2017, 07:44 PM RE: Little explorations with the HP calculators - pier4r - 06-01-2017, 10:19 PM RE: Little explorations with the HP calculators - pier4r - 06-13-2017, 09:49 AM RE: Little explorations with the HP calculators - pier4r - 06-15-2017, 01:09 PM RE: Little explorations with the HP calculators - pier4r - 09-09-2017, 09:21 PM RE: Little explorations with the HP calculators - Joe Horn - 09-09-2017, 10:03 PM RE: Little explorations with the HP calculators - pier4r - 09-10-2017, 09:55 AM RE: Little explorations with the HP calculators - AlexFekken - 09-10-2017, 10:46 AM RE: Little explorations with the HP calculators - Gilles59 - 09-11-2017, 09:21 PM RE: Little explorations with HP calculators (pre Prime) - pier4r - 09-24-2017, 08:17 PM RE: Little explorations with HP calculators (pre Prime) - brickviking - 09-25-2017, 12:12 AM RE: Little explorations with HP calculators (pre Prime) - rprosperi - 09-25-2017, 12:23 AM RE: Little explorations with HP calculators (pre Prime) - pier4r - 10-17-2017, 08:18 AM RE: Little explorations with HP calculators (pre Prime) - pier4r - 10-17-2017, 01:34 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-17-2017, 04:46 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-22-2017, 08:50 AM RE: Little explorations with HP calculators (no Prime) - John Keith - 10-22-2017, 01:52 PM RE: Little explorations with HP calculators (no Prime) - brickviking - 10-22-2017, 08:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-22-2017, 03:51 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 10-22-2017, 11:05 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-30-2017, 08:46 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-31-2017, 09:54 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-31-2017, 02:22 PM RE: Little explorations with HP calculators (no Prime) - Claudio L. - 10-31-2017, 03:12 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-31-2017, 05:50 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 10-31-2017, 06:23 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 11-05-2017, 04:05 PM RE: Little explorations with HP calculators (no Prime) - Joe Horn - 11-05-2017, 10:28 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 11-07-2017, 07:33 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 11-07-2017, 09:26 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 11-07-2017, 09:33 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-12-2017, 10:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-17-2017, 10:05 AM RE: Little explorations with HP calculators (no Prime) - Joe Horn - 12-18-2017, 02:26 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-18-2017, 07:08 AM RE: Little explorations with HP calculators (no Prime) - Joe Horn - 12-19-2017, 04:20 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-19-2017, 09:15 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-19-2017, 09:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 02-19-2018, 08:03 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 02-19-2018, 09:25 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 02-19-2018, 09:40 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 02-19-2018, 10:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-10-2018, 11:36 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 03-10-2018, 05:14 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-11-2018, 12:07 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 03-11-2018, 06:25 PM RE: Little explorations with HP calculators (no Prime) - Neve - 03-11-2018, 06:47 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-11-2018, 07:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-12-2018, 09:03 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-16-2018, 07:48 PM RE: Little explorations with HP calculators (no Prime) - Thomas Okken - 03-16-2018, 08:24 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-16-2018, 09:10 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-17-2018, 12:27 PM RE: Little explorations with HP calculators (no Prime) - Csaba Tizedes - 03-17-2018, 01:22 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 03-17-2018, 08:00 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 03-17-2018, 02:37 PM RE: Little explorations with HP calculators (no Prime) - Dieter - 03-16-2018, 08:29 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-17-2018, 08:55 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-18-2018, 12:14 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 03-18-2018, 12:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 08:55 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-18-2018, 09:54 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-18-2018, 12:50 PM RE: Little explorations with Casio 9750/9860 calculators - brickviking - 03-20-2018, 11:59 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-21-2018, 10:45 AM RE: Little explorations with Casio 9750/9860 calculators - brickviking - 03-22-2018, 02:16 AM RE: Little explorations with HP calculators (no Prime) - 3298 - 03-22-2018, 11:32 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-21-2018, 11:23 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-21-2018, 08:37 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-21-2018, 11:27 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-22-2018, 11:15 AM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-23-2018, 02:02 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-23-2018, 08:20 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 09:29 AM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 01:50 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 02:24 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 05:34 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 06:17 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 04-02-2018, 08:53 PM RE: Little explorations with HP calculators (no Prime) - Zaphod - 04-02-2018, 09:16 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-25-2018, 05:01 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-25-2018, 07:34 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 03-25-2018, 08:00 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 03-25-2018, 08:45 PM RE: Little explorations with HP calculators (no Prime) - Dave Frederickson - 03-25-2018, 09:40 PM RE: Little explorations with HP calculators (no Prime) - Dave Frederickson - 03-25-2018, 09:43 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 04-02-2018, 01:53 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 04-02-2018, 09:11 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-05-2018, 02:22 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 05-21-2018, 12:53 PM RE: Little explorations with HP calculators (no Prime) - grsbanks - 05-21-2018, 01:04 PM RE: Little explorations with HP calculators (no Prime) - rprosperi - 05-21-2018, 06:02 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-26-2018, 09:34 PM RE: Little explorations with HP calculators (no Prime) - ijabbott - 12-26-2018, 11:41 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 03:22 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 04:59 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-31-2018, 07:59 PM RE: Little explorations with HP calculators (no Prime) - brickviking - 12-28-2018, 09:42 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-28-2018, 09:53 PM RE: Little explorations with HP calculators (no Prime) - 3298 - 12-28-2018, 11:21 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-29-2018, 12:10 AM RE: Little explorations with HP calculators (no Prime) - 3298 - 12-29-2018, 02:05 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-27-2018, 06:52 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 08:31 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-27-2018, 08:47 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-27-2018, 10:20 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018, 11:58 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-28-2018, 01:38 AM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-28-2018, 04:04 AM RE: Little explorations with HP calculators (no Prime) - ijabbott - 12-28-2018, 09:06 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-28-2018, 02:00 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 12-28-2018, 03:41 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-29-2018, 04:06 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 12-29-2018, 07:35 AM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-29-2018, 10:02 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-30-2018, 01:26 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-31-2018, 04:30 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 12-31-2018, 08:20 PM RE: Little explorations with HP calculators (no Prime) - DavidM - 12-31-2018, 08:36 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-13-2019, 10:48 AM RE: Little explorations with HP calculators (no Prime) - DavidM - 01-13-2019, 05:05 PM RE: Little explorations with HP calculators (no Prime) - John Keith - 01-13-2019, 08:27 PM RE: Little explorations with HP calculators (no Prime) - Thomas Klemm - 01-13-2019, 01:44 PM RE: Little explorations with HP calculators (no Prime) - Albert Chan - 01-13-2019, 01:48 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-13-2019, 07:00 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-28-2019, 01:38 PM RE: Little explorations with HP calculators (no Prime) - pier4r - 01-31-2019, 04:07 PM
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# A Fractional Sum
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Given a positive integer n, which integers can arise as the following sum
1/x1 + 2/x2 + 3/x3 + ... + n/xn
where xi is a strictly increasing sequence of positive integers?
Source: 13th Irish Olympiad, Crux Mathematicorum May 2006. | 107 | 363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | longest | en | 0.750896 |
http://mathhelpforum.com/calculus/152655-world-population-problem-using-integrals.html | 1,480,784,631,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00153-ip-10-31-129-80.ec2.internal.warc.gz | 180,965,951 | 10,253 | # Thread: World Population Problem Using Integrals
1. ## World Population Problem Using Integrals
Here is the factoids from the original problem:
F(0) = 6
lim(x -> infinity) F(x) = 30
rate of population growth = $(A*e^t)/((.02*A+e^t)^2)$
where t is a continuous variable and t >= 0
I need to figure out at what t will the population reach 10.
I managed to figure out that A = 46.15, but after that I get lost.
The book says I should be solving for t with the equation $30-((46.15)/(.923+e^t)) = 10$, but I keep coming up with the equation $6+((46.15)/(.923+e^t)) = 10$ and I don't know where I am going wrong.
If anyone has some insight for me, that would be super. I've been trying to figure this out for two days now and I need some help.
Thanks.
2. Originally Posted by Kristen
Here is the factoids from the original problem:
F(0) = 6
lim(x -> infinity) F(x) = 30
rate of population growth = $(A*e^t)/((.02*A+e^t)^2)$
where t is a continuous variable and t >= 0
I need to figure out at what t will the population reach 10.
I managed to figure out that A = 46.15, but after that I get lost.
The book says I should be solving for t with the equation $30-((46.15)/(.923+e^t)) = 10$, but I keep coming up with the equation $6+((46.15)/(.923+e^t)) = 10$ and I don't know where I am going wrong.
If anyone has some insight for me, that would be super. I've been trying to figure this out for two days now and I need some help.
Thanks.
You are given:
$F'(t)=\dfrac{Ae^t}{(0.02A+e^t)^2}$
so now integrate this to find $F(t)$ and use that together with the given data to solve the problem.
$F(t)=30-\dfrac{A}{0.02A+e^t}$
the $30$ is there because that is the value of $F(\infty)$ where the second term on the right is $0$
CB | 537 | 1,736 | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.950139 |
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Find Final Velocity when given initial velocity and Force in terms.In other words, velocity measures not only speed, but the direction that an object is traveling as mentioned in the study guide on motion.Avail a Free Physics Tutoring Session and get Help with Physics Concepts from the Physics Problem Solver.How to compute the average velocity of an object moving in one dimension from a graph of its position vs. time.
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They both reach the end points at the same time, though one went positive, the other went negative.Give an example where two objects could have the same average velocity but different average speeds.The velocity unit would include the speed traveled as well as the direction traveled in.Just as there is a speed formula and an acceleration formula there is a formula for velocity. | 470 | 2,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-47 | latest | en | 0.901313 |
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textbook from the previous versions is the addition of more exercises with answers or hints in theappendix. Although not part of the textbook, there are now 107 online videos with about 14 hours of content that span the first seven chapters of this book. These videos are freely availableonlineat Grand Valley’s Department of Mathematics YouTube | 3,981 | 17,855 | {"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} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.861744 |
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A173953 a(n) = numerator of (Zeta(2, 3/4) - Zeta(2, n-1/4)), where Zeta is the Hurwitz Zeta function. 11
0, 16, 928, 119344, 3078464, 1132669904, 606887707616, 49610806397296, 48006150564413056, 48265162121607952, 8192066749392160288, 15200753287254377716912, 33677610844789597790454208 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS All numbers in this sequence are divisible by 16. For A173953/16 see A173955. a(n+2)/A173954(n+2) is, for n >= 0, the partial sum Sum_{k=0..n} 1/(k + 3/4)^2 = 16*Sum_{k=0..n} 1/(4*k + 3)^2. The limit n -> infinity is given in A282824 as Zeta(2, 3/4) = Psi(1, 3/4) = Pi^2 - 8*Catalan, with the trigamma function Psi(1, z) and the Catalan constant A006752. LINKS G. C. Greubel, Table of n, a(n) for n = 1..250 Eric Weisstein's World of Mathematics, Hurwitz Zeta Function Eric Weisstein's World of Mathematics, Trigamma Function FORMULA a(n) = Numerator of (Pi^2 - 8*Catalan - Zeta(2, (4 n - 1)/4)). Numerator of 128*n*Sum_{k>=1} (4*k - 1 + 2*n) / ((4*k - 1)^2 * (4*k - 1 + 4*n)^2). - Vaclav Kotesovec, Nov 14 2017 Numerator of 16*Sum_{k=0..n-2} 1/(4*k + 3)^2, n >= 2, with a(1) = 0. See a comment above. - Wolfdieter Lang, Nov 14 2017 EXAMPLE The rationals r(n) = Zeta(2, 3/4) - Zeta(2, n-1/4) begin: 0/1, 16/9, 928/441, 119344/53361, 3078464/1334025, 1132669904/481583025, 606887707616/254757420225, 49610806397296/20635351038225, ... - Wolfdieter Lang, Nov 14 2017 MAPLE r := n -> Zeta(0, 2, 3/4) - Zeta(0, 2, n-1/4): seq(numer(simplify(r(n))), n=1..13); # Peter Luschny, Nov 14 2017 MATHEMATICA Table[Numerator[FunctionExpand[Pi^2 - 8*Catalan - Zeta[2, (4*n - 1)/4]]], {n, 1, 20}] (* Vaclav Kotesovec, Nov 14 2017 *) Numerator[Table[128*n*Sum[(4*k - 1 + 2*n) / ((4*k - 1)^2 * (4*k - 1 + 4*n)^2), {k, 1, Infinity}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *) Numerator[Table[16*Sum[1/(4*k + 3)^2, {k, 0, n-1}], {n, 1, 20}]] (* Vaclav Kotesovec, Nov 15 2017 *) PROG (PARI) for(n=1, 20, print1(numerator(16*sum(k=0, n-2, 1/(4*k+3)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018 (MAGMA) [0] cat [Numerator((&+[16/(4*k+3)^2: k in [0..n-2]])): n in [2..20]]; // G. C. Greubel, Aug 23 2018 CROSSREFS Denominators are in A173954. Cf. A006752, A120268, A173945, A173947, A173948, A173949, A173955. Sequence in context: A006089 A260620 A290940 * A211105 A276637 A211081 Adjacent sequences: A173950 A173951 A173952 * A173954 A173955 A173956 KEYWORD frac,nonn,easy AUTHOR Artur Jasinski, Mar 03 2010 EXTENSIONS Name simplified by Peter Luschny, Nov 14 2017 STATUS approved
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Last modified January 22 01:28 EST 2022. Contains 350481 sequences. (Running on oeis4.) | 1,233 | 3,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.521353 |
https://hindimaintutorial.in/gcd-sort-of-an-array-solution-leetcode-you-are-given-an-integer-array-nums-and-you-can-perform-the-following-operation-any-number-of-times-on-nums-swap-the-positions-of-two-ele/ | 1,656,726,878,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00689.warc.gz | 352,591,041 | 18,898 | For Solution
You are given an integer array `nums`, and you can perform the following operation any number of times on `nums`:
• Swap the positions of two elements `nums[i]` and `nums[j]` if `gcd(nums[i], nums[j]) > 1` where `gcd(nums[i], nums[j])` is the greatest common divisor of `nums[i]` and `nums[j]`.
Return `true` if it is possible to sort `nums` in non-decreasing order using the above swap method, or `false` otherwise.
Example 1:
```Input: nums = [7,21,3]
Output: true
Explanation: We can sort [7,21,3] by performing the following operations:
- Swap 7 and 21 because gcd(7,21) = 7. nums = [21,7,3]
- Swap 21 and 3 because gcd(21,3) = 3. nums = [3,7,21]
```
Example 2:
```Input: nums = [5,2,6,2]
Output: false
Explanation: It is impossible to sort the array because 5 cannot be swapped with any other element.
```
Example 3:
```Input: nums = [10,5,9,3,15]
Output: true
We can sort [10,5,9,3,15] by performing the following operations:
- Swap 10 and 15 because gcd(10,15) = 5. nums = [15,5,9,3,10]
- Swap 15 and 3 because gcd(15,3) = 3. nums = [3,5,9,15,10]
- Swap 10 and 15 because gcd(10,15) = 5. nums = [3,5,9,10,15]
```
Constraints:
• `1 <= nums.length <= 3 * 104`
• `2 <= nums[i] <= 105` | 445 | 1,212 | {"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} | 4.28125 | 4 | CC-MAIN-2022-27 | latest | en | 0.661566 |
https://en-academic.com/dic.nsf/enwiki/2055294 | 1,679,886,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00184.warc.gz | 282,138,304 | 11,273 | # Formula (mathematical logic)
Formula (mathematical logic)
In mathematical logic, a formula is a type of abstract object a token of which is a symbol or string of symbols which may be interpreted as any meaningful unit (i.e. a name, an adjective, a proposition, a phrase, a string of names, a string of phrases, etcetera) in a formal language. Two different strings of symbols may be tokens of the same formula. It is not necessary for the existence of a formula that there be any tokens of it. The exact definition of a formula depends on the particular formal language in question. [Hunter, Geoffrey, Metalogic: An Introduction to the Metatheory of Standard First-Order Logic]
A fairly typical definition (specific to first-order logic) goes as follows: Formulas are defined relative to a particular formal language and "relation symbols", where each of the function and relation symbols comes supplied with an arity that indicates the number of arguments it takes.
Then a term is defined recursively as
#A variable,
#A constant, or
#"f"("t"1,...,"t""n"), where "f" is an "n"-ary function symbol, and "t"1,...,"t""n" are terms.
An atomic formula is one of the form:
#"t"1="t"2, where "t"1 and "t"2 are terms, or
#"R"("t"1,...,"t""n"), where "R" is an "n"-ary relation symbol, and "t"1,...,"t""n" are terms.
Finally, the set of formulae is defined to be the smallest set containing the set of atomic formulae such that the following holds:
#$egphi$ is a formula when $phi$ is a formula;
#$\left(phi land psi\right)$ and $\left(phi lor psi\right)$ are formulae when $phi$ and $psi$ are formulae;
#$exists x, phi$ is a formula when "x" is a variable and $phi$ is a formula;
#$forall x, phi$ is a formula when $x$ is a variable and $phi$ is a formula (alternatively, $forall x, phi$ could be defined as an abbreviation for $egexists x, egphi$).
If a formula has no occurrences of $exists x$ or $forall x$, for any variable $x$, then it is called "quantifier-free". An "existential formula" is a string of existential quantification followed by a quantifier-free formula.
ee also
*Well-formed formula
*Theorem
References
*cite book | author = Hinman, P. | title = Fundamentals of Mathematical Logic | publisher = A K Peters | year = 2005 | id = ISBN 1-568-81262-0
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### Look at other dictionaries:
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# Capacitive Vs Resistive Strain Stretch Measurement Figure 6 A Sensor That Is Stretched To 30 Strain Held At That Strain For 10 Seconds And Then Relaxed Over 2 Seconds We Calculate Capacitance For The Anderson Bridge Circuit Diagram #9318
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How do You Wire a Switch?
One of the most popular electrical wiring thoughts is on how to wire a switch. Though using switches at your house is fairly uncomplicated, wiring one particular may not be that uncomplicated for everybody. An ON-OFF switch is in fact quite simple to wire. You will discover several types of switches, but for this instance, let us say you happen to be putting in a single-pole toggle switch, an exceedingly popular switch (as well as easiest).
You will find three hues of wires within a typical single-pole toggle switch: black, white, and green. Splice the black wire in two and hook up them on the terminal screws - a person on major and the other about the base screw with the swap. The white wire serves to be a supply of uninterrupted electricity and is particularly usually connected to the light-weight colored terminal screw (e.g., silver). Connect the inexperienced wire for the ground screw of the swap.
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How do You Replace a Breaker and just how Would you Incorporate a Sub Panel?
When a lot of endeavor to try and do these responsibilities by themselves, a lot of people are suggested to rent an electrician as an alternative. It's a lot more complicated and so harmful for most people to test to switch a breaker or increase a panel. To give you an thought about such popular electrical wiring inquiries, you should should operate on the warm electrical panel. In case you you should not even understand what this means, you're just not geared up to accomplish the task by yourself. Even though you really have to devote extra by hiring knowledgeable electrician, it can be considerably safer plus more wise to carry out this rather. - Capacitive Vs Resistive Strain Stretch Measurement Figure 6 A Sensor That Is Stretched To 30 Strain Held At That Strain For 10 Seconds And Then Relaxed Over 2 Seconds We Calculate Capacitance For The Anderson Bridge Circuit Diagram #9318
You will discover factors why these are one of the most commonly requested electrical wiring queries. Just one, a lot of imagine it's basic to try and do, and two, these are typically the typical electrical duties in your house. But you then mustn't set your protection in danger as part of your goal to save cash. The stakes could even be considerably increased when you make an effort to lower your expenses and do an electrical wiring job with no ample information or expertise.
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MitoPedia Terms and abbreviations Preprints and history MiP and biochemistry Concepts and methods MitoPedia: SUIT MitoPedia: O2k
Height of humans
## Description
The height of humans, h, is given in SI units in meters [m]. Humans are countable objects, and the symbol and unit of the number of objects is N [x]. The average height of N objects is, H = h/N [m/x], where h is the heights of all N objects measured on top of each other. Therefore, the height per human has the unit [m·x-1] (compare body mass [kg·x-1]). Without further identifyer, H is considered as the standing height of a human, measured without shoes, hair ornaments and heavy outer garments.
Abbreviation: h [m]; H [m·x-1]
Reference: Tipton 2012 Nursing
Healthy reference population Body mass excess BFE BME cutoffs BMI H M VO2max mitObesity drugs
```Communicated by Gnaiger Erich 2020-02-12
in: Catastrophe XXX XXX-mass Carol on BME and mitObesity of A X-mass Carol
```
## Units of height
### From height to area and volume
Any power function of H, such as H2 (dimension of area) and H3 (dimension of volume), has the corresponding units [m2·x-1] (area per object) and [m3·x-1] (volume per object), respectively. The unit is not, e.g., [m2·x-2], since the quantities are bound to a single object.
### When distinguishing H from h is not an issue
Height, H, is measured as a distance [m] from feet to head of a (single) individual [x]. The unit [x] is discussed in the context of number of entities. It is useful to consider the unit [x], if the quantities h [m] and H [m·x-1] have to be distinguished in a particular context. The same argument applies for distinguishing the body mass of an object, M [kg·x-1], from the mass of a tissue sample, m [kg]. Then the unit [x] cancels in the ratio of M·H-1 [kg·m-1] or in the BMI = M·H-2 [kg·m-2]. It is reasonable, to accept some level of inconsistency, when omitting the unit [x] from the unit for H, in a context that does not introduce any ambiguity between H and h.
## Measurement of height
The person is standing upright on a firm horizontally leveled surface without shoes, hair ornaments and heavy outer garments. A small gap of 0.1 m (10 cm) is maintained between the heels of the feet which face straight ahead and arms at sides. The back of the head, shoulder blades, buttocks and heels are touching the wall-mounted statiometer. For facing straingt, the ear canal and cheek bone are level. The 90° head of the statiometer is lowered to press the hair flat. This SOP applies to mobile persons who can stand steadily for the measurement.
## Self-reported measurements
'Men overestimated their height by 1.3 to 1.9 cm and the women by 0.5 to 1.3 cm. Men overestimated their weight by up to 0.45 kg and women underestimated their weight by up to 1.4 kg' (Tipton 2012 Nursing).
## References: BMI and height
- >>>>>>> - Click on [Expand] or [Collapse] - >>>>>>>
Reference
Bonthuis 2013 PLOS ONEBonthuis M, Jager KJ, Abu-Hanna A, Verrina E, Schaefer F, van Stralen KJ (2013) Application of body mass index according to height-age in short and tall children. PLOS ONE 8:e72068.
Bosy-Westphal 2009 Br J NutrBosy-Westphal A, Plachta-Danielzik S, Dörhöfer RP, Müller MJ (2009) Short stature and obesity: positive association in adults but inverse association in children and adolescents. Br J Nutr 102:453-61.
Cohen 2008 Am J Clin NutrCohen DA, Sturm R (2008) Body mass index is increasing faster among taller persons. Am J Clin Nutr 87:445-8.
De Onis 2007 Bull World Health Organizationde Onis M, Onyango AW, Borghi E, Siyam A, Nishida C, Siekmann J (2007) Development of a WHO growth reference for school-aged children and adolescents. Bull World Health Organization 85:660-7.
De Onis 2019 Public Health Nutritionde Onis M, Borghi E, Arimond M, Webb P, Croft T, Saha K, De-Regil LM, Thuita F, Heidkamp R, Krasevec J, Hayashi C, Flores-Ayala R (2019) Prevalence thresholds for wasting, overweight and stunting in children under 5 years. Public Health Nutrition 22:175-9.
Diverse Populations Collaborative Group 2005 Am J Phys AnthropolDiverse Populations Collaborative Group (2005) Weight-height relationships and body mass index: some observations from the Diverse Populations Collaboration. Am J Phys Anthropol 128:220-9.
Gnaiger 2019 MiP2019
OXPHOS capacity in human muscle tissue and body mass excess – the MitoEAGLE mission towards an integrative database (Version 6; 2020-01-12).
