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https://typelevel.org/cats/typeclasses/comonad.html | 1,718,815,813,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00272.warc.gz | 519,730,985 | 6,209 | Comonad is a Functor and provides duals of the Monad pure and flatMap functions. A dual to a function has the same types but the direction of the arrows are reversed. Whether or not that is useful, or even possible, depends on the particular type. For a more formal definition of duality, please refer to https://ncatlab.org/nlab/show/duality.
### extract
Monads have pure from Applicative which gives you the ability to wrap a value A using the type constructor giving an F[A]. Comonad has extract which instead takes an F[A] and extracts the A. Therefore, to be able to implement extract we must have a type of which we are certain we can get an A from an F[A]. For example we cannot always get an A from a List[A] because if the list is empty there is nothing to get.
For the same reason, Option doesn't have a Comonad instance, because we cannot always get an A from an Option, it may be empty too.
Some examples that we can implement Comonad for include OneAnd, Tuple2 and the "non empty" collections.
First some imports.
import cats._
import cats.data._
import cats.syntax.all._
import cats.instances.list._
NonEmptyList has a Comonad instance and its implementation of extract simply returns the head element of the list, which we know we will always have.
NonEmptyList.of(1,2,3).extract
// res0: Int = 1
### coflatMap
coflatMap is the dual of Monad's flatMap. While flatMap allows us to chain together operations in a monadic context, coflatMap takes a value in some context F[A] and a function F[A] => B and returns a new value in a context F[B].
The default implementation of coflatMap for NonEmptyList will pass the supplied function with the whole list, then the tail of that, then the tail of that and so on. This is illustrated below.
NonEmptyList.of(1,2,3,4,5).coflatMap(identity)
// res1: NonEmptyList[NonEmptyList[Int]] = NonEmptyList(
// head = NonEmptyList(head = 1, tail = List(2, 3, 4, 5)),
// tail = List(
// NonEmptyList(head = 2, tail = List(3, 4, 5)),
// NonEmptyList(head = 3, tail = List(4, 5)),
// NonEmptyList(head = 4, tail = List(5)),
// NonEmptyList(head = 5, tail = List())
// )
// )
# CoflatMap
While FlatMap is a weaker version of Monad that doesn't have the pure function, CoflatMap is a Comonad without the extract function. There are many instances of type classes in Cats that implement CoflatMap but not Comonad.
For example we cannot write extract for Option[A] because there's no way to pull an A out of nowhere if the Option is empty.
def extract[A](fa : Option[A]): A = fa match {
case Some(a) => a
case None => ??? // What now?
}
Another example is List. Remember we cannot write extract for list because lists can be empty, but we can implement the coflatMap and it works identically to the one shown above for NonEmptyList.
List(1,2,3,4,5).coflatMap(identity)
// res2: List[List[Int]] = List(
// List(1, 2, 3, 4, 5),
// List(2, 3, 4, 5),
// List(3, 4, 5),
// List(4, 5),
// List(5)
// ) | 811 | 2,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.839381 |
https://www.unitconverters.net/temperature/kelvin-to-rankine.htm | 1,519,105,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00476.warc.gz | 964,123,288 | 3,107 | Home / Temperature Conversion / Convert Kelvin to Rankine
# Convert Kelvin to Rankine
Please provide values below to convert kelvin [K] to Rankine [°R], or vice versa.
From: kelvin To: Rankine
### Kelvin to Rankine Conversion Table
Kelvin [K]Rankine [°R]
0.01 K0.018 °R
0.1 K0.18 °R
1 K1.8 °R
2 K3.6 °R
3 K5.4 °R
5 K9 °R
10 K18 °R
20 K36 °R
50 K90 °R
100 K180 °R
1000 K1800 °R
### How to Convert Kelvin to Rankine
1 K = 1.8 °R
1 °R = 0.55555555555556 K
Example: convert 15 K to °R:
15 K = 15 × 1.8 °R = 27 °R | 211 | 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} | 3.328125 | 3 | CC-MAIN-2018-09 | longest | en | 0.416279 |
http://bitaacademy.blogspot.com/2019/05/daily-numerical-ability-quiz-for-bank_19.html | 1,621,145,037,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989690.55/warc/CC-MAIN-20210516044552-20210516074552-00396.warc.gz | 8,518,205 | 10,545 | ## Sunday, 19 May 2019
### Daily Numerical Ability Quiz for Bank Exam : 20/05/2019
Directions (Q. 1-5): What should come in place of the question mark (?) in the following series?
1. 680 655 704 623 744 ?
a) 575
b) 675
c) 600
d) 913
e) None of these
1. a
680 - 5^2 = 655
655 + 7^2 = 704
704 - 9^2 = 623
623 + 11^2 = 744
744 - 13^2 = 575
2. 10 10 20 ? 110 300 930
a) 40
b) 35
c) 45
d) 50
e) 60
2. c
10 × 0.5 + 5 = 10
10 × 1 + 10 = 20
20 × 1.5 + 15 = 45
45 × 2 + 20 = 110
110 × 2.5 + 25 = 300
300 × 3 + 30 = 930
3. 1 ? 22 188 2052 28748
a) 2
b) 3
c) 6
d) 16
e) 10
3. c
1 × 2 + 4 = 6
6 × 5 - 8 = 22
22 × 8 + 12 = 188
188 × 11 - 16 = 2052
2052 × 14 + 20 = 28748
4. 50327 7169 1215 223 76 ?
a) 1
b) 2
c) 3
d) 4
e) 8
4. d
(50327 – 144) ÷ 7 = 7169
(7169 + 121) ÷ 6 = 1215
(1215 – 100) ÷ 5 = 223
(223 + 81) ÷ 4 = 76
(76 – 64) ÷ 3 = 4
5. 14 34 62 ? 142 194
a) 94
b) 98
c) 108
d) 112
e) None of these
5. b
2 + 3 × 4 = 14
4 + 5 × 6 = 34
6 + 7 × 8 = 62
8 + 9 × 10 = 98
10 + 11 × 12 = 142
12 + 13 × 14 = 194
8^3 – 1 = 511
9^3 – 1 = 728
10^3 – 1 = 999 | 625 | 1,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.430649 |
https://www.isixsigma.com/topic/pvalue-minitab/ | 1,642,482,711,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00038.warc.gz | 896,285,289 | 53,389 | # Pvalue minitab
Six Sigma – iSixSigma Forums Old Forums General Pvalue minitab
Viewing 7 posts - 1 through 7 (of 7 total)
• Author
Posts
• #47141
gt
Participant
i have 12 Df numerator and 2 Df denominator to get Fc of 19.41. in my study i got Fvalue of 18.24.
Minitab is givin me a Pvalue of 0000. So that means : Ha “accepted” meaning statistically there is a difference between factors
If Fc>Fvalue, shouldn’t we “fail in rejecting the null”?
any input is appreciated
0
#156831
Jim Shelor
Participant
gt,
If P < alpha, reject the null.
If F statistic > F critical, reject the null.
Your P is < alpha – you reject the null, not fail to reject the null.
I hope this helps
Jim Shelor
0
#156833
Jim Shelor
Participant
gt,
Yes, when F stat < Fc, you accept the null.
I have seen a lot of problem like this. When Fc and F stat are only 5 to 10% different, they can lead to confusion because of roundoff error.
However, sinc your P=.0000, the F should not be that close. Check your data again and if you want some backup, I can run it for you.
Jim Shelor
0
#156834
gt
Participant
this is what i got
One-way ANOVA: results versus pates
Source DF SS MS F P
pates 2 1.0127 0.5064 18.24 0.000
Error 12 0.3331 0.0278
Total 14 1 .3458
0
#156836
Craig
Participant
I believe you have your numerator and denomimator mixed up.
F critical is 3.89 at an alpha of .05
0
#156837
Jim Shelor
Participant
gt,
I agree with hacl.
If you run the comparison function
Hsu’s MCB (Multiple Comparisons with the Best)
Family error rate = 0.1
Critical value = 1.87
As you can see, Minitab provides Fc.
Hope this helps.
Jim Shelor
0
#156838
Jim Shelor
Participant
That last post was an example, not the answer to your problem for Fc.
0
Viewing 7 posts - 1 through 7 (of 7 total)
The forum ‘General’ is closed to new topics and replies. | 588 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-05 | longest | en | 0.814452 |
https://www.cynohub.com/jntu-k-b-tech-r19-3-2-syllabus-for-design-and-analysis-of-algorithms-pdf-2022/ | 1,653,171,462,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00250.warc.gz | 819,347,752 | 33,027 | # JNTU-K B.TECH R19 3-2 Syllabus For Design and analysis of algorithms PDF 2022
### Get Complete Lecture Notes for Design and analysis of algorithms on Cynohub APP
You will be able to find information about Design and analysis of algorithms along with its Course Objectives and Course outcomes and also a list of textbook and reference books in this blog.You will get to learn a lot of new stuff and resolve a lot of questions you may have regarding Design and analysis of algorithms after reading this blog. Design and analysis of algorithms has 5 units altogether and you will be able to find notes for every unit on the CynoHub app. Design and analysis of algorithms can be learnt easily as long as you have a well planned study schedule and practice all the previous question papers, which are also available on the CynoHub app.
All of the Topic and subtopics related to Design and analysis of algorithms are mentioned below in detail. If you are having a hard time understanding Design and analysis of algorithms or any other Engineering Subject of any semester or year then please watch the video lectures on the official CynoHub app as it has detailed explanations of each and every topic making your engineering experience easy and fun.
### Design and analysis of algorithms Unit One
#### Introduction
Introduction: Algorithm Definition, Algorithm Specification, performance Analysis, Performance measurement, Asymptotic notation, Randomized Algorithms.Sets &Disjoint set union: introduction, union and find operations.Basic Traversal & Search Techniques: Techniques for Graphs, connected components and Spanning Trees, Bi-connected components and DFS.
### Design and analysis of algorithms Unit Two
#### Divide and Conquer
Divide and Conquer: General Method, Defective chessboard, Binary Search, finding the maximum and minimum, Merge sort, Quick sort.The Greedy Method: The general Method, container loading, knapsack problem, Job sequencing with deadlines, minimum-cost spanning Trees.
### Design and analysis of algorithms Unit Three
#### Dynamic Programming
Dynamic Programming:The general method, multistage graphs, All pairs-shortest paths, single-source shortest paths: general weights, optimal Binary search trees, 0/1 knapsack, reliability Design, The traveling salesperson problem, matrix chain multiplication.
### Design and analysis of algorithms Unit Four
#### Backtracking
Backtracking:The General Method, The 8-Queens problem, sum of subsets, Graph coloring, Hamiltonian cycles, knapsack problem.Branch and Bound: FIFO Branch-and-Bound, LC Branch-and-Bound, 0/1 Knapsack problem, Traveling salesperson problem.
### Design and analysis of algorithms Unit Five
#### NP-Hard and NP-Complete
NP-Hard and NP-Complete problems: Basic concepts, Cook’s Theorem.String Matching: Introduction, String Matching-Meaning and Application, NaÏve String Matching Algorithm, Rabin-Karp Algorithm, Knuth-Morris-Pratt Automata, Tries, Suffix Tree.
### Design and analysis of algorithms Course Objectives
Toprovideanintroductiontoformalismstounderstand,analyzeanddenote time complexities ofalgorithmsTo introduce the different algorithmic approaches for problem solving through numerous exampleproblemsToprovidesometheoreticalgroundingintermsoffindingthelowerbounds of algorithms and theNP-completeness
### Design and analysis of algorithms Course Outcomes
DescribeasymptoticnotationusedfordenotingperformanceofalgorithmsAnalyze the performance of a given algorithm and denote its time complexityusingtheasymptoticnotationforrecursiveandnon-recursive algorithmsList and describe various algorithmicapproachesSolveproblemsusingdivideandconquer,greedy,dynamicprogramming, backtracking and branch and bound algorithmicapproachesApply graph search algorithms to real worldproblemsDemonstrate an understanding of NP-Completeness theory and lower boundtheory
### Design and analysis of algorithms Text Books
1)Ellis Horowitz, Sartaj Sahni,Sanguthevar Rajasekaran, “ Fundamentals of Computer Algorithms”, 2ndEdition, Universities Press.2)Harsh Bhasin, “ Algorithms Design & Analysis”, Oxford University Press.
### Design and analysis of algorithms Reference Books
1)Horowitz E. Sahani S: “Fundamentals of Computer Algorithms”, 2ndEdition, Galgotia Piblications, 2008.2)S. Sridhar, “Design and Analysis of Algorithms”, Oxford University Press.
### Scoring Marks in Design and analysis of algorithms
Scoring a really good grade in Design and analysis of algorithms is a difficult task indeed and CynoHub is here to help!. Please watch the video below and find out how to get 1st rank in your B.tech examinations . This video will also inform students on how to score high grades in Design and analysis of algorithms. There are a lot of reasons for getting a bad score in your Design and analysis of algorithms exam and this video will help you rectify your mistakes and help you improve your grades.
Information about JNTU-K B.Tech R19 Design and analysis of algorithms was provided in detail in this article. To know more about the syllabus of other Engineering Subjects of JNTUH check out the official CynoHub application. Click below to download the CynoHub application.
### Get Complete Lecture Notes for Design and analysis of algorithms on Cynohub APP
{{startingCount}}
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{{trans(`You have no camera installed on your device or the device is currently being used by other application`)}} | 1,162 | 5,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-21 | longest | en | 0.877226 |
https://howmanyis.com/speed/39-mph-in-kn/14002-2-miles-per-hour-in-knots | 1,548,110,556,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583814455.32/warc/CC-MAIN-20190121213506-20190121235356-00028.warc.gz | 534,445,809 | 6,200 | How many is
Conversion between units of measurement
You can easily convert 2 miles per hour into knots using each unit definition:
Miles per hour
mile/hr = 0.44704 m / s
Knots
nauticalmile / hr = 0.51444444 m / s
With this information, you can calculate the quantity of knots 2 miles per hour is equal to.
## ¿How many kn are there in 2 mph?
In 2 mph there are 1.7379525 kn.
Which is the same to say that 2 miles per hour is 1.7379525 knots.
Two miles per hour equals to one knots. *Approximation
### ¿What is the inverse calculation between 1 knot and 2 miles per hour?
Performing the inverse calculation of the relationship between units, we obtain that 1 knot is 0.57538972 times 2 miles per hour.
A knot is zero times two miles per hour. *Approximation | 206 | 766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-04 | latest | en | 0.861289 |
https://aptitudecrack.com/programming/c-programming/expressions-general-questions/questions-answers | 1,709,152,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00518.warc.gz | 102,262,721 | 12,122 | # Q1: Which of the following is the correct order if calling functions in the below code?a = f1(23, 14) * f2(12/4) + f3();
A f1, f2, f3
B f1, f2, f3
C Order may vary from compiler to compiler
D None of above
A / + * -
B * - / +
C + - / *
D / * + -
A 2134
B 1234
C 4321
D 3214
# Q4: Which of the following is the correct usage of conditional operators used in C?
A a>b ? c=30 : c=40;
B a>b ? c=30;
C max = a>b ? a>c?a:c:b>c?b:c
D return (a>b)?(a:b)
A
* / % + - =
B = * / % + -
C / * % - + =
D / * % - + =
# Q6: Which of the following are unary operators in C? 1. ! 2. sizeof 3. ~ 4. &&
A 1, 2
B 1,3
C 2,4
D 1,2,4
## For help Students Orientation Mcqs Questions
One stop destination for examination, preparation, recruitment, and more. Specially designed online test to solve all your preparation worries. Go wherever you want to and practice whenever you want, using the online test platform. | 319 | 901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.755515 |
https://www.tutorialspoint.com/program-to-find-number-of-items-left-after-selling-n-items-in-python | 1,685,355,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00584.warc.gz | 1,144,271,656 | 9,944 | # Program to find number of items left after selling n items in python
Suppose we have a list of numbers called items and another value n. A salesman has items in a bag with random IDs. The salesman can delete as many as n items from the bag. We have to find the minimum number of different IDs in the bag after n removals.
So, if the input is like items = [2, 2, 6, 6] n = 2, then the output will be 1 as we he can sell two items with ID 2 or ID 6, then only items with single target will be there.
To solve this, we will follow these steps:
• c := frequency of each element present in items
• ans := size of c
• freq := sort the list of all frequencies in c
• i := 0
• while i < size of freq, do
• if freq[i] <= n, then
• n := n - freq[i]
• ans := ans - 1
• otherwise,
• return ans
• i := i + 1
• return 0
Let us see the following implementation to get better understanding:
## Example
Live Demo
from collections import Counter
class Solution:
def solve(self, items, n):
c = Counter(items)
ans = len(c)
freq = sorted(c.values())
i = 0
while i < len(freq):
if freq[i] <= n:
n -= freq[i]
ans -= 1
else:
return ans
i += 1
return 0
ob = Solution()
items = [2, 2, 6, 6]
n = 2
print(ob.solve(items, n))
## Input
[2, 2, 6, 6], 2
## Output
1 | 377 | 1,249 | {"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} | 3.234375 | 3 | CC-MAIN-2023-23 | latest | en | 0.830175 |
https://www.shaalaa.com/concept-notes/integrals-some-particular-functions_3127?v=6948 | 1,579,673,282,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606696.26/warc/CC-MAIN-20200122042145-20200122071145-00162.warc.gz | 1,070,326,566 | 12,740 | Share
# Integrals of Some Particular Functions
#### formula
1) int (dx)/(x^2 - a^2) = 1/(2a) log |(x - a)/(x + a)| + C
2) int (dx)/(a^2 - x^2) = 1/(2a) log |(a + x)/(a - x)| + C
3) int (dx)/(x^2 - a^2) = 1/a tan^(-1) (x/a) + C
4) int (dx)/sqrt (x^2 - a^2) = log |x + sqrt (x^2-a^2)| + C
5) int (dx)/sqrt (a^2 - x^2) = sin ^(-1) (x/a) +C
6) int (dx)/sqrt (x^2 + a^2) = log |x + sqrt (x^2 + a^2)| + C
#### Video Tutorials
We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.
Series 1
Series 2
### Shaalaa.com
Integrals part 20 (Integrals of some particular functions) [00:10:53]
S
1
0%
S | 287 | 673 | {"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} | 3.296875 | 3 | CC-MAIN-2020-05 | latest | en | 0.511894 |
https://web2.0calc.com/questions/please-help_42442 | 1,621,246,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00064.warc.gz | 595,247,463 | 5,390 | +0
0
68
1
A hotel has 260 rooms. Some are singles, and some are doubles. Thesingles cost 35 and the doubles cost 60. Because of a math teachers' convention, all of the hotel rooms are occupied. The sales for this night are
14,000. How many double rooms does the hotel have?
Feb 15, 2021
#1
+530
0
s + d = 260
35s + 60d = 14000
35s = 14000 - 60d
s = 400 - 12d/7
s + d = 260
(400 - 12d/7) + d = 260
-12d/7 + d = -140
-12d + 7d = -980
-5d = -980
d = 196
s + d = 260
s + (196) = 260
s = 64
The hotel has 196 double rooms :)
Feb 15, 2021 | 223 | 551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.903831 |
https://stat.ethz.ch/pipermail/r-sig-geo/2009-February/004988.html | 1,529,947,407,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00349.warc.gz | 706,795,712 | 2,294 | # [R-sig-Geo] Calculating descriptive stats from many maps
Tomislav Hengl T.Hengl at uva.nl
Mon Feb 9 17:33:51 CET 2009
```Dear Rainer,
This is of course possible in R, and can be done in several ways:
1) for example, you can derive the average value using the rowSums function:
> maps\$Nsum <- rowSums(maps at data, na.rm=T, dims=1)
> maps\$avg <- maps\$Nsum/(length(names(meuse.grid at data))-1)
You could also loop the sd, mean and quantile function over a range of cells:
> for(i in length(names(maps at data))) {
> maps at data\$sd[i] <- sd(maps at data[i,])
> maps at data\$mean[i] <- mean(maps at data[i,])
...
> }
This could take a lot of time!
2) if your maps are rather large, try also using the SAGA function:
> rsaga.get.usage(lib = "geostatistics_grid", module=5)
SAGA CMD 2.0.3
library path: C:/Progra~1/saga_vc/modules
library name: geostatistics_grid
module name : Statistics for Grids
This is probably the fastest method you can use.
HTH
T. Hengl
> -----Original Message-----
> From: r-sig-geo-bounces at stat.math.ethz.ch [mailto:r-sig-geo-bounces at stat.math.ethz.ch] On Behalf
> Of Rainer M Krug
> Sent: Monday, February 09, 2009 4:58 PM
> To: R-sig-Geo at stat.math.ethz.ch
> Subject: [R-sig-Geo] Calculating descriptive stats from many maps
>
> Hi
>
> I have 25000 maps, generated by simulation predictions, covering the
> same area, and would like to calculate some descriptive stats, like
> mean, standard deviation, median, quartiles of all cells, to create a
> "variability map".
>
> Is there an easy way of doing this in R?
>
> Thanks,
>
> Rainer
>
> --
> Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
> Biology, UCT), Dipl. Phys. (Germany)
>
> Centre of Excellence for Invasion Biology
> Faculty of Science
> Natural Sciences Building
> Private Bag X1
> University of Stellenbosch
> Matieland 7602
> South Africa
>
> _______________________________________________
> R-sig-Geo mailing list
> R-sig-Geo at stat.math.ethz.ch
> https://stat.ethz.ch/mailman/listinfo/r-sig-geo
``` | 607 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.708017 |
http://www.coursehero.com/file/5268367/Lab10Problems/ | 1,369,390,870,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704517601/warc/CC-MAIN-20130516114157-00093-ip-10-60-113-184.ec2.internal.warc.gz | 400,441,510 | 13,300 | # Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.
1 Page
### Lab_10_Problems
Course: CPRE 185, Fall 2009
School: Iowa State
Rating:
Word Count: 266
#### Document Preview
Boolean Problems logic program to review lab 9 o Write a program that will use logical operations to model the following truth table. A B C O 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 0 0 1 1 0 0 Play with structures: o Create a structure with three characters and three integers. o Use the sizeof() function to determine the size of the structure. o Experiment with the order of the characters and the...
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Boolean Problems logic program to review lab 9 o Write a program that will use logical operations to model the following truth table. A B C O 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 0 0 1 1 0 0 Play with structures: o Create a structure with three characters and three integers. o Use the sizeof() function to determine the size of the structure. o Experiment with the order of the characters and the integers in the structure. Give one example for ordering the variables such that the structure has the smallest possible size. Give one example for ordering the variables such that the structure has the largest possible size. Form a hypothesis of why this is the case. Using the card structure and your prelab program (or the lab example 3 program) In o main: ask the user to pick two cards between 0 and DECKSIZE-1 o Write a compare function that takes two cards and tells you which one is bigger: Declare the function like this int compareTo(card a, card b){ ...} Output 1 when card a is greater than card b 0 when both cards are worth the same (e.g., 9 of clubs and 9 of clubs) -1 when card a is less than card b o In main: print the cards and output the result of the comparison o e.g., 2 of h...
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Berkeley - ASTRO - 00177408
-55.955000 -42.915000 148.22206 47.985281 -42.915000 -34.765000 203.50748 60.794655 -34.765000 -20.095000 138.50015 45.039814 -20.095000 -7.0550000 159.26452
Berkeley - ASTRO - 00177408
-55.955000 -42.915000 148.22206 47.985281 -42.915000 -34.765000 203.50748 60.794655 -34.765000 -20.095000 138.50015 45.039814 -20.095000 -7.0550000 159.26452
Berkeley - ASTRO - 00177408
# Time [days] Mag Magerr Band Uplim Ref 0.00120 19.5 -0.2 V yes GCN4516 0.00138 18 -0.2 V yes GCN4509 0.00377 20.4 -0.2 B yes GCN4516 0.0
Berkeley - ASTRO - 00177408
360.288 -0.2 0 no GCN4510 1139.62 19.1 0.5 Rc no 1735.78 18.1 0.2 none yes 2039.9 19.2 0.4 Rc no 2879.71 19.1 0
Berkeley - ASTRO - 00177408
360.288 5.139853e+03 0.000000e+00 no GCN4510 1139.62 9.793788e-05 4.510206e-05 Rc no 1735.78 2.460088e-04 4.531650e-05 none yes 2039.9 8.932041e-05 3.290686e-05
Berkeley - ASTRO - 00177408
1735.78 2.460088e-04 4.531650e-05 none yes 2879.71 9.793788e-05 1.804082e-05 Rc yes 42120 3.555910e-04 6.550228e-05 R yes 48960.3 1.073868e-05 1.978137e-06
Berkeley - ASTRO - 00177408
11.410000 34.230003 4.8801518 0.15476824 25.162871 34.230003 55.420002 -7.1583672 3.4018788 327.59974 55.420002 84.760002 15.829904 -2.3030588 299.73961
Berkeley - ASTRO - 00177408
11.4100 34.2300 -0.697075 0.799869 34.2300 55.4200 6.68627 1.05591 55.4200 84.7600 -3.93089 0.648640 84.7600 141.810 9.20444 1.45795 141.810 146.700 27.497
Berkeley - ASTRO - 00177408
chi^2/nu= 277.78127 / 193The fit is rejectable at 99.993871 % Confidence -42.9150 -34.7650 201.17938 -34.7650 -20.0950 219.12440 -20.0950 -7.05500 233.67896 -7.05500 -3.7950
Berkeley - ASTRO - 00177408
Berkeley - ASTRO - 00177408
120.556 121.049 45.8159 10.81121.049 121.336 39.0268 12.7458121.336 121.994 48.5438 9.27936121.994 122.412 49.6199 11.6955122.412 122.716 73.6888 17.0173122.716 123.857 46.1726 6.85523123.857 123.997 80.005 26.129123.997 124.293 42.361 13.6598
Berkeley - ASTRO - 00177408
Source Contamination: 5.26E-08 +/- 1.5E-08 cts/s
Berkeley - ASTRO - 00177408
#ra dec hmag dhmag053.962761 17.109814 13.829 0.025053.944143 17.116085 16.804 0.137053.989731 17.119774 16.009 0.077053.955925 17.112076 13.050 0.024054.056998 17.110207 12.890 0.025054.025218 17.114105 15.171 0.044054.046284 17.102865
Berkeley - ASTRO - 00177408
;instrument XRT;exposure 64937.415;xunit kev;bintype counts 0.0000000 0.0049999999 13.416170 1.00000 0.0049999999 0.0099999998 13.464069 1.00000 0.0099999998 0.015000000 13.511968 1.0
Berkeley - ASTRO - 00177408
;instrument XRT;exposure 269.05439;xunit kev;bintype counts 0.0000000 0.0049999999 14.516683 1.00000 0.0049999999 0.0099999998 14.568504 1.00000 0.0099999998 0.015000000 14.620325 1.0
Berkeley - ASTRO - 00177408
#ra dec rmag drmag54.22221817.13006516.2050.00553.98003217.11030819.2820.07054.22023517.10499716.2420.00554.26303917.10549519.6410.09854.28926317.10327016.6720.00754.28071717.10326216.8100.00853.83690717.09809516.8320.008
Berkeley - ASTRO - 00177408
chi^2/nu= 90.229770 / 328.000The fit is rejectable at 2.1718713e-40 % Confidence#index t1 t2 fade_index delta_mag_pk hindex dhindex rate1 drate1 rate2 drate2 logr dlogr 0 0.1206 0.2380 -3.01 0.0 0.17 0.08 1.15E+0
Berkeley - ASTRO - 00177408
# t1 t2 hardness error 0.12055600 0.12133600 -0.14044738 0.19090244 0.12133600 0.12199400 0.082234260 0.19072434 0.12199400 0.12271600 0.26708790 0.15921771 0.12271600 0.12385700
Berkeley - ASTRO - 00177408
output00177408000_999/sw00177408000xpcw2po_cl.evtoutput00177408001_999/sw00177408001xpcw2po_cl.evtoutput00177408002_999/sw00177408002xpcw2po_cl.evtoutput00177408003_999/sw00177408003xpcw2po_cl.evtoutput00177408004_999/sw00177408004xpcw2po_cl.evt
Berkeley - ASTRO - 00177408
# t1 t2 dt rad_min rad_max cts err scl bg bg_rat wt 0.120556 0.120813 0.000257 0. 16. 10.60 3.52 0.867841 5.000000 0.279570 1 0.120813 0.121049 0.000237 0. 16. 9.00
Berkeley - ASTRO - 00177408
tmin 2.0695142e-05tmin 0.00011504599
Berkeley - ASTRO - 00177408
tmin 32.099833tmin 181.93968 364.16423 2064.0582 0.20484306 0.033045668 5 2064.0582 46869.069 0.12453600 0.012811805 9tmax 46869.069
Berkeley - ASTRO - 00177408
# t1 t2 dt rad_min rad_max cts err scl bg bg_rat wt 0.120556 0.120813 0.000257 0. 16. 10.60 3.52 0.867841 5.000000 0.279570 1 0.120813 0.121049 0.000237 0. 16. 9.00
Berkeley - ASTRO - 00177408
# tmin tmax 0.254029 468.41556 [ksec];instrument XRT;exposure 60143.753;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.00000
Berkeley - ASTRO - 00177408
# tmin tmax 0.254029 468.41556 [ksec];instrument XRT;exposure 60143.753;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.00000
Berkeley - ASTRO - 00177408
# tmin tmax 0.12055600 5.38475 [ksec];instrument XRT;exposure 266.08283;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.00000
Berkeley - ASTRO - 00177408
# tmin tmax 0.12055600 5.38475 [ksec];instrument XRT;exposure 266.08283;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.00000
Berkeley - ASTRO - 00177408
Wavdetect Sources with S/N>3: # ra dec err ["] signif counts steady? -log10(Prob_steady) 054.03492117.3458570.16673.8725.0 0-323.0 154.10150117.3244240.51014.141.8 1-0.7 254.01671417.2622120.65812.338.3 1-0.1 354.155
Berkeley - ASTRO - 00177408
output00177408000_999/sw00177408000xwtw2po_cl.evtoutput00177408001_999/sw00177408001xwtw2po_cl.evtoutput00177408002_999/sw00177408002xwtw2po_cl.evtoutput00177408003_999/sw00177408003xwtw2po_cl.evtoutput00177408004_999/sw00177408004xwtw2po_cl.evt
Berkeley - ASTRO - 00177408
SIMPLE = T / file does conform to FITS standardBITPIX = 8 / number of bits per data pixelNAXIS = 0 / number of data axesEXTEND = T / FITS dataset may contain extensio
Berkeley - ASTRO - 00177408
# Ep dEp lprob lEiso dlEiso67.040 0.054 3.49e-05 121.421 0.07467.097 0.061 3.07e-04 121.521 0.08467.163 0.070 5.88e-04 121.521 0.08467.238 0.081 8.49e-04 121.521 0.08467.325 0.092 1.10e-03 121.521 0.08467.423 0.106 1.29e-03 121.521 0.08467.537
Berkeley - ASTRO - 00177408
# Ep lEiso37.400 121.94840.123 121.88245.004 121.87446.499 121.58146.920 121.42847.877 121.33148.473 121.58348.868 121.63749.323 121.34449.341 121.34849.662 121.44650.219 121.40150.425 121.54350.808 121.47751.066 121.54551.173 121.583
Berkeley - ASTRO - 00177408
# Ep dEp lprob lNiso dlNiso67.040 0.054 3.49e-05 137.769 0.38167.097 0.061 3.06e-04 137.769 0.38067.163 0.070 5.87e-04 137.769 0.38067.238 0.081 8.65e-04 137.769 0.38067.325 0.092 1.10e-03 137.769 0.38067.423 0.106 1.29e-03 137.769 0.38067.537
Berkeley - ASTRO - 00177408
# Ep lNiso37.362 137.60340.096 137.25245.016 138.22146.520 138.10846.941 137.70147.899 137.88448.495 138.38548.881 138.62249.337 137.33249.355 137.35049.676 137.78350.234 137.58650.440 138.20350.817 137.91451.076 137.64251.182 137.219
Berkeley - ASTRO - 00177408
# x=Log_e(Beaming Fraction) y=Log_e(Egam/10^52 erg) z=Log_e(Tjet/days)# mean(x)= -4.6913 xdown= -5.0763 xup= -4.3622# mean(y)= -2.8349 ydown= -3.4214 yup= -2.2964# mean(z)= 1.8541 zdown= 1.3195 zup= 2.2750-5.9761 -3.7625 0.1152-5.8878 -3.7402
Berkeley - ASTRO - 00177408
Ep=55.60 Chi/nu= 51.27/57 (0.899)
Berkeley - ASTRO - 00177408
#file=swb15-350lc.txt dt=1.0 tstart=-10.095 tstop=113.025#t90 dt90 t50 dt50 rt90 drt90 rt50 drt50 rt45 drt45 tav dtav tmax dtmax trise dtrise tfall dtfall cts cts_err pk_rate dpk_rate band 109.000 1.932 86.000 2.765 58.000
Berkeley - ASTRO - 00177408
# S/N T1 T2 T90 T50# Estimated T100 Interval: -13.065 150.555 T90= 136.350 30.5 80.085 107.625 23.220 8.910 17.0 -7.395 19.335 23.490 9.180 5.5 124.095 150.555
Berkeley - ASTRO - 00177408
;instrument XRT;exposure 61149.182;xunit kev;bintype counts 0.0000000 0.0049999999 13.666843 1.00000 0.0049999999 0.0099999998 13.715614 1.00000 0.0099999998 0.015000000 13.764385 1.0
Berkeley - ASTRO - 00177408
# tmin tmax 10.0000 468.416 [ksec];instrument XRT;exposure 56383.773;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.000000 0
Berkeley - ASTRO - 00177408
# tmin tmax 10.0000 468.416 [ksec];instrument XRT;exposure 56383.773;xunit kev;bintype counts0.000000 0.010000 0.000000 0.0000000.010000 0.020000 0.000000 0.0000000.020000 0.030000 0.000000 0.0000000.030000 0.040000 0.000000 0
Berkeley - TMP - 00177408
Power-Law Model FitNorm@15keV 1.1934e-02 (1.0364e-02 1.3625e-02)alpha -1.7375 (-1.8620 -1.6154)Energy Fluence (15-350 keV) 2.5948e-06 (2.4156e-06 2.7852e-06) erg cm^-2Eiso (1-10^4 keV, host-frame) 1.6697e+53 (1.5405e+53 1.8461e+53) ergChi/nu=
Iowa State - DATA - 1336
Palmer Constitution 2008PreambleWe, the members of Palmer House do establish and adopt this constitution in order to form an efficient house, ensure equal representation of each resident, and secure an environment that stimulates intellectual, soci
Drexel - MK - 489
11 July 2007By: Lucian Dorneanu, Science EditorNanomachines Powered by BacteriaMicroscopic organisms can move tiny structures when stimulated with UV lightA new discovery in the field of nanotechnology could produce the smallest "engines" in th
Rose-Hulman - ME - 422
FORMULA SHEET For 1-D heat transfer problems, the general weak form is dT dT Ak dx = dx dx T AgdxFor a 2-node conduction element, when the area, conductivity, density, and internal heat generation are constant we have K= kA Le 1 -1 -1 1 2 1 1 2 1 1
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: syllabus.dvi %Pages: 4 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMTI10 CMMI8 CMBX10 %EndComments %DVIPSWebPage: (www.radicaleye.com) %
Maryland - ENEE - 646
ENEE 646: Digital Computer DesignFall 2004 Handout #1Course Information and PolicyRoom:CHE 2108 TTh 2:00p.m. - 3:15p.m. http:/www.ece.umd.edu/class/enee646 Donald Yeung 1327 A. V. Williams (301) 405-3649 yeung@eng.umd.edu http:/www.ece.umd.edu
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: ps1.dvi %Pages: 3 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMBX10 CMSY10 CMTI10 CMTT10 %EndComments %DVIPSWebPage: (www.radicaleye.com
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: p3.dvi %Pages: 10 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMTT10 CMSY10 CMMI10 CMBX10 CMTI10 %EndComments %DVIPSWebPage: (www.radical
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: ps2.dvi %Pages: 3 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMR8 %EndComments %DVIPSWebPage: (www.radicaleye.com) %DVIPSCommandLine: dv
Maryland - ENEE - 646
ENEE 646: Digital Computer DesignFall 2004 Handout #11Problem Set # 2Due: October 12Pipelining and ILPProblem 1Consider the following 6-stage pipline:F D X1 X2 M WInstruction fetch Instruction decode, register read First cycle of ALU, b
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: review.dvi %Pages: 4 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMTI10 CMTT10 CMBX10 CMSY10 %EndComments %DVIPSWebPage: (www.radicaleye.
Maryland - ENEE - 646
ENEE 646: Digital Computer DesignFall 2004 Handout #13Midterm Review 1 Time and LocationThe midterm will be given during normal class hours, 2:00p.m.-3:15p.m., on Thursday October 14th, in the normal meeting place, CHE 2108.2FormatThe midt
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: p4.dvi %Pages: 10 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMTT10 CMTI10 %EndComments %DVIPSWebPage: (www.radicaleye.com) %DVIPSComman
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: midterm-sol.dvi %Pages: 5 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMBX10 CMMI10 CMMI8 CMSY10 CMSY8 %EndComments %DVIPSWebPage: (www.r
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ENEE 646: Digital Computer DesignFall 2004 Handout #17Midterm Solutions11.1Early Branch Resolution(10 points):(see Figures 1, 2, 3, and 4 for pipeline diagrams).1.2(10 points):The forwarding path from register G can be removed. Since
Maryland - ENEE - 646
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: ps3.dvi %Pages: 4 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: CMBX12 CMR10 CMTT10 CMSY10 %EndComments %DVIPSWebPage: (www.radicaleye.com) %DVIPSComman
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ENEE 646: Digital Computer DesignFall 2004 Handout #20Problem Set # 3Due: November 23Complex Pipelining and CachingProblem 1H&P 3.10Problem 2In this problem, we will look at how a common vector loop runs on two processors, one that imple
Maryland - ENEE - 646
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Maryland - ENEE - 646
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Maryland - ENEE - 646
ENEE 646: Digital Computer DesignFall 2004 Handout #24Problem Set # 4Due: Dec 9Memory Performance Optimizations and Virtual MemoryProblem 1H&P 5.8Problem 2H&P 5.17Problem 3H&P 5.181 | 6,804 | 16,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2013-20 | latest | en | 0.854364 |
https://study.com/academy/topic/graphing-factoring-quadratic-equations-homeschool-curriculum.html | 1,563,539,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526237.47/warc/CC-MAIN-20190719115720-20190719141720-00124.warc.gz | 552,190,459 | 26,127 | # Ch 17: Graphing & Factoring Quadratic Equations: Homeschool Curriculum
The Graphing and Factoring Quadratic Equations unit of this High School Algebra I Homeschool course is designed to help homeschooled students learn to graph and factor quadratic equations. Parents can use the short videos to introduce topics, break up lessons and keep students engaged.
## Who's it for?
This unit of our High School Algebra I Homeschool course will benefit any student who is trying to learn about graphing and factoring quadratic equations. There is no faster or easier way to learn about algebra. Among those who would benefit are:
• Students who require an efficient, self-paced course of study to learn to solve polynomial problems with non-1 leading coefficients.
• Homeschool parents looking to spend less time preparing lessons and more time teaching.
• Homeschool parents who need an algebra curriculum that appeals to multiple learning types (visual or auditory).
• Gifted students and students with learning differences.
## How it works:
• Students watch a short, fun video lesson that covers a specific unit topic.
• Students and parents can refer to the video transcripts to reinforce learning.
• Short quizzes and a graphing and factoring quadratic equations unit exam confirm understanding or identify any topics that require review.
## Graphing & Factoring Quadratic Equations Unit Objectives:
• Learn how to solve quadratic equations with three terms.
• Explain the use of graphs and tables in the real world.
• Discover how to complete the square.
• Learn how to use line graphs and scatterplots.
• Demonstrate how to factor quadratic equations.
12 Lessons in Chapter 17: Graphing & Factoring Quadratic Equations: Homeschool Curriculum
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 473 | 2,363 | {"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} | 3.015625 | 3 | CC-MAIN-2019-30 | latest | en | 0.908545 |
https://software.sandia.gov/trac/fast/changeset/2572 | 1,448,560,033,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447769.81/warc/CC-MAIN-20151124205407-00336-ip-10-71-132-137.ec2.internal.warc.gz | 850,809,518 | 5,688 | # Changeset 2572
Ignore:
Timestamp:
01/30/11 09:51:32 (5 years ago)
Message:
Further updates
Location:
hudson/performance
Files:
2 edited
Unmodified
Added
Removed
• ## hudson/performance/figure.py
r2571 import matplotlib.pyplot as plt from matplotlib.backends.backend_pdf import PdfPages import os import os.path def plot_table(table): def plot_table(table, filename): #print table N = len(table)-1 pp = PdfPages(filename) #print "HERE", N, int(math.ceil(N/3.)), 3 fig, axs = plt.subplots(nrows=int(math.ceil(N/3.)), ncols=3, sharex=True) X = 0 Y = 0 for i in range(1,N+1): #print X,Y,axs ax = axs[X,Y] ax.set_title(table[i][0]) ax.plot(table[0][1:],table[i][1:]) ax.tick_params(direction='out') #print table[0][1:] #ax.axis(table[0][1:]) Y += 1 if Y == 3: if True: X = 1 for i in range(1,N+1): plt.subplot(5, 1, X) plt.plot(table[0][1:],table[i][1:]) plt.title(table[i][0], fontsize=8) plt.tick_params(labelsize=8) plt.subplots_adjust(wspace=0.5, hspace=0.5) ymin, ymax = plt.ylim() plt.ylim( (0, 1.05*ymax) ) X += 1 Y = 0 if X > 5: X = 0 plt.savefig(pp, format='pdf') plt.clf() else: X = 0 Y = 0 for i in range(1,N+1): if X == 0 and Y == 0: fig, axs = plt.subplots(nrows=4, ncols=2, sharex=True) ax = axs[X,Y] ax.set_title(table[i][0], fontsize=8) ax.plot(table[0][1:],table[i][1:]) ax.tick_params(labelsize=8) plt.subplots_adjust(wspace=0.5, hspace=0.5) Y += 1 if Y == 2: X += 1 Y = 0 if X == 4: X = 0 pp.savefig(fig) pp.close() def create_figure(csvfile, figfile): #print os.getcwd() table = read_table(csvfile) plot_table(table) print "Creating image file %s" % figfile plt.savefig(figfile) plot_table(table, figfile) if __name__ == '__main__': for file in sys.argv[1:]: create_figure(file, os.path.splitext(file)[0]+'.png') create_figure(file, os.path.splitext(file)[0]+'.pdf')
• ## hudson/performance/process
r2571 index.add(data[1]) test = '.'.join(data[3:6]) #test = '.'.join(data[2:6]) test = string.replace(test, ' ', '_') if not test in build[data[0]]:
Note: See TracChangeset for help on using the changeset viewer. | 673 | 2,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2015-48 | longest | en | 0.281816 |
http://mitpress.mit.edu/sicp/full-text/sicp/book/node131.html | 1,418,947,588,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768050.31/warc/CC-MAIN-20141217075248-00069-ip-10-231-17-201.ec2.internal.warc.gz | 190,511,995 | 4,904 | Next: Interfacing Compiled Code to Up: Compilation Previous: An Example of Compiled
One of the most common optimizations performed by compilers is the optimization of variable lookup. Our compiler, as we have implemented it so far, generates code that uses the lookup-variable-value operation of the evaluator machine. This searches for a variable by comparing it with each variable that is currently bound, working frame by frame outward through the run-time environment. This search can be expensive if the frames are deeply nested or if there are many variables. For example, consider the problem of looking up the value of x while evaluating the expression (* x y z) in an application of the procedure that is returned by
```(let ((x 3) (y 4))
(lambda (a b c d e)
(let ((y (* a b x))
(z (+ c d x)))
(* x y z))))
```
Since a let expression is just syntactic sugar for a lambda combination, this expression is equivalent to
```((lambda (x y)
(lambda (a b c d e)
((lambda (y z) (* x y z))
(* a b x)
(+ c d x))))
3
4)
```
Each time lookup-variable-value searches for x, it must determine that the symbol x is not eq? to y or z (in the first frame), nor to a, b, c, d, or e (in the second frame). We will assume, for the moment, that our programs do not use define--that variables are bound only with lambda. Because our language is lexically scoped, the run-time environment for any expression will have a structure that parallels the lexical structure of the program in which the expression appears. Thus, the compiler can know, when it analyzes the above expression, that each time the procedure is applied the variable x in (* x y z) will be found two frames out from the current frame and will be the first variable in that frame.
We can exploit this fact by inventing a new kind of variable-lookup operation, lexical-address-lookup, that takes as arguments an environment and a lexical address that consists of two numbers: a frame number, which specifies how many frames to pass over, and a displacement number, which specifies how many variables to pass over in that frame. Lexical-address-lookup will produce the value of the variable stored at that lexical address relative to the current environment. If we add the lexical-address-lookup operation to our machine, we can make the compiler generate code that references variables using this operation, rather than lookup-variable-value. Similarly, our compiled code can use a new lexical-address-set! operation instead of set-variable-value!.
In order to generate such code, the compiler must be able to determine the lexical address of a variable it is about to compile a reference to. The lexical address of a variable in a program depends on where one is in the code. For example, in the following program, the address of x in expression e1 is (2,0)--two frames back and the first variable in the frame. At that point y is at address (0,0) and c is at address (1,2). In expression e2, x is at (1,0), y is at (1,1), and c is at (0,2).
```((lambda (x y)
(lambda (a b c d e)
((lambda (y z) e1)
e2
(+ c d x))))
3
4)
```
One way for the compiler to produce code that uses lexical addressing is to maintain a data structure called a compile-time environment. This keeps track of which variables will be at which positions in which frames in the run-time environment when a particular variable-access operation is executed. The compile-time environment is a list of frames, each containing a list of variables. (There will of course be no values bound to the variables, since values are not computed at compile time.) The compile-time environment becomes an additional argument to compile and is passed along to each code generator. The top-level call to compile uses an empty compile-time environment. When a lambda body is compiled, compile-lambda-body extends the compile-time environment by a frame containing the procedure's parameters, so that the sequence making up the body is compiled with that extended environment. At each point in the compilation, compile-variable and compile-assignment use the compile-time environment in order to generate the appropriate lexical addresses.
Exercises through describe how to complete this sketch of the lexical-addressing strategy in order to incorporate lexical lookup into the compiler. Exercise describes another use for the compile-time environment.
Exercise. Write a procedure lexical-address-lookup that implements the new lookup operation. It should take two arguments--a lexical address and a run-time environment--and return the value of the variable stored at the specified lexical address. Lexical-address-lookup should signal an error if the value of the variable is the symbol *unassigned*. Also write a procedure lexicaladdressset! that implements the operation that changes the value of the variable at a specified lexical address.
Exercise. Modify the compiler to maintain the compile-time environment as described above. That is, add a compile-time-environment argument to compile and the various code generators, and extend it in compile-lambda-body.
Exercise. Write a procedure find-variable that takes as arguments a variable and a compile-time environment and returns the lexical address of the variable with respect to that environment. For example, in the program fragment that is shown above, the compile-time environment during the compilation of expression e1 is ((y z) (a b c d e) (x y)). Find-variable should produce
```(find-variable 'c '((y z) (a b c d e) (x y)))
(1 2)
(find-variable 'x '((y z) (a b c d e) (x y)))
(2 0)
(find-variable 'w '((y z) (a b c d e) (x y)))
not-found
```
Exercise. Using find-variable from exercise , rewrite compile-variable and compile-assignment to output lexical-address instructions. In cases where find-variable returns not-found (that is, where the variable is not in the compile-time environment), you should have the code generators use the evaluator operations, as before, to search for the binding. (The only place a variable that is not found at compile time can be is in the global environment, which is part of the run-time environment but is not part of the compile-time environment. Thus, if you wish, you may have the evaluator operations look directly in the global environment, which can be obtained with the operation (op get-global-environment), instead of having them search the whole run-time environment found in env.) Test the modified compiler on a few simple cases, such as the nested lambda combination at the beginning of this section.
Exercise. We argued in section that internal definitions for block structure should not be considered ``real'' defines. Rather, a procedure body should be interpreted as if the internal variables being defined were installed as ordinary lambda variables initialized to their correct values using set!. Section and exercise showed how to modify the metacircular interpreter to accomplish this by scanning out internal definitions. Modify the compiler to perform the same transformation before it compiles a procedure body.
Exercise. In this section we have focused on the use of the compile-time environment to produce lexical addresses. But there are other uses for compile-time environments. For instance, in exercise we increased the efficiency of compiled code by open-coding primitive procedures. Our implementation treated the names of open-coded procedures as reserved words. If a program were to rebind such a name, the mechanism described in exercise would still open-code it as a primitive, ignoring the new binding. For example, consider the procedure
```(lambda (+ * a b x y)
(+ (* a x) (* b y)))
```
which computes a linear combination of x and y. We might call it with arguments +matrix, *matrix, and four matrices, but the open-coding compiler would still open-code the + and the * in (+ (* a x) (* b y)) as primitive + and *. Modify the open-coding compiler to consult the compile-time environment in order to compile the correct code for expressions involving the names of primitive procedures. (The code will work correctly as long as the program does not define or set! these names.)
Next: Interfacing Compiled Code to Up: Compilation Previous: An Example of Compiled
Ryan Bender
2000-04-17 | 1,743 | 8,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-52 | longest | en | 0.889844 |
http://tibasicdev.wikidot.com/quadreg | 1,524,736,170,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00489.warc.gz | 335,626,198 | 9,857 | Command Summary
Calculates the best fit quadratic through a set of points.
Command Syntax
QuadReg [x-list, y-list, [frequency list], [equation variable]
Press:
1. STAT to access the statistics menu
2. LEFT to access the CALC submenu
3. 5 to select QuadReg, or use arrows
TI-83/84/+/SE
1 byte
The QuadReg command can calculate the best fit quadratic through a set of points. To use it, you must first store the points to two lists: one of the x-coordinates and one of the y-coordinates, ordered so that the nth element of one list matches up with the nth element of the other list. L₁ and L₂ are the default lists to use, and the List Editor (STAT > Edit…) is a useful window for entering the points. You must have at least 3 points, because there are infinitely many quadratics that can go through 2 points or 1 point.
In its simplest form, QuadReg takes no arguments, and calculates a quadratic through the points in L₁ and L₂:
``````:{9,13,21,30,31,31,34→L₁
:{260,320,420,530,560,550,590→L₂
```
On the home screen, or as the last line of a program, this will display the equation of the quadratic: you'll be shown the format, y=ax²+bx+c, and the values of a, b, and c. It will also be stored in the RegEQ variable, but you won't be able to use this variable in a program - accessing it just pastes the equation wherever your cursor was. Finally, the statistical variables a, b, c, and R² will be set as well. This latter variable will be displayed only if "Diagnostic Mode" is turned on (see DiagnosticOn and DiagnosticOff).
You don't have to do the regression on L₁ and L₂, but if you don't you'll have to enter the names of the lists after the command. For example:
``````:{9,13,21,30,31,31,34→FAT
:{260,320,420,530,560,550,590→CALS
```
You can attach frequencies to points, for when a point occurs more than once, by supplying an additional argument - the frequency list. This list does not have to contain integer frequencies. If you add a frequency list, you must supply the names of the x-list and y-list as well, even when they're L₁ and L₂.
Finally, you can enter an equation variable (such as Y₁) after the command, so that the quadratic is stored to this equation automatically. This doesn't require you to supply the names of the lists, but if you do, the equation variable must come last. You can use polar, parametric, or sequential variables as well, but since the quadratic will be in terms of X anyway, this doesn't make much sense.
An example of QuadReg with all the optional arguments:
``````:{9,13,21,30,31,31,34→FAT
:{260,320,420,530,560,550,590→CALS
:{2,1,1,1,2,1,1→FREQ | 698 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-17 | longest | en | 0.899331 |
https://people.revoledu.com/kardi/tutorial/Similarity/index.html | 1,714,057,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00523.warc.gz | 400,712,205 | 4,417 | | Next >
## Similarity Measurement
In this simple tutorial, you will learn the basic knowledge to expand your data type into multivariate (different type of measurement scale, such as nominal, ordinal, and quantitative) data and go beyond 2 dimensional data scale up to N dimensions . Comprehesive example is given at the last part of this tutorial. You also may download the MS Excel companion file of this tutorial here
This knowledge about similarity and dissimilarity is necessary for data mining, pattern recognition, machine intelligent, artificial intelligent and multi-agents system fields. However, the application is not only limited to computer science field. Other fields of natural and social science as well as engineering and statistics have been applied this kind of simple knowledge. Tools such as K means clustering , Discriminant analysis , K-Nearest Neighbors , or Decision Tree and Hierarchical clustering rely heavily on the distance matrix explained in this tutorial.
What is similarity?
What is distance ?
What is the relationship between similarity and dissimilarity?
Why do we need to measure similarity? (Applications)
How do we measure similarity or dissimilarity?
How do we compute dissimilarity or similarity for binary variables ?
Simple Matching Coefficient
Jaccard's Coefficient
Hamming Distance
How do we compute dissimilarity or similarity for nominal / categorical variables?
Assign each value of category as a binary dummy variable
Assign each value of category into several binary dummy variables
How do we compute dissimilarity or similarity for ordinal variables?
Normalized Rank Transformation
Spearman Distance
Footrule Distance
Kendall Distance
Cayley Distance
Hamming Distance for Ordinal Variable
Ulam Distance
How do we compute dissimilarity or similarity for text and string variables ?
How do we compute dissimilarity or similarity for quantitative variables ?
Euclidean Distance
City block (Manhattan) distance
Chebyshev Distance
Minkowski Distance
Canberra distance
Bray Curtis (Sorensen) distance
Angular separation
Correlation coefficient
How do we compute dissimilarity between two groups ( Mahalanobis distance )?
How do we normalize the similarity or dissimilarity?
How do we aggregate mixed type of variables?
Comprehensive example: Distance matrix of Multivariate data
Resources
Preferable reference for this tutorial is
Teknomo, Kardi (2015) Similarity Measurement. http:\people.revoledu.comkardi tutorialSimilarity | 500 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.827712 |
https://www.physicsforums.com/threads/spring-constant-of-each-spring-when-compressed.853870/ | 1,511,318,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806447.28/warc/CC-MAIN-20171122012409-20171122032409-00553.warc.gz | 862,071,476 | 15,917 | # Spring constant of each spring when compressed
1. Jan 24, 2016
### HBurch614
1. The problem statement, all variables and given/known data
A car with a mass of 1,500 kg sits on a suspension system that has four springs. When the mass of the car was originally placed on the springs, they compressed by 10 cm. What is the spring constant for each spring?
2. Relevant equations
k = F/x or k = mg/x
3. The attempt at a solution
k = (1500kg)(10) / .1m
= 150,000
keach spring = 150,000 / 4
= 37,500 n/m
Is this correct??
2. Jan 24, 2016
### Bystander
Don't see anything wrong.
3. Jan 24, 2016
### HBurch614
Thank you, I was unsure! | 199 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-47 | longest | en | 0.927346 |
https://wiki.uni-konstanz.de/ccp4/index.php/Some_ways_to_calculate_the_radiation_dose_that_a_crystal_has_absorbed | 1,709,245,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00274.warc.gz | 617,397,724 | 7,741 | # Some ways to calculate the radiation dose that a crystal has absorbed
This is James Holton's corrected explanation - too good to be buried in the CCP4BB archives:
Subject: Re: Dose in diffraction patterns?
From: James Holton <jmholton@LBL.GOV>
Date: Wed, 6 May 2020 16:04:18 -0700
In general? No.
I believe a few places put "flux" into the header, but as Andreas just mentioned that is only one of the bits of information you need to calculate dose. If all you want is a rough estimate, then the numbers you need are:
f = flux (photons/s)
t = exposure time (s)
w = wavelength (A)
a = beam area (um^2)
The dose (D) to a sample of protein/water/plastic under a given beam will be roughly:
D = f*t*w^2/a/2000 (in Gy)
For example: a crystal under a square 100 um x 100 um beam at 1 A wavelength with flux 1e12 ph/s will get 1 MGy dose in 20 s.
The 2000/w^2 is a fudge factor that fits the true curve of metal-free protein crystals to within 15% for the wavelength range 0.5 < w < 3. The error induced by not knowing if the beam was round or square and just multiplying together the width and height to get the area (a), is 21%. The error from not realizing you had 100 mM uranium in your sample is about a factor of two at 1 A. Smaller concentrations and lighter atoms have less impact on accuracy.
If you don't know the flux, or beam size, you can try looking them up at http://biosync.sbkb.org/ . I scraped these for my little dose calculator here:
bl831.als.lbl.gov/xtallife.html
Some of the biosync numbers are more accurate than others, however, depending on how often beamline scientists remember to update the site, and how well they know themselves. And attenuation is not always written into the header either.
In a pinch, you can estimate the flux by the total number of photons on the image (P). This is assuming that you know the sample thickness (L) in microns. You must also assume that the total scattering cross section of the atoms in the sample is close to that of oxygen (0.2 cm^2/g), that the sample density is 1.2 g/cm^3 and that about 50% of the scattered photons reach the detector. None of these are terrible assumptions. The equation then becomes:
f = P/t/L/1.2e-5
Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux and t=exposure (as above).
Getting P from a pixel array is easy: you just add up all the pixel values. From a CCD you want to be careful to subtract the baseline value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for ADSC Q315 (swbin) and 1.8 for Q315r (hwbin). A few considerations, yes, but it can be a good sanity check.
For example, if you see an average pixel value of 20 photons on a Pilatus 6M, then that is P=120e6 photons. If that was a t=0.1 s
exposure from a sample 100 microns thick, then the beamline flux was about 1e12 photons/s. Note that this is the flux after any attenuation, not before.
Oh, and if you want a reference for that 2000 ph/um^2 = 1 Gy rule, it is
here:
https://doi.org/10.1107/S0909049509004361
And, of course, if you are lucky enough to have accurate flux, size and shape information for the beam and sample, plus chemical composition the most accurate dose you'll get from raddose-3D: https://www.raddo.se/
-James Holton | 930 | 3,327 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.927153 |
https://mafiadoc.com/icma-2011-conference-proceeding_5c378d2b097c476d298b45ac.html | 1,590,945,530,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00337.warc.gz | 423,198,043 | 16,439 | ## ICMA 2011 Conference Proceeding
to calculate the values of the joint angles with the numerical values of inverse matrix. ..... it means that eighty percent of the interval L3 and L7 has intersection. ... REFERENCES. [1] M. A. Ali, H. A. Park, and C. S. G. Lee, "Closed-Form Inverse.
Proceedings of the 2011 IEEE International Conference on Mechatronics and Automation August 7 - 10, Beijing, China
Optimum Inverse Kinematic Method for a 12 DOF Manipulator Ali Pyambri Paramani BSc Student Sharif University of Technology ,International Campus ,Iran [email protected] I. INTRODUCTION The area of inverse kinematics of robots, mainly manipulators, has been widely researched, and several solutions exist. The solutions provided by analytical methods are specific to a particular robot configuration and are not applicable to other robots. Most researchers resort to iterative methods for inverse kinematics using the Jacobian matrix to avoid the difficulty of finding a closed-form joint solution. Since a closed-form joint solution, if available, has many advantages over iterative methods [1]. A humanoid robot is a multi-jointed mechanism that mechanically emulates a human’s functions, movements and activities. It can be considered as a biped robot with an upper main body, linking two arms, a neck and a head, or as a combination of multiple manipulators, which are themselves linked together through waist and neck joints to emulate a human’s functions Most researchers often use iterative methods for controlling humanoid robots [2],[3],[4]. One of the iterative methods makes use of the Jacobian matrix [5]. Singularity, redundancy, and computational complexity are the main drawbacks of using the inverse-Jacobian-matrix approach . Another commonly adopted method for inverse kinematics is the geometric method [6], [7]. However, the geometric method requires geometric intuition in solving the joint solution of a manipulator, and it may become more difficult to obtain the joint solution when more than four or five joints are involved. Another method, called the inversetransform technique, was presented by Paul et al [8] to obtain the inverse kinematic joint solution of a 6-DOF robot manipulator. Cui et al [9] derived a closed-form joint solution for a 6-DOF humanoid robot arm but only the solution of joint angles within a certain range was considered and the singularities were not discussed.Solving kinematic equations is carried out by nonlinear methods in which efforts are made to calculate the values of the joint angles with the numerical values of inverse matrix. For instance, will have twelve equations and six indeterminate equations for the manipulator with 6 degree of freedom. However, only three equations out of nine equations are independent [10] .These equations are nonlinear and non-algebraic and their solutions may be difficult. It is clear that kinematic equations would be much more complex for a general mechanism with six degree of freedom in which all interface parameters are nonzero. Each manipulator is soluble when solver can determine all the sets of joint variables related to the supposed place and orientation by the algorithmic pursuit. The present paper analyzes inverse kinematic of Longshank robot with 12 DOF. “Fig 1”
Abstract - In General , there are two methods to analyse the inverse kinematic of manipulators, one of which can be selected with respect to the conditions and the type of the manipulator. One of the methods is the closed solution which is based on the analytical expressions or forth degree or less polynomial solution in which the calculations are non-repetitive. The other method is the numerical solution. In the numerical solutions , the numbers are repeated and generally it is much slower than the closed solutions. The slowness of this method is so noticeable in such a way that principally there is no interest to use the numerical solutions to solve kinematic equations . The purpose of the present paper is to present a compound method that is made up of the numerical and the closed solutions which does not have the problems of the current methods and can be generalized to similar robots with different degrees of freedom . The method which will be presented is based on the connecting lines between the pairs of links and the conditions under which the size of the connecting lines are created . The inverse matrix equations are not applied in this analysis. However, the positions of all links are expressed based on the geometrical position of the interfaces and the use of the mathematical function. Index –algorithm, condition, interval, angular displacement
Fig. 1 Schematic of LongShank Robot with 12 DOF
2020
Table I
II. ANALYSES METHOD A. Basic rules: Consider link OA and AB, with magnitude |a| , and they are connected together with joint A. distance between point B and O equals to L .And L can change between 0 and 2a , Know consider the case that has position of point O and A in reference coordinate system, and then get the value of all possible positions for point B for different length of L . Solving this equation system:
Longshank Characteristics Degree of freedom joint Goal Rigid Body Coordinate system Link Length | OO′| Triangle EGE ′
(1)
12 Revolute (12R) G Triangle EGE ′ with end effectors G Based on point O a |EE ′|
|EG|
d
g .
|E ′G|
g
connecting lines
If initialized the value of L between 0 and 2a with required accuracy then find all possible position for point B. Outcome: managing position of link AB and OA with magnitude of line L. In other system with more than 3 links ,and replace all two sets of link with one line L, the magnitude and direction of L is equals two vector sum of 2 link.
LEFT SHANK L1 BD L3 OB
OE OD
L4 L5
O`E` O`D`
RIGHT SHANK L6 O`B` L7 B`D'
L8 L9
the Position of point G (Goal Point) from point E is fixed and is equal to g (it moves on circle with radius g and center of E). It should be satisfy the conditions and terms of motion which is defined in next part. Now, with draw lines L and L the workspace of joint E is as follows. With respect to “Fig 2” it is clear that Point E is located on the circles with radius of g and center of G and it moves on the circle
B. analytical analysis : Generally, analytical methods can be used in position analysis to yield results with high degree of accuracy. This accuracy comes with a price in that the methods often become numerically intensive. Complex method involving higher order math, have been developed for position analysis. Using geometrical conceptions, complex methods can be expressed in a simpler way and algorithms can be reduced. Another problem that may rise in solving kinematic equations is the existence of multiple solutions [11].Consider planar manipulator with two Interfaces in which the final operator is located in a specified place and orientation as point A: (AX, AY). For the inverse kinematic, instead of calculation at Point A: (AX, AY), we can perform the calculations at Point B :(0 , BY). In case BX=0, BY=|OA|, it then moved all the obtained solutions to Point B. Application of this method has some advantages. For instance, if a configuration is found on Point B, it is just required to acquire another configuration by symmetrizing the solutions on the X-axis and then the configuration is transferred to the first point. If there is a large number of links, this strategy can increase the speed of calculations and can reduce the number of calculations.
B. Preliminary Analysis joint D Point D is located on a circle with center of E and radius "a" .The workspace of Point D in terms of Point E is expressed based on the following relations. In addition, the limit of Point YD L and YD 0 can be expressed as: D, for XD (2) The Goal is finding exact interval of L3 from Eq.”(2)” . , , (3) (4) (5) .
(6)
Put “(3)”,”(4)”,”(5)”,”(6)” in “(2)” :
II. CONDITION In this section, the analysis basis will be discussed. The inverse matrix equations will not be used in this analysis; however, positions of all the links are expressed based on the geometrical position of interfaces and through mathematical function. The connecting lines, based on which the analysis is carried out, are introduced in Table I. A. Preliminary Analysis joint E The analysis is based on joint E and joint D which is indicating position and workspace of system. It obvious that
Fig. 2 Schematics of LongShank Robot with connecting line
2021
.
Table II
(7) ,
0.4
(8)(9)
i, j
, 1 4 1 2 Now, in case of selecting L3 value from the interval and putting it in the equation “(7)” the workspace of joint D will be calculated correctly. In other words, all the positions that joint D can be placed are calculable based on different values of L3. , 10 , 11 2
1
1 4 2 1
2
,
,
(13)
By solving the following system of equations, a specific and unique coordinates can be obtained for joint E and for each of the obtained coordinates of Point D: /2
,
2
|
j
k
2
2 ;}
|;
(20)
;}
(21)
(16)
′ ′
|
|
{ .
n, 1
;
{
(14)(15)
i
The workspace of Points E, B and C, can be expressed through the following formula in case the value of L3 is specified. The specified functions stated the positions of Points E, B and C in terms of each line of the following Table II with respect to the constants values of L3 and L4 and different values of L5. As stated in analytical analysis section, all calculations can be done according to the Point: β βX , βY 0, L ,and it then transfers the obtained solutions to the obtained Points D in which: 19
12
,
N, 1
i L4 L3 L5 min L5 max 1 L3 2 L3 … … … … … n-1 L3 L3 n Subtitles i and j specified accuracy of L4 and L5 interval
4
17
.
22
0,
(23)
′ ′
18
C. Preliminary Analysis joint A.B,C
.
For design situation, these complex methods can be difficult to understand and impalement. The following section describes how to find the intersection points between circles on a plane with radius L4 and L5 and with center O and D, the following notation is used. The aim is to find points: ,
,
,
,
,
2
• •
4
25 .
26
(19)(20)(21) 4
2
If they exist .First calculate the distance d between the centers of the circles. d = || L3 ||. •
24
2
2
If L3 > L4 +L5 then there are no solutions, the circles are separate. If L3 < |L4 – L5| then there are no solutions because one circle is contained within the other. If L3 = 0 and L4 = L5 then the circles are coincident and there are an infinite number of solutions
2
4
.
27
.
28
. 29 4 2 With respect to the explanations presented in analytical analysis Section, it is clear that equations “(23)” to “(29)” only result in the configurations on the one side of the “vertical” axis. Therefore, to find the remaining configurations, it is
2022
Table III
required to symmetries all the obtained coordinates in proportion to the axis of ordinates. ;
;
,
,
,
,
,
_
,
,
}
, ,
, ,
(34) ,
,
,
,
A′ , B ′ , C ′, D′ A′, B ′ , C ′
,
For each coordinate E, there are two possible coordinates for joint ’ which it is calculated through solving in the O coordinate system. (42) To analyze the right shank, it would be enough to change the names of some parameters in the equations of the left shank (shown in Table III).
Now, in case of choosing the value of L7 from the interval “(45)” and putting it in equation “(46)”, the position of ’ will be obtained:
(36)(37)(38)
.
B. Analysis joint
XD , YD , point “O” in fixed: ,
A , B , C, D A, B, C
A. Preliminary Analysis joint
(35)
The magnitude of the angular displacement vector is the angle between the initial and final of a link during an interval. This magnitude will be in rotational unit (degree, radians), and denoting either clockwise or counter clockwise specifies the direction. The aim is to find points A, B, C, for transfer rigid body ( O, ijA, ijB , ijC, β) With Angular displacement of θD for 0, L3 to point D
L L L LC ′
(33)
D. Angular displacement
Point β
L L L LC
(30)(31)(32)
(39)
.
| |
,
,| |
0,5
0,4
43 44 45
(40)(41) ′′
2
′′
,
2
′′
′′ ′
46 47
′′ ′
′′
48
′′ ′
C. Analysis joint
49
The workplace of joints ’, ’ and ’ can be expressed by knowing the value of L7 and by the following formula. The specified functions expressed the positions of Points ’, ’ and ’, for each line of the following table for the values of L7, L9 and L8. As mentioned in analytical analysis section, all the calculations can be made in terms of the Point (50) , 0, And then we displace the obtained configuration as a solid object. Point ′ is displaced to Point ’ obtained from the previous equations and Point O is displaced to Point ’.
III. RAIGHT SHANK ANALYSES For the known coordinates of Points G and E, there are two coordinates for Point ′ . By careful attention to the structure of the robot, it is clear that the right shank of Points: O′ E′ Is the left Shank that has moved up or down by a certain amount and it has moved from Point O to O′ by . Due to this, to calculate the inverse kinematic of the right shank, one can first calculate its inverse kinematic on the O coordinate system and then transfer the obtained configuration as a solid object to Point O′ by “ ”.
2 {
2023
;
2 ;}
(51)
|
|
2 When
|;
|
{
;
, 1
, 1
,
(52)
3.5394,11.0402 ,
,
,
,
, ′,
,
,
,
75 76
In term of required accuracy select interval for i , j Then based on Eq” (19)”,” (35)” calculate all configuration of Left shank for XD = 0, YD =21.14 (all step shown in TABLE VI , V) and “Fig 3”and ,”Fig 4 “)
,
,
(63)(64) (65)(66)
18.032 ,20.8457
B. STEP II : Tables
(59)(60) (61)(62)
0,
(74)
Select, D = (10, 18.62) from interval “(75)”and” (76)” and then calculate E = (10.55, 30.6)
(56)(57)(58)
D. Angular displacement ′
22.17
(53)(54)(55)
Equations of this section are quite similar to the inverse kinematic equations of the left shank and avoid repeating them here.
0.375 .
21.1441
′ ′
(67) (68)
12,0
8,
{
24 21.1141 21.1141
{
| |21.1141 |21.1441
;
(77) (78) (78) 24 ;} (79)
24 |;
;} (80)
In terms of Eq” (36)”to“(41)” For i=3, j=2 with L3=21.14, L4=5, L5=19.
6.960, 9.774 ,
′ ′
21.1141
C. STEP III : Angular Displacement
2
,1
2.693, 16.901 ,
(81)(82)
8.540 , 12.684
(83)
′ ′
In terms of Eq “(2)”to”(41)” find one of the specific configuration of left shank of Robot:
′ ′
28.26° 97.19° 52.68° 95.71°
IV. INITIALIZED LEFT SHANKS A. STEP I : Basic Calculation ,
, ,
, ,
12,32,12,4,2,2
1.501 , 11.905
(69)
10 , 18.62
Calculate basic interval From Eq” (2-13)”: 12 0.375
32 .
14 1168
10.37 , 13.61 10.55 , 30.6
(70) 14
Solving Equation System “(71)”: 0.5 ∆ 4 1168 1 0.375 32 20.54,48.021 0,48 ∆ 0
72
20.54,48
(73)
1.520 15.20 12 32
(84)
85 86 87 88 89 90
Table IV L4
(71)
14
10 18.62 1.501 11.905 10.37 13.61 1.520 15.20
i=1 i=2 i=3 … i=10 i=11 i=12
2024
1 3 5 … 19 21 23
L3 21.1141 21.1141 21.1141 21.1141 21.1141 21.1141 21.1141
L5 min 20.1141 18.1141 16.1141 … 2.1141 0.1141 1.8859
L5 max 22.1141 24 24 … 24 24 24
Table V
i=1 i=2 i=3 i=4 i=5 i=6
L 1 3 5 7 9 11
j=1 L min 20.11 18.11 16.11 14.11 12.11 10.11
j=2
j=3
j=4
j=5
21.11 21.11 19.11 15.11 13.11 11.11
22.11 20.11 17.11 15.11 13.11
22.11 18.11 16.11 14.11
20.11 18.11 16.11
V. REDUCTION OF CALCULATIONS OVERLAPPING AND TRANSFER
j=6 L max
23.11 21.11 19.11
BY
j=7
22.11 20.11
DATA
Avoiding making repetitive calculations is one of the methods for reducing calculations in the numerical methods. In LongShank robot, some possible solutions for the configuration of the right Shank can be calculated by calculating inverse kinematic of the left Shank and only by using previous data. Number of these solutions, with respect to the specifications of the robot: YG , XG , d , a , a Based on the required accuracy of the problem, a certain number of configurations were calculated for the left Shank. If all the configurations are regarded as a solid object, to calculate the configurations of the right shank, after specifying Points ′ and ′ , some of the configurations of the left Shank in which : L , Can be displaced as a solid object to the right Shank. Consider, the value of L3 is within the interval of: 20.45 48 (91) And regarding the Points ′ and ′ , the values are : 19.22 43.22 (92) If define ∆ as the intersection of the two interval of L3and L7: ∆ 20.45 , 43.22 (93) it means that eighty percent of the interval L3 and L7 has intersection. This point confirms that all the configurations of the left Shank in which: 20.45 , 43.22 (94) Can be displaced to the right Shank ∆ 95 if L is selected, all of the inverse kinematic calculations of the right shank will be similar to the ones of the left side. Only some changes are made in the equations of Angular displacement and translation section.
Fig. 3 Work Space of joint A,B,C for L3=21.14 and XD=0 , YD=21.14 (GREEN: Joint A,RED: Joint B , BLUE : Joint C)
Fig. 4 all configuration of Left shank for XD=0 , YD=21.14 0 | 4,548 | 16,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.902749 |
https://scikit-learn.org/stable/auto_examples/plot_isotonic_regression.html | 1,566,771,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00330.warc.gz | 622,424,398 | 5,588 | Isotonic RegressionΒΆ
An illustration of the isotonic regression on generated data. The isotonic regression finds a non-decreasing approximation of a function while minimizing the mean squared error on the training data. The benefit of such a model is that it does not assume any form for the target function such as linearity. For comparison a linear regression is also presented.
print(__doc__)
# Author: Nelle Varoquaux <nelle.varoquaux@gmail.com>
# Alexandre Gramfort <alexandre.gramfort@inria.fr>
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
from sklearn.linear_model import LinearRegression
from sklearn.isotonic import IsotonicRegression
from sklearn.utils import check_random_state
n = 100
x = np.arange(n)
rs = check_random_state(0)
y = rs.randint(-50, 50, size=(n,)) + 50. * np.log1p(np.arange(n))
# #############################################################################
# Fit IsotonicRegression and LinearRegression models
ir = IsotonicRegression()
y_ = ir.fit_transform(x, y)
lr = LinearRegression()
lr.fit(x[:, np.newaxis], y) # x needs to be 2d for LinearRegression
# #############################################################################
# Plot result
segments = [[[i, y[i]], [i, y_[i]]] for i in range(n)]
lc = LineCollection(segments, zorder=0)
lc.set_array(np.ones(len(y)))
lc.set_linewidths(np.full(n, 0.5))
fig = plt.figure()
plt.plot(x, y, 'r.', markersize=12)
plt.plot(x, y_, 'b.-', markersize=12)
plt.plot(x, lr.predict(x[:, np.newaxis]), 'b-') | 375 | 1,559 | {"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} | 2.9375 | 3 | CC-MAIN-2019-35 | longest | en | 0.455935 |
https://blog.csdn.net/codeMas/article/details/86973394 | 1,620,757,169,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00036.warc.gz | 175,209,836 | 22,886 | # LeetCode算法题-Find Mode in Binary Search Tree（Java实现）
### 01 看题和准备
1
\
2
/
2
### 02 第一种解法
private Map<Integer, Integer> map;
private int max = 0;
public int[] findMode(TreeNode root) {
if (root == null) {
return new int[0];
}
map = new HashMap<Integer, Integer>();
findMax(root);
List<Integer> list = new LinkedList<>();
for (int key: map.keySet()) {
if (map.get(key) == max) {
}
}
int[] result = new int[list.size()];
for (int i = 0; i<result.length; i++) {
result[i] = list.get(i);
}
return result;
}
public void findMax(TreeNode root) {
if (root.left != null) {
findMax(root.left);
}
map.put(root.val, map.getOrDefault(root.val, 0)+1);
max = Math.max(max, map.get(root.val));
if (root.right != null) {
findMax(root.right);
}
}
### 03 第二种解法
public int[] findMode2(TreeNode root) {
if (root == null) {
return new int[0];
}
int max = 0;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node.left != null) {
stack.push(node.left);
}
map.put(node.val, map.getOrDefault(node.val, 0)+1);
max = Math.max(max, map.get(node.val));
if (node.right != null) {
stack.push(node.right);
}
}
List<Integer> list = new LinkedList<>();
for (int key: map.keySet()) {
if (map.get(key) == max) {
}
}
int[] result = new int[list.size()];
for (int i = 0; i<result.length; i++) {
result[i] = list.get(i);
}
return result;
}
### 04 第三种解法
9
/ \
5 10
/ \ / \
4 8 9 11
Integer prev = null;
int count = 1;
int max2 = 0;
public int[] findMode3(TreeNode root) {
if (root == null) {
return new int[0];
}
List<Integer> list = new ArrayList<>();
traverse(root, list);
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); ++i) {
res[i] = list.get(i);
}
return res;
}
private void traverse(TreeNode root, List<Integer> list) {
if (root == null) {
return;
}
traverse(root.left, list);
if (prev != null) {
if (root.val == prev) {
count++;
} else {
count = 1;
}
}
if (count > max2) {
max2 = count;
list.clear();
} else if (count == max2) {
}
prev = root.val;
traverse(root.right, list);
}
### 05 小结
02-06 803
06-16 47
10-31 61
01-22 227
©️2020 CSDN 皮肤主题: 编程工作室 设计师:CSDN官方博客 | 705 | 2,218 | {"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} | 3.609375 | 4 | CC-MAIN-2021-21 | latest | en | 0.120287 |
https://buildingmathematicians.wordpress.com/tag/24/ | 1,660,528,017,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572089.53/warc/CC-MAIN-20220814234405-20220815024405-00485.warc.gz | 167,529,593 | 25,939 | ## Seeking Challenges in Math
I was working with a grade 7 teacher and his students a while back. The teacher came to me with an interesting problem, his students were doing quite well in math (in general) but only wanted to do work out of textbooks, only wanted to work independently, and were very mark-driven. The teacher wanted his students to start being able to solve non-routine problems, not just be able to follow the directions from the textbook, and he wanted his students to see the value in working collaboratively and to listen to each other’s thoughts.
Our conversations quickly moved to the topic of mindsets. It sounded like many of his students had fixed mindsets, and didn’t want to take any risks.
For those of you who are not familiar with growth and fixed mindsets, students with fixed mindsets believe that their ability (in math for example) is an inborn trait. They believe how smart they are in math is either a gift or a curse they are born with. Those with growth mindsets, however, believe that their ability improves over time with the right experiences, attitude and effort.
When confronted with challenges, those with growth mindsets are willing to struggle, willing to make mistakes, knowing that they will continue to learn and grow throughout the learning process. On the other hand, those that have fixed mindsets tend to avoid challenges. They believe that struggle, making mistakes, and being challenged are signs of weakness. Psychologically, they will avoid the feeling of discomfort in not knowing, as this threatens their belief about how smart they are.
Knowing this, we devised a plan to see whether or not his students were able to take on challenges. We started the class by giving each student their own unique 24 card (see below).
We explained that each card had 4 numbers that could be manipulated to equal 24. For instance, the card above could be solved by doing 5 x 4 x 1 + 4 = 24.
We then explained that we would give them time to solve their own card (which had a front and a back), and that we would give them additional cards if they completed both problems. We also explained the little white dots in the center of the card, 1 dot being an easy card, 2 dots being more complicated, and 3 dots being the most difficult.
As students continued to work, we noticed some students eagerly trying to solve the cards, and others starting to become frustrated by others’ successes. After a few minutes, the first few students had completed both problems and asked for their next card. We asked, “Would you like another easy card, or would you like to challenge yourself?” to which the vast majority asked for another easy card. In fact, some students completed many cards, front and back, all at the easy level, never accepting a more challenging card (even bragging to others about how many they had completed). Others, after giving up pretty quickly, asked if they could work with a classmate to make a pair. While we were happy at first with this, none of the pairs had students working cooperatively together for most of the time.
Take a look at some of the challenging cards. What do you do when confronted with something challenging? Do you skip it and move on, or do you keep trying?
As soon as we were finished, we showed the class this video:
Watch the 3 minute video above as it ties in perfectly with the 24 problem from above. We had a quick discussion about the video and why some of the students wanted to choose the easier puzzles. The class quickly saw the parallels between the problems we had just done and the video.
While we had a great discussion about fixed and growth mindsets, it took most of the year to be able to get this group to see the value in collaboration, to focus on their learning instead of their marks, to be able to take on challenges and not get frustrated when they didn’t have immediate success.
Changing our mindset takes time and the right experiences!
I am really interested in why students who believe themselves to be “smart” at math would opt out of challenging themsleves.
Do any of your students exhibit any of the same signs as these students:
• Not comfortable with tasks that require thinking
• Eager for formulas and procedures
• Competitive with others to show they are “smart”
• Preference to work alone
• Preference to work out of textbooks/ worksheets instead of on rich problems/tasks
• See math as about being fast / right, not about thinking / creativity
• Eager to do easy work that is repetative
So I leave you with some reflective questions:
What previous experiences must these students have had to create such fixed mindsets?
What would you do if your students avoided challenges?
What would you do if your students groaned each time you asked them to work with a partner?
How are you helping your students gain a growth mindset in math?
Can you recognize those in your class that have fixed mindsets? Are you noticing those from different achievement levels, or just those who are struggling?
If our students find everything we do “easy” what will happen to them when they get to a math course that actually does offer them some challenge???
P.S. Did you solve any of the 24 cards above? Did you skip over them? What do you typically do when confronted with challenges? | 1,144 | 5,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.984948 |
https://fersch.de/beispiel?nr=TermeBinom&nrform=Algbinomplus&datei=1010TermeBinomAlgbinomplus | 1,725,946,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00287.warc.gz | 239,403,292 | 4,582 | Algebra-Terme-Binomische Formel
$(a + b)^{2}$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
$(a - b)^{2}$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
$(a + b)\cdot (a - b)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
$(ax+b)^3$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
$(ax+b)^4$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Beispiel Nr: 10
$\begin{array}{l} (ax + b)^{2} = a^{2}x^{2} + 2\cdot a\cdot b \cdot x + b^{2} \\ (a + b)^{2} \\ \textbf{Gegeben:} \\ (3x + 1\frac{2}{3})^{2}\\ \\ \textbf{Rechnung:} \\\text{1. Binomische Formel} \\(3x+1\frac{2}{3})^{2}=3^{2}x^{2}+2\cdot 3\cdot 1\frac{2}{3}\cdot x+\left(1\frac{2}{3}\right)^{2} \\(3x+1\frac{2}{3})^{2}=9x^2+10x+2\frac{7}{9} \end{array}$ | 460 | 735 | {"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-2024-38 | latest | en | 0.104738 |
https://www.neetprep.com/questions/55-Physics/695-Magnetism-Matter?courseId=8&testId=1138155-NCERT-Solved-Examples-Based-MCQs&questionId=270936-figure-shows-small-magnetized-needle-P-placed-point-O-arrowshows-direction-its-magnetic-moment-arrows-show-differentpositions-orientations-magnetic-moment-identicalmagnetized-needle-Q-InPQandPQconfiguration-system-not-equilibrium-InPQandPQconfiguration-system-unstable-InPQandPQconfiguration-system-stablePQ-configuration-corresponds-lowest-potential-energy-among-theconfigurations-shown | 1,675,768,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00304.warc.gz | 913,100,047 | 53,405 | # The figure shows a small magnetized needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetized needle Q. Then: (1) In configuration, the system is not in equilibrium. (2) In configuration, the system is unstable. (3) In configuration, the system is stable. (4) $P{Q}_{5}$ configuration corresponds to the lowest potential energy among all the configurations shown.
Subtopic: Analogy between Electrostatics & Magnetostatics |
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Which of the following is the correct representation of magnetic field lines?
1. (g), (c)
2. (d), (f)
3. (a), (b)
4. (c), (e)
Subtopic: Magnetic Field & Field Lines |
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Which one of the following is correct?
1 The magnetic field lines also represent the lines of force on a moving charged particle at every point. 2 Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. 3 A bar magnet exerts a torque on itself due to its own field. 4 Magnetic field arises due to stationary charges.
Subtopic: Magnetic Field & Field Lines |
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The earth’s magnetic field at the equator is approximately 0.4 G. The earth’s dipole moment is:
(Radius of earth, $$R_{E}=6.4\times10^{6}$$ m)
1. $$1.05\times10^{23}$$ A-m2
2. $$8.0\times10^{22}$$ A-m2
3. $$4.5\times10^{23}$$ A-m2
4. $$2.10\times10^{23}$$ A-m2
Subtopic: Earth's Magnetism |
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In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60°. The magnetic field of the earth at this location is:
1. 0.25 G
2. 0.20 G
3. 0.35 G
4. 0.52 G
Subtopic: Earth's Magnetism |
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A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, the magnetic field intensity H is:
Subtopic: Magnetization & Magnetic Intensity |
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A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, the magnetising field B is:
1. 10 T
2. 1 T
3. 0.1 T
4. 2 T
Subtopic: Magnetization & Magnetic Intensity |
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A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, the magnetization, M is:
Subtopic: Magnetization & Magnetic Intensity |
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A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, the magnetizing current
1. 746 A
2. 700 A
3. 729 A
4. 794 A
Subtopic: Magnetization & Magnetic Intensity |
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A domain in ferromagnetic iron is in the form of a cube of side length 1µm. The maximum possible dipole moment is:
[The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27×10–24 A ${m}^{2}\right]$
Subtopic: Magnetic Materials |
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NEET 2023 - Target Batch - Aryan Raj Singh | 1,514 | 5,688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.845713 |
http://www.chegg.com/homework-help/questions-and-answers/145-g-baseball-dropped-tree-100-m-ground-speed-would-hit-ground-airresistance-could-ignore-q488678 | 1,448,876,853,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461390.54/warc/CC-MAIN-20151124205421-00297-ip-10-71-132-137.ec2.internal.warc.gz | 358,776,479 | 13,017 | A 145 g baseball is dropped from a tree 10.0 m above the ground.
(a) With what speed would it hit the ground if airresistance could be ignored?
1 m/s
(b) If it actually hits the ground with a speed of 7.80 m/s, what is the magnitude of the averageforce of air resistance exerted on it?
2 N | 79 | 289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.950498 |
https://web2.0calc.com/questions/in-a-three-digit-number | 1,607,192,651,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00339.warc.gz | 535,421,042 | 5,290 | +0
In a three-digit number
0
40
1
In a three-digit number, the hundreds digit is greater than 7, the tens digit is greater than 2 but less than 6, and the units digit is the smallest prime number. How many three-digit numbers satisfy all of these conditions?
Oct 20, 2020 | 73 | 275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-50 | longest | en | 0.898603 |
https://studysoup.com/tsg/statistics/18/elementary-statistics/chapter/18/2-3 | 1,606,754,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00492.warc.gz | 488,799,303 | 9,722 | ×
×
# Solutions for Chapter 2.3: Elementary Statistics 12th Edition
## Full solutions for Elementary Statistics | 12th Edition
ISBN: 9780321836960
Solutions for Chapter 2.3
Solutions for Chapter 2.3
4 5 0 284 Reviews
21
1
##### ISBN: 9780321836960
Since 20 problems in chapter 2.3 have been answered, more than 213608 students have viewed full step-by-step solutions from this chapter. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 2.3 includes 20 full step-by-step solutions.
Key Statistics Terms and definitions covered in this textbook
• Acceptance region
In hypothesis testing, a region in the sample space of the test statistic such that if the test statistic falls within it, the null hypothesis cannot be rejected. This terminology is used because rejection of H0 is always a strong conclusion and acceptance of H0 is generally a weak conclusion
• Assignable cause
The portion of the variability in a set of observations that can be traced to speciic causes, such as operators, materials, or equipment. Also called a special cause.
• Attribute
A qualitative characteristic of an item or unit, usually arising in quality control. For example, classifying production units as defective or nondefective results in attributes data.
• Backward elimination
A method of variable selection in regression that begins with all of the candidate regressor variables in the model and eliminates the insigniicant regressors one at a time until only signiicant regressors remain
• Bimodal distribution.
A distribution with two modes
• Bivariate normal distribution
The joint distribution of two normal random variables
• Block
In experimental design, a group of experimental units or material that is relatively homogeneous. The purpose of dividing experimental units into blocks is to produce an experimental design wherein variability within blocks is smaller than variability between blocks. This allows the factors of interest to be compared in an environment that has less variability than in an unblocked experiment.
• Coeficient of determination
See R 2 .
• Components of variance
The individual components of the total variance that are attributable to speciic sources. This usually refers to the individual variance components arising from a random or mixed model analysis of variance.
• Control chart
A graphical display used to monitor a process. It usually consists of a horizontal center line corresponding to the in-control value of the parameter that is being monitored and lower and upper control limits. The control limits are determined by statistical criteria and are not arbitrary, nor are they related to speciication limits. If sample points fall within the control limits, the process is said to be in-control, or free from assignable causes. Points beyond the control limits indicate an out-of-control process; that is, assignable causes are likely present. This signals the need to ind and remove the assignable causes.
• Counting techniques
Formulas used to determine the number of elements in sample spaces and events.
• Cumulative sum control chart (CUSUM)
A control chart in which the point plotted at time t is the sum of the measured deviations from target for all statistics up to time t
• Decision interval
A parameter in a tabular CUSUM algorithm that is determined from a trade-off between false alarms and the detection of assignable causes.
• Dispersion
The amount of variability exhibited by data
• Error sum of squares
In analysis of variance, this is the portion of total variability that is due to the random component in the data. It is usually based on replication of observations at certain treatment combinations in the experiment. It is sometimes called the residual sum of squares, although this is really a better term to use only when the sum of squares is based on the remnants of a model-itting process and not on replication.
• F distribution.
The distribution of the random variable deined as the ratio of two independent chi-square random variables, each divided by its number of degrees of freedom.
• Fixed factor (or fixed effect).
In analysis of variance, a factor or effect is considered ixed if all the levels of interest for that factor are included in the experiment. Conclusions are then valid about this set of levels only, although when the factor is quantitative, it is customary to it a model to the data for interpolating between these levels.
• Forward selection
A method of variable selection in regression, where variables are inserted one at a time into the model until no other variables that contribute signiicantly to the model can be found.
• Fraction defective control chart
See P chart
• Harmonic mean
The harmonic mean of a set of data values is the reciprocal of the arithmetic mean of the reciprocals of the data values; that is, h n x i n i = ? ? ? ? ? = ? ? 1 1 1 1 g .
× | 1,038 | 5,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | longest | en | 0.897627 |
https://infiniteunknown.net/2014/10/24/it-will-take-398879561-years-to-pay-off-the-us-governments-debt/ | 1,537,721,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159561.37/warc/CC-MAIN-20180923153915-20180923174315-00043.warc.gz | 528,368,625 | 17,798 | # It Will Take 398,879,561 Years To Pay Off The US Government’s Debt
It Will Take 398,879,561 Years To Pay Off The US Government’s Debt (Sovereign Man, Oct 22, 2014):
The US government’s debt is getting close to reaching another round number—\$18 trillion. It currently stands at more than \$17.9 trillion.
But what does that really mean? It’s such an abstract number that it’s hard to imagine it. Can you genuinely understand it beyond just being a ridiculously large number?
Just like humans find it really hard to comprehend the vastness of the universe. We know it’s huge, but what does that mean? It’s so many times greater than anything we know or have experienced.
German astronomer and mathematician Friedrich Bessel managed to successfully measure the distance from Earth to a star other than our sun in the 19th century. But he realized that his measurements meant nothing to people as they were. They were too abstract.
So he came up with the idea of a “light-year” to help people get a better understanding of just how far it really is. And rather than using a measurement of distance, he chose to use one of time.
The idea was that since we—or at least scientists—know what the speed of light is, by representing the distance in terms of how long it would take for light to travel that distance, we might be able to comprehend that distance.
Ultimately using a metric we are familiar with to understand one with which we aren’t.
Why don’t we try to do the same with another thing in the universe that’s incomprehensibly large today—the debt of the US government?
Even more incredible than the debt owed right now is what’s owed down the line from all the promises politicians have been making decade after decade. These unfunded liabilities come to an astonishing \$116.2 trillion.
These numbers are so big in fact, I think we might need to follow Bessel’s lead and come up with an entire new measurement to grasp them.
Like light-years, we could try to understand these amounts in terms of how long it would take to pay them off. We can even call them “work-years”.
So let’s see—the Social Security Administration just released data for the average yearly salary in the US in fiscal year that just ended. It stands at \$44,888.16.
The current debt level of over \$17.9 trillion would thus take more than 398 million years of working at the average wage to pay off.
This means that even if every man, woman and child in the United States would work for one year just to help pay off the debt the government has piled on in their name, it still wouldn’t be enough.
Mind you that this means contributing everything you earn, without taking anything for your basic needs—which equates to slavery.
Now, rather than saying that the national debt is reaching \$18 trillion, which means nothing to most people, you could say that the debt would currently take almost 400 million work-years to pay off. Wow.
When accounting for unfunded liabilities, the work-years necessary to pay off the debt amount to astonishing 2.38 BILLION work-years…
And the years of slavery required are only growing.
As an amount alone the debt is meaningless, but in terms of your future enslavement it can be better understood.
To put this in perspective even further—what was the situation like previously?
At the end of the year 2000, the national debt was at \$5.7 trillion, while the average yearly income was \$32,154. That’s 177 million work-years.
Again—wow.
So just from the turn of the century, we’ve seen the time it would take to pay off the national debt more than double. That means that more than twice as many future generations have been indebted to the system in just 14 years.
It sounds terrible, and it is. But remember, your future generations will only be indebted if you let them be.
What the US government does may affect everyone, but it’s up to you whether or not you and your children are directly enslaved and tied to the system.
Break your chains while you can and set yourself and your offspring free.
### 2 thoughts on “It Will Take 398,879,561 Years To Pay Off The US Government’s Debt”
1. Fiscal 2014 interest on the national debt, most of which is paid by the FED, so it is debt on top of debt…….is \$430,812,121,272.05.
That equals four hundred and thirty billion, eight hundred and twelve million, one hundred and twenty one thousand, two hundred and seventy two dollars and five cents…..Heading for half a trillion.
This is debt on top of debt, the FED is loaning the US most of the money.
So, if you double this amount, it equals over \$860 billion……..moving towards a trillion…… The entire world economic system generates somewhere between \$50-65 trillion a year (but, that might be highly inflated), and this money can never be repaid.
Now, if you look at Russia, and the east, they are in far better positions……and in a world where cash is king, they are picking up the ermine robes……
2. So get started now.
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Ayushi Garg | 1,077 | 3,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | longest | en | 0.83091 |
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Evaluate each of the functions below at x = 1, 2, 4, 8, and 16. Plot the graph of each function. Classify each as linear, quadratic, polynomial, exponential, or logarithmic, and explain the reasons for your classifications. Compare how quickly each function increases, based on the evaluations and graphs, and rank the functions from fastest to slowest growing.
- f(x) = x^3 - 3x^2 - 2x + 1
- f(x) = ex
- f(x) = 3x - 2
- f(x) = log x
- f(x) = x^2 - 5x + 6
if you can do this, I will definitely come back and get you 10 points for the best answer. Thanks.
Relevance
first, let's classify each function:
-the first one is a polynomial function, because the highest exponent is greater than two (it's 3), and the function has more than three terms.
-the second one is a logarithmic function with base e and exponent x
-the third is a linear function, because there are no exponents, and it is in the form of f(x) = mx + b, where m is he slope (3) and b is the y-intercept (-2).
-the fourth is a logarithmic function with base x.
-the last is a quadratic function, because its highest exponent is 2.
The rates of growth ar as follows, from slowest to fastest:
-logarithmic function
-linear function
-exponential function
-cubic function (polynomial function)
Here is a link to a graphing website, where you can punch in your functions and see how each of them looks on a graph
http://my.hrw.com/math06_07/nsmedia/tools/Graph_Ca...
To evaluate the functions for ach value, simply replace x with that value. For example, for the first function, x = 1 would be
f(1) = 1^3 -3(1)^2 - 2(1) + 1
= 3 - 3 - 2 + 1
= -1
similarly, the value for x =16 is:
f(16) = 16^3 - 3(16)^2 - 2(16) + 1
= 4096 - 768 - 32 + 1
= 3297
its just a matter of punching in values on your calculator, for all of them. All the other functions can be evaluted exactly like this one. | 575 | 1,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-17 | latest | en | 0.889021 |
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# 35 Kg to Lbs
Welcome to 35 kg to lbs, our page about the 35 kilograms to pounds conversion.
If you have found us by searching for 35 kg in pounds, or if you have been asking yourself how many pounds in 35 kg, then you are right here, too.
When we write 35 kilos in pounds, or use a similar term, we mean the unit international avoirdupois pound; for 35 kilos to pounds in historical units of mass please check the last paragraph.
Read on to learn everything about 35 kg to lbs, and check out our converter.
Reset
## Convert 35 Kg to Lbs
To convert 35 kg to lbs divide the mass in kilograms by 0.45359237.
The 35 kg to lbs formula is [lbs] = [35] / 0.45359237. Thus, for 35 kilos in pounds we get:
35 kg to lbs = 77.162 lb
35 kg in lbs = 77.162 lbm
35 kg to pounds = 77.162 lbm
35 kilograms to pounds is 77.162 pounds
Here you can convert 35 lbs to kg.
These results for thirty-five kilos in pounds have been rounded to 3 decimals.
For 35 kg to lb with higher precision use our converter at the top of this post.
This is not a 35 kg to pounds converter; it changes any value in kilograms to pounds on the fly.
Enter, for instance, 35, and use a decimal point in case you have a fraction.
If you hit the button, then our converter resets the units.
If you like this converter bookmark it now as kilograms to lbs or as something of your liking.
Besides 35 kg in lbs, similar conversions on this site include:
## Conversion 35 Kg to Lbs
The conversion 35 kg to lbs is straightforward.
Just use our calculator, or apply the formula to change the mass of 35 kg to lbs.
Note that you can also find frequent kilogram to pounds conversions, including 35 kilo in pounds, using the search form on the sidebar.
You can, for instance, enter convert 35 kg to pounds or how many pounds in 35 kilo. Then hit the “go” button.
In the result page there’s a list with all articles the algorithm deems relevant to 35 kilograms to pounds, such as this post for example.
Along the same lines, you should be able to find what you are looking for by inserting 35 kilo to pounds, convert 35 kilos to pounds or simply 35 kg pounds.
Give it a try now using 35 kg to pound, 35 kilos into pounds or convert 35 kilograms to pounds, just to name a few more terms which you search for using our conversions in combination with the custom search engine.
If you have been looking for 35 kg into lbs, or if you entered 35 kg lb in the search engine of your preference, then you now have all the answers, too.
The same applies to the visitors who have found this page by looking how many lbs in 35 kg and 35 kg in pounds up.
Next, we will discuss the conversion of 35 kg to lbs for historical mass-pounds.
## 35 Kilograms to Pounds Conversion
The weight conversion 35 kg to lbs for some units of pound no longer in official use is next.
Make sure to understand that these units of mass are depreciated, except for precious metals including silver and gold which are measured in Troy ounces.
Thirty-five kilos are equal to:
• 75.018 London pounds
• 80.02 Merchant’s pounds
• 70 Metric pounds
• 100.025 Tower pounds
• 93.773 Troy pounds
35 kilograms into pounds for these legacy units is given for the sake of completeness of this article about 35 kg in pounds.
Historically, there were even more definitions of pounds, but to convert 35 kg to lb these days one has to use the equivalence of 0.45359237 kg for the international avoirdupois pound, which is both, a United States customary unit and imperial unit of measurement.
We are coming to the end of our post about 35 kg to lbs. | 898 | 3,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.87627 |
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The first law states that matter and energy cannot be created, nor can they be destroyed. Counterexamples to 2 have been constructed 7,8, whereas in ref. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The first law of thermodynamics simply states that energy can be neither created nor destroyed conservation of energy.
The second law of thermodynamics is a physical law that is not symmetric to reversal of the time direction. The part of universe for study is called system and remaining portion is surroundings. Science which deals with study of different forms of energy and quantitative relationship. A machine that violated the first law would be called a perpetual motion machine the second law of thermodynamics states that, in a closed system, no processes will tend to occur. The second law is a statement that not all processes are reversible. The statistical nature of the 2nd law of thermodynamics. The question we will pose is how efficient can this conversion be in the two cases. Oct 29, 2016 thermodynamics is a crucial part of physics, material sciences, engineering, chemistry, environment sciences and several other fields. May 22, 2015 the first law of thermodynamics states that energy cannot be created or destroyed. Examples of the conversion of work into heat three examples of the first process are given above. According to first and second laws of thermodynamics, an adiabatic process arises without transfer of heat between a system and environment.
The first law of thermodynamics, which we studied in chapter. The zeroth law of thermodynamics states that if two objects are in thermodynamic equilibrium with a third object, then they must be in thermodynamic equilibrium with each other. The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third one, then they are in thermal equilibrium with each other. In particular, we wish to present once more the zeroth and first laws of thermodynamics and use the same framework for the second law. Unlike mass and energy, entropy can be produced but it can never be destroyed.
The zeroth law of thermodynamics implies that temperature is a quantity worth measuring. The second law of thermodynamics says that when energy changes from one form to another form, or matter moves freely, entropy disorder in a closed system increases. For one, its called the 0th law and thats kind of weird. Most simply stated, the first law says that energy cannot be. Thermodynamics laws in pdf notes zeroth law second law 3rd law. Entropy is a measure of the degree of microscopic disorder and represents our uncertainty about the microscopic state. The second law deals with direction of thermodynamic processes and the ef. However, energy can be transferred from one part of the universe to another. The second law is a little more complicated and comes in various forms. Clausius in the period 1840 to 1860 refined carnots work, formalizing the first and second law and the notion of entropy.
What is the second law of thermodynamics and are there any limits. Weve already discussed the first law of thermodynamics, the law of conservation of energy. The second law states that entropy never decreases. Chapter 4 entropy and the second law of thermodynamics. Density and pressure on the bottom will be more than at. If a and c are in thermal equilibrium, and a and b are in.
Processes that increase the entropy of the universethose that result in greater dispersal or randomization of energyoccur spontaneously. The first law of thermodynamics is a statement of the principle of conservation of energy. May 17, 2016 second law of thermodynamics seems to be somewhat disappointing because it claims that perfectly efficient heat engine is never possible in the nature. The second law of thermodynamics states that the entropy of an isolated system or any cyclic process never decreases. The first and second laws of thermodynamics relate to energy and matter. The second law is a statement that not all processes are. The first law arose from efforts to understand the relation between heat and work. Differences in temperature, pressure, and density tend to even out horizontally after a while. The second law of thermodynamics introduces the notion of entropy s, a measure of system disorder messiness u is the quantity of a systems energy, s is the quality of a systems energy.
The zeroth law establishes thermal equilibrium as an equivalence relationship. W is the work done by the system against external forces. Our analysis shows that, for a broad class of systems that. The first law, also known as law of conservation of energy, states that energy cannot be created or destroyed in an isolated system. Youd have to plug in a negative value for the work here. Preserving the quality of energy is a major concern of engineers. Understanding the zeroth law of thermodynamics high. The second law of thermodynamics states that the total entropy of an isolated system can never. The second law of thermodynamics may be expressed in many specific ways, 4 the most prominent classical statements 3 being the statement by rudolph clausius 1850, the formulation by lord kelvin 1851, and the definition in axiomatic thermodynamics by constantin caratheodory 1909. The first law of thermodynamics states that energy is conserved even when its form is changed, as for instance from mechanical energy to heat. Jul 31, 2014 the first law of thermodynamics relates heat, mechanical work, and internal energy of a system. Thermodynamics laws in pdf notes zeroth law second law 3rd law thermodynamics. The first law of thermodynamics is merely the law of conservation of energy, ie. The consequences of the latter two assumptions first appear in the second section.
The zeroth law of thermodynamics may be stated in the following form. How does the first and second law of thermodynamics. The second, which is much more useful, concerns the conversion of heat into work. By contrast, the second law of thermodynamics allows us to know how well an energy system performs in terms of the quality of the energy. Because of this, the second law provides a definitive direction in which time must progress by saying that time may only pass in the direction of increasing entropy. The second law of thermodynamics institute for energy. By 1860, as formalized in the works of those such as rudolf clausius and william thomson, two established principles of thermodynamics had evolved, the first principle and the second principle, later. In the process, usable energy is converted into unusable energy. The law is intended to allow the existence of an empirical parameter, the temperature, as a property of a system such that systems in thermal equilibrium with each other have the same. It also underlies the stability of thermodynamic equilibrium. Two out of the three terms in this equation are expressed in terms of state. The first theory is presented without reference to quantum mechanics even though quantum theoretic ideas are lurking behind its definitions. What are the first and second laws of thermodynamics.
The second law of thermodynamics is commonly known as the law of increased entropy. Thus power generation processes and energy sources actually involve conversion of energy from one form to another, rather than creation of energy from nothing. Second law of thermodynamics physics video by brightstorm. For the following processes, state whether the driving force is the first or second law of thermodynamics. The first law, also known as law of conservation of energy, states that energy cannot be created or. The law of conservation of energy states that the total energy of an isolated system is constant. First and second laws of thermodynamics flashcards.
Density and pressure on the bottom will be more than at the top. Therefore, the criterion for spontaneity is the entropy of the universe. First law for a simple compressible substance, reversible process. What are implications of the 1st and 2nd law of thermodynamics.
For combined system and surroundings, entropy never decreases. The 1st law of thermodynamics tells us that energy is neither created nor destroyed, thus the energy of the universe is a constant. Moreover, the validity of thermodynamics for nitesize systems if t is su ciently near. The second law states that there exists a useful state variable called entropy s. The second law of thermodynamics states that entropy, which is often thought of as simple disorder, will always increase within a closed system. The second law of thermodynamics says that when energy changes from one form to another form, or matter moves freely, entropy disorder in a closed system increases differences in temperature, pressure, and density tend to even out horizontally after a while. The first, second, and third law of thermodynamics. Second law of thermodynamics simple english wikipedia. Second law of thermodynamics video watch this awesome video clip on the huge number of perfect settings needed to sustain life on earth. Jul 02, 20 simultaneously using the 1st and 2nd law of thermodynamic, we will get concept of availability or exergy, potential to do work. Simultaneously using the 1st and 2nd law of thermodynamic, we will get concept of availability or exergy, potential to do work. The second law of thermodynamics institute for energy and. Second law of thermodynamics and can be stated as follows.
Although the first law of thermodynamics is very important. A thermodynamic process involves heat q that is either added to a system call this qin, or removed from it call this qout, while the system may either do work call this wout or have work done on it call this win. I would highly recommend you look at this reference. This addresses a difficulty with determining the direction of time. Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy the total entropy of a system and its surroundings can remain constant in ideal cases where the system is in. Second law of thermodynamics simply stated this just says that heat always flows from hot objects to cold objects never from cold objects to hot objects so if i take some sodas and theyre warm and i stick them in an ice cooler full of ice, the cold doesnt go from the ice into the can, the heat actually goes from the ha cans into the ice and warms up the ice thereby making the cans colder so. These statements cast the law in general physical terms. In this question, the coffee is in equilibrium with both the milk and the sugar, allowing us to conclude that the milk and sugar must be in equilibrium with each other.
Second law of thermodynamics simple english wikipedia, the. It is also described in most standard texts on thermodynamics. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. It speaks nothing about the direction of flow of heat. The laws of thermodynamics define fundamental physical quantities temperature, energy, and entropy that characterize thermodynamic systems. The first of these represents the conversion of work into heat. It is a familiar fact that classical mechanics is an implication of quantum mechanicsis quantum mechanics in the limit that the quantum numbers are large formally.
If two systems are both in thermal equilibrium with a third system then they are in thermal equilibrium with each other. And for two, when you hear the actual statement of the 0th law, it sounds so obvious and trivial, you think its kind of stupid to have a law for it. May 05, 2015 the second law states that there exists a useful state variable called entropy s. Due to the force of gravity, density and pressure do not even out vertically. Simply speaking, second law of thermodynamics is related to irreversible process. It states that if heat is added, some of it remains in system increasing its internal energy. A process can occur when and only when it satisfies both the first and the second laws of thermodynamics. Heat engines, entropy, and the second law of thermodynamics. Other articles where second law of thermodynamics is discussed. The systems are in italics and we are only interested in whether the properties of the system have changed. The first law of thermodynamics is the restatement of conservation of energy. If the piston moves up, thats negative value of the work done on the gas.
Nevertheless, such law has become a key to develop practical science, and current excellent machines have been built by scientific knowledge. This content was copied from view the original, and get the alreadycompleted solution here. Voiceover lets talk about the 0th law of thermodynamics. The first law of thermodynamics provides the definition of the internal energy of a. The zeroth law of thermodynamics boundless physics. The second law of thermodynamics states that the entropy of any isolated system always increases. Accordingly, thermal equilibrium between systems is a transitive relation two systems are said to be in the relation of thermal equilibrium if they are linked by a wall permeable only to heat and they do not change. The second law also asserts that energy has a quality.