Heymsfield 2014 Am J Clin NutrHeymsfield SB, Peterson CM, Thomas DM, Heo M, Schuna JM Jr, Hong S, Choi W (2014) Scaling of adult body weight to height across sex and race/ethnic groups: relevance to BMI. Am J Clin Nutr 100:1455-61.
Hood 2019 Nutr DiabetesHood K, Ashcraft J, Watts K, Hong S, Choi W, Heymsfield SB, Gautam RK, Thomas D (2019) Allometric scaling of weight to height and resulting body mass index thresholds in two Asian populations. Nutr Diabetes 9:2. doi: 10.1038/s41387-018-0068-3.
Indian Academy of Pediatrics Growth Charts Committee 2015 Indian PediatrIndian Academy of Pediatrics Growth Charts Committee, Khadilkar V, Yadav S, Agrawal KK, Tamboli S, Banerjee M, Cherian A, Goyal JP, Khadilkar A, Kumaravel V, Mohan V, Narayanappa D, Ray I, Yewale V (2015) Revised IAP growth charts for height, weight and body mass index for 5- to 18-year-old Indian children. Indian Pediatr 52:47-55.
Sperrin 2016 J Public Health (Oxf)Sperrin M, Marshall AD, Higgins V, Renehan AG, Buchan IE (2016) Body mass index relates weight to height differently in women and older adults: serial cross-sectional surveys in England (1992-2011). J Public Health (Oxf) 38:607-613.
WHO 2006 Acta PaediatrWHO Multicentre Growth Reference Study Group (2006) WHO Child Growth Standards based on length/height, weight and age. Acta Paediatr Suppl 450:76-85.
WHO 2006 Geneva: World Health OrganizationWHO Multicentre Growth Reference Study Group (2006) WHO child growth standards: length/height-for-age, weight-for-age, weight-for-length, weight-for-height and body mass index-for-age: Methods and development. Geneva: World Health Organization 312 pp.
Zucker 1962 Committee on Biological Handbooks, Fed Amer Soc Exp BiolZucker TF (1962) Regression of standing and sitting weights on body weight: man. In: Growth including reproduction and morphological development. Altman PL, Dittmer DS, eds: Committee on Biological Handbooks, Fed Amer Soc Exp Biol:336-7.
## Publications: BME and height
» Height of humans
Reference
Bosy-Westphal 2009 Br J NutrBosy-Westphal A, Plachta-Danielzik S, Dörhöfer RP, Müller MJ (2009) Short stature and obesity: positive association in adults but inverse association in children and adolescents. Br J Nutr 102:453-61.
De Onis 2007 Bull World Health Organizationde Onis M, Onyango AW, Borghi E, Siyam A, Nishida C, Siekmann J (2007) Development of a WHO growth reference for school-aged children and adolescents. Bull World Health Organization 85:660-7.
Gnaiger 2019 MiP2019
OXPHOS capacity in human muscle tissue and body mass excess – the MitoEAGLE mission towards an integrative database (Version 6; 2020-01-12).
Hood 2019 Nutr DiabetesHood K, Ashcraft J, Watts K, Hong S, Choi W, Heymsfield SB, Gautam RK, Thomas D (2019) Allometric scaling of weight to height and resulting body mass index thresholds in two Asian populations. Nutr Diabetes 9:2. doi: 10.1038/s41387-018-0068-3.
Indian Academy of Pediatrics Growth Charts Committee 2015 Indian PediatrIndian Academy of Pediatrics Growth Charts Committee, Khadilkar V, Yadav S, Agrawal KK, Tamboli S, Banerjee M, Cherian A, Goyal JP, Khadilkar A, Kumaravel V, Mohan V, Narayanappa D, Ray I, Yewale V (2015) Revised IAP growth charts for height, weight and body mass index for 5- to 18-year-old Indian children. Indian Pediatr 52:47-55.
Zucker 1962 Committee on Biological Handbooks, Fed Amer Soc Exp BiolZucker TF (1962) Regression of standing and sitting weights on body weight: man. In: Growth including reproduction and morphological development. Altman PL, Dittmer DS, eds: Committee on Biological Handbooks, Fed Amer Soc Exp Biol:336-7.
## MitoPedia: BME and mitObesity
» Body mass excess and mitObesity | BME and mitObesity news | Summary |
TermAbbreviationDescription
BME cutoff pointsBME cutoffObesity is defined as a disease associated with an excess of body fat with respect to a healthy reference condition. Cutoff points for body mass excess, BME cutoff points, define the critical values for underweight (-0.1 and -0.2), overweight (0.2), and various degrees of obesity (0.4, 0.6, 0.8, and above). BME cutoffs are calibrated by crossover-points of BME with established BMI cutoffs.
Body fat excessBFEIn the healthy reference population (HRP), there is zero body fat excess, BFE, and the fraction of excess body fat in the HRP is expressed - by definition - relative to the reference body mass, M°, at any given height. Importantly, body fat excess, BFE, and body mass excess, BME, are linearly related, which is not the case for the body mass index, BMI.
Body massm [kg]; M [kg·x-1]The body mass, M, is the mass (kilogram [kg]) of an individual (object) [x] and is expressed in units [kg/x]. Whereas the body weight changes as a function of gravitational force (you are weightless at zero gravity; your floating weight in water is different from your weight in air), your mass is independent of gravitational force, and it is the same in air and water.
Body mass excessBMEThe body mass excess, BME, is an index of obesity and as such BME is a lifestyle metric. The BME is a measure of the extent to which your actual body mass, M [kg/x], deviates from M° [kg/x], which is the reference body mass [kg] per individual [x] without excess body fat in the healthy reference population, HRP. A balanced BME is BME° = 0.0 with a band width of -0.1 towards underweight and +0.2 towards overweight. The BME is linearly related to the body fat excess.
Body mass indexBMIThe body mass index, BMI, is the ratio of body mass to height squared (BMI=M·H-2), recommended by the WHO as a general indicator of underweight (BMI<18.5 kg·m-2), overweight (BMI>25 kg·m-2) and obesity (BMI>30 kg·m-2). Keys et al (1972; see 2014) emphasized that 'the prime criterion must be the relative independence of the index from height'. It is exactly the dependence of the BMI on height - from children to adults, women to men, Caucasians to Asians -, which requires adjustments of BMI-cutoff points. This deficiency is resolved by the body mass excess relative to the healthy reference population.
ComorbidityComorbidities are common in obesogenic lifestyle-induced early aging. These are preventable, non-communicable diseases with strong associations to obesity. In many studies, cause and effect in the sequence of onset of comorbidities remain elusive. Chronic degenerative diseases are commonly obesity-induced. The search for the link between obesity and the etiology of diverse preventable diseases lead to the hypothesis, that mitochondrial dysfunction is the common mechanism, summarized in the term 'mitObesity'.
Healthy reference populationHRPA healthy reference population, HRP, establishes the baseline for the relation between body mass and height in healthy people of zero underweight or overweight, providing a reference for evaluation of deviations towards underweight or overweight and obesity. The WHO Child Growth Standards (WHO-CGS) on height and body mass refer to healthy girls and boys from Brazil, Ghana, India, Norway, Oman and the USA. The Committee on Biological Handbooks compiled data on height and body mass of healthy males from infancy to old age (USA), published before emergence of the fast-food and soft-drink epidemic. Four allometric phases are distinguished with distinct allometric exponents. At heights above 1.26 m/x the allometric exponent is 2.9, equal in women and men, and significantly different from the exponent of 2.0 implicated in the body mass index, BMI [kg/m2].
Height of humansh [m]; H [m·x-1]The height of humans, h, is given in SI units in meters [m]. Humans are countable objects, and the symbol and unit of the number of objects is N [x]. The average height of N objects is, H = h/N [m/x], where h is the heights of all N objects measured on top of each other. Therefore, the height per human has the unit [m·x-1] (compare body mass [kg·x-1]). Without further identifyer, H is considered as the standing height of a human, measured without shoes, hair ornaments and heavy outer garments.
Lengthl [m]Length l is an SI base quantity with SI base unit meter m. Quantities derived from length are area A [m2] and volume V [m3]. Length is an extensive quantity, increasing additively with the number of objects. The term 'height' h is used for length in cases of vertical position (see height of humans). Length of height per object, LUX [m·x-1] is length per unit-entity UX, in contrast to lentgth of a system, which may contain one or many entities, such as the length of a pipeline assembled from a number NX of individual pipes. Length is a quantity linked to direct sensory, practical experience, as reflected in terms related to length: long/short (height: tall/small). Terms such as 'long/short distance' are then used by analogy in the context of the more abstract quantity time (long/short duration).
MitObesity drugsBioactive mitObesity compounds are drugs and nutraceuticals with more or less reproducible beneficial effects in the treatment of diverse preventable degenerative diseases implicated in comorbidities linked to obesity, characterized by common mechanisms of action targeting mitochondria.
ObesityObesity is a disease resulting from excessive accumulation of body fat. In common obesity (non-syndromic obesity) excessive body fat is due to an obesogenic lifestyle with lack of physical exercise ('couch') and caloric surplus of food consumption ('potato'), causing several comorbidities which are characterized as preventable non-communicable diseases. Persistent body fat excess associated with deficits of physical activity induces a weight-lifting effect on increasing muscle mass with decreasing mitochondrial capacity. Body fat excess, therefore, correlates with body mass excess up to a critical stage of obesogenic lifestyle-induced sarcopenia, when loss of muscle mass results in further deterioration of physical performance particularly at older age.
VO2maxVO2max; VO2max/MMaximum oxygen consumption, VO2max, is and index of cardiorespiratory fitness, measured by spiroergometry on human and animal organisms capable of controlled physical exercise performance on a treadmill or cycle ergometer. VO2max is the maximum respiration of an organism, expressed as the volume of O2 at STPD consumed per unit of time per individual object [mL.min-1.x-1]. If normalized per body mass of the individual object, M [kg.x-1], mass specific maximum oxygen consumption, VO2max/M, is expressed in units [mL.min-1.kg-1].
MitoPedia concepts: MiP concept
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MitoPedia:BME, Height | 3,832 | 15,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.898559 |
http://mathoverflow.net/questions/54371?sort=votes | 1,369,487,883,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705955434/warc/CC-MAIN-20130516120555-00042-ip-10-60-113-184.ec2.internal.warc.gz | 168,580,172 | 10,902 | ## Signatures on the infinite symmetric group
I have a question about the symmetric group. Taking signatures of permutations defines a surjective homomorphism $S_n \rightarrow \mathbb{Z}/2$. This is compatible with the natural inclusions $S_n \hookrightarrow S_{n+1}$, so we get a surjection $S_{\infty} \rightarrow \mathbb{z}/2$. Here $S_{\infty}$ is the direct limit of the $S_n$. In other words, $S_{\infty}$ is the group of finitely supported permutations of a countable set.
This brings me to my question. let $S_{\infty}'$ be the set of all permutations of a countable set. We have an inclusion $S_{\infty} \hookrightarrow S_{\infty}'$. Does the signature map $S_{\infty} \rightarrow \mathbb{Z}/2$ extend to $S_{\infty}'$?
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See also mathoverflow.net/questions/12291/…. this question actually starts with this observation here and tries to replace Z/2 by another group. – Martin Brandenburg Feb 5 2011 at 9:00
The answer to the question is "no". In fact, $S'_{\infty}$ is a perfect group, so there are no maps from it to an abelian group. Even more is true -- every element of $S'_{\infty}$ can be expressed as a commutator! This is much stronger than simply saying that $[S'_{\infty},S'_{\infty}] = S'_{\infty}$.
For these results, see Theorem 6 of the following paper.
MR0040298 (12,671e) Ore, Oystein Some remarks on commutators. Proc. Amer. Math. Soc. 2, (1951). 307–314.
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Thank you very much! – Ben S Feb 4 2011 at 22:36 For EVEN MORE, see the paper of M. Droste and I. Rivin (on arxiv.org, though it has now appeared online in Bulletin of the AMS). – Igor Rivin Feb 5 2011 at 3:35 Sorry, that's bulletin of the LMS above. – Igor Rivin Feb 5 2011 at 16:10
Here is a direct way to see the answer is 'no'.
Let our countable set be the set of all integers $\mathbb{Z}$.
What is the sign of $(1,2)(3,4)(5,6)(7,8)\dots$? (Note that we're fixing all nonpositive integers here.)
Whatever it is, you can multiply by the transposition $(1,2)$ to get a permutation with the opposite sign, then you can conjugate, which doesn't affect the sign, by $(\dots,-3,-2,-1,0,1,2,3,\dots)^{-2}$ giving back the element you started with. Contradiction.
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Another way to look at it: the above argument expresses $(1,2)$ as a commutator. – ndkrempel Feb 5 2011 at 3:38 | 696 | 2,272 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2013-20 | latest | en | 0.811523 |
https://inches-to-mm.appspot.com/95.7-inches-to-mm.html | 1,716,747,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00626.warc.gz | 272,608,985 | 6,393 | Inches To Mm
# 95.7 in to mm95.7 Inch to Millimeters
in
=
mm
## How to convert 95.7 inch to millimeters?
95.7 in * 25.4 mm = 2430.78 mm 1 in
A common question is How many inch in 95.7 millimeter? And the answer is 3.7677165354 in in 95.7 mm. Likewise the question how many millimeter in 95.7 inch has the answer of 2430.78 mm in 95.7 in.
## How much are 95.7 inches in millimeters?
95.7 inches equal 2430.78 millimeters (95.7in = 2430.78mm). Converting 95.7 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 95.7 in to mm.
## Convert 95.7 in to common lengths
UnitLengths
Nanometer2430780000.0 nm
Micrometer2430780.0 µm
Millimeter2430.78 mm
Centimeter243.078 cm
Inch95.7 in
Foot7.975 ft
Yard2.6583333333 yd
Meter2.43078 m
Kilometer0.00243078 km
Mile0.0015104167 mi
Nautical mile0.0013125162 nmi
## What is 95.7 inches in mm?
To convert 95.7 in to mm multiply the length in inches by 25.4. The 95.7 in in mm formula is [mm] = 95.7 * 25.4. Thus, for 95.7 inches in millimeter we get 2430.78 mm.
## Alternative spelling
95.7 in in mm, 95.7 Inch to Millimeters, 95.7 Inch in Millimeters, 95.7 Inches in mm, 95.7 Inch to mm, 95.7 Inch to Millimeter, 95.7 Inch in Millimeter, 95.7 Inches to Millimeters, 95.7 Inches in Millimeters, | 474 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.755717 |
https://documen.tv/what-is-the-sum-of-the-absolute-value-of-15-and-the-absolute-value-of-20-17656992-23/ | 1,686,177,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00520.warc.gz | 254,560,590 | 14,964 | Question
What is the sum of the absolute value of -15 and the absolute value of -20?
Step-by-step explanation:
By absolute value it means how far from 0 so all negatives will become positive | 43 | 193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.87322 |
https://answerprime.com/which-function-below-has-the-smallest-slope/ | 1,702,249,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00421.warc.gz | 117,083,749 | 39,299 | # Which function below has the smallest slope?
Which function below has the smallest slope? f(x) f(x) = 2x − 3 g(x) graph going through (0, negative 5) and (4, 3) h(x) x h(x) −1 3 1 7 3 11
We have three functions: FIRST. The slope of f(x) is 2 SECOND. g(x) graph going through (0, -5) and (4, 3), so: We can find the slope of g(x) as follows: The slope of g(x) is 2 THIRD. h(x) passes through (-1, 3), (1, 7) and (3, 11) We can find the slope of h(x) as follows: The slope of h(x) is 2 So, it is obvious that the three functions have the same slope 2.
The functions all have the same slope Step-by-step explanation: just took the assessment and I got it right.
The functions all have different slopes
The slope of the other functions can be found by finding ∆y/∆x, the change in function value divided by the change in x value. Slope of g(x) = (3 – (-5))/(4 – 0) = 8/4 = 2. Slope of h(x) = (7 – 3)/(1 – (-1)) = 4/2 = 2. The appropriate choice is The functions all have the same slope
Scroll to Top | 325 | 1,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.941311 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-15-differentiation-in-several-variables-15-4-differentiability-and-tangent-planes-preliminary-questions-page-789/3 | 1,580,307,561,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00347.warc.gz | 909,181,511 | 12,357 | ## Calculus (3rd Edition)
Since the linearization of $f$ at $(2,3)$ is $$L(x,y)=f(2,3)+f_x(2,3)(x-2)+f_y(2,3)(y-3)$$ then the right answer is (b), because (a) has a different form. | 75 | 181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-05 | latest | en | 0.859619 |
https://www.techwhiff.com/issue/in-1986-several-robotic-spacecrafts-were-sent-into--421772 | 1,675,914,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00796.warc.gz | 1,018,548,677 | 12,232 | # In 1986 several robotic spacecrafts were sent into space to study the halleys comet. Which of these statements best explains why humans should continue to send robotic spacecrafts into space?A) they absorb harmful radiations in spaceB) they help change weather conditions on earthC) they help discard some myths about objects in spaceD) they destroy meteors and comets which might strike earth
###### Question:
in 1986 several robotic spacecrafts were sent into space to study the halleys comet. Which of these statements best explains why humans should continue to send robotic spacecrafts into space?
A) they absorb harmful radiations in space
B) they help change weather conditions on earth
D) they destroy meteors and comets which might strike earth
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### Rights wich are based upon laws of nature and witch cannot be taken away are called
Rights wich are based upon laws of nature and witch cannot be taken away are called... | 1,249 | 4,590 | {"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} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.651921 |
https://www.mathnasium.com/ca/math-centres/reddeer/news/6820221020-the-zero-one-relationship-one-of-the-bases-for-a-complete-sense-of-numbers | 1,718,651,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00181.warc.gz | 801,644,458 | 14,229 | # THE ZERO-ONE RELATIONSHIP: ONE OF THE BASES FOR A COMPLETE SENSE OF NUMBERS
Oct 20, 2022 | Red Deer
Understanding proper fractions: the zero-one relationship that a student must know before introducing the algorithm related to fractions.
One of our students, a 6th grader, said lightly that it doesn’t matter if your math is bad in elementary and middle school, as long as it’s good in high-school. Well .. really? Foundational gaps do not happen overnight so it is very unlikely you can close the gaps really quick. There are no silver bullets for curing long-term problems with math, especially for middle and high school students.
Just after our last blog “FLUENCY IN FRACTIONS IS NECESSARY FOR SUCCESS IN ALGEBRA .. AND BEYOND” where it says that the typical mistake students make when ordering fractions is to put proper fractions before 0, it happened again. Just like most kids who did our assessment, a highschooler did the same mistake; a sign that they missed to understand this concept back in upper elementary. It’s not uncommon to see students who are able to do fractions operations but actually they missed the basic understanding of the value of proper fractions.
Let’s talk about the zero-one relationship then. Understanding this concept is part of an important math learning process for the primary grades, and also the foundation of a focused remedial program for middle school and high school.
The Zero-One Relationship
The zero-one relationship defines all other relationships in mathematics. When the distance from 0 to 1 is known, all other relationships are locked-in. When you know how big 1 is, then the size of 2, 3, and even the fractions 1/2, 3/4, and so on, are automatically determined and easily demonstrated.
The number 1, in addition to being part of the sequence “0, 1, 2...,” also represents a whole, one whole thing. This notion of 1 representing the whole is the basis on which Proportional Thinking, including the important skills of computing ratio and percent, is built.
The numbers between 0 and 1 (the proper fractions) have some often-overlooked characteristics. Here’s a good example:
Multiplication – Proper Fractions
You know that when we multiply by a number bigger than 1, the answer is larger than the number being multiplied. For instance:
2 x 6 = 12, where the answer (12) is larger than the number being multiplied (6).
(This question says, “How much is 6, two times?”)
But when we multiply by a number smaller than 1, the answer is smaller than the number being multiplied. For example:
1/2 x 6 = 3, where the answer (3) is smaller than the number being multiplied (6).
(This question says, “How much is half of 6?” Obviously, half of any number is always smaller than the number itself.)
So, multiplying does not always make things bigger. It depends on what the multiplier is.
Division – Proper Fractions
Similarly, dividing does not always make things smaller.
When we divide by a number bigger than 1, the answer is smaller than the number being divided. For instance:
12 : 2 = 6, where the answer (6) is smaller than the number being divided (12).
(This question says, “How many 2s are there in 12?”)
But be aware: when we divide by a number smaller than 1, the answer is bigger than the number being divided. Notice that:
6 : 1/2 = 12, where the answer (12) is larger than the number being divided (6).
(This question says, “How many halves are there in 6 wholes?” When you think about it, there must be more halves than wholes in any number.)
Please pay close attention, the language of “how many of these are in this?” is so important to build students’ understanding before introducing the algorithm related to fractions.
The nature and behavior of the numbers between 0 and 1 should be studied as a continuing strand throughout the math learning. | 861 | 3,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.947509 |
http://www.evi.com/q/14_oz_in_kg | 1,419,139,723,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770718.139/warc/CC-MAIN-20141217075250-00068-ip-10-231-17-201.ec2.internal.warc.gz | 508,401,135 | 6,068 | # 14 oz in kg
• 14 ounces is 0.4 kilograms.
• tk10npubl tk10ncanl
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• 14 oz in kgs | 1,530 | 4,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-52 | latest | en | 0.870346 |
https://www.numbersaplenty.com/2259440 | 1,653,063,017,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00216.warc.gz | 1,075,254,982 | 3,747 | Search a number
2259440 = 24561463
BaseRepresentation
bin1000100111100111110000
311020210100222
420213213300
51034300230
6120232212
725130201
oct10474760
94223328
102259440
111303607
1290b668
13611561
1442b5a8
152e96e5
hex2279f0
2259440 has 40 divisors (see below), whose sum is σ = 5350848. Its totient is φ = 887040.
The previous prime is 2259437. The next prime is 2259449. The reversal of 2259440 is 449522.
2259440 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a tau number, because it is divible by the number of its divisors (40).
It is a super-2 number, since 2×22594402 = 10210138227200, which contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (2259449) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 4649 + ... + 5111.
22259440 is an apocalyptic number.
2259440 is a gapful number since it is divisible by the number (20) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 2259440, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (2675424).
2259440 is an abundant number, since it is smaller than the sum of its proper divisors (3091408).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
2259440 is a wasteful number, since it uses less digits than its factorization.
2259440 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 537 (or 531 counting only the distinct ones).
The product of its (nonzero) digits is 2880, while the sum is 26.
The square root of 2259440 is about 1503.1433730686. The cubic root of 2259440 is about 131.2200717854.
The spelling of 2259440 in words is "two million, two hundred fifty-nine thousand, four hundred forty". | 608 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | latest | en | 0.913144 |
https://edurev.in/question/477198/A-consumer-spends-Rs--80-on-purchasing-a-commodity | 1,606,233,804,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00092.warc.gz | 279,417,846 | 41,329 | Courses
# A consumer spends Rs. 80 on purchasing a commodity when its price is Re. 1 per unit and spends Rs. 96 when the price is Rs. 2 per unit. Calculate the price elasticity of demand. a)0.2b)0.3c)0.4d)0.5Correct answer is option 'C'. Can you explain this answer? Related Test: Test: Theory Of Demand- 3
## CA Foundation Question
By Dhanush Surya · Sep 24, 2018 ·CA Foundation
Chandna Arora answered Jun 05, 2018
Initial Total Expenditure (TEo)=Rs 80
Final Total Expenditure (TE1)=Rs 96
Initial Price (Po)=Rs 1
Final Price (P1)=Rs 2
Now, Quantity Q = TE/P
Qo = 80/1 = 80
Q1 = 96/2 = 48
Now,
Ed=(−)[Po/Qo] x [ΔQ/ΔP]
Ed=(−)1/80 x [48−80]/(2−1)
Ed=(−)1/80 x (−32/1)
Ed=(−)−0.4
Ed=0.4
Thus, the price elasticity of demand is 0.4.
Goplar Ajay answered Dec 23, 2018
Shaheen Fatma answered Sep 24, 2018
80/96×2 =0.4
Dhanush Surya answered Jun 05, 2018
Any body explain this and how | 361 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-50 | latest | en | 0.807789 |
kinhdientamquoc.vn | 1,675,720,382,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00600.warc.gz | 365,932,104 | 5,283 | # Probability là gì
Until yesterday, the project was just a possibility, but now it has become a real probability (= it is likely to happen).