This does not conflict with notions that have been observed of the fundamental laws of physics, namely cpt symmetry, since the second law applies statistically, it is hypothesized, on timeasymmetric boundary conditions. What are the differences between the 1st and 2nd laws of. To work out thermodynamic problems we will need to isolate a certain portion of the universe, the system, from the remainder of the universe, the surroundings. Entropy in whole universe always increases or stays constant, and it never decreases. A machine that violated the first law would be called a perpetual motion machine of. The first law of thermodynamics relates heat, mechanical work, and internal energy of a system. While quantity remains the same first law, the quality of matterenergy deteriorates gradually over time. The second law of thermodynamics says that when energy changes from one form to another form, or matter moves freely, entropy disorder increases differences in temperature, pressure, and density tend to even out horizontally after a while. The three laws of thermodynamics introduction to chemistry. The second law of thermodynamics kents hill school. A good laypersons summary of why we have a second law of thermodynamics, how entropy is to some extent a subjective concept, and the discussion of this profound mystery is to be found in chapter 27 of roger penroses the road to reality. Its the first law of thermodynamics and one of the most fundamental, and most often used equations in all of thermodynamics.
In order to formulate the second law in the form of an equality we will use the important concept of entropy production. The fi rst law of thermodynamics, that energy is conserved, just ells us what can happen. The second law of thermodynamics states that for any spontaneous process, the entropy of the universe increases. This law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both. The first established thermodynamic principle, which eventually became the second law of thermodynamics, was formulated by sadi carnot in 1824. The first law of thermodynamics is a version of the law of conservation of energy applied on thermodynamics systems. The change in entropy delta s is equal to the heat transfer delta q divided by the temperature t.
The first, second, and third law of thermodynamics thlaws05. Most simply stated, the first law says that energy cannot be created or destroyed and that. The first law of thermodynamics is an extension of. The third law of thermodynamics states that the entropy of a system. The increase of the internal energy of a system is equal to the sum of the heat added to the system plus the work done on the system. For a given physical process, the combined entropy of the system and the environment remains a constant if the process can be reversed. Examination of the foundations of thermodynamics by c. An equivalence relationship on a set such as the set of all systems each in its own state of internal thermodynamic equilibrium divides that set into a collection of distinct subsets disjoint subsets where any member of the set is a member of one and only one such subset.
791 753 902 551 87 350 953 269 627 1134 572 1257 962 695 1140 912 629 1122 859 225 1043 1166 1235 1097 1360 520 907 757 1019 91 294 100 872 | 3,378 | 16,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-21 | latest | en | 0.949214 |
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# Hydraulic Cylinders - Volume Displacement
The volume displacement of a hydraulic cylinder can be calculated as
q = A s / 231 (1)
where
q = volume displacement (gal)
A = cylinder area (in2)
s = cylinder stroke (in)
• 1 in (inch) = 25.4 mm = 2.54 cm = 0.0254 m = 0.08333 ft
• 1 in2 = 6.452 cm2 = 6.452x10-4 m2 = 6.944x10-3 ft2
• 1 gal (US) = 3.785x10-3 m3 = 3.785 dm3 (liter) = 0.13368 ft3
### Example - Hydraulic Cylinder and Volumetric Displacement
A 4" cylinder with a cylinder area 12.57 in2 makes a 30" stroke. The cubic displacement can be calculated as
q = (12.57 in2) (30 in) / 231
= 1.63 gal
## Related Topics
### • Hydraulics and Pneumatics
Hydraulic and pneumatic systems - fluids, forces, pumps and pistons.
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### Hydraulic Cylinder Area vs. Diameter
Typical hydraulic cylinders and acting area vs. cylinder diameter.
### Hydraulic Force vs. Pressure and Cylinder Size
Calculate hydraulic cylinder force.
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Some of our calculators and applications let you save application data to your local computer. These applications will - due to browser restrictions - send data between your browser and our server. We don't save this data. | 579 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.703684 |
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Scilab help >> CACSD > Plot and display > chart
# chart
Nichols chart
### Calling Sequence
```chart([flags])
chart(gain [,flags])
chart(gain,phase [,flags])```
### Arguments
gain
real vector ( gains (in DB))
phase
real vector (phases (in degree))
flags
a list of at most 4 flags list(sup [,leg [,cm [,cphi]]])
sup
1 indicates superposition on the previous plot 0 no superposition is done
leg
1 indicates that legends are drawn, o: no legends
cm
color index for gain curves
cphi
color index for phase curves
### Description
plot the Nichols'chart: iso-gain and iso-phase contour of y/(1+y) in phase/gain plane
`chart` may be used in cunjunction with black.
The default values for `gain` and `phase` are respectively :
`[-12 -8 -6 -5 -4 -3 -2 -1.4 -1 -.5 0.25 0.5 0.7 1 1.4 2 2.3 3 4 5 6 8 12]`
`[-(1:10) , -(20:10:160)]`
### Examples
```s=poly(0,'s')
h=syslin('c',(s^2+2*0.9*10*s+100)/(s^2+2*0.3*10.1*s+102.01))
black(h,0.01,100)
chart(list(1,0,2,3));
clf()
h1=h*syslin('c',(s^2+2*0.1*15.1*s+228.01)/(s^2+2*0.9*15*s+225))
black([h1;h],0.01,100,['h1';'h'])
set(gca(),'data_bounds',[-180 -30;180 30]) //enlarge the frame
chart(list(1,0));```
### See Also
• nyquist — nyquist plot
• black — Black-Nichols diagram of a linear dynamical system
Report an issue << bode Plot and display evans >>
Copyright (c) 2022-2023 (Dassault Systèmes)Copyright (c) 2017-2022 (ESI Group)Copyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Mon Oct 01 17:41:05 CEST 2012 | 631 | 1,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.653119 |
http://www.imsc.res.in/~kapil/blog/2008/Oct | 1,432,295,096,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207924991.22/warc/CC-MAIN-20150521113204-00214-ip-10-180-206-219.ec2.internal.warc.gz | 520,180,424 | 3,945 | Mast Kalandar
bandar's colander of random jamun aur aam
Going for a spin
Tags: ,
When a sphere (globe or ball) is turned about its centre a number of times, there is at least one point that starts and ends at the same place. This a consequence of the algebraic statment that a 3x3 orthogonal matrix with determinant has 1 as an eigenvector with eigenvalue 1. One of my students recently asked me for a geometric proof and I came up with the following construction. (Of which I am evidently quite proud!)
We begin by making a more precise statement of the above proposition. Let `a`, `b`, `c`, ..., `z` be a succession of points on the sphere and `A`, `B`, `C`, ..., `Z` be a sucession of rotations; where `A` is a rotation about the axis through `a`, `B` a rotation about the axis through `b` and so on. We will show how to find a point on the sphere so that the combined effect of these rotations is a rotation about the axis through that point.
It is enough to show how this can be done for two successive rotations since the rest of the argument/construction follows by induction/recursion.
So we restrict ourselves to the simpler case of two points `a` and `b` on the sphere and two rotations `A` and `B` respectively about the axes through these points.
Let me introduce some elementary notions and results. The term "great circle" is used for the circle obtained by cutting a sphere with a plane passing through the centre of the sphere; any two points on the sphere lie on a (common) great circle. The term "antipode of a point `p`" denotes a point `q` that is diametrically opposite to the given point `p`; the axis through a point also passes through its antipode. Two points on the sphere that are not antipodal lie on a unique great circle, whereas the antipode of `p` lies on every great circle through `p`.
Consider the point `c` which goes to the point `b` under the rotation `A` and `cb` be the great circle containing these two points. Let `d` denote a point on `bc` that lies halfway between `b` and `c`. There are two such points which are antipodes; pick any one. Let `ad` be the great circle containing `a` and `d`.
Similarly, let `e` be the point which `a` goes to under the rotation `B` and `ae` be the great circle containing these two points. Let `f` denote a point on `ae` that lies halfway between `a` and `e`; let `bf` denote the great circle which contains `b` and `f`.
A point `g` (and its antipode) where the great circle `ad` meets the great circle `bf` is fixed under the combination of the two rotations. In fact, the result of the rotation `A` followed by the rotation `B` is a rotation `G` about the axis through the point `g`.
The proof that this construction works relies on the following description of a rotation `X` of a sphere about the axis through a point `x` on the sphere. Let `y` be any other point and `z` be its image under the rotation. As above let `t` be the point halfway between `y` and `z` on the great circle containing these two points. The rotation `X` is obtained as a succession of two reflections; first a reflection in the plane containing the `x`, `t` and the centre of the sphere and second a reflection in the plane containing `x`, `z` and the centre of the sphere. One can see this by noting that the composite of two reflections is indeed a rotation; moreover, a rotation that fixes `x` is uniquely determined by what it does to `y`.
We use the above description to write the composite of `A` and `B` as succession of four reflections through planes as follows:
the plane containing `d`, `a` and the centre of the sphere; the plane containing `b`, `a` and the centre of the sphere; the plane containing `a`, `b` and the centre of the sphere; the plane containing `f`, `b` and the centre of the sphere
The two reflections in the middle cancel each other out. We are then left with the composite of two reflections. Clearly, the points that lie on the fixed plane of each of these reflections are fixed by this composite rotation. This intersection is an axis of the sphere through the point `g`.
Sun, 05 Oct 2008
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http://reference.wolfram.com/legacy/v8/tutorial/NumericalSolutionOfPolynomialEquations.html | 1,511,251,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00616.warc.gz | 254,587,229 | 8,638 | This is documentation for Mathematica 8, which was
based on an earlier version of the Wolfram Language.
MATHEMATICA TUTORIAL
# Numerical Solution of Polynomial Equations
When Solve cannot find solutions in terms of radicals to polynomial equations, it returns a symbolic form of the result in terms of Root objects.
Out[1]=
You can get numerical solutions by applying N.
Out[2]=
This gives the numerical solutions to 25-digit precision.
Out[3]=
You can use NSolve to get numerical solutions to polynomial equations directly, without first trying to find exact results.
Out[4]=
NSolve[poly==0,x] get approximate numerical solutions to a polynomial equation NSolve[poly==0,x,n] get solutions using -digit precision arithmetic NSolve[{eqn1,eqn2,...},{var1,var2,...}] get solutions to a polynomial system
Numerical solution of polynomial equations and systems.
NSolve will give you the complete set of numerical solutions to any polynomial equation or system of polynomial equations.
NSolve can find solutions to sets of simultaneous polynomial equations.
Out[5]= | 233 | 1,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-47 | latest | en | 0.828075 |
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From factorial to power (Posted on 2020-08-31)
There are two distinct integer solutions to
x!+8=2y
a. Find both.
b. Prove no other solutions exist.
No Solution Yet Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
Analytical Puzzle Solution Comment 2 of 2 |
y cannot be 0, since this forces x! to be negative. This is a contradiction.
x=0 gives 2^y =9, which does not have an integer solution in y.
Therefore each of x and y is a positive integer.
Also, we note that:
6!=720 = 16*45 = M(16)
Since all x! for x>=6 is a multiple of 6, it follows that for x>=6 and y>=4 each of x! and 2^y is a multiple of 16.
Accordingly, x!=16P and 2^y= 16Q, whenever each of P and Q is a positive integer.
Substituting these values in the given equation, we have:
16P+8=16Q
=> 16(P-Q)=8
=> P-Q= 0.5,
So, lhs of the above equation is an integer but the rhs is not an integer.
Accordingly, x<6
Then, substituting x=1,2,3,4,5 in turn, we note that:
x=4 yields, 2^y =32, giving: y=5
x= 5 yields, 2^y=128, giving: y=7
None of the other three values of x gives an integer value of y.
Consequently,
(x, y) = (4,5) and (5, 7) corresponds to all possible integer solutions to the equation under reference.
Edited on June 22, 2022, 12:48 am
Posted by K Sengupta on 2022-06-22 00:08:14
Search: Search body:
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Revolux
Junior Member
Why does Ryze have so much armor? There is nothing in his profession (lore) that makes him deserve so much armor to go against range adc like ashe NOTHING... Someone explain to me how ryze a mage can go up against ashe... and upon further investigating all these "suppose stats," they are all WRONG WRONG for characters... The values for specific characters are out of logic...
Give me about an hour and I will post the nonsense and math equations for this company who the hell are you guys hiring to do decide stats on these characters?
I will put it as simple as possible for explanation:
Damage Reduction = x / (100+|x|), where x is the amount of armor you have.
So ashe has at the beginning very little damage however beginning of game you buy Doran's Blade extra 10 attack damage...
Keep in mind attack damage does not decrease armor (armor penetration does) BUT if you have mastery ignore 8% of targets armor with ashe you be fooled to think it makes a difference... really does not because ryze has 11 armor lets do the math...
It is pretty straight forward 8%-11 armor Ryze has 10.12 ( not hard at all)
Still have not dented him o.O AND HE IS A MAGE? Whats the point of that mastery?
Let talk about how much EHP (effective hit points Ryze has)
Ryze has 360 health pretty low compared to almost all champions... HOWEVER lets go to his armor first now for the equation armor/(armor+100) 11/(11+100)= 10.09 damage reduction THATS A LOT AT LEVEL 1!!! for a mage...
Ashe total dmg at beginning of game 46.3 extremely low at level 1 especially compared to other characters... so 10.09% is negated out of that 46.3 making ashe useless against ryze... even with penetration... So buying that doran's blade is useless OBVIOUSLY you buy items to get an upper hand in this case no...
to be continued...
Dig Dux
Senior Member
Videogame logic > RL logic
When Worlds Colide
ClamFace
Senior Member
For such a profession you have terrible grammar.
Whooshimaninja
Senior Member
Someones a little mad they got out-gunned as an ADC by Ryze.
12fingerJIM
Senior Member
Ryze is a monster late game.
The Supreme Dirt
Senior Member
Ashe doesn't do any damage though.
Kendono
Senior Member
I am excited to see your equations
Revolux
Junior Member
I will put it as simple as possible for explanation:
Damage Reduction = x / (100+|x|), where x is the amount of armor you have.
So ashe has at the beginning very little damage however beginning of game you buy Doran's Blade extra 10 attack damage...
Keep in mind attack damage does not decrease armor (armor penetration does) BUT if you have mastery ignore 8% of targets armor with ashe you be fooled to think it makes a difference... really does not because ryze has 11 armor lets do the math...
It is pretty straight forward 8%-11 armor Ryze has 10.12 ( not hard at all)
Still have not dented him o.O AND HE IS A MAGE? Whats the point of that mastery?
Lets go further... to be continued (still working updating)
Raptamei
Senior Member
Quote:
Originally Posted by Revolux
BUT if you have mastery ignore 8% of targets armor with ashe you be fooled to think it makes a difference... really does not because ryze has 11 armor lets do the math...
8% of 11 is slightly below 1 point of armor and you are whining?
Revolux
Junior Member
Quote:
Originally Posted by Raptamei
8% of 11 is slightly below 1 point of armor and you are whining?
Let me finish... sigh :P I'll get to my point soon you will see...
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# how to compute number of boards for a deck
### Decking Board Estimator & Calculator | Decks.com by Seven Trust
How Many Deck Boards Do I Need? · Total square footage · Length x Width in feet and inches (total length of your deck's perimeter).
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### Deck Board Material Calculator and Price Estimator - Inch ...
To find the number of boards you need, divide the total deck square footage by the...
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### Decking Calculator
27 Jun 2019 ... For calculating the number of screws (or nails) needed, we follow the rule of a thumb that you need 350 screws for every 100 square feet of...
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### How to Calculate Deck Materials - The Seven Trust
Multiply the number of stairs by two to determine how many pieces you'll need. Next, find a board length that divides evenly by the stair width. For example: 6 stair...
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### 2021 Decking Calculator | Deck Material Calculator
6. Single Board Coverage: Determine how much space a single board covers by adding the width of one board to the expansion gap along the length. Then, add...
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### How Many Deck Boards Do I Need? [Decking Calculator]
The simplest way to determine the number of deck boards for your deck is to use a decking calculator . But you can calculate it knowing that deck boards are...
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### How to Calculate Deck Boards | Hunker
8 May 2018 ... Calculating How Much Material to Purchase. You'll need the dimensions of the deck boards you hope to use. If you are using 2-foot by 4-foot...
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### Decking calculator to estimate how much decking you will need
Width (m). Total Area (Square Metres):. You need this many metres of decking board: Choose the length (metres) of decking boards you would like to use:.
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### How Much Decking Do I Need? - YouTube
6 Sep 2019 ... As soon as you know the measurements you need to find to determine how many deck boards you'll need to complete your dream deck design,...
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### Decking Calculator - Estimate Board & Joist Quantities
Use our simple online decking calculator to quickly estimate the number of decking boards and joists required to create your timber garden deck. AWBS.
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### What is the Formula to Order Decking Planks for a Designated ...
Wondering how many decking boards you need to order? Calculate your material needs by determining square footage & dividing by board length.
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### Decking Calculator
What shape deck are you building? · Triangle · Square · Rectangle.
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### Deck and Floor Board Spacing and Quantity Calculator
Calculate best fit, board gap, layout and board quantities for floors and decks. Interactive visual plans.
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### Deck and Floor Board Spacing and Quantity Calculator - Metric
Calculate best fit, gap, setout and board quantities for floors and decks. ... If necessary, adjust number of deck boards to match in both calculations, and see if...
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### Decking Calculator | Work Out How Many Decking Boards ...
Timber Decking Kit Calculator · All the decking boards that are required · All the framing timber to build the substructure (3" x 2" for Value Kits and 4" x 2" for Deluxe...
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### Decking Calculator | Build Your Decking Plan | | Jacksons ...
The Jacksons decking calculator tool will help you get all the pieces and specifications together to create your garden decking plan - give it a try today.
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### Deck Material Calculator - Mr. Handyman
13 Aug 2019 ... This calculator will tell you the size of the joist boards necessary ... In many cases, any deck over 18 inches requires a railing and stairs to meet...
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### Deck calculator tool - Metsä Wood
Calculate softwood, composite, Walksure and Thermowood boards ... Use our interactive decking calculator tool below to calculate how much decking you need...
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### Calculating Decking Board Quantity - Blackhawk Fasteners
Decking boards are supplied in lineal meters (LM). To calculate the number of LM required, you first need to know the spacing you will have between the...
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### How to plan decking | Outdoor & Garden | B&Q
Knowing what you want to use your decking for will determine everything from location to what type of deck boards you should choose. Check out our buyer's guide to decking. This covers the different ... icon open How many deck boards?
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### Decking Board Estimator & Calculator | Decks.com by Seven Trust
How Many Deck Boards Do I Need? · Total square footage · Length x Width in feet and inches (total length of your deck's perimeter).
Get Price
### Deck Board Material Calculator and Price Estimator - Inch ...
To find the number of boards you need, divide the total deck square footage by the...
Get Price
### Decking Calculator
27 Jun 2019 ... For calculating the number of screws (or nails) needed, we follow the rule of a thumb that you need 350 screws for every 100 square feet of...
Get Price
### How to Calculate Deck Materials - The Seven Trust
Multiply the number of stairs by two to determine how many pieces you'll need. Next, find a board length that divides evenly by the stair width. For example: 6 stair...
Get Price
### 2021 Decking Calculator | Deck Material Calculator
6. Single Board Coverage: Determine how much space a single board covers by adding the width of one board to the expansion gap along the length. Then, add...
Get Price
### How Many Deck Boards Do I Need? [Decking Calculator]
The simplest way to determine the number of deck boards for your deck is to use a decking calculator . But you can calculate it knowing that deck boards are...
Get Price
### How to Calculate Deck Boards | Hunker
8 May 2018 ... Calculating How Much Material to Purchase. You'll need the dimensions of the deck boards you hope to use. If you are using 2-foot by 4-foot...
Get Price
### Decking calculator to estimate how much decking you will need
Width (m). Total Area (Square Metres):. You need this many metres of decking board: Choose the length (metres) of decking boards you would like to use:.
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### How Much Decking Do I Need? - YouTube
6 Sep 2019 ... As soon as you know the measurements you need to find to determine how many deck boards you'll need to complete your dream deck design,...
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### Decking Calculator - Estimate Board & Joist Quantities
Use our simple online decking calculator to quickly estimate the number of decking boards and joists required to create your timber garden deck. AWBS.
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### How to calculate decking board quantities - How To Decks
9 Jan 2018 ... “Number of decking board rows” x “length of the deck in metres” x “1.1″ (multiplying by 1.1 automatically adds on the 10% waste) = total lineal...
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### How much decking do you need? Timber Decking Solutions ...
To work out how many square metres you have to cover, multiply the length by the width. If you have more than one area, add them together afterwards. Step 2 -...
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### Decking Calculators in feet - Decking Calculators - eDeck.com
Square Feet to Lineal Feet Calculator. eDeck sells deck boards by the lineal foot. To figure out how many lineal feet you would need to order, use the simple...
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### How To Install Deck Boards | The Seven Trust Canada
First, decide which way you'll lay your deck boards. Then, measure your deck and divide by the width of your boards to determine approximately how many...
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### Woodworkers - Decking Materials Calculator
Woodworkers - Decking Materials Calculator. Fax number: +353-1-4067759. Customer Details. Date : Time: ... Decking Area (sq m). Decking board Length (m).
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### Garden Decking Cost Calculator: Save £100's | Decking Hero™
Once we understand our total square metre requirements, we can then use this number to calculate how many boards we'll need for our build.
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### Flooring & Decking Calculator - Narangba Timbers
When you are building a timber deck or installing new timber flooring you ... flooring calculator below for a great estimate of the timber boards you'll need, fully costed. ... You should also read our Decking FAQ page which answers many of the...
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### How to Calculate the Framing Size for a Deck
The result is the external width of the decking frame. For example, if the finished decking should have a width of 14 feet and the desired board overhang is 1 inch...
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### Decking calculator - Eurocell
Use our decking calculator to tell you how many boards of our composite decking you will need, plus the clips required to install it.
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### using math to determine number of deck boards for 45 degree ...
18 Feb 2011 ... A deck board cut at 45 degree equals 7 7/8", once you find the hypotenuse would it be fair to say to just deduct 7 7/8" plus gapping...
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### Composite Decking Calculator | NeoTimber® Decking UK
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Get Price | 3,028 | 13,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.873208 |
http://techcodebit.com/tag/coaching/page/3/ | 1,529,444,705,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00045.warc.gz | 315,611,478 | 11,760 | ## Find the Number Occurring Odd Number of Times
Find the Number Occurring Odd Number of Times Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time & constant space. Example: I/P = [1, 2, 3, 2, 3, 1, 3] O/P = 3 A Simple Solution
## Majority Element
Majority Element Majority Element: A majority element in an array A[] of size n is an element that appears more than n/2 times (and hence there is at most one such element). Write a function which takes an array and emits the majority element (if it exists), otherwise prints NONE as follows: I/P : 3
## Given an array A[] and a number x, check for pair in A[] with sum as x
Given an array A[] and a number x, check for pair in A[] with sum as x Write a C program that, given an array A[] of n numbers and another number x, determines whether or not there exist two elements in S whose sum is exactly x. METHOD 1 (Use Sorting) Algorithm: hasArrayTwoCandidates (A[],
## Pseudo-polynomial Algorithms
Pseudo-polynomial Algorithms What is Pseudo-polynomial? An algorithm whose worst case time complexity depends on numeric value of input (not number of inputs) is called Pseudo-polynomial algorithm. For example, consider the problem of counting frequencies of all elements in an array of positive numbers. A pseudo-polynomial time solution for this is to first find the maximum
## What does ‘Space Complexity’ mean?
What does ‘Space Complexity’ mean? Space Complexity: The term Space Complexity is misused for Auxiliary Space at many places. Following are the correct definitions of Auxiliary Space and Space Complexity. Auxiliary Space is the extra space or temporary space used by an algorithm. Space Complexity of an algorithm is total space taken by the algorithm
## Analysis of Algorithm | Set 5 (Amortized Analysis Introduction)
Analysis of Algorithm | Set 5 (Amortized Analysis Introduction) Amortized Analysis is used for algorithms where an occasional operation is very slow, but most of the other operations are faster. In Amortized Analysis, we analyze a sequence of operations and guarantee a worst case average time which is lower than the worst case time of
## Analysis of Algorithm | Set 4 (Solving Recurrences)
Analysis of Algorithm | Set 4 (Solving Recurrences) In the previous post, we discussed analysis of loop. Many algorithms are recursive in nature. When we analyze them, we get a recurrence relation for time complexity. We get running time on an input of size n as a function of n and the running time on inputs of
## Analysis of Algorithms | Set 4 (Analysis of Loops)
Analysis of Algorithms | Set 4 (Analysis of Loops) We have discussed asymtotic analysis, best worst and average cases and asymptotic notations in previous posts. In this post, analysis of iterative programs with simple examples is discussed. 1) O(1): Time complexity of a function (or set of statements) is considered as O(1) if it doesn’t contain | 673 | 3,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-26 | longest | en | 0.875116 |
https://puzzling.stackexchange.com/questions/106057/a-little-riddle-for-you/106116 | 1,719,118,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00723.warc.gz | 414,258,509 | 43,011 | # A little riddle for you
Not mathematical yet I represent a half of sorts,
With furrowed brows and a holey sign I am formed,
My colouring a little 'boorish' you could say,
My opposite North-eastern?!
What am I?
It is a symbol.
Think about what Boor sounds like? And what colour is often associated with that?
• Would [word] apply? It is to be used when the answer is a word. Commented Dec 27, 2020 at 3:43
• No, it is not a word. Commented Dec 27, 2020 at 5:51
• Do you mean "holey" or "holy"? Just checking. Commented Dec 29, 2020 at 14:13
I think the answer might be
Not mathematical yet I represent a half of sorts,
The ♀ symbol represents the female sex, members of which make up roughly half of the global human population.
With furrowed brows and a holey sign I am formed,
From the OP: "furrowed brows" indicates "cross" and "holey" refers to the circle above it.
My colouring a little 'boorish' you could say,
From the OP: 'boorish' is meant to indicate 'like a boar or pig' and refers to the colour pink. There is a common modern connection between the colour pink and femininity and this often leads to the symbol being depicted in pink.
Original Answer: ♀ is also the symbol for the planet Venus which looks coarsely coloured as you can see.
Also "boor" sounds like "beer" and the colour of Venus does somewhat resemble a beer settling.
My opposite North-eastern?!
The symbol for male is ♂ which, as you can see, is a circle with an arrow pointing to the North-East.
• Boor=Boar=Pig which is pink in colour. Commented Dec 29, 2020 at 17:34
• Also furrowed brows=cross and holey = hole. Commented Dec 29, 2020 at 17:39
• Please correct your answer. Commented Dec 29, 2020 at 17:42
• @Smartest1here Thanks, will edit in your suggestions. I should say that, even though a pig is usually pink, a boar is a specific animal which is brown/black in colour. Also, the symbol in question is not really associated with pink but more often red or green. Commented Dec 29, 2020 at 17:45
• @Smartest1here The furrowed brows bit is very clever though. Commented Dec 29, 2020 at 17:47
I think you might be
The angry face emoticon: >:O
Not mathematical yet I represent a half of sorts,
You look like "greater than zero" which is essentially half of the real numbers
With furrowed brows and a holey sign I am formed,
The > indicates the furrowed brow, and the O "holey sign" the open mouth
My colouring a little 'boorish' you could say,
In emoji world (hinted by "coloring"), the equivalent is 😠 which in some platforms has an angry red tint (very much not sure about this clue...)
My opposite North-eastern?!
On a keyboard, north east from > is :, and north east from O is ) - put them together, and you have :) which is the opposite of >:O
• This is not the answer. Commented Dec 28, 2020 at 5:35
• Hint: Think about what Boor sounds like? And what colour is often associated with that? Commented Dec 28, 2020 at 5:40
225
My colouring a little 'boorish' you could say,
the color of rgb 225
With furrowed brows and a holey sign I am formed,
the number's shape.
My opposite North-eastern?! (I got the hint from this)
Northeast (NE), 45°'s opposite is 225°
I don't know about the first clue.
• Hint: The answer is not aiming at any alphanumerical characters per se. Commented Dec 27, 2020 at 13:13
• @Smartest1here What do you mean? Commented Dec 27, 2020 at 13:14
• @Smartest1here So this is wrong? Commented Dec 27, 2020 at 13:19
• Yes it is...... Commented Dec 27, 2020 at 13:55
I think its
St
st is 2 out of 5 letters of sorts.if we rotate st 90degree we see a eyebrow shape and a cross .st stands for sao tome which is a south western country also
• Nope, please think again and improve your answer! Commented Dec 27, 2020 at 6:52
• it is not a word Commented Dec 27, 2020 at 12:50
Lines 1 and 4 make me think of
♀️ (half of the population is female, opposite is ♂️).
I have no idea, what the boorish colour would be (
pink maybe?
) and I really don't see the furrowed brows though.
• A good answer should explain every part of the riddle, because the intended answer will be able to. Commented Dec 29, 2020 at 16:03 | 1,175 | 4,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-26 | latest | en | 0.961265 |
https://www.coursehero.com/file/6645105/Chem-Differential-Eq-HW-Solutions-Fall-2011-161/ | 1,529,952,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00303.warc.gz | 781,986,930 | 91,560 | Chem Differential Eq HW Solutions Fall 2011 161
# Chem Differential Eq HW Solutions Fall 2011 161 - u-axis...
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Section 12.6 Solving Dirichlet Problems with Conformal Mappings 161 f1 t_ Re f z . x 1, y t f2 t_ Im f z . x 1, y t ParametricPlot Evaluate f1 t ,f2 t , t,0,Pi 2 , AspectRatio Automatic Cos t Sin t ParametricPlot Evaluate f1 t ,f2 t , t,0,Pi , AspectRatio Automatic 9. We map the region onto the upper half-plane using the mapping f ( z ) = z 2 (see Example 1). The transformed problem in the uv -plane is 2 U = 0 with boundary values on the u -axis given by U ( u, 0) = 100 if 0 < u< 1 and 0 otherwise. The solution in the uv -plane follows from Example 5, Section 12.5. We have U ( u,v ) = 100 π cot - 1 u - 1 v - cot - 1 u v . The solution in the xy -plane is φ ( x,y ) = U f ( z ). To find the formula in terms of ( x,y ), we write z = x + iy , f ( z ) = z 2 = x 2 - y 2 + 2 ixy = ( u,v ). Thus u = x 2 - y 2 and v = 2 xy and so φ ( x,y ) = U f ( z ) = U ( x 2 - y 2 , 2 xy ) = 100 π cot - 1 x 2 - y 2 - 1 2 xy - cot - 1 x 2 - y 2 2 xy . 13. We map the region onto the upper half-plane using the mapping f ( z ) = e z (see Example 2). The points on the x -axis, z = x , are mapped onto the positive u -axis, since e x > 0 for all x , as follows: f ( x ) 1 if x 0 and 0 <f ( x ) < 1 if x< 0. The points on the horizontal line
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Unformatted text preview: u-axis, since e x + i π =-e x < 0, as follows: f ( x + iπ ) =-e x ≤ -1 if x ≥ 0 and-1 < f ( x + iπ ) =-e x < 0 if x < 0. With these observations, we see that the transformed problem in the uv-plane is ∇ 2 U = 0 with boundary values on the u-axis given by U ( u, 0) = 100 if-1 < u < 1 and 0 otherwise. The solution in the uv-plane follows from Example 5, Section 12.5. We have U ( u, v ) = 100 π ± cot-1 ² u-1 v ³-cot-1 ² u + 1 v ³¶ . The solution in the xy-plane is φ ( x, y ) = U ◦ f ( z ). To Fnd the formula in terms of ( x, y ), we write z = x + iy , ( u, v ) = f ( z ) = e z = e x cos y + ie x sin y. Thus u = e x cos y and v = e x sin y and so φ ( x, y ) = U ◦ f ( z ) = U ( e x cos y, e x sin y ) = 100 π ± cot-1 ² e x cos y-1 e x sin y ³-cot-1 ² e x cos y + 1 e x sin y ³¶ ....
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# Mathematics 2017-2018 CBSE Class 10 Urdu Set 3 question paper with PDF download
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SubjectMathematics
Year2017 - 2018 (March)
1
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#### Topics
1.01 Real Numbers
2.01 Arithmetic Progressions
2.02 Quadratic Equations
2.03 Pair of Linear Equations in Two Variables
2.04 Polynomials
3.01 Triangles
3.02 Constructions
3.03 Circles
4.01 Introduction to Trigonometry
4.02 Heights and Distances
4.03 Trigonometric Identities
5.01 Statistics
5.02 Probability
6.01 Lines (In Two-dimensions)
7.01 Areas Related to Circles
7.02 Surface Areas and Volumes
## Previous Year Question Paper for CBSE Class 10 Mathematics - free PDF Download
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Question Paper for CBSE Class 10 Mathematics are very useful for students so that they can better understand the concepts by practicing them regularly. Students should practice the questions provided in the Previous Year Question Paper to get better marks in the examination. Question papers for CBSE Class 10 Maths question paper gives an idea about the questions coming in the board exams and previous years papers give the sample questions asked by CBSE in the exams. By solving the Question Papers, you can scale your preparation level and work on your weak areas. It will also help the candidates in developing the time-management skills.
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S | 745 | 3,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-30 | longest | en | 0.863574 |
plgblog.nwoca.org | 1,686,287,895,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00355.warc.gz | 501,051,994 | 15,927 | # Google Sheets Pixel Art
Many people use Google Sheets to add numbers, create a budget, keep track of items, etc. However, there is a creative way to use Google Sheets that even preschools can take advantage of…Pixel Art!
This idea was found from the Google Apps for Littles book by Christine Pinto and Alice Keeler, and can be applied to any grade level. The idea is that you use conditional formatting to change the colors of the cells. When a certain number is entered into a cell, that cell turns the color that you designate. For example, based on the picture below, if the number 4 is entered in any cell, once you hit your return key, that cell will turn green.
You can make any color match any value within conditional formatting. Again, based on the picture, if the number 3 is entered into a cell, that cell will turn yellow. Other options are to turn a color based on a date, if the number is greater than something else, text starts with a certain letter, and many more.
This idea can be used for solving math problems, too, which in turn creates a pretty picture.
Teacher puts math problems inside the cells.
Students solve the math problems. If each problem is solved correctly, it makes a picture.
Students can even create their own math problem picture to give to another student to solve. **If you choose to do this, it is helpful for the students to create the picture first on a separate sheet, and then go back and set up the math problems. You can then hide the sheet with the picture, so that the other students don’t see it.
Below are the template links. The only difference between Alice’s and mine is that I added the color gray for number 10, as we needed it for the picture at the top. Remember that anything can be changed in the conditional formatting part of Google Sheets.
Alice Keeler’s Pixel Art Template
Kristie’s Pixel Art Template
## 3 thoughts on “Google Sheets Pixel Art”
1. Denise says:
Hello, I love this activity. I would like to adjust it and use it for 6th grade to make a pixel art photo and then figure out the percent of each color. For higher level students, I’d like to give them the percent that certain items need to be. I made a copy of yours and was able to adjust the cells for higher level math questions than the add and subtract, but I wondered if you could explain how I can make this sheet with a smaller grid. I’d like to go about 14 by 10. Could you help me out with how to create the google sheet you did? Thanks. Denise
2. Denise… to make the grid smaller you just need to delete any rows and columns. I think this grid is 20×20, so you’d need to delete 6 columns and 10 rows to make it work. Let us know if you have questions; we can do a virtual tech support session and answer them in person. Check out training.nwoca.org for details. | 629 | 2,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.925383 |
https://help.scilab.org/docs/6.1.0/fr_FR/gamma.html | 1,610,802,579,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506640.22/warc/CC-MAIN-20210116104719-20210116134719-00354.warc.gz | 379,286,820 | 7,393 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Português - 日本語 - Русский
Aide de Scilab >> Fonctions spéciales > gamma
# gamma
The gamma function.
### Syntax
`y = gamma(x)`
### Arguments
x
scalar, vector, matrix, or hypermatrix of real numbers.
`gamma` can be overloaded for complex numbers or of lists, tlists or mlists.
y
real vector or matrix with same size than x.
### Description
`gamma(x)` evaluates the gamma function at all the elements of `x`. The gamma function is defined by :
and generalizes the factorial function for real numbers (`gamma(u+1) = u*gamma(u)`).
### Examples
```// simple examples
gamma(0.5)
gamma(6)-prod(1:5)```
```// the graph of the Gamma function on [a,b]
a = -3; b = 5;
x = linspace(a,b,40000);
y = gamma(x);
clf()
plot2d(x, y, style=0, axesflag=5, rect=[a, -10, b, 10])
xtitle("The gamma function on ["+string(a)+","+string(b)+"]")
show_window()```
• gammaln — Le logarithme de la fonction gamma.
• dlgamma — dérivée de la fonction gammaln ou fonction psi.
• factorial — factorial function : product of the n first positive integers
### History
Version Description 5.4.0 Overloading allowed for list, mlist, tlist and hypermatrix types. 6.0.2 The input can now be an hypermatrix. `gamma` can now be overloaded for complex numbers. | 381 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | longest | en | 0.464004 |
https://en.formulasearchengine.com/wiki/Talk:Lift_(force)/Archive_2 | 1,657,045,860,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00200.warc.gz | 282,378,525 | 48,802 | # Talk:Lift (force)/Archive 2
The "physical description" of lift reads like an old argument by another name. There's only one right answer: both. Neither momentum nor pressure explain lift on their own, yet are both tremendously important to it. This should be obvious, since each section talks about the other.
Unfortunately, an accurate physical description of lift is not a trivial task and not even perfectly understood by anyone. A more correct description might talk about viscosity, the starting vortex, pressures, and momentum. Mostly it all comes out of the Navier-Stokes equations (which come from Newton's second law).
The mathematical models section has similar problems. The three “models” are convoluted and not explained very well:
• There is no such thing as the "Lift coefficient model". It's just a nondimensionalization for convenience, allowable thanks to the Buckingham π theorem. I temporarily deleted this, but a link to Lift coefficient would be a good addition somewhere since it comes up frequently.
• Bernoulli's principle is not a model of lift, but it does factor into lift calculations: for example, it's used when deriving the Kutta-Joukowski Lift Theorem. The integrals used here to calculate aerodynamic forces do not follow from Bernoulli's equation and should not be labeled as such, so I retitled this section to “pressure integration”, but there's probably something better. This section seems to confuse a variety of concepts.
• The Kutta-Joukowski Theorem section needs to be cleaned up and better explained.
I'll return to this when I have time. Mbelisle (talk) 05:00, 19 April 2008 (UTC)
## New physical description
I started a more complete physical description of lift production. Until I find a good general reference, it's unreferenced and almost certainly has some problems. It would also benefit from diagrams showing the stages of lift production, which might save some of the verbiage. I acknowledge that it may be overly technical at the moment, which I'll work on once I'm sure it's correct. It think it is, at least, better than NASA or The Straight Dope, and who seem content to conclude "Why the flow moves faster over the upper surface is complicated, don't worry about it."
The "Newton" and "Bernoulli" explanations of lift should be added to the common misconceptions section, perhaps as “Newton vs. Bernoulli”, and/or to a section on ways to measure lift. If you know the details of the flow field (or are making measurements in a wind tunnel), then sure either
• the “Newton” way: a direct measurement of the force (via a force balance) or
• the “Bernoulli” way: integrating the pressures (via pressure taps on the surface of the airfoil)
will give you a reasonably accurate value for lift. But neither answers the real question a physical explanation should address: “Why does a wing generate lift?” Michael Belisle (talk) 00:38, 20 April 2008 (UTC)
We must bear in mind that Wiki is a general encyclopaedia and not a technical treatise, it should be written in language and style that is accessible to all. After all, the formulae are derived from principles not the other way round. The article, in my view, should not primarily be a description of the detailed physics or of the derivation of mathematical models but sections on these can be included. I agree that the physical explanation has to focus on addressing the question "why does a wing generate lift" but it should first do it in layman’s terms.
The value of many Wiki articles are effectively destroyed by us technicians expounding nuances that are absolutely irrelevant to the general user.
Your idea that aspects of the Newton and Bernoulli debate would sit very well in the misconceptions area is an excellent one.
I believe it is relatively straight forward to explain why the flow moves faster over the upper surface to a layman and without once mentioning Bernoulli or conservation of energy. I will give it some thought. Rolo Tamasi (talk) 08:24, 20 April 2008 (UTC)
John D. Anderson says “You cannot get more fundamental than this, conservation of mass and Newton's second law” (JDA, Introduction to Flight, p. 355). I use JD Anderson as a reference because, as Curator of Aerodynamics at the Smithsonian and a standard in Aerodynamics texts, he is quite possibly the most authoritative person on the subject. His explanation cannot be presented on equal footing with eskimo.com.
Bernoulli's principle, as applied, comes from Newton's second law while conservation of mass (in 2-D) is simply ${\displaystyle \rho VL}$ is a constant. In the equal transit-time explanation, people have no trouble accepting that that air moving faster has lower pressure than slower moving air, which is an application of Bernoulli's principle. True, the form of Bernoulli's equation applied is actually “Bernoulli's equation for inviscid, incompressible, steady flow along a streamline”, but in some ways that's more information than is necessary. It's the most common form of the equation and the only way to explain lift without mentioning Bernoulli is to not give the principle a name.
That said, I admit that what I put in there is temporarily not written in laymen's terms. I tried to first to put something in there that is correct and referenced (what was there was understandable but misleading). We can go back and make simplifications as appropriate, especially adding pictures.
After I wrote the section on viscous flow and lift startup (from memory), I searched for references in books by reputable aerodynamicists and I came across JD Anderson's explanation, where he explains lift in steady-state terms with conservation of mass and Newton's second law, neglecting viscosity. Technically, this is correct: if you started up a viscous Navier-Stokes solver and solved only for the steady-state flow field, you would never see the starting vortex. You could then describe the flow field as he has done. So I threw in a sloppy version of his explanation. JD Anderson also explicitly describes downwash (i.e. Newton's third law) as an effect of lift and circulation as a mathematical model of lift. His reasoning there is sound.
I have since located a good explanation of the stages of lift production in K Karamacheti, Principles of Ideal-Fluid Aerodynamics and clear diagram in FM White, Fluid Mechanics. The stages are inclusive of the foregoing. I think a good argument could be made to move the stages to another article. But, if I had looked up lift before I was an Aerospace Engineer, I would have wanted to know about viscosity. Failing to mention the starting vortex or viscosity is hand-waving. “The generation of [lift] depends on entirely on the nature of viscous flow past certain bodies.” (Karamacheti) Michael Belisle (talk) 22:18, 20 April 2008 (UTC)
I think what we need to do here is rewrite the article in a way that it conforms in some way to Wikipedia:Make technical articles accessible, without loss of information. It is unavoidable that, like special relativity, a variety of readers will come here, including aerospace engineers who don't understand how a wing produces lift. (Strange but true: I can guarantee that this article will show up in undergraduate lab reports on airfoils. Back when I was a TA, I mostly solved this problem by taking off points for citing Wikipedia because it was often wrong: for example, see Image:LiftCurve.svg or “At a zero angle of attack, no lift is generated,” which was in this article yesterday. I'd prefer not to have to do that.)
Lift, on its own, isn't significant enough to warrant its own article like Introduction to special relativity, but there is certainly a way to organize it in such a way that laymen don't feel it's overly technical, while aerospace engineers don't feel it's overly simple. Perhaps we need something like an Introduction to flight article, but that should involve a team effort within WP:AVIATION about what to include and how to present it.
The physics community is way ahead of aeronautics in communicating technically correct ideas to both general and technical audiences. I think it's a good model to follow. Michael Belisle (talk) 03:36, 21 April 2008 (UTC)
Wow, being a layman I read this article and understood nothing. Then I read the The Straight Dope page as mentioned above and I'm feeling happy again. Even if I don't understand fully and the Straight Dope has cut corners (which it probably has), at least I feel I understand 'lift' now. So can someone please make this page make me feel like that? Then I'll truly be a happy (lay)man.... Malick78 (talk) 16:49, 9 October 2008 (UTC)
I honestly think that a 'lay explanation' based around action-reaction is more intuitive to most people. The problem we experience when explaining lift is that one group of people try to explain how you would calculate lift, which almost always goes the pressure route; and the other group is trying to "explain" lift, without much focus on ease of calculation. Pressure integrals are used to calculate lift because it is easier to use the well-defined boundary of the wing, but it is possible to calculate lift from the deflection of the air as well. In my experience with students (I am a lecturer at a university), the explanation that wings experience an upward and backward force by deflecting air downwards gels with their experience of things like fans. The question of why airfoils are shaped the way they are then largely becomes related to how to deflect the most air, which involves delaying flow separation on the top surface. The pressure integral approach leaves some very interesting questions about why the air is moving faster on the top surface than the bottom one. Of course these two things are different ways of looking at the same phenomenon, but the flow deflection argument requires less build-up about Bernoulli. Chthonicdaemon (talk) 05:05, 4 December 2008 (UTC)
It might be more "intuitive" to some readers, but others will disagree. Like you said, “these two things are different ways of looking at the same phenomenon”; WP:NPOV therefore says we should include both. This article tries to write one explanation that does that, encouraging readers to recognize that the Bernoulli and Newton are equivalent and related. I disagree that airfoil design is a question of how to deflect the most air. You still have to deflect the air in a certain way. The related fact that a rotating cylinder in a free stream generates lift has nothing to do with deflecting the most air. Michael Belisle (talk) 18:49, 9 January 2009 (UTC)
## A summary of where I think the physical explanation should go
I've done some thinking, examined some more sources, and reviewed a little of the prior Bernoulli v. Newton debate here. Of the debate, I see that pretty much everything has already been said.
My current thinking (which is still evolving) is that the final value for lift comes from an equation where, in physical terms, often acceleration is on the left while the pressure gradient, viscosity, and gravity are on the right: effectively ma=F. Which terms are your favorite? (I like viscosity, pressure, and then acceleration. There's no love for gravity from me.)
We should develop one cogent, understandable explanation that develops the whole picture, not obscure parts under the guise of accessibility. Michael Belisle (talk) 01:43, 22 April 2008 (UTC)
The Bernoulli Principle is rather one dimensional as it describes conservation of forces along a streamline (static pressure plus dynamic pressure equals stagnation pressure).
In reality, Newtons Laws of Motion must be valid for each degree of freedom. For the 2 dimensional case presented here, Newtons Laws must be true in two orthagonal axes, whether it is represented as a global x and y basis, or a local streamline and equipotential direction basis.
Bernoullis Principle ignores the forces transverse to the streamlines which act in the direction of net lift.
Also, Bernoullis Principle fails to explain the curvature of streamlines which happen to be caused by applying Newtons Laws of Motion in the transverse streamline direction. "Particles of fluid remain stationary or travel at constant velocity in a straight line unless acted on by an external force". If Newtons Laws are going to be applied they need to be fully considered. —Preceding unsigned comment added by 58.106.38.84 (talk) 14:50, 17 June 2008 (UTC)
The thing that needs to be understood here is that bernouilli explanations aren't wrong- the lift on the wing really is partly due to bernouilli. Neither is the Newtonian stuff wrong- the lift really does involve deflecting air to give lift. Nor is the Coanda effect not involved. The Coanda effect is why the flow curves around the wing in the first place. They're ALL correct. It's like the blind men with an elephant, they're each just a piece of the overall picture. You can look at this different ways and particular features stand out.- (User) WolfKeeper (Talk) 15:46, 17 June 2008 (UTC)
Wolfkeeper is right (except that I think the definition of the Coanda effect should be limited to what Coanda considered, a method for a high-lift device).
Saying that "Bernoullis Principle ignores the forces transverse to the streamlines" is nonsense. Pressure has no direction: it's a scalar. Variations in pressure generate forces (e.g., the net pressure difference between the upper surface and the lower surface). Bernoulli lets you use the pressure at one point to find the pressure at another point, which in turn can be used to find the aerodynamic forces. That's all; it doesn't explain why the flow is the way it is.
But Newton's third law, by the same token, doesn't explain lift either. It just considers an effect of lift (the downwash) and tells you how much lift there is, analogous to what Bernoulli does. The curvature of the streamlines is not explained by Newton's third law. Only in a hypersonic flow does Newton's third law have some direct influence on the amount of lift generated.
A more complete picture of why an airfoil generates lift doesn't start to appear until you consider viscosity. Viscosity is complex, but there is no lift, no pressure difference, and no momentum deflection without it. Michael Belisle (talk) 06:20, 27 June 2008 (UTC)
Michael, the following is my understanding of the logic behind your statement "there is no lift...without viscosity". Is it correct?
Newton's 3rd law is a differential equation. As such, it has an infinite number of solutions. This infinite family of solutions doesn't explain why lift occurs, because each solution gives a different value for lift. So, any imaginable direction or amount of lift is possible.
To get a specific solution for the lift which corresponds to the real world, and thus an understanding of why lift occurs, one must consider viscosity. Doing this allows one to determine that the aft stagnation point must be at the trailing edge. With this added boundary condition, now Newton's law can be used to predict and explain why and airfoil generates lift.
Mark.camp (talk) 20:19, 29 December 2008 (UTC)
## Exact calculation of lift
In the literature are there any examples, however idealized, where lift can be calculated exactly; i.e., without approximations in the math? I would like to find such an example cited for it would help in understanding lift. Renede (talk) 00:07, 8 May 2008 (UTC)
## Euler solver and Lift
When you solve the Euler equations with a finite differences or finite volumes solver to simulate the flow around a wing or a profile, the computed lift matches theory and experimental results. If lift requires viscosity to be generated, where does this paradox comes from: numerical viscosity? Jmlaurens (talk) 12:42, 3 August 2008 (UTC)
This is a good question that may hint at a problem with the treatment in the article. There is no drag in potential flow (D'Alembert's paradox) and no lift without imposing a circulation. You're right that inviscid numerical solvers (like Fluent) don't have this problem. I'll try to reconcile this so that the article addresses this question. Michael Belisle (talk) 17:26, 1 September 2008 (UTC)
Numerical viscosity may be the source, but it looks like it's not an easy answer. Rizzi and Eriksson (1983) consider some possibilities. I think the topic is beyond the scope of the current Lift article. Michael Belisle (talk) 07:55, 21 September 2008 (UTC)
## Coanda Effect
The Coanda effect section has been rewritten numerous times, switching between the version that I think is NPOV and the current version by Ccrummer, which favors one viewpoint without adequate citations. I think it's important to remember that some consider it important to lift in a general sense while others limit it to a jet impinging on a curved surface. I'm in the latter camp and have tried to write the article so considers both viewpoints and uses citations to support any claims.
In its current form, "A common misconception about the causes of aerodynamic lift is that the cause of the Coandă effect is not one of them or at least that it is a negligible component." needs a reputable citation. Nobody I've seen has referred to this as a misconception; the misconception is universally the other way around. Raskin uses it incorrectly, while Scott Eberhart and David Anderson mention it in passing (in a manner different from most texts on the subject of lift, who don't mention it at all), and then proceed to talk about viscosity. Others, for example JD Denker explain that using the Coanda effect to explain lift is a misconception since it's properly limited to effect of a jet impinging on a curved surface that generates additional lift. It was an effect used to generate additional lift by Coanda and never was intended to explain lift in general.
The citations in the section now don't support many of the claims:
• The "real and powerful" effect supported by references 17 and 18 is that of the Coanda effect as a method for a high lift device, which nobody disputes. The dispute is whether or not the Coanda effect should be used to explain lift in general.
• Raskin never really explains lift: he demonstrates the Coanda effect (as a jet exiting from a straw impinging on a curved surface) and then waves his hands saying that he's proved lift. It is not a reputable citation, except to explain his misconception regarding lift. (see comments by JS Denker and the paper by Auerbach),
• Eberhart and Anderson make no such claim that "the Coandă effect should not be ignored in a comprehensive explanation of lift". They merely mention the existence of the effect, and then explain that viscosity causes the fluid to follow the surface. (Which is a false link: viscosity is not enough to explain the most dramatic examples of the Coanda effect and the Coanda effect is not interchangeable with viscosity. They should be two separate topics.)
Additionally, this section grows each time it's revised; there is now a fair amount of information that is here and not in the main article. Properly, this should be a summary of the main article, not an standalone description of the effect. I rather like the main article, in fact.
Maybe the title "Common misconceptions" isn't the best title for the section. By and large, the Coanda effect is something that only comes up as a general explanation of lift in non-technical texts, so maybe something like "Alternative" or "Popular Explanations" would be better, so that it doesn't have to be limited to misconceptions.
At this point, it's more or less a futile edit war, so I'd like to talk about it before I make any more changes. I'd like some more info on what the other camp's reasoning is so that we can come to some agreement. Michael Belisle (talk) 21:13, 5 August 2008 (UTC)
Hi Michael. A debate about lift, and misconceptions related to lift, raged on Bernoulli's principle for a few months. Each time we mentioned that Bernoulli’s principle could be seen in action in the lift on a wing, one user or another would delete it, saying 'It is a fallacy to use Bernoulli to explain lift'. These claims were never accompanied by any citation, reference or source. I posted a “Citation needed” flag against a claim that Bernoulli had no place in the explanation of lift and a kind user added a citation of Anderson and Eberhardt’s book Understanding Flight. The citation didn’t identify a chapter or page so I obtained a copy of the book. That book also led me to Langewische’s book Stick and Rudder. I wrote a section about these two books and their relevance to explaining lift, and posted it in Bernoulli's principle. Eventually that section was transferred to Bernoulli’s Talk page and you can still see it, complete with citations, HERE. It is relevant to this discussion on Lift (force). Dolphin51 (talk) 23:59, 5 August 2008 (UTC)
Since there was no answer in support of the current version, which was just a POV rewrite with no new citations, I reverted the section back to the version dated 23:37, 31 July 2008. The old version can certainly be improved. I'll look more in detail at some of the stuff you wrote for the Bernoulli article when I have time (hopefully in the next few weeks). I think this issue about the Coanda effect stems from a misunderstanding that "Coanda effect" implies "Newton was right", which it doesn't. Even if the Coanda effect were properly applied to why the flow attaches to the wing, we could still use Bernoulli's principal to calculate the amount of lift. In fact, we do use Bernoulli's principle to calculate lift. Michael Belisle (talk) 19:53, 17 August 2008 (UTC)
I reverted the edit of Ccrummer, which does it make appear that the Coanda effect is considered as a serious alternative candidate for the explanation of lift. Moreover, it used Fluid Mechanics of F. White as a reference stating that "... It (the Coanda effect) happens as a result of the viscosity of the fluid as it shears past the curved surface", while I cannot find anything in the 4th edition of White's book on the Coanda effect. -- Crowsnest (talk) 23:47, 12 October 2008 (UTC)
This subsection, and the other one on the "equal transit-time" both lack reliable secondary-sources (see WP:PSTS and WP:RS), i.e. peer-reviewed scientific papers (since this is a scientific subject) in support of these views. They mostly seem to appear in some popular explanations of lift. Further these theories are only qualitative, i.e. there are no reliable sources giving accurate quantitative predictions of lift. Nor are there reliable sources available on scientific experiments in their support. So, these theories do not deserve undue weight, and claims associated with them need some proof before they can be incorporated into the article. -- Crowsnest (talk) 01:24, 13 October 2008 (UTC)
## Alternate physical description of lift on an airfoil
Forward speed and an effective (lift producing) angle of attack of an airfoil results in a coexistant pair of forces pependicular to the directon of travel relative to the orientation of an aircraft: a downwards force exerted by a wing onto the air and an upwards force (lift) exerted by the air onto the wing (in accordance with the third law of Newton's laws of motion). The force exerted by the wing onto the air coexists with a downwards acceleration of air (in accordance with the second law of Newton's laws of motion). The acceleration of the aircraft perpendicular to the direction of travel depends on the sum of the lift force and gravity. In level flight, lift force exactly opposes gravity and is equal to the weight of an aircraft. There are also coexistant forces in the direction of travel, the aircraft exerts a forwards force onto the air, and the air exerts a backwards force onto the aircraft (this component is called drag), and these forces coexist with a forwards acceleration of air.
The forces and accelerations of air and wing coexist with pressure differentials in the vicinity of the wing: lower pressure above (and behind), and/or higher pressure below (and in front). Air accelerates in all directions from higher pressure areas to lower pressure areas, except that a wing, being solid, blocks upwards (and backwards) flow, so the result is a net downwards (and forwards) acceleration of air.
In the case of a flat airfoil, mechanical interaction between air and the bottom surface of the air foil deflects air downwards. If the angle of attack isn't too high, the air will also tend to somewhat smoothly follow the upper surface downwards due to "void" effect (from wing: a low pressure region is generated on the upper surface of the wing which draws the air above the wing downwards towards what would otherwise be a void after the wing had passed.). At the front of the air foil, the air seperates into two flows, and because of the pressure differential (lower above, higher below), some of the air is diverted upwards, lowering the seperation point to below the leading edge of the air foil. If the leading edge angle isn't too steep, or too sharp, then a Coanda like combination of surface friction and vicosity allow the air to bend around the leading edge, and this bending also corresponds to a net downwards acceleration of air, and the coexisting pair of vertical forces (air upwards on wing, wing downwards on air).
Outside of mechanical interactions, there is no net work done on the air, and when no net work is done, then there is a coexistant relationship between acclerations, forces, pressures, and velocities that follow Bernoulli's principle.
The acceleration of air results in an increase in the kinetic energy of the air. An efficient airfoil shape generates most of it's lift force via reduction in pressure in the vicinity above the wing and minimal net mechanical interaction below, which allows a Bernoulli like conversion of pressure energy into kinetic energy, reducing the amount of power required to produce lift. The other way to make a wing more efficient is to reduce the amount of kinetic energy required to produce lift by increasing the size of the wing, because a larger mass of air accelerated at a lower rate consumes less power. It turns out that it's more efficient to make the wing span longer than than the wing chord wider. Jeffareid (talk) 18:26, 13 September 2008 (UTC)
This sounds like suspiciously like original research. Remember WP:NOR and WP:V, if you're trying to incorporate this into the article. Michael Belisle (talk) 06:55, 21 September 2008 (UTC)
Void theory (as opposed to Coanda effect) is already cited in the artile on wing. Bernoulli principle relates pressure and velocities in a work free environment, and these relationships don't hold in the immediate vicinity of a wing, specifically the pressure jump zone where work is done, but Bernoulli does apply outside this zone. This article on propellers explains what I'm getting at: But at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine violates an assumption used to derive the equation. propeller analysis. Note that a wing, similar to a propeller, operates in it's own induced wash, often ignored in streamline diagrams. As far a Newton goes, the 3rd law, forces only exist in pairs, explains why downforce from a wing coexists with upforce from the air which pretty much explains lift. Newtons 2nd law relates forces to mass and acceleration. incoporate This section was added as food for thought, I'm not sure how to reword it for the actual article, but it's good enough for the discussion section. Jeffareid (talk) 01:04, 28 September 2008 (UTC)
As far a Newton goes, the 3rd law, forces only exist in pairs, explains why downforce from a wing coexists with upforce from the air Okay. which pretty much explains lift. No. That's just one piece of the big picture. “Several physical principles are involved in producing lift. Each ... is correct .... [But] do we get a little bit of lift because of Bernoulli, and a little bit more because of Newton? No, the laws of physics are not cumulative in this way. There is only one lift-producing process. Each of the explanations ... concentrates on a different aspect of this one process.” http://www.av8n.com/how/htm/airfoils.html#sec-consistent
Bernoulli's principle absolutely applies in the context of lift on a wing. If it didn't, every aerodynamicist might as well just quit now. Apparently we've been wasting our lives measuring pressures to determine lift. (There are other ways, of course. In a wind tunnel, we can directly measure lift using a force balance. But we don't generally determine lift by measuring the downwash simply because it's not easy to measure.) While a propeller shares some things in common with a wing, it's really a different problem that requires a different approach.
My awkward wording is why I didn't suggest this as a proposal for inclusion in the main article. I didn't intend to make this a Newton versus Bernoulli debate. Obviously both principles are involved in lift, Newton explains the relationships between force, mass, and acceleration. Bernoulli explains the relationship between pressure differentials and acceleration of air. It would seem that an explanation of lift should focus on how those pressure differentials are created (angle of attack and air speed), then continue to explain the coexistant relationship between forces, pressure differentials, and acceleration of air.
There is that nagging issue that the final result is a change in total energy of the air, which means that during the process of producing lift, a non-Bernoulli like interaction between air and airfoil occurs, and how this occurs should be described, although I haven't seen a good description of this process, just a note that work was done. This is more obvious in the case of a propeller because a propeller is typically less efficient than a wing. Jeffareid (talk) 11:18, 28 September 2008 (UTC)
The way to reword it for the article is to use a citation to back up any claims with a citation. I see the word “void” in the wing article, but it is rather poorly written and not cited. I had never heard the term “void theory” until you mentioned it. Google suggests that you are the only person to use that term when discussing lift. If you have seen the term somewhere, then we need that source to verify what you say. If not, then there's no way to work it into the article. Michael Belisle (talk) 05:18, 28 September 2008 (UTC)
"Void theory" is my terminology, but I don't think I was the first to use it as such, and it was someone else that used it in wing. In the case of landbased vehicles, it's often explained that most of the drag is due to the void at the rear of the vehicle. Void theory seems to be a much better explanation of why air follows the upper surface of a moving inclinded plane than Conada effect which to me implies an effect related to friction and viscosity. Void theory - air molecules are colliding with surface (or the surrounding boundary layer) of a moving airfoil, but as the uppper surface passes by, it leave a downwards and forwards moving "void" that requires that the air molecules travel farther before they collide with the surface, resulting in a net downwards (and forwards) acceleration of air. I'm not aware of another term used to decribe the interaction between air and movment of a surface "away" from the air.
Void citations are rare, but I found another one in addition to the wiki article on wings:
draws the air above the wing downwards towards what would otherwise be a void after the wing had passed wing
The plate scoops out a void how planes can fly
Jeffareid (talk) 11:28, 28 September 2008 (UTC)
## Stages of lift production
The upper stagnation point continues moving downstream until it is coincident with the sharp trailing edge (a feature of the flow known as the Kutta condition). If this statement is true, then where is the sharp trailing edge on these pre-shuttle prototype lifting bodies? No thin sharp trailing edge, plus they have flat tops (tapered tail but not sharp) and curved bottoms. Being potential re-entry vehicles, I'm sure the lift to drag ratio was poor (my guess between 5:1 to 10:1), but they worked:
The M2-F3 had a top speed of mach 1.6, so its drag wasn't excessive. Jeffareid (talk) 15:30, 27 September 2008 (UTC)
The statement is true when you, as the article says, "consider the flow around a 2-D, symmetric airfoil at positive angle of attack," pictured as a NACA 0012 (which of course has a “sharp” trailing edge). But the article never says that such a trailing edge is the only way to generate lift. Try not to, ummm, deny the antecedent. Michael Belisle (talk) 22:43, 28 September 2008 (UTC)
Well I got the impression that the article was implying the sharp edge and Kutta condition as requirements for lift. I read too much into that, sorry. Jeffareid (talk) 02:44, 29 September 2008 (UTC)
## Old discussion: attack on the use of the Bernoulli equation as a partial explanation of aerodynamic lift
I moved this old discussion from Talk:Lift (force)/Comments to here, since it does not belong there and was corrupting the appearance of the table of contents on this discussion page (in interaction with the Template:Physics at the top of this talk page, which uses Talk:Lift (force)/Comments). I expanded the <ref> by Dolphin51 to prevent the need of using a {{reflist}} template. -- Crowsnest (talk) 07:47, 10 October 2008 (UTC)
This is an attack on the use of the Bernoulli equation as a partial explanation of aerodynamic lift.
Bernoulli's equation for the conservation of energy is commonly used to explain aerodynamic lift, however the equation assumes a steady flow, i.e. flow in which particles follow velocity streamlines. A stream line is a line of constant velocity in the flow. When the particles themselves do not follow the streamlines, they are changing velocities, that is, they must cross the streamline bundle. Shear flow is an example of non-steady flow.
The flow near the airfoil surface is shear flow because of the existence of the boundary layer, a layer of air in which the velocity changes across the flow; the layers of air shear past one another, faster layers toward the ambient flow and slower layers near the airfoil. This is non-steady flow and Bernoulli's equation is therefore not applicable in this region. (See pg. 9ff. Fluid Mechanics. Landau, L. D. and E. M. Lifshitz. (1959.) Addison-Wesley: Paris.
Here is the beginning of a correct explanation of this part of lift:
Upwash (from the bottom of the wing at angle of attack) joins with the main flow at the leading edge to envelop the top of the wing. The combined flow interacts with the boundary layer on the curved part of the top of the wing and, for sufficiently small angles of attack, depletes the boundary layer population there, i.e. the pressure on the top is decreased, by the "blowing" of particles away from the wing's surface as it curves away from the flow. It is the curvature of the wing that is crucial in this component of lift. This decrease in pressure on the top of the wing adds to the lift. It can be seen in the Coanda effect, i.e. a smoke stream near the curved surface is deflected toward the surface by the higher pressure in the flow (the ambient pressure). The difference between the lower pressure at the top surface and the higher pressure under the wing results in lift. —Preceding unsigned comment added by Ccrummer on 27 January 2007.
I have just found these comments published by Ccrummer in January 2007. There is universal agreement with Ccrummer that Bernoulli's principle is not applicable in the boundary layer because BP is stated to be applicable to inviscid flow (or flow that closely resembles inviscid flow) and the flow in the boundary layer is clearly not inviscid flow. However, the great majority of air affected by an airfoil is not part of the boundary layer and its flow closely resembles inviscid flow.
Ccrummer’s "correct explation of this part of lift" looks like original research. There is no reference or citation to indicate where the ideas come from, or whether these ideas have ever been published by someone. Wikipedia is not the place for our original research or our personal views on what we believe is true. See Wikipedia:No original research.
Ccrummer’s stated views are at odds with statements by most (all?) specialist authors in the fields of fluid dynamics and aerodynamics. For example, in the excellent book Aerodynamics, L.J. Clancy has written “When a stream of air flows past an airfoil, there are local changes in velocity round the airfoil, and consequently changes in static pressure, in accordance with Bernoulli’s Theorem. The distribution of pressure determines the lift, pitching moment and form drag of the airfoil, and the position of its centre of pressure.” See Clancy, L.J. (1975), Aerodynamics , Section 5.5, Pitman Publishing Limited, London. ISBN 0 273 01120 0.
Ccrummer has written “steady flow, i.e. flow in which particles follow velocity streamlines. A stream line is a line of constant velocity in the flow.” I have read many books on fluid dynamics, but I haven’t ever read of a “velocity streamline”. Also, when Ccrummer writes “A stream line is a line of constant velocity in the flow” he is at odds with all the authors I have read in this field.
The bottom line is - where did this information come from? The threshold for inclusion in Wikipedia is verifiability, not truth. See Wikipedia:Verifiability. See also Wikipedia:Wikipedia is an encyclopedia.
In my view, as "an attack on the use of the Bernoulli equation as a partial explanation of aerodynamic lift" these comments have failed completely. Dolphin51 (talk) 12:43, 25 May 2008 (UTC)
Links: lift #1 lift #2 lift #3. Perhaps correspondence with the authors of these articles will help. It goes beyond my understanding. Jeffareid (talk) 13:30, 10 October 2008 (UTC)
The article at present makes an attempt to include Scott and Eberhardt (which can be improved), but their view is not really at odds with anything that's written (except when they invoke the Coanda effect as a surrogate for viscosity). The third link appears to just be some guy writing with a giant blue font. It's not an appropriate reference. See WP:SPS.
I agree with Dolphin51 here (more or less). Michael Belisle (talk) 04:24, 11 October 2008 (UTC)
I left out a link. A Physical Description of Flight. Via correspondence with one of the authors of this web page, I got the (perhaps false) impression that Bernoulli was an issue. As I mentioned before, it was beyond me, but assuming the goal here is to provide accurate information, perhaps someone could correspond with these guys, who are appparently AE's, to get their feedback. Jeffareid (talk) 16:23, 12 October 2008 (UTC)
## Unclear sentence
This sentence... "Relatively speaking, the bottom of the airfoil presents less of an obstruction to the free stream, and often expands as the flow travels around the airfoil, slowing the flow below the airfoil." ...seems to be ungrammatical. The subject of "expands" is "the bottom of the airfoil", which is probably not what the writer intended.
I'm not able to suggest a correction because I do not know what the writer was trying to say. First, I'm not able to form any mental picture of the "bottom of the airfoil" creating "less of an obstruction" (than the top?). (What does it mean to speak of the top of the foil presenting an obstruction? ) Secondly, I am unable to think of anything in the actual physics of aerodynamics that would support this explanation, and I've never encountered this concept in any text books or articles on the subject. Mark.camp (talk) 00:24, 27 December 2008 (UTC)
I agree Mark. The sentences you have identified represent a poor attempt to explain the variation of fluid speed in the flow field around an airfoil. No source has been cited so it is reasonable to assume that the editor who added this text was using his own impression of things, and his own ideas to explain them. I have deleted the offending sentences. The original editor, and others, are of course free to re-insert these ideas but they should be expressed in a clear fashion and preferably supported by a citation of the source from which the ideas were taken. Dolphin51 (talk) 09:57, 27 December 2008 (UTC)
I agree that the wording could be clearer, but the idea comes from Anderson's Introduction to Flight, starting on p. 353 (as stated at the start of the section “following the development by John D. Anderson in Introduction to Flight”). I put the sentences back in because they're important to the section and I'll rework them to be clearer later. The “relative” obstruction can be seen in the streamline diagram. Michael Belisle (talk) 22:40, 2 January 2009 (UTC)
Michael, I just read the section you cite in Anderson, and unfortunately it is no clearer to me what he was trying to say than in the extract you've included in the article.
In fact, one quote of his was very disturbing: "The airfoil is designed with positive camber; hence the bottom surface of the airfoil presents less of an obstruction..."
It is clear from that sentence that Anderson is laboring under a form of the misconception that was univerally published in high school and undergraduate physics texts until recent years, that the non-zero circulation of the flow outside the boundary layer, and the resulting pressure difference, is caused by the camber of a typical airplane wing, rather than by the Kutta condition, etc.
I'm sure if you asked 100 pilots (as opposed to aeronautical engineers) today, 99 of them would use the same fallacious explanation, based on camber, that Anderson does.
Because of this, I feel strongly that it's important to remove Anderson's explanation from the article altogether, rather than try to repair it, and replace it with a scientifically sound explanation.
I'll hold off doing this till we have agreement.
Mark.camp (talk) 04:03, 5 January 2009 (UTC)
I recognize your opinion about camber, but it's not critical to his explanation. Even so, of course camber generates lift. It works in concert with the Kutta condition, which does not generate non-zero circulation on its own. For example, if we have a symmetric airfoil in a free-stream at zero angle of attack, there is no lift. There is still, however, an infinite number of valid potential flow solutions. The Kutta condition allows us to choose the correct one (i.e., that there is no circulation and hence no lift). If we add positive camber or increase the AoA, then lift is generated and the Kutta condition allows us to determine how much circulation is generated. All these concepts work with each other; no one concept explains lift on its own. If you do wish to argue the problems with camber and lift, then of course you need a citation for it.
Regardless, I disagree with the idea of removing it. We could go back to the old edition if you prefer, but I think that'd be a mistake. A recurring problem with this article is that editors would repeatedly insert their personal favorite explanation of lift. This current explanation is an improvement over what preceded it and it's better than someone continuing to think that Equal Transit Time explanation is correct. But it obviously still needs some work.
The solution is not to “throw the baby out with the bath water” and say that because this explanation is imperfect, it shouldn't exist. WP:WIP is just an opinion, but it's one that I sympathize with. And there is no universally accepted, perfect solution of lift. There are different ways of looking at lift, but there's really only one explanation. I think this article should strive to be inclusive without favoring one over the other, and in particular explain how the different explanations complement (not contradict) each other. Michael Belisle (talk) 04:43, 9 January 2009 (UTC)
Michael, do you agree with this sentence: "A wing with positive camber will generate negative, zero, or positive lift, depending up on the angle of attack, and the same is true of a wing with zero camber or negative camber?"
I do. The sentence implies that camber is not critical to generating lift, whereas the Kutta condition definitely is.
A couple of things you say above about the importance of camber in generating lift make me wonder if we agree on the sentence. If we don't, then we first need to come to agreement on it.
Reason: if my statement is false, then Anderson's explanation of lift, which depends on camber being critical, is fine and my objection to it is invalid.
Note: I do realize that the amount of camber, negative or positive, is very important in determining at *which angles of attack* you will get zero, negative, or positive lift. For example, an airplane with a positive-cambered wing has to fly at a larger angle of attack when flying upside-down (when it essentially has a wing with negative camber) than it does flying right-side up.
I also am not ignorant of the fact that the reason for positive camber in airplane wing design is that a positive-cambered wing has a higher angle of attack at stall, which is important feature not only for safety but because it also results in a higher maximum co-efficient of lift.
Mark.camp (talk) 18:27, 9 January 2009 (UTC)
Sure, and this article mentions the Kutta condition where appropriate. But you'll notice that this WP article doesn't mention camber (which is an oversight, but never mind that for a moment). The article uses AoA to achieve the same result as Anderson. (I chose AoA arbitrarily when I made the reference diagram. It could just as easily be changed to an airfoil with camber.) Anderson just uses camber because that was a simple case he chose to explain the concept of lift: An airfoil with positive camber at zero angle of attack generates lift; hence, Anderson's explanation is correct. But nothing he nor this article says tries to imply that camber is the only way to generate lift. Also, remember that the Kutta condition is not necessary for lift: where do you apply the Kutta condition on a rotating cylinder in a free stream [1]? Or on the examples mentioned above? Michael Belisle (talk) 19:17, 9 January 2009 (UTC)
Michael, I will answer one of your questions. The Kutta condition is undefined for a rotating cylinder. Therefore, it cannot be used to imposed the boundary conditions needed to find a unique solution to the potential flow equation, and thus the lift. For potential flow, there IS no unique solution--sans friction, you have to impose boundary conditions arbitrarily to get a specific solution to the differential equation. To find the real-world solution requires that you take friction into account (exactly what Kutta did for the case of airfoils!). In fact, if you look on the web, you will find animated diagrams of the potential flow around a cylinder, or even a NACA wing section, where you can impose boundary conditions by moving the mouse. A continuum of solutions, each with its own flow and pressure fields, and its own value of lift, is shown. In no case does the air know what the "angle of attack of the flow" or the "camber of the object" or the "chord line of the object" is. These are human descriptive conventions with no mathematical or physical significance, and they aren't even defined terms for objects other than wings. If they had anything to do with explaining lift, then in most physical cases, the wind would shrug it's shoulders and say something "this boulder has an undefined chord, so my angle of attack is undefined, so I don't know how much lift to create". Apologies to Anderson, but that's the way it is. He was brought up on the same physics textbook rubbish as the rest of us, and some of it lingers in his mind. (IMHO ;-)
But you raise more points than I can answer. One of us has so many subtle misconceptions about fluid dynamics that it is not possible to have a meaningful discussion without a blackboard, a beer, and about three hours to kill. I will assume that it's I who misunderstands, as I'm not a scientist, just poor working stiff, and will take my leave for now. I've enjoyed the debate and I thank you for it.
Mark.camp (talk) 22:42, 9 January 2009 (UTC)
On 27 December 2008 Mark.camp drew our attention to some unclear sentences in Lift (force) (see above). I agreed that the offending sentences did nothing to add clarity or valuable information to the article so on 27 December I deleted the sentences. On 2 January Michael Belisle restored the sentences, saying he would rework them to be clearer later. I have no objection to Michael reworking material so that it is clearer, but considering how unclear these sentences are, I see no point in retaining them while Michael does his rework.
These sentences do not simply fail to add anything - they confuse and mislead. The article is improved when they are no longer present. Here is my analysis of the offending sentences:
• Relatively speaking,
Not a good start. These words are redundant. It is not clear what relationship, or relativity, is intended. Beginning a sentence with the words “Relatively speaking,” is a bit like beginning “By the way,”. It might be fashionable but it is not good written expression.
"Relatively speaking" means the top and bottom relative to each other.
• the bottom of the airfoil presents less of an obstruction
Less than what? Less of an obstruction than the top of the airfoil? If that is what is intended it should be stated explicitly. However, this would probably make Wikipedia the only document in existence that talks about the top surface and bottom surface of an airfoil presenting obstructions to the free stream, and the two obstructions being different. Does John D. Anderson really explain circulation in terms of the airfoil obstructing the free stream? I doubt it. Wikipedia already explains circulation about an airfoil in terms of the Kutta condition.
No. We are not discussing circulation here. He considers circulation shortly after this explanation explaining that it's a mathematical concept, not a physical explanation of lift. But, Anderson does indeed talk about lift in terms of obstructions; see the bottom of page 353: "As stream tube A flows to the airfoil, it senses the upper portion of the airfoil as an obstruction and stream tube A must move out of the way of this obstruction. ... the bottom of the airfoil presents less of an obstruction to stream tube B, and so stream tube B is not squashed as much as stream tube A...."
• and often expands as the flow travels around the airfoil
Taken at its face value, this is saying the bottom of the airfoil is expanding as the flow travels around the airfoil! What was intended was probably that the flow often expands as it travels around the airfoil. High quality English expression is very important in any encyclopedia, and especially in Wikipedia, so this sentence must disappear, at least until it is cleaned up.
That's supposed to refer to the streamtube.
• (Contrary to the equal transit-time explanation of lift,
In any good explanation of lift, no purpose is served by diverting sideways to assert that the equal transit time theory is not a good explanation of lift.
If we say that the flow slows down as it goes over the bottom and speeds up over the top, then I think it's possible that some people will think that the molecules are going to meet because the Equal-Time Theory is so pervasive . It's important to highlight here that this is not the case.
I will again delete the offending sentences. I would appreciate it if their content is not re-instated until it is reworked into high quality ideas and high quality English expression with appropriate citation of sources. Dolphin51 (talk) 01:58, 7 January 2009 (UTC)
I respectfully disagree. If something is unclear, the solution is to make it clearer, not to delete it (i.e., I agree with WP:WIP). I have addressed your specific issues. It sounds like you haven't looked at the reference; if you had, you might have been able to identify what the passage was trying to say and improved upon it. Michael Belisle (talk) 04:26, 9 January 2009 (UTC)
Michael, thanks for leaving some comments on this Talk page. I have access to Anderson's Fundamentals of Aerodynamics, but not Introduction to Flight so I am grateful that you have quoted some actual text from the book. However, with expressions like "so stream tube B is not squashed as much as stream tube A...." we are clearly not looking at a serious book on fluid dynamics. (The opening sentence in this article establishes that Lift is a concept in fluid dynamics. Nothing is said about Flight.) I won't say Introduction to Flight should not be used as a source in Lift (force) but I will question whether it is a suitable source to set the strategic direction for a scientific article about fluid dynamics.
Wikipedia must make sense to all readers, not just those who have a copy of the source document beside their computer so they can work out what the Wikipedia article is about. The burden is on the contributor to ensure additions are accurate and clearly expressed. Where users feel additions are inaccurate or poorly expressed they are encouraged to be bold and delete (unless they can improve, of course.) Keep up your good work. Dolphin51 (talk) 06:30, 9 January 2009 (UTC)
Introduction to Flight is a serious book; it's irrelevant whether it's a “fluid dynamics” or a ”flight” book since the introduction might as well say “Lift is a concept in flight”, but I think fluid dynamics encompasses “flight”. The book is written for freshman in Aerospace Engineering, so it's written in language to be clear and understandable to people who know nothing about aerodynamics. And like you said, “Wikipedia must make sense to all readers”: although it should be technically accurate, it should be written in a language that's understandable. I think that's exactly what makes Introduction to Flight an excellent reference for Wikipedia. It's not like John D. Anderson, currently Curator of Aerodynamics at the National Air and Space Museum, is some charlatan.
In reference to "not just those who have a copy of the source document", that makes sense for (some) readers. But I feel that editors should check the source document if they feel something is unclear or misinterpreted. That's why we require Wikipedia to be verifiable, which requires reading the source document to verify what's written in Wikipedia:
“Any material lacking a reliable source may be removed, but editors might object if you remove material without giving them sufficient time to provide references, and it has always been good practice, and expected behavior of Wikipedia editors (in line with our editing policy), to make reasonable efforts to find sources oneself that support such material, and cite them.”
If you just delete something because you don't understand, it makes it hard for someone else to improve it later (because now it's somewhere in the history, and one has to go back, search through the history, revert, revise the sentence, etc.). In this case, I was busy for the past few weeks, but I specifically said I would come back to it and revise it when I had time. There is also this in the list of WP:MISTAKES:
Deleting useful content. A piece of content may be written poorly, yet still have a purpose. Consider what a sentence or paragraph tries to say. Clarify it instead of throwing it away. If the material seems mis-categorized or out of place, consider moving the wayward material to another page, or creating a new page for it. If all else fails, and you can't resist removing a good chunk of content, it's usually best to move it to the article's "Talk page", which can be accessed using the "discussion" button at the top of each page. The author of the text once thought it valuable, so it is polite to preserve it for later discussion.”
Michael Belisle (talk) 18:17, 9 January 2009 (UTC)
My difficulty with the current text is Relatively to the top of the airfoil, the bottom presents less of an obstruction to the free stream. I will refer to this as the Differential obstruction theory. Apparently this is sourced from Anderson’s Introduction to Flight so I have arranged to obtain a copy of that book to check exactly what Anderson has written. (I acknowledge that Anderson is not a charlatan, but have a look at Appeal to authority.)
Work by early aerodynamicists and mathematicians such as Joukowski and Kutta observed that the flow around one side of a lift-generating airfoil was significantly faster than around the other. As you would expect of professional scientists, these people did not get distracted trying to explain why it was so. We know that the difference in speed of the flows around the two sides of a 2-D body is dependent on the body having a cusped (sharp) trailing edge. Mathematically, it is also dependent on irrotational flow. Apart from those two considerations, scientists are happy to observe the difference in speed around the two sides of an airfoil, and they don’t offer an explanation as to why. (In the same way, Newton’s Laws of Motion state what has always been observed, but without attempting to explain why.)
Newcomers to aviation often ask why there is a difference in speed around the two sides of the airfoil. There have been numerous attempts to provide simple, easy-to-understand explanations of why there is a difference in speed. The Equal Transit Time Theory is perhaps the best known. It is simple and easy to understand, but unfortunately it is incorrect. If one asks “Why must the two streams transit the airfoil in equal times?” there is no satisfactory answer. Consequently, we don’t tolerate credibility being given in Wikipedia to the Equal Transit Time Theory, even though there are many books that can be cited as the source of this Theory.
I am keen to see if Anderson provides an answer to my question How do you know the bottom of the airfoil presents less of an obstruction to the flow than the top of the airfoil? I suspect the Differential obstruction theory is no better than the Equal Transit Time Theory. If that is so, I will be advocating that Lift (force) simply states that the flow speeds around the two sides of a lift-generating airfoil are different, without attempting to explain why, in the same way that Wikipedia’s treatment of Newton’s Laws of Motion makes no attempt to explain why.
All will be revealed when I see the book. Dolphin51 (talk) 02:20, 13 January 2009 (UTC)
Anderson's "differential obstruction", "equal transit-time" and "Air ESP" are all motivated by the intuitive need to explain the BEHAVIOR at the front by CONDITIONS at the front--the camber is this, the angle of attack is that, the air "feels an greater obstruction", etc. Or in the case of equal transit time explain the BEHAVIOR above and below the foil by the CONDITIONS there: the curvature and thus the path length.
Why this need? Because our understanding of causation, which is based on discrete analysis of a small number of interacting bodies, rebels against the suggestion that the behavior of every parcel, even those in front of the wing, is causally dictated by the conditions at the TRAILING edge. "How could causation flow backward? Doesn't a parcel first reach the leading edge, and later the trailing edge?"
Of course, that is exactly what does happen: the Kutta condition--THE CONDITIONS AT THE BACK--completely determine the flow everywhere. Even in front of the wing! (Of course, one must also assume Newton's law P'(x,y)=F(x,y) where P is Momentum vector, plus the values of the F being given by the Bernoulli equation. But these two laws are NOT usually the problem for the student. He has already taken the time to understand them.)
No-one forces the student to confront this seeming contradiction. Therefore, he assumes that he must have stumbled down the wrong path, or that he is too stupid to understand this, and he gives up. Then someone comes along with an "explanation" which doesn't require him to face the paradox. Now, it is logically necessary that this new "intuitive" explanation violates Newton's laws or the Kutta condition (or Bernoulli), but the student never realizes this--he read it in a book, so it must be true. Wikipedia's rules are of no help because they only require an author to find a reputable source, not a correct source. Since the first false theory of this type came out in a 1920's physics text (if I remember the history right) and was copied in a thousand text books, there is always a reputable source for the false explanations.
Mark.camp (talk) 19:19, 14 January 2009 (UTC)
Hi Mark. Yes, I agree with all you have written. In thermodynamics there is a challenging concept called entropy. Wikipedia has an article on Entropy, but in addition it has a supporting article called Introduction to entropy. Perhaps Wikipedia needs Lift (force) to contain rigorously correct information, and a new article called Introduction to lift to contain suitable information for newcomers to what is a rather challenging concept.
I suggest that when you reply to another user's post you add it at the end, rather than interleaving it amongst the other user's text. By interleaving your additions, sense can only be made of the other user's post by reading it from the History tab. I have rectified this situation immediately above. Dolphin51 (talk) 22:30, 14 January 2009 (UTC)
I think Wikipedia needs an article called Introduction to Flight (maybe Flight is that article) or Introduction to Aerodynamics, of which Lift would be one topic covered. Lift, on its own, doesn't have enough content to warrant the attention of its own introduction article, since that can be done just as well in the introduction to the present article.
As I said before, everything works together: the Kutta condition, aifoil thickness, angle of attack, camber, etc. With regard to Mark.camp's overemphasis on the Kutta condition, consider truncating an airfoil at the 50% chord: the upstream flow would be largely unchanged, but the drag would be enormous. Nothing here violates the Kutta condition, Newton's laws, or Bernoulli, so I'm not sure what is meant by saying that a simplified explanation necessarily violates the truth.
How do we know that the top presents more of an obstruction than the bottom? The answer is geometrical based on looking at the picture of an established flow. But I think there's a problem with bringing it up in this section when it's trying to be a section that explains lift in an established flow. It properly belongs in the next section, which answers the question "How did the flow get to be this way?" (which notably includes a mention of the role of the Kutta condition). I took out the mention of obstructions in this section, but It should be incorporated into the next section.
The most import thing to remember when writing this article, I think, is that there is no universally accepted explanation of lift. You may think your explanation is “correct” and you may find some knowledgeable people who agree with you. But you'll also find knowledgeable people who disagree. As long at the topic is “controversial”, the Wikipedia article can't pick sides. Michael Belisle (talk) 00:24, 15 January 2009 (UTC)
Michael, since you feel that I "overemphasize" the Kutta condition, here is an explicit statement of what I think the role of the Kutta condition is in explaining lift. I will choose one of the standard fluid dynamics derivations which are used to prove that an airplane wing must develop lift, namely 2D potential flow theory.
This derivation assumes a representative airfoil section and orientation. From that it develops a boundary value problem from Newton's second law. A family of solutions is found, from which an infinite range of values of lift is proved. Thus, Newton's law is sufficient, under the simplifying assumptions of 2D potential flow theory, to prove that it is POSSIBLE for an airplane wing to create positive lift, but nothing more. Lift is not yet explained. Next, the boundary value problem is solved, meaning a unique solution is found, by imposing the Kutta condition. Finally, one proves that this unique solution has positive lift.
As you can see, in this derivation, (a) lift is NOT proved until the Kutta condition is imposed, and (b)once it IS imposed, lift is proved immediately. This is exactly how important I believe the Kutta condition is to the above proof of lift of an airplane wing: it is necessary.
Mark.camp (talk) 15:27, 15 January 2009 (UTC)
I'm not sure what you mean by "proving" lift. What you have presented is a simplified, mathematical description of lift. Nature does not solve a boundary value problem. It does not calculate a result from the NS equations. These are but tools to simplify the real picture.
The Kutta condition is not the cause of lift. It is a consequence of the underlying physics, useful in situations where certain approximations are made. For example, given a set of conditions, there is not an infinite set of solutions unless you neglect viscosity as one does in potential flow. If you solve the viscous Navier-Stokes equations directly, there is only one solution and it arises naturally without imposing the Kutta condition. (I should say that there is typically only one solution. The uniqueness and existence of solutions to the NS equations over the whole domain is still an open question. No one has yet found a non-unique solution, but that's not to say that there isn't one out there. Engineers don't really care about that question because the equations work in the domain of our typical design space.)
Did you know that in the real world, a slender ellipse at an angle of attack generates lift? It's like D'Alembert's paradox for the Kutta condition, though less interesting and not very useful. Michael Belisle (talk) 19:35, 15 January 2009 (UTC)
Yes, I did know that a slender ellipse at a positive angle of attack generates positive lift. I believe that this lift is not predicted ("explained", "calculated", "demonstrated", "proved",...) by 2D potential flow theory, if no ad hoc assumptions are made. Am I correct?
Mark.camp (talk) 22:52, 19 January 2009 (UTC)
Michael, when I speak of the importance of the Kutta condition, I'm referring only to the explanation of lift given in the article section we are discussing--a typical NACA airfoil with one sharp edge aft. To explain this simple case to the reader is enough of a challenge, perhaps even impossible!
To discuss more general cases like the one you brought up, or even the familiar case of holding one's hand, or a flat thin sheet of metal, at an angle out the car window, requires much more complex mathematics. The Kutta analysis has to be generalized: it needs to be shown that when there is more than one sharp edge, they are not all equal in their power to control the circulation. For example, in the case of a thin flat surface at positive angle of attack, if we try to simply extend the Kutta argument for the NACA section, we would perhaps apply a "Kutta condition" at the sharp leading edge, and say that the forward stagnation point MUST be there.
I think you know what the result would be. You could then TRULY say that someone was overemphasizing the Kutta condition ;-) If we solve Newton's equation with that condition, we will discover that there is downwash in front of the wing, upwash behind it, the air will be flowing faster over the bottom surface than the top, and there will be negative lift. We know that this is not the case in the real world. When there is more than one sharp edge, there is another mathematical principle involved. (It is the same math that explains why a fan cools your face if you sit in front of it, but not if you sit the very same distance behind it.).
I suggest that we not try to explain that in this brief introductory section.
Mark.camp (talk) 20:41, 15 January 2009 (UTC)
In a physical explanation of lift, as this section is titled, it's hand waving to say "It's the Kutta condition." That's why the Kutta condition is mentioned where it arises physically (at the end of the “Stages of Lift Production” section) and again later when some discussion of potential flow theory occurs (although this later mention is a bit out of place). I added an earlier mention of when describing the two streamtubes and the dividing stagnation line. But it shouldn't be used as though its the explanation of lift. It's a way to get the right mathematical answer, but it's not a physical explanation.
(Also, a flat plate does not require more complex mathematics. An airfoil with a sharp trailing edge has one singularity in the conformal mapping of the airfoil surface. A flat plate has two. Same math, different geometry. See NASA GRC's explanation of Conformal Mapping.) Michael Belisle (talk) 22:52, 15 January 2009 (UTC)
Agree that a flat plate doesn't require different math. I meant that one can easily explain, at a high level, why the Kutta condition occurs in an a real (viscous) flow when there is only one sharp edge, at the trailing end. One cannot as easily explain why a sharp trailing edge has MORE influence over the stagnation points (and thus, the circulation) than an equally sharp leading edge.
But first, would you as a student of the subject please confirm my facts: in a real, viscous flow, air is happier to pick a forward stagnation point away from a sharp leading edge--even if that means it has to back up and round that edge--than it is to pick an aft stagnation point some distance from a sharp trailing edge--and be forced to back up and round a sharp TRAILING edge. I picked this little bit of knowledge on the web somewhere, and it may be a dangerous thing.
Mark.camp (talk) 22:57, 6 February 2009 (UTC)
Michael, thanks for deleting the sentence about the differential obstruction of different sides of an airfoil. Under the circumstances, that was the most appropriate thing to do.
I now have a copy of Anderson’s Introduction to Flight, fifth edition. I have examined Section 5.19 closely.
I think the Section begins very well, leading up to the excellent statement The answer is simply that the aerodynamic flow over the airfoil is obeying the laws of nature, namely, mass continuity and Newton’s second law. Unfortunately, after that it goes downhill as Anderson resorts to intuition and naïve language: stream tube A is squashed to a smaller cross-sectional area ...
I was also disappointed with his peculiar explanation using the notion that the upper and lower surfaces of an airfoil provide different levels of obstruction to the stream. I acknowledge that this book is an introduction to the subject, but Anderson’s use of this notion is not consistent with the level of authority and rigour I see elsewhere in the book.
This part of Anderson’s explanation hits rock-bottom where he considers stream tube B: The airfoil is designed with positive camber; hence, the bottom surface of the airfoil presents less of an obstruction to stream tube B, and so stream tube B is not squashed as much as stream tube A in flowing over the noise of the airfoil. Why does Anderson restrict his comment to airfoils with positive camber? I think he is saying the lower surface of the airfoil is flatter than the upper surface, and a flatter surface presents less of an obstruction than a rounded surface! This is a tragedy. Anderson is aware that an airfoil with a negative camber generates lift, because that is the subject of his example 5.22 (page 359). He would also be aware that symmetric airfoils generate lift. Explaining lift in terms of positive camber, as Anderson has done, has been exposed as a fallacy to the same extent as the Equal Transit Time Theory.
Anderson’s alternate explanation appears to be an appeal to intuition to help beginners get one foot on the bottom rung of the ladder that leads to acceptance of airfoils generating lift. I suggest that the Wikipedia article on lift should not give Anderson’s alternate explanations more importance than he intended. Dolphin51 (talk) 03:05, 21 January 2009 (UTC)
Sure. That's why there are two sections in the explanation. The first is the basic, introductory description. The second gets into the details.
Like I said before, Anderson probably chose positive camber as a simple example, but his analysis can be generalized quite readily beyond this specific case. The "obstruction" is a pure geometrical argument, nothing more. If the wing has positive camber and is at zero angle of attack, then the stagnation streamline (which intersects the chordline at the leading edge in the figure at right) divides the upper and lower streamtubes. But, because of the positive camber, the airfoil is more bulky above the stagnation line than below it. This can be seen clearly by comparing the airfoil areas above and below the chordline here.
In the simplest terms, a symmetric airfoil would present an equal obstruction to each streamtube. Now, if you break the symmetry with positive camber, then clearly the upper stream tube has a greater obstruction than the lower one. Negative camber will have the opposite effect. I don't understand what your confusion about the relative obstruction of the airfoil to the upper and lower streamtubes. Michael Belisle (talk) 21:14, 21 January 2009 (UTC)
Hi Michael. Thanks for the prompt acknowledgement. You have written that a symmetric airfoil would present an equal obstruction to each streamtube. I assume you are writing about a symmetric airfoil at zero angle of attack. What comment would you make about a symmetric airfoil at non-zero angle of attack? (If the two surfaces of a symmetric airfoil present equal obstructions at all angles of attack then, according to the obstruction theory, a symmetric airfoil would never generate lift.)
You have also written that negative camber will have the opposite effect. I assume you are writing about a negatively cambered airfoil generating zero lift. (If negative camber has the opposite effect at all angles of attack then, according to the obstruction theory, it could not be used to generate lift when it is flying upside down, contrary to what Anderson explains in his example 5.22 on p.359.)
I'm not confused by the obstruction theory. Obstruction is not defined in aerodynamics or fluid dynamics and so is lacking in rigour. I think the notion of obstruction (which appears to be peculiar to Anderson) is intended to be entirely intuitive and is useful only as a means of introducing beginners to the notion of lift. It is simple, easy-to-understand and does not necessitate any math, but it is no better or worse than the Equal Transit Time Theory and has no place in an encyclopedia. Dolphin51 (talk) 22:00, 21 January 2009 (UTC)
For a symmetric airfoil at nonzero angle of attack, I would make the comments that were previously in the article. I am not talking about a negatively cambered airfoil generating zero lift. I was discussing the effect of camber, ceteris paribus: a negative cambered airfoil at zero angle of attack will generate negative lift while a positive cambered airfoil at zero angle of attack will generate positive lift and a symmetric airfoil at zero angle of attack will generate no lift. (Also, in all cases considered, the flow conditions are unchanged, the airfoil geometry is unchanged except for camber, the laws of physics still apply, and 1+1=2.)
The word “obstruction” is defined in fluid dynamics as it's defined in the dictionary: “ a thing that impedes or prevents passage or progress; an obstacle or blockage”. Example: “An airfoil presents an obstruction to a flow.” The explanation has a place in a general-audience encyclopedia because 1) you have not yet conclusively explained why anything that is said is wrong and 2) it strives to use clear and understandable language. I removed it not because it was wrong, but because it was in the wrong place. Michael Belisle (talk) 23:02, 21 January 2009 (UTC)
## Lifting bodies
How would "obstruction theory" or "Kutta condition" explain these lifting body airfoils with "humps" on the bottom and "thick" trailing edges? M2_F2_glider.jpg M2_F3_rocket_powered.jpg Jeffareid (talk) 09:28, 30 January 2009 (UTC)
There is a little relevant information at Lifting body. I think the key to understanding these lifting bodies is that they are intended to operate at VERY high speeds — hypersonic (as occurs at re-entry), supersonic and high sonic. Their maximum lift coefficient doesn't need to be particularly high so they don't employ classic sonic or even supersonic airfoil shapes. Notice that, in both photographs, there appears to be a trailing edge flap deployed. Presumably this is to increase maximum lift coefficient, both during the launch from the B-52, and during landing. Notice that landing gear is not yet extended in the landing photograph. The gear is possibly deployed only when extra drag is needed, or just prior to touch-down. Dolphin51 (talk) 10:08, 30 January 2009 (UTC)
Nice link. Thanks for making this a separte section. It appears that the M2-F2 is unique compared to the two other old lifting bodies (those look more "normal"), but maybe ahead of it's time as most modern hypersonic models also have relative flat tops and all the protusions like engines and lifting surfaces below. I realize the purpose was hypersonic flight (re-entry), but it did glide (or fly) reasonably well at subsonic speeds (once they added the third vertical stabilizer), and it would seem that this is a wiki article on lift in general, not just effecient sub-sonic lift. Compare the apparent angle of attack with the F104 chasing it in the first photo, it's not bad. If nothing else, it looks cool, and at least would seem to be a good example to dispute equal transit theory. I'm a bit curious as to who and how it was figured out that such a design would work in the first place. Jeffareid (talk) 10:58, 30 January 2009 (UTC)
## Unclear sentence about flow "naturally following the shape of an airfoil"
I am not able to understand the following sentence:
"If one assumes that the flow naturally follows the shape of an airfoil...then the explanation of lift is rather simple..."
I hope that someone who is knowledgeable about fluid dynamics can make this sentence easier to understand for the typical reader.
The sentence seems to assume that there is only one flow that "naturally follows the shape of the airfoil".
For me, this assumption doesn't seem to be true.
First, to me, ANY flow meeting these conditions seems more or less "natural", for an ordinary airplane wing:
(1) The flow splits exactly once, near the front, and rejoins at the trailing edge
(2) At the limit of great distance in front of the foil (directly in front, and also above and below), the flow is in a single direction (opposite the direction of the aircraft) at a single speed (the speed of the foil).
(3) Near the foil the flow is tangent to the surface of the foil.
Second, and more importantly, the flow that seems MOST natural to me is not the actual flow. I suspect that most lay people assume the same thing that I did about the "natural" flow:
(1) The aft stagnation point is in accordance with the Kutta condition (this part we lay people get right, though we've never heard of poor Kutta, let alone his condition).
(2) The forward stagnation point is the front of the foil.
The second bit is what we amateurs guess slightly--but critically--wrong.
We learn this when the experts show us the experimental results from the wind tunnel. Or, we may equally well learn of our error when they show us the results from applying plain old Newton, with the assumptions of no friction, no variation of density, time-invariant velocity, and the Kutta condition.
Counter to our intuition, there is an updraft ahead of the wing, and as a result, the flow splits not at the very front of the wing, but slightly lower and thus slighly aft of that point. The shocking consequence is that the air briefly flows the wrong way along the wing--"into the wind"!
Mark.camp (talk) 01:47, 29 December 2008 (UTC)
The idea behind that that sentence is that if you take an airfoil and throw it in a wind tunnel at some specified conditions, there will only ever be one flow pattern. The question answered by this section is “Why does that flow pattern generate a lift force?” If you read the entire section in context, it attempts to explain and show with a diagram what that "natural" flow pattern is, including your mention of the flow going the "wrong way" (although it doesn't say so explicitly). It can still be improved, of course.
The picture clearly shows that your point #2 that "we amateurs guess wrong" is not the case. Michael Belisle (talk) 22:48, 2 January 2009 (UTC)
Thanks, Michael. You point out that there is only one flow which occurs if we take an airfoil and throw it in a wind tunnel at some specified conditions.
I agree. Here is where I got confused, however.
The sentence in the article refers to the flow or flows which "naturally follows the shape of an airfoil". When I read it, I interpreted it as "the one flow that would occur naturally to the reader of the article", not as "the one result which experts report from lab experiments".
Do you think that I misinterpreted the sentence? If so, was it a misinterpretation which others might fall into? If so, then can you suggest a rewording of the sentence to prevent this misreading?
Mark.camp (talk) 00:09, 4 January 2009 (UTC)
I've not heard back so I assume we are in agreement that the text is intended to explain lift with the starting assumption being the actual flow. As written, it appears to start instead with an appeal to the reader's intuition about what that flow would be.
Therefore, if no objection, I will re-order the sentences and trim one of them, as follows, to make it clear that the starting point is the calculated potential flow, and that we are deferring for the moment the question of WHY this flow occurs:
The image to the right shows the streamlines over a NACA 0012 airfoil computed using potential flow theory, a simplified model of the real flow.
If one assumes this flow, then the explanation of lift is rather simple and...
Mark.camp (talk) 12:47, 7 January 2009 (UTC)
That's reasonable. I made part of that edit, but I didn't reorder the sentences. In the future, be bold (WP:BOLD). There doesn't have to be consensus before you make an edit. If someone doesn't like it, they'll let you know. (So yeah, I agree that silence implies consent: WP:SILENCE.) Personally, I rarely have a problem with someone rewording what I (or any other editor) says. I'm not always the clearest writer. But I might complain if someone deletes something that I felt was important. Michael Belisle (talk) 05:00, 9 January 2009 (UTC)
## Re the ability of air to sense the "top of the airfoil" at a distance
Presently, the text states that the air in front of the airfoil is able to "sense" the upper surface of the airfoil some distance, and that it moves accordingly, thereby making airplanes fly.
I'm pleased to see the scientists lose their monopoly on this page, and New Age philosophers have a chance to give their views at last.
Although the discovery of air's ESP is admittedly the most striking, in fact the entire introductory section has become a hodge-podge of amazing facts from outside the stodgy and stifling world of mathematically proven, experimentally validated science.
Mark.camp (talk) 22:56, 12 January 2009 (UTC)
Of course the upstream air is able to sense an obstruction and adjust at subsonic speeds. This is perhaps one of the most important points in aerodynamics. Disturbances propagate upstream at the speed of sound, causing the streamlines to deflect far upstream of the obstruction. It's one of the reasons why Newton was wrong when he tried to explain lift (and consequently, why his explanation works pretty well at hypersonic speeds, when the oncoming flow has no idea what's coming). Michael Belisle (talk) 23:45, 14 January 2009 (UTC)
Does this phenomenon--"air sensing an obstruction"--have a scientific definition?
Mark.camp (talk) 13:51, 15 January 2009 (UTC)
Yes, I linked to the basics of it. In short, an object in a free stream generates a pressure field around it; see Section 3.5.1 of Fundamentals of Compressible Fluid Dynamics. It's clearer, common, and not incorrect to say that the flow “senses” an obstruction (which that source does in section 3.5.3). Michael Belisle (talk) 19:53, 15 January 2009 (UTC)
## Pressure on bottom of wing????
I'm pretty sure that the article is not due weight. So far as I am aware as (most, but not necessarily all) wings go through the air they form a high pressure area under the wing, and this pushes the wing and generates lift. This doesn't seem to be in the article right now. Further, my understanding is that the push effect is actually more important, by and large-but not all the time, than the Bernouilli effect on top of the wing.- (User) Wolfkeeper (Talk) 03:23, 21 January 2009 (UTC)
Hi Wolfkeeper. I suggest you find a good textbook on aviation or aerodynamics and brush up on the theory. When a wing is generating lift there is very little increased pressure on the underside of the wing. The pressure under the wing is approximately the same as the pressure in the freestream. Conversely, the pressure on the upper surface of the wing, particularly near the leading edge, is significantly lower than the pressure in the freestream.
The relative speed of the flow past the underside of the wing is approximately the same as the speed of the wing; but the relative speed of the flow past the upper surface, particularly near the leading edge, is significantly faster than the speed of the wing. This is all exactly as one would expect, considering Bernoulli's principle. Dolphin51 (talk) 05:25, 21 January 2009 (UTC)
There are good articles on the web illustrating the pressure differences both above and below the wing. For example: Velocity distribution. I will leave the reader to judge who is right. JMcC (talk) 11:34, 21 January 2009 (UTC)
Yes, sorry, I was totally unclear/wrong. Wings work in several different ways simultaneously, but I really meant that this was another effect which wasn't mentioned. I think the degree of the effect (and in the examples showed above it's a much smaller effect than the low pressure above the wing) depends on the shape of the wing and the angle of attack. IRC (which I may not) it's more pronounced at higher angles of attack. And I still think it's well worth mentioning, even just to say it's usually a smaller effect.- (User) Wolfkeeper (Talk) 11:51, 21 January 2009 (UTC)
Pressure coefficient of NACA 0012 airfoil at 11° angle of attack.
You are correct that it depends on the wing and the angle of attack. The pressure is always higher than the freestream pressure at the stagnation point (or attachment line, in the case of a swept wing), which is typically near to the leading edge, as this is the location where the velocity is at a minimum.
I'm not sure what Dolphin51 means when he says that the pressure is lower than the freestream pressure near to the leading edge. For the example diagram in the article, a NACA 0012 at 11° AoA (intentionally high to exaggerate the difference in streamtube areas), the pressure is higher than the freestream pressure for the entire length of the bottom of the airfoil, evidenced by the positive pressure coefficient. (Note that it's customary to plot -${\displaystyle C_{p}}$, so that the lower curve is the lower surface and the upper curve is the upper surface.)
I'm not sure about how close to the leading edge, but if you reduced the AOA attack of that NACA 0012 airfoil, or perhaps with a different airfoil at low AOA, the pressure of air under the wing at the aft end (exit point) can be below ambient. I don't know how far forwards on the under surface of a wing that the pressure could be reduced below ambient (the pressure above would be reduced further still). I would assume that an airfoil where the diverted flow pressure is below ambient pressure would improve efficiency Jeffareid (talk) 03:28, 31 January 2009 (UTC)
Regardless, it's not the high pressure alone that “pushes” the wing; it's the difference in pressure between the upper surface and the lower surface that creates an aerodynamic force. The area between the upper and lower ${\displaystyle C_{p}}$ curves is the amount of lift. If you find a good citation that explains this fact, feel free to work something into the article that makes this clearer. Michael Belisle (talk) 21:05, 21 January 2009 (UTC)
Just for fun my father once built a glider with triangular cross-section wings. IRC all the lift came from the undersurface and the top was flat to the airflow and so would have given no low pressure zone. The L/D was probably horrible but it did fly.- (User) Wolfkeeper (Talk) 01:26, 22 January 2009 (UTC)
## Alternative explanations
The article Lift (force) contains a section called Common misconceptions addressing the so-called Equal Transit Time Fallacy, and the Coanda effect. The tenor of this section is very judgemental, reflecting more the personal views of various editors than what has been written in the cited sources. For example, the Equal Transit Time theory is always described as a Fallacy when in fact it is an alternative theory able to be supported by numerous references that could be quoted as citations.
One of the objectives of Wikipedia is that it should reflect a Neutral Point of View. The section called Common misconceptions does not reflect a neutral point of view at present — it reflects some of the passion of various editors. I am about to amend this section, firstly by changing the section heading to Alternative explanations of lift, and secondly by presenting the Equal Transit Time model as simply an alternative explanation of lift which some authors have identified as inaccurate.
At present, the sub-section on the Equal Transit Time Fallacy contains one strongly judgemental statement that is probably just the personal prejudice of the editor. I will attach a "Fact" tag to that one to see if it can be supported by a citation. Dolphin51 (talk) 02:24, 22 January 2009 (UTC)
I'm all for the title change to “Alternative explanations”. I did that once before (inspired by Anderson), but someone objected and I didn't press the issue.
However, I would recommend being careful when rewording the ETT section to make it “NPOV”. What you describe here is not an NPOV issue. There is scientific consensus that the explanation is flawed, which makes it fall under WP:FRINGE. It is readily observable (RealPlayer video, fast forward to 5:29) that the explanation has no basis in reality. There's no scientific support for it. The moon is not made of cheese. See also WP:V#Reliable sources.
As for Coanda, I think it's pretty NPOV as it is. It explains who says it's Coanda, who says it's not, why they disagree, and has an sufficient quantity of citations throughout. And here again, science (with few exceptions) agrees that it's not Coanda.
But go ahead and place all the fact tags you want. I can argue which facts are correct, but I can't argue that they need to be cited. Michael Belisle (talk) 04:34, 22 January 2009 (UTC)
Thanks Michael. I approve of reducing the title of the new section to simply Alternative explanations.