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Probability is the study of the mathematics of calculating the likelihood that particular events will happen.
high/low/strong probability There is a strong probability that the country will fall into recession again.
assess/calculate/estimate the probabilities They also estimated the probabilities of dividend increases & decreases under certain circumstances.
Transition probabilities were calculated directly from clinical trial data for the first 3 years và then extrapolated to 10 years.
We assume that the age differences within a cohort are small enough to consider these subgroups as being homogeneous with respect lớn transition probabilities.
The main interest is in characterizing how precisely we can "bound" the probabilities associated with various atoms.
His treatment is probabilistic, where probabilities are obtained from statistical data, rather than being subjective probabilities.
The probabilities here can be derived from statistical data on the relative frequency of prescriptions of drugs under given conditions.
At each cycle of the process, the cohort is reallocated lớn health states according to specified transition probabilities.
A better position, surely, would be to lớn consider the objective probabilities of the unrealized alternatives.
I hold that cases in which (only) numbers are different are structurally similar to lớn cases in which (only) probabilities are different.
Conversely, canopy trees show slower dynamics, with probabilities of mortality and recruitment lowered by a mean factor of about 0.5 with respect to other species.
All probabilities are appropriately conditioned on the observed data, and users can find any probabilities of interest. | 405 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-06 | latest | en | 0.904891 |
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# Loan-to-Value (LTV) Ratio
## What Is the Loan-to-Value (LTV) Ratio?
The loan-to-value (LTV) ratio is an assessment of lending risk that financial institutions and other lenders examine before approving a mortgage. Typically, loan assessments with high LTV ratios are considered higher risk loans. Therefore, if the mortgage is approved, the loan has a higher interest rate.
Additionally, a loan with a high LTV ratio may require the borrower to purchase mortgage insurance to offset the risk to the lender. This type of insurance is called private mortgage insurance (PMI).
### Key Takeaways
• Loan-to-value (LTV) is an often used ratio in mortgage lending to determine the amount necessary to put in a down-payment and whether a lender will extend credit to a borrower.
• Most lenders offer mortgage and home-equity applicants the lowest possible interest rate when the loan-to-value ratio is at or below 80%.
• Fannie Mae's HomeReady and Freddie Mac's Home Possible mortgage programs for low-income borrowers allow an LTV ratio of 97% (3% down payment) but require mortgage insurance until the ratio falls to 80%.
## Understanding the Loan-to-Value (LTV) Ratio
Interested homebuyers can easily calculate the LTV ratio of a home. This is the formula:
\begin{aligned} <V ratio=\frac{MA}{APV}\\ &\textbf{where:}\\ &MA = \text{Mortgage Amount}\\ &APV = \text{Appraised Property Value}\\ \end{aligned}
An LTV ratio is calculated by dividing the amount borrowed by the appraised value of the property, expressed as a percentage. For example, if you buy a home appraised at $100,000 for its appraised value, and make a$10,000 down payment, you will borrow $90,000. This results in an LTV ratio of 90% (i.e., 90,000/100,000). Determing an LTV ratio is a critical component of mortgage underwriting. It may be used in the process of buying a home, refinancing a current mortgage into a new loan, or borrowing against accumulated equity within a property. Lenders assess the LTV ratio to determine the level of exposure to risk they take on when underwriting a mortgage. When borrowers request a loan for an amount that is at or near the appraised value (and therefore has a higher LTV ratio), lenders perceive that there is a greater chance of the loan going into default. This is because there is very little equity built up within the property. As a result, in the event of a foreclosure, the lender may find it difficult to sell the home for enough to cover the outstanding mortgage balance and still make a profit from the transaction. The main factors that impact LTV ratios are the amount of the down payment, sales price, and the appraised value of a property. The lowest LTV ratio is achieved with a higher down payment and a lower sales price. ## How LTV is Used by Lenders A LTV ratio is only one factor in determining eligibility for securing a mortgage, a home-equity loan, or a line of credit. However, it can play a substantial role in the interest rate that a borrower is able to secure. Most lenders offer mortgage and home-equity applicants the lowest possible interest rate when their LTV ratio is at or below 80%. A higher LTV ratio does not exclude borrowers from being approved for a mortgage, although the interest on the loan may rise as the LTV ratio increases. For example, a borrower with an LTV ratio of 95% may be approved for a mortgage. However, their interest rate may be a full percentage point higher than the interest rate given to a borrower with an LTV ratio of 75%. If the LTV ratio is higher than 80%, a borrower may be required to purchase private mortgage insurance (PMI). This can add anywhere from 0.5% to 1% to the total amount of the loan on an annual basis. For example, PMI with a rate of 1% on a$100,000 loan would add an additional $1,000 to the total amount paid per year (or$83.33 per month). PMI payments are required until the LTV ratio is 80% or lower. The LTV ratio will decrease as you pay down your loan and as the value of your home increases over time.
In general, the lower the LTV ratio, the greater the chance that the loan will be approved and the lower the interest rate is likely to be. In addition, as a borrower, it's less likely that you will be required to purchase private mortgage insurance (PMI).
While it is not a law that lenders require an 80% LTV ratio in order for borrowers to avoid the additional cost of PMI, it is the practice of nearly all lenders. Exceptions to this requirement are sometimes made for borrowers who have a high income, lower debt, or have a large investment portfolio.
## Example of LTV
For example, suppose you buy a home that appraises for $100,000. However, the owner is willing to sell it for$90,000. If you make a $10,000 down payment, your loan is for$80,000, which results in an LTV ratio of 80% (i.e., 80,000/100,000). If you were to increase the amount of your down payment to $15,000, your mortgage loan is now$75,000. This would make your LTV ratio 75% (i.e., 75,000/100,000).
## Variations on Loan-to-Value Ratio Rules
Different loan types may have different rules when it comes to LTV ratio requirements.
### FHA Loans
FHA loans are mortgages designed for low-to-moderate-income borrowers. They are issued by an FHA-approved lender and insured by the Federal Housing Administration (FHA). FHA loans require a lower minimum down payment and credit scores than many conventional loans. FHA loans allow an initial LTV ratio of up to 96.5%, but they require a mortgage insurance premium (MIP) that lasts for as long as you have that loan (no matter how low the LTV ratio eventually goes). Many people decide to refinance their FHA loans once their LTV ratio reaches 80% in order to eliminate the MIP requirement.
### VA and USDA Loans
VA and USDA loans—available to current and former military or those in rural areas—do not require private mortgage insurance even though the LTV ratio can be as high as 100%. However, both VA and USDA loans do have additional fees.
### Fannie Mae and Freddie Mac
Fannie Mae's HomeReady and Freddie Mac's Home Possible mortgage programs for low-income borrowers allow an LTV ratio of 97%. However, they require mortgage insurance until the ratio falls to 80%.
For FHA, VA, and USDA loans, there are streamline refinancing options available. These waive appraisal requirements so the home's LTV ratio doesn't affect the loan. For borrowers with an LTV ratio over 100%—also known as being "underwater" or "upside down"—Fannie Mae's High Loan-to-Value Refinance Option and Freddie Mac's Enhanced Relief Refinance are also available options.
## LTV vs. Combined LTV (CLTV)
While the LTV ratio looks at the impact of a single mortgage loan when purchasing a property, the combined loan-to-value (CLTV) ratio is the ratio of all secured loans on a property to the value of a property. This includes not only the primary mortgage used in LTV but also any second mortgages, home equity loans or lines of credit, or other liens. Lenders use the CLTV ratio to determine a prospective home buyer's risk of default when more than one loan is used—for example, if they will have two or more mortgages, or a mortgage plus a home equity loan or line of credit (HELOC). In general, lenders are willing to lend at CLTV ratios of 80% and above and to borrowers with high credit ratings. Primary lenders tend to be more generous with CLTV requirements since it is a more thorough measure.
Let's look a little closer at the difference. The LTV ratio only considers the primary mortgage balance on a home. Therefore, if the primary mortgage balance is $100,000 and the home value is$200,000, LTV = 50%.
Consider, however, the example if it also has a second mortgage in the amount of $30,000 and a HELOC of$20,000. The combined loan to value now becomes ($100,000 +$30,000 + $20,000 /$200,000) = 75%; a much higher ratio.
These combined considerations are especially important if the mortgagee defaults and foes into foreclosure.
The main drawback of the information that a LTV provides is that it only includes the primary mortgage that a homeowner owes, and does not include in its calculations other obligations of the borrower, such as a second mortgage or home equity loan. Therefore, the CLTV is a more inclusive measure of a borrower's ability to repay a home loan.
### Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
1. Experian. "What Is Loan-to-Value Ratio (LTV)."
2. Freddie Mac. "Home Possible." | 2,043 | 8,940 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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} | 3.46875 | 3 | CC-MAIN-2022-21 | latest | en | 0.908305 |
https://brilliant.org/problems/quiet-complex-1/ | 1,524,776,953,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948549.21/warc/CC-MAIN-20180426203132-20180426223132-00605.warc.gz | 582,170,579 | 11,170 | # Quiet Complex -1
Algebra Level 4
Z is a complex number satisfying the equation (Z)^2 - (3+i)(Z) + (m+2i)=0 , where "m" is a real number. suppose the equation has a real root .then the additive inverse of the non real root ,is
NOTE : i=square root of (-1)
× | 81 | 262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-17 | latest | en | 0.827458 |
http://mathematica.stackexchange.com/questions/18044/findshortestpath-with-negative-weights-isnt-working/18047 | 1,464,337,860,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276564.72/warc/CC-MAIN-20160524002116-00239-ip-10-185-217-139.ec2.internal.warc.gz | 179,269,852 | 18,481 | # FindShortestPath with negative weights isn't working
I have the following functions defined:
RandomTree[n_, opts:OptionsPattern[]] :=
TreeGraph[UndirectedEdge[RandomInteger[{1,#}], # + 1] & /@ Range[1, n - 1], opts]
RandomCycleTree[n_, opts:OptionsPattern[]] := Module[{tree, e},
tree = RandomTree[n];
e = RandomChoice[EdgeList[GraphComplement[tree]]];
GraphUnion[tree, Graph[{e}], opts]]
RandomCycleTreeWeighted[n_, opts:OptionsPattern[]] :=
RandomCycleTree[n, EdgeWeight -> RandomReal[{-1, 1}, n], opts]
If I execute:
FindShortestPath[RandomCycleTreeWeighted[10,
VertexLabels -> "Name", ImagePadding -> 10], 3, 7]
it returns unevaluated. Can anyone reproduce this and explain why it is happening?
Note: I am using Mathematica 9. I also tried executing this in Mathematica 8.0 and it also didn't work.
-
The RandomTree function is from the documentation (up to minor variations). See the Applications section here – becko Jan 19 '13 at 2:44
Seems the problem is with the negative weights. I tried BellmanFord to no avail – Dr. belisarius Jan 19 '13 at 2:48
@belisarius Yes. Putting EdgeWeight -> RandomReal[{0, 1}, n] works fine. Weird, since in the documentation it is claimed that the "BellmanFord" method option should support negative weights. Is this a bug? – becko Jan 19 '13 at 2:53
FindShortestPath works for your graphs with Method->"BellmanFord" but ... your graphs should be Directed.
Bellman-Ford's algo works for graphs with negative edge weights, but only if they are free of negative weight cycles. Think of it as if you could get a -Infinity path: if your graph is undirected, you can always get a -Infinity valued path by going again and again forth and back over the same edge.
-
You can also try this: – becko Jan 20 '13 at 1:42
RandomTreeWeighted[n_, opts:OptionsPattern[]] := RandomTree[n, EdgeWeight -> RandomReal[{-1, 1}, n - 1], opts] – becko Jan 20 '13 at 1:43
Then FindShortestPath[RandomTreeWeighted[10], 3, 7,Method -> "BellmanFord"] also returns unevaluated. Please tell me if you can reproduce this too. – becko Jan 20 '13 at 1:44
It should work, because there are no cycles now. – becko Jan 20 '13 at 1:45
@becko No, probably I haven't explained it clearly. An undirected graph is never cycle-free because a->b->a is a cycle and is always possible – Dr. belisarius Jan 20 '13 at 3:24
RandomCycleTreeWeighted may generate graphs with negative cycles, whose edge sum to a negative value, and there may not be a shortest path. See Wikipedia Bellman-Ford page [1].
- | 704 | 2,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-22 | longest | en | 0.747139 |
https://powerpointmaniac.com/autocad/how-do-i-set-relative-coordinates-in-autocad.html | 1,695,793,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00807.warc.gz | 485,473,102 | 31,455 | # How do i set relative coordinates in autocad?
Contents
Use relative coordinates when you know the location of a point in relation to the previous point. To specify relative coordinates, precede the coordinate values with an @ sign. For example, entering @3,4 specifies a point 3 units along the X axis and 4 units along the Y axis from the last point specified.
## How do I set absolute coordinates in AutoCAD?
Solution:
1. On the command line, enter DSETTINGS.
2. In the Dynamic Settings dialog box, click the Dynamic Input tab.
3. Under Pointer Input, click Settings.
4. In the Pointer Input Settings dialog box, select either Relative Coordinates or Absolute Coordinates, as desired.
24 мая 2017 г.
## What is relative coordinate system?
Relative coordinates are locations defined by their distance from a reference point. An example of a relative coordinate is the distance from your computer monitor to your printer.
## What is the difference between relative and absolute coordinates?
Absolute coordinates: is distance or angle of axes relative to the origin of the coordinates. When entering absolute coordinates, enter the coordinates of points in the command bar. … Relative coordinates: the distance or angle of axes relative to the last point.5 мая 2019 г.
## How do I set coordinates in AutoCAD?
Using Specific Coordinates
1. Click Home tab > Draw panel > Line. Find.
2. Type the coordinate value for the first point by typing the X value, a comma, then the Y value, for example 1.65,4.25.
3. Press the Spacebar or Enter.
4. Do one of the following: …
5. Press the Spacebar or Enter.
## How do I find coordinates in AutoCAD?
Help
1. Click Home tab Utilities panel ID Point. Find.
2. Click the location that you want to identify. The X,Y,Z coordinate values are displayed at the Command prompt. With object snaps turned on, you can select an object and see the coordinates for a feature such as an endpoint, midpoint, or center.
## What do you mean by absolute polar coordinates?
Absolute polar coordinates are measured from the UCS origin (0,0), which is the intersection of the X and Y axes. Use absolute polar coordinates when you know the precise distance and angle coordinates of the point. With dynamic input, you can specify absolute coordinates with the # prefix.
## What are the two types of coordinate systems?
Data is defined in both horizontal and vertical coordinate systems. Horizontal coordinate systems locate data across the surface of the earth, and vertical coordinate systems locate the relative height or depth of data.
## What are the coordinate systems in AutoCAD?
World Coordinate System (WCS), User Coordinate System (UCS).
There is 4 AutoCAD coordinates system you should know. Absolute coordinate system, Relative Rectangular coordinate system, Relative Polar coordinate system and Interactive system(Direct coordinate system).
## What is an absolute coordinate system?
Absolute coordinates refers to a Cartesian System that uses x-axis, y-axis, and sometimes a z-axis to establish a point some distance from a common origin. For example, the picture’s origin point is ‘0,0’ and the absolute coordinate from that point is ‘8,7’ making it 8 along the x-axis and 7 along the y-axis.
IT IS INTERESTING: How do i weight in autocad?
## What is absolute point?
Absolute location refers to a specific, fixed point on the Earth’s surface as expressed by a scientific coordinate system. It is more precise than the relative location, which describes where a place is located using other places nearby.
## What is the difference between absolute and incremental coordinates?
Absolute coordinates are defined as each position on the work piece is unique. … With incremental coordinates, the last point traveled to becomes the new reference point on which the operator bases his next move. | 788 | 3,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.714891 |
http://www.stata.com/statalist/archive/2012-07/msg01227.html | 1,500,986,979,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425193.20/warc/CC-MAIN-20170725122451-20170725142451-00498.warc.gz | 568,856,389 | 4,856 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: how to get monthly values from Bimonthly data
From Nick Cox To statalist@hsphsun2.harvard.edu Subject Re: st: how to get monthly values from Bimonthly data Date Tue, 31 Jul 2012 13:00:11 -0500
```That wasn't clear.
So, you will need to double up. If monthly dates go up in steps of 2
expand 2
bysort date : replace date = date + 1 if _n == 2
On Tue, Jul 31, 2012 at 12:45 PM, sabbas gidarokostas
> thanks Nick for your reply and the useful code. But I need to convert
> these values to monthly data because the rest of my data are only
> monthly values.
>
> cheers
>
> On 7/31/12, Nick Cox <njcoxstata@gmail.com> wrote:
>> I would leave these data almost as they are. What you can do is
>>
>> gen year = real(substr(dates, -4, 4))
>> gen period = substr(dates, 1, 2)
>> gen month = cond(per == "JF", 1, cond(per == "MA", 3, cond(per ==
>> "MJ", 5, cond(per == "JA", 7, cond(per == "SO", 9, cond(per == "ND",
>> 11, .))))))
>>
>> gen date = ym(year, month)
>> tsset id date , delta(2)
>>
>> The key point is that you have bimonthly data and that's fine. No
>> doubling up or interpolation is needed; in fact that can only produce
>> a spurious increase in your sample size.
>>
>> Some people would undoubtedly prefer -recode- for the mapping "JF" -> 1,
>> etc.
>>
>> Nick
>>
>> On Fri, Jul 27, 2012 at 3:48 AM, sabbas gidarokostas
>>> Dear all,
>>>
>>> I have the following panel data set
>>> clear all
>>> input str8 (id text dates values1
>>> values2 )
>>>
>>> "1" "ot" " " "" " "
>>> "1" "wt" " " "" " "
>>> "1" "rt" "MA 2005" "32.4" "2.5"
>>> "1" "gh" "MJ 2005" "82.3" "12.5"
>>> "1" "bg" "JA 2005" "2.5" "24.5"
>>> "1" "se" "SO 2005" "82.5" " "
>>> "1" "cv" "ND 2005" "62.3" "26.3"
>>> "1" "cv" "JF 2006" "12.2" "22.3"
>>> "1" "cv" "MA 2006" "22.1" "2.4"
>>> "2" "ot" " " " " " "
>>> "2" "wt" " " " " " "
>>> "2" "rt" "MA 2005"
>>> "2" "gh" "MJ 2005" "32.3" "23.5"
>>> "2" "bg" "JA 2005" "52.4" "22.3"
>>> "2" "se" "SO 2005" "82.9" "22.5"
>>> "2" "cv" "ND 2005" " " "2.5"
>>> "2" "cv" "JF 2006" "332.4" " "
>>> "2" "cv" "MA 2006" "312.4" "21.5"
>>> end
>>> The above data set is a sample.
>>> Basically,I have 30000 individuals and instead of the last 2 columns
>>> that contain the numbers I have 29 such columns.
>>>
>>> The data set refers to bimontly values.
>>>
>>> Is there a way to convert the above data to montly data?
>>> Specifically, What I think is that first I have to split the dates as
>>> follows
>>>
>>> M/2005
>>> A/2005
>>> M/2005
>>> J/2005
>>>
>>> and so forth and then to convert them to
>>>
>>> 3/2005
>>> 4/2005
>>> 6/2005
>>> and so forth.
>>>
>>> So regarding the first 3 columns I should get
>>>
>>> "1" "ot" " "
>>> "1" "wt" " "
>>> "1" "rt" "3/2005"
>>> "1" "rt" "4 2005"
>>> "1" "gh" "5/2005"
>>> "1" "gh" "6/2005"
>>>
>>> and so forth.I am not exactly sure how to handle the last two columns,
>>> though(numerical values)
>>>
>>> I think that one way would be to divide the bimonthly values by 2. Then I
>>> get
>>>
>>> "1" "ot" " "
>>> "1" "wt" " "
>>> "1" "rt" "3/2005" "32.4/2" "2.5/2"
>>> "1" "rt" "4 2005" "32.4/2" "2.5/2"
>>> "1" "gh" "5/2005" "82.3/5" "12.5/5"
>>> "1" "gh" "6/2005" "82.3/5" "12.5/5"
>>>
>>> I am not sure that this makes sense as I have repeated values for two
>>> consequtive months.
>>>
>>> Another way would be interpolation?
>>>
>>> Any code provided is greatly appreciated.
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 1,604 | 4,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.836031 |
https://www.jiskha.com/display.cgi?id=1330049049 | 1,502,951,305,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102967.65/warc/CC-MAIN-20170817053725-20170817073725-00566.warc.gz | 934,555,730 | 3,613 | # GeOmEtRy!!!
posted by .
How many x-intercepts does the graph of y=x^2-4x+4?
• GeOmEtRy!!! -
Hint:
x^2-4x+4=(x-2)²
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## Ccscambridge.org
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### Microsoft word - cosmetic headline - maplewood aesthetics 1.doc
The Cosmetic Headline Published by MFY Inc. All rights reserved An update with Dr. Janette McDermett of Maplewood Aesthetics. CH - Dr. McDermott – what’s new at Maplewood Aesthetics and what do you now offer? DM – Microdermabrasion, Botox, Juvéderm and other dermal fillers. Also, teeth whitening and specific chemical peels for acne, sports peels, massages etc. CH – And
### neuroscience.ubc.ca
Advances in psychiatric treatment (2009), vol. 15, 242–249 doi: 10.1192/apt.bp.105.001834Antiglucocorticoids in psychiatry† Sean A. McIsaac, Åsa Westrin & Allan H. Young Sean McIsaac studied for his endocrine tissue: the hypothalamus, pituitary and adrenal cortices are its major components. The Significant evidence has accrued suggesting that HPA axis is regulated by external i | 1,555 | 5,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-34 | longest | en | 0.866899 |
https://nl.mathworks.com/matlabcentral/cody/problems/30-sort-a-list-of-complex-numbers-based-on-far-they-are-from-the-origin/solutions/1911482 | 1,601,012,070,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400222515.48/warc/CC-MAIN-20200925053037-20200925083037-00299.warc.gz | 544,095,587 | 16,576 | Cody
Problem 30. Sort a list of complex numbers based on far they are from the origin.
Solution 1911482
Submitted on 25 Aug 2019 by Ethan Zhang
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
j = sqrt(-1); z = [-4 6 3+4*j 1+j 0]; zSorted_correct = [6 3+4*j -4 1+j 0]; assert(isequal(complexSort(z),zSorted_correct))
2 Pass
z = 1:10; zSorted_correct = 10:-1:1; assert(isequal(complexSort(z),zSorted_correct)) | 168 | 529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-40 | latest | en | 0.614697 |
https://math.stackexchange.com/questions/1712866/magnitude-of-floating-point-rounding-errors-in-operations | 1,571,112,744,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655864.19/warc/CC-MAIN-20191015032537-20191015060037-00121.warc.gz | 593,936,087 | 32,831 | # Magnitude of floating point rounding errors in operations
I'm doing the following exercise:
Consider $$f(x)= \begin{cases} \displaystyle\frac{1-(\cos(x))^3}{x^2}, \quad if\ x\neq 0\\ \displaystyle\frac{3}{2}, \quad if\ x= 0\\ \end{cases}$$
Calculate $f(x_0)$, with $x_0=0.000011$, with a C program (simple and double precision) and with a pocket calculator. Identify the cause of the error. Justify the differences and the magnitude of the errors.
I know that there's cancellation on the numerator and the rounding errors on every operation ($\cos(x)$, $(\cos(x))^3$, $1-(\cos(x))^3$ and dividing that quantity by $x^2$). The calculator gives me $0$, and the correct answer is very near to $3/2$. This is because a pocket calculator can store only 9 significant digits, and $1-(\cos(x_0))^3$ is $0$ for a calculator. But how can I measure the magnitude of the errors when there're not rounding errors on the input? The quantity given by $x_0$ is exact and the computer write this quantity correctly in simple and double precision, so there's only rounding errors inside every operation. Can I measure the magnitude by using the propagation of errors' formula? This formula says: $$|\Delta f(x)|=|f'(x)|\cdot|\Delta x|.$$
Thanks!
• The correct answer is not close to $3/2$. Check your function definition. – Yves Daoust Mar 25 '16 at 11:40
• @YvesDaoust The exact value is like 1.499999999894125000... – Relure Mar 25 '16 at 11:44
• Oooops, sorry, I was cubing the whole numerator. – Yves Daoust Mar 25 '16 at 11:54
• (1) The C program will not represent $0.000011$ exactly, since it's not an integer divided by an integer power of $2$. But that's not the main cause of error, as you know. (2) Don't assume everyone knows which propagation of errors formula you mean. What's the formula? – David K Mar 25 '16 at 12:04
The error of the denominator can be estimated using relative error formulas for floating point evaluations as $$(1-(\cos(x)·(1+δ_1))^3·(1+δ_2))·(1+δ_3) \\ = (1-\cos(x)^3)·(1+δ_3)-\cos(x)^3·(3δ_1+δ_2)+O(δ^2)$$ where $|δ_k|<\mu=2^{-52}$ (the last for double). Thus the relative error of the denominator computation is $$δ_4=δ_3+\frac{\cos(x)^3}{1-\cos(x)^3}·(3δ_1+δ_2)$$ which for $x=1.1·10^{-5}$ is bounded by $|δ_4|<5.51·10^9·\mu$. For the full quotient this relative error transfers and using a more exact evaluation formula, this error is confirmed as
In [1]: raw = (1-cos(x)**3)/x**2
In [2]: raw
Out[2]: 1.5000012251581842
In [3]: refined = 2*(sin(x/2)/x)**2*(1+cos(x)+cos(x)**2)
In [4]: refined
Out[4]: 1.4999999998941247
In [5]: (raw-refined)/refined
Out[5]: 8.168427064213852e-07
In [6]: (raw-refined)/refined*2**52
Out[6]: 3678732508.259658
which gives the relative error as $3.68·10^9⋅μ$.
• Thanks for the answer! Where does the error of applying $\frac{1}{x^2}$ appear in your solution? Thanks! – Relure Mar 25 '16 at 15:43
• Nowhere. Like the error $δ_3$ of the subtraction, the squaring and division will add an additional $2$ or $3$ to the factor $5.51·10^9$, so I left out this trivial amount. For $x\gg 0$ these would play a greater role. – LutzL Mar 25 '16 at 15:48
• I've been trying to figure it out by myself, but I can't see why you turn this $(1-(\cos(x)·(1+δ_1))^3·(1+δ_2))·(1+δ_3) \\ = (1-\cos(x)^3)·(1+δ_3)-\cos(x)^3·(3δ_1+δ_2)$ into this $δ_4=δ_3+\frac{\cos(x)^3}{1-\cos(x)^3}·(3δ_1+δ_2)$. Where does the $δ_4$ come from? – Relure Mar 25 '16 at 17:47
• $δ_4=$(approxvalue-exactvalue)/exactvalue for the computation of $1-\cos(x)^3$, omitting the higher-order error terms. Which means it is the first expression minus $1-\cos(x)^3$ divided by $1-\cos(x)^3$. – LutzL Mar 25 '16 at 18:17 | 1,243 | 3,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-43 | latest | en | 0.833457 |
https://play.google.com/store/apps/details?id=integral.calculator.stepbystep | 1,721,259,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00076.warc.gz | 398,143,589 | 150,250 | # Integral Calculator with Steps
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Data can’t be deleted | 789 | 4,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.899544 |
http://stackoverflow.com/questions/4976997/vhdl-how-to-convert-a-floating-point-number-to-integer?answertab=oldest | 1,455,385,965,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701167113.3/warc/CC-MAIN-20160205193927-00232-ip-10-236-182-209.ec2.internal.warc.gz | 215,020,259 | 18,190 | # VHDL: How to convert a floating point number to integer
I want to pass a from a floating point number to a integer number. Basically I have a floating point number between 1 and 0, with three decimal places and I want to pass it to a integer number as if I multiply it by 1000. I suspect there should be a more optimal way to do it than using the arithmetic mult operation x1000. I´m looking for a piece of code preferably. Thank you to all possible pices of code, references, articles or comments.
-
Are you using float (real) in behavioral code for simulation or in RTL code that should go to hardware? – Philippe Feb 12 '11 at 8:49
I want to sinthesize floating point operations. – Peterstone Feb 12 '11 at 14:10
If you have a binary floating point number which is not a constant, there is no way to convert it to an integer in [0,1000] without actually performing multiplication and integral conversion.
Try doing the binary arithmetic on paper. The base-10 scaling doesn't work out to a shift or anything special… it's just a generic scaling operation.
Since the number is in the range [0,1], consider doing all the arithmetic in fixed point, perhaps using 1/1024 as the unit, or even in fixed point decimal with 1/1000 as the unit.
You should only use floating point math if you need more precision closer to zero.
-
If you want this for simulations only you can do something like:
i <= integer(r * 1000);
If you want to synthesize to hardware, you might consider a power-of two to make the logic more compact.
i <= integer(r * 1024);
However, @Potatoswatter has a valid point: you should consider if you really need a real (floating point).
- | 395 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-07 | latest | en | 0.902283 |
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# OCR (non mei) M1 Friday 17th June 2016
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1. it's not 97.18 because the triangle you are using isn't the resolved force and you could draw the forces with that angle and resolve them to find it's incorrect, drawing a line connecting the 2 ends and finding it is equal to 6 is not a resolved force
2. (Original post by physicskid123)
QQ
QQ
3. (Original post by ajamesg)
Attachment 551872this is the proof that it is 97.18 degrees
You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.
4. (Original post by sandpitturtle)
was P 12 on q5?
Yes
5. (Original post by Furthermaths100)
You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.
As I said previously, I agree now that I am wrong. Do you remember how many marks the whole question was worth?
6. * unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
7. (Original post by Parallex)
agree w/ 6.76 for the last one
tension was 1.81N or something for the pulleys
can't remember much lol
yes thats what i got too
8. (Original post by czj1997)
* unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
definitely
9. (Original post by czj1997)
* unrelated question.. but
IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
Yes but you will need higher As to make up for the Bs. You just need to average 80 UMS overall to get an A (480+ total UMS points).
10. Any idea what 68 would be UMS wise?
11. its angle to the vertical not horizontal that's why its 88 degrees. (180-97)
12. (Original post by Buymoria)
Any idea what 68 would be UMS wise?
68 marks could get you 100 ums, depends on how everyone did overall. Doubt that will get you less than 96 ums though.
13. I also did the cosine rule but youre forgetting to minus the value of 97 whatever from 180
14. (Original post by Arnold12345)
its angle to the vertical not horizontal that's why its 88 degrees. (180-97)
82.7 degrees
15. forgot to work on time on question 1 , 2 marks off?
(also what was the question on the contact force asking?)
16. I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?
17. also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??
18. (Original post by G_8)
I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?
No chance they will mark you down, I was thinking of doing it the complicated way but realised before I started, both are legitemate ways to do it and will both be on the mark scheme
19. (Original post by G_8)
also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??
This was quite hard for an M1 paper I would say - papers like this are usually about 63 for 90 ums and this will be no exception.
20. (Original post by Snicherz)
forgot to work on time on question 1 , 2 marks off?
(also what was the question on the contact force asking?)
Yes 2 marks off. Not too sure on the contact force question.
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https://cs50.stackexchange.com/questions/tagged/linear-search?tab=Votes | 1,606,560,930,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00709.warc.gz | 256,069,746 | 28,506 | # Questions tagged [linear-search]
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### PSET3 LINEAR SEARCH
Please help me with this: My code for linear search: // To check n is a positive int. if (n<0) { return false; } // To check if value can be found in values array. if values is found return ...
50 views
### How to create an input for the search function?
I am trying to test my search function by creating a separate function that takes an input [some integer] and then checks whether that integer fits within the given array [numbers 10-19]. I made the ...
161 views
### Pset3 Linear Search (control may reach end of non-void function)
This seems pretty simple, but I'm not sure what I'm doing wrong. Thanks in advance for helping! bool search(int value, int values[], int n) { if (n >= 0) { for (int i = 0; i <= n; i++...
292 views
I'm doing pset3 and I seem to have a functioning linear search. But I can't find my "needle" if it's in a higher array position than 331, which is where 127 is (I've copied the output of "generate" to ...
52 views
### Caesar: Validating the key - stuck
I am stuck with Caesar, validating the key. Before i use the atoi function, I want to check if argv consists of digits. I watched the linear search movie but is is to cryptic for me. I have to ...
49 views
### Linear find function always finding needle in hay stack and generating multiple numbers
I wanted to start simple with a linear search in the helper.c function: bool search(int value, int values[], int n) { // TODO: implement a searching algorithm for (int i = 0; value < n; i++) { ...
64 views
### Pset 3 Helper: Linear search function implem/ header file won't compile
I just tried to compile this file(helpers.c), as a function implementation that is already included as a header for the main programme(find.c). I am not sure what else to do anymore, I have tried ...
16 views
### Why am I generating the following errors with this code if it says it's compiled correctly?
The code compiles, and based off what I've seen in the linear search video, this should work. Tad bit confused why with this code below this error, I'm getting this error. In function `_start': (....
59 views
### pset3 linear search not working
So I've started working on pset3 after a long break, I couldnt even start making a binary search function. Tried to do a linear search function but it doesnt seem to work when i run "./generate 1000 ...
366 views
### Cannot find the needle 127 in the haystack of 1000 random numbers using linear search . Any ideas on what I may have done wrong?
I'm having a bit of difficulty in compiling the helpers.c program. Specifically when I try to check the correctness of the program, it seems it cannot find the needle 127 in the haystack of 1000 ...
512 views
### Linear Search not working
Alright, so, I'm not quite sure what's going wrong here, but maybe you people could help. My implementation of linear search in pset3 will return false no matter what (well, specifically I've only ...
42 views
### I have written a linear searh algorithm. How do I test it to see that it works?
This is my linear search code. How do I test it to see if it works? /** * helpers.c * * Computer Science 50 * Problem Set 3 * * Helper functions for Problem Set 3. */ #include <cs50.h> ...
115 views
### PSet 3 Helpers/Find Compile Error
I started changing "helpers.h" to implement a linear search. I have not changed generate.c or find.c. #include <cs50.h> bool search(int value, int values[], int n); { for(int i = 0; i < n;...
54 views
### PSET 3 Linear Search Error
Keep getting the error of "control may reach end of non-void function" This is for the linear search, can someone please give me some help/tips into getting this error to go away? This is my code: ...
34 views
### search not compiling -
I want to just see if the search part of helpers.c works, but upon compiling, I get a really long error: this goes on for about 20 lines. Not sure what to do. I read through other posts, tried make ...
795 views
### pset3 helpers.c linear search confusion
bool search(int value, int values[], int n) // TODO: implement a searching algorithm if (n >= 0) { for (int i = 0; i < n; i++) { if (values[i] == value) ...
91 views
### Why isn't my linear search working correctly?
I cannot get my linear search programm to work and it seems to be a problem with the return of false if the needle is not found. My coding for the search is bool search(int value, int values[], int n)...
86 views
### My linear search can't find any values. Why?
Here's the code for my pset 3, which fails to find the '127' in the stack. Any hints will be massively appreciated: bool search(int value, int values[], int n) { for(int i = 0; i < n; i++) ...
126 views
### Pset3 - working on linear search
I'm currently working on adding the linear search and receive the following error when I try to compile... Also, when changing the code for helpers.c, is it necessary to compile that file or just ...
74 views
### Control may reach end of non-void function [duplicate]
I have read the following and am still unable to answer my question: How to Fix ācontrol may reach end of non-void functionā and āNo such file or directoryā Errors? Error: control may reach end of ...
249 views
### pset3 - Linear search wont work at all
So, I'm currently stuck on pset3. I'm doing the linear search coding on helpers.c but it doesn't seem to search for the needle in the hay stack. It is behaving quite weird. If i run ./generate 1000 ...
806 views
### linear search compilation error
I've updated search function in helpers.c with code below. I had to remove reference to helpers.h since that was causing even more errors. #include <cs50.h> // #include "helpers.h" bool search ... | 1,429 | 5,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | longest | en | 0.843091 |
http://www.topcodingsolutions.net/2015/02/shoppingsurveydiv2.html | 1,653,063,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00473.warc.gz | 109,033,069 | 12,255 | ## Monday, February 16, 2015
### ShoppingSurveyDiv2
Problem:
TopCoder Problem Statement - ShoppingSurveyDiv2 Single Round Match 634 Round 1 - Division II, Level Two
Overview:
Given a count of how many of each item was purchased, and the number of customers, determine the smallest number of customers that could have purchased one of each item.
Java Source:
```01: public class ShoppingSurveyDiv2 {
02:
03: public int minValue(int N, int[] s) {
04:
05: int numItemsPurchased = 0;
06:
07: for (int i : s) {
08: numItemsPurchased += i;
09: }
10:
11: int maxItemsPurchasedWithNoBigShoppers = (s.length - 1) * N;
12:
13: int numBigShoppers =
14: numItemsPurchased - maxItemsPurchasedWithNoBigShoppers;
15:
16: return (numBigShoppers < 0) ? 0 : numBigShoppers;
17:
18: }
19: }
```
Notes:
First, we'll count the total number of items that have been purchased. This is stored in the variable numItemsPurchased.
Next, we calculate the number of items that could be purchased if every customer stopped just short of buying one of each. That is each customer, buys (s.length -1) items. We'll call that maxItemsPurchasedWithNoBigShoppers.
The number of BigShoppers that we'll need is now just the difference between numItemsPurchased and maxItemsPurchasedWithNoBigShoppers. This value may be negative, in which case we'll return zero.
This problem is a bit difficult to understand, but once you get it, the code is really easy. | 385 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.8855 |
http://stackoverflow.com/questions/3407563/haskell-numeric-type-heirarchy-in-sicp-exercises?answertab=active | 1,394,216,618,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999650252/warc/CC-MAIN-20140305060730-00014-ip-10-183-142-35.ec2.internal.warc.gz | 172,608,873 | 16,217 | # Haskell numeric type heirarchy in SICP exercises
I've been learning Haskell recently and was talking to a friend who is working through SICP. We were curious to compare Common Lisp and Scheme and so I decided as an exercise to try to translate exercise 1.29 into Haskell.
This exercise uses a function sigma which represents the mathematical Summation function Sigma. This function takes a function f to apply to each term, a lower bound, a function to apply to each term to get the next term, and an upper bound. It returns the sum of f applied to each term.
simpsonIntegral is supposed to use Simpson's rule to approximate the integral of the function f over the range [a, b] using an "accuracy" n. I'm having trouble getting this function to work because there seems to be something I don't understand about the types involved.
This code will compile with version 6.12.1 of ghc but simpsonIntegral will be given a type context (Integral a, Fractional a) which doesn't make any sense and the function blows up as soon as you call it. I've got this working at one point but what I did was so obviously a hack that I wanted to ask here how this would be handled idiomatically.
How does one idiomatically handle the Integral -> Fractional/Real conversion needed in h? I read a number of things but nothing seemed obvious and clean.
sigma :: (Ord a, Num b) => (a -> b) -> a -> (a -> a) -> a -> b
sigma f a next b = iter a 0
where
iter current acc | current > b = acc
| otherwise = iter (next current) (acc + f current)
simpsonIntegral f a b n = 1.0 * (h / 3) * (sigma simTerm 0 (1+) n)
where
h = (b - a) / n
simTerm k = (yk k) * term
where
yk k = f (a + h * k)
term =
case k of
0 -> 1
1 -> 1
otherwise -> if odd k then 4 else 2
-
To follow up on Justice's answer: if you're curious about where to put the fromIntegrals, the following compiles:
simpsonIntegral :: (Integral a, Fractional b) => (b -> b) -> a -> a -> a -> b
simpsonIntegral f a b n = 1.0 * (h / 3) * (sigma simTerm 0 (1+) n)
where
h = fromIntegral (b - a) / fromIntegral n
simTerm k = (yk k) * term
where
yk k = f (fromIntegral a + h * fromIntegral k)
term =
case k of
0 -> 1
1 -> 1
otherwise -> if odd k then 4 else 2
And seems to work:
*Main> simpsonIntegral (^3) 0 1 100
0.2533333233333334
*Main> simpsonIntegral (^3) 0 1 1000
0.2503333333323334
-
The problem is that the function "odd" expects it's argument to be an Integral type. The compiler then infers that your variable "k" is of type Integral. But by using the operation "/", the compiler infers "k" to be also of type Fractional. The solution can be as simple as converting "k" to Integer where it's really needed:
if odd (round k) then 4 else 2 | 756 | 2,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.927254 |
http://mymathforum.com/complex-analysis/335959-write-complex-number-bi-form-b-real.html | 1,544,681,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00483.warc.gz | 195,088,136 | 9,238 | My Math Forum write complex number in a+bi form with a and b real
Complex Analysis Complex Analysis Math Forum
September 17th, 2016, 12:06 PM #1 Newbie Joined: Jun 2012 Posts: 11 Thanks: 0 write complex number in a+bi form with a and b real $\frac{z}{(z+1)^2}$, with $z=\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}i$ I know the answer is $1-\frac{\sqrt{2}}{2}$, but I don't know how to get there. Anybody know which steps I should take? Last edited by skipjack; October 31st, 2016 at 09:30 PM.
September 17th, 2016, 01:21 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond \displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{e^{i\pi/4}}{(e^{i\pi/4}+1)^2} \\ &=\frac{e^{i\pi/4}}{e^{i\pi/2}+2e^{i\pi/4}+1} \\ &=\frac12\frac{\sqrt2+\sqrt2i}{i+\sqrt2+\sqrt2i +1} \\ &=\frac{\sqrt2}{2}\frac{1+i}{(\sqrt2+1)(1+i)} \\ &=\frac{\sqrt2}{2}\frac{1}{\sqrt2+1} \\ &=\frac{\sqrt2}{2}(\sqrt2-1) \\ &=\frac{2-\sqrt2}{2} \\ &=1-\frac{\sqrt2}{2}\end{align*} Thanks from wortel
October 31st, 2016, 09:51 PM #3 Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850 \displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{1}{z + 2 + 1/z} \\ &= \frac{1}{2 + \sqrt2} \\ &= \frac{2 - \sqrt2}{2} \\ &= 1 - \frac{\sqrt2}{2}\end{align*} Thanks from Joppy
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Contact - Home - Forums - Cryptocurrency Forum - Top | 677 | 1,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-51 | latest | en | 0.724233 |
http://7puzzleblog.com/229/ | 1,521,699,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647777.59/warc/CC-MAIN-20180322053608-20180322073608-00108.warc.gz | 5,488,136 | 8,605 | # DAY 229:
Today’s Challenge
There are five different 24game challenges below.
For each group of four numbers, your task is to arrive at the target answer of 24 by using each of the four digits exactly once, with + – × ÷ available:
• 1 2 3 4
• 2 3 4 5
• 3 4 5 6
• 4 5 6 7
• 5 6 7 8
All five challenges are possible, can you do it? Have a go!
The 7puzzle Challenge
The playing board of the 7puzzle game is a 7-by-7 grid containing 49 different numbers, ranging from up to 84.
The 1st & 7th rows of the playing board contain the following fourteen numbers:
2 4 9 11 14 15 22 24 27 30 40 70 72 77
What is the total when adding together all the odd numbers?
Make 229 Challenge
Can you arrive at 229 by inserting 2, 3, 4, 5 and 7 into the gaps below?
◯²×◯+◯²×◯+◯ = 229
Answers can be found here.
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This entry was posted in 7puzzleblog.com. Bookmark the permalink. | 360 | 1,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-13 | longest | en | 0.702217 |
http://codeforces.com/blog/entry/45258 | 1,585,407,645,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00544.warc.gz | 45,114,489 | 22,669 | Rating changes for the last round are temporarily rolled back. They will be returned soon. ×
### MikhailRubinchik.ru's blog
By MikhailRubinchik.ru, 4 years ago, translation, ,
Hello everyone
I created and prepared all problems for the first round of Yandex.Algorithm . Below you will find editorial. I am sorry if something in solutions discriptions is not clear, if you have questions — please ask them in comments.
## Problem A. Cheese
If potato takes less time to heat, than cheese to melt, then it is impossible. Otherwise, we either heat the potato a little and then put cheese on it, or heat the cheese to some degree, put it on the potato and continue to heat them together.
## Problem В. Travelling in the Square
After leaving the central square there are four possible directions to go to, on the next step there is a choice from two turns and after that the route is fully determined. Thus, there are only 8 different routes. Each of them can be represented by a single string of length 9. Let’s put these strings in a constant array. While reading the input data let’s ignore all line breaks, so that we have a single string of length 9. Now we just need to find a string in the constant array, that equals to the input string (given that “?” is equal to any symbol).
## Problem С. Opencup
Note, that if your team is not in top 30, it doesn’t matter which place it took. So let’s think that the only places available are the places from 1 to 31 (the last place means that we should output 1000). For every place from 1 to 30 let’s check whether we can stay in top 10. The worst case scenario is top 10 teams (excluding yours) take top 10 places in the last Grand Prix. Let’s assign a place to each of top 10 teams in such a way, so they would rank higher than you team. That we can do using simple brute-force. The total amount of operations will then be 31 * 10! * 10, which is about a billion. 31 can be replaced with log 31, and that takes us to about 200 million operations, which will probably be enough to get an OK for the solution, if it is implemented efficiently. Otherwise, several straight-forward optimizations will lead to a success. For example, you can check correctness of a permutation while building it. To eliminate a risk of exceeding the Time Limit, you can use a greedy approach when assigning points to top 10 teams.
## Problem D. Cold countries
Let t be the number of pairs (boy + girl) that will eventually sleep next to each other. We are going to solve the problem for each t separately and then sum the answers with coefficients C(p, t). There are 2t! satisfying permutations if we are solving the problem for pairs only. To every such permutation we should now add w — t girls and m — t boys. Between some pairs (or between a pair and a wall) we can put only girls, and between other pairs — only boys. If t is odd, there is x ways to do it (for both girls and boys), where t = 2x — 1. The number of ways to put boys into permutation then looks like this: m! * C(m + x, x). First multiplier describes the order of boys and the second solves a standard combinatorics problem of putting m objects between t walls. We can then simplify the expression and get (m + x)! / x! (the same for girls). So, for an odd t we have an answer 2 * (t!) * (m + x)! * (w + x)! / x! / x!. We will not compute the answer for an even t here, but note, that in this case one should consider two possibilities: when the leftmost person in a first pair a boy and when it is a girl. This solution has O(p * log MOD) complexity, and this is too slow. The second factor appears because of the necessity of performing a division (or finding modular multiplicative inverse). Let’s notice, that we need to find it only for factorials of numbers from 1 to p and 1 / fact[y] = 1/ fact[y — 1] * 1 / y. We can find multiplicative inverse for numbers 1 .. p in O(p log log p) time, using Sieve of Eratosthenes, and then calculate the inverse for composite numbers in O(1), and for primes in log MOD. Also, you could have gotten an OK using only caсhing of already computed values. There is also a linear algorithm for finding multiplicative inverse elements, you can find its’ description in comments to this discussion: http://apps.topcoder.com/forums/?module=Thread&threadID=680416&mc=26&view=threaded
## Problem E. Lillebror and Karlsson
Let’s represent a chocolate as a binary string of length m which contains n ones. We now need to split the string into a minimal amount of substrings, that fit our criteria. The resulting partition will contain n strings, of which some consist of ones and zeroes, and some only of zeroes (let’s call them zero-strings). To solve the problem we must minimize the amount of zero-strings. Let f[i] be a minimal number of zero-strings in a partition of a prefix of length i. We are going to initialize f[i] = f[j] if (i + j) / 2 — integer, s[(i + j) / 2] = ‘1’ and there are no other ones between positions i and j. If there is no such j, we initialize f[i] with infinity. Now we need to update f[i] = min(f[i], f[k] + 1]), using all k such that s[k..i] is a zero-string. Using the fact that all such k form a single substring, we can maintain minimum of them and update f[i] using O(1) time. We now have a solution with complexity O(m). Let’s transform it into O(n) solution. Initialization: f[0] = 0; f[i] = 1, if there are no ones to the left from i and s[i] = ‘0’. Now let’s find initial values of f[i] for all positions between the leftmost (first) ‘1’ and the second one. To do that we’ll need to reverse the array f. If the right end of array “covers” the right “1”, then we just need to delete everything after it. If the right end didn’t reach the right “1”, we need to add some values into array (let’s take a minimal value in the array + 1). After that step we have initial values for all positions between first two ones. Now let’s run through array from left to right and update a value on each position to a (minimum on the prefix with the end in this position) + 1. Similarly, we can calculate f[i] for all positions between second and third “1” and so on. Our array is too long, so we need to compress it. If there are x consecutive values y we will store corresponding part of array as a pair (x, y). Now we can add a bunch of equal values (and in this algorithm we only add equal values) in O(1) time. The amount of deleted values is not greater than the amount of added. Finally, let’s notice, that the difference between maximal and minimal values in the array is 1 or less. c — position of the minimal element in the array. To the right from it there can’t be a number greater than c + 1. Similarly, on a previous step we could not have had numbers greater than c + 1 to the left from it. Thus, our array in compressed form never has more than 3 elements, and we can reverse and run through it in O(1) time.