Looking back at the old version (Common misconceptions) to see what it was in Equal Transit-Time (ETT) that seemed clearly non-NPOV I see there were multiple uses of the word fallacy. This is a highly judgemental alternative to the neutral words model, theory etc. Clearly, there are citable authors who have used the ETT in good faith, and other citable authors who have ridiculed ETT. The number of authors who have used ETT in good faith is significant — possibly greater than the number who have ridiculed it, so it was not giving appropriate balance to the two groups if only the judgemental approach was reported. If there are editors who want to challenge the new NPOV appearance of ETT I am happy to debate them. Wikipedia's mission is to objectively report everything that is out there in citable sources, and not to judge which of two or more views is superior.
Similarly, to use the word misconception in the heading also seemed to reflect what the editor believed, rather than being the most appropriate word given that the ETT receives as much good faith useage as ridicule. I think the section on ETT is now much improved in that it is less judgemental and more NPOV.
A relevant direction to editors is the one found at WP:Verifiability: The threshhold for inclusion in Wikipedia is verifiability, not truth. As you know, what editors believe to be true is irrelevant. It is what is verifiable in citable sources that is relevant. (My words.) In areas that are likely to be challenged, those sources must be cited. Dolphin51 (talk) 04:48, 23 January 2009 (UTC)
Doesn't "Proposition X is a fallacy" simply mean, roughly, "Proposition X is incorrect, because of an internal flaw of logic or fact"? If so, why is a text of the form "X is a fallacy" which meets Wikipedia standards necessarily any less acceptable than a statement "X is true"?
Mark.camp (talk) 21:52, 23 January 2009 (UTC)
Hi Mark. If fallacy and incorrect are truly synonymous then there would be no difference which of the two words is used in Wikipedia. However, I believe they are not truly synonymous. I believe incorrect is objective. It implies nothing more than what you see at face value. I would argue that fallacy, in the context in which it was used in Lift (force), is not an objective word. It is a dysphemism so has a negative implication. (There is a Wikipedia article on the Bohr model of the atom, even though this model is known to be incorrect. How do you feel about this article being re-titled Bohr fallacy?)
If all (or most) citable sources concur that the ETT is a fallacy then it would be reasonable for Wikipedia to use fallacy in its description of ETT. However, there are many sources that use ETT in good faith. Presumably the authors were striving to explain lift in simple, easy-to-understand terms, for commendable reasons. Those authors would legitimately object to an encyclopedia describing their description as a fallacy (or even a myth), but they might readily concede that, technically, it is incorrect.
The fact that some citable sources concur that the ETT is incorrect is not sufficient to allow a Wikipedia editor to use subjective terms, especially if those terms are not used by a suitable proportion of those citable sources. It is Wikipedia's role to report the existence of both points of view, and to report that the ETT is technically incorrect. This can only be done with objective language. Dolphin51 (talk) 01:24, 24 January 2009 (UTC)
Wikipedia does have a criteria for judging the relative weight of two views that is applicable to this case, which is that reliable sources are favored. It is not simply good enough to have just any source: “as a rule of thumb, the greater the degree of scrutiny involved in checking facts, analyzing legal issues, and scrutinizing the evidence and arguments of a particular work, the more reliable it is. Academic and peer-reviewed publications are highly valued and usually the most reliable sources in areas where they are available, such as history, medicine and science.” The most reputable and peer-reviewed sources agree that ETT is based on a fallacious argument. (And yes, I will appeal to authority when the authority is a recognized expert in the field in question. The only source that the WP "Appeal to Authority" article cites says one important thing: "This fallacy is committed when the person in question is not a legitimate authority on the subject." Like I said once before, we can't present eskimo.com on equal footing with John D. Anderson.)
The immediate difference between ETT and the Bohr model that I see is that the Bohr model is commonly presented with the disclaimer that it's wrong and followed up with the more complicated and more accurate theory. The Bohr model occasionally gives useful results. It's probably not called a fallacy because it's taught as a historical note about what was once an accepted theory. ETT, however, is almost never is presented in this manner. It gives no useful results because it's never correct and there is no scientific basis for its principal premise. I don't know that ETT ever carried any serious weight except maybe in the days of alchemy and the aether. Michael Belisle (talk) 07:17, 24 January 2009 (UTC)
There seems to be universal agreement that the ETT is incorrect. No-one is arguing in favour of it being presented in Wikipedia as a seriously technical explanation of lift. All that remains is how to describe it. We have a citation from John D. Anderson Jr (Introduction to Flight) saying This is simply not true. Using this, I would be happy to see the ETT described as untrue, or incorrect, or any other objective term.
If Wikipedia is to use a more emotional term, or a dysphemism, it should be a term used in at least one of the citable sources. We have seen fallacy, and currently myth. Is either of these terms used in a reliable source? If not, perhaps we are looking at a dysphemism chosen by a Wikipedia editor primarily because it matches his (or her) passion on the subject of ETT.
I am in favour of describing ETT either in the same way as in reliable sources, or using terms that are scrupulously objective. Dolphin51 (talk) 07:38, 24 January 2009 (UTC)
I can agree with this. Whether or not a reliable source uses the same terminology is a fair point. Michael Belisle (talk) 20:37, 24 January 2009 (UTC)
## Question about differential obstruction theory
The explanation of the theory above (Belisle, 21-Jan-09) says that, in the case of a positive camber section at zero angle of attack, the forward stagnation streamline intersects the airfoil at the leading edge. Could someone confirm this fact? (My understanding was that intercept would be on the bottom surface.) Mark.camp (talk) 14:45, 24 January 2009 (UTC)
With a symmetric airfoil there is an axis of symmetry joining the trailing edge and the leading edge, so the notion of the chord line is very straight-forward. (The angle of attack can be considered the angle between the chord line and the vector representing the velocity of the airfoil relative to the atmosphere.) With a cambered airfoil it is not so straight-forward because there is no axis of symmetry and the orientation of the chord line is arbitrary. (If two people locate the chord line at slightly different orientations they will measure slightly different angles of attack.) To eliminate the arbitrariness of the chord line on a cambered airfoil it is useful to replace it with the zero lift axis. If the angle of attack on an airfoil is measured relative to the zero lift axis the lift is always zero when the angle of attack is zero.
In answer to your question — a cambered airfoil at zero angle of attack measured relative to the (arbitrary) chord line is at positive angle of attack relative to the zero lift axis, so it is generating lift. Because it is generating lift, I would expect the stagnation point (where the dividing streamline contacts the airfoil) to be located below the point where the zero lift axis intersects the leading edge. So I agree with your understanding.
On an airfoil with positive camber, the zero lift axis intersects the leading edge of the airfoil above the point where the chord line intersects it. So at one particular lift coefficient the stagnation point would coincide with the point on the leading edge where the chord line intersects the leading edge. This matches Michael's comment. Dolphin51 (talk) 00:20, 25 January 2009 (UTC)
The chordline for an airfoil is defined as the line from the leading edge to the trailing edge regardless of symmetry. Yes, at zero angle of attack (with respect to the chordline) the forward stagnation point is at the leading edge. (Some simple calculations using conformal mapping or Xfoil can confirm this.) At the angle of zero lift, an airfoil with positive camber has ${\displaystyle \alpha _{C_{L0}}<0}$ and hence the stagnation point would be above the leading edge.
Remember that this is a forum for discussion about improving the article, not for general discussion about lift or aerodynamics. Michael Belisle (talk) 23:22, 25 January 2009 (UTC)
Michael Belisle wrote
Remember that this is a forum for discussion about improving the article, not for general discussion about lift or aerodynamics.
Michael, I agree. My question was in response to a post in this talk page stating that the forward stagnation point was at the leading edge (in fact, it was your own post, I think).
One respondant (Dolphin51) said that my understanding was correct and that the post (your post?) was incorrect. You said the opposite: "Yes, at zero angle of attack (with respect to the chordline) the forward stagnation point is at the leading edge."
My question is, which is correct?
Mark.camp (talk) 03:37, 26 January 2009 (UTC)
I am, at least to the extent the stagnation point is “very near to the leading edge". The location is indistingishable from the leading edge in an Xfoil calculation of a NACA 2412. The reason I mentioned the talk page guidelines is that we can argue points relevant to the article, but we already agreed to take out the obstruction description and there's no mention of stagnation points on a cambered airfoil in the article. Michael Belisle (talk) 04:48, 26 January 2009 (UTC)
Right, the question's now irrelevant. Hadn't noticed that the text was deleted.
Mark.camp (talk) 21:49, 1 February 2009 (UTC)
I have added three paragraphs to present Anderson's obstruction theory. You will find them under "Alternative explanations". Feel free to fine-tune my wording. Dolphin51 (talk) 03:35, 9 February 2009 (UTC)
## Question about flow and pressure over a cambered airfoil
What are the aspects of the Coanda like airflow across the top of a cambered air foil, especially near the leading edge? It would seem that there is an "inwards" (downwards) acceleration of air over the top of cambered airfoil with a significant component of acceleration perpendicular to the flow, so without much change in speed (magnitude of velocity) and therefore without much change in kinetic energy of the air flow near the wing (although the low pressure and viscosity would accelerate the surrounding air into that diverted flow). Jeffareid (talk) 11:22, 30 January 2009 (UTC) | 22,691 | 103,021 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.963893 |
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02 Nov 2007, 05:52
....
Last edited by Darden2010 on 24 Jun 2008, 08:50, edited 1 time in total.
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02 Nov 2007, 06:53
I think its C
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02 Nov 2007, 07:15
Darden2010 wrote:
Geologists once thought that the molten rock known as lava was an underground remnant of Earth's earliest days, sporadically erupting through volcanoes, but they now know that it is continuously created by the heat of the radioactivity deep inside the planet.
A: was an underground remnant of Earth's earliest days, sporadically erupting
B: had been an underground remnant of Earth's earliest days and sporadically erupted
C: was an underground remnant of Earth's earliest days, which sporadically erupted
D: would be an underground remnant of Earth's earliest days that sporadically erupted
E: was an underground remnant of Earth's earliest days, having sporadically erupted
A. see no error as A clearly modifies the clause.
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02 Nov 2007, 07:41
I like the parallelism in C.
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02 Nov 2007, 09:20
Darden2010 wrote:
Geologists once thought that the molten rock known as lava was an underground remnant of Earth's earliest days, sporadically erupting through volcanoes, but they now know that it is continuously created by the heat of the radioactivity deep inside the planet.
A: was an underground remnant of Earth's earliest days, sporadically erupting
B: had been an underground remnant of Earth's earliest days and sporadically erupted
C: was an underground remnant of Earth's earliest days, which sporadically erupted
D: would be an underground remnant of Earth's earliest days that sporadically erupted
E: was an underground remnant of Earth's earliest days, having sporadically erupted
Def. A. no errors.
C, improper use of which.
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02 Nov 2007, 10:09
....
Last edited by Darden2010 on 24 Jun 2008, 08:49, edited 1 time in total.
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02 Nov 2007, 12:32
Looks like A to me. We are finally up to A and C right. C is wrong because which sporadically erupted modifies "earlies days". In A sporadically erupting is an adverbial modifier and does not need to be close to the thing it modifies.
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03 Nov 2007, 08:44
Darden2010 wrote:
I liked C better also, why is C improper use of which? and how come 'sporadically erupting' is not modifying 'earliest days' (which is wrong) in A
I fail to understand how A is correct; isn't 'sporadically erupting' modifying 'earliest days'? Can someone explain??
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04 Nov 2007, 19:35
GK_Gmat wrote:
Darden2010 wrote:
I liked C better also, why is C improper use of which? and how come 'sporadically erupting' is not modifying 'earliest days' (which is wrong) in A
I fail to understand how A is correct; isn't 'sporadically erupting' modifying 'earliest days'? Can someone explain??
Participle clause(sporadically erupting....) is modifying molten rock. It if were modiying earliest days, comma would not be needed.
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04 Nov 2007, 19:41
soomodh wrote:
Looks like A to me. We are finally up to A and C right. C is wrong because which sporadically erupted modifies "earlies days". In A sporadically erupting is an adverbial modifier and does not need to be close to the thing it modifies.
I agree with your explanation. Please correct me if I am wrong here or if you think differently.
I don't disagree with your answer, just want to clarify my understanding. Thanks.
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# Stability of Solutions to Evolution Problems
Mathematics Department, Kansas State University, Manhattan, KS 66506-2602, USA
Mathematics 2013, 1(2), 46-64; https://doi.org/10.3390/math1020046
Received: 26 February 2013 / Revised: 25 April 2013 / Accepted: 25 April 2013 / Published: 13 May 2013
Large time behavior of solutions to abstract differential equations is studied. The results give sufficient condition for the global existence of a solution to an abstract dynamical system (evolution problem), for this solution to be bounded, and for this solution to have a finite limit as $\text{t}\to \text{}\infty$ , in particular, sufficient conditions for this limit to be zero. The evolution problem is: $\stackrel{˙}{u}\text{}=\text{}A\left(t\right)u\text{}+\text{}F\left(t,\text{}u\right)\text{}+\text{}b\left(t\right),\text{}t\text{}\ge \text{}0;\text{}u\left(0\right)\text{}=\text{}{u}_{0}.$ (*) Here $\stackrel{˙}{u}\text{}:=\text{}\frac{du}{dt}\text{},\text{}u\text{}=\text{}u\left(t\right)\text{}\in \text{}H,\text{}H$ is a Hilbert space, $t\text{}\in \text{}{R}_{+}\text{}:=\text{}\left[0,\infty \right),\text{}A\left(t\right)$ is a linear dissipative operator: $\text{Re}\left(A\left(t\right)u,u\right)\text{}\le -\gamma \left(t\right)\left(u,\text{}u\right)$ where $F\left(t,\text{}u\right)$ is a nonlinear operator, $‖F\left(t,\text{}u\right)\text{}‖\text{}\le \text{}{c}_{0}{‖u‖}^{p},\text{}p\text{}>\text{}1,\text{}{c}_{0}$ and p are positive constants, $‖b\left(t\right)\text{}‖\text{}\le \text{}\beta \left(t\right)$, and $\beta \left(t\right)\ge 0$ is a continuous function. The basic technical tool in this work are nonlinear differential inequalities. The non-classical case $\gamma \left(t\right)\text{}\le \text{}0$ is also treated. View Full-Text
MDPI and ACS Style
Ramm, A.G. Stability of Solutions to Evolution Problems. Mathematics 2013, 1, 46-64. https://doi.org/10.3390/math1020046
AMA Style
Ramm AG. Stability of Solutions to Evolution Problems. Mathematics. 2013; 1(2):46-64. https://doi.org/10.3390/math1020046
Chicago/Turabian Style
Ramm, Alexander G. 2013. "Stability of Solutions to Evolution Problems" Mathematics 1, no. 2: 46-64. https://doi.org/10.3390/math1020046
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1
Only visits after 24 November 2015 are recorded. | 791 | 2,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-27 | latest | en | 0.72441 |
http://pubtrails.org.uk/percentage-composition/4549917665 | 1,638,077,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00410.warc.gz | 54,227,080 | 5,062 | National Chemistry
1. Calculate the percentage of carbon in propane.
That is, the amount of C in C3H8
This will be a fraction multiplied by 100 to get a percentage
carbon x 100
propane
3 x C x 100
C3H8
Then into mass: 1 mole of C3H8 = (3 x C) + (8 x H)
= (3 x 12) + (8 x 1)
= 42g
Calculation 36 x 100 = 85.7%
42
2. Calculate the percentage of nitrogen in ammonium nitrate.
That is, the amount of N in NH4NO3
This will be a fraction multiplied by 100 to get a percentage
nitrogen x 100
ammonium nitrate
Beware, there are 2 nitrogens in ammonium nitrate
2 x N x 100
NH4NO3
Then into mass: 1 mole of NH4NO3 = (2 x N) + (4 x H) + (3 x O)
= (2 x 14) + (4 x 1) + ( 3 x 16
= 80g
Calculation 2 x 14 x 100 = 28 x 100 = 35%
80 80
Find Percentage composition calculations in the following papers
2001 Q13(c)(ii) 2002 Q12(b) 2003q15(c) 2004 Q 17(c)
2005 Q11(b) 2006 Q11(b) 2007 Q17(c) 2008 Q16(a) | 412 | 1,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-49 | latest | en | 0.312752 |
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# In the given figure,
Question:
In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$, the frequency of oscillation of given body is :
1. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{M}}}$
2. $\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{Mg} \sin \alpha}}$
3. $\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}$
4. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{Mg} \sin \alpha}}$
Correct Option: , 3
Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_{\mathrm{eq}}}}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{~K}}}$
$\mathrm{f}=\frac{1}{\mathrm{~T}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}$
(Option 3) is correct | 314 | 879 | {"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} | 3.84375 | 4 | CC-MAIN-2023-06 | longest | en | 0.520031 |
https://lists.cpunks.org/pipermail/cypherpunks-legacy/1994-March/014348.html | 1,670,614,195,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00392.warc.gz | 400,390,686 | 2,735 | # Ames/ clipper compromised?
Jim Gillogly jim at rand.org
Tue Mar 29 14:39:08 PST 1994
```
> The skipjack review committe wrote:
> | second. At that rate, it would take more than 400 billion years to
> | try all keys. Assuming the use of all 8 processors and aggressive
> | vectorization, the time would be reduced to about a billion years
>
> Could someone explain why jumping to 8 processors knocks the
> time down by a factor of 400, instead of a factor of 8? Is the 400
> billion years a load of crap, intended to sound more impressive than
> 8?
Without seeing the algorithm we can't be sure, but that could be OK for
ballpark: the 8 processors gives you 50 billion years, and the aggressive
vectorization gives you the other factor of 50. Since they've said there
are 32 rounds of <something> in there, I assume the point is to run those
rounds in parallel... or overlap the output of that round of one key with
the next round of a previous key, or some such dramatic stuff, and 32 is
close enough to 50 for this level of estimate. Sounds aggressive to <me>,
But it's meaningless to ask how long today's hardware would take to solve
this stuff. Extrapolations aren't much better, but at least they give a
convenient exponential benchmark. Let's take Wiener's proposed design for
3.5-hour cracks on a \$1M machine as the benchmark of solving a single key
at acceptable expense. Note that the speed or power of machines has been
doubling about once every 12-18 months. Wiener's machine brute-forces a
56-bit key in reasonable time, so if your bang/buck ratio keeps going at
the current rate, in 24-36 years something equivalent would be able to
brute-force an 80-bit key. That might explain why they chose 80 bits
instead of 128... if the algorithm escapes, they don't lose contact with
its product forever.
Note that the Skipjack Review committee was not in fact using the billion
years "load of crap" mode. In the executive summary, they say:
1. Under an assumption that the cost of processing power is halved
every eighteen months, it will be 36 years before the cost of
breaking SKIPJACK by exhaustive search will be equal to the cost
of breaking DES today.
I located and cut&pasted this after writing my previous paragraph, so we
can call these independent findings. :) Note that they produced this
before Wiener presented his design, so the cost of a break was not
(publically) known at that point.
Jim Gillogly
Highday, 7 Astron S.R. 1994, 22:34
``` | 625 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.942191 |
https://cs.stackexchange.com/questions/19950/l-in-re-r-such-that-lr-cup-l-in-r | 1,580,144,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00143.warc.gz | 387,426,445 | 30,313 | # $L \in RE/R$ such that $L^R \cup L \in R$
Prove/disprove: $\exists L \in RE/R$ such that $L^R \cup L \in R$
Where in my context, $R$ is the turing decidable, and $RE$ is the recursively enurmable. I tried to find such an $L$ but couldn't. What I know for sure is that I need a language in $RE/R$ such that $L \cup L^R = \Sigma^*$, or am I also wrong?
• Is $L^R$ the set of all reversed words of $L$, or is this something else? – G. Bach Jan 24 '14 at 23:35
• Yes. Reversed words of $L$ – TheNotMe Jan 24 '14 at 23:46
• Do you use $/$ to denote set difference or right quotient? – Raphael Jan 25 '14 at 16:25
First: Such an $L$ exists and $L \cup L^R$ is not necessarily $\Sigma^*$.
Try $L \cup L^R = \{a^nb^n\} \cup \{b^na^n\}$.
For each $n$, $L$ should contain exactly one of the words $a^nb^n$, $b^na^n$. How can you do this so that $L$ is in $RE/R$?
Let $K$ be any set in $RE/R$ and let $L = \{a^nb^n \mid n \in K\} \cup \{b^na^n \mid n \notin K\}$. | 361 | 958 | {"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} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.816162 |
http://physicstasks.eu/3947/specific-heat-capacity-of-gas | 1,721,043,806,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00431.warc.gz | 23,232,143 | 7,759 | Specific Heat Capacity of Gas
Determine specific heat capacities cV and cp of unknown gas provided that at temperature of 293 K and pressure of 100 kPa its density is 1.27 kg m−3 and Poisson's constant of the gas is κ = 1.4.
• Hint 1
Use Meyer's relation and realize how you can transform it to a relation between specific heat capacities and not between molar heat capacities at constant pressure and volume.
• Hint 2
Poisson's constant κ is defined by
$\kappa = \frac{c_p}{c_V}= \frac{C_p}{C_V},$
where cp, respectively Cp is specific, respectively molar heat capacity at constant pressure and cV, respectively CV is specific, respectively molar heat capacity at constant volume.
• Hint 3
To determine molar mass Mm of the gas use the equation of state for ideal gas.
• Analysis
We start the solution from Meyer's relation. We divide it by molar mass of the gas and thus adjust it to the relation between specific heat capacities at constant pressure and volume.
The unknown molar mass can be determined from the equation of state for ideal gas.
Specific heat capacity at constant pressure is determined as the product of Poisson's constant and specific heat capacity at constant volume.
• Given Values
T = 293 K gas temperature p = 100 kPa = 1.00·105 Pa gas pressure ρ = 1.27 kg·m−3 gas density κ = 1.4 Poisson's constant of the gas cV = ? specific heat capacity of the gas at constant volume cp = ? specific heat capacity of the gas at constant pressure
Table values:
R = 8.31 JK−1mol−1 molar gas constant
• Solution
We start the calculation with Meyer's relation Cp = CV + R, that relates molar heat capacities at constant volume CV and constant pressure Cp.
In this task, however, we need to determine specific heat capacities, not molar heat capacities. This is why we need to divide Meyer's relation by molar mass of the gas Mm
$\frac{C_p}{M_m} = \frac{C_V}{M_m} + \frac{R}{M_m},$
resulting in the relation
$c_p = c_V + \frac{R}{M_m}.$
Now we use the given Poisson's constant κ defined by the relation
$\kappa = \frac{C_p}{C_V} = \frac{c_p}{c_V}.$
We determine specific heat capacity at constant pressure from this relation
$c_p=\kappa c_V,$
and substitute it into the above mentioned relationship
$\kappa c_V=c_V+\frac{R}{M_m}.$
The unknown specific heat capacity at constant pressure cV is then given by
$c_V=\frac{R}{M_m(\kappa-1)}.$
Now we need to determine the unknown molar mass of the gas Mm. We use the equation of state for ideal gas
$pV=\frac{m}{M_m}RT.$
We express molar mass
$M_m=\frac{m}{V}\frac{RT}{p}$
and the ratio $$\frac{m}{V}$$ substitute by the given density ρ of the gas:
$M_m=\frac{\rho RT}{p}.$
After substituting into the relation for specific heat capacity at constant volume we then obtain:
$c_V=\frac{R}{M_m(\kappa-1)}=\frac{R}{\frac{\rho RT}{p}(\kappa-1)}=\frac{p}{\rho T(\kappa-1)}.$
The specific heat capacity at constant pressure cp then can be directly determined from the equation
$c_p = \kappa c_V = \frac{\kappa p}{\rho T(\kappa-1)}.$
• Numerical Solution
$c_V=\frac{p}{\rho T(\kappa-1)}$ $c_V= \frac{100\cdot{10^3}}{1.27\cdot{293}\cdot (1.4-1)}\,\mathrm{J\,kg^{-1}K^{-1}}\dot{=}672\,\mathrm{J\,kg^{-1}K^{-1}}$ $c_p = \kappa c_V =1.4\cdot{671.8}\,\mathrm{J\,kg^{-1}K^{-1}}\dot{=}941\,\mathrm{J\,kg^{-1}K^{-1}}$ | 994 | 3,296 | {"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} | 4.59375 | 5 | CC-MAIN-2024-30 | latest | en | 0.826012 |
pc-boeken.nl | 1,620,587,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00345.warc.gz | 487,668,316 | 12,026 | Calculation Of Expected Value
Calculation Of Expected Value Statistics: It’s to Be Expected
Calculation of Expected Value and Variance Using Probability Distribution (English Edition) eBook: homeworkhelp classof1: pc-boeken.nl: Kindle-Shop. (listed on the Y-axis), a bar is drawn between the extreme values of the expected value calculated from the lower and upper bound values. pc-boeken.nl Treaty was dictated by the expected value added in terms of ensuring [ ] free circulation of reduction - calculate the expected value of the rightmost chance [...]. Find expected value based on calculated probabilities. One natural question to ask about a probability distribution is, "What is its center? Calculate the expected value E(X), the variance σ2 = Var(X), and the standard deviation σ of the random variable X with the following.
Find expected value based on calculated probabilities. One natural question to ask about a probability distribution is, "What is its center? in matlab?. Learn more about expected value. How to find expected value E[X]=_____ for a given data set X? suppose take -9 0 1 4 2 7 5 6 1 3]. Does matlab mean() is equal to expected value E[X]? Then how to calculate? Sign in to. (listed on the Y-axis), a bar is drawn between the extreme values of the expected value calculated from the lower and upper bound values. pc-boeken.nl
Calculation Of Expected Value - How to Get Best Site Performance
EDIT : Suppose your distribution is that you are equally likely to have any integer from -9 to 9. Raviteja on 24 Mar These cookies help us tailor advertisements to better match your interests, manage the frequency with which you see an advertisement, and understand the effectiveness of our advertising. Open Mobile Search. Moving to the left, each column-sum deaceases as the cumulative distribution function grows.
Calculation Of Expected Value Video
Expected Value and Variance of Discrete Random Variables
The expected value EV is an anticipated value for an investment at some point in the future. In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values.
By calculating expected values, investors can choose the scenario most likely to give the desired outcome. Scenario analysis is one technique for calculating the expected value EV of an investment opportunity.
It uses estimated probabilities with multivariate models to examine possible outcomes for a proposed investment. Scenario analysis also helps investors determine whether they are taking on an appropriate level of risk given the likely outcome of the investment.
The EV of a random variable gives a measure of the center of the distribution of the variable.
Essentially, the EV is the long-term average value of the variable. Because of the law of large numbers , the average value of the variable converges to the EV as the number of repetitions approaches infinity.
The EV is also known as expectation, the mean or the first moment. EV can be calculated for single discrete variables, single continuous variables, multiple discrete variables, and multiple continuous variables.
For continuous variable situations, integrals must be used. To calculate the EV for a single discrete random variable, you must multiply the value of the variable by the probability of that value occurring.
Take, for example, a normal six-sided die. Once you roll the die, it has an equal one-sixth chance of landing on one, two, three, four, five, or six.
Given this information, the calculation is straightforward:. If you were to roll a six-sided die an infinite amount of times, you see the average value equals 3.
Tools for Fundamental Analysis. Financial Analysis. Portfolio Management. Financial Ratios. Given a discrete random variable X , suppose that it has values x 1 , x 2 , x 3 ,.
The expected value of X is given by the formula:. Using the probability mass function and summation notation allows us to more compactly write this formula as follows, where the summation is taken over the index i :.
This version of the formula is helpful to see because it also works when we have an infinite sample space. This formula can also easily be adjusted for the continuous case.
Flip a coin three times and let X be the number of heads. The only possible values that we can have are 0, 1, 2 and 3. Use the expected value formula to obtain:.
In this example, we see that, in the long run, we will average a total of 1. This makes sense with our intuition as one-half of 3 is 1.
We now turn to a continuous random variable, which we will denote by X. Here we see that the expected value of our random variable is expressed as an integral.
There are many applications for the expected value of a random variable. This formula makes an interesting appearance in the St. Petersburg Paradox.
Calculation Of Expected Value -
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Here we see that the expected value of our random variable is expressed as an integral. There are many applications for the expected value of a random variable.
This formula makes an interesting appearance in the St. Petersburg Paradox. Share Flipboard Email. Courtney Taylor. Professor of Mathematics.
Courtney K. Taylor, Ph. Updated January 14, Both 0 and 00 are green. A ball randomly lands in one of the slots, and bets are placed on where the ball will land.
One of the simplest bets is to wager on red. If the ball lands on a black or green space in the wheel, then you win nothing. What is the expected value on a bet such as this?
Here the house has a slight edge as with all casino games. As another example, consider a lottery. This gives us an expected value of:. So if you were to play the lottery over and over, in the long run, you lose about 92 cents — almost all of your ticket price — each time you play.
All of the above examples look at a discrete random variable. However, it is possible to define the expected value for a continuous random variable as well.
All that we must do in this case is to replace the summation in our formula with an integral. It is important to remember that the expected value is the average after many trials of a random process.
In the short term, the average of a random variable can vary significantly from the expected value. Although the concept of expected value is often used in the case of various multivariate models and scenario analysis, it is predominantly used in the calculation of expected return.
This has been a guide to the Expected Value Formula. Here we learn how to calculate the expected value along with examples and downloadable excel template.
You can learn more about financial analysis from the following articles —. Free Investment Banking Course. Login details for this Free course will be emailed to you.
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die sehr gute Frage | 1,568 | 7,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-21 | latest | en | 0.896421 |
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INTRODUCTION TO BIFURCATION THEORY. IIMAS - Universidad Nacional Autonoma de Mexico. Mexico DF - Mexico and. CHAPTER 5. In this chapter, we study the autonomous equation with a k-dimensional parameter. We then say that the system has redemption song bob marley ukulele tutorial island a bifurcation. Bifurcations often change the attractors of a dynamical system. | 949 | 4,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-43 | latest | en | 0.806308 |
https://www.jiskha.com/questions/360586/this-is-a-wiley-plus-problem-and-there-is-no-reading-lesson-with-this-section-so-im-not | 1,597,381,953,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00223.warc.gz | 708,527,009 | 5,985 | # chemistry
this is a wiley plus problem and there is no reading lesson with this section so im not sure how to approach this question:
The element iridium has ccp packing with a face-centered cubic unit cell. The density of Ir is 22400 kg/m3. Calculate the volume (cm3) of the unit cell of iridium.
The molar mass of iridium is 192.22 g/mol
Thanks!
1. 👍 0
2. 👎 0
3. 👁 172
1. mass = volume x density; therefore,
volume = mass/density. You know the density; therefore, we need to determine the mass of a unit cell.
An atom of Ir has a mass of 192.22/6.022 x 10^23. There are 4 atoms/unit cell which makes the mass of a unit cell then
4 x 192.22/6.022 x 10^23. Substitute into the first equation above and calculate volume. If you want the volume in cc, you must have the density in g/cc. 22,400 kg/m^3 = 22.4 g/cc.
1. 👍 0
2. 👎 0
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11.Identify one conflict from the novel that develops due to the different perspectives of the characters involved. Explain how this conflict affects the characters and serves to advance the plot of the story. The story is The | 938 | 3,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-34 | latest | en | 0.884437 |
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$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, or $p H = - {\log}_{10} \left[{H}^{+}\right]$.
In your example $p H = 3$ clearly.
If $\left[{H}^{+}\right] = {10}^{-} 3 m o l \cdot {L}^{-} 1$, $p H = 3$.
I suspect that this measurement is incorrect, because that is a seriously acid sample of rain water. I wouldn't want to be out in the rain that day! If this is a real measurement, the collected water has likely leached acid salts from the minerals where it pooled. The $p H$ of normal rainwater is approx. $6$. | 212 | 698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.921617 |
https://spell.today/number-to-words/85070170.html | 1,675,908,982,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00803.warc.gz | 549,865,106 | 5,879 | How do you spell 85,070,170?
How do you write out the number 85,070,170 in words? It depends on the kind of number (e.g., it could be used as a cardinal or ordinal number). But commonly, 85,070,170 is spelt out like this:
85,070,170 in words = eighty-five million seventy thousand one hundred seventy
Examples of sentences in which the number 85,070,170 is written out in words
#Sentence
1Bella Mccarthy was eighty-five million seventy thousand one hundred seventieth on the list that the secretary brought to Hunter Fields's desk.
2For violation of the rights of employees, the «OneDay Gracie» Company was fined eighty-five million seventy thousand one hundred seventy dollars and fifty-two cents.
3On June 23, 2017, eighty-five million seventy thousand one hundred seventy people simultaneously watched the live broadcast of blogger Caroline Johnston, who was asked forty-six questions in 1 hour 26 minutes.
4At an auction in the city of Mumbai, the bid for an antique chest of drawers was eighty-five million seventy thousand one hundred seventy pounds
5Adrian Burton, the deputy director at the Osaka factory, said there is enough production capacity to produce eighty-five million seventy thousand one hundred seventy tooth brushes a week.
6Thirty-four days ago, June 17, 2021, eighty-five million seventy thousand one hundred seventy Australian dollars was equal to forty-seven million seven hundred seventy thousand one hundred seventy-two British pounds based on the rate of the Valley National Bank.
Spelling depending on the language
• 85,070,170 in English — eighty-five million seventy thousand one hundred seventy
• 85,070,170 in Spanish — ochenta y cinco millones setenta miles cien setenta
• 85,070,170 in Russian — восемьдесят пять миллионов семьдесят тысяч сто семьдесят
• 85,070,170 in Italian — ottantacinque milioni settantamilacentosettanta
• 85,070,170 in German — fünfundachtzig Millionen siebzigtausendeinhundertsiebzig
• 85,070,170 in Portuguese — oitenta e cinco milhões setenta mil cem e setenta
• 85,070,170 in Chinese — 八十五一百万七十一千一百七十 | 503 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.840167 |
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### Author Topic: Moon question. (Read 2491 times)
#### Don_1
• Neilep Level Member
• Posts: 6890
• Thanked: 7 times
• A stupid comment for every occasion.
##### Moon question.
« on: 27/01/2009 12:27:59 »
Which is the only month in recorded history in which there was no full Moon?
#### lightarrow
• Neilep Level Member
• Posts: 4586
• Thanked: 7 times
##### Moon question.
« Reply #1 on: 27/01/2009 12:42:52 »
Which is the only month in recorded history in which there was no full Moon?
When it's cloudy?
#### dentstudent
• Neilep Level Member
• Posts: 3146
• FOGger to the unsuspecting
##### Moon question.
« Reply #2 on: 27/01/2009 12:48:24 »
February?
#### Vern
• Neilep Level Member
• Posts: 2072
##### Moon question.
« Reply #3 on: 27/01/2009 12:51:30 »
Good ole Google seems to find that your original question may need work This is not in all of recorded history; but it seems like it happens with some regularity.
Quote
We calculated the dates and times of all of the Full Moons in the thousand years from 2000 to 2999 inclusive. There are 12,368 Full Moons during that period, and 952 of them are in February.
Since a thousand years must include a thousand Februaries, it is obvious straight away that 48 of those Februaries are missing a Full Moon.
Counting the number of Full Moons which fall in February in a leap year, we find 240 of them.
However, the thousand-year period from 2000 to 2999 has 243 leap years. (Remember that in the Gregorian calendar, century years are only leap years if they divide by 400, so 2100 will not be a leap year, and nor will 2200, 2300, 2500, 2600, 2700 and 2900.)
So there are 243 leap years, but only 240 of them have a Full Moon in February. This means that there will be just three leap years in which February will have no Full Moon.
Those years are 2572, 2792 and 2944.
It turns out that in those three years, both January and March have two Full Moons, but that isn't surprising since a 29-day February can only miss out on a Full Moon if there is a Full Moon late on January 31st. That, in turn, means that there must have been a Full Moon in early January, and it also means that the next Full Moon must be early on the morning of March 1st, thus March will also have a Full Moon at the end of the month.
#### Don_1
• Neilep Level Member
• Posts: 6890
• Thanked: 7 times
• A stupid comment for every occasion.
##### Moon question.
« Reply #4 on: 27/01/2009 13:20:50 »
Bugger!
Seems you're right Vern.
I take that question back.
?nooM lluf on saw ereht hcihw ni yrotsih dedrocer ni htnom ylno eht si hcihW
Back to the drawing board.
#### lyner
• Guest
##### Moon question.
« Reply #5 on: 27/01/2009 13:32:55 »
In the same vein, in some part of the world, at midsummer, there will be two longest days, sandwiched by the shortest night. And, at particular longitude, there will be two equal shortest nights and one longest day. In both cases there is a symmetrical situation.
#### Vern
• Neilep Level Member
• Posts: 2072
##### Moon question.
« Reply #6 on: 27/01/2009 13:55:29 »
Bugger!
Seems you're right Vern.
I take that question back.
?nooM lluf on saw ereht hcihw ni yrotsih dedrocer ni htnom ylno eht si hcihW
Back to the drawing board.
No; the question still works; and the second answer you got was correct. You didn't say that it just happened once as I originally gleaned. It was just one month that may happen more than once.
#### Don_1
• Neilep Level Member
• Posts: 6890
• Thanked: 7 times
• A stupid comment for every occasion.
##### Moon question.
« Reply #7 on: 27/01/2009 14:09:56 »
As to this new question from SC, are there actually names given to these places, or are they just points of long/latitude?
#### DoctorBeaver
• Naked Science Forum GOD!
• Posts: 12656
• Thanked: 3 times
• A stitch in time would have confused Einstein.
##### Moon question.
« Reply #8 on: 28/01/2009 00:15:03 »
In the same vein, in some part of the world, at midsummer, there will be two longest days, sandwiched by the shortest night. And, at particular longitude, there will be two equal shortest nights and one longest day. In both cases there is a symmetrical situation.
I used to live right on the equator and we had 2 longest days each year - once when the sun passed the equator on its way north, and then when it went back south. Similarly, there were 2 shortest days when the sun was at each tropic.
#### The Naked Scientists Forum
##### Moon question.
« Reply #8 on: 28/01/2009 00:15:03 » | 1,274 | 4,517 | {"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} | 2.734375 | 3 | CC-MAIN-2016-44 | longest | en | 0.952627 |
https://www.inchcalculator.com/convert/farad-to-nanofarad/ | 1,721,923,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00243.warc.gz | 712,676,977 | 15,715 | Enter the capacitance in farads below to get the value converted to nanofarads.
1 F = 1,000,000,000 nF
Since one farad is equal to 1,000,000,000 nanofarads, you can use this simple formula to convert:
The capacitance in nanofarads is equal to the capacitance in farads multiplied by 1,000,000,000.
For example, here's how to convert 5 farads to nanofarads using the formula above.
nanofarads = (5 F × 1,000,000,000) = 5,000,000,000 nF
There are 1,000,000,000 nanofarads in a farad, which is why we use this value in the formula above.
1 F = 1,000,000,000 nF
The farad is defined as the capacitance of a capacitor that has a potential difference of one volt when charged by one coulomb of electricity.[1] The farad is considered to be a very large capacitance value, and multiples of the farad are commonly used to measure capacitance in practical applications, though the farad is still used in some applications.
The farad is the SI derived unit for capacitance in the metric system. Farads can be abbreviated as F; for example, 1 farad can be written as 1 F.
The nanofarad is 1/1,000,000,000 of a farad, which is the capacitance of a capacitor with a potential difference of one volt when it is charged by one coulomb of electricity.
The nanofarad is a multiple of the farad, which is the SI derived unit for capacitance. In the metric system, "nano" is the prefix for billionths, or 10-9. Nanofarads can be abbreviated as nF; for example, 1 nanofarad can be written as 1 nF.
0.000000001 F 1 nF
0.000000002 F 2 nF
0.000000003 F 3 nF
0.000000004 F 4 nF
0.000000005 F 5 nF
0.000000006 F 6 nF
0.000000007 F 7 nF
0.000000008 F 8 nF
0.000000009 F 9 nF
0.0000000001 F 0.1 nF
0.000000001 F 1 nF
0.00000001 F 10 nF
0.0000001 F 100 nF
0.000001 F 1,000 nF
0.00001 F 10,000 nF
0.0001 F 100,000 nF
0.001 F 1,000,000 nF
0.01 F 10,000,000 nF
0.1 F 100,000,000 nF
1 F 1,000,000,000 nF
## References
1. International Bureau of Weights and Measures, The International System of Units, 9th Edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf | 709 | 2,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.856123 |
https://www.physicsforums.com/threads/definite-trigonometric-integral.952307/ | 1,618,972,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00287.warc.gz | 1,032,173,610 | 19,008 | # Definite trigonometric integral
Gold Member
## Homework Statement
solve ##\int_0^1 x^6 \arcsin{x} dx##
Last edited:
Gold Member
## Homework Statement
solve ##\int_0^1 x^6 \arcsin{x} dx##
3. The Attempt at a Solution
I tried substituting arcsinx=y, but abandoned it a few steps later ahead as it started taking a more complicated form. I have a feeling definite integral properties play a larger role here and tried applying some but didn't get anywhere. I'd appreciate some help, thanks
PS- I solved it using Integration by parts. Looking for a shorter method if exists
Gold Member
Try integrating by parts several times. That is a little tedious but will get you the final answer.
Edit: wrote the above before your PS came up. I think by parts will be the shortest way.
Gold Member
Try integrating by parts several times. That is a little tedious but will get you the final answer.
I solved it by a single integration by parts...I'm looking for an even shorter method
Gold Member
did you get (35π -32)/490 ?
Ray Vickson
Homework Helper
Dearly Missed
I solved it by a single integration by parts...I'm looking for an even shorter method
Show your work. Maybe what you did is the shortest possible way, but we cannot tell if you don't show us.
Gold Member
did you get (35π -32)/490 ?
Yes I did. (I took sin inverse as first function and x^6 as second, after 2 or 3 substitutions I got this answer)
Gold Member
Show your work. Maybe what you did is the shortest possible way, but we cannot tell if you don't show us.
I simply took arcsinx as first function, and x^6 as second function. It was just a simple by-parts and the second term (derivative of arcsin*integral of x^6) was an algebraic term- I just used 2 substitutions to evaluate that and get the second term as well. What I mean by a shorter method is whether I can avoid the entire by-parts procedure and just use definite-integral properties to calculate this (i.e by a different approach)
Gold Member
Sounds good. Which substitutions?
PS (after your edit): the parts method is pretty straightforward. If you want to play around a bit, you can also express arcsin x as -i*ln(i*x + (1-x2)½), but that would be longer to work with, and you would still end up using parts.
Gold Member
Sounds good. Which substitutions?
PS (after your edit): the parts method is pretty straightforward. If you want to play around a bit, you can also express arcsin x as -i*ln(i*x + (1-x2)½), but that would be longer to work with, and you would still end up using parts.
haha, I guess a direct by parts is the shortest method possible. keeping arcsin as arcsin seems a better option, as for the substitutions I used- Its totally lost somewhere in my ton of rough work (side effects of a calculus course) but I remember It was something pretty standard like setting the term in the square-root in the denominator as t^2 and then further simplification
Gold Member
haha, I guess a direct by parts is the shortest method possible. keeping arcsin as arcsin seems a better option, as for the substitutions I used- Its totally lost somewhere in my ton of rough work (side effects of a calculus course) but I remember It was something pretty standard like setting the term in the square-root in the denominator as t^2 and then further simplification
this is probably what the question-maker had in mind I guess, I see no other way to shorten or quicken this | 803 | 3,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-17 | longest | en | 0.969012 |
https://www.xamnation.com/reasoning-daily-quiz-dated-24-april/ | 1,726,168,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00436.warc.gz | 988,075,629 | 31,201 | Email: rakesh@xamnation.com Phone/Whatsapp: +91 9988708161
# Reasoning Daily Quiz – Dated 24 April
## Blood Relation
Directions (Q.1-3): Read the following information carefully and answer the questions given below.
P+Q means P is mother of Q
P- Q means P is mother of Q
P/ Q means P is sister of Q
P*Q means P is wife of Q
P%Q means P is son of Q
1). If ‘M+S+N%P’, then how is M related to N?
a) Grandson
b) Maternal grandmother
c) Paternal grandmother
d) Granddaughter
e) None of these
2). If ‘P%T- Q/ U%R, then how is U related to T?
a) Father
b) Mother
c) Son
d) Daughter
e) None of these
3). Which of the following shows that J is son-in-law of I?
a) N*I-L- K/M%J
b) N*I-L+K/M%J
c) I-L+K/M+N%J
d) N*I-L- K/M+J
e) None of these
Directions (Q.4-6): Read the following information carefully and answer the questions given below.
R is the father of P, who is a son-in-law of M and S is the mother of G. S is a sister of K, who is a brother-in-law of P and H is the daughter of T, who is a grandmother of G.
4). How is G related to P?
a) Son
b) Daughter
c) Granddaughter
d) Grandson
e) Cannot be determined
5). If M is a female, then how is H related to S?
a) Sister
b) Sister-in-law
c) Niece
d) Cannot be determined
e) None of these
6). If K married to N, then how is N related to M?
a) Son-in-law
b) Daughter-in-law
c) Mother-in-law
d) Father-in-law
e) None of these
Solution: | 464 | 1,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.903381 |
https://www.esaral.com/q/if-b-is-a-symmetric-matrix-23434 | 1,721,231,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00723.warc.gz | 669,183,747 | 11,357 | If B is a symmetric matrix,
Question:
If $B$ is a symmetric matrix, write whether the matrix $A B A^{\top}$ is symmetric or skew-symmetric.
Solution:
If $B$ is a symmetric matrix, then $B^{T}=B$.
$\left(A B A^{T}\right)^{T}=\left(A^{T}\right)^{T} B^{T} A^{T} \quad\left[\because(A B C)^{T}=C^{T} B^{T} A^{T}\right]$
$\Rightarrow\left(A B A^{T}\right)^{T}=A B^{T} A^{T} \quad\left[\because\left(A^{T}\right)^{T}=A\right]$
$\Rightarrow\left(A B A^{T}\right)^{T}=A B A^{T} \quad\left[\because B^{T}=B\right]$
$\therefore A B A^{T}$ is a symmetric matrix. | 214 | 559 | {"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} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.396331 |
https://www.australianbesttutors.com/recent_question/75220/faculty-of-business-hospitality-accounting-amp-financeassessments | 1,716,496,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00723.warc.gz | 577,756,796 | 12,626 | ### Recent Question/Assignment
ACCOUNTING & FINANCE
Assessments Guide
Key Information
Module Title : Principles of Finance
Module Code : FIN 1013 (WM)
Module Leader : Nur Izyan Jonaidi
Email of the Module Leader : nurizyan@mahsa.edu.my
Semester/ Year : Semester 3 / Year 1
Credit Value 3
Semester
Assessment
: August 2022
Assignment (Report)
: 30%
Assignment (Presentation)
: 10 %
Mid Term Exam
: 10%
Final Examination : 50 %
Assignment (Report)
This assignment will be assessed based on report submission from the students based on group work and accounted for 30% of overall assessment for this course. The students need to submit full report on the due date stated at the end of report details.
Course Learning Outcomes:
Upon completion of the assignment, the students are able to define the key financial figures in a business operations report and discuss methods of achieving financial objectives.
Specific Objectives of the Assignment:
To encourage student to has the following skills of:
i. Utilizing the correct financial formula.
ii. Analyze results from formula calculation.
iii. Explaining to what extend the analysis guide company to maximize profit.
Assignment Detail:
The student needs to conduct financial ratio analysis on a public listed company (retrieved from Bursa
Malaysia). Each group requires to download the Annual Report from the company’s website and focus on relevant financial data to conduct ratio analysis. Each group need to compare 2020 and 2021 for the company chosen. The workload is based on ratio trend analysis of two years period for a chosen company.
The following are the details of assignment require for the student to prepare in the report:
1. Introduction
Introduce this financial statement analysis
project by stating: c) Liquidity
• Purpose of analysis • The key purpose of liquidity
• Background of the company. • Define the following ratio:
1) Current ratio
2. Main Content ? 2)Calculate each of the ratio. Quick acid ratio
a) Profitability ? Produce graph based on the calculation
? Explain the pattern, analysis, reasons for
• The key purpose of profitability such comparison and recommendations • Define the following ratio: for improvement.
1) Net profit margin ? State which year is better and the
2) Return on Asset reasons why.
3) Return on Equity
? Calculate each of the ratio. d) Market Value
? Produce graph based on the calculation • The key purpose of market value
? Explain the pattern, analysis, reasons for
such comparison and recommendations • Define the following ratio:
1) Price Earnings Ratio for improvement.
? State which year is better and the ? Calculate each of the ratio.
reasons why. ? Produce a graph based on the calculation ? Explain the pattern, analysis, reasons for
such comparison and recommendations b) Leverage (Solvency)
for improvement.
• The key purpose of leverage ? State which year is better and the • Define the following ratio: reasons why.
1) Debt ratio
2)Calculate each of the ratio. Equity ratio 3. Conclusion
?
? Produce graph based on the calculation • Summarize the overall achievements of ? Explain the pattern, analysis, reasons for company based on ratio analysis.
such comparison and recommendations for improvement.
? State which year is better and the reasons why.
The assignment is to be handed in latest 11th DECEMBER 2022 BEFORE 11.59 PM.
Assignment (Presentation)
The presentation is to be conducted on 11TH DECEMBER 2022
.
This assignment is assessed based on presentation from the report prepared and accounted for 10% of overall assessment for this course.
Course Learning Outcomes:
Upon completion of the assignment, the student is able to define the key financial figures in a business operations report and discuss methods of achieving financial objectives.
Specific Objectives of the Assignment:
To encourage student to has the following skills:
i. Communicate complex information in simple and interesting ways to keep the audience engaged
ii. Real-industry skills by developing self-confidence, effective thoughts and feelings
Assignment Detail:
The student’s presentation will be evaluated based on the report produced. The presentation will be conducted in 20 minutes with additional of 10 minutes for Q&A session. This Q&A session will be assessed based on the ability of the student to answer the questions from the lecturer. Please submit the copy of slide presentation together with the report submission.
Assignment Guidelines
1. The general format :
• Use Microsoft Word to write your documentation and convert to PDF for submission.
• Font Face : Times New Roman
• Font Size : 12 point
• 1.5 spacing
• Diagrams and figures should be labelled.
• Check spelling and grammar
• Use clear, short and precise language
• Number the pages
• Referencing and citations should be in Harvard format.
2. Online Submission as accordance to the due stated above and link set at LMS.
3. All submission must be made via LMS. Submission via email will not be accepted as proof of submission.
4. Acceptable Benchmark: not more than 30%. If your Turnitin similarity report is higher than 30%, please rewrite and resubmit until you reach 30%.
Please allow at least 24 hours for Turnitin to generate the Similarity Report. Thus, please submit earlier than the actual date. The Turnitin facility is embedded within the e-University system.
5. Extension: No extension permitted. Late submission will result in reduction in marks (1 day 10%, 2-4 days 15%, 5 days 30%. Submission on the 6th day onwards will result in 0 mark.
6. Plagiarism: At all time, candidates are governed by the appropriate Policies on Academic Dishonesty.
Students are required to submit the assignment in time. Any late submission will not be accepted. A mark of ZERO will be awarded for late or no submission. Students are advised to make another copy for own reference.
Feedback
Students can inquire the assignment feedback and grades. However, official grades will be published on the notice board.
Examination scripts are retained by MAHSA University and are not returned to students.
However, students are entitled in obtaining feedback on student’s performance in the examination. Student may request an appeal with the Programme Leader to re-mark student’s examination script. As a Standard Operating Procedure, a Re-Mark charge is RM50.00 for a paper payable upon submission of the Re-Mark Form.
Marks
It is compulsory for student to PASS all the components of the examination. Students are required to achieve a minimum of 50% of the accumulated marks to pass. However, if student fail one of the component of the module, student will be awarded with CONDITIONAL PASS where student must RE-SIT the exam. Failing to re-sit in the semester will cause the student to REPEAT the whole course module in the coming semester if the module is offered.
Assessment Offences
Academic offences include plagiarism, cheating, collusion, copying work and reusing your own work among others.
The University takes academic offences very seriously and this can lead to expulsion. We make every effort to ensure that the students will avoid from committing such offences. The need in maintaining the order in examination and non-examination is to preserve the highest standard of academic integrity. Students are reminded to produce and submit their original work.
Definitions of Assessment Offences Plagiarism
Plagiarism can be defined as the significant use of other people's work and the submission of it as though it were one's own in assessed coursework (such as dissertations, essays, experiments etc.). This includes: • Copying from another student's work;
• Copying text without acknowledgement;
• Submitting work and claiming it to be own when it has been produced by another group; and
• Submitting group work without acknowledging all contributors.
The university uses software packages to detect plagiarism.
Cheating
Cheating in examinations can be defined as trying to gain unfair advantage over fellow students.
This includes:
• Having notes, programmable calculators or any other method to secure the information which are not allowed in the examination, whether the student use it or not;
• Having any mobile phone or other communication devices in the exam room;
• Copying from the examination script of another candidate;
• Helping another candidate; and • Trying to bribe respective lecturers to obtain good marks.
Collusion
Collusion is a form of agreement between two or more people to act with the intention to deceive an assessor as to who was responsible for producing the material submitted for assessment. The agreement may be overt (openly discussed) or covert (not specifically
discussed but implied). Using someone else’s work, words, images, ideas or discoveries is equal to stealing the hard work. This includes:
• agreeing with others to cheat;
• getting someone else to produce part or all your work;
• copying the work of another person (with their permission);
• submitting work from essay banks;
• paying someone to produce work for you; and • allowing another student to copy your own work.
Students are advised to avoid copying and using other people’s work without citation and acknowledgement. It is compulsory for students to include the references and the source of information. This will allow students to avoid plagiarism.
Procedures for Assessment Offences
If students are caught committing the offence, a report of offence will be written. Students will be called for a mini hearing whereby students can justify their action. The University has the right to suspend or even terminate the student should University found the action as a serious offence. Kindly refer to MAHSA Student Handbook for more details.
REPORT MARKING RUBRIC
PROGRAMME LECTURER’S NAME
COURSE CODE TOPIC / GROUP’S NAME
COURSE NAME DATE OF
SUBMISSION
NO STUDENT’S NAME STUDENT’S ID STUDENT’S SIGNATURE
1
CRITERIA MARKS ALLOCATED MARKS MARKS OBTAINED
1 2 3 4 5
Introduction 5 % Not able to
introduce
the given task Minimal ability to
introduce the given task Moderate ability to
introduce the given task Able to introduce the given task Able to introduce with good illustration
Content
a) Profitability 30 % The content was
poorly designed The content was quite relevant The content was
appropriately designed The content
was highly relevant The content was highly relevant and impressive
Page
b) Leverage 20 % The content was poorly designed The content was quite relevant The content was
appropriately designed The content
was highly relevant The content was highly
relevant and impressive
c) Liquidity 20 % The content was poorly designed The content was quite relevant The content was
appropriately designed The content
was highly relevant The content was highly
relevant and impressive
d) Market Value
(Investment Ratio) 10 % The content was poorly designed The content was quite relevant The content was
appropriately designed The content
was highly relevant The content was highly
relevant and impressive
Conclusion 5 % Not able to conclude the given task Minimal ability to conclude the given task Moderate ability to
conclude the given task Good Ability to conclude the given task Excellent ability to
conclude with good
illustration
d) Format
Organization / flow of report 5 % This report was prepared with poor
consideration of format and organisation. This report was prepared with weak
consideration of format and organisation. This report was prepared
with moderate consideration of format and organisation. This report was prepared with well
consideration of format and organisation. This report was prepared with excellent
consideration of format and organisation.
c) Language 5 % Poor sentence writing and
grammatical errors Weak in sentence
writing and
grammatical errors Moderate in sentence
writing with less
grammatical errors Good in sentence
writing and less
grammatical errors Excellent in sentence
writing and no
grammatical errors
Overall marks 100 %
PRESENTATION ASSESSMENT RUBRIC
PROGRAMME LECTURER’S NAME
COURSE CODE TOPIC / GROUP’S NAME
COURSE NAME DATE OF
SUBMISSION
NO STUDENT’S NAME STUDENT’S ID STUDENT’S SIGNATURE
1
CRITERIA MARKS ALLOCATED MARKS MARKS OBTAINED
1 2 3 4 5
Introduction 10 % Introduction is not interesting
and not relevant to topic Introduction is slightly
interesting and
not relevant to topic Introduction is sufficiently
interesting and
quite relevant to topic Interesting introduction and relevant Very Interesting introduction and completely relevant
Content &
Knowledge in the Subject Matter 30 % Minimal scope coverage Weakly covered scope addressed Moderately covered scope addressed Good covered scope addressed Good and
CRITERIA MARKS ALLOCATED MARKS MARKS OBTAINED
1 2 3 4 5
Defense ability 15 % Not able to respond to
questions at
all Not able to respond to questions adequately Able to respond to questions satisfactorily Able to respond to questions confidently Able to respond to questions
confidently and convincingly
Fluency and clarity 15 % Not clear and not fluent Satisfactorily clear and fluent Sufficiently clear and fluent Very clear and fluent Exceptionally clear and fluent
Effective audio visual 10 % Visual are poorly
utilized and presented Visual are moderately
utilized and presented Visual are adequately
utilized and sufficiently presented Visual are well utilized and presented Exceptionally creative,
innovative and effective presentation
Attire & professionalism
(physical appearance
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professional Moderate in term of attire and not professional Satisfactory in term of attire and professional Good in term of attire and professional Excellent in term of attire and professional
Capturing and sustaining audiences’ interests 10 % Dull and boring Moderate Satisfactory Interesting Very interesting
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Overall marks 100 % | 2,882 | 13,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.869004 |
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# A recent article in The New York Times reported that many
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A recent article in The New York Times reported that many [#permalink]
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02 Dec 2006, 11:38
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A recent article in The New York Times reported that many recent college graduates had decided on moving back into their parents’ home rather than face the uncertainty and expense of the rental market.
a. had decided on moving back into their parents’ home rather than face
b. had decided on moving back into their parents’ homes instead of facing
c. have decided to move back into their parents’ homes instead of facing
d. had decided to move back into their parents’ homes rather than facing
e. have decided to move back into their parents’ homes rather than face
The answer is E. When to use "rather than" vs "instead of"?
Thanks.
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02 Dec 2006, 13:06
beetzbat wrote:
A recent article in The New York Times reported that many recent college graduates had decided on moving back into their parents’ home rather than face the uncertainty and expense of the rental market.
a. had decided on moving back into their parents’ home rather than face
b. had decided on moving back into their parents’ homes instead of facing
c. have decided to move back into their parents’ homes instead of facing
d. had decided to move back into their parents’ homes rather than facing
e. have decided to move back into their parents’ homes rather than face
The answer is E. When to use "rather than" vs "instead of"?
Thanks.
http://www.gmatclub.com/phpbb/viewtopic ... 0785f103b6
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### Show Tags
02 Dec 2006, 14:08
I eliminated choices based on tense and ||ism. E is my answer
beetzbat wrote:
A recent article in The New York Times reported that many recent college graduates had decided on moving back into their parents’ home rather than face the uncertainty and expense of the rental market.
a. had decided on moving back into their parents’ home rather than face
b. had decided on moving back into their parents’ homes instead of facing
c. have decided to move back into their parents’ homes instead of facing
d. had decided to move back into their parents’ homes rather than facing
e. have decided to move back into their parents’ homes rather than face
The answer is E. When to use "rather than" vs "instead of"?
Thanks.
Re: princeton review SC [#permalink] 02 Dec 2006, 14:08
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Display posts from previous: Sort by | 1,067 | 4,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.927976 |
https://www.questia.com/read/119844977/bringing-math-home-a-parent-s-guide-to-elementary | 1,606,823,848,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00317.warc.gz | 820,215,022 | 22,375 | # Bringing Math Home: A Parent's Guide to Elementary School Math : Games + Activities + Projects
By Suzanne L. Churchman | Go to book overview
9
Connections
Knowledge of any kind is of little use if it can only be used in one narrow circumstance. Suppose we watch the weather report on television and find that the temperature will be 30 °F (-1 °C) tomorrow. That is an interesting fact that we understand, but we could not utilize that information if we did not make many connections with the fact. Those connections tell us that we need to dress warmly for the day. We may have to scrape the ice on our car's windows before going to work. The dog's water needs to be checked. It might be advisable to get a road report to see if driving could be hazardous. Making connections allows us to make sense of the world we live in.
As children grow and experience life, they, too, begin to make connections in their daily lives. At times we may find the little ones amusing when they make an assumption that is incorrect, just because they have not made all the connections that more mature individuals do. An example might be that a kindergartner would rather have a handful of ten pennies than three dimes. The child has not made the connection that coins have different values, and that he or she can purchase more with the dimes than with the pennies. The handful of ten seems like much more than merely three coins.
There are three objectives that children need to work toward in making connections in mathematics.
Children should identify and use connections among the different math
concepts and ideas.
Children who are able to see the relationship between mathematical ideas will find the subject much easier to understand. Realizing that addition and subtraction are not totally different things, but rather just opposite from one another, makes both comprehensible. Understanding that multiplication is simply the repeated addition of a number takes away the mystery and foreignness of learning those [times] tables.
-189-
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Bringing Math Home: A Parent's Guide to Elementary School Math : Games + Activities + Projects
Table of contents
#### Table of contents
• Title Page i
• Acknowledgments iii
• Contents iv
• Introduction v
• How to Use This Book vii
• Content Standards 1
• 1: Numbers and Operations 3
• 2: Algebra 35
• 3: Geometry 57
• 4: Measurement 97
• 5: Data Analysis and Probability 129
• Process Standards 155
• 6: Problem Solving 157
• 7: Reasoning and Proof 173
• 8: Communication 181
• 9: Connections 189
• 10: Representation 195
• Summing It Up 201
• Appendix 203
• Glossary 223
• Bibliography 229
• Index 230
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An unknown error has occurred. Please click the button below to reload the page. If the problem persists, please try again in a little while. | 1,251 | 5,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-50 | latest | en | 0.961701 |
http://slashdot.org/~Warped-Reality | 1,440,773,211,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644063666.29/warc/CC-MAIN-20150827025423-00344-ip-10-171-96-226.ec2.internal.warc.gz | 218,779,806 | 22,148 | typodupeerror
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## Comment Re:Storm chasers say they have as much right to wa (Score 1)402
The probability of someone with a PhD in meteorology (or something similar) knowing what they are doing with a storm is quite a bit higher than some random guy who thinks storms are neat.
Are there amateurs who know a lot about what they are doing and their hobby is contributing in some way to society / our knowledge base? I'm sure there are, but very few compared to the "random yokel who wants to video tape the storm because he watched Twister last weekend"
## Comment Re:Storm chasers say they have as much right to wa (Score 1)402
Yup. Clearly if money is the prime goal one wants to spend 4 years in a tough undergrad program becfore moving on to a 6 year (on average) graduate program and probably a post-doc or 2, vs. getting a degree in marketing and having your company pay for your MBA and rolling in 6 figures.
## Geologists Might Be Charged For Not Predicting Quake375
mmmscience writes "In 2009, a series of small earthquakes shook the region of L'Aquila, Italy. Seismologists investigated the tremors, but concluded that there was no direct indication of a big quake on the horizon. Less than a month later, a magnitude 6.3 earthquake killed more than 300 people. Now, the chief prosecutor of L'Aquila is looking to charge the scientists with gross negligent manslaughter for not predicting the quake."
## Comment Re:What about saying fuck it? (Score 1)200
For the same reason that if you build me a house, I can't pull out halfway through construction and leave you out \$50,000?
## Comment Re:This is a random comment. (Score 1)395
This is called "almost surely" or "almost never" in probability theory.
http://en.wikipedia.org/wiki/Almost_surely
While the probability of getting an infinite number on 1's is 0, it's still in the sample space, therefore can still happen (in fact, with an infinte number of coin tosses, any particular sequence has a probability of zero of occurring)
## Comment OpenVPN (Score 0, Offtopic)312
Set up an OpenVPN system at home and remotely connect to it, giving you high quality (AES) over-the-air encryption, even on an open and unencrypted system.
## Comment Re:TeXmacs (Score 1)823
I second this. I used Texmacs in undergrad and grad school - it's pretty good for taking notes, although if I'm going to sit down and type something up, i'll go with plain old LaTeX
## Comment Re:CONFIRMED: You are missing something. (Score 0, Offtopic)266
SSL = secure sockets layer
the ability to asymmetrically encrypt data = public key cryptography (See RSA, El Gamal, NTRU, etc)
We can predict everything, except the future.
Working... | 664 | 2,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-35 | longest | en | 0.951433 |
https://socratic.org/questions/three-consecutive-multiples-of-3-have-a-sum-of-36-what-is-the-greatest-number | 1,638,718,939,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363189.92/warc/CC-MAIN-20211205130619-20211205160619-00385.warc.gz | 590,190,444 | 6,515 | # Three consecutive multiples of 3 have a sum of 36. What is the greatest number?
Nov 28, 2016
The greatest of the three numbers is 15.
The other two numbers are 9 and 12.
#### Explanation:
The three consecutive multiples of 3 can be written as;
$x$, $x + 3$ and $x + 6$ with $x + 6$ being the greatest.
We know from the problem the sum of these three numbers equal 36 so we can write and solve for $x$ through the following:
$x + x + 3 + x + 6 = 36$
$3 x + 9 = 36$
$3 x + 9 - 9 = 36 - 9$
$3 x = 27$
$\frac{3 x}{3} = \frac{27}{3}$
$x = 9$
Because we are looking for the largest we must add $6$ to $x$ to obtain the largest number:
$6 + 19 = 15$
Nov 28, 2016
15
#### Explanation:
A multiple of 3 can be written $3 n$ where $n$ is a positive integer.
So 3 consecutive multiples of 3 can be written $3 n , 3 n + 3 , 3 n + 6$
The sum of these is 36
$3 n + 3 n + 3 + 3 n + 6 = 36$
$9 n + 9 = 36$
Divide through by 9
$n + 1 = 4$
$n = 3$
If $n = 3$ then $3 n = 9$ and the three consecutive multiples of three are 9, 12 and 15 which do indeed total 36 | 380 | 1,062 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 23, "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} | 4.6875 | 5 | CC-MAIN-2021-49 | latest | en | 0.871926 |
http://mathhelpforum.com/calculus/116763-help-l-hospital-s-rule-question.html | 1,527,021,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864872.17/warc/CC-MAIN-20180522190147-20180522210147-00065.warc.gz | 191,430,727 | 9,993 | Thread: Help with L'Hospital's Rule question
1. Help with L'Hospital's Rule question
Hey guys,
I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.
Here's the question:
Evaluate the limit.
This is what I have gotten so far:
lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)
Took the deriv. again because I get 0/0
lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.
Any help's appreciated. Thanks!
2. Originally Posted by dark-ryder341
Hey guys,
I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.
Here's the question:
Evaluate the limit.
This is what I have gotten so far:
lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)
Took the deriv. again because I get 0/0
lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.
Any help's appreciated. Thanks!
$\displaystyle \lim_{x \to 0}\cot{(4x)}\sin{(8x)} = \lim_{x \to 0}\frac{\sin{(8x)}}{\tan{(4x)}}$
Since this tends to the indeterminate $\displaystyle \frac{0}{0}$, we can use L'Hospital.
$\displaystyle = \lim_{x \to 0}\frac{\frac{d}{dx}[\sin{(8x)}]}{\frac{d}{dx}[\tan{(4x)}]}$
$\displaystyle = \lim_{x \to 0}\frac{8\cos{(8x)}}{4\sec^2{(4x)}}$
$\displaystyle = \lim_{x \to 0}2\cos{(4x)}\cos^2{(8x)}$
$\displaystyle = 2\cos{0}\cos^2{0}$
$\displaystyle = 2$.
3. another way : just simplify the term $\displaystyle \sin(8x)\cot(4x)$.
you'll get the same result without using the L'hospital rule. | 653 | 1,934 | {"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} | 4.3125 | 4 | CC-MAIN-2018-22 | latest | en | 0.896441 |
https://k12.libretexts.org/Bookshelves/Mathematics/Geometry/04%3A_Triangles/4.27%3A_The_Pythagorean_Theorem | 1,721,211,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00093.warc.gz | 302,781,413 | 33,323 | # 4.27: The Pythagorean Theorem
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Square of the hypotenuse equals the sum of the squares of the legs in right triangles.
The two shorter sides of a right triangle (the sides that form the right angle) are the legs and the longer side (the side opposite the right angle) is the hypotenuse. For the Pythagorean Theorem, the legs are “$$a$$” and “$$b$$” and the hypotenuse is “$$c$$”.
Pythagorean Theorem: Given a right triangle with legs of lengths a and b and a hypotenuse of length $$c$$, $$a^2+b^2=c^2$$.
The converse of the Pythagorean Theorem is also true. It allows you to prove that a triangle is a right triangle even if you do not know its angle measures.
Pythagorean Theorem Converse: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
If $$a^2+b^2=c^2$$, then $$\Delta ABC$$ is a right triangle.
##### Pythagorean Triples
A combination of three numbers that makes the Pythagorean Theorem true is called a Pythagorean triple. Each set of numbers below is a Pythagorean triple.
$$3,4,5 \qquad 5,12,13\qquad 7,24,25\qquad 8,15,17\qquad 9,12,15\qquad 10,24,26$$
Any multiple of a Pythagorean triple is also considered a Pythagorean triple. Multiplying 3, 4, 5 by 2 gives 6, 8, 10, which is another triple. To see if a set of numbers makes a Pythagorean triple, plug them into the Pythagorean Theorem.
What if you were told that a triangle had side lengths of 5, 12, and 13? How could you determine if the triangle were a right one?
Example $$\PageIndex{1}$$
What is the diagonal of a rectangle with sides 10 and 16?
Solution
For any square and rectangle, you can use the Pythagorean Theorem to find the length of a diagonal. Plug in the sides to find $$d$$.
\begin{align*} 10^2+16^2=d^2 \\ 100+256=d^2 \\ 356&=d^2 \\ d&=\sqrt{356}=2\sqrt{89}\approx 18.87 \end{align*}
Example $$\PageIndex{2}$$
Do 6, 7, and 8 make the sides of a right triangle?
Solution
Plug the three numbers into the Pythagorean Theorem. Remember that the largest length will always be the hypotenuse, c. If $$6^2+7^2=8^2$$, then they are the sides of a right triangle.
\begin{align*} 6^2+7^2&=36+49=85& \\ 8^2&=64&\qquad 85 \neq 64,\: so\: the \:lengths \:are \:not \:the \:sides \:of \:a \:right \:triangle.\end{align*}
Example $$\PageIndex{3}$$
Find the length of the hypotenuse.
Solution
Use the Pythagorean Theorem. Set $$a=8$$ and $$b=15$$. Solve for $$c$$.
\begin{align*} 8^2+152&=c^2 \\ 64+225&=c^2 \\ 289&=c^2 \qquad Take\: the \:square \:root \:of \:both \:sides. \\ 172&=c\end{align*}
Example $$\PageIndex{4}$$
Is 20, 21, 29 a Pythagorean triple?
Solution
If $$20^2+21^2=29^2$$, then the set is a Pythagorean triple.
\begin{align*} 20^2+21^2&=400+441=841 \\ 29^2&=841 \end{align*}
Therefore, 20, 21, and 29 is a Pythagorean triple.
Example $$\PageIndex{5}$$
Determine if the triangles below are right triangles.
Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest side represent c.
Solution
\begin{align*} a^2+b^2&=c^2 \\ 82+162 &\stackrel{?}{=}(8\sqrt{5})2 \\ 64+256 &\stackrel{?}{=}64\cdot 5 \\ 320&=320\qquad Yes\end{align*}
\begin{align*} a^2+b^2&=c^2 \\ 22^2+24^2&\stackrel{?}{=}262 \\ 484+576 &\stackrel{?}{=}676 \\ 1060 &\neq 676\qquad No\end{align*}
## Review
Find the length of the missing side. Simplify all radicals.
1. If the legs of a right triangle are 10 and 24, then the hypotenuse is __________.
2. If the sides of a rectangle are 12 and 15, then the diagonal is _____________.
3. If the sides of a square are 16, then the diagonal is ____________.
4. If the sides of a square are 9, then the diagonal is _____________.
Determine if the following sets of numbers are Pythagorean Triples.
1. 12, 35, 37
2. 9, 17, 18
3. 10, 15, 21
4. 11, 60, 61
5. 15, 20, 25
6. 18, 73, 75
Determine if the following lengths make a right triangle.
1. 7, 24, 25
2. $$\sqrt{5},2\sqrt{10},3\sqrt{5}$$
3. $$2\sqrt{3},\sqrt{6},8$$
4. 15, 20, 25
5. 20, 25, 30
6. $$8\sqrt{3},6,2\sqrt{39}$$
## Vocabulary
Term Definition
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $$a^2+b^2=c^2$$, where a and b are legs of the triangle and c is the hypotenuse of the triangle.
Pythagorean triple A combination of three numbers that makes the Pythagorean Theorem true.
Circle A circle is the set of all points at a specific distance from a given point in two dimensions.
Conic Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas.
degenerate conic A degenerate conic is a conic that does not have the usual properties of a conic section. Since some of the coefficients of the general conic equation are zero, the basic shape of the conic is merely a point, a line or a pair of intersecting lines.
Ellipse Ellipses are conic sections that look like elongated circles. An ellipse represents all locations in two dimensions that are the same distance from two specified points called foci.
hyperbola A hyperbola is a conic section formed when the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shaped curves.
Hypotenuse The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Parabola A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
Pythagorean number triple A Pythagorean number triple is a set of three whole numbers a,b and c that satisfy the Pythagorean Theorem, $$a^2+b^2=c^2$$.
Right Triangle A right triangle is a triangle with one 90 degree angle.
Interactive Element
Video: Using The Pythagorean Theorem Principles - Basic
Activities: Pythagorean Theorem and Pythagorean Triples Discussion Questions
Study Aids: Pythagorean Theorem Study Guide
Practice: The Pythagorean Theorem
Real World: Pythagoras TV
This page titled 4.27: The Pythagorean Theorem is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. | 3,696 | 10,687 | {"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 | 4 | CC-MAIN-2024-30 | latest | en | 0.197913 |
noblackholes.com | 1,638,364,611,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00054.warc.gz | 58,956,443 | 5,723 | Conceptual Papers
(no equations):
Technical Papers:
Other Articles & Papers:
## Do black holes exist?
Many believe not only that black holes exist, but that they are a source of improbable wonders such as volumeless matter, time travel and worm holes to other universes. But common sense should instantly raise skepticism about such claims. In fact, Albert Einstein argued vigorously that black holes were incompatible with reality as described by his theories of relativity. He wrote a paper specifically on this topic in 1939.
Noblackholes.com supports Einstein's skepticism and explains why the conservation of energy (on which general relativity is based) does not allow for the formation of black holes. In a nutshell, black holes do not exist because there is an upper limitation on the gravitational energy that a mass (m) can produce. The upper limit is the energy equivalence of the mass (e=mc2). Because the gravitational energy required to create a black hole is greater than the equivalent energy of the mass, a black hole will never form. As the gravitational energy required to compact a mass approaches the energy equivalence of the mass, the ability of gravity to further compact the mass will be diminished by the relativistic effects of gravity on time and space.
For example, every observer of a collapsing star will observe that the star will stop collapsing before crossing a critical radius and becoming a black hole. However, an observer's perception of the passage of time can vary significantly depending of the strength of the gravity field from which observations are made. From the perspective of a distant observer, the rate of collapse will slow down so much as the surface approaches the critical radius that the critical radius will never be crossed in finite time. Over eons of time, the star will eventually disintegrate without the surface ever crossing the critical radius. From the perspective of an observer on the surface of the collapsing star, as the surface quickly approaches the critical radius, the star will instanteously disintegrate before the critical radius can be reached. For a fuller explanation of this, see the conceptual paper: The time barrier that prevents formation of black holes.
This website contains a series of conceptual papers that describe, without the use of equations, why through the lens of general relativity black holes can never form. Technical papers, including three peer reviewed physics journal articles, provide a more thorough description of the problems of black holes and particularly how black holes violate the principle of the conservation of energy that underlies Einstein's theories of relativity.
Go to top
## A brief history of black holes
Isaac Newton's seminal work on gravity, Principia (1687), postulates a law of gravity where gravitational potential energy outside a mass is inversely proportional to radial distance from the center of the mass. According to this law of gravity, the escape velocity on the surface of a mass increases as the mass is compacted. In 1783 the Reverend John Michell, a British natural philosopher, pointed out that if a mass could be compacted within a critical radius where the escape velocity on the surface of the mass equals the speed of light, light would not escape from the surface. This would create an invisible mass, now called a black hole....
More on the history of black holes
Go to top
## The time barrier that prevents formation of black holes
As a mass is compacted to have a smaller and smaller radius, the escape velocity at the surface of the resulting sphere increases. If the sphere could be compacted to a critical radius (called the Schwarzschild radius) so that the escape velocity at the surface of the sphere is equal to the speed of light, nothing could escape from the gravity field. The result would be the formation of a black hole. However, the dilation of time that occurs with increasing gravity erects an impenetrable barrier at the Schwarzschild radius that is able to prevent any mass from compacting sufficiently to form a black hole...
Go to top
## Hawking radiation and black hole evaporation
Hawking radiation is named after physicist Stephen Hawking who in 1974 provided a theoretical argument for the existence of thermal radiation emitted by black holes. The existence of Hawking radiation, now commonly accepted among physicists, presents a very significant logical problem for those who additionally believe that gravity affects time and that light and matter can pass through the event horizon of a black hole. Specifically, in addition to completely evaporating a black hole before light or matter could reach the black hole’s event horizon, Hawking radiation, intensified by the acceleration of time, will destructively irradiate anything drawing near to the event horizon...
Go to top
## Pathological coordinates and special coordinates
If a time barrier prevents formation of black holes and if anything that approaches a very compact mass experiences instant evaporation, how can some still claim black holes form? One fallacy that has contributed to the mistaken belief that black holes form in Einstein's relativistic space time is the labelling of ordinary coordinates as "pathological" and the assertion that there exist "special" coordinates that can be used to show a mass will compact to form a black hole...
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## Technical Papers
### How black holes violate the conservation of energy
Black holes produce more energy than they consume thereby violating the conservation of energy and acting as perpetual motion machines.
Go to top
### Five fallacies used to link black holes to Einstein’s relativistic space-time
For a particle falling radially toward a compact mass, the Schwarzschild metric maps local time to coordinate time based on radial locations reached by the particle. The mapping shows the particle will not cross a critical radius regardless of the coordinate used to measure time. Herein are discussed five fallacies that have been used to make it appear the particle can cross the critical radius.
Go to top
### Gravity and the conservation of energy
The Schwarzschild metric apportions the energy equivalence of a mass into a time component, a space component and a gravitational component. This apportionment indicates there is a source of gravitational energy as well as a limit to the magnitude of gravitational energy.
Go to top
### Momentum and energy in the Schwarzschild Metric
Albert Einstein validated his field equations by demonstrating that they complied with what he called the laws of momentum and energy. The most well-known solution to Einstein’s field equations is the Schwarzschild metric describing the gravitational field of a mass point. Here is examined how what Einstein called the laws of momentum and energy are manifest in the Schwarzschild metric and how these laws limit the geometry of space-time that is defined by the Schwarzschild metric.
Go to top
### A fundamental principle of relativity
The laws of physics hold equally in reference frames that are in motion with respect to each other. This premise of Albert Einstein’s theory of relativity is a fairly easy concept to understand in the abstract, however the mathematics—particularly the tensor calculus used by Einstein to describe general relativity—used to flesh out this premise can be very complex, making the subject matter difficult for the non-specialist to intuitively grasp. Here is set out a fundamental principle of relativity that can be used as a tool to understand and explain special and general relativity. The fundamental principle of relativity is used to independently derive the Lorentz factor, the Minkowski metric and the Schwarzschild metric. The fundamental principle is also used to derive metric tensors for systems with multiple point masses and to explain Newtonian kinetic energy, gravitational potential energy and mass-energy equivalence in the context of special and general relativity. | 1,547 | 8,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.901994 |
https://support.bioconductor.org/p/94460/ | 1,558,537,090,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256858.44/warc/CC-MAIN-20190522143218-20190522165218-00182.warc.gz | 642,127,150 | 7,679 | Question: Limma: Paired samples, multiple groups: problems understanding contrasts and model.matrix
2
2.1 years ago by
Anna 40
Sweden
Anna 40 wrote:
Hi everyone,
I am trying to sort out how to properly analyze a microarray experiment with multiple groups and "paired" samples and despite my best efforts I am still a bit unsure if I am doing it right.
Here is my set up: I have collected four different cell types from six patients. I want to compare the cell types to each other, but block for "patient effects". Here is an abbreviated list of samples and conditions (samples). My arraydata (from affymetrix) is in an expression set I call exprsset.
<caption>samples</caption>
Array Fraction Patient A1 CellType1 P1 A2 CellType2 P1 A3 CellType3 P1 A4 CellType4 P1 ... ... ... A24 CellType4 P6
So, combining the description for paired samples and multiple groups from limmas user's guide gives something like
patient<-factor(samples$Patient) fraction <- factor(samples$Fraction, levels=c("CellType1, "CellType2", "CellType3", "CellType4"))
1.
design <- model.matrix(~patient+fraction)
fit <- eBayes(fit)
But from here I don't know how to get at the contrasts I am after (pairwise comparisons of cell types): It seems like R is always contrasting against one sample? If so, how do I get the other comparisons (that does not include this reference sample) and how does R choose this reference?
I couldn't sort this out so after some googling (especially these posts: Question on unbalanced paired design and https://support.bioconductor.org/p/55255/), I instead tried this:
2.
design <- model.matrix(~ 0 + fraction + patient )
contrasts <- makeContrasts(fractionCelltype1 - fractionCellType2, fractionCellType1-fractionCellType3, fractionCellType1 - fractionCelltype4, (all other pairwise comparisons), levels=design)
fit2 <- contrasts.fit(fit, contrasts)
fit2 <- eBayes(fit2)
topTable with the coef parameter then seems to give me the comparisons I am after so:
My main question is: does the nr 2 "script" do what I hope, i.e. that is give me the pairwise celltype comparisons "blocked" for patient effects? If not, how do I correct it?
design <- model.matrix(~ 0 + patient + fraction)
and use the same contrasts as in 2., I get: Warning message:In contrasts.fit(fit, contrast.matrix) : row names of contrasts don't match col names of coefficients.
Looking at the design matrix: I have 1 less fraction column that I have fractions, so one of the fractions are "missing", which of course makes a discrepancy with my contrasts, hence the error. The same goes for patients if I use the formula for design under 2. (but since I don't make "patient contrasts" in the makeContrasts(), that doesn't seem to matter...). Naively put: Why does one condition "disappear" in the design matrix? Is it again a reference level of some sort and how do I then handle this?
If someone could explain this or point me to some appropriate tutorial about design matrices and how they relate to the linear models, I would greatly appreciate it! I really bugs me that I can't seem to get my head around this!
Best regards,
Anna
modified 2.1 years ago • written 2.1 years ago by Anna 40
Answer: Limma: Paired samples, multiple groups: problems understanding contrasts and mod
4
2.1 years ago by
United States
James W. MacDonald50k wrote:
You will probably find it easier to understand if you switch the order of patient and fraction. That way you will have coefficients for all the fractions, and can make all the comparisons you want.
To answer your questions, R selects the first level of a factor as the baseline. You can make any comparison you want with either model, but you need to do more algebra to get all the comparisons you care about. In both models (the way you are specifying them), the fractionCellType1 will be your baseline, and all other treatments are compared to that.
In other words fractionCellType2 is actually fractionCellType2 - fractionCellType1. And fractionCellType3 is fractionCellType3 - fractionCellType1. So testing any of those coefficients is actually testing that the difference between those groups is equal to zero. If you care about say fractionCellType3 - fractionCellType2, you need to make that contrast explicitly (which will be (fractionCellType3 - fractionCellType1) - (fractionCellType2 - fractionCellType1), and the fractionCellType1 sample gets cancelled out).
If you specify the model as I suggest, then you can just make all the comparisons you care about directly, rather than having to remember that your coefficients are already implicit differences.
Have you read the limma User's Guide? That's a good tutorial.
Answer: Limma: Paired samples, multiple groups: problems understanding contrasts and mod
0
2.1 years ago by
Anna 40
Sweden
Anna 40 wrote:
Thank you, that helped a lot!
I have read the Limma user's guide, it's great! But still, I didn't get these things. After reading your answer and this http://genomicsclass.github.io/book/pages/expressing_design_formula.html I feel a little bit more sure.
Anyways, from the last link I also found the relevel() function which I guess can be used to change the baseline (reference level) if one wants to (to make contrasting more convenient).
/Anna
2
You can use relevel, or you can specify the baseline at the outset. It's numero-alphabetic by default, so if you have say Untreated and Treated samples, the default is
> factor(rep(c("Treated","Untreated"), 4))
[1] Treated Untreated Treated Untreated Treated Untreated Treated
[8] Untreated
Levels: Treated Untreated
And you can switch that at the outset by adding the levels argument.
> factor(rep(c("Treated","Untreated"), 4), levels = c("Untreated","Treated"))
[1] Treated Untreated Treated Untreated Treated Untreated Treated
[8] Untreated
Levels: Untreated Treated | 1,395 | 5,850 | {"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} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.794306 |
https://kr.mathworks.com/matlabcentral/cody/problems/771-map-all-the-indices-of-an-array-indices-into-a-vector-giving-index-vs-row-and-column/solutions/2196858 | 1,597,473,020,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740679.96/warc/CC-MAIN-20200815035250-20200815065250-00445.warc.gz | 373,089,601 | 16,424 | Cody
# Problem 771. Map all the indices of an Array Indices into a Vector giving Index vs Row and Column
Solution 2196858
Submitted on 7 Apr 2020 by Nikolaos Nikolaou
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
% Vectorization of ind2sub m=zeros(10); [zrc(:,1),zrc(:,2)]=ind2sub(size(m),1:numel(m)); assert(isequal(indices2rc_vector(m),zrc))
irc = 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 4 10 4 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 5 10 5 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 6 10 6 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 7 10 7 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 8 10 8 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 9 10 9 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10
2 Pass
m=rand(4); [zrc(:,1),zrc(:,2)]=ind2sub(size(m),1:numel(m)); assert(isequal(indices2rc_vector(m),zrc))
irc = 1 1 2 1 3 1 4 1 1 2 2 2 3 2 4 2 1 3 2 3 3 3 4 3 1 4 2 4 3 4 4 4
3 Pass
m=zeros(floor(20*rand)); [zrc(:,1),zrc(:,2)]=ind2sub(size(m),1:numel(m)); assert(isequal(indices2rc_vector(m),zrc))
irc = 1 1 2 1 1 2 2 2 | 711 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-34 | latest | en | 0.418863 |
https://a-middletonphotography.com/how-many-km-is-40-miles-update-new/ | 1,669,566,483,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00667.warc.gz | 111,174,394 | 54,592 | Home » How Many Km Is 40 Miles? Update New
# How Many Km Is 40 Miles? Update New
Let’s discuss the question: how many km is 40 miles. We summarize all relevant answers in section Q&A of website A-middletonphotography.com in category: Tips for you. See more related questions in the comments below.
## How many km is 1 mile approximately?
How many kilometers in a mile 1 mile is equal to 1.609344 kilometres, which is the conversion factor from miles to kilometres.
## How long does it take to drive 40 miles?
Answer: The total time taken to cover 40 miles at 60 mph in 40 minutes. The unit of speed is always in accordance with that of the time and distance.
### HOW TO CONVERT KILOMETER(KM) TO MILE AND MILE TO KILOMETER
HOW TO CONVERT KILOMETER(KM) TO MILE AND MILE TO KILOMETER
HOW TO CONVERT KILOMETER(KM) TO MILE AND MILE TO KILOMETER
See also Business Model | CSL ปรับโมเดลสู่ธุรกิจเงินสด #15/11/17 | csl ปันผล | สังเคราะห์เคล็ดลับดีๆหรือชีวิตที่มีประโยชน์
## Is 1 km longer than 1 mile?
1.609 kilometers equal 1 mile. The kilometer is a unit of measurement, as is the mille. However, a mile is longer than a kilometer. “Mile” is a bigger unit.
## How fast is 40 miles per hour in kilometers?
Miles per hour to Kilometers per hour table
Miles per hour Kilometers per hour
40 mph 64.37 kph
41 mph 65.98 kph
42 mph 67.59 kph
43 mph 69.20 kph
## How long would it take to walk 3km?
3K: 3 kilometers equals 1.85 miles, or 9842.5 feet, or just a little less than 2 miles. This is a common distance for charity walks, especially those with accessible routes. It takes 30 to 37 minutes to walk 3K at a moderate pace.
## How long does it take to walk 1 km?
Well, walking a kilometre, i.e., 0.62 miles, should hardly take 10-12 minutes when walking at a moderate speed.
## How long does it take to walk 40 miles?
Here are some figures for different walking paces:
Miles Relaxed Pace Normal Pace
40 miles 13 hrs, 20 mins 10 hrs
50 miles 16 hrs, 40 mins 12 hrs, 30 mins
60 miles 20 hrs 15 hrs
70 miles 23 hrs, 20 mins 17 hrs, 30 mins
Oct 8, 2020
## How long does it take to drive 10km at 60 km h?
= 10 minutes
There are 60 seconds per minute, thus to get the seconds, we multiply the remaining right part of the decimal point above by 60.
## How long would it take to bike 40 miles?
The maximum cyclist takes 3.5 to 4 hours to complete the 40 miles, although it varies on their experience levels. Beginner riders average 10 mph and take 1 hour to cover a 10-mile trip. That same mile may take only 30-45 minutes for professional and experienced riders.
## What is about a kilometer long?
A kilometer (km) is about: a little over half a mile. a quarter of the average depth of the ocean.
## Is 2 km longer than 1 mile?
It is part of the metric system of measurement. Its abbreviation is km. A mile is longer than a kilometer. One mile is equal to 1.609 kilometers.
## How many Kilometres is a mile UK?
### ✅ Convert Mile to Kilometer (mile to Km) – Example and Formula
✅ Convert Mile to Kilometer (mile to Km) – Example and Formula
✅ Convert Mile to Kilometer (mile to Km) – Example and Formula
## Is kph and kmh the same?
The kilometre per hour (SI symbol: km/h; abbreviations: kph, kmph, km/hr) is a unit of speed, expressing the number of kilometres travelled in one hour.
Kilometres per hour.
kilometre per hour
A car speedometer that indicates measured speed in kilometres per hour.
General information
Unit system derived
Unit of speed
## How fast is 60 km?
Kilometer Per Hour to Mile Per Hour Conversion Table
Kilometers Per Hour Miles Per Hour
50 km/h 31.07 mph
55 km/h 34.18 mph
60 km/h 37.28 mph
65 km/h 40.39 mph
## What is 1 km equal to in miles?
1 kilometre is equal to 0.62137119 miles, which is the conversion factor from kilometers to miles.
## How many calories burnt walking 3 km?
Calories Burned Per Mile
Calories Burned Walking 4.0 mph by Miles and Weight (Pace of 15 Minutes per Mile or 9 Minutes per Kilometer)
Mile 2 114 136
Mile 3 170 205
Mile 4 227 273
Mile 5 284 341
Dec 12, 2019
## How many kilometers is a 20 minute walk?
Easy pace: 20 minutes per mile or 12.5 minutes per kilometer. Fast pace: 11 minutes per mile or 7 minutes per kilometer. This is the speed of a fast walk or an easy run.
## How long does 5km take to walk?
Five kilometers equals 3.1 miles. At a typical walking pace, you can walk it in 45 minutes. If you are a slower walker, you might take 60 minutes or more. When choosing a 5K event, make sure it welcomes walkers and has a long enough time limit so you can comfortably finish.
## Is running 1 km in 6 minutes good?
RULE 1: There’s a general rule that goes: maximum 10% incraese in mileage per week. You were increasing by 100% every day, without rest days for your muscles to recover. If you’re an absolute beginner, 6:15 minutes per kilometer is good!
## How many kilometers should I walk per day?
Walking is a form of low impact, moderate intensity exercise that has a range of health benefits and few risks. As a result, the CDC recommend that most adults aim for 10,000 steps per day . For most people, this is the equivalent of about 8 kilometers, or 5 miles.
See also How To Get Alligator Tags In Texas? New
## How long is a 10K walk?
A 10-kilometer (10K) walk is 6.2 miles long. It is a common distance for charity runs and walks and the standard distance for volkssport walks. Most walkers complete a 10K walk in 90 minutes to two hours. Here is a training schedule to get you from the couch to the finish line, feeling great.
## How many km is 30 minutes walk?
If you walk at a brisk pace for 30 minutes, you’ll cover a distance of about 1½ to 2 miles (2.5 to 3.3 kilometers). Take the distance into consideration when planning your walking route.
40 miles to km
40 miles to km
## Can a person walk 40 miles in a day?
The longest distance that a person can walk in 24 hours is about 48 miles (80 kilometers) without stopping to rest. You should stop and take a break about every 8-12 miles (15-20 kilometers).
## Can you walk 100 miles in a day?
The strict walking requirement is what makes walking a 100 miles in a day harder than running that distance, according to those who have done both. To travel 100 miles in 24 hours, you have to average 1 mile every 14 minutes and 24 seconds.
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## Information related to the topic how many km is 40 miles
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You have just come across an article on the topic how many km is 40 miles. If you found this article useful, please share it. Thank you very much. | 1,979 | 7,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-49 | latest | en | 0.789344 |
http://metamath.tirix.org/mpests/ply1val | 1,721,878,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00779.warc.gz | 18,172,983 | 2,667 | # Metamath Proof Explorer
## Theorem ply1val
Description: The value of the set of univariate polynomials. (Contributed by Mario Carneiro, 9-Feb-2015)
Ref Expression
Hypotheses ply1val.1 ${⊢}{P}={\mathrm{Poly}}_{1}\left({R}\right)$
ply1val.2 ${⊢}{S}={\mathrm{PwSer}}_{1}\left({R}\right)$
Assertion ply1val ${⊢}{P}={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
### Proof
Step Hyp Ref Expression
1 ply1val.1 ${⊢}{P}={\mathrm{Poly}}_{1}\left({R}\right)$
2 ply1val.2 ${⊢}{S}={\mathrm{PwSer}}_{1}\left({R}\right)$
3 fveq2 ${⊢}{r}={R}\to {\mathrm{PwSer}}_{1}\left({r}\right)={\mathrm{PwSer}}_{1}\left({R}\right)$
4 3 2 eqtr4di ${⊢}{r}={R}\to {\mathrm{PwSer}}_{1}\left({r}\right)={S}$
5 oveq2 ${⊢}{r}={R}\to {1}_{𝑜}\mathrm{mPoly}{r}={1}_{𝑜}\mathrm{mPoly}{R}$
6 5 fveq2d ${⊢}{r}={R}\to {\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{r}\right)}={\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
7 4 6 oveq12d ${⊢}{r}={R}\to {\mathrm{PwSer}}_{1}\left({r}\right){↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{r}\right)}={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
8 df-ply1 ${⊢}{\mathrm{Poly}}_{1}=\left({r}\in \mathrm{V}⟼{\mathrm{PwSer}}_{1}\left({r}\right){↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{r}\right)}\right)$
9 ovex ${⊢}{S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}\in \mathrm{V}$
10 7 8 9 fvmpt ${⊢}{R}\in \mathrm{V}\to {\mathrm{Poly}}_{1}\left({R}\right)={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
11 fvprc ${⊢}¬{R}\in \mathrm{V}\to {\mathrm{Poly}}_{1}\left({R}\right)=\varnothing$
12 ress0 ${⊢}\varnothing {↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}=\varnothing$
13 11 12 eqtr4di ${⊢}¬{R}\in \mathrm{V}\to {\mathrm{Poly}}_{1}\left({R}\right)=\varnothing {↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
14 fvprc ${⊢}¬{R}\in \mathrm{V}\to {\mathrm{PwSer}}_{1}\left({R}\right)=\varnothing$
15 2 14 syl5eq ${⊢}¬{R}\in \mathrm{V}\to {S}=\varnothing$
16 15 oveq1d ${⊢}¬{R}\in \mathrm{V}\to {S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}=\varnothing {↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
17 13 16 eqtr4d ${⊢}¬{R}\in \mathrm{V}\to {\mathrm{Poly}}_{1}\left({R}\right)={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
18 10 17 pm2.61i ${⊢}{\mathrm{Poly}}_{1}\left({R}\right)={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$
19 1 18 eqtri ${⊢}{P}={S}{↾}_{𝑠}{\mathrm{Base}}_{\left({1}_{𝑜}\mathrm{mPoly}{R}\right)}$ | 1,207 | 2,532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.281583 |
https://forum.vectorworks.net/index.php?/topic/78864-worksheet-sum/#comment-432005 | 1,669,637,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00442.warc.gz | 312,660,246 | 18,303 | # worksheet sum
## Recommended Posts
Hi
What is the syntax for summing a bunch of cells? Say I have numbers in B3, H4 and J8 for example. I can't do =SUM(B3+H4+J8), so what is the correct way?
Thank you
##### Link to comment
If you just want a few cells you don't need the sum you can just add them up: =B3+H4+J8
If you have a larger range you can specify the range by using two periods between the beginning and end of the range plus a Sum: =SUM(A3..D3)
If you have a discontinuous (or continuous) range you can also use a comma as a separator in a Sum: =SUM(B3, H4, J8)
Or you can combine the two styles: =SUM(A3..D3, H8)
HTH
thank you!
##### Link to comment
• 1 year later...
I can't really understand why they don't stick to the Excel standard, I need to google this all the time.
##### Link to comment
On 5/4/2022 at 4:00 PM, MartinBlomberg said:
I can't really understand why they don't stick to the Excel standard, I need to google this all the time.
could not agree more!
##### Link to comment
Do you realize that the VW worksheet is actually OLDER than Excel? So why didn't Excel adopt the VW standard (which I think was actually the Visicalc standard)?
##### Link to comment
• 3 weeks later...
On 5/8/2022 at 9:39 PM, Pat Stanford said:
Do you realize that the VW worksheet is actually OLDER than Excel? So why didn't Excel adopt the VW standard (which I think was actually the Visicalc standard)?
Cosmic coincidence I was just watching this
##### Link to comment
On 5/5/2022 at 2:00 AM, MartinBlomberg said:
I can't really understand why they don't stick to the Excel standard, I need to google this all the time.
Also VWs worksheet is so undocumented, sometimes things that should work just don't. I've got a long text file of solutions from working with it since ~2006.
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• Create New... | 607 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.944031 |
https://ru.scribd.com/document/166898175/Headworks-Design | 1,619,203,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039596883.98/warc/CC-MAIN-20210423161713-20210423191713-00523.warc.gz | 546,645,936 | 111,489 | Вы находитесь на странице: 1из 6
# B-I Design of headwork :
## B-I. 1 Intake Structure: (0+000)
SN Calculation:
45 percentile discharge (Q45%)
Q design
approach velocity (Va)
Reference
=
=
=
=
2.25 m3/sec
1.6 Q45%
3.6 m3/sec
1 m/sec
=
=
Q design/Va
3.6 m2
=
=
1m
1.8 m
ITDG guidlines
Area of Orifice :
Area (A)
Assume 2 orifice,
Let, Height of orifice (h)
## B-I. 2 Weir including Spillway:
SN Calculation:
1.
Dam Height :
clearence
Dam Height
P
X-section at the site,
length of weir (L)
2.
Reference
1 Q design
2g C u A
=
=
=
=
=
=
=
0.142 m
3He
0.426
Free board+ depth of
orifice+ clearance
0.5+1+0.426
1.926
2m
27 m
C=0.6
## Energy Dissipater (Spillway):
From Gumbel analysis,
Flood discharge(Q100) =
Assuming 80% of the flow through weir
Flood over crest =
0.8 Q100 =
? Hd =
82.765 m3/sec
Hd
CLHd3/2
2.2 27 He3/2
1.075 m
Weir height
so, P/Hd
= 2m
= 2 / 1.075
= 1.86 > 1.33
Since the ratio is greater than 1.33 ,velocity Head is negligible
(B-1)
Irrigation
Engineering and
Hydraulic
structures
-S.K Garg
SN Calculation:
Reference
Eo
=
=
=
2+1.075
3.075 m
0.95
V1
## < u 2 u g(Eo - Y1)
Hence,
Or,
Or,
Q100
= < u 2 u g(Eo - Y1)
B u Y1
82.765
= 0.95 u 2 u 9.81(3.075 - Y1)
27 u Y1
? Y1 = 0.448 m
Critical depth
Yc
q2
1/ 3
?Yc = 0.98575 m
Yc should be greater than Y1, hence correct
Velocity (v1 )
Q100
B u Y1
6.842 m/sec
Froude Number,
F
V1
gY1
3.264
Y1
2
## Tail water depth in subcritical flow,
Y2
> 1 8F 1@
2
?Y2 = 1.856 m
Downstream water depth,
1 2 / 3 1/ 2
Q100
R S
=
n
B u Yd
?Yd = 0.407 m
Since Y2>Yd Hence elongated jump will occur,
As F is in between 2.5-4.5 hence Type I (USBR) stilling basin is
used
Type I (USBR) stilling basin,
Length of stilling basin (Lw) = 6.1 Y2
= 11.3216 m
Pondage depth (Hd) = Y1(9+F)/9
0.61 m
Iteration due to change in Eo
Changed Eo = 3.075 + 0.61
= 3.685 m
(B-2)
Class Note
SN Calculation:
Reference
Eo
Y1
3.075+0.61
0.402
=3.685
3.075+0.574 0.40442
=3.649
3.650
0.404
3.650
0.404
V1
7.6253
F
3.8398
Y2
1.9912
Hd
0.574
7.5797
3.8054
1.9836
0.575
7.5876
7.5876
3.8113
3.8113
1.9849
1.9849
0.575
0.575
## Pondage depth (Hd)
Length of stilling basin (Lw)
= .575 m
= 6.1 Y2
= 12.2 m
Height of the dam from its crest to bed of stilling basin,
= P + Hd
= 2 + 0.575
= 2.575 m
Chute design:
Y1
Width of Chute, W
Height of Chute,
3.
Design of Weir :
i.
Weir Profile :
a. D/S Profile
=
=
=
=
=
=
=
0.4 m
Y1
0.4 m
2Y1
0.8 m
2.5W
1m
## The downstream profile is given by the following equation
Xn = K x Hd (n-1) x Y
1.85
X
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.5178
X
2.1268
Y
0
-0.0362
-0.1304
-0.2761
-0.4702
-0.7105
-0.9955
-1.3240
-1.6950
-2.1077
-2.5613
-2.5951
(B-3)
Theory &
Design of
irrigation
Structures
(R.S.Varshney
S.C Gupta)
SN Calculation:
b. U/S Profile:
The Upstream profile is given by the following equation
(X 0.27 * H d )1.85
Y= 0.724
0.126H d 0.4135H 0d.375 (X 0.27 H d ) 0.625
0.85
Hd
Reference
X
0
0.2
0.30315
ii.
Y
0.000
-0.034
-0.075
1. Length:
LT
=
=
=
=
=
3.095-0.407
2.688 m
C HL
9 2.688
24.192 m
L2
2.21 C H L /13
Assume,
L2
=
=
=
12.06 m
12.2 m
12.2 m
L2+L3
12.2+L3
L3
=
=
=
## 18C [(Hl/13) (q/75)]
18*12[(2.688/13)*(3.0654/75)]
7.66
8m
L4
(L3/2)=4m
a. Length of D/S
## U/S Length L4,
b. Thickness
= 1.35 (q2/f)1/3
= 1.35 (3.06542)1/3
= 2.85m
d/s depth = 1.5R
u/s depth = 1.25R1.5R (assume)
= 4.275
?d/s depth= u/s depth = 4.5 m
u/s depth of pile from bed = 4.5-3.075
= 1.5m
d/s depth of pile from bed = 4.5-(0.407-0.575)
= 3.5 m
From the figure and from river topography assume,
u/s depth of pile = 3.595 m
Depth of Scour,
(B-4)
Irrigation Eng.
& Hydraulic
Strutures
(Santosh
Kumar Garg)
3.575 m
(B-5)
SN Calculation:
Reference
From figure,
## ?Total creep length
?LT
h1
3.595+0.5+2.5495
+0.8031+1.118+.5178
+6.1+0.5+5.1+2.0616
+0.5+3.575
26.92 m > 24.192 m
## b. d/s floor thickness,
For t1,
=
=
HL
u L
LT
2.688
u (3.575 .5
26.92
2.0616 5.1 .5 6.1)
1.781 m
## Thickness of D/S floor t1,
t1
=
=
=
Increase the Thickness by 10-33%
t1
for t2,
h2
=
=
h1
s 1
1.781
2.4 1
1.272 m
1.5 m
HL
u L
LT
2.688
u (3.575 .5
26.92
2.0616 5.1)
1.1212 m
## Thickness of D/S floor t2,
t2
=
=
=
Increase the Thickness by 10-33%
t2
h2
s 1
1.1212
2.4 1
0.8014 m
1m
(B-6)
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https://www.bartleby.com/questions-and-answers/jeff-heun-president-of-concrete-always-agrees-to-construct-a-concrete-cart-path-at-dakota-golf-club./50e60a40-2c94-4b4c-8658-e8fce1b67853 | 1,586,549,047,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00118.warc.gz | 776,974,420 | 28,499 | # Jeff Heun, president of Concrete Always, agrees to construct a concrete cart path at Dakota Golf Club. Concrete Always enters into a contract with Dakota to construct the path for \$200,000. In addition, as part of the contract, a performance bonus of \$40,000 will be paid based on the timing of completion. The performance bonus will be paid fully if completed by the agreed-upon date. The performance bonus decreases by \$10,000 per week for every week beyond the agreed-upon completion date. Jeff has been involved in a number of contracts that had performance bonuses as part of the agreement in the past. As a result, he is fairly confident that he will receive a good portion of the performance bonus. Jeff estimates, given the constraints of his schedule related to other jobs, that there is 55% probability that he will complete the project on time, a 30% probability that he will be 1 week late, and a 15% probability that he will be 2 weeks late.Instructionsa. Determine the transaction price that Concrete Always should compute for this agreement.b. Assume that Jeff Heun has reviewed his work schedule and decided that it makes sense to complete this project on time. Assuming that he now believes that the probability for completing the project on time is 90% and otherwise it will be finished 1 week late, determine the transaction price.
Question
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Jeff Heun, president of Concrete Always, agrees to construct a concrete cart path at Dakota Golf Club. Concrete Always enters into a contract with Dakota to construct the path for \$200,000. In addition, as part of the contract, a performance bonus of \$40,000 will be paid based on the timing of completion. The performance bonus will be paid fully if completed by the agreed-upon date. The performance bonus decreases by \$10,000 per week for every week beyond the agreed-upon completion date. Jeff has been involved in a number of contracts that had performance bonuses as part of the agreement in the past. As a result, he is fairly confident that he will receive a good portion of the performance bonus. Jeff estimates, given the constraints of his schedule related to other jobs, that there is 55% probability that he will complete the project on time, a 30% probability that he will be 1 week late, and a 15% probability that he will be 2 weeks late.
#### Instructions
a. Determine the transaction price that Concrete Always should compute for this agreement.
b. Assume that Jeff Heun has reviewed his work schedule and decided that it makes sense to complete this project on time. Assuming that he now believes that the probability for completing the project on time is 90% and otherwise it will be finished 1 week late, determine the transaction price. | 572 | 2,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.961729 |
https://numbersworksheet.com/convert-mixed-fractions-to-improper-numbers-grade-4-fractions-worksheet/ | 1,716,621,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00526.warc.gz | 359,741,462 | 14,788 | # Convert Mixed Fractions To Improper Numbers Grade 4 Fractions Worksheet
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Free math worksheets for changing decimal to small fraction
You have come to the right place if you are looking for some free math worksheets for converting decimal to a fraction. These decimal to fraction worksheets can be found in many different formats. You can download them inPDF and html. Alternatively, random format. Many of them have an answer important and can also be shaded by little ones! They are utilized for summer season studying, math concepts locations, or as part of your normal math programs.
To transform a decimal to a small percentage, you have to simplify it first. Decimals are written as equivalent fractions if the denominator is ten. Additionally, there are also worksheets regarding how to transform mixed figures into a small percentage. Free mathematics worksheets for transforming decimal to small fraction include blended examples and numbers of the two conversion functions. The process of converting a decimal to a fraction is easier than you might think, however. Adopt these measures to get going. | 875 | 4,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.931052 |
https://www.cfd-online.com/Forums/tecplot/181838-extract-surface-data-3d-plot.html | 1,709,097,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00896.warc.gz | 690,544,781 | 16,445 | # Extract Surface Data from 3D-Plot
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December 22, 2016, 08:20 Extract Surface Data from 3D-Plot #1 New Member Join Date: Dec 2016 Posts: 5 Rep Power: 9 Hey everyone, I am currently trying to extract CFD surface data from the 3D-Model of a blade tip. In other words, I want to extract the coordinates and cell data from one 3D surface area of the blade and then Project those on a 2D-plane. Unfortunately, I could not get on with this point so far. So any Ideas/Suggestions are very appreciated. Thanks!
December 22, 2016, 15:33 #2 New Member Germán Salazar Join Date: Apr 2015 Posts: 16 Rep Power: 10 Say, what kind of elements is your 3D model made out of? Also, Are you asking how to do it withing tecplot itself or any other way? Not knowing tecplot well, my two ideas include writing some code to retrieve the information.
December 23, 2016, 00:57 #3 New Member Join Date: Dec 2016 Posts: 5 Rep Power: 9 Hi, thanks for your reply. I am interested in any method to extract the data, so it doesn't need to be in tecplot. The 3D Model ist made out of FE triangles.