## Problem F. Dynamic Complexity of String
For each border of length g there is a period of length T such that g + T = |S|. Because of that, we can find a minimal period of a string by subtracting from its’ length the length of maximal border. This is derived directly from the definitions of period and border (for more information search the Internet for “border array and period”). Furthermore, if the borders are scanned in decreasing order, the prefix function of each next border will be greater, than the previous one, and that means that minimal period will decrease (or stay the same). After calculating a prefix-function we can in O(1) find complexity for each prefix. To do that take the answer for the maximal border and add 1 if its’ period is different from the period of the next border (in decreasing order). We now know how to find an answer and how to write a checker for the problem. What left is to learn how to find the right string. First, it’s useful to look at known “interesting“ strings, for example “random string”, Zimin strings, Thue–Morse strings, Fibonacci strings, string from identical letters and so on (for more information search the Internet for “combinatorics on words”). In this case we can use Fibonacci strings. Note, that solution to this problem has a practical value. Some string algorithms have complexity O(dynamic complexity of string), and solution described above generates worst-case tests for such algorithms. Dynamic complexity of random strings is O(n), so if your problem can potentially be solved using borders’ analysis, you definitely should add such tests to a testset to increase the perfomance time of those algorithms by log n.
• +37
» 4 years ago, # | +46 The editorial is good and all, except for permutations in C. It's like talking about problem of adding two numbers as:You can generate a random number and check if it's the answer. Choosing from interval [min(a, b), 2·max(a, b)] and repeating 105 times should be enough to get AC. Though, you can just print a + b to avoid any risk.And nice problems btw., I enjoyed D and E a lot.
» 4 years ago, # | +16 Problem C can also be solved with maximum bipartite matching. Iterate over resulting place (from 31 down to 1) and check if it's possible. How to check: build graph with two parts — places and top10. Add an edge between i-th place and j-th team if j-th team will have >= our points. If |max matching| < 10 then we found the answer.
» 4 years ago, # | ← Rev. 2 → 0 int main() { int p,a,b,res=-1; cin>>p>>a>>b;if(p>=a) { res=max(p,a+b); } cout<
• » » 4 years ago, # ^ | +5 must be p ≥ b.
• » » » 4 years ago, # ^ | 0 thanks it worked | 2,360 | 9,557 | {"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} | 3.5 | 4 | CC-MAIN-2020-16 | latest | en | 0.920071 |
http://oeis.org/A045483 | 1,571,366,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00039.warc.gz | 147,587,708 | 4,102 | This site is supported by donations to The OEIS Foundation.
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A045483 McKay-Thompson series of class 5B for the Monster group with a(0) = 1. 3
1, 1, 9, 10, -30, 6, -25, 96, 60, -250, 45, -150, 544, 360, -1230, 184, -675, 2310, 1410, -4830, 750, -2450, 8196, 4920, -16180, 2376, -7875, 25644, 15000, -48720, 7126, -22800, 73221, 42310 (list; graph; refs; listen; history; text; internal format)
OFFSET -1,3 LINKS Seiichi Manyama, Table of n, a(n) for n = -1..1000 J. H. Conway and S. P. Norton, Monstrous Moonshine, Bull. Lond. Math. Soc. 11 (1979) 308-339. D. Ford, J. McKay and S. P. Norton, More on replicable functions, Commun. Algebra 22, No. 13, 5175-5193 (1994). J. McKay and H. Strauss, The q-series of monstrous moonshine and the decomposition of the head characters, Comm. Algebra 18 (1990), no. 1, 253-278. FORMULA Expansion of 7 + (eta(q) / eta(q^5))^6 in powers of q. - Michael Somos, May 22 2013 a(n) = A007252(n) = A106248(n) unless n=0. a(n) = A229793(n) - A078905(n) for n > 0. - Seiichi Manyama, Jan 01 2017 EXAMPLE 1/q + 1 + 9*q + 10*q^2 - 30*q^3 + 6*q^4 - 25*q^5 + 96*q^6 + 60*q^7 - ... MATHEMATICA a[ n_] := SeriesCoefficient[ 7 + 1/q (QPochhammer[ q] / QPochhammer[ q^5])^6, {q, 0, n}] (* Michael Somos, May 22 2013 *) PROG (PARI) q='q+O('q^30); a= 7 + (eta(q)/eta(q^5))^6/q; Vec(a) \\ G. C. Greubel, Jun 02 2018 CROSSREFS Cf. A007252, A106248 (same except for initial terms). Sequence in context: A041170 A041168 A042635 * A007252 A322653 A322465 Adjacent sequences: A045480 A045481 A045482 * A045484 A045485 A045486 KEYWORD sign AUTHOR STATUS approved
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Last modified October 17 22:40 EDT 2019. Contains 328134 sequences. (Running on oeis4.) | 745 | 1,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-43 | latest | en | 0.553363 |
http://www.jiskha.com/display.cgi?id=1320175295 | 1,493,341,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122720.81/warc/CC-MAIN-20170423031202-00295-ip-10-145-167-34.ec2.internal.warc.gz | 560,461,085 | 3,558 | # physics
posted by on .
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls up next to the spy’s boat, both boats reach the edge of a 5.2 m waterfall. The spy’s speed is 15 m/s and the officials’ speed is 25 m/s.
How far apart will the two vessels be when they land below the waterfall? The accelera- tion of gravity is 9.81 m/s2 .
Answer in units of m
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Post a New Question | 139 | 565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.90165 |
http://slideplayer.com/slide/4052310/ | 1,521,950,916,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00527.warc.gz | 271,187,625 | 17,880 | Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting Reactant Reactions don’t always use the correct proportions of reactants.
Presentation on theme: "Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting Reactant Reactions don’t always use the correct proportions of reactants."— Presentation transcript:
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting Reactant Reactions don’t always use the correct proportions of reactants. Many reactions are carried out with an excess amount of one reactant …more than is actually needed. A chemical reaction depends on the reactant that is present in limiting amount - limiting reactant.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Ham sandwich example 10 slices bread + 5 slices ham 5 ham sandwiches 2 slices of bread + 1 slice ham = 1 sandwich
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Ham sandwich 14 slices bread + 5 slices ham excess reactantlimiting reactant 5 ham sandwiches + 4 slices of bread unreacted
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant example 2Sb (s) + 3I 2 (s)2SbI 3 (s) Determine the limiting reactant and theoretical yield when: 1. 1.20 moles Sb and 2.40 moles I 2 are mixed. 2. 1.20g Sb and 2.40g of I 2 are mixed.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Determing limiting reactant 1. Calculate the amount of product that would be formed if the first reactant were completely consumed. 2. Repeat this calculation for the second reactant. 3. Choose the smaller of the two amounts. This is the theoretical yield. The reactant that produces the smaller amount - limiting reactant.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant cont… 2Sb (s) + 3I 2 (s)2SbI 3 (s) 1.20mol Sb x 2mol SbI 3 = 1.20mol SbI 3 2mol Sb 2.40mol I 2 x 2mol SbI 3 = 1.60mol SbI 3 3mol I 2 1.20 mol SbI 3 is the theoretical yield and Sb is the limiting reactant.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant cont… 2Sb (s) + 3I 2 (s)2SbI 3 (s) 1.20g Sb x 1mol Sb x 2mol SbI 3 x 502.5g SbI 3 = 4.95g SbI 3 121.8g Sb 2mol Sb 1mol SbI 3 2.40g I 2 x 1mol I 2 x 2mol SbI 3 x 502.5g SbI 3 = 3.17g SbI 3 253.8g I 2 3mol I 2 1mol SbI 3 3.17g SbI 3 is the theoretical yield and I 2 is the limiting reactant.
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https://cupsix.com/4012/motherboards-inc-manufactures-computer-parts-the-companys-total-revenue-is/ | 1,680,313,444,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00534.warc.gz | 228,794,522 | 9,570 | # Motherboards inc manufactures computer parts the company's total revenue is
Posted By Admin @ September 06, 2022
#### Question:
Motherboards, Inc., manufactures computer parts. The company’s total revenue is the money the company earns after paying all of its production costs. the total wages the company pays workers in its factories and stores. the amount of money the company earns from selling an individual part. the amount the company receives from the sale of all of its computer parts.a. the money the company earns after paying all of its production costs.
b. the total wages the company pays workers in its factories and stores.
c. the amount of money the company earns from selling an individual part.
d. the amount the company receives from the sale of all of its computer parts.
The company’s total revenue is the amount the company receives for the sale of all its computer parts (d)
Explanation:
The money the company earns after paying production costs is known as gross profit, not total revenue (a)
The salary that the company pays to its workers in factories and stores is considered an expense, therefore it cannot be called total revenue (b).
To know the total revenue of a company, all units sold, not just an individual part (c), must be considered, therefore the total revenue is the amount Motherboards Inc., receives for the total sale of its products (d)
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Fractions on the Board
Fractions on the Board
# Comparing Fractions
Unit 1: Crazy About Fractions
Lesson 4 of 10
## Big Idea: Fraction strips and concrete models can be used to help students understand the size of the parts of a fraction
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Environment: Urban | 296 | 1,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-09 | latest | en | 0.858368 |
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9th Class Physics Chapter 7 PROPERTIES OF MATTER MCQs Related To According To Hooke’s Law, The Density Of A Substance Can Be Found With The Help Of, According To Archimedes, Upthrust Is Equal To, What Should Be The Approximate Length Of A Glass Tube To Construct A Water Barometer, SI Unit Of Pressure Is Pascal, Which Is Equal To, Which Of The Substances Is The Lightest One, In Which Of The Following State, Molecules Do Not Leave Their Position:
## Class 9 Physics Chapter 7 PROPERTIES OF MATTER MCQs TEST:
12
Created on By Quizllc
Chapter 7 – Properties of Matter
1 / 7
According to Hooke's law?
2 / 7
The density of a substance can be found with the help of?
3 / 7
According to Archimedes, upthrust is equal to?
4 / 7
What should be the approximate length of a glass tube to construct a water barometer?
5 / 7
SI unit of pressure is Pascal, which is equal to?
6 / 7
Which of the substances is the lightest one?
7 / 7
In which of the following state, molecules do not leave their position?
The average score is 55%
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https://ebrainanswer.com/mathematics/question1876472 | 1,621,374,249,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00301.warc.gz | 232,731,165 | 31,059 | , 08.11.2019 12:31, demigod0701
# What is the definition to absolute value
What is the definition to absolute value
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Mathematics, 21.06.2019 19:30, angie249
Bob is putting books on a shelf and takes a break after filling 18 1/2 inches on the book shelf. the book shelf is now 2/3 full. how long is the entire bookshelf? | 269 | 828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-21 | latest | en | 0.840251 |
http://math.stackexchange.com/questions/250388/maximal-positive-invariant-set-some-fine-print | 1,469,627,472,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826907.66/warc/CC-MAIN-20160723071026-00124-ip-10-185-27-174.ec2.internal.warc.gz | 154,110,011 | 16,828 | Maximal Positive Invariant Set — Some fine print
I would like to share something I noticed on the definition of Maximal Positively Invariant Sets.
Definition 1. For a discrete-time system of the form $x_{k+1}=f(x_{k})$ (and $x_{k}\in \mathbb{X}\subseteq \mathbb{R}^n$ and $\mathbb{X}$ is a closed set), a set $\mathcal{O}\subseteq\mathbb{X}$ is called a Positively Invariant set if $x_{k+1}\in\mathcal{O}$ whenever $x_{k}\in\mathcal{O}$.
Definition 2a. The set $\mathcal{O}_\infty$ is called a Maximal Positively Invariant set is it is a Positively Invariant set and it contains all other positively invariant sets.
Definition 2b. On the other hand, the term maximal appears in partially ordered spaces like $(\mathbb{X},\subseteq)$ in a slightly different context. From that point of view, $\mathcal{O}_\infty$ is a Maximal Positively Invariant set if it is positively invariant and for every positively invariant set $\mathcal{O}_\infty'$ with $\mathcal{O}_\infty'\supseteq \mathcal{O}_\infty$, it is $\mathcal{O}_\infty'=\mathcal{O}_\infty$.
Are the two definitions equivalent?
-
Is there some discrepancy between the two definitions after all? What if one may find two sets $\mathcal{O}_\infty'$ & $\mathcal{O}_\infty$ that are both maximal positively invariant (according to definition 2b this is feasible). Of course these two sets are neither a subset of the other one. It turns out that this is not the case and in fact this (pathological) pair of sets does not exist since unions of positively invariant sets are positively invariant. As a result, $\mathcal{O}_\infty\cup\mathcal{O}_\infty'$ would be a positively invariant set (hence, none of $\mathcal{O}_\infty$, $\mathcal{O}_\infty'$ is maximal). | 493 | 1,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2016-30 | latest | en | 0.759162 |
https://de.mathworks.com/matlabcentral/cody/problems/44457-triangle-of-numbers/solutions/1542164 | 1,603,514,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00712.warc.gz | 295,021,043 | 17,132 | Cody
# Problem 44457. Triangle of numbers
Solution 1542164
Submitted on 28 May 2018 by omer lugasi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('triangle.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
n = 0; mat_correct = []; assert(isequal(triangle(n),mat_correct))
3 Pass
n = 1; mat_correct = 1; assert(isequal(triangle(n),mat_correct))
4 Pass
n = 6; mat_correct = [1 0 0; 2 3 0; 4 5 6]; assert(isequal(triangle(n),mat_correct))
5 Pass
n = 12; mat_correct = [1 0 0 0; 2 3 0 0; 4 5 6 0; 7 8 9 10; 11 12 0 0]; assert(isequal(triangle(n),mat_correct))
6 Pass
n = 50; mat_correct = [1,zeros(1,8); 2:3,zeros(1,7); 4:6,zeros(1,6); 7:10,zeros(1,5); 11:15,zeros(1,4); 16:21,zeros(1,3); 22:28,0,0; ; 29:36,0; 37:45; 46:50,zeros(1,4)]; assert(isequal(triangle(n),mat_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 404 | 1,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.465861 |
https://www.html5gamedevs.com/topic/41090-reversing-an-object-in-babylon-js/ | 1,680,086,228,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00526.warc.gz | 897,584,853 | 20,647 | # Reversing an object in babylon js
## Recommended Posts
I am trying to reverse the direction of a sphere as soon as it touches the ground. I have tried every possibilities to try reverse the direction and I want it to bounce up and down starting from the ground to go upward and then coming back down. I am not understanding how to do it. Below is the code for that part. Please help me.
Thank You
var originalVelocity = -.005;
scene.registerAfterRender(function () {
sphere3.position.y = sphere3.position.y + originalVelocity;
if (sphere3.intersectsMesh(ground, true)) {
sphere3.material.diffuseColor = new BABYLON.Color3(1, 0, 0);
sphere3.position.y = sphere3.position.y - 4*originalVelocity; //this line is not working
}
});
##### Share on other sites
hi @Eisha
welcome to forum
try this
bx.scaling.y = -1;
miss understand
##### Share on other sites
``````var velocity = 0.005;
var dir = -1.0;
scene.registerBeforeRender(function () {
sphere.position.y += velocity * dir
if (sphere.intersectsMesh(ground, false)) {
console.log('intersected');
dir = 1.0;
}
if (sphere.position.y > 3) {
dir = -1.0;
}
});``````
Explanation:
Sphere is moving in the direction of velocity * -1 (negative y direction is downwards).
When the sphere touches the ground, the direction is flipped to the positive direction.
When the sphere reaches a height of y = 3, it returns to the negative direction.
Going Further:
More like gravity: Increase the velocity a little each step when traveling in the negative y direction (like accelerating downwards via gravity). Decrease the velocity while traveling in the positive y direction (like a ball reaching the top of an arc and then falling). No need to flip the direction manually at the top anymore, this fake gravity will have done it already.
More like a rubber ball: if the ball's scale is compressed in the y direction when it hits the ground, it'll look like a bouncy ball. There are other little details depending how bouncy/squishy one would like things to look.
Demo:
##### Share on other sites
10 hours ago, NasimiAsl said:
hi @Eisha
welcome to forum
try this
bx.scaling.y = -1;
miss understand
Hello @NasimiAsl
Thank You ?
##### Share on other sites
9 hours ago, timetocode said:
``````var velocity = 0.005;
var dir = -1.0;
scene.registerBeforeRender(function () {
sphere.position.y += velocity * dir
if (sphere.intersectsMesh(ground, false)) {
console.log('intersected');
dir = 1.0;
}
if (sphere.position.y > 3) {
dir = -1.0;
}
});``````
Explanation:
Sphere is moving in the direction of velocity * -1 (negative y direction is downwards).
When the sphere touches the ground, the direction is flipped to the positive direction.
When the sphere reaches a height of y = 3, it returns to the negative direction.
Going Further:
More like gravity: Increase the velocity a little each step when traveling in the negative y direction (like accelerating downwards via gravity). Decrease the velocity while traveling in the positive y direction (like a ball reaching the top of an arc and then falling). No need to flip the direction manually at the top anymore, this fake gravity will have done it already.
More like a rubber ball: if the ball's scale is compressed in the y direction when it hits the ground, it'll look like a bouncy ball. There are other little details depending how bouncy/squishy one would like things to look.
Demo:
Hello @timetocode
Thank you so much for your help its a great explanation ?
Very well explained
Thank You
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× | 932 | 3,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-14 | longest | en | 0.823704 |
http://mathhelpforum.com/calculus/53196-uniform-continuous.html | 1,526,826,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863570.21/warc/CC-MAIN-20180520131723-20180520151723-00100.warc.gz | 195,785,867 | 9,734 | 1. ## Uniform Continuous?
Prove that $\displaystyle f(x) = \frac {1}{x^2+1}$ is either uniform continuous or not.
Proof:
I claim that this function is uniform continuous, since its derivative is bounded, here is my proof.
Let $\displaystyle \epsilon > 0$ be given, and pick $\displaystyle 0 < \delta < \epsilon$, then for $\displaystyle x,y \in \mathbb {R} , |x-y|< \delta$, we have $\displaystyle | f(x) - f(y) | = | \frac {1}{x^2+1} - \frac {1}{y^2+1} | = | \frac {y^2+1-x^2-1}{(x^2+1)(y^2+1)} |$ $\displaystyle \frac { | x^2 - y^2 | }{|(x^2+1)(y^2+1)|} = \frac {|x-y||x+y|}{|(x^2+1)(y^2+1)|} < \frac { \delta |x+y|}{|(x^2+1)(y^2+1)|} < \delta < \epsilon$
Is this right? Thanks.
2. Hi,
this is right, but I think you should explain the inequality $\displaystyle \frac { \delta |x+y|}{|(x^2+1)(y^2+1)|} < \delta$, using like $\displaystyle \frac { |x+y|}{|(x^2+1)(y^2+1)|} \leq \frac{ |x|+|y|}{|x^2+1||y^2+1|}\leq \frac{ \frac{1}{2}(x^2+1)+\frac{1}{2}(y^2+1)}{|x^2+1||y^2 +1|}$ $\displaystyle =\frac{1}{2}\frac{1}{y^2+1}+\frac{1}{2}\frac{1}{x^ 2+1}\leq \frac{1}{2}+\frac{1}{2}=1$ (where I used $\displaystyle |a|\leq \frac{a^2+1}{2}$, which results from $\displaystyle a^2+1-2a=(a-1)^2\geq 0$). There may be a quicker proof though, that I didn't notice. | 542 | 1,260 | {"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} | 4.21875 | 4 | CC-MAIN-2018-22 | latest | en | 0.652872 |
https://www.teachoo.com/1656/518/Ex-6.2--4---In-figure--DE----AC-and-DF----AE.-Prove-that/category/Ex-6.2/ | 1,723,364,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00301.warc.gz | 782,393,004 | 21,509 | Ex 6.2
Chapter 6 Class 10 Triangles
Serial order wise
### Transcript
Ex 6.2, 4 In figure, DE || AC and DF || AE. Prove that, prove that π΅πΉ/πΉπΈ = π΅πΈ/πΈπΆ Given: DE II AC and DF II AE To prove: π΅πΉ/πΉπΈ=π΅πΈ/πΈπΆ Proof: From (1) and (2) π΅πΉ/πΉπΈ=π΅πΈ/πΈπΆ Hence proved | 189 | 324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.627701 |
http://www.amosweb.com/cgi-bin/awb_nav.pl?s=wpd&c=dsp&k=elasticity%20and%20supply%20intercept | 1,718,772,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00365.warc.gz | 40,873,796 | 8,987 | Tuesday June 18, 2024
AmosWEB means Economics with a Touch of Whimsy!
ELASTICITY AND SUPPLY INTERCEPT:
The intersection of a straight-line supply curve with vertical price axis and/or horizontal quantity axis reveals the relative price elasticity of supply. Intersection with the horizontal quantity axis means inelastic and intersection with the vertical price axis means elastic. Intersection with the origin means unit elastic supply.
The positioning of a supply curve relative to the price and quantity axes indicates the price elasticity of supply. Unlike demand, a straight-line supply curve does not contain all five elasticity alternatives. However, the elastic or inelastic nature of a supply curve is revealed by the intersection with the vertical price axis and/or horizontal quantity axis.
### Five Elasticity Alternatives
Five Supply Curves
The elasticity of a supply curve depends on whether the supply curve intersects the price axis, the quantity axis, or the origin. The exhibit to the right can be used to illustrate the five elasticity alternatives.
• Perfectly Elastic: First up is perfectly elastic supply, which means an infinitesimally small change in price results in an infinitely large change in quantity supplied. The coefficient of elasticity for this alternative is E = ∞. A click of the [Perfectly E] button reveals a perfectly elastic supply curve, which is horizontal and intersects the vertical price axis.
• Relatively Elastic: Next up is relatively elastic supply, which means a relatively small change in price results in a relatively large change in quantity supplied. The coefficient of elasticity for this alternative is in the range of 1 < E < ∞. A click of the [Relatively E] button reveals a relatively elastic supply curve, which is flat but not horizontal and intersects the price axis.
• Unit Elastic: Third on the list is unit elastic supply, which means that any change in price is matched by an equal relative change in quantity. The coefficient of elasticity for this alternative is in the range of E = 1. A click of the [Unit Elastic] button reveals a unit elastic supply curve, which is a straight line extending from the origin.
• Relatively Inelastic: Next is relatively inelastic supply, which means a relatively large change in price is needed to induce a relatively small change in quantity supplied. The coefficient of elasticity for this alternative is in the range of 0 < E < 1. A click of the [Relatively In] button reveals a relatively inelastic supply curve, which is steep but not vertical and intersects the quantity axis.
• Perfectly Inelastic: Lastly is perfectly inelastic supply, means that quantity supplied is unaffected by any change in price. The coefficient of elasticity for this alternative is E = 0. A click of the [Perfectly In] button reveals a perfectly inelastic supply curve, which is vertical and intersects the horizontal quantity axis.
An explanation of this connection between elasticity and intercept can be had with a closer look at the elasticity concept. Elasticity is comparison between the percentage changes in price and quantity. A percentage change depends on both amount of the change, or the unit change, and the starting point, or base value, of the change. Because slope is constant for a straight-line supply curve, unit changes are always proportional for any of the supply curves displayed here.