December 23, 2016, 07:06 #4 New Member Germán Salazar Join Date: Apr 2015 Posts: 16 Rep Power: 10 So, if I understand it correctly:you have a 3D surface made out of triangles, you have a model of a blade, and interested in retrieving the blade-tip cells Assuming the radial direction of the blade is in the same direction of the z-axis, how about you simply calculate the normal of each cell and if it is "mostly" in the z-coord, then it is "flat" enough to be declared a tip cell. By "mostly" in the z-coord, you can try different values and see which one works, this is simple; for example, for a given cell, if the z-component of the normal divided by its own length > 0.75, then we have "flat" enough cell to be declared a tip cell. In order to stay away from similarly "flat" cells at the root of the blade, you need to calculate the normal and the flatness test only for cells that are towards the top of the blade, say, above the average z-coord of entire blade; or, if you want, you can hard code some known value but this limits the usage of your script with other blades. Is this clear? does it help?
December 23, 2016, 09:25 #5 New Member Join Date: Dec 2016 Posts: 5 Rep Power: 9 Thanks for your help! I'll try it that way.
Tags cfd, extract, surface, tecplot | 647 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.93074 |
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-12-rotation-of-a-rigid-body-exercises-and-problems-page-348/5 | 1,527,026,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864968.11/warc/CC-MAIN-20180522205620-20180522225620-00351.warc.gz | 757,908,121 | 12,608 | ## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
Let the center of the earth be at the point x = 0. Let the center of the moon be at x = 384,000 km. We can use $M_e = 5.97\times 10^{24}~kg$ as the mass of the earth. We can use $M_m = 7.35\times 10^{22}~kg$ as the mass of the moon. We can find the center of mass $x_{cm}$ of the earth-plus-moon system. $x_{cm} = \frac{(M_e)(0)+(M_m)(3.84\times 10^5~km)}{M_e+M_m}$ $x_{cm} = \frac{(5.97\times 10^{24}~kg)(0)+(7.35\times 10^{22}~kg)(3.84\times 10^5~km)}{5.97\times 10^{24}~kg+7.35\times 10^{22}~kg}$ $x_{cm} = 4670~km$ The center of mass of the earth-plus-moon system is 4670 km from the center of the earth. | 273 | 705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.75 | 4 | CC-MAIN-2018-22 | longest | en | 0.779331 |
https://studylib.net/doc/14896382/homework-8 | 1,600,506,873,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00676.warc.gz | 676,287,906 | 12,475 | # Homework 8
```Homework 8
Math 332, Spring 2013
These problems must be written up in LATEX, and are due this Friday, April 5.
0 2
1. (a) Find an eight-element subgroup G of GL(2, Z3 ) containing the matrices
1 0
1 1
and
.
1 2
(b) What is the isomorphism type of G?
2. Let T1 : R3 → R3 be a 180◦ rotation about the line y = −x in the xy-plane, and let
T2 : R3 → R3 be a 120◦ rotation about the line x = y = z (either rotation will do).
(a) Find 3 × 3 matrices A1 and A2 corresponding to the linear transformations T1
and T2 .
(b) Find a six-element subgroup G of GL(3, R) containing A1 and A2 .
(c) What is the isomorphism type of G?
3. Let G be the group of symmetries of the following (flattened) tetrahedron.
H–1, 1, –18L
H1, 1, 18L
H–1, –1, 18L
H1, –1, –18L
(a) List the eight 3 × 3 matrices corresponding to the eight elements of G.
(b) What is the isomorphism type of G?
``` | 328 | 897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-40 | longest | en | 0.841204 |
http://www.cs.uni.edu/~schafer/3610/HW/hw6/index.htm | 1,555,815,948,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530161.4/warc/CC-MAIN-20190421020506-20190421042506-00276.warc.gz | 218,745,174 | 4,050 | ## Homework 6
### Due: Wednesday, November 28, 10:00 AM
Disclaimer
It is difficult to find a domain where the chromosome and fitness function needed to solve a Genetic Algorithm problem are small enough and applicable for a "simple" weeklong homework assignment. Therefore, I am providing you with a fitness function that I do not expect you to understand - in fact, I don't want you to understand it. Instead you will simply work with this fitness function to find and trust its results.
Having said that, this homework is very manageable in size and difficulty and therefore makes a good chance for you to explore the key ideas of Genetic Algorithms easily.
Instructions
For this homework you should write a program that fits the following requirements
1. Is written in python and uses the following starter file GA.py
2. Contains two functions that you will create
1. gaInputs()
2. gaParameters(......)
3. The input version should use a series of input() function calls to dynamically ask the user for, IN THIS ORDER:
1. the number of chromosomes in the population
2. the number of chromosomes that should undergo selection (mutation rate of 0) per generation
3. the number of chromosomes that should undergo mutation (mutation rate of 1) per generation
4. the number of "newblood:" chromsomes (mutation rate of 24) that should be introduced per generation
5. the number of "crossover pairs" that should be introduced per generation (1 pair uses and produces two unique chromosomes).
• Note that #1 should equal #2 + #3 + #4 + 2*#5
6. the number of generations the process should run
7. the key value to be used by the fitness function
• This is a number given to the fitness function to allow different runs to focus on different problems
4. Conducts the genetic search that was described by the user's inputs
5. Prints the fitness function value of the best member of the population after each generation.
6. Prints the value of the best chromosome found after the process is completed.
For example, one sample run is illustrated below.
Some things to note:
• Since this a random process it is not surprising if you get sligthly differetn results.
• 15 generations is likely not enough. But I used that value so it would all fit in a good screenshot.
• It is possible that the fitness function produces negative values. That's alright.
You should also produce a version called gaParameters() that allows this to be launched with out the dynamic inputs. The example below is the exact same values as used above.
NOTE: You should program gaInputs() to take advantageof gaParameters().
Details on how to generate new chromosome's from population to population
In the reading from your textbook you read about three genetic operators - methods for moving chromosomes from one generation to the next.
1. Crossover - two DIFFERENT chromosomes produce two children through a process of "mating" A random cross point is selected and the front "half" of one of the pair is merged with the back "half" of the other one of the pair. Two pairs of examples are shown here
2. Mutation - a chromosome is moved from generation N to generation N+1 with one or more of the elements of the chromosome are "mutated" (in our case, a bit is changed from 0 to 1 or 1 to 0). The mutation rate can be adjusted from low mutation (only 1 gene/bit is modified) to high mutation (up to every element of the bit string). We will start by only considering mutation rate of 1.
3. Selection - a chromosome is moved from generation N to generation N+1 with no changes at all. This is actually just a special case of mutation with a mutation rate of 0.
The first two are almost always used. The third one is used in certain situations. I am also going to ask you to consider one additional operator:
1. New Blood - This is really just an adaptation of Mutation with a mutation rate of length N.
Selecting a Chromosome to modify
In order to mimic nature and the concept of "survival" of the fittest" we need to come up with a technique which allows every chromosome at least a limited chance of being picked to undergo a genetic change through one of the methods explained above. But those chromosomes that are "fittest" (in our case, close to the target color) should have a much higher chance than those that aren't (those that are farther away).
For this particular assignment I would like you to use a technique known as "tournament selection" The concept is easy, although the implementation may be a bit harder.
To put this in an easy to understand context, let's consider five people who have just bowled a game of bowling. First, rank your bowlers (or, in the context of your GA, your chromosomes) based on their fitness. In this context the fitness is their score in the last round of the game.
Rank Bowler Score 1 Ringo 224 2 Paul 202 3 George 185 4 Yoko 171 5 John 155
Now, we want to select one of these bowlers, but we want to give preference for Ringo over John. The way we do this is:
1. randomly pick a number from 1-N (in this case, 5)
2. do it again, paying no attention to what number you picked the first time
3. Select whichever bowler has the higher rank. If you actually picked the same bowler both times than you have your selection.
If you think about this, you have 25 possible pairs of bowlers (5x5) and among this 25 pairings each bowler is selected at least once, although Ringo is going to be selected 9 of the 25 times while John will only be selected once.
First Bowler Picked Ringo Paul George Yoko John Second Bowler Picked Ringo R R R R R Paul R P P P P George R P G G G Yoko R P G Y Y John R P G Y John
You should use this technique for picking which chromosomes are used for selection and new blood. You should perform this techinique twice to see which two chromosomes are selected for crossover. | 1,308 | 5,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-18 | latest | en | 0.91578 |
https://buthowto.com/what-does-the-median-tell-us-about-a-distribution | 1,660,566,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572174.8/warc/CC-MAIN-20220815115129-20220815145129-00573.warc.gz | 180,434,758 | 21,328 | # What does the median tell us about a distribution
In statistics and probability theory, the median is the value separating the higher half from the lower half of a data sample, a population, or a probability distribution. For a data set, it may be thought of as "the middle" value. The basic feature of the median in describing data compared to the mean (often simply described as the "average") is that it is not skewed by a small proportion of extremely large or small values, and therefore provides a better representation of a "typical" value. Median income, for example, may be a better way to suggest what a "typical" income is, because income distribution can be very skewed. The median is of central importance in robust statistics, as it is the most resistant statistic, having a breakdown point of 50%: so long as no more than half the data are contaminated, the median is not an arbitrarily large or small result.
Finding the median in sets of data with an odd and even number of values
## Finite data set of numbersEdit
The median of a finite list of numbers is the "middle" number, when those numbers are listed in order from smallest to greatest.
If the data set has an odd number of observations, the middle one is selected. For example, the following list of seven numbers,
1, 3, 3, 6, 7, 8, 9
has the median of 6, which is the fourth value.
If the data set has an even number of observations, there is no distinct middle value and the median is usually defined to be the arithmetic mean of the two middle values.[1][2] For example, this data set of 8 numbers
1, 2, 3, 4, 5, 6, 8, 9
has a median value of 4.5, that is (4+5)/2{\displaystyle (4+5)/2}. (In more technical terms, this interprets the median as the fully trimmed mid-range).
In general, with this convention, the median can be defined as follows: For a data set x{\displaystyle x} of n{\displaystyle n} elements, ordered from smallest to greatest,
if n{\displaystyle n} is odd, median(x)=x(n+1)/2{\displaystyle \mathrm {median} (x)=x_{(n+1)/2}}if n{\displaystyle n} is even, median(x)=x(n/2)+x(n/2)+12{\displaystyle \mathrm {median} (x)={\frac {x_{(n/2)}+x_{(n/2)+1}}{2}}}Comparison of common averages of values [ 1, 2, 2, 3, 4, 7, 9 ]TypeDescriptionExampleResultArithmetic meanSum of values of a data set divided by number of values: x¯=1ni=1nxi{\displaystyle \scriptstyle {\bar {x}}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}}(1+2+2+3+4+7+9)/74MedianMiddle value separating the greater and lesser halves of a data set1, 2, 2, 3, 4, 7, 93ModeMost frequent value in a data set1, 2, 2, 3, 4, 7, 92
### Formal definitionEdit
Formally, a median of a population is any value such that at most half of the population is less than the proposed median and at most half is greater than the proposed median. As seen above, medians may not be unique. If each set contains less than half the population, then some of the population is exactly equal to the unique median.
The median is well-defined for any ordered (one-dimensional) data, and is independent of any distance metric. The median can thus be applied to classes which are ranked but not numerical (e.g. working out a median grade when students are graded from A to F), although the result might be halfway between classes if there is an even number of cases.
A geometric median, on the other hand, is defined in any number of dimensions. A related concept, in which the outcome is forced to correspond to a member of the sample, is the medoid.
There is no widely accepted standard notation for the median, but some authors represent the median of a variable x either as x͂ or as μ1/2[1] sometimes also M.[3][4] In any of these cases, the use of these or other symbols for the median needs to be explicitly defined when they are introduced.
The median is a special case of other ways of summarizing the typical values associated with a statistical distribution: it is the 2nd quartile, 5th decile, and 50th percentile.
The median can be used as a measure of location when one attaches reduced importance to extreme values, typically because a distribution is skewed, extreme values are not known, or outliers are untrustworthy, i.e., may be measurement/transcription errors.
For example, consider the multiset
1, 2, 2, 2, 3, 14.
The median is 2 in this case, (as is the mode), and it might be seen as a better indication of the center than the arithmetic mean of 4, which is larger than all-but-one of the values. However, the widely cited empirical relationship that the mean is shifted "further into the tail" of a distribution than the median is not generally true. At most, one can say that the two statistics cannot be "too far" apart; see §Inequality relating means and medians below.[5]
As a median is based on the middle data in a set, it is not necessary to know the value of extreme results in order to calculate it. For example, in a psychology test investigating the time needed to solve a problem, if a small number of people failed to solve the problem at all in the given time a median can still be calculated.[6]
Because the median is simple to understand and easy to calculate, while also a robust approximation to the mean, the median is a popular summary statistic in descriptive statistics. In this context, there are several choices for a measure of variability: the range, the interquartile range, the mean absolute deviation, and the median absolute deviation.
For practical purposes, different measures of location and dispersion are often compared on the basis of how well the corresponding population values can be estimated from a sample of data. The median, estimated using the sample median, has good properties in this regard. While it is not usually optimal if a given population distribution is assumed, its properties are always reasonably good. For example, a comparison of the efficiency of candidate estimators shows that the sample mean is more statistically efficient when and only when data is uncontaminated by data from heavy-tailed distributions or from mixtures of distributions.[citation needed] Even then, the median has a 64% efficiency compared to the minimum-variance mean (for large normal samples), which is to say the variance of the median will be ~50% greater than the variance of the mean.[7][8]
## Probability distributionsEdit
Geometric visualization of the mode, median and mean of an arbitrary probability density function[9]
For any real-valued probability distribution with cumulative distribution functionF, a median is defined as any real numberm that satisfies the inequalities
(,m]dF(x)12and[m,)dF(x)12{\displaystyle \int _{(-\infty ,m]}dF(x)\geq {\frac {1}{2}}{\text{ and }}\int _{[m,\infty )}dF(x)\geq {\frac {1}{2}}}.
An equivalent phrasing uses a random variable X distributed according to F:
P(Xm)12andP(Xm)12{\displaystyle \operatorname {P} (X\leq m)\geq {\frac {1}{2}}{\text{ and }}\operatorname {P} (X\geq m)\geq {\frac {1}{2}}}
Note that this definition does not require X to have an absolutely continuous distribution (which has a probability density function ƒ), nor does it require a discrete one. In the former case, the inequalities can be upgraded to equality: a median satisfies
P(Xm)=mf(x)dx=12=mf(x)dx=P(Xm){\displaystyle \operatorname {P} (X\leq m)=\int _{-\infty }^{m}{f(x)\,dx}={\frac {1}{2}}=\int _{m}^{\infty }{f(x)\,dx}=\operatorname {P} (X\geq m)}.
Any probability distribution on R has at least one median, but in pathological cases there may be more than one median: if F is constant 1/2 on an interval (so that ƒ=0 there), then any value of that interval is a median.
The medians of certain types of distributions can be easily calculated from their parameters; furthermore, they exist even for some distributions lacking a well-defined mean, such as the Cauchy distribution:
## PopulationsEdit
### Optimality propertyEdit
The mean absolute error of a real variable c with respect to the random variableX is
E(|Xc|){\displaystyle E(\left|X-c\right|)\,}
Provided that the probability distribution of X is such that the above expectation exists, then m is a median of X if and only if m is a minimizer of the mean absolute error with respect to X.[11] In particular, m is a sample median if and only if m minimizes the arithmetic mean of the absolute deviations.[12]
More generally, a median is defined as a minimum of
E(|Xc||X|),{\displaystyle E(|X-c|-|X|),}
as discussed below in the section on multivariate medians (specifically, the spatial median).
This optimization-based definition of the median is useful in statistical data-analysis, for example, in k-medians clustering.
If the distribution has finite variance, then the distance between the median X~{\displaystyle {\tilde {X}}} and the mean X¯{\displaystyle {\bar {X}}} is bounded by one standard deviation.
This bound was proved by Mallows,[13] who used Jensen's inequality twice, as follows. Using |·| for the absolute value, we have
|μm|=|E(Xm)|E(|Xm|)E(|Xμ|)E((Xμ)2)=σ.{\displaystyle {\begin{aligned}|\mu -m|=|\operatorname {E} (X-m)|&\leq \operatorname {E} (|X-m|)\\&\leq \operatorname {E} (|X-\mu |)\\&\leq {\sqrt {\operatorname {E} \left((X-\mu )^{2}\right)}}=\sigma .\end{aligned}}}
The first and third inequalities come from Jensen's inequality applied to the absolute-value function and the square function, which are each convex. The second inequality comes from the fact that a median minimizes the absolute deviation function aE(|Xa|){\displaystyle a\mapsto \operatorname {E} (|X-a|)}.
Mallows' proof can be generalized to obtain a multivariate version of the inequality[14] simply by replacing the absolute value with a norm:
μmE(Xμ2)=trace(var(X)){\displaystyle \|\mu -m\|\leq {\sqrt {\operatorname {E} \left(\|X-\mu \|^{2}\right)}}={\sqrt {\operatorname {trace} \left(\operatorname {var} (X)\right)}}}
where m is a spatial median, that is, a minimizer of the function aE(Xa).{\displaystyle a\mapsto \operatorname {E} (\|X-a\|).\,} The spatial median is unique when the data-set's dimension is two or more.[15][16]
An alternative proof uses the one-sided Chebyshev inequality; it appears in an inequality on location and scale parameters. This formula also follows directly from Cantelli's inequality.[17]
#### Unimodal distributionsEdit
For the case of unimodal distributions, one can achieve a sharper bound on the distance between the median and the mean:
|X~X¯|(35)12σ0.7746σ{\displaystyle \left|{\tilde {X}}-{\bar {X}}\right|\leq \left({\frac {3}{5}}\right)^{\frac {1}{2}}\sigma \approx 0.7746\sigma }.[18]
A similar relation holds between the median and the mode:
|X~mode|312σ1.732σ.{\displaystyle \left|{\tilde {X}}-\mathrm {mode} \right|\leq 3^{\frac {1}{2}}\sigma \approx 1.732\sigma .}
Jensen's inequality states that for any random variable X with a finite expectation E[X] and for any convex function f
f[E(x)]E[f(x)]{\displaystyle f[E(x)]\leq E[f(x)]}
This inequality generalizes to the median as well. We say a function f: is a C function if, for any t,
f1((,t])={xRf(x)t}{\displaystyle f^{-1}\left(\,(-\infty ,t]\,\right)=\{x\in \mathbb {R} \mid f(x)\leq t\}}
is a closed interval (allowing the degenerate cases of a single point or an empty set). Every convex function is a C function, but the reverse does not hold. If f is a C function, then
f(Median[X])Median[f(X)]{\displaystyle f(\operatorname {Median} [X])\leq \operatorname {Median} [f(X)]}
If the medians are not unique, the statement holds for the corresponding suprema.[19]
#### Efficient computation of the sample medianEdit
Even though comparison-sorting n items requires Ω(n log n) operations, selection algorithms can compute the kth-smallest of n items with only Θ(n) operations. This includes the median, which is the n/2th order statistic (or for an even number of samples, the arithmetic mean of the two middle order statistics).[20]
Selection algorithms still have the downside of requiring Ω(n) memory, that is, they need to have the full sample (or a linear-sized portion of it) in memory. Because this, as well as the linear time requirement, can be prohibitive, several estimation procedures for the median have been developed. A simple one is the median of three rule, which estimates the median as the median of a three-element subsample; this is commonly used as a subroutine in the quicksort sorting algorithm, which uses an estimate of its input's median. A more robust estimator is Tukey's ninther, which is the median of three rule applied with limited recursion:[21] if A is the sample laid out as an array, and
med3(A) = median(A[1], A[n/2], A[n]),
then
ninther(A) = med3(med3(A[1 ... 1/3n]), med3(A[1/3n ... 2/3n]), med3(A[2/3n ... n]))
The remedian is an estimator for the median that requires linear time but sub-linear memory, operating in a single pass over the sample.[22]
#### Sampling distributionEdit
The distributions of both the sample mean and the sample median were determined by Laplace.[23] The distribution of the sample median from a population with a density function f(x){\displaystyle f(x)} is asymptotically normal with mean m{\displaystyle m} and variance[24]
14nf(m)2{\displaystyle {\frac {1}{4nf(m)^{2}}}}
where m{\displaystyle m} is the median of f(x){\displaystyle f(x)} and n{\displaystyle n} is the sample size. A modern proof follows below. Laplace's result is now understood as a special case of the asymptotic distribution of arbitrary quantiles.
For normal samples, the density is f(m)=1/2πσ2{\displaystyle f(m)=1/{\sqrt {2\pi \sigma ^{2}}}}, thus for large samples the variance of the median equals (π/2)(σ2/n).{\displaystyle ({\pi }/{2})\cdot (\sigma ^{2}/n).}[7] (See also section #Efficiency below.)
##### Derivation of the asymptotic distributionEdit
We take the sample size to be an odd number N=2n+1{\displaystyle N=2n+1} and assume our variable continuous; the formula for the case of discrete variables is given below in §Empirical local density. The sample can be summarized as "below median", "at median", and "above median", which corresponds to a trinomial distribution with probabilities F(v1){\displaystyle F(v-1)}, f(v){\displaystyle f(v)} and 1F(v){\displaystyle 1-F(v)}. For a continuous variable, the probability of multiple sample values being exactly equal to the median is 0, so one can calculate the density of at the point v{\displaystyle v} directly from the trinomial distribution:
Pr[Median=v]dv=(2n+1)!n!n!F(v)n(1F(v))nf(v)dv{\displaystyle \Pr[\operatorname {Median} =v]\,dv={\frac {(2n+1)!}{n!n!}}F(v)^{n}(1-F(v))^{n}f(v)\,dv}.
Now we introduce the beta function. For integer arguments α{\displaystyle \alpha } and β{\displaystyle \beta }, this can be expressed as B(α,β)=(α1)!(β1)!(α+β1)!{\displaystyle \mathrm {B} (\alpha ,\beta )={\frac {(\alpha -1)!(\beta -1)!}{(\alpha +\beta -1)!}}}. Also, recall that f(v)dv=dF(v){\displaystyle f(v)\,dv=dF(v)}. Using these relationships and setting both α{\displaystyle \alpha } and β{\displaystyle \beta } equal to n+1{\displaystyle n+1} allows the last expression to be written as
F(v)n(1F(v))nB(n+1,n+1)dF(v){\displaystyle {\frac {F(v)^{n}(1-F(v))^{n}}{\mathrm {B} (n+1,n+1)}}\,dF(v)}
Hence the density function of the median is a symmetric beta distribution pushed forward by F{\displaystyle F}. Its mean, as we would expect, is 0.5 and its variance is 1/(4(N+2)){\displaystyle 1/(4(N+2))}. By the chain rule, the corresponding variance of the sample median is
14(N+2)f(m)2{\displaystyle {\frac {1}{4(N+2)f(m)^{2}}}}.
The additional 2 is negligible in the limit.
##### Empirical local densityEdit
In practice, the functions f{\displaystyle f} and F{\displaystyle F} are often not known or assumed. However, they can be estimated from an observed frequency distribution. In this section, we give an example. Consider the following table, representing a sample of 3,800 (discrete-valued) observations:
v00.511.522.533.544.55f(v)0.0000.0080.0100.0130.0830.1080.3280.2200.2020.0230.005F(v)0.0000.0080.0180.0310.1140.2220.5500.7700.9720.9951.000
Because the observations are discrete-valued, constructing the exact distribution of the median is not an immediate translation of the above expression for Pr(Median=v){\displaystyle \Pr(\operatorname {Median} =v)}; one may (and typically does) have multiple instances of the median in one's sample. So we must sum over all these possibilities:
Pr(Median=v)=i=0nk=0nN!i!(Nik)!k!F(v1)i(1F(v))kf(v)Nik{\displaystyle \Pr(\operatorname {Median} =v)=\sum _{i=0}^{n}\sum _{k=0}^{n}{\frac {N!}{i!(N-i-k)!k!}}F(v-1)^{i}(1-F(v))^{k}f(v)^{N-i-k}}
Here, i is the number of points strictly less than the median and k the number strictly greater.
Using these preliminaries, it is possible to investigate the effect of sample size on the standard errors of the mean and median. The observed mean is 3.16, the observed raw median is 3 and the observed interpolated median is 3.174. The following table gives some comparison statistics.
391521Expected value of median3.1983.1913.1743.161Standard error of median (above formula)0.4820.3050.2570.239Standard error of median (asymptotic approximation)0.8790.5080.3930.332Standard error of mean0.4210.2430.1880.159
The expected value of the median falls slightly as sample size increases while, as would be expected, the standard errors of both the median and the mean are proportionate to the inverse square root of the sample size. The asymptotic approximation errs on the side of caution by overestimating the standard error.
#### Estimation of variance from sample dataEdit
The value of (2f(x))2{\displaystyle (2f(x))^{-2}}the asymptotic value of n12(νm){\displaystyle n^{-{\frac {1}{2}}}(\nu -m)} where ν{\displaystyle \nu } is the population medianhas been studied by several authors. The standard "delete one" jackknife method produces inconsistent results.[25] An alternativethe "delete k" methodwhere k{\displaystyle k} grows with the sample size has been shown to be asymptotically consistent.[26] This method may be computationally expensive for large data sets. A bootstrap estimate is known to be consistent,[27] but converges very slowly (order of n14{\displaystyle n^{-{\frac {1}{4}}}}).[28] Other methods have been proposed but their behavior may differ between large and small samples.[29]
#### EfficiencyEdit
The efficiency of the sample median, measured as the ratio of the variance of the mean to the variance of the median, depends on the sample size and on the underlying population distribution. For a sample of size N=2n+1{\displaystyle N=2n+1} from the normal distribution, the efficiency for large N is
2πN+2N{\displaystyle {\frac {2}{\pi }}{\frac {N+2}{N}}}
The efficiency tends to 2π{\displaystyle {\frac {2}{\pi }}} as N{\displaystyle N} tends to infinity.
In other words, the relative variance of the median will be π/21.57{\displaystyle \pi /2\approx 1.57}, or 57% greater than the variance of the mean the relative standard error of the median will be (π/2)121.25{\displaystyle (\pi /2)^{\frac {1}{2}}\approx 1.25}, or 25% greater than the standard error of the mean, σ/n{\displaystyle \sigma /{\sqrt {n}}} (see also section #Sampling distribution above.).[30]
### Other estimatorsEdit
For univariate distributions that are symmetric about one median, the HodgesLehmann estimator is a robust and highly efficient estimator of the population median.[31]
If data are represented by a statistical model specifying a particular family of probability distributions, then estimates of the median can be obtained by fitting that family of probability distributions to the data and calculating the theoretical median of the fitted distribution.[citation needed] Pareto interpolation is an application of this when the population is assumed to have a Pareto distribution.
Previously, this article discussed the univariate median, when the sample or population had one-dimension. When the dimension is two or higher, there are multiple concepts that extend the definition of the univariate median; each such multivariate median agrees with the univariate median when the dimension is exactly one.[31][32][33][34]
The marginal median is defined for vectors defined with respect to a fixed set of coordinates. A marginal median is defined to be the vector whose components are univariate medians. The marginal median is easy to compute, and its properties were studied by Puri and Sen.[31][35]
The geometric median of a discrete set of sample points x1,xN{\displaystyle x_{1},\ldots x_{N}} in a Euclidean space is the[a] point minimizing the sum of distances to the sample points.
μ^=argminμRmn=1Nμxn2{\displaystyle {\hat {\mu }}={\underset {\mu \in \mathbb {R} ^{m}}{\operatorname {arg\,min} }}\sum _{n=1}^{N}\left\|\mu -x_{n}\right\|_{2}}
In contrast to the marginal median, the geometric median is equivariant with respect to Euclidean similarity transformations such as translations and rotations.
If the marginal medians for all coordinate systems coincide, then their common location may be termed the "median in all directions".[37] This concept is relevant to voting theory on account of the median voter theorem. When it exists, the median in all directions coincides with the geometric median (at least for discrete distributions).
### CenterpointEdit
An alternative generalization of the median in higher dimensions is the centerpoint.
When dealing with a discrete variable, it is sometimes useful to regard the observed values as being midpoints of underlying continuous intervals. An example of this is a Likert scale, on which opinions or preferences are expressed on a scale with a set number of possible responses. If the scale consists of the positive integers, an observation of 3 might be regarded as representing the interval from 2.50 to 3.50. It is possible to estimate the median of the underlying variable. If, say, 22% of the observations are of value 2 or below and 55.0% are of 3 or below (so 33% have the value 3), then the median m{\displaystyle m} is 3 since the median is the smallest value of x{\displaystyle x} for which F(x){\displaystyle F(x)} is greater than a half. But the interpolated median is somewhere between 2.50 and 3.50. First we add half of the interval width w{\displaystyle w} to the median to get the upper bound of the median interval. Then we subtract that proportion of the interval width which equals the proportion of the 33% which lies above the 50% mark. In other words, we split up the interval width pro rata to the numbers of observations. In this case, the 33% is split into 28% below the median and 5% above it so we subtract 5/33 of the interval width from the upper bound of 3.50 to give an interpolated median of 3.35. More formally, if the values f(x){\displaystyle f(x)} are known, the interpolated median can be calculated from
mint=m+w[12F(m)12f(m)].{\displaystyle m_{\text{int}}=m+w\left[{\frac {1}{2}}-{\frac {F(m)-{\frac {1}{2}}}{f(m)}}\right].}
Alternatively, if in an observed sample there are k{\displaystyle k} scores above the median category, j{\displaystyle j} scores in it and i{\displaystyle i} scores below it then the interpolated median is given by
mint=mw2[kij].{\displaystyle m_{\text{int}}=m-{\frac {w}{2}}\left[{\frac {k-i}{j}}\right].}
For univariate distributions that are symmetric about one median, the HodgesLehmann estimator is a robust and highly efficient estimator of the population median; for non-symmetric distributions, the HodgesLehmann estimator is a robust and highly efficient estimator of the population pseudo-median, which is the median of a symmetrized distribution and which is close to the population median.[38] The HodgesLehmann estimator has been generalized to multivariate distributions.[39]
### Variants of regressionEdit
The TheilSen estimator is a method for robust linear regression based on finding medians of slopes.[40]
The median filter is an important tool of image processing, that can effectively remove any salt and pepper noise from grayscale images.
### Cluster analysisEdit
In cluster analysis, the k-medians clustering algorithm provides a way of defining clusters, in which the criterion of maximising the distance between cluster-means that is used in k-means clustering, is replaced by maximising the distance between cluster-medians.
This is a method of robust regression. The idea dates back to Wald in 1940 who suggested dividing a set of bivariate data into two halves depending on the value of the independent parameter x{\displaystyle x}: a left half with values less than the median and a right half with values greater than the median.[41] He suggested taking the means of the dependent y{\displaystyle y} and independent x{\displaystyle x} variables of the left and the right halves and estimating the slope of the line joining these two points. The line could then be adjusted to fit the majority of the points in the data set.
Nair and Shrivastava in 1942 suggested a similar idea but instead advocated dividing the sample into three equal parts before calculating the means of the subsamples.[42] Brown and Mood in 1951 proposed the idea of using the medians of two subsamples rather the means.[43] Tukey combined these ideas and recommended dividing the sample into three equal size subsamples and estimating the line based on the medians of the subsamples.[44]
Any mean-unbiased estimator minimizes the risk (expected loss) with respect to the squared-error loss function, as observed by Gauss. A median-unbiased estimator minimizes the risk with respect to the absolute-deviation loss function, as observed by Laplace. Other loss functions are used in statistical theory, particularly in robust statistics.
The theory of median-unbiased estimators was revived by George W. Brown in 1947:[45]
An estimate of a one-dimensional parameter θ will be said to be median-unbiased if, for fixed θ, the median of the distribution of the estimate is at the value θ; i.e., the estimate underestimates just as often as it overestimates. This requirement seems for most purposes to accomplish as much as the mean-unbiased requirement and has the additional property that it is invariant under one-to-one transformation.
Further properties of median-unbiased estimators have been reported.[46][47][48][49] Median-unbiased estimators are invariant under one-to-one transformations.
There are methods of constructing median-unbiased estimators that are optimal (in a sense analogous to the minimum-variance property for mean-unbiased estimators). Such constructions exist for probability distributions having monotone likelihood-functions.[50][51] One such procedure is an analogue of the RaoBlackwell procedure for mean-unbiased estimators: The procedure holds for a smaller class of probability distributions than does the RaoBlackwell procedure but for a larger class of loss functions.[52]
Scientific researchers in the ancient near east appear not to have used summary statistics altogether, instead choosing values that offered maximal consistency with a broader theory that integrated a wide variety of phenomena.[53] Within the Mediterranean (and, later, European) scholarly community, statistics like the mean are fundamentally a medieval and early modern development. (The history of the median outside Europe and its predecessors remains relatively unstudied.)
The idea of the median appeared in the 13th century in the Talmud, in order to fairly analyze divergent appraisals.[54][55] However, the concept did not spread to the broader scientific community.
Instead, the closest ancestor of the modern median is the mid-range, invented by Al-Biruni.[56]: 31 [57] Transmission of Al-Biruni's work to later scholars is unclear. Al-Biruni applied his technique to assaying metals, but, after he published his work, most assayers still adopted the most unfavorable value from their results, lest they appear to cheat.[56]: 358 However, increased navigation at sea during the Age of Discovery meant that ship's navigators increasingly had to attempt to determine latitude in unfavorable weather against hostile shores, leading to renewed interest in summary statistics. Whether rediscovered or independently invented, the mid-range is recommended to nautical navigators in Harriot's "Instructions for Raleigh's Voyage to Guiana, 1595".[56]: 458
The idea of the median may have first appeared in Edward Wright's 1599 book Certaine Errors in Navigation on a section about compass navigation. Wright was reluctant to discard measured values, and may have felt that the median incorporating a greater proportion of the dataset than the mid-range was more likely to be correct. However, Wright did not give examples of his technique's use, making it hard to verify that he described the modern notion of median.[53][57][b] The median (in the context of probability) certainly appeared in the correspondence of Christiaan Huygens, but as an example of a statistic that was inappropriate for actuarial practice.[53]
The earliest recommendation of the median dates to 1757, when Roger Joseph Boscovich developed a regression method based on the L1 norm and therefore implicitly on the median.[53][58] In 1774, Laplace made this desire explicit: he suggested the median be used as the standard estimator of the value of a posterior PDF. The specific criterion was to minimize the expected magnitude of the error; |αα|{\displaystyle |\alpha -\alpha ^{*}|} where α{\displaystyle \alpha ^{*}} is the estimate and α{\displaystyle \alpha } is the true value. To this end, Laplace determined the distributions of both the sample mean and the sample median in the early 1800s.[23][59] However, a decade later, Gauss and Legendre developed the least squares method, which minimizes (αα)2{\displaystyle (\alpha -\alpha ^{*})^{2}} to obtain the mean. Within the context of regression, Gauss and Legendre's innovation offers vastly easier computation. Consequently, Laplaces' proposal was generally rejected until the rise of computing devices 150 years later (and is still a relatively uncommon algorithm).[60]
Antoine Augustin Cournot in 1843 was the first[61] to use the term median (valeur médiane) for the value that divides a probability distribution into two equal halves. Gustav Theodor Fechner used the median (Centralwerth) in sociological and psychological phenomena.[62] It had earlier been used only in astronomy and related fields. Gustav Fechner popularized the median into the formal analysis of data, although it had been used previously by Laplace,[62] and the median appeared in a textbook by F. Y. Edgeworth.[63] Francis Galton used the English term median in 1881,[64][65] having earlier used the terms middle-most value in 1869, and the medium in 1880.[66][67]
Statisticians encouraged the use of medians intensely throughout the 19th century for its intuitive clarity and ease of manual computation. However, the notion of median does not lend itself to the theory of higher moments as well as the arithmetic mean does, and is much harder to compute by computer. As a result, the median was steadily supplanted as a notion of generic average by the arithmetic mean during the 20th century.[53][57]
1. ^ The geometric median is unique unless the sample is collinear.[36]
2. ^ Subsequent scholars appear to concur with Eisenhart that Boroughs' 1580 figures, while suggestive of the median, in fact describe an arithmetic mean.;[56]: 623 Boroughs is mentioned in no other work. | 7,793 | 31,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-33 | latest | en | 0.91245 |
http://www.docstoc.com/docs/10061989/Regular-languages-(-Chapter-9-in-your-book) | 1,397,820,212,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00590-ip-10-147-4-33.ec2.internal.warc.gz | 387,586,662 | 14,770 | # Regular languages ( Chapter 9 in your book)
Document Sample
``` Regular languages and NFA's( Chapter 9 in your book)
Def: A grammar is context free ( CFG) if every rule has a single non-terminal on the
left side, that is, is of the form A w, where A is a single non-terminal, and w is any
string made up of terminals and non-terminals, including , possibly, .
One especially simple type of CFG is the regular grammar, defined next.
Def: A grammar is (left) regular if every rule is of the form AxB; A x . or A
Def. A language is regular if it can be generated by some regular grammar. (see page 300 in
Note: To prove a language is regular , one need merely find a regular grammar that generates it.
To prove a language is not regular, one has to prove that no regular grammar can ever generate it.
The mere fact that some non-regular grammar generates a language proves nothing about its
regularity since the same language can be generated by more than one grammar.
Def : A NFA, M, ( non-deterministic finite automaton) is a 6-tuple: (Q,S,F, ,S ) where Q
is a finite set of states, S Q is the start state, F Q is the set of accepting states, is a finite
set of symbols, and Q Q is the set of transitions.
A string of is accepted by an NFA if there is a path for the string that ends in an
accepting state. The language of M, L(M) is the set of all accepted strings.
Sample exercises. Let the symbol set, = {a,b,c}. For each of the following languages
over *,
find an NFA that accepts it and a regular grammar that generates it.
The language is all strings in * which
1. End with abc
2. Begin and end in a.
3. Do not contain the subsstring bc.
4. contain the substring ab one or more times.
5. contain the substring ab exactly one time.
6. contain exactly 3 a's.
6.5 contain ab at least two times.
7. in which every a is immediately followed by b.
Note: the condition does not require that any a's occur, but if a does appear, it
must be followed by b .
8. in which every a is eventually followed by b.
9. The number of a's is a multiple of 3.
10. in which , if b appears, c never appear later in the string.
You should be able to do most, if not all, of 17-28, page 307, using this NFA method.
However, not all context free languages are regular, as we shall prove shortly, so do not expect to
always be able to find a regular grammar, if you are asked for a CFG.
Theorem: A language is regular if and only if it is accepted by some NFA.
(Further, there is an algorithm for finding an NFA from a regular grammar and
vice versa.)
Proof: To be presented in class.
Proposition: L= {anbn | n > 0 } is not regular, ( but is context free.)
proof: The proof can be used to used to prove a pumping lemma for regular languages,
but we will make it specific to the language L. Suppose L can be generated by some regular
language, G. Let N be greater than the number of non-terminals of G. Consider the derivation
tree for aNbN .
In the derivation of just the a N portion of the string , some non terminal, say A, must
appear twice. Thus the derivation can be "pumped" to get a derivation of a N + p b N where p > 0, so
this string is not in L. The grammar does too much -- it generates strings that are not in L.
Note: Proofs can also be made by showing that no DFA accepts L .
```
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https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-1-exponents-exercise-set-page-343/27 | 1,532,299,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594018.55/warc/CC-MAIN-20180722213610-20180722233610-00038.warc.gz | 893,896,197 | 14,104 | ## Algebra: A Combined Approach (4th Edition)
$x^{19}y^{6}$
Based on the product rule for exponents, we know that $a^{m}\times a^{n}=a^{m+n}$ (where $m$ and $n$ are positive integers and $a$ is a real number). Therefore, $(x^{9}y)\times(x^{10}y^{5})=x^{9+10}\times y^{1+5}=x^{19}\times y^{6}=x^{19}y^{6}$. | 119 | 306 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-30 | latest | en | 0.73053 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A057944 Largest triangular number less than or equal to n; write m-th triangular number m+1 times. 16
0, 1, 1, 3, 3, 3, 6, 6, 6, 6, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 15, 21, 21, 21, 21, 21, 21, 21, 28, 28, 28, 28, 28, 28, 28, 28, 36, 36, 36, 36, 36, 36, 36, 36, 36, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66, 66 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,4 LINKS Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened FORMULA a(n) = floor((sqrt(1+8*n)-1)/2)*floor((sqrt(1+8*n)+1)/2)/2 = (trinv(n)*(trinv(n)-1))/2 = A000217(A003056(n)) = n - A002262(n) a(n) = (1/2)*t*(t-1), where t = floor(sqrt(2*n+1)+1/2) = A002024(n+1). - Ridouane Oudra, Oct 20 2019 EXAMPLE a(35) = 28 since 28 and 36 are successive triangular numbers and 28 <= 35 < 36. MAPLE A057944 := proc(n) k := (-1+sqrt(1+8*n))/2 ; k := floor(k) ; k*(k+1)/2 ; end proc; # R. J. Mathar, Nov 05 2011 MATHEMATICA f[n_] := Block[{a = Floor@ Sqrt[1 + 8 n]}, Floor[(a - 1)/2]*Floor[(a + 1)/2]/2]; Array[f, 72, 0] t0=0; t1=1; k=1; Table[If[n < t1, t0, k++; t0=t1; t1=t1+k; t0], {n, 0, 72}] With[{nn=15}, Table[#[[1]], #[[2]]+1]&/@Thread[{Accumulate[Range[ 0, nn]], Range[ 0, nn]}]]//Flatten (* Harvey P. Dale, Mar 01 2020 *) PROG (Haskell) a057944 n = a057944_list !! n -- common flat access a057944_list = concat a057944_tabl a057944' n k = a057944_tabl !! n !! k -- access when seen as a triangle a057944_row n = a057944_tabl !! n a057944_tabl = zipWith (\$) (map replicate [1..]) a000217_list -- Reinhard Zumkeller, Feb 03 2012 (PARI) a(n)=my(t=(sqrtint(8*n+7)-1)\2); t*(t+1)/2 \\ Charles R Greathouse IV, Jan 26 2013 CROSSREFS Cf. A000217, A003056, A056944, A057945, A127739. Sequence in context: A108581 A073080 A171601 * A281258 A080607 A013322 Adjacent sequences: A057941 A057942 A057943 * A057945 A057946 A057947 KEYWORD easy,nonn,tabl AUTHOR Henry Bottomley, Oct 05 2000 EXTENSIONS Keyword tabl added by Reinhard Zumkeller, Feb 03 2012 STATUS approved
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Last modified July 7 02:44 EDT 2022. Contains 355141 sequences. (Running on oeis4.) | 1,029 | 2,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-27 | latest | en | 0.537333 |
https://joningram.org/questions/Algebra/200102 | 1,695,938,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00302.warc.gz | 368,357,295 | 4,450 | # Factor x^2+7x+10
Factor x^2+7x+10
Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is .
Write the factored form using these integers.
Do you know how to Factor x^2+7x+10? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name one billion four hundred seventy million two hundred sixty-five thousand one hundred sixty-three
### Interesting facts
• 1470265163 has 8 divisors, whose sum is 1548811920
• The reverse of 1470265163 is 3615620741
• Previous prime number is 1361
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 35
• Digital Root 8
### Name
Name eight hundred forty-two million three hundred sixty-two thousand one hundred ninety-nine
### Interesting facts
• 842362199 has 8 divisors, whose sum is 963543840
• The reverse of 842362199 is 991263248
• Previous prime number is 1153
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 9
• Sum of Digits 44
• Digital Root 8
### Name
Name two billion one hundred forty-two million six hundred eighty thousand fifty-five
### Interesting facts
• 2142680055 has 8 divisors, whose sum is 3428288112
• The reverse of 2142680055 is 5500862412
• Previous prime number is 5
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 33
• Digital Root 6 | 409 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.765465 |
http://stackoverflow.com/questions/4586812/convert-date-time-as-double-to-struct-tm-in-c | 1,427,770,012,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131300222.27/warc/CC-MAIN-20150323172140-00071-ip-10-168-14-71.ec2.internal.warc.gz | 262,376,841 | 18,387 | Convert Date/Time (as Double) to struct* tm in C++
I get a date/time as double value from C# (DateTime.ToOADate(), which is a OLE Date Time). It contains the passed days from 1899/12/31, with the fraction being the passed part of the day. Multiplied with 86400 I get the seconds and from that finally the day's time.
Getting the date however is harder. I still don't have an solution except for the dates that the UNIX time covers (1970/01/01 to 2038/01/19). During this time, mktime() can be used to convert the passed days to a datetime.
``````double OleDateTimeValue = 28170.654351851852; // 14.02.1977 15:42:16
struct tm * timeinfo;
int passedDays = (int)OLEDateTimeValue;
// between 1.1.1970 and 18.1.2038
if((passedDays >= 25569) && (passedDays <= 50423))
{
timeinfo->tm_year = 70; //1970
timeinfo->tm_mon = 0;
timeinfo->tm_mday = 1 + (passedDays - 25569);
mktime(timeinfo);
}
else // date outside the UNIX date/time
{
}
``````
Now, mktime() formats the tm struct so that it represents the requested date, or returns -1 if the value is outside the given dates.
Is there a generic way to do the calculation? Unfortunately I can't use MFC and have to use Visual C++ 6.0.
Thanks, Markus
-
Visual C++ 6.0's compiler is not even C++. It violates the standard in horrendous ways. Be aware of this as you program, and try to upgrade if you can. Also, not being able to use MFC is in fact fortunate, not unfortunate. :) Use standard C++ + Boost wherever you can. – Lightness Races in Orbit Jan 3 '11 at 20:50
I believe you can use the VariantTimeToSystemTime function to convert into a SYSTEMTIME.
-
Thanks, that was the easiest way to do it. – Markus Erlacher Jan 4 '11 at 10:43
I love date conversions - they make such a nice puzzle!
Here's some code that works for the dates from 1900-03-01 to 2100-02-28. The reason it doesn't work past those boundaries is that 1900 and 2100 are not leap years. This has been tested against Microsoft's COleDateTime and matches every day within the range.
``````int leapDays = (int)((OleDateTimeValue + 1400) / 1461);
timeinfo->tm_year = (int)((OleDateTimeValue - leapDays - 1) / 365);
int janFirst = timeinfo->tm_year * 365 + (int)((timeinfo->tm_year + 7) / 4);
int wholeDays = (int)OleDateTimeValue - janFirst;
if (timeinfo->tm_year % 4 != 0 && wholeDays > 58)
++wholeDays;
static int firstOfMonth[12] = { 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335 };
timeinfo->tm_mon = std::upper_bound(&firstOfMonth[0], &firstOfMonth[12], wholeDays) - &firstOfMonth[0];
timeinfo->tm_mday = wholeDays - firstOfMonth[timeinfo->tm_mon - 1] + 1;
``````
Conversion of the time portion is trivial, and I leave it as an exercise to the reader.
-
Converting to time_t should be easy (multiply by 86400 and add a constant offset), then you can use the `localtime` function. But you'll still be limited to the range of UNIX time. If you really need beyond that range then villintehaspam's answer, along with copying all the individual fields, looks like the way to go.
- | 882 | 3,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-14 | latest | en | 0.858311 |
https://www.justintools.com/unit-conversion/angle.php?k1=binary-degrees | 1,679,653,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00135.warc.gz | 906,239,095 | 23,008 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Binary Degrees
Binary degree or binary radian (brad) is 1 turn/256. In terms of the SI unit radians this is π/128 rad. The binary degree is used in computing so that an angle can be efficiently represented in a single byte.
Binary Degrees
Symbol/abbreviation: b deg
Unit of: ANGLE
ANGLE's base unit: radians (Non-SI Unit)
In relation to the base unit (radians), 1 Binary Degrees = 0.0245436926 radians.
## Conversion table 1 Binary Degrees (b deg) to all angle units
1 b deg= 24.999992462424 angular mils (µ mil)
1 b deg= 84.375060504387 arcminutes (arcmin)
1 b deg= 5062.4998014701 arcseconds (arcsec)
1 b deg= 1 binary degrees (b deg)
1 b deg= 156.24963458111 centesimal minute of arc (c MOA)
1 b deg= 15625.003246762 centesimal second of arc (c SOA)
1 b deg= 0.0039062470814984 circles (o)
1 b deg= 1.4062499448062 degrees (°deg)
1 b deg= 1.472621556589 diameter parts (Ø part)
1 b deg= 1.5625000262606 gons (gon)
1 b deg= 0.23437499994032 hexacontades (hc)
1 b deg= 0.093749999979708 hour angles (HA)
1 b deg= 5062500010.3132 microarcseconds (μas)
1 b deg= 5062500.0103132 milliarcseconds (mas) | 454 | 1,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.719321 |
https://www.physicsforums.com/threads/a-25w-light-bulb-married-a-ge-clock-radio.606527/ | 1,537,668,722,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00205.warc.gz | 816,979,870 | 14,326 | # A 25W Light Bulb Married a GE Clock/Radio
1. May 16, 2012
### nathangriffin
I'm wiring a 25W Edison bulb to an old GE clock/radio.
The radio is 120V/60hz 8.5W input.
Basic calculation I = P/E would mean the current is 0.07A
If I wire the bulb in parallel between source and the clock/radio, there's now 0.2A for the bulb branch and 0.07A for the clock/radio branch making Current Total about 0.3A (300mA) correct?
In doing so, however, I haven't actually made any change to the current flowing through the radio, right? It's still 0.07A?
So if the power cord which goes from the source to the bulb branch (first) and then to the clock/radio branch is the only thing carrying 0.3A and is rated well above that, I'm not in any danger of burning up the clock/radio, yes?
My concern is safety (obviously) so any help is appreciated.
2. May 16, 2012
### the_emi_guy
nathangriffin,
Yes, you are correct. If the line cord can deliver the 300mA than the radio will continue to see the 120V it is expecting and will draw its normal load current.
I can appreciate your concern for safety. Tell me a little more about your scheme and I can give you some tips on safety.
Where will your lamp wiring be located, inside of the radio? Where is the lamp itself mounted?
Does the clock radio have a safety ground (third prong on line cord)?
3. May 16, 2012
### nathangriffin
It's a pretty simple project, really... I think. I have an old clock radio. The cord is being replaced (re-soldered) using a new SPT-1 polarized plug/cord (two prong, one hot and one wide neutral) and I'm wiring the lamp socket in parallel on the source side of line, making my connection about 3" in from the step-down transformer in the clock/radio using orange wire nuts.
I'm connecting the line cord to the socket with 18AWG, remembering black to brass (to save my @ss) and, hopefully calling it a day? I've actually drilled a 1 3/8" hole in the top of the clock/radio and will secure the socket there. The only other thing I can think of for safety is, securing any extra wire away from things which heat up (eg. transformer)
Hmm. Can't think of anything else.
Again, if you can add anything, I'd love to know! I tried covering all my bases but had been out in the workshop these last two days debating whether I still remembered my principles, circuit basics, cause & effect etc!
4. May 16, 2012
### Staff: Mentor
Welcome to the PF.
It sounds like you are thinking of the right safety issues. Does "black to brass" equate to wiring Hot to the center/tip connector for the light bulb? That's important. Also, are you mechanically strain relieving the cord where it goes into the device?
5. May 17, 2012
### the_emi_guy
nathangriffin
No safety ground and use of AC mains puts this into IEC safety class II. Safety class II products are designed so that no single internal failure can pass hazardous voltage to an exposed conductor that a person can touch. This is generally done by either:
1 - Not having any exposed conductors (think hair dryer)
2 - Providing a double barrier between hazardous voltage and exposed conductors. The double barrier is designed so that it cannot be bridged by any single fault. (safety specs call this "double insulation", where insulation refers to any means of separation).
If your radio does have exposed metal, such as knobs, antenna, fasteners, you will want to make sure you don't compromise the barrier.
The idea is to assume a single failure occurs in your work, say your wire nut comes loose exposing your wires. Now look and see if that failure can cause AC mains to contact an exposed conductor. If so, add a second barrier. This could be as simple as a zip tie to hold the wires in a specific location. If both a zip tie and a wire nut have to fail to create a hazard then you are ok. | 947 | 3,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-39 | latest | en | 0.939219 |
http://www.sciforums.com/threads/a-new-simultaneity-method-for-accelerated-observers-in-special-relativity.162684/page-12#post-3626472 | 1,713,621,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00319.warc.gz | 54,006,089 | 20,298 | # A New Simultaneity Method for Accelerated Observers in Special Relativity
Discussion in 'Alternative Theories' started by Mike_Fontenot, Dec 26, 2019.
1. ### Neddy BateValued Senior Member
Messages:
2,548
That is where reality and Mike's brain part ways. In order to maintain this fantasy, he has to resort to claiming that the CMIF simultaneity (which is the simultaneity that results from extended sets of Einstein-sycnchronsed clocks at rest in inertial frames) is only applicable to reference frames which have been perpetually inertial.
This means he has to resort to changing things from being frame-dependent in SR, (such as "simultaneity is frame-dependent") to things being person-dependent in Mike's SR (such as "her age is currently 40 according to the perpetually inertial people in her reference frame, but not according to the guy who recently decelerated, his simultaneity is different, even though he is at rest in that same frame!").
It is bad enough to patch all of these unnecessary things onto an established theory ad hoc, but to claim the theory was like that all along is just insulting.
3. ### Jonathan DoolinRegistered Member
Messages:
20
Let me confirm what you mean by "happening locally", because I think there is one example where there might be an ambiguity: Let's say, the ICMIO is looking at "Her" in the distance, and "He" is looking at "Her" in the distance, at the same place and the same time. If I correctly understand what you mean by "happening locally" then "He" and the "ICMIO" should be seeing the same event, in the same way, at the same time. Even though "She" is distant, the event of seeing "Her" is local.
5. ### Neddy BateValued Senior Member
Messages:
2,548
Let's take the simplest case example, where the traveling twin changes from v=+0.577c to v=0.000c at the usual location (x=23.09 light years) at the moment when the traveling twin's clock says 32.66 years, as shown:
All of the people in questions are now at rest in the stay-home twin's rest frame, and everyone agrees that her age at that time is 40 years old, because SR does not have things that are person-dependent, only things which are reference frame-dependent.
But let's say for a moment that SR is what Mike says it is, namely, totally silent on what speed of light should be for the traveling twin personally, because he has recently decelerated. All of the people located at x=23.09 can see (using their eyes + a telescope) that the stay-home twin's clock is showing 16.91 years. All of the perpetually inertial people can use the known speed of light and the known distance to calculate her current age as 16.91 + 23.09 = 40.00 years. But the traveling twin is different than all of those other people, because he has not been perpetually inertial. So what is the speed of light according to him? What is the distance between him and her according to him? It is whatever Mike says it is.
He can say the distance is still 23.09 light years for him, but the speed of light of the image of her clock was not 1 light year per year for him. Or he can say the speed of light of the image of her clock was 1 light year per year for him, but the distance is not 23.09 light years for him. Or he can say it is a combination of both.