The difference in elasticity rests with the base values. For a supply curve that intersects the vertical price axis, price starts with a larger base value than quantity. As such, any subsequent price change is relatively smaller, compared to this larger base, than the quantity change. This results in a relatively elastic supply.
For a supply curve that intersects the horizontal quantity axis, quantity starts with a larger base value than the price. As such, any subsequent quantity change is relatively smaller, compared to this larger base, than the price change. This results in a relatively inelastic supply.
For a supply curve that goes through the origin, price and quantity start with equal zero values, and equal bases. As such, all subsequent price and quantity changes are proportional. This results in a unit elastic supply.
<= ELASTICITY AND DEMAND SLOPE ELASTICITY DETERMINANTS =>
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Thanks for visiting AmosWEB | 1,079 | 5,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.910807 |
http://stackoverflow.com/questions/5681194/orbital-animation | 1,436,285,783,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375099755.63/warc/CC-MAIN-20150627031819-00190-ip-10-179-60-89.ec2.internal.warc.gz | 246,230,884 | 18,177 | # Orbital animation
first of all I'll try to explain what I'm trying to do. The task looks quite simple at first look, but it took me some time to realize its very complex. What I'm trying to do is simple animation with easing at the beginning and at the end - I know how to use easing, the hard part is that I'm trying to do something like Orbit - Elliptical orbit with lets say 5 rectangles attached to it. I want to move the rectangles along the Elliptical orbit (elliptical path) without changing the rotation angle of each rectangle.
I've tried with Path animation, but it seems that motion path animations don't support easing ? Am I wrong ? Second solution was to group the path and the rectangles and rotate the whole group, but this changes rectangles rotation angle too. Is there a simple way to do this ? Please point me to an article or something, or if you have a similar scenario please share the solution. Thanks.
-
Here's another approach...
What you want to do is not actually a rotation, but a translation along an elliptic path. The trouble is, a `TranslateTransform` is defined by an X and Y, not an angle and radius... But it's easier to animate an angle, so you have to convert polar coordinates to cartesian coordinates.
To do that, let's define two converters: `SinConverter` and `CosConverter`:
``````public class SinConverter : IValueConverter
{
#region Implementation of IValueConverter
public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
{
try
{
double angle = System.Convert.ToDouble(value);
double angleRad = Math.PI * angle / 180;
}
catch
{
return Binding.DoNothing;
}
}
public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
{
throw new NotImplementedException();
}
#endregion
}
public class CosConverter : IValueConverter
{
#region Implementation of IValueConverter
public object Convert(object value, Type targetType, object parameter, CultureInfo culture)
{
try
{
double angle = System.Convert.ToDouble(value);
double angleRad = Math.PI * angle / 180;
}
catch
{
return Binding.DoNothing;
}
}
public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
{
throw new NotImplementedException();
}
#endregion
}
``````
Now, we need an angle property to animate: so we define a dummy `RotateTransform` in the resources, that will be the target of the animation.
Next, we apply a `TranslateTransform` on the "satellite", and we bind the X and Y to the angle, using our converters.
Eventually, we just need to create the animation itself that will animate the angle.
Here's the complete XAML:
``````<Window x:Class="WpfCS.Orbit"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:my="clr-namespace:WpfCS"
Title="Orbit"
Height="300" Width="300">
<Window.Resources>
<my:SinConverter x:Key="sinConverter" />
<my:CosConverter x:Key="cosConverter" />
<RotateTransform x:Key="rotate" Angle="0" />
</Window.Resources>
<Grid>
<Rectangle Width="30" Height="30" Fill="Blue">
<Rectangle.RenderTransform>
<TranslateTransform X="{Binding Path=Angle,
Source={StaticResource rotate},
Converter={StaticResource cosConverter},
ConverterParameter=100}"
Y="{Binding Path=Angle,
Source={StaticResource rotate},
Converter={StaticResource sinConverter},
ConverterParameter=60}"/>
</Rectangle.RenderTransform>
</Rectangle>
<Ellipse Width="5" Height="5" Fill="White" Stroke="Black" StrokeThickness="1" />
</Grid>
<Window.Style>
<Style TargetType="Window">
<Style.Triggers>
<EventTrigger.Actions>
<BeginStoryboard>
<Storyboard>
<DoubleAnimation Storyboard.Target="{StaticResource rotate}"
Storyboard.TargetProperty="Angle"
From="0" To="360" Duration="0:0:5"
RepeatBehavior="Forever" />
</Storyboard>
</BeginStoryboard>
</EventTrigger.Actions>
</EventTrigger>
</Style.Triggers>
</Style>
</Window.Style>
</Window>
``````
-
I thought about doing it via X & Y animations too, nice job getting it done this way. – H.B. Apr 15 '11 at 20:38
Thanks Thomas, that's the answer I've been looking for. – zlat Apr 18 '11 at 12:00
Try this, it's great fun:
``````<Canvas Height="100" Width="100" RenderTransformOrigin="0.5,0.5">
<Canvas.Triggers>
<BeginStoryboard>
<Storyboard>
<DoubleAnimation From="0" To="360"
RepeatBehavior="Forever"
Duration="0:0:1"
Storyboard.TargetProperty="RenderTransform.Angle">
<DoubleAnimation.EasingFunction>
<CubicEase EasingMode="EaseInOut"/>
</DoubleAnimation.EasingFunction>
</DoubleAnimation>
</Storyboard>
</BeginStoryboard>
</EventTrigger>
</Canvas.Triggers>
<Canvas.RenderTransform>
<RotateTransform />
</Canvas.RenderTransform>
<Canvas.Resources>
<Style TargetType="{x:Type Rectangle}">
<Setter Property="RenderTransformOrigin" Value="0.5,0.5"/>
<Setter Property="RenderTransform" Value="{Binding RenderTransform.Inverse, RelativeSource={RelativeSource AncestorType=Canvas}}"/>
</Style>
</Canvas.Resources>
<Rectangle Fill="Red" Height="20" Width="20" Canvas.Left="40" Canvas.Top="0"/>
<Rectangle Fill="Green" Height="20" Width="20" Canvas.Left="40" Canvas.Top="80"/>
</Canvas>
``````
Key Points:
1. `RotateTransform` as `RenderTransform` on the Canvas which is animated.
2. Implicit style via `Canvas.Resources` binds the `RenderTransform` of Rectangles to the inverse RotateTransform of the parent canvas.
3. ???
4. Profit!
-
Very clever, I never noticed the Transform class had an Inverse property... It's simpler than my approach, since you don't need any code – Thomas Levesque Apr 15 '11 at 20:29 | 1,320 | 5,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.847707 |
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# Determining Gravity with a Pendulum
Extracts from this document...
Introduction
SPH141 Practical 2
Practical Experiment 2
Determining Gravity with a Pendulum
Aim
To determine the local acceleration due to gravity using Galileo pendulum technique.
Theory
Gravity is a force that acts on Earth every day. Sir Isaac Newton was first to underline the principles of gravity when an apple fell on his head (Ashbacher 2002). He stated that each particle with a mass attracts all other particles with mass with a gravitational force that is directly proportional to the product of their masses and inversely proportional to their distance of separation squared (Ashbacher 2002).
This is due to that gravity acts between objects (Ashbacher 2002), consequently causing a force of attraction which pulls the two object together, such as that an object with a mass will fall down towards earth ground. The Earth’s mass creates a gravitational force, which pulls the object down towards Earth.
This theory is also supported by Newton’s three law of motions, particularly the first law stating that, ‘an object in motion or at rest will remain in motion or at rest unless acted upon by an external fore‘. An object will remain at rest floating in the air, however since an external force, gravity, acts upon it, the object falls towards Earth.
Theoretically, the acceleration due to gravity on Earth is 9.8ms-2
Middle
Average
0.30
10.9
11.3
10.2
10.8
0.60
15.8
15.7
15.7
15.7
0.90
19.1
19.0
18.9
19.0
Resolution Ruler – 0.1cm Stop Watch – 0.01s
Calculations
Calculating the gravitational acceleration
T = 2π
T = 2π
g =
Calculating Gravitational Acceleration for 0.30m
10.8s per 10 pendulum swing cycle = 1.08s per pendulum swing cycle
L = 0.30m and T = 1.08s
g =
g = 10.2ms-2
Calculating Gravitational Acceleration for 0.60m
15.7s per 10 pendulum swing cycle = 1.57s per pendulum swing cycle
L = 0.60m and T = 1.08s
g =
g = 9.6ms-2
Calculating Gravitational Acceleration for 0.90m
19.0s per 10 pendulum swing cycle = 1.90s per pendulum swing cycle
L = 0.90m and T = 1.90s
g =
g = 9.8ms-2
Calculating Uncertainties for the gravitational acceleration
0.30m Pendulum
Since T = 10.8 and L = 0.30, the uncertainty for T = 10.8s ± 0.05s and L = 0.30m ± 0.05m
Highest value for the gravitation acceleration using 0.30m pendulum is;
L = 0.30m + 0.05m
= 0.35m
T = 10.8s – 0.05
=10.75s per 10 cycles
g =
where L = 0.35 and T = 1.075s per cycle
g =
g = 11.9ms-2
Lowest value for the gravitation acceleration using 0.30m pendulum is;
L = 0.30m - 0.05m
= 0.25m
T = 10.8s + 0.05
=10.85s per 10 cycles
g =
where L = 0.25 and T = 1.085s per cycle
g =
g = 8.4ms-2
0.60m Pendulum
Since T = 15.7 and L = 0.60, the uncertainty for T = 15.7s ± 0.05s and L = 0.6m ± 0.05m
Highest value for the gravitation acceleration using 0.60m pendulum is;
L = 0.60m + 0.05m
= 0.65m
T = 15.7s – 0.05
=15.65s per 10 cycles
g =
where L = 0.65 and T = 1.565s per cycle
g =
g = 10.5ms-2
Lowest value for the gravitation acceleration using 0.
Conclusion
Conclusion
The acceleration due to gravitation was determined to be 10.2ms-2, 9.6ms-2 and 9.8ms-2 for the pendulum measurements of 0.30m, 0.60m and 0.90m. This shows that the aim f the experiment was achieved through the conduction of the experiment. Though, the theoretical acceleration due to gravitation on Earth is determined to be 9.8ms-2, in which it was found that by using the 0.90m, the exact value could be calculated. However there were some errors involved such as the parallax error, but within all trials, the acceleration due to gravity of each individual was within the highest and lowest uncertainty range. An improvement was suggested in regards to the errors and that was to use a longer pendulum to reduce the pendulum cycle time. Overall the experiment was followed according to the method, and the result obtained had a percentage error less than 10%, hence the results are considered acceptable.
References
Ashbacher, C 2002, ‘Sir Isaac Newton: The Gravity of Genius’, Mathematics & Computer Education, vol. 36, no. 3, pp. 302-310, viewed 5 September, via Education Research Complete
Houston, K 2012, ‘The Simple Pendulum’, College Physics, vol. 1, no.1, pp.1-4, viewed 5 September, <http://cnx.org/content/m42243/latest/?collection=col11406/latest>
Appendix
Diagram 1.1
Experiment Set Up
This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.
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# Related GCSE Forces and Motion essays
1. ## The Simple Pendulum Experiment
4 star(s)
length, then I can find the acceleration of gravity by taking the gradient of this line to be m in the above equation () L (cm) T2 (s2) (3 s.f.) 80 3.53 90 3.90 100 4.33 110 4.78 120 5.06 130 5.24 140 5.62 150 6.00 T2 = time for one oscillation2 (s2)
2. ## length of a simple pendulum affects the time
4 star(s)
and the length is adjusted by pulling the cotton through the two blocks. Gravity may be considered to act through the centre of gravity of the bob. For this reason the length of the cotton is measured from the wooden blocks to the centre of the bob.
1. ## Determination of the acceleration due to gravity using a simple pendulum.
could be affected by human error in recording or calculation of the results. Also drawing a T2 graph illustrates that the graph is exponential as it produces a straight line. This also shows that length is directly proportional to T2.
2. ## Determining the acceleration due to gravity by using simple pendulum.
of pendulum. Divide this time by 20 to get a value for the average periodic time (T) of the motion. By using these averaging techniques it minimizes random errors.
1. ## Investigating the period of a simple pendulum and measuring acceleration due to gravity.
Time for 20 oscillations(s) Mean time(s) Time� (s�) 0.10 13.37 13.13 13.25 0.44 0.30 23.09 23.02 23.06 1.33 0.50 28.38 28.30 28.34 2.01 0.70 33.38 33.94 33.66 2.83 0.90 37.78 37.69 37.74 3.56 1.10 41.53 41.56 41.55 4.32 1.30 44.78 44.44 44.61 4.98 1.50 48.26 48.59 48.43 5.86 Gradient of the graph = ?y = ?T� ?x ?L = 5.20 1.30 = 4 ms��.
2. ## Period of Oscillation of a Simple Pendulum
To check that this is correct, I shall apply it to an answer, which I already know. I shall work out the time for a 60 cm pendulum as follows: - L ? t� L = kt� 0.6 = 0.253 * t� 0.6 ??????????t� t� = 2.372 seconds t =
1. ## Measuring Acceleration due to Gravity using a simple Pendulum.
References Advanced physics by M. Nelkon & M. Dethereidge "S.H.M. is acceleration is directly proportional to displacement and directed (acts towards) a fixed point." Evaluation From my graph I have realised that I only have 2 anomalous results, these results are at the points: length = 0.900 and 1.000 These anomalous results have occurred because of the errors.
2. ## The determination of the acceleration due to gravity at the surface of the earth, ...
The gradient of this graph is 4?� / g. Therefore the value of g can be calculated by 4?� / gradient. This equation doesn't involve mass, amplitude of displacement. It doesn't take into account air resistance as this is negligible.
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# GMAT CR 700 level question bank
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GMAT CR 700 level question bank [#permalink] 09 Aug 2012, 08:47
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Hi ,
I have collected a set of 60 questions from the forum and prepared a doc out of it. The answers are put as hyperlink , which leads to the relevant thread on gmatclub. Thought it would be useful for guys who want to get practice on some tough CR questions.
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CR Question Bank 700 level.doc [131 KiB]
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Re: GMAT CR 700 level question bank [#permalink] 10 Aug 2012, 03:11
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Posting the updated document with OA , will try to put in OE , meanwhile go through the hyperlinks
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Re: GMAT CR 700 level question bank [#permalink] 09 Aug 2012, 12:52
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Expert's post
At least OA, better with OE
+1 kudos
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Re: GMAT CR 700 level question bank [#permalink] 09 Aug 2012, 22:07
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Interesting collection. Pls provide the OA for all 60.
My answers below for first 10 -
1. D
2. E
3.D
4.E
5.C
6.C
7.A
8.D
9.C
10.B
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Re: GMAT CR 700 level question bank [#permalink] 09 Aug 2012, 09:47
enlist the OA and OE as well.
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Re: GMAT CR 700 level question bank [#permalink] 10 Aug 2012, 10:19
+1 kudos
thanks for sharing.
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Re: GMAT CR 700 level question bank [#permalink] 15 Aug 2012, 02:31
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Re: GMAT CR 700 level question bank [#permalink] 23 Sep 2012, 20:30
crackHSW wrote:
Posting the updated document with OA , will try to put in OE , meanwhile go through the hyperlinks
Hello,
Any luck with the OE?
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Re: GMAT CR 700 level question bank [#permalink] 20 Sep 2013, 06:22
What are the sources?
Thanks!
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Re: GMAT CR 700 level question bank [#permalink] 20 Sep 2013, 06:22
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### Eureka Math Grade 7 Module 3 Lesson 11 Example Answer Key
Example 1.
The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.
The angles 86°, 68°, and the angle between them, which is vertically opposite and equal in measure to x, are angles on a line and have a sum of 180°.
86 + x + 68 = 180
x + 154 = 180
x + 154 – 154 = 180 – 154
x = 26
Example 2.
In a complete sentence, describe the angle relationships in the diagram. You may label the diagram to help describe the angle relationships. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.
The angle formed by adjacent angles a° and b° is vertically opposite to the 77° angle. The angles x°, a°, and b° are adjacent angles that have a sum of 90° (since the adjacent angle is a right angle and together the angles are on a line).
x + 77 = 90
x + 77 – 77 = 90 – 77
x = 13
Example 3.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measures of ∠JAH and ∠GAF. Confirm your answers by measuring the angle with a protractor.
The sum of the degree measurements of ∠JAH, ∠GAH, ∠GAF, and the arc that subtends ∠JAF is 360°.
225 + 2x + 90 + 3x = 360
315 + 5x = 360
315 – 315 + 5x = 360 – 315
5x = 45
($$\frac{1}{5}$$)5x = ($$\frac{1}{5}$$)45
x = 9
m∠JAH = 2(9°) = 18° m∠GAF = 3(9°) = 27°
Example 4.
In the accompanying diagram, the measure of ∠DBE is four times the measure of ∠FBG.
a. Label ∠DBE as y° and ∠FBG as x°. Write an equation that describes the relationship between ∠DBE and ∠FBG.
y = 4x
b. Find the value of x.
50 + x + 4x = 180
50 + 5x = 180
5x + 50 – 50 = 180 – 50
5x = 130
($$\frac{1}{5}$$)(5x) = ($$\frac{1}{5}$$)(130)
x = 26
c. Find the measures of ∠FBG, ∠CBD, ∠ABF, ∠GBE, and ∠DBE.
m∠FBG = 26°
m∠CBD = 26°
m∠ABF = 4(26°) = 104°
m∠GBE = 50°
m∠DBE = 104°
d. What is the measure of ∠ABG? Identify the angle relationship used to get your answer.
∠ABG = ∠ABF + ∠FBG
∠ABG = 104 + 26
∠ABG = 130
m∠ABG = 130°
To determine the measure of ∠ABG, you need to add the measures of adjacent angles ∠ABF and ∠FBG.
### Eureka Math Grade 7 Module 3 Lesson 11 Exercise Answer Key
Opening Exercise
a. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.
The angles marked by x°, 90°, and 14° are angles on a line and have a sum of 180°.
x + 90 + 14 = 180
x + 104 = 180
x + 104 – 104 = 180 – 104
x = 76
b. $$\overleftrightarrow{C D}$$ and $$\overleftrightarrow{E F}$$ are intersecting lines. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for y. Confirm your answers by measuring the angle with a protractor.
The adjacent angles marked by y° and 51° together form the angle that is vertically opposite and equal to the angle measuring 147°.
y + 51 = 147
y + 51 – 51 = 147 – 51
y = 96
c. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for b. Confirm your answers by measuring the angle with a protractor.
The adjacent angles marked by 59°, 41°, b°, 65°, and 90° are angles at a point and together have a sum of 360°.
59 + 41 + b + 65 + 90 = 360
b + 255 = 360 – 255
b = 105
d. The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for z. Confirm your answers by measuring the angle with a protractor.
The angles marked by z°, 158°, and z° are angles on a line and have a sum of 180°.
z + 158 + z = 180
2z + 158 = 180
2z + 158 – 158 = 180 – 158
2z = 22
z = 11
e. Write an equation for the angle relationship shown in the figure and solve for x. In a complete sentence, describe the angle relationship in the diagram. Find the measurements of ∠EPB and ∠CPA. Confirm your answers by measuring the angle with a protractor.
∠CPA, ∠CPE, and ∠EPB are angles on a line and their measures have a sum of 180°.
5x + 90 + x = 180
6x + 90 = 180
6x + 90 – 90 = 180 – 90
6x = 90
($$\frac{1}{6}$$)6x = ($$\frac{1}{6}$$)90
x = 15
∠EPB = 15°
∠CPA = 5(15°) = 75°
Exercise 1.
The following figure shows four lines intersecting at a point. In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angle with a protractor.
The angles x°, 25°, y°, and 40° are angles on a line and have a sum of 180°; the angle marked y° is vertically opposite and equal to 96°.
y = 96, vert. ∠s
x + 25 + (96) + 40 = 180
x + 161 = 180
x + 161 – 161 = 180 – 161
x = 19
Exercise 2.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angles with a protractor.
The measures of angles x and y have a sum of 90°; the measures of angles x and 27 have a sum of 90°.
x + 27 = 90
x + 27 – 27 = 90 – 27
x = 63
(63) + y = 90
63 – 63 + y = 90 – 63
y = 27
Exercise 3.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measure of ∠JKG. Confirm your answer by measuring the angle with a protractor.
The sum of the degree measurements of ∠LKJ, ∠JKG, ∠GKM, and the arc that subtends ∠LKM is 360°.
5x + 24 + x + 90 = 360
6x + 114 = 360
6x + 114 – 114 = 360 – 114
6x = 246
($$\frac{1}{6}$$)6x = ($$\frac{1}{6}$$)246
x = 41
m∠JKG = 41°
### Eureka Math Grade 7 Module 3 Lesson 11 Problem Set Answer Key
In a complete sentence, describe the angle relationships in each diagram. Write an equation for the angle relationship(s) shown in the figure, and solve for the indicated unknown angle. You can check your answers by measuring each angle with a protractor.
Question 1.
Find the measures of ∠EAF, ∠DAE, and ∠CAD.
∠GAF, ∠EAF, ∠DAE, and ∠CAD are angles on a line and their measures have a sum of 180°.
6x + 4x + 2x + 30 = 180
12x + 30 = 180
12x + 30 – 30 = 180 – 30
12x = 150
x = 12.5
m∠EAF = 2(12.5°) = 25°
m∠DAE = 4(12.5°) = 50°
Question 2.
Find the measure of a.
Angles a°, 26°, a°, and 126° are angles at a point and have a sum of 360°.
a + 126 + a + 26 = 360
2a + 152 = 360
2a + 152 – 152 = 360 – 152
2a = 208
($$\frac{1}{2}$$)2a = ($$\frac{1}{2}$$)208
a = 104
Question 3.
Find the measures of x and y.
Angles y° and 65° and angles 25° and x° have a sum of 90°.
x + 25 = 90
x + 25 – 25 = 90 – 25
x = 65
65 + y = 90
65 + y = 90
65 – 65 + y = 90 – 65
y = 25
Question 4.
Find the measure of ∠HAJ.
Adjacent angles x° and 15° together are vertically opposite from and are equal to angle 81°.
x + 15 = 81
x + 15 – 15 = 81 – 15
x = 66
m∠HAJ = 66°
Question 5.
Find the measures of ∠HAB and ∠CAB.
The measures of adjacent angles ∠CAB and ∠HAB have a sum of the measure of ∠CAH, which is vertically opposite from and equal to the measurement of ∠DAE.
2x + 3x + 70 = 180
5x = 110
($$\frac{1}{5}$$)5x = ($$\frac{1}{5}$$)110
x = 22
m∠HAB = 3(22°) = 66°
m∠CAB = 2(22°) = 44°
Question 6.
The measure of ∠SPT is b°. The measure of ∠TPR is five more than two times ∠SPT. The measure of ∠QPS is twelve less than eight times the measure of ∠SPT. Find the measures of ∠SPT, ∠TPR, and ∠QPS.
∠QPS, ∠SPT, and ∠TPR are angles on a line and their measures have a sum of 180°.
(8b – 12) + b + (2b + 5) = 180
11b – 7 = 180
11b – 7 + 7 = 180 + 7
11b = 187
($$\frac{1}{11}$$)11b = ($$\frac{1}{11}$$)187
b = 17
m∠SPT = (17°) = 17°
m∠TPR = 2(17°) + 5° = 39°
m∠QPS = 8(17°) – 12° = 124°
Question 7.
Find the measures of ∠HQE and ∠AQG.
∠AQG, ∠AQH, and ∠HQE are adjacent angles whose measures have a sum of 90°.
2y + 21 + y = 90
3y + 21 = 90
3y + 21 – 21 = 90 – 21
3y = 69
($$\frac{1}{3}$$)3y = ($$\frac{1}{3}$$)69
y = 23
m∠HQE = 2(23°) = 46°
m∠AQG = (23°) = 23°
Question 8.
The measures of three angles at a point are in the ratio of 2:3:5. Find the measures of the angles.
∠A = 2x, ∠B = 3x, ∠C = 5x
2x + 3x + 5x = 360
10x = 360
($$\frac{1}{10}$$)10x = ($$\frac{1}{10}$$)360
x = 36
∠A = 2(36°) = 72°
∠B = 3(36°) = 108°
∠C = 5(36°) = 180°
Question 9.