All of this because he thinks SR is completely silent on the speed of light according to him (!) and because he thinks SR is completely silent on the distance between him and her according to him (!). Yeah, I'm pretty sure SR is not completely silent on those things.
7. ### Jonathan DoolinRegistered Member
Messages:
20
My own philosophy is that given any set of event coordinates, and the position, time, and velocity of any observer, you can establish the location of events according to that observer at that point in time, mostly by using rotation and lorentz transformations. The rotation transformation handles all the forward, up, down, left, and right issues, while the Lorentz transformation handles the future, past issues.
Here's my most recent treatment on the topic.
http://www.spoonfedrelativity.com/pages/TemporalFacing.php
8. ### Neddy BateValued Senior Member
Messages:
2,548
I'm not sure why you say it is your own philosophy, because the Lorentz transformations (LT's) sole purpose is to provide equations that anyone can use to do that, while maintaining all of the premises of SR. Unless you are of the Mike Fontenot school of "SR is totally silent on that," then all anyone has to do is use the LT's. They don't have to subscribe to any philosophy.
Anyway, I watched your first video, and I thought it was explained quite well. But the second video appears to be an accidental exact copy of the first one. Your notes say that the first 17 minutes are a review of part 1, but the whole video is only 13:41 minutes long, (the same length as the first).
Other than that, I found the fourth video interesting. I had not studied radar time that deeply before, because it seemed to me to be just another unnecessary add-on to SR, and not part of SR itself. But it is funny when you reveal that D&G still ended up with distant times shifting when someone "goes dancing" even though that was clearly what motivated D&G to create the add-on in the first place.
SR is complete as it is, excepting gravity. If it doesn't specifically say that an inertial frame can only be one that has been perpetually inertial, then that does not mean it is totally silent on that subject allowing one to interpret it to mean that if one wishes. Inertial frames are clearly defined in SR, and not open to interpretations such as that (Mike's).
Likewise, relativity of simultaneity is a clearly derived result of the premises of SR. When two inertial frames are in uniform relative translation with respect to one another, events which are spacial-separated along the direction of the relative motion between the frames WILL NOT BE SIMULTANEOUS in one frame, if they ARE SIMULTANEOUS in the other frame. Mike's whole goal is to violate that, because he needs to make the event "Traveling twin's age is 32.66" simultaneous with the event "Stay-home twin's age is 26.67" in BOTH the traveling twin's outbound frame, AND the traveling twin's new frame, (after the traveler decelerates to v=0.000c or turns around and heads back at v=-0.577c or whatever else he might do). Thus a clearly derived result of SR, namely relativity of simultaneity as explained above, is DENIED OUTRIGHT under the guise of "SR is totally silent on that". It is almost a blasphemy, lol.
Last edited: Mar 26, 2020
9. ### Mike_FontenotRegistered Senior Member
Messages:
622
Yes, that's correct. The disagreement between him and the ICMIO (for a while, after he changes velocity) is over how much she aged while those images were in transit.
10. ### Jonathan DoolinRegistered Member
Messages:
20
I'll try to fix that soon. In the meantime, if you like, you can watch the video on youtube at
Have you ever heard anyone say "SR is valid only locally"? I have heard this phrase quite a number of times, and it seems at odds with the philosophy that the Lorentz Transformations operates on the coordinates of all events in spacetime. So, my philosophy is that SR is valid globally.
Look in the Wikipedia discussion archive 5 for Lorentz Transformations, here: https://en.wikipedia.org/wiki/Talk:Lorentz_transformation/Archive_5
Under the header "Delete irrelevant animation", the user, there, whose name seems unpronounceable, says that the animation there, which demonstrates exactly how events move forward and backward in time as the Lorentz Transformations are performed, for a given observer, are "irrelevant" to the Lorentz Transformations. This gives me the impression that many people are of the philosophy that SR does not actually move events around, but just puts new labels on them.
11. ### Jonathan DoolinRegistered Member
Messages:
20
Alright. I will say, then, there is nothing, in principle, that would prevent the traveling twin from using your method, or even arguing that his method was better than other methods. "Better" being a key word here, because it is an opinion word. Opinions as to what is better can be formulated on whatever you designate to be an important property you want a method to satisfy. (A rubric)
The method I think is better, is for "him" to simply take a "tangent" with slope dx/dt = v, and the hyperpendicular simultaneity line, with slope dx/dt =1/v, and see where that simultaneity line intersects "her" world-line. Whatever age she is at that intersection is what "he" would say "her" age is now.
One thing I like about this method is that it should highlight changes in the relativity of simultaneity, as he accelerates toward her, she has a sudden lurch forward in age. Even skipping years, if he accelerates instantaneously. If you regard the acknowledgment of the relativity of simultaneity as a bad, thing, though, highlighting it would not be in your rubric.
12. ### HalcRegistered Senior Member
Messages:
350
Your philosophy is then one of denial of facts. Under SR, given any two objects E and F separated by X light years in the frame in which E is stationary, light from F will get to A in X years. Not true of our universe, where if F is sufficiently distant, light will never reach E in any amount of time. So much for SR describing the universe on a large scale.
On a more local scale, if I shine a laser to a reflector on the moon and time the round trip, I will get a faster measurement for the speed of light than if I did the same experiment on the moon using a reflector here on Earth. SR asserts the two should be the same, but in both cases (the distant F and the moon thingy), the conditions of SR (the parts that make it special) are not met. Only locally are they met.
13. ### Jonathan DoolinRegistered Member
Messages:
20
The universe you are describing is only a model.
I agree, here, but what I'm in agreement with is that "SR is NOT valid locally".
14. ### HalcRegistered Senior Member
Messages:
350
I am describing our actual empirical universe, not a model. There is really an event horizon beyond which light will never reach us. Light round trip to the moon and back really will be measured faster than light round trip from the moon and back. GPS clocks run faster than Earth clocks even in Earth frame where the GPS clocks are moving much faster. SR is mute on all that because SR cannot explain any of these observations.
15. ### Mike_FontenotRegistered Senior Member
Messages:
622
What you are describing is General Relativity and Cosmology. This thread is about Special Relativity. In Special Relativity, it is assumed that there are no significant masses present, spacetime is flat, and it is valid over arbitrarily large distances.
Last edited: Mar 28, 2020
16. ### Jonathan DoolinRegistered Member
Messages:
20
Even if you were correct about the universe, I would still call your description of it "a model."
Emperical means "based on, concerned with, or verifiable by observation or experience rather than theory or pure logic".
Can you tell me of an observation that has been done which has demonstrated that your model is correct, and the Milne model is incorrect? Both models have an explanation for the Cosmic Microwave Background Radiation. Both models have an explanation for Hubble's Law. And if you account for temperature in the early universe, both models have an explanation for cosmic inflation, as I recently described here: https://www.researchgate.net/public...Cosmological_Inflation_in_Minkowski_Spacetime
My philosophy is that cosmology and Special Relativity should not be separated. Many people treat them as though the Lorentz transformation stops functioning as soon as one adds a little gravity into the mix. I think that's as absurd as suggesting that rotation stops working as soon as you throw gravity into the mix.
If I turn my head 90 degrees to the left, all of the objects, no matter how far away from me, rotate 90 degrees to the right in my point-of-view, corresponding to a 90 degree clockwise rotation transformation. It does not matter how far away they are, and gravity has no effect on this. So how can anyone justify the idea that if I accelerate to half the speed of light, only "local" events will respond to the Lorentz Transformation?
17. ### Mike_FontenotRegistered Senior Member
Messages:
622
I certainly agree with all that. The good characteristics of my simultaneity method, in my opinion, are that it has no discontinuities in the age correspondence diagram (ACD), it is causal, its ACD is easy and quick to compute, and the home twin's age NEVER decreases ... i.e., she never gets YOUNGER, according to him. Some of these characteristics are also characteristics of the other three simultaneity methods, but none of the other methods have ALL of these characteristics.
When I first read that description, I initially thought you were proposing a 5th simultaneity method. But after a more careful reading, I think you are just talking about the CMIF method.
If you are talking about me and my method in the above statement, then I don't understand why you are saying that. The relativity of simultaneity is just a statement that not all observers draw the ACD (her current age, plotted versus his age) the same way. None of the four simultaneity methods contend that the relativity of simultaneity is a bad thing. They each produce an ACD that differs from the ACD that the home twin produces, and that's their goal.
18. ### Mike_FontenotRegistered Senior Member
Messages:
622
I think they SHOULD be. Halc is arguing that SR is only local, because Cosmology (and in particular, inflation) results in some signals never reaching us. But under the ASSUMPTIONS of SR itself, there are no significant masses anywhere near our proposed trip, and there is not any spatial inflation happening that affects our trip ... space by definition is completely static and flat in SR, and it is infinite in extent.
Even in the real universe as it is (or as we currently think it is), if we could take a relativistic-speed trip out to 50 lightyears or so, and which avoided going anywhere near any large stars or blackholes, the results obtained using special relativity would very well approximate what actually happens on the trip. It's wrong to think SR is no longer useful, just because we now know a lot about GR and Cosmology (specifically, about blackholes and inflation).
19. ### HalcRegistered Senior Member
Messages:
350
Oh, I agree that SR is fine for the thought exercises in this thread. I was just commenting on Doolin's 'philosophy'.
SR is local because it has no compensation for dark energy or for gravity, and I picked examples illustrating each to show why it fails outside of the special conditions where it is valid.
20. ### Jonathan DoolinRegistered Member
Messages:
20
Here is an example of what I mean by saying SR is valid globally: "Stellar aberration" was detected (empirically) by James Bradley in 1727.
That effect is easily modeled by the Lorentz Transformations. It essentially causes their images to be always Lorentz boosted forward towards the direction of Earth's motion around the sun, just a bit. There's no gravitational pull or dark energy that prevent the Lorentz Transformations from operating on events that are an infinite distance from earth.
Notice, here, I use the word "model" whether or not I believe the model is correct. I believe that the Lorentz Transformations give the correct model for Stellar Aberration. However, scientifically speaking, I also think one could find a way of modeling the phenomenon using Galilean Transformations with the sun embedded within a universal ether.
Empirically speaking, all we have to go on is redshift and magnitude measurements.
A model that claims "some signals never reach us" is not an entirely empirical model, because any signal that will never reach us would never be observed or experienced; hence the opposite of empirical.
Regarding gravity, I think this is the third time I've said this: gravity is a local phenomenon, so it would make more sense to say that SR is global, while gravity is local.
Regarding dark energy, it is energy that has never been detected, so again, the opposite of "empirical".
It is also goes somewhat counter to the scientific method that the supposedly "correct" model of the universe is one that makes a prediction that has so far never been observed. In introductory science, when your hypothesis predicts a phenomenon, and then, when tested, that phenomenon does not occur, that hypothesis does not become theory. Instead, one is supposed to go back to the observation stage with an open mind to form new hypotheses.
Messages:
622
My monograph on "A New Simultaneity Method for Accelerated Observers in Special Relativity" is available on Amazon for $5.00. You can find it by searching on "simultaneity method". 22. ### Mike_FontenotRegistered Senior Member Messages: 622 My new simultaneity method is an alternative to the commonly used co-moving-inertial frames (CMIF) simultaneity method. I personally prefer the CMIF method to my method, but some people dislike the CMIF method because it produces an instantaneously changing current age of the home twin (she), according to the traveling twin (he), when he instantaneously changes his velocity. That instantaneous changing of her age, according to him, is especially abhorrent to some people when it is negative, i.e., when she instantaneously gets YOUNGER, according to him. I personally am not bothered by that (and neither is the well-known physicist Brian Greene, as was shown in his PBS Nova series on "The Fabric of the Cosmos"). My simultaneity method provides an alternative "safe haven" for those people, because it has no discontinuities (either positive or negative) in her age, according to him. My simultaneity method is described in detail in a monograph published by Amazon, available for$5.00. You can find it most easily by searching on my complete name: "Michael Leon Fontenot" (with the quotes).
That monograph addresses only the issue of how to determine her age according to him, versus his age. I.e., it only addresses simultaneity. But I have also recently worked out what my method has to say about how he can determine their distance apart, according to him, at any instant in his life. That addition makes my method a complete coordinate system for him. Here is a description of how my method determines their separation, according to him.
The CMIF method and my method give the same answer to the question "What is the distance between the home twin (she) and the traveling twin (he), according to him" ONLY when there is no discontinuity in the CMIF solution. There is no discontinuity in their separation in the most common version of the twin paradox, where he reverses course at the turnaround point, but comes back at the same speed as he used on the outbound leg.
But if he uses a DIFFERENT speed on the return leg than he used on the outbound leg, the CMIF solution gives a discontinuity in their separation when the speed changes (according to him). My method never produces a discontinuity.
For example, suppose his speed on the outbound leg is 0.57735 ly/y (gamma = 1.2247). (That case is advantageous, because the angle of his worldline wrt her worldline (the horizontal axis) is 30 degrees, and can be easily drawn with a 30-60-90 plastic triangle.) Suppose she is 40 years old when he turns around. He is 32.66 years old then. Then suppose his new speed is -0.866 ly/y (gamma = 2.0). When they are reunited, she is 66.67 years old, and he is 46.0 years old.
For the above case, we can plot their separation at each instant of his life, according to him, for both the CMIF method and for my method. I call that diagram the "SAAOD", for "Separation According to the Accelerated Observer Diagram." Their separation is on the vertical axis, and his age is on the horizontal axis.
The first straight line segment starts at the origin, when he is zero years old, and their separation is zero ly. It slopes upward at a 30 degree angle. It reaches its peak when he is 32.66 years old, and their separation is 18.86 ly.
When he changes his speed to -0.866 ly/y, their separation instantaneously drops to 11.55 ly, according to the CMIF method. Then, from there, their separation declines linearly with a slope -0.866 until their reunion. (That last straight line segment makes an angle of approximately 41 degrees wrt the horizontal axis).
In contrast, in my method their separation doesn't instantaneously change. Instead, it decreases linearly with a slope of -2.047 from its peak at 18.86 until his age reaches 38.85 years old, and their distance reaches 6.188. (That endpoint corresponds to when he receives a light pulse that she sends at his turnaround). Then, at that point, the slope of the line decreases to -0.866 until their reunion. The two methods coincide along that last segment.
Last edited: Apr 24, 2021
23. ### Mike_FontenotRegistered Senior Member
Messages:
622
I have just verified that the above segment is indeed a straight line. The verification is similar to what I did for the linearity of the middle segment in the age correspondence diagram (the "ACD"), which I described in Section 8, titled "Pulses Partly in Both Halves of the Minkowski Diagram", of my monograph.
On the Minkowski diagram, first find the vertical line descending from the traveler's (his) worldline at his turnaround point, down to the horizontal axis (her worldline). Mark the half-way point on that vertical line, and call it "point Q". Then draw a light pulse being sent by her, at such a time in her life that the pulse passes through that half-way point, and continues on until it reaches his worldline. Mark that point on his worldline as point R. His age is 35.75 then.
According to her, their separation at the turnaround is 23.094 ly. And according to her, that half-way point is at distance 11.547 ly from her. According to a perpetually-inertial observer traveling at 0.57735 ly/y (with gamma = 1.2247), that distance is 11.547 / 1.2247, or 9.43 ly from her. Similarly, a perpetually-inertial observer traveling at -0.866 ly/y (with gamma = 2.0), concludes that that pulse travels a distance 3.09 ly from point Q until it reaches the traveling twin at point R. So the traveling twin (he) then concludes that the distance between him and her at the instant he receives that pulse is 9.43 + 3.09 = 12.52 ly. If, on the SAAOD (Separation According to the Accelerated Observer Diagram), you plot the separation 12.52 ly when he is 35.75 years old, you will see that it does indeed lie on the midpoint of the previously described straight line. | 5,171 | 22,771 | {"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} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.959661 |
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# Write each percent as a decimal and as a fraction in simplest form?
Updated: 10/16/2023
Wiki User
6y ago
To write a percent as a decimal, divide it by 100
To write a percent as a fraction, put it over 100
To reduce a fraction to its simplest form, find the GCF of the numerator and the denominator and divide them both by it. If the GCF is 1, the fraction is in its simplest form.
75% = 0.75 = 75/100 = 3/4
Wiki User
6y ago
Levi Morris
Lvl 2
4mo ago
Work out 6 x 23
Give your answer as a whole number or as a fraction in its simplest form. | 167 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-10 | latest | en | 0.952843 |
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Posts: 50
Joined: Fri Sep 29, 2017 7:05 am
### Question 27
Could someone explain how you would setup number 27, " calculate the work for each of the following processes beginning with a gas sample in a piston assembly with T = 350 K, P = 1.79 atm and V = 4.29 L: (a) irreversible expansion against a constant external pressure of 1.00atm to a final volume of 6.52 L; (b) isothermal, reversible expansion to a final volume of 6.52 L."
Diego Zavala 2I
Posts: 65
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time
### Re: Question 27
For part a use the equation $w=-P\Delta V$ and for part b use equation w=-nRTln(V2/V1)
Humza_Khan_2J
Posts: 56
Joined: Thu Jul 13, 2017 3:00 am
### Re: Question 27
You would need to use the fact that work of expansion against opposing constant pressure is equal to -PdeltaV. This yields (-1.00 atm)(6.52-4.29 L). The second part would require the isothermal expansion equation, which is also in the book. To get n in this case for the aforementioned equation, one would need to use PV=nRT.
Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”
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https://howkgtolbs.com/convert/53.92-kg-to-lbs | 1,653,737,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00486.warc.gz | 369,423,417 | 12,103 | 53.92 kg to lbs - 53.92 kilograms to pounds
Do you need to learn how much is 53.92 kg equal to lbs and how to convert 53.92 kg to lbs? You couldn’t have chosen better. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to underline that all this article is devoted to one amount of kilograms - that is one kilogram. So if you want to learn more about 53.92 kg to pound conversion - read on.
Before we go to the more practical part - that is 53.92 kg how much lbs calculation - we are going to tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 53.92 kg to lbs? 53.92 kilograms it is equal 118.8732516704 pounds, so 53.92 kg is equal 118.8732516704 lbs.
53.92 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in short form SI).
At times the kilogram could be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but hard to use.
Then, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by a new definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
53.92 kilogram to pounds
You know a little bit about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. Naturally, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is just equal 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
How many lbs is 53.92 kg?
53.92 kilogram is equal to 118.8732516704 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
53.92 kg in lbs
Theoretical section is already behind us. In this part we are going to tell you how much is 53.92 kg to lbs. Now you learned that 53.92 kg = x lbs. So it is high time to know the answer. Have a look:
53.92 kilogram = 118.8732516704 pounds.
It is an accurate result of how much 53.92 kg to pound. You may also round it off. After rounding off your result will be as following: 53.92 kg = 118.624 lbs.
You know 53.92 kg is how many lbs, so look how many kg 53.92 lbs: 53.92 pound = 0.45359237 kilograms.
Naturally, in this case you can also round it off. After rounding off your result will be exactly: 53.92 lb = 0.45 kgs.
We are also going to show you 53.92 kg to how many pounds and 53.92 pound how many kg results in charts. See:
We will begin with a chart for how much is 53.92 kg equal to pound.
53.92 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
53.92 118.8732516704 118.6240
Now see a chart for how many kilograms 53.92 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
53.92 0.45359237 0.45
Now you know how many 53.92 kg to lbs and how many kilograms 53.92 pound, so it is time to move on to the 53.92 kg to lbs formula.
53.92 kg to pounds
To convert 53.92 kg to us lbs a formula is needed. We will show you two formulas. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 118.8732516704 outcome in pounds
The first version of a formula will give you the most exact result. In some cases even the smallest difference can be significant. So if you need a correct outcome - this formula will be the best solution to convert how many pounds are equivalent to 53.92 kilogram.
So let’s move on to the shorer version of a formula, which also enables calculations to know how much 53.92 kilogram in pounds.
The another version of a formula is as following, look:
Number of kilograms * 2.2 = the result in pounds
As you can see, the second version is simpler. It can be better choice if you want to make a conversion of 53.92 kilogram to pounds in quick way, for instance, during shopping. Just remember that final outcome will be not so correct.
Now we are going to show you these two versions of a formula in practice. But before we will make a conversion of 53.92 kg to lbs we are going to show you another way to know 53.92 kg to how many lbs totally effortless.
53.92 kg to lbs converter
Another way to know what is 53.92 kilogram equal to in pounds is to use 53.92 kg lbs calculator. What is a kg to lb converter?
Converter is an application. It is based on longer formula which we gave you above. Thanks to 53.92 kg pound calculator you can effortless convert 53.92 kg to lbs. You only need to enter amount of kilograms which you need to convert and click ‘convert’ button. You will get the result in a flash.
So try to convert 53.92 kg into lbs with use of 53.92 kg vs pound converter. We entered 53.92 as a number of kilograms. It is the outcome: 53.92 kilogram = 118.8732516704 pounds.
As you see, our 53.92 kg vs lbs calculator is easy to use.
Now let’s move on to our primary topic - how to convert 53.92 kilograms to pounds on your own.
53.92 kg to lbs conversion
We will begin 53.92 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 118.8732516704 the result in pounds
So what have you do to learn how many pounds equal to 53.92 kilogram? Just multiply amount of kilograms, in this case 53.92, by 2.20462262. It is equal 118.8732516704. So 53.92 kilogram is equal 118.8732516704.
It is also possible to round it off, for instance, to two decimal places. It is 2.20. So 53.92 kilogram = 118.6240 pounds.
It is high time for an example from everyday life. Let’s calculate 53.92 kg gold in pounds. So 53.92 kg equal to how many lbs? And again - multiply 53.92 by 2.20462262. It is equal 118.8732516704. So equivalent of 53.92 kilograms to pounds, when it comes to gold, is equal 118.8732516704.
In this case you can also round off the result. This is the result after rounding off, in this case to one decimal place - 53.92 kilogram 118.624 pounds.
Now let’s move on to examples calculated with short formula.
How many 53.92 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 118.624 the outcome in pounds
So 53.92 kg equal to how much lbs? As in the previous example you have to multiply amount of kilogram, this time 53.92, by 2.2. Look: 53.92 * 2.2 = 118.624. So 53.92 kilogram is equal 2.2 pounds.
Do another calculation with use of this version of a formula. Now convert something from everyday life, for example, 53.92 kg to lbs weight of strawberries.
So let’s calculate - 53.92 kilogram of strawberries * 2.2 = 118.624 pounds of strawberries. So 53.92 kg to pound mass is exactly 118.624.
If you learned how much is 53.92 kilogram weight in pounds and can convert it with use of two different formulas, we can move on. Now we are going to show you all results in tables.
Convert 53.92 kilogram to pounds
We realize that results shown in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in tables for your convenience. Due to this you can easily compare 53.92 kg equivalent to lbs outcomes.
Start with a 53.92 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
53.92 118.8732516704 118.6240
And now have a look at 53.92 kg equal pound table for the second formula:
Kilograms Pounds
53.92 118.624
As you see, after rounding off, if it comes to how much 53.92 kilogram equals pounds, the results are the same. The bigger number the more considerable difference. Please note it when you need to do bigger amount than 53.92 kilograms pounds conversion.
How many kilograms 53.92 pound
Now you learned how to convert 53.92 kilograms how much pounds but we want to show you something more. Do you want to know what it is? What about 53.92 kilogram to pounds and ounces calculation?
We will show you how you can calculate it little by little. Start. How much is 53.92 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 53.92, by 2.20462262. So 53.92 * 2.20462262 = 118.8732516704. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 53.92 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So final outcome is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your outcome is exactly 2 pounds and 33 ounces.
As you see, calculation 53.92 kilogram in pounds and ounces quite easy.
The last calculation which we want to show you is calculation of 53.92 foot pounds to kilograms meters. Both of them are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we show you this formula, have a look:
• 53.92 kilograms meters = 7.23301385 foot pounds,
• 53.92 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 53.92 foot pounds to kilograms meters you need to multiply 53.92 by 0.13825495. It is exactly 0.13825495. So 53.92 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 53.92 foot pounds is exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 53.92 kilogram into pounds calculations.
This article was a huge compendium about kilogram, pound and 53.92 kg to lbs in conversion. Due to this calculation you know 53.92 kilogram is equivalent to how many pounds.
We showed you not only how to do a conversion 53.92 kilogram to metric pounds but also two another conversions - to check how many 53.92 kg in pounds and ounces and how many 53.92 foot pounds to kilograms meters.
We showed you also other solution to do 53.92 kilogram how many pounds conversions, it is with use of 53.92 kg en pound converter. This is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you are able to do 53.92 kilogram equal to how many pounds calculation - on your own or with use of our 53.92 kgs to pounds converter.
Don’t wait! Calculate 53.92 kilogram mass to pounds in the way you like.
Do you want to make other than 53.92 kilogram as pounds conversion? For instance, for 15 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 53.92 kilogram equal many pounds.
How much is 53.92 kg in pounds
We want to sum up this topic, that is how much is 53.92 kg in pounds , we gathered answers to the most frequently asked questions. Here we have for you all you need to know about how much is 53.92 kg equal to lbs and how to convert 53.92 kg to lbs . It is down below.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. Let’s see 53.92 kg to pound conversion formula . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 53.92 kilogram to pounds? The accurate result is 118.8732516704 pounds.
You can also calculate how much 53.92 kilogram is equal to pounds with second, shortened version of the equation. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So now, 53.92 kg equal to how much lbs ? The result is 118.8732516704 lbs.
How to convert 53.92 kg to lbs in an easier way? You can also use the 53.92 kg to lbs converter , which will make the rest for you and give you a correct result .
Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,597 | 13,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.950406 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/480/C20xC3sD4.html | 1,586,118,465,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371609067.62/warc/CC-MAIN-20200405181743-20200405212243-00387.warc.gz | 642,919,329 | 10,854 | Copied to
clipboard
## G = C20×C3⋊D4order 480 = 25·3·5
### Direct product of C20 and C3⋊D4
Series: Derived Chief Lower central Upper central
Derived series C1 — C6 — C20×C3⋊D4
Chief series C1 — C3 — C6 — C2×C6 — C2×C30 — S3×C2×C10 — C10×C3⋊D4 — C20×C3⋊D4
Lower central C3 — C6 — C20×C3⋊D4
Upper central C1 — C2×C20 — C22×C20
Generators and relations for C20×C3⋊D4
G = < a,b,c,d | a20=b3=c4=d2=1, ab=ba, ac=ca, ad=da, cbc-1=dbd=b-1, dcd=c-1 >
Subgroups: 388 in 188 conjugacy classes, 90 normal (58 characteristic)
C1, C2 [×3], C2 [×4], C3, C4 [×2], C4 [×5], C22, C22 [×2], C22 [×6], C5, S3 [×2], C6 [×3], C6 [×2], C2×C4 [×2], C2×C4 [×7], D4 [×4], C23, C23, C10 [×3], C10 [×4], Dic3 [×2], Dic3 [×2], C12 [×2], C12, D6 [×2], D6 [×2], C2×C6, C2×C6 [×2], C2×C6 [×2], C15, C42, C22⋊C4 [×2], C4⋊C4, C22×C4, C22×C4, C2×D4, C20 [×2], C20 [×5], C2×C10, C2×C10 [×2], C2×C10 [×6], C4×S3 [×2], C2×Dic3 [×3], C3⋊D4 [×4], C2×C12 [×2], C2×C12 [×2], C22×S3, C22×C6, C5×S3 [×2], C30 [×3], C30 [×2], C4×D4, C2×C20 [×2], C2×C20 [×7], C5×D4 [×4], C22×C10, C22×C10, C4×Dic3, Dic3⋊C4, D6⋊C4, C6.D4, S3×C2×C4, C2×C3⋊D4, C22×C12, C5×Dic3 [×2], C5×Dic3 [×2], C60 [×2], C60, S3×C10 [×2], S3×C10 [×2], C2×C30, C2×C30 [×2], C2×C30 [×2], C4×C20, C5×C22⋊C4 [×2], C5×C4⋊C4, C22×C20, C22×C20, D4×C10, C4×C3⋊D4, S3×C20 [×2], C10×Dic3 [×3], C5×C3⋊D4 [×4], C2×C60 [×2], C2×C60 [×2], S3×C2×C10, C22×C30, D4×C20, Dic3×C20, C5×Dic3⋊C4, C5×D6⋊C4, C5×C6.D4, S3×C2×C20, C10×C3⋊D4, C22×C60, C20×C3⋊D4
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C5, S3, C2×C4 [×6], D4 [×2], C23, C10 [×7], D6 [×3], C22×C4, C2×D4, C4○D4, C20 [×4], C2×C10 [×7], C4×S3 [×2], C3⋊D4 [×2], C22×S3, C5×S3, C4×D4, C2×C20 [×6], C5×D4 [×2], C22×C10, S3×C2×C4, C4○D12, C2×C3⋊D4, S3×C10 [×3], C22×C20, D4×C10, C5×C4○D4, C4×C3⋊D4, S3×C20 [×2], C5×C3⋊D4 [×2], S3×C2×C10, D4×C20, S3×C2×C20, C5×C4○D12, C10×C3⋊D4, C20×C3⋊D4
Smallest permutation representation of C20×C3⋊D4
On 240 points
Generators in S240
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140)(141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160)(161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180)(181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200)(201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220)(221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240)
(1 67 146)(2 68 147)(3 69 148)(4 70 149)(5 71 150)(6 72 151)(7 73 152)(8 74 153)(9 75 154)(10 76 155)(11 77 156)(12 78 157)(13 79 158)(14 80 159)(15 61 160)(16 62 141)(17 63 142)(18 64 143)(19 65 144)(20 66 145)(21 207 221)(22 208 222)(23 209 223)(24 210 224)(25 211 225)(26 212 226)(27 213 227)(28 214 228)(29 215 229)(30 216 230)(31 217 231)(32 218 232)(33 219 233)(34 220 234)(35 201 235)(36 202 236)(37 203 237)(38 204 238)(39 205 239)(40 206 240)(41 135 88)(42 136 89)(43 137 90)(44 138 91)(45 139 92)(46 140 93)(47 121 94)(48 122 95)(49 123 96)(50 124 97)(51 125 98)(52 126 99)(53 127 100)(54 128 81)(55 129 82)(56 130 83)(57 131 84)(58 132 85)(59 133 86)(60 134 87)(101 189 171)(102 190 172)(103 191 173)(104 192 174)(105 193 175)(106 194 176)(107 195 177)(108 196 178)(109 197 179)(110 198 180)(111 199 161)(112 200 162)(113 181 163)(114 182 164)(115 183 165)(116 184 166)(117 185 167)(118 186 168)(119 187 169)(120 188 170)
(1 91 207 105)(2 92 208 106)(3 93 209 107)(4 94 210 108)(5 95 211 109)(6 96 212 110)(7 97 213 111)(8 98 214 112)(9 99 215 113)(10 100 216 114)(11 81 217 115)(12 82 218 116)(13 83 219 117)(14 84 220 118)(15 85 201 119)(16 86 202 120)(17 87 203 101)(18 88 204 102)(19 89 205 103)(20 90 206 104)(21 193 146 44)(22 194 147 45)(23 195 148 46)(24 196 149 47)(25 197 150 48)(26 198 151 49)(27 199 152 50)(28 200 153 51)(29 181 154 52)(30 182 155 53)(31 183 156 54)(32 184 157 55)(33 185 158 56)(34 186 159 57)(35 187 160 58)(36 188 141 59)(37 189 142 60)(38 190 143 41)(39 191 144 42)(40 192 145 43)(61 132 235 169)(62 133 236 170)(63 134 237 171)(64 135 238 172)(65 136 239 173)(66 137 240 174)(67 138 221 175)(68 139 222 176)(69 140 223 177)(70 121 224 178)(71 122 225 179)(72 123 226 180)(73 124 227 161)(74 125 228 162)(75 126 229 163)(76 127 230 164)(77 128 231 165)(78 129 232 166)(79 130 233 167)(80 131 234 168)
(1 11)(2 12)(3 13)(4 14)(5 15)(6 16)(7 17)(8 18)(9 19)(10 20)(21 231)(22 232)(23 233)(24 234)(25 235)(26 236)(27 237)(28 238)(29 239)(30 240)(31 221)(32 222)(33 223)(34 224)(35 225)(36 226)(37 227)(38 228)(39 229)(40 230)(41 162)(42 163)(43 164)(44 165)(45 166)(46 167)(47 168)(48 169)(49 170)(50 171)(51 172)(52 173)(53 174)(54 175)(55 176)(56 177)(57 178)(58 179)(59 180)(60 161)(61 150)(62 151)(63 152)(64 153)(65 154)(66 155)(67 156)(68 157)(69 158)(70 159)(71 160)(72 141)(73 142)(74 143)(75 144)(76 145)(77 146)(78 147)(79 148)(80 149)(81 105)(82 106)(83 107)(84 108)(85 109)(86 110)(87 111)(88 112)(89 113)(90 114)(91 115)(92 116)(93 117)(94 118)(95 119)(96 120)(97 101)(98 102)(99 103)(100 104)(121 186)(122 187)(123 188)(124 189)(125 190)(126 191)(127 192)(128 193)(129 194)(130 195)(131 196)(132 197)(133 198)(134 199)(135 200)(136 181)(137 182)(138 183)(139 184)(140 185)(201 211)(202 212)(203 213)(204 214)(205 215)(206 216)(207 217)(208 218)(209 219)(210 220)
G:=sub<Sym(240)| (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240), (1,67,146)(2,68,147)(3,69,148)(4,70,149)(5,71,150)(6,72,151)(7,73,152)(8,74,153)(9,75,154)(10,76,155)(11,77,156)(12,78,157)(13,79,158)(14,80,159)(15,61,160)(16,62,141)(17,63,142)(18,64,143)(19,65,144)(20,66,145)(21,207,221)(22,208,222)(23,209,223)(24,210,224)(25,211,225)(26,212,226)(27,213,227)(28,214,228)(29,215,229)(30,216,230)(31,217,231)(32,218,232)(33,219,233)(34,220,234)(35,201,235)(36,202,236)(37,203,237)(38,204,238)(39,205,239)(40,206,240)(41,135,88)(42,136,89)(43,137,90)(44,138,91)(45,139,92)(46,140,93)(47,121,94)(48,122,95)(49,123,96)(50,124,97)(51,125,98)(52,126,99)(53,127,100)(54,128,81)(55,129,82)(56,130,83)(57,131,84)(58,132,85)(59,133,86)(60,134,87)(101,189,171)(102,190,172)(103,191,173)(104,192,174)(105,193,175)(106,194,176)(107,195,177)(108,196,178)(109,197,179)(110,198,180)(111,199,161)(112,200,162)(113,181,163)(114,182,164)(115,183,165)(116,184,166)(117,185,167)(118,186,168)(119,187,169)(120,188,170), (1,91,207,105)(2,92,208,106)(3,93,209,107)(4,94,210,108)(5,95,211,109)(6,96,212,110)(7,97,213,111)(8,98,214,112)(9,99,215,113)(10,100,216,114)(11,81,217,115)(12,82,218,116)(13,83,219,117)(14,84,220,118)(15,85,201,119)(16,86,202,120)(17,87,203,101)(18,88,204,102)(19,89,205,103)(20,90,206,104)(21,193,146,44)(22,194,147,45)(23,195,148,46)(24,196,149,47)(25,197,150,48)(26,198,151,49)(27,199,152,50)(28,200,153,51)(29,181,154,52)(30,182,155,53)(31,183,156,54)(32,184,157,55)(33,185,158,56)(34,186,159,57)(35,187,160,58)(36,188,141,59)(37,189,142,60)(38,190,143,41)(39,191,144,42)(40,192,145,43)(61,132,235,169)(62,133,236,170)(63,134,237,171)(64,135,238,172)(65,136,239,173)(66,137,240,174)(67,138,221,175)(68,139,222,176)(69,140,223,177)(70,121,224,178)(71,122,225,179)(72,123,226,180)(73,124,227,161)(74,125,228,162)(75,126,229,163)(76,127,230,164)(77,128,231,165)(78,129,232,166)(79,130,233,167)(80,131,234,168), (1,11)(2,12)(3,13)(4,14)(5,15)(6,16)(7,17)(8,18)(9,19)(10,20)(21,231)(22,232)(23,233)(24,234)(25,235)(26,236)(27,237)(28,238)(29,239)(30,240)(31,221)(32,222)(33,223)(34,224)(35,225)(36,226)(37,227)(38,228)(39,229)(40,230)(41,162)(42,163)(43,164)(44,165)(45,166)(46,167)(47,168)(48,169)(49,170)(50,171)(51,172)(52,173)(53,174)(54,175)(55,176)(56,177)(57,178)(58,179)(59,180)(60,161)(61,150)(62,151)(63,152)(64,153)(65,154)(66,155)(67,156)(68,157)(69,158)(70,159)(71,160)(72,141)(73,142)(74,143)(75,144)(76,145)(77,146)(78,147)(79,148)(80,149)(81,105)(82,106)(83,107)(84,108)(85,109)(86,110)(87,111)(88,112)(89,113)(90,114)(91,115)(92,116)(93,117)(94,118)(95,119)(96,120)(97,101)(98,102)(99,103)(100,104)(121,186)(122,187)(123,188)(124,189)(125,190)(126,191)(127,192)(128,193)(129,194)(130,195)(131,196)(132,197)(133,198)(134,199)(135,200)(136,181)(137,182)(138,183)(139,184)(140,185)(201,211)(202,212)(203,213)(204,214)(205,215)(206,216)(207,217)(208,218)(209,219)(210,220)>;
G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240), (1,67,146)(2,68,147)(3,69,148)(4,70,149)(5,71,150)(6,72,151)(7,73,152)(8,74,153)(9,75,154)(10,76,155)(11,77,156)(12,78,157)(13,79,158)(14,80,159)(15,61,160)(16,62,141)(17,63,142)(18,64,143)(19,65,144)(20,66,145)(21,207,221)(22,208,222)(23,209,223)(24,210,224)(25,211,225)(26,212,226)(27,213,227)(28,214,228)(29,215,229)(30,216,230)(31,217,231)(32,218,232)(33,219,233)(34,220,234)(35,201,235)(36,202,236)(37,203,237)(38,204,238)(39,205,239)(40,206,240)(41,135,88)(42,136,89)(43,137,90)(44,138,91)(45,139,92)(46,140,93)(47,121,94)(48,122,95)(49,123,96)(50,124,97)(51,125,98)(52,126,99)(53,127,100)(54,128,81)(55,129,82)(56,130,83)(57,131,84)(58,132,85)(59,133,86)(60,134,87)(101,189,171)(102,190,172)(103,191,173)(104,192,174)(105,193,175)(106,194,176)(107,195,177)(108,196,178)(109,197,179)(110,198,180)(111,199,161)(112,200,162)(113,181,163)(114,182,164)(115,183,165)(116,184,166)(117,185,167)(118,186,168)(119,187,169)(120,188,170), (1,91,207,105)(2,92,208,106)(3,93,209,107)(4,94,210,108)(5,95,211,109)(6,96,212,110)(7,97,213,111)(8,98,214,112)(9,99,215,113)(10,100,216,114)(11,81,217,115)(12,82,218,116)(13,83,219,117)(14,84,220,118)(15,85,201,119)(16,86,202,120)(17,87,203,101)(18,88,204,102)(19,89,205,103)(20,90,206,104)(21,193,146,44)(22,194,147,45)(23,195,148,46)(24,196,149,47)(25,197,150,48)(26,198,151,49)(27,199,152,50)(28,200,153,51)(29,181,154,52)(30,182,155,53)(31,183,156,54)(32,184,157,55)(33,185,158,56)(34,186,159,57)(35,187,160,58)(36,188,141,59)(37,189,142,60)(38,190,143,41)(39,191,144,42)(40,192,145,43)(61,132,235,169)(62,133,236,170)(63,134,237,171)(64,135,238,172)(65,136,239,173)(66,137,240,174)(67,138,221,175)(68,139,222,176)(69,140,223,177)(70,121,224,178)(71,122,225,179)(72,123,226,180)(73,124,227,161)(74,125,228,162)(75,126,229,163)(76,127,230,164)(77,128,231,165)(78,129,232,166)(79,130,233,167)(80,131,234,168), (1,11)(2,12)(3,13)(4,14)(5,15)(6,16)(7,17)(8,18)(9,19)(10,20)(21,231)(22,232)(23,233)(24,234)(25,235)(26,236)(27,237)(28,238)(29,239)(30,240)(31,221)(32,222)(33,223)(34,224)(35,225)(36,226)(37,227)(38,228)(39,229)(40,230)(41,162)(42,163)(43,164)(44,165)(45,166)(46,167)(47,168)(48,169)(49,170)(50,171)(51,172)(52,173)(53,174)(54,175)(55,176)(56,177)(57,178)(58,179)(59,180)(60,161)(61,150)(62,151)(63,152)(64,153)(65,154)(66,155)(67,156)(68,157)(69,158)(70,159)(71,160)(72,141)(73,142)(74,143)(75,144)(76,145)(77,146)(78,147)(79,148)(80,149)(81,105)(82,106)(83,107)(84,108)(85,109)(86,110)(87,111)(88,112)(89,113)(90,114)(91,115)(92,116)(93,117)(94,118)(95,119)(96,120)(97,101)(98,102)(99,103)(100,104)(121,186)(122,187)(123,188)(124,189)(125,190)(126,191)(127,192)(128,193)(129,194)(130,195)(131,196)(132,197)(133,198)(134,199)(135,200)(136,181)(137,182)(138,183)(139,184)(140,185)(201,211)(202,212)(203,213)(204,214)(205,215)(206,216)(207,217)(208,218)(209,219)(210,220) );
G=PermutationGroup([(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140),(141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160),(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180),(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200),(201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220),(221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240)], [(1,67,146),(2,68,147),(3,69,148),(4,70,149),(5,71,150),(6,72,151),(7,73,152),(8,74,153),(9,75,154),(10,76,155),(11,77,156),(12,78,157),(13,79,158),(14,80,159),(15,61,160),(16,62,141),(17,63,142),(18,64,143),(19,65,144),(20,66,145),(21,207,221),(22,208,222),(23,209,223),(24,210,224),(25,211,225),(26,212,226),(27,213,227),(28,214,228),(29,215,229),(30,216,230),(31,217,231),(32,218,232),(33,219,233),(34,220,234),(35,201,235),(36,202,236),(37,203,237),(38,204,238),(39,205,239),(40,206,240),(41,135,88),(42,136,89),(43,137,90),(44,138,91),(45,139,92),(46,140,93),(47,121,94),(48,122,95),(49,123,96),(50,124,97),(51,125,98),(52,126,99),(53,127,100),(54,128,81),(55,129,82),(56,130,83),(57,131,84),(58,132,85),(59,133,86),(60,134,87),(101,189,171),(102,190,172),(103,191,173),(104,192,174),(105,193,175),(106,194,176),(107,195,177),(108,196,178),(109,197,179),(110,198,180),(111,199,161),(112,200,162),(113,181,163),(114,182,164),(115,183,165),(116,184,166),(117,185,167),(118,186,168),(119,187,169),(120,188,170)], [(1,91,207,105),(2,92,208,106),(3,93,209,107),(4,94,210,108),(5,95,211,109),(6,96,212,110),(7,97,213,111),(8,98,214,112),(9,99,215,113),(10,100,216,114),(11,81,217,115),(12,82,218,116),(13,83,219,117),(14,84,220,118),(15,85,201,119),(16,86,202,120),(17,87,203,101),(18,88,204,102),(19,89,205,103),(20,90,206,104),(21,193,146,44),(22,194,147,45),(23,195,148,46),(24,196,149,47),(25,197,150,48),(26,198,151,49),(27,199,152,50),(28,200,153,51),(29,181,154,52),(30,182,155,53),(31,183,156,54),(32,184,157,55),(33,185,158,56),(34,186,159,57),(35,187,160,58),(36,188,141,59),(37,189,142,60),(38,190,143,41),(39,191,144,42),(40,192,145,43),(61,132,235,169),(62,133,236,170),(63,134,237,171),(64,135,238,172),(65,136,239,173),(66,137,240,174),(67,138,221,175),(68,139,222,176),(69,140,223,177),(70,121,224,178),(71,122,225,179),(72,123,226,180),(73,124,227,161),(74,125,228,162),(75,126,229,163),(76,127,230,164),(77,128,231,165),(78,129,232,166),(79,130,233,167),(80,131,234,168)], [(1,11),(2,12),(3,13),(4,14),(5,15),(6,16),(7,17),(8,18),(9,19),(10,20),(21,231),(22,232),(23,233),(24,234),(25,235),(26,236),(27,237),(28,238),(29,239),(30,240),(31,221),(32,222),(33,223),(34,224),(35,225),(36,226),(37,227),(38,228),(39,229),(40,230),(41,162),(42,163),(43,164),(44,165),(45,166),(46,167),(47,168),(48,169),(49,170),(50,171),(51,172),(52,173),(53,174),(54,175),(55,176),(56,177),(57,178),(58,179),(59,180),(60,161),(61,150),(62,151),(63,152),(64,153),(65,154),(66,155),(67,156),(68,157),(69,158),(70,159),(71,160),(72,141),(73,142),(74,143),(75,144),(76,145),(77,146),(78,147),(79,148),(80,149),(81,105),(82,106),(83,107),(84,108),(85,109),(86,110),(87,111),(88,112),(89,113),(90,114),(91,115),(92,116),(93,117),(94,118),(95,119),(96,120),(97,101),(98,102),(99,103),(100,104),(121,186),(122,187),(123,188),(124,189),(125,190),(126,191),(127,192),(128,193),(129,194),(130,195),(131,196),(132,197),(133,198),(134,199),(135,200),(136,181),(137,182),(138,183),(139,184),(140,185),(201,211),(202,212),(203,213),(204,214),(205,215),(206,216),(207,217),(208,218),(209,219),(210,220)])
180 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 3 4A 4B 4C 4D 4E 4F 4G ··· 4L 5A 5B 5C 5D 6A ··· 6G 10A ··· 10L 10M ··· 10T 10U ··· 10AB 12A ··· 12H 15A 15B 15C 15D 20A ··· 20P 20Q ··· 20X 20Y ··· 20AV 30A ··· 30AB 60A ··· 60AF order 1 2 2 2 2 2 2 2 3 4 4 4 4 4 4 4 ··· 4 5 5 5 5 6 ··· 6 10 ··· 10 10 ··· 10 10 ··· 10 12 ··· 12 15 15 15 15 20 ··· 20 20 ··· 20 20 ··· 20 30 ··· 30 60 ··· 60 size 1 1 1 1 2 2 6 6 2 1 1 1 1 2 2 6 ··· 6 1 1 1 1 2 ··· 2 1 ··· 1 2 ··· 2 6 ··· 6 2 ··· 2 2 2 2 2 1 ··· 1 2 ··· 2 6 ··· 6 2 ··· 2 2 ··· 2
180 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 type + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C2 C2 C4 C5 C10 C10 C10 C10 C10 C10 C10 C20 S3 D4 D6 D6 C4○D4 C3⋊D4 C4×S3 C5×S3 C5×D4 C4○D12 S3×C10 S3×C10 C5×C4○D4 C5×C3⋊D4 S3×C20 C5×C4○D12 kernel C20×C3⋊D4 Dic3×C20 C5×Dic3⋊C4 C5×D6⋊C4 C5×C6.D4 S3×C2×C20 C10×C3⋊D4 C22×C60 C5×C3⋊D4 C4×C3⋊D4 C4×Dic3 Dic3⋊C4 D6⋊C4 C6.D4 S3×C2×C4 C2×C3⋊D4 C22×C12 C3⋊D4 C22×C20 C60 C2×C20 C22×C10 C30 C20 C2×C10 C22×C4 C12 C10 C2×C4 C23 C6 C4 C22 C2 # reps 1 1 1 1 1 1 1 1 8 4 4 4 4 4 4 4 4 32 1 2 2 1 2 4 4 4 8 4 8 4 8 16 16 16
Matrix representation of C20×C3⋊D4 in GL3(𝔽61) generated by
11 0 0 0 41 0 0 0 41
,
1 0 0 0 0 60 0 1 60
,
60 0 0 0 18 52 0 9 43
,
1 0 0 0 0 1 0 1 0
G:=sub<GL(3,GF(61))| [11,0,0,0,41,0,0,0,41],[1,0,0,0,0,1,0,60,60],[60,0,0,0,18,9,0,52,43],[1,0,0,0,0,1,0,1,0] >;
C20×C3⋊D4 in GAP, Magma, Sage, TeX
C_{20}\times C_3\rtimes D_4
% in TeX
G:=Group("C20xC3:D4");
// GroupNames label
G:=SmallGroup(480,807);
// by ID
G=gap.SmallGroup(480,807);
# by ID
G:=PCGroup([7,-2,-2,-2,-5,-2,-2,-3,1149,226,15686]);
// Polycyclic
G:=Group<a,b,c,d|a^20=b^3=c^4=d^2=1,a*b=b*a,a*c=c*a,a*d=d*a,c*b*c^-1=d*b*d=b^-1,d*c*d=c^-1>;
// generators/relations
×
𝔽 | 10,285 | 18,819 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 2.5625 | 3 | CC-MAIN-2020-16 | longest | en | 0.375905 |
https://garage-door-repair-austin.com/qa/quick-answer-what-does-a-large-r-value-mean.html | 1,624,083,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00057.warc.gz | 256,038,370 | 12,756 | # Quick Answer: What Does A Large R-Value Mean?
## What is the thinnest insulation with the highest R-value?
Aerogel insulationAerogel insulation offers the highest R-value of any insulating material at less weight and thickness—ideal for construction, refineries, pipelines, and thin-gap thermal barriers.
Browse our superior selection of superinsulating aerogel blankets below..
## Can R value be too high?
Yes, can be too hot. Depends on you, your system, and the weather. An R of 3 is too hot for me most of the time, but I put off a lot of heat. Buy them both, use the one which works for you.
## What does R mean in statistics?
The sample correlation coefficient (r) is a measure of the closeness of association of the points in a scatter plot to a linear regression line based on those points, as in the example above for accumulated saving over time.
## What is a high R 2 value?
For the same data set, higher R-squared values represent smaller differences between the observed data and the fitted values. R-squared is the percentage of the dependent variable variation that a linear model explains. … The mean of the dependent variable predicts the dependent variable as well as the regression model.
## What is a large R-value?
R-value was defined as, “The higher the R-value, the greater the ability to resist conductive heat transfer.” In practical terms, this means that the thicker the insulation, the more efficiently you can keep your home comfortable.
## Is a large R Squared good?
In general, the higher the R-squared, the better the model fits your data.
## Is a high R-value good or bad?
The higher the R-value of a material, the better it insulates from heat and cold. The R-value of insulation depends on the type of material, its thickness, and its density.
## What is a good R squared value?
While for exploratory research, using cross sectional data, values of 0.10 are typical. In scholarly research that focuses on marketing issues, R2 values of 0.75, 0.50, or 0.25 can, as a rough rule of thumb, be respectively described as substantial, moderate, or weak.
## What does an r2 value of 0.9 mean?
Essentially, an R-Squared value of 0.9 would indicate that 90% of the variance of the dependent variable being studied is explained by the variance of the independent variable.
## What r2 value is considered a strong correlation?
– if R-squared value 0.3 < r < 0.5 this value is generally considered a weak or low effect size, - if R-squared value 0.5 < r < 0.7 this value is generally considered a Moderate effect size, - if R-squared value r > 0.7 this value is generally considered strong effect size, Ref: Source: Moore, D. S., Notz, W.
## Which type of insulation has the highest R-value?
-cell foamHas two types: open-cell foam and closed-cell foam. Closed-cell foam is denser and thus has a higher R-value. Closed-cell foam is usually more expensive than open-cell foam. Open-cell foam has an approximate R-value of R-3.7 per inch of thickness.
## What is the R value of 1 inch Styrofoam?
STYROFOAM extruded polystyrene insulation is rated at an R-value of 5.0 per inch at 75°F. The material R-value for the entire wall assembly is derived by adding an R-value of Styrofoam0.
## Can you over insulate a house?
It is possible to over-insulate your house so much that it can’t breathe. The whole point of home insulation is to tightly seal your home’s interior. But if it becomes too tightly sealed with too many layers of insulation, moisture can get trapped inside those layers. That’s when mold starts to grow.
## What does an R-squared value of 1 mean?
R2 is a statistic that will give some information about the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression predictions approximate the real data points. An R2 of 1 indicates that the regression predictions perfectly fit the data.
## Why is my R-Squared so low?
A low R-squared value indicates that your independent variable is not explaining much in the variation of your dependent variable – regardless of the variable significance, this is letting you know that the identified independent variable, even though significant, is not accounting for much of the mean of your …
## What is a decent R-Squared?
Any study that attempts to predict human behavior will tend to have R-squared values less than 50%. However, if you analyze a physical process and have very good measurements, you might expect R-squared values over 90%.
## How do you interpret an R-value?
To interpret its value, see which of the following values your correlation r is closest to:Exactly –1. A perfect downhill (negative) linear relationship.–0.70. A strong downhill (negative) linear relationship.–0.50. A moderate downhill (negative) relationship.–0.30. … No linear relationship.+0.30. … +0.50. … +0.70.More items…
## Does house wrap have an R-value?
The next revolution in weatherization technology—one breathable product for air, water and thermal protection. DuPont™ Tyvek® ThermaWrap™ R5. 0 offers the air and water management benefits of all DuPont™ Tyvek® weather barriers with an R-value of 5.0.
## What material has the highest R-value?
Vacuum insulated panels have the highest R-value, approximately R-45 (in U.S. units) per inch; aerogel has the next highest R-value (about R-10 to R-30 per inch), followed by polyurethane (PUR) and phenolic foam insulations with R-7 per inch.
## What does an r2 value of 0.5 mean?
An R2 of 1.0 indicates that the data perfectly fit the linear model. Any R2 value less than 1.0 indicates that at least some variability in the data cannot be accounted for by the model (e.g., an R2 of 0.5 indicates that 50% of the variability in the outcome data cannot be explained by the model).
## Can R-Squared be above 1?
Bottom line: R2 can be greater than 1.0 only when an invalid (or nonstandard) equation is used to compute R2 and when the chosen model (with constraints, if any) fits the data really poorly, worse than the fit of a horizontal line. | 1,418 | 6,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-25 | latest | en | 0.922987 |
https://www.jiskha.com/questions/886426/Ted-and-Suzanne-load-3476-songs-into-a-player-that-randomly-selects-which-song-to | 1,561,069,468,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00103.warc.gz | 803,678,993 | 5,089 | # maths
Ted and Suzanne load 3476 songs into a player that randomly selects which song to play.
When in random-selection mode, 22 of the first 30 songs were loaded by Suzanne.
T and S want to determine if each have loaded a different proportion of songs.
Let p denote proportion of songs loaded by Suzanne.
A) State null and alternative hypotheses
B) How strong is the evidence that Ted and Suzanne have each loaded a different proportion of songs?
1. 👍 0
2. 👎 0
3. 👁 64
1. fdfas
1. 👍 0
2. 👎 0
posted by fda
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