The sum of the measures of two adjacent angles is 72°. The ratio of the smaller angle to the larger angle is 1∶3. Find the measures of each angle.
∠A = x, ∠B = 3x
x + 3x = 72
4x = 72
($$\frac{1}{4}$$)(4x) = ($$\frac{1}{4}$$)(72)
x = 18
∠A = (18°) = 18°
∠B = 3(18°) = 54°
Question 10.
Find the measures of ∠CQA and ∠EQB.
4x + 5x = 108
9x = 108
($$\frac{1}{9}$$)9x = ($$\frac{1}{9}$$)108
x = 12
m∠CQA = 5(12°) = 60°
m∠EQB = 4(12°) = 48°
### Eureka Math Grade 7 Module 3 Lesson 11 Exit Ticket Answer Key
Question 1.
Write an equation for the angle relationship shown in the figure and solve for x. Find the measures of ∠RQS and ∠TQU.
($$\frac{1}{7}$$)7x = ($$\frac{1}{7}$$)49 | 3,445 | 9,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.8125 | 5 | CC-MAIN-2024-26 | latest | en | 0.890349 |
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Question description
Resources: Appendix A1 at the end of Basic Statistics for Business and Economics
Answer question 68 in the Data Set Exercises from Ch. 3 of Basic Statistics for Business and Economics.
Answer question 34 in the Data Set Exercises from Ch. 4 of Basic Statistics for Business and Economics.
68. Refer to the Real Estate data, which reports information on homes sold in the Phoenix,
Arizona, area last year.
a. Select the variable selling price.
1. Find the mean, median, and the standard deviation.
2. Write a brief summary of the distribution of selling prices.
b. Select the variable referring to the area of the home in square feet.
1. Find the mean, median, and the standard deviation.
2. Write a brief summary of the distribution of the area of homes.
Refer to the Real Estate data, which report information on homes sold in the Phoenix,
Arizona, area last year. Select the variable selling price.
a. Develop a box plot. Estimate the first and the third quartiles. Are there any outliers?
b. Develop a scatter diagram with price on the vertical axis and the size of the home on
the horizontal. Does there seem to be a relationship between these variables? Is the
relationship direct or inverse?
(Top Tutor) Daniel C.
(997)
School: Boston College
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Here’s a simple solution to the classic “For a certain race, 3 teams were allowed to enter 3 members each” GMAT question. First, let’s look at the question text so you know what we’re talking about: For a certain race, 3 teams were allowed to enter 3 members each. A team earned 6 – n … Read More
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I need to evaluate the hypergeometric function $$_2F_1$$ with $$|z| > 1$$ as in Wolfram Language with GSL but the GSL documentation says the $$_2F_1$$ needs $$|z| < 1$$.
Is there any way I can use GSL and $$_2F_1$$ with $$|z| > 1$$? Any reference about the subject?
• Have a look at DLMF 15.8.2, for example. Nov 27, 2021 at 23:35
The reason GSL doesn't do this is that you need an analytic continuation. That's what Mathematica does. For example, see here and here. A very similar question was asked here, but they propose a roll-your-own solution.
Edit: Useful paper reference here per my link above (thanks for pulling it out @JoseM)
Please, consider this as a toy or educational code... If you want to do it yourself consider the use of a variable precision library like GMP.
In my case z > 2, so reading the paper on the forum suggested by @spencer-bryngelson, Computation of Hypergeometric Functions you get Eq.4.20 (use table 13) or you can use DLMF 15.8.2 as suggested by @a-rural-reader (for me this's more clear because you can see a discontinuity if $$a - b \in \mathbb{Z}$$ see Hypergeometric2F1 too)
#include <stdio.h>
#include <stdlib.h>
#include <gsl_sf.h>
#include <gsl_math.h>
double _2F1(double a,
double b,
double c,
double z)
{
/* a - b !€ Z */
double bd = b - 1e-15;
double A = (gsl_sf_gamma(bd - a)*gsl_sf_gamma(c)*pow(-z, -a))/(gsl_sf_gamma(b)*gsl_sf_gamma(c - a));
double Fa = gsl_sf_hyperg_2F1(a, a - c + 1.0, a - bd + 1.0, 1.0/z);
double B = (gsl_sf_gamma(a - bd)*gsl_sf_gamma(c)*pow(-z, -b))/(gsl_sf_gamma(a)*gsl_sf_gamma(c - bd));
double Fb = gsl_sf_hyperg_2F1(bd, bd - c + 1.0, bd - a + 1.0, 1.0/z);
return A*Fa + B*Fb;
}
int main()
{
/* Mathematica
Hypergeometric2F1[2., 3., 4., 5.0]
= 0.156542 + 0.150796i
*/
double a = 2.0;
double b = 3.0;
double c = 4.0;
double z = 5.0;
double result = _2F1(a, b, c, z);
printf("2F1(%.1f, %.1f, %.1f, %.1f): %f\n", a, b, c, z, result);
/* result = 0.15625*/
/*
Hypergeometric2F1[2., 3., 4., 10.0]
Out[30]= 0.03985 + 0.0188496 I
*/
z = 10.0;
result = _2F1(a, b, c, z);
printf("2F1(%.1f, %.1f, %.1f, %.1f): %f\n", a, b, c, z, result);
/* result = 0.0341796875*/
return 0;
}
As you can see the results are far from accurate.
If you don't want to do it yourself, I think Arb seems a good choice.
– user20857
Dec 1, 2021 at 19:17
You can use a transformation formula. Here is how I do in R:
library(gsl)
Gauss2F1 <- function(a,b,c,x){
if(x>=0 & x<1){
hyperg_2F1(a,b,c,x)
}else{
hyperg_2F1(c-a,b,c,1-1/(1-x))/(1-x)^b
}
} | 944 | 2,562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | longest | en | 0.723145 |
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Cumulative probability function is
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MIS can be used in controlling
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Top Drawer Teachers: Multipliction by zero
1. 1. Multiplying by zeroBeing able to generalise that‘multiplying any number by zerogives a product of zero’ helpsstudents apply this knowledge tosituations such as 3 × 6 × 0.Multiplying by zero
2. 2. 2 packets of 3 chocolates2 × 3 = 6Multiplying by zero
3. 3. 1 packet of 3 chocolates1 × 3 = 3Multiplying by zero
4. 4. No packets of 3 chocolatesJust 3 empty chocolate wrappers!0 × 3 = 0Multiplying by zero
5. 5. Real life storyI had 3 bags and I put zero lollies in each bag.How many lollies did I have?Zero lollies!3 × 0 = 0Multiplying by zero
6. 6. Real life storyI had 3 bags and I put zero lollies in each bag.How many lollies did I have?Zero lollies!3 × 0 = 0Multiplying by zero | 456 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-45 | latest | en | 0.537729 |
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Title: Today we will compute simple division of fractions.
1
Today we will compute simple division of
fractions.
Compute calculate or work out But First lets
2
Remember that when multiplying fractions
• When multiplying fractions, they do NOT need to
have a common denominator.
• To multiply two (or more) fractions, multiply
across, numerator by numerator and denominator by
denominator.
• If the answer can be simplified, then simplify
it.
• Example
• Example
3
Equivalent Fractions
• Name the same amount but have different
numerators and denominators.
1
2
1
1
4
2
4
2
1
4
4
Equivalent Fractions
• Are sometimes called equal fractions two
or more fractions that name the same number.
1
2
1
1
4
2
4
2
1
4
5
Equivalent Fraction Models
1
2
3
2
4
6
6
Simplifying
Reduce
10 Greatest Common Factor
Write the numerator and denominator as a product
of factors, then cancel common factors and obtain
the result.
7
Simplifying
Write the numerator and denominator as a product
of common factors then cancel the common factors
and obtain the result.
Repeat the simplification process until all
common factors are removed.
8
Converting Mixed Numbers
1
1
5
5
3
3
3
3
16
3
Multiply the whole number by the denominator
Add the numerator to the result
Place over the initial denominator
Compute the final result
9
Converting Mixed Numbers
2
2
6
6
9
9
9
9
56
9
Multiply the whole number by the denominator
Add the numerator to the result
Place over the initial denominator
Compute the final result
10
Lets think about what dividing fractions means.
In this expression it means How many one
eighths are in three fourths?
For example, how many one eighth slices of pizza
are in three fourths of a pizza?
11
How many one eighths are in three fourths?
To find this we must first find 3/4 of the pizza.
We then cut each fourth into halves to make
eighths.
We can see there are 6 eighths in three fourths.
12
EXAMPLE 1
How many one eighths are in one half?
Using a fraction manipulative, we show one half
of a circle.
To find how many one eighths are in one half, we
cover the one half with eighths and count how
many we use.
We find there are 4. There are four one eighths
in one half.
13
Dividing Fractions
• When dividing fractions, they do NOT need to have
a common denominator.
• To divide two fractions, change the operation to
multiply and take the reciprocal of the second
fraction (flip the second fraction).
Keep-Change-Change.
14
Dividing Fractions
• Finish the problem by following the rules for
multiplying fractions.
15
Dividing Fractions
Ex) Divide.
16
Dividing Fractions
Ex) Divide.
17
Dividing Fractions
Ex) Divide. Invert and Multiply !
18
Dividing Fractions
Ex) Divide. 15 and 5 have a common
factor.
19
Dividing Fractions
Ex) Divide. Divide them both by 5.
20
Dividing Fractions
Ex) Divide.
21
Dividing Fractions
Ex) Divide. 22 and 6 have a
common factor.
22
Dividing Fractions
Ex) Divide. Divide them both by
2.
23
Dividing Fractions
Ex) Divide. Divide them both by
2.
24
Dividing Fractions
Ex) Divide.
25
Dividing Fractions
Ex) Divide.
26
It is important to know how to compute simple
division of fractions because...
• You will need to know it for 7th grade math!
• You will need it for the CSTs!
• Why else is it important?
27
What weve learned!
• How do you reduced fractions? Do you use the LCM
or the GCF?
• What does compute mean?
• What happens if you need to multiply or divide
fractions and the number is mixed? | 1,078 | 4,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-45 | latest | en | 0.791921 |
http://mathhelpforum.com/math-topics/9756-pully-question-print.html | 1,506,327,017,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690376.61/warc/CC-MAIN-20170925074036-20170925094036-00711.warc.gz | 218,397,170 | 2,926 | # pully question
• Jan 9th 2007, 12:03 PM
math619
pully question
This is another confusing word problem where you need to have a good diagram to work off of.
A pully is suspended 13.5 m above a small bucket of cement on the ground. A rope is put over the pully. One end of the rope is tied to the bucket and the other end dangles loosly to the ground. A construction worker holds the end of the rope at a constant height (1.5 m) and walks away from beneath the pully at 1.6 m/s. How fast is the bucket rising when he is 9 m away from the path of the rising cement bucket? :confused:
Can anyone help out with this one? It's really bothering me.
Thx
• Jan 9th 2007, 12:52 PM
galactus
Like many of these related rates problems, it involves ol' Pythagoras.
Let z=the hypoteneuse of the triangle (the length of the rope).
By Pythagoras:
$x^{2}+(13.5-1.5)^{2}=z^{2}$
$x^{2}+144=z^{2}$
Differentiate:
$2x\frac{dx}{dt}=2z\frac{dz}{dt}$
When x=9, we can see that z=15.
$\frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}$
$\frac{dz}{dt}=\frac{9}{15}(\frac{8}{5})=\frac{24}{ 25} \;\ m/sec$ | 355 | 1,079 | {"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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | longest | en | 0.87518 |
https://wikimili.com/en/Tilting_theory | 1,701,390,501,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00186.warc.gz | 699,847,893 | 21,157 | # Tilting theory
Last updated
It turns out that there are applications of our functors which make use of the analogous transformations which we like to think of as a change of basis for a fixed root-system — a tilting of the axes relative to the roots which results in a different subset of roots lying in the positive cone. For this reason, and because the word 'tilt' inflects easily, we call our functors tilting functors or simply tilts.
## Contents
Brenner & Butler (1980, p. 103)
In mathematics, specifically representation theory, tilting theory describes a way to relate the module categories of two algebras using so-called tilting modules and associated tilting functors. Here, the second algebra is the endomorphism algebra of a tilting module over the first algebra.
Tilting theory was motivated by the introduction of reflection functors by JosephBernšteĭn , Israel Gelfand ,andV. A. Ponomarev ( 1973 ); these functors were used to relate representations of two quivers. These functors were reformulated by MauriceAuslander , María Inés Platzeck ,and Idun Reiten ( 1979 ), and generalized by SheilaBrennerandMichael C. R. Butler ( 1980 ) who introduced tilting functors. DieterHappelandClaus Michael Ringel ( 1982 ) defined tilted algebras and tilting modules as further generalizations of this.
## Definitions
Suppose that A is a finite-dimensional unital associative algebra over some field. A finitely-generated right A-module T is called a tilting module if it has the following three properties:
Given such a tilting module, we define the endomorphism algebra B = EndA(T). This is another finite-dimensional algebra, and T is a finitely-generated left B-module. The tilting functors HomA(T,), Ext1
A
(T,), BT and Tor B
1
(,T) relate the category mod-A of finitely-generated right A-modules to the category mod-B of finitely-generated right B-modules.
In practice one often considers hereditary finite-dimensional algebras A because the module categories over such algebras are fairly well understood. The endomorphism algebra of a tilting module over a hereditary finite-dimensional algebra is called a tilted algebra.
## Facts
Suppose A is a finite-dimensional algebra, T is a tilting module over A, and B = EndA(T). Write F = HomA(T,), F = Ext1
A
(T,), G = BT, and G = TorB
1
(,T). F is right adjoint to G and F is right adjoint to G.
Brenner & Butler (1980) showed that tilting functors give equivalences between certain subcategories of mod-A and mod-B. Specifically, if we define the two subcategories ${\displaystyle {\mathcal {F}}=\ker(F)}$ and ${\displaystyle {\mathcal {T}}=\ker(F')}$ of A-mod, and the two subcategories ${\displaystyle {\mathcal {X}}=\ker(G)}$ and ${\displaystyle {\mathcal {Y}}=\ker(G')}$ of B-mod, then ${\displaystyle ({\mathcal {T}},{\mathcal {F}})}$ is a torsion pair in A-mod (i.e. ${\displaystyle {\mathcal {T}}}$ and ${\displaystyle {\mathcal {F}}}$ are maximal subcategories with the property ${\displaystyle \operatorname {Hom} ({\mathcal {T}},{\mathcal {F}})=0}$; this implies that every M in A-mod admits a natural short exact sequence ${\displaystyle 0\to U\to M\to V\to 0}$ with U in ${\displaystyle {\mathcal {T}}}$ and V in ${\displaystyle {\mathcal {F}}}$) and ${\displaystyle ({\mathcal {X}},{\mathcal {Y}})}$ is a torsion pair in B-mod. Further, the restrictions of the functors F and G yield inverse equivalences between ${\displaystyle {\mathcal {T}}}$ and ${\displaystyle {\mathcal {Y}}}$, while the restrictions of F and G yield inverse equivalences between ${\displaystyle {\mathcal {F}}}$ and ${\displaystyle {\mathcal {X}}}$. (Note that these equivalences switch the order of the torsion pairs ${\displaystyle ({\mathcal {T}},{\mathcal {F}})}$ and ${\displaystyle ({\mathcal {X}},{\mathcal {Y}})}$.)
Tilting theory may be seen as a generalization of Morita equivalence which is recovered if T is a projective generator; in that case ${\displaystyle {\mathcal {T}}=\operatorname {mod} -A}$ and ${\displaystyle {\mathcal {Y}}=\operatorname {mod} -B}$.
If A has finite global dimension, then B also has finite global dimension, and the difference of F and F' induces an isometry between the Grothendieck groups K0(A) and K0(B).
In case A is hereditary (i.e. B is a tilted algebra), the global dimension of B is at most 2, and the torsion pair ${\displaystyle ({\mathcal {X}},{\mathcal {Y}})}$ splits, i.e. every indecomposable object of B-mod is either in ${\displaystyle {\mathcal {X}}}$ or in ${\displaystyle {\mathcal {Y}}}$.
Happel (1988) and Cline, Parshall & Scott (1986) showed that in general A and B are derived equivalent (i.e. the derived categories Db(A-mod) and Db(B-mod) are equivalent as triangulated categories).
## Generalizations and extensions
A generalized tilting module over the finite-dimensional algebra A is a right A-module T with the following three properties:
• T has finite projective dimension.
• Ext i
A
(T,T) = 0 for all i > 0.
• There is an exact sequence ${\displaystyle 0\to A\to T_{1}\to \dots \to T_{n}\to 0}$ where the Ti are finite direct sums of direct summands of T.
These generalized tilting modules also yield derived equivalences between A and B, where B = EndA(T).
Rickard (1989) extended the results on derived equivalence by proving that two finite-dimensional algebras R and S are derived equivalent if and only if S is the endomorphism algebra of a "tilting complex" over R. Tilting complexes are generalizations of generalized tilting modules. A version of this theorem is valid for arbitrary rings R and S.
Happel, Reiten & Smalø (1996) defined tilting objects in hereditary abelian categories in which all Hom- and Ext-spaces are finite-dimensional over some algebraically closed field k. The endomorphism algebras of these tilting objects are the quasi-tilted algebras, a generalization of tilted algebras. The quasi-tilted algebras over k are precisely the finite-dimensional algebras over k of global dimension ≤2 such that every indecomposable module either has projective dimension ≤1 or injective dimension1. Happel (2001) classified the hereditary abelian categories that can appear in the above construction.
Colpi & Fuller (2007) defined tilting objects T in an arbitrary abelian category C; their definition requires that C contain the direct sums of arbitrary (possibly infinite) numbers of copies of T, so this is not a direct generalization of the finite-dimensional situation considered above. Given such a tilting object with endomorphism ring R, they establish tilting functors that provide equivalences between a torsion pair in C and a torsion pair in R-Mod, the category of allR-modules.
From the theory of cluster algebras came the definition of cluster category (from Buan et al. (2006)) and cluster tilted algebra (Buan, Marsh & Reiten (2007)) associated to a hereditary algebra A. A cluster tilted algebra arises from a tilted algebra as a certain semidirect product, and the cluster category of A summarizes all the module categories of cluster tilted algebras arising from A.
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The Beilinson–Bernstein localization theorem is a foundational result of geometric representation theory, a part of mathematics studying the representation theory of e.g. Lie algebras using geometry. | 3,295 | 14,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.852132 |
https://yutsumura.com/tag/gaussian-integers/ | 1,611,475,978,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00312.warc.gz | 1,095,878,031 | 20,200 | # Tagged: Gaussian integers
## Problem 534
Let $I$ be a nonzero ideal of the ring of Gaussian integers $\Z[i]$.
Prove that the quotient ring $\Z[i]/I$ is finite.
## Problem 188
Denote by $i$ the square root of $-1$.
Let
$R=\Z[i]=\{a+ib \mid a, b \in \Z \}$ be the ring of Gaussian integers.
We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to
$N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.$
Here $\bar{\alpha}$ is the complex conjugate of $\alpha$.
Then show that an element $\alpha \in R$ is a unit if and only if the norm $N(\alpha)=\pm 1$.
Also, determine all the units of the ring $R=\Z[i]$ of Gaussian integers. | 209 | 627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-04 | longest | en | 0.667107 |
http://wiki.fool.com/wiki/index.php?title=How_to_Calculate_the_Dividend_Payout_Ratio_Based_on_Profit_Margin&oldid=24447 | 1,448,514,272,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446500.34/warc/CC-MAIN-20151124205406-00289-ip-10-71-132-137.ec2.internal.warc.gz | 247,513,850 | 11,794 | What is Foolsaurus?
It's a glossary of investing terms edited and maintained by our analysts, writers and YOU, our Foolish community. Get Started Now!
# How to Calculate the Dividend Payout Ratio Based on Profit Margin
Original post by Bryan Keythman of Demand Media
A dividend payout ratio shows the percentage of a company's net income that it pays out as cash dividends to its shareholders. Profit margin shows the percentage of a company's sales that it keeps as net income. If you know a company's profit margin, you can determine its dividend payout ratio using some of its other financial information. While there is no specific acceptable payout ratio, one that is close to 100 percent means a company is paying out nearly all of its earnings as dividends.
## Contents
### Step 1
Find the amount of a company's revenue on each of its last four quarterly income statements. A public company reports its first quarter through third quarter income statements in its 10-Q quarterly reports and its fourth-quarter income statement in its 10-K annual report. You can obtain these for free from the U.S. Securities and Exchange Commission's EDGAR online database. For example, assume a company generated \$5 million, \$4 million, \$6 million and \$5 million in revenue during the past four quarters.
### Step 2
Calculate the sum of its revenue in each of its past four quarters to determine the revenue it generated over the past 12 months, or its trailing 12-month revenue. In this example, calculate the sum of the four quarters to get \$20 million in trailing 12-month revenue.
### Step 3
Multiply the company's trailing 12-month revenue by its known profit margin to calculate its trailing 12-month net income. In this example, if the company's profit margin is 10 percent, so multiply \$20 million by 10 percent to get \$2 million in trailing 12-month net income.
### Step 4
Find a company's trailing 12-month dividend per share and its number of shares outstanding in the company's stock quote section. In this example, assume the company's trailing 12-month dividend is \$1 per share and that the company has 600,000 shares outstanding.
### Step 5
Multiply the company's trailing 12-month dividend per share by the total number of shares outstanding to calculate its total trailing 12-month dividend. In this example, multiply \$1 by 600,000 to get a trailing 12-month dividend of \$600,000.
### Step 6
Divide the company's trailing 12-month dividend by its trailing 12-month net income to calculate its dividend payout ratio. In this example, divide \$600,000 by \$2 million to get a dividend payout ratio of 0.3, or 30 percent. This suggests the company generates sufficient earnings to likely continue its dividend payment.
```
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``` | 613 | 2,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2015-48 | latest | en | 0.931259 |
https://howtofunda.com/quadrilateral-maths-working-model/ | 1,721,553,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00099.warc.gz | 269,061,566 | 13,781 | # how to make quadrilateral math’s working model
In this post we write about making of the quadrilateral maths working model – tlm for class 9 | class 10
Creating working models of different types of quadrilaterals using color paper and cardboard can be a fantastic way to explore their properties visually.
Let’s build models for four common types of quadrilaterals: square, rectangle, parallelogram, and trapezoid and so on
Materials needed:
1. Different colored papers (for example, red, blue, green, and yellow)
2. Cardboard or thick paperboard
3. Scissors
4. Glue or tape
5. Ruler
6. Pencil
Step-by-step instructions:
1. Make the different quadrilateral types such as square, rectangle, parallelogram, and trapezoid and so on :
2. This model can be helpful in understanding the properties of each quadrilateral type, such as the number of parallel sides, angles, and side lengths. You can label the sides, angles, and properties to make it more educational and interactive. | 227 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.861176 |
https://www.radfordmathematics.com/probabilities-and-statistics/continuous-probability-distributions/continuous-random-variables-probability-ditributions/probability-density-functions-continuous-random-variables.html | 1,555,869,815,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00483.warc.gz | 814,902,521 | 8,694 | # Continuous Probability Distributions
## (Continuous Random Variables)
Continuous random variables are used to model continuous phenomena or quantities, such as time, length, mass, ... that depend on chance.
We refer to continuous random variables with capital letters, typically $$X$$, $$Y$$, $$Z$$, ... .
For instance the heights of people selected at ranom would correspond to possible values of the continuous random variable $$X$$ defined as:
$$X$$ : height, in cm, of a person selected at random
### Probability Distributions
When working with continuous random variables, such as $$X$$, we only calculate the probability that $$X$$ lie within a certain interval; like $$P\begin{pmatrix}X \leq k\end{pmatrix}$$ or $$P\begin{pmatrix}a\leq X \leq b \end{pmatrix}$$.
We don't calculate the probability of $$X$$ being equal to a specific value $$k$$. In fact that following result will always be true: $P\begin{pmatrix}X = k \end{pmatrix} = 0$ This can be explained by the fact that the total number of possible values of a continuous random variable $$X$$ is infinite, so the likelihood of any one single outcome tends towards $$0$$.
### Calculating Probabilities
To calculate probabilities we'll need two functions:
• The probability density function (PDF)
• The cumulative density function (CDF) a.k.a the cumulative distribution function
Each of these is defined, further down, but the idea is to integrate the probability density function $$f(x)$$ to define a new function $$F(x)$$, known as the cumulative density function.
To calculate the probability that $$X$$ be within a certain range, say $$a \leq X \leq b$$, we calculate $$F(b) - F(a)$$, using the cumulative density function.
Put "simply" we calculate probabilities as: $P\begin{pmatrix}a\leq X \leq b \end{pmatrix} = \int_a^b f(x) dx$ where $$f(x)$$ is the variable's probability density function.
### Tutorial
In the following tutorial we learn about continuous random variables and how to calculate probabilities using probability density functions.
## Probability Density Functions (PDF)
Given a continuous random variable $$X$$, its probability density function $$f(x)$$ is the function whose integral allows us to calculate the probability that $$X$$ lie within a certain range, $$P\begin{pmatrix}a\leq X \leq b\end{pmatrix}$$.
The curve $$y=f(x)$$ serves as the "envelope", or contour, of the probability distribution.
#### Exam Hint
We'll often be given a pdf with an unknown parameter that we'll need to find using the second property (see question 2.a below).
### Properties of the PDF
Probability density functions are always greater than or equal to $$0$$: $f(x) \geq 0, \quad x \in \mathbb{R}$ The area enclosed by a probability density function and the horizontal axis equals to $$1$$: $\int_{-\infty}^{+\infty}f(x)dx = 1$ If the area isn't equal to $$1$$ then $$X$$ is not a continuous random variable.
Note: these properties are often used in exam questions.
### Examples
#### Example 1
The piecewise function defined as: $f(x) = \begin{cases} 3x^2,\quad 0\leq x \leq 1 \\ 0, \quad \text{elsewhere} \end{cases}$ could be the probability density function for some continuous random variable $$X$$.
Indeed, we can see from its graph that $$f(x)\geq 0$$. Furthermore we can check that the area enclosed by the curve and the $$x$$-axis equals to $$1$$: \begin{aligned} \int_{-\infty}^{+\infty}f(x)dx & = \int_0^13x^2dx \\ & = \begin{bmatrix}x^3 \end{bmatrix}_0^1 \\ & = 1^3 - 0^3 \\ \int_{-\infty}^{+\infty}f(x)dx & = 1 \end{aligned}
#### Example 2
Another example, that we'll learn about with normal distributions, could be the function defined as: $f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, \quad x \in \mathbb{R}$
We can see from its graph that $$f(x)\geq 0$$. And although we cannot integrate this by hand, use numerical methods and our calculator we find: $\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx = 1$
#### Terminology
Keep in mind that the Cumulative density function is frequently called the cumulative distribution function or simply the distribution function.
It's important to know that all $$3$$ names refer to the same thing: the CDF.
## Cumulative Density Function (CDF)
### (Calculating Probabilities)
Given a continuous random variable $$X$$ and its probability density function $$f(x)$$, the cumulative density function, written $$F(x)$$, allows us to calculate the probability that $$X$$ be less than, or equal to, any value of $$x$$, in other words: $$P\begin{pmatrix}X \leq x \end{pmatrix} = F(x)$$.
Where: $F(x) =\int_{-\infty}^x f(t)dt$ There are three "types" of probability calculations that we'll need to be comfortable with. Each is shown here:
#### Calculating $$P\begin{pmatrix}X \leq k \end{pmatrix}$$
Since $$F(x) = P\begin{pmatrix}X \leq x \end{pmatrix}$$ we write: $P\begin{pmatrix}X\leq k \end{pmatrix} = \int_{-\infty}^k f(x)dx$ This "tells us" that the probability that the continuous random variable $$X$$ be less than or equal to some value $$k$$ equals to the area enclosed by the probability density function and the horizontal axis, between $$-\infty$$ and $$k$$.
#### Calculating $$P\begin{pmatrix}a \leq X \leq b \end{pmatrix}$$
To calculate the probability that a continuous random variable $$X$$, lie between two values say $$a$$ and $$b$$ we use the following result: $P\begin{pmatrix}a \leq X \leq b \end{pmatrix} = \int_a^b f(x)dx$
#### Calculating $$P\begin{pmatrix} X \geq k \end{pmatrix}$$
To calculate the probability that a continuous random variable $$X$$ be greater than some value $$k$$ we use the following result: $P\begin{pmatrix} X \geq k \end{pmatrix} = \int_k^{+\infty} f(x)dx$
## Example
A continuous random variable $$X$$ has probability density function defined as: $f(x) = \begin{cases} -\frac{3}{4}x(x-2), \quad 0\leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases}$
1. Find $$P\begin{pmatrix}X \leq 1.5\end{pmatrix}$$.
2. Find $$P\begin{pmatrix}0.5 \leq X \leq 1 \end{pmatrix}$$.
3. Find $$P\begin{pmatrix}X \geq 1 \end{pmatrix}$$.
### Solution
1. To find $$P\begin{pmatrix}X \leq 1.5\end{pmatrix}$$, we use write: \begin{aligned} P\begin{pmatrix} X \leq 1.5 \end{pmatrix} & = \int_{-\infty}^{1.5} f(x)dx \\ & = \int_{-\infty}^{\frac{3}{2}}f(x)dx \\ & = \int_{-\infty}^{\frac{3}{2}}-\frac{3}{4}x(x-2)dx \\ & = -\frac{3}{4} \int_{-\infty}^{\frac{3}{2}}x(x-2)dx \\ & = -\frac{3}{4} \int_{-\infty}^{\frac{3}{2}}\begin{pmatrix} x^2 - 2x \end{pmatrix}dx \\ & = -\frac{3}{4} \begin{bmatrix}\frac{x^3}{3} - x^2 \end{bmatrix}_0^{\frac{3}{2}} \\ & = - \frac{1}{4}\begin{bmatrix}x^3 - 3x^2 \end{bmatrix}_0^{\frac{3}{2}} \\ & = - \frac{1}{4}\begin{bmatrix}\begin{pmatrix} \frac{3}{2}\end{pmatrix}^3 - 3 \times \begin{pmatrix} \frac{3}{2} \end{pmatrix}^2 - 0 \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \frac{27}{8} - \frac{27}{4} \end{bmatrix} \\ & = - \frac{1}{4} \begin{bmatrix} -\frac{27}{8} \end{bmatrix} \\ & = \frac{27}{32} \\ P\begin{pmatrix} X \leq 1.5 \end{pmatrix} & = 0.844 \end{aligned}
Graphically, this result can be interpreted as follows:
The area enclosed by the probability density function's curve and the horizontal axis, from $$-\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures).
The probability is equal to the area so: $$P\begin{pmatrix} X \leq 1.5\end{pmatrix} = 0.844$$
2. To find $$P\begin{pmatrix}0.5 \leq X \leq 1\end{pmatrix}$$ we write: \begin{aligned} P\begin{pmatrix}0.5 \leq X \leq 1\end{pmatrix} & = \int_{0.5}^1f(x)dx \\ & = \int_{0.5}^1 -\frac{3}{4}x(x-2)dx \\ & = -\frac{3}{4} \int_{0.5}^1 x(x-2)dx \\ & = -\frac{3}{4} \int_{\frac{1}{2}}^1 \begin{pmatrix} x^2-2x \end{pmatrix} dx \\ & = -\frac{3}{4}\begin{bmatrix} \frac{x^3}{3} - x^2 \end{bmatrix}_{\frac{1}{2}}^1 \\ & = -\frac{3}{4}\times \frac{1}{3}\begin{bmatrix} x^3 - 3x^2 \end{bmatrix}_{\frac{1}{2}}^1 \\ & = -\frac{1}{4}\begin{bmatrix} x^3 - 3x^2 \end{bmatrix}_{\frac{1}{2}}^1 \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix} 1^3 - 3\times 1^2\end{pmatrix} - \begin{pmatrix} \begin{pmatrix}\frac{1}{2} \end{pmatrix}^3 - 3\begin{pmatrix}\frac{1}{2} \end{pmatrix}^2\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix} 1 - 3\end{pmatrix} - \begin{pmatrix} \frac{1}{8} - 3\times \frac{1}{4}\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix} -2\end{pmatrix} - \begin{pmatrix} \frac{1}{8} - \frac{3}{4}\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} -2 - \begin{pmatrix} \frac{1}{8} - \frac{6}{8}\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} -2 - \begin{pmatrix} - \frac{5}{8}\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} -2 + \frac{5}{8} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} - \frac{11}{8} \end{bmatrix} \\ & = \frac{11}{32} \\ P\begin{pmatrix}0.5 \leq X \leq 1\end{pmatrix} & = 0.344 \end{aligned}
Graphically, this result can be interpreted as follows:
The area enclosed by the probability density function's curve and the horizontal axis, between $$x=0.5$$ and $$x=1$$ is equal to $$0.344$$ (rounded to 3 significant figures).
The probability is equal to the area so: $$P\begin{pmatrix}0.5 \leq X \leq 1\end{pmatrix} = 0.344$$
3. To find $$P\begin{pmatrix}X \geq 1\end{pmatrix}$$ we write: \begin{aligned} P\begin{pmatrix}X \geq 1\end{pmatrix} & = \int_1^{+\infty}f(x)dx \\ & = \int_1^2 -\frac{3}{4}x(x-2)dx \\ & = -\frac{3}{4}\int_1^2 x(x-2)dx \\ & = -\frac{3}{4}\int_1^2 \begin{pmatrix} x^2 - 2x \end{pmatrix} dx \\ & = -\frac{3}{4}\begin{bmatrix}\frac{x^3}{3}-x^2 \end{bmatrix}_1^2 \\ & = -\frac{3}{4}\times \frac{1}{3}\begin{bmatrix}x^3-3x^2 \end{bmatrix}_1^2 \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix}2^3-3\times 2^2\end{pmatrix} - \begin{pmatrix}1^3-3\times 1^2\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix}8-12\end{pmatrix} - \begin{pmatrix}1-3\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix}-4\end{pmatrix} - \begin{pmatrix}-2\end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} \begin{pmatrix}-4\end{pmatrix} +2 \end{bmatrix} \\ & = -\frac{1}{4}\begin{bmatrix} -2 \end{bmatrix} \\ & = \frac{1}{2} \\ P\begin{pmatrix}X \geq 1\end{pmatrix} & = 0.5 \end{aligned}
Graphically, this result can be interpreted as follows:
The area enclosed by the probability density function's curve and the horizontal axis, between $$x=1$$ and beyond is equal to $$0.5$$.
The probability is equal to the area so: $$P\begin{pmatrix}X \geq 1\end{pmatrix} = 0.5$$.
Note: we could have stated this result directly, without integrating, as $$x=1$$ is the axis of symmetry of the parabola $$y=-\frac{3}{4}x(x-2)$$.
### Important Result
When working with continuous random variables the following results will always be true: $P\begin{pmatrix}X\leq k \end{pmatrix} = P\begin{pmatrix}X < k \end{pmatrix}$
and
$P\begin{pmatrix}X\geq k \end{pmatrix} = P\begin{pmatrix}X > k \end{pmatrix}$ Consequently, the following four probabilities are equal and calculated in exactly the same way: \begin{aligned} P\begin{pmatrix}a \leq X \leq b \end{pmatrix} & = P\begin{pmatrix}a \leq X < b \end{pmatrix} \\ & = P\begin{pmatrix}a < X \leq b \end{pmatrix} \\ & = P\begin{pmatrix}a < X < b \end{pmatrix} \\ & = \int_a^b f(x)dx \end{aligned}
## Exercise
1. A continuous random variable $$X$$ has pdf: $f(x) = \begin{cases} \frac{sin(x)}{2}, \quad 0 \leq x \leq \pi \\ 0, \quad \text{elsewhere} \end{cases}$
1. Calculate $$P\begin{pmatrix}X \leq \frac{\pi}{3}\end{pmatrix}$$
2. Calculate $$P\begin{pmatrix} \frac{\pi}{3} \leq X \leq \frac{2\pi}{3}\end{pmatrix}$$
3. Show that $$X$$ is indeed a continuous random variable.
2. A continuous random variable has probability density function defined as: $f(x) = \begin{cases} kx^2, \quad 0 \leq x \leq 2\\ 0, \quad \text{elsewhere} \end{cases}$
1. Find the value of $$k$$.
2. Find $$P \begin{pmatrix}X \leq 1 \end{pmatrix}$$.
3. Find $$P \begin{pmatrix}1 < X < 1.5 \end{pmatrix}$$.
1. $$P\begin{pmatrix}X \leq \frac{\pi}{3}\end{pmatrix} = \frac{1}{4} = 0.25$$
2. $$P\begin{pmatrix} \frac{\pi}{3} \leq X \leq \frac{2\pi}{3}\end{pmatrix} = \frac{1}{2} = 0.5$$
3. We show that $$X$$ is a coontinuous random variable by showing that $$\int_{-\infty}^{+\infty}f(x)dx = 1$$: \begin{aligned} \int_{-\infty}^{+\infty}f(x)dx & = \int_0^{\pi}\frac{sin(x)}{2}dx \\ & = \begin{bmatrix}- \frac{cos(x)}{2} \end{bmatrix}_0^{\pi} \\ & = -\frac{1}{2}\begin{bmatrix}cos(x) \end{bmatrix}_0^{\pi} \\ & = -\frac{1}{2}\begin{bmatrix}cos\begin{pmatrix}\pi \end{pmatrix} - cos\begin{pmatrix}0 \end{pmatrix} \end{bmatrix} \\ & = -\frac{1}{2}\begin{bmatrix}-1 - 1 \end{bmatrix} \\ & = -\frac{1}{2}\begin{bmatrix}-2 \end{bmatrix} \\ \int_{-\infty}^{+\infty}f(x)dx & = 1 \end{aligned} Since $$\int_{-\infty}^{+\infty}f(x)dx = 1$$ we can confirm that $$X$$ is a continuous random variable.
1. $$k = \frac{3}{8}$$
2. $$P\begin{pmatrix}X \leq 1 \end{pmatrix} = 0.125$$
3. $$P\begin{pmatrix}1 \leq X \leq 1.5 \end{pmatrix} = \frac{19}{64}=0.297$$ | 4,727 | 12,978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-18 | latest | en | 0.811461 |
https://www.tutorialspoint.com/program-to-find-minimum-length-of-lossy-run-length-encoding-in-python | 1,716,659,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00433.warc.gz | 909,328,549 | 25,821 | # Program to find minimum length of lossy Run-Length Encoding in Python
Suppose we have a lowercase string s and another value k. Now consider an operation where we perform a run-length encoding on a string by putting repeated successive characters as a count and character. So if the string is like "aaabbc" would be encoded as "3a2bc". Here we do not put "1c" for "c" since it only appears once successively. So we can first remove any k consecutive characters in s, then find the minimum length possible of the resulting run-length encoding.
So, if the input is like s = "xxxxxyyxxxxxzzxxx", k = 2, then the output will be 6, as the two obvious choices are to remove the "yy"s or the "zz"s. If we remove the "yy"s, then we will get "10x2z3x" which has length of 7. If we remove the "zz"s, then will get "5x2y8x" which has length of 6, this is the smallest.
To solve this, we will follow these steps −
• Define a function calc_cost() . This will take l
• if l is same as 0, then
• return 0
• if l is same as 1, then
• return 1
• otherwise,
• return size of str(l) + 1
• Define a function prefix() . This will take s
• pre := a list initially with pair [0, 0]
• last := null
• for each c in s, do
• if c is same as last, then
• insert a pair (0th element of last item of pre, 1 + 1st element of last item of pre) into pre
• otherwise,
• insert (0th element of last item of pre) + calc_cost(1st element of last item of pre, 1) into pre
• last := c
• return pre
• From the main method do the following:
• pre := prefix(s)
• suf := reverse of prefix(s in reverse order)
• ans := infinity
• for i in range 0 to size of s - k + 1, do
• j := i + k
• pair (left, midl) := pre[i]
• pair (right, midr) := suf[j]
• cost := left + right
• c1 := s[i - 1] if i > 0 otherwise null
• c2 := s[j] if j < size of s otherwise null
• if c1 is same as c2, then
• cost := cost + calc_cost(midl + midr)
• otherwise,
• cost := cost + calc_cost(midl) + calc_cost(midr)
• ans := minimum of ans and cost
• return ans
Let us see the following implementation to get better understanding −
## Example
Live Demo
def calc_cost(l):
if l == 0:
return 0
if l == 1:
return 1
else:
return len(str(l)) + 1
class Solution:
def solve(self, s, k):
def prefix(s):
pre = [[0, 0]]
last = None
for c in s:
if c == last:
pre.append([pre[-1][0], pre[-1][1] + 1])
else:
pre.append([pre[-1][0] + calc_cost(pre[-1][1]),1])
last = c
return pre
pre = prefix(s)
suf = prefix(s[::-1])[::-1]
ans = float("inf")
for i in range(len(s) - k + 1):
j = i + k
left, midl = pre[i]
right, midr = suf[j]
cost = left + right
c1 = s[i - 1] if i > 0 else None
c2 = s[j] if j < len(s) else None
if c1 == c2:
cost += calc_cost(midl + midr)
else:
cost += calc_cost(midl) + calc_cost(midr)
ans = min(ans, cost)
return ans
ob = Solution()
s = "xxxxxyyxxxxxzzxxx"
print(ob.solve(s, 2))
## Input
s = "xxxxxyyxxxxxzzxxx"
## Output
6
Updated on: 10-Oct-2020
171 Views | 930 | 2,938 | {"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": 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} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.784825 |
https://metanumbers.com/22621 | 1,628,127,830,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00430.warc.gz | 404,663,746 | 10,777 | 22621
22,621 (twenty-two thousand six hundred twenty-one) is an odd five-digits prime number following 22620 and preceding 22622. In scientific notation, it is written as 2.2621 × 104. The sum of its digits is 13. It has a total of 1 prime factor and 2 positive divisors. There are 22,620 positive integers (up to 22621) that are relatively prime to 22621.
Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 13
• Digital Root 4
Name
Short name 22 thousand 621 twenty-two thousand six hundred twenty-one
Notation
Scientific notation 2.2621 × 104 22.621 × 103
Prime Factorization of 22621
Prime Factorization 22621
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 22621 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.0266 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 22,621 is 22621. Since it has a total of 1 prime factor, 22,621 is a prime number.
Divisors of 22621
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 22622 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 11311 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 150.403 Returns the nth root of the product of n divisors H(n) 1.99991 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 22,621 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 22,621) is 22,622, the average is 11,311.
Other Arithmetic Functions (n = 22621)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 22620 Total number of positive integers not greater than n that are coprime to n λ(n) 22620 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2527 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 22,620 positive integers (less than 22,621) that are coprime with 22,621. And there are approximately 2,527 prime numbers less than or equal to 22,621.
Divisibility of 22621
m n mod m 2 3 4 5 6 7 8 9 1 1 1 1 1 4 5 4
22,621 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
Base conversion (22621)
Base System Value
2 Binary 101100001011101
3 Ternary 1011000211
4 Quaternary 11201131
5 Quinary 1210441
6 Senary 252421
8 Octal 54135
10 Decimal 22621
12 Duodecimal 11111
20 Vigesimal 2gb1
36 Base36 hgd
Basic calculations (n = 22621)
Multiplication
n×i
n×2 45242 67863 90484 113105
Division
ni
n⁄2 11310.5 7540.33 5655.25 4524.2
Exponentiation
ni
n2 511709641 11575383789061 261846756692348881 5923235483137624037101
Nth Root
i√n
2√n 150.403 28.2816 12.2639 7.42852
22621 as geometric shapes
Circle
Diameter 45242 142132 1.60758e+09
Sphere
Volume 4.84869e+13 6.43033e+09 142132
Square
Length = n
Perimeter 90484 5.1171e+08 31990.9
Cube
Length = n
Surface area 3.07026e+09 1.15754e+13 39180.7
Equilateral Triangle
Length = n
Perimeter 67863 2.21577e+08 19590.4
Triangular Pyramid
Length = n
Surface area 8.86307e+08 1.36417e+12 18470
Cryptographic Hash Functions
md5 3fb5bb321738f743b4dc240f7aa5ea23 e9021d47cdaf81da1937b4c00c6cc7f78c934a51 36b212ef149cc79e63eb109177f831812969fa8e86f9c09b405448276b0ec76f 3aa0b11e07f88d05384e29de0c4677adf201b9df84b9d1ca9fd884a16d0700cd887081d211ac2e47acbf0a92f4938f1ecbfba1df7beea44b8db252702c920fb8 e4709fdef54f9842d2a35df6d77c1aa9bcf7e489 | 1,397 | 3,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-31 | latest | en | 0.832442 |
https://familycouncil.org/?p=9811 | 1,716,747,849,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00359.warc.gz | 208,610,854 | 15,583 | An old myth keeps resurfacing about the Arkansas Lottery: The idea that by not allocating a large percentage of its revenue for scholarships each year, the Arkansas Lottery is somehow more profitable and able to “generate” more scholarship money in the long-run.
Currently, about 19% – 20% of the Lottery’s revenue goes to scholarship. Lottery proponents say if the Arkansas Lottery were required to allocate 30% of its gross revenue for scholarships (like many state lotteries), it would have less money to spend on promotional activity and prizes, which drive lottery ticket sales. By letting the Arkansas Lottery allocate a smaller percentage of revenue for sales, the Lottery is able to spend more money on prizes, which leads to more lottery ticket sales and–ultimately–more college scholarships.
To put it bluntly, this idea is complete nonsense. To prove it, let’s do the math.
Right now, the Arkansas Lottery projects total sales for 2014 at around \$460 million. If everything goes according to plan, they will sell close to half-a-billion dollars worth of lottery tickets and they will put somewhere around \$90 million in the state’s college scholarship fund. Sales and scholarship funding for 2014 are not on track, but for the sake of example let’s crunch these numbers.
Take a look at the table below.
Gross Lottery Revenue 20% for Scholarships 30% for Scholarships \$ 460,000,000.00 \$ 92,000,000.00 \$ 138,000,000.00 \$ 440,000,000.00 \$ 88,000,000.00 \$ 132,000,000.00 \$ 420,000,000.00 \$ 84,000,000.00 \$ 126,000,000.00 \$ 400,000,000.00 \$ 80,000,000.00 \$ 120,000,000.00 \$ 380,000,000.00 \$ 76,000,000.00 \$ 114,000,000.00 \$ 360,000,000.00 \$ 72,000,000.00 \$ 108,000,000.00 \$ 340,000,000.00 \$ 68,000,000.00 \$ 102,000,000.00 \$ 320,000,000.00 \$ 64,000,000.00 \$ 96,000,000.00 \$ 310,000,000.00 \$ 62,000,000.00 \$ 93,000,000.00
The column on the left shows gross lottery revenue; in the middle is the scholarship money if the Arkansas Lottery allocates 20% of that gross revenue for scholarships; and on the right is scholarship money if they allocate 30% of that gross revenue for scholarships.
Doing the math, we see that if the Arkansas Lottery raised its scholarship allocation from 20% of gross revenue to 30%, the Arkansas Lottery could pay out more for college scholarships than it currently does even if its ticket sales plummeted by \$150 million!
The math is pretty clear. But what can the experiences of other states tell us?
Well, a few years ago we published an article highlighting how Louisiana’s lottery paid out \$30 million more for education than Arkansas’ lottery did even though Louisiana’s lottery took in \$100 million less. How was the Louisiana Lottery able to take in less money but pay out more? Louisiana earmarked 35% of its gross lottery revenue for education while Arkansas only earmarked 20%.
It really is that simple.
So let’s recap: If the Arkansas Lottery raised its scholarship allocation from 20 cents on the dollar to 30 cents on the dollar:
1. Scholarship funding would almost certainly go up.
2. Even if lottery ticket sales dropped as a result, they would have to drop by more than \$150 million before scholarship funding would be any worse than it currently is.
Does anyone really think putting more money toward scholarships is somehow going to cost the Arkansas Lottery \$150 million a year? If the Arkansas Lottery is serious about funding college scholarship, raising the percentage they allocate from 20% to 30% is the obvious thing to do. | 863 | 3,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.86811 |
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