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https://www.coursehero.com/file/6659835/elemprob-fall2010-page32/ | 1,481,153,313,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542250.48/warc/CC-MAIN-20161202170902-00137-ip-10-31-129-80.ec2.internal.warc.gz | 941,252,151 | 67,499 | elemprob-fall2010-page32
# elemprob-fall2010-page32 - e-x ) = 1-F X ( e-x ) . Taking...
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An exponential is the time for something to occur. A gamma is the time for t events to occur. A gamma with parameters 1 2 and n 2 is known as a χ 2 n , a chi-squared random variable with n degrees of freedom. Gammas and chi- squared’s come up frequently in statistics. Another distribution that arises in statistics is the beta: f ( x ) = 1 B ( a,b ) x a - 1 (1 - x ) b - 1 , 0 < x < 1 , where B ( a,b ) = R 1 0 x a - 1 (1 - x ) b - 1 . Cauchy . Here f ( x ) = 1 π 1 1 + ( x - θ ) 2 . What is interesting about the Cauchy is that it does not have finite mean, that is, E | X | = . Often it is important to be able to compute the density of Y = g ( X ). Let us give a couple of examples. If X is uniform on (0 , 1] and Y = - log X , then Y > 0. If x > 0, F Y ( x ) = P ( Y x ) = P ( - log X x ) = P (log X ≥ - x ) = P ( X e - x ) = 1 - P ( X
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Unformatted text preview: e-x ) = 1-F X ( e-x ) . Taking the derivative, f Y ( x ) = d dx F Y ( x ) =-f X ( e-x )(-e-x ) , using the chain rule. Since f X = 1, this gives f Y ( x ) = e-x , or Y is exponential with parameter 1. For another example, suppose X is N (0 , 1) and Y = X 2 . Then F Y ( x ) = P ( Y x ) = P ( X 2 x ) = P (- x X x ) = P ( X x )-P ( X - x ) = F X ( x )-F X (- x ) . Taking the derivative and using the chain rule, f Y ( x ) = d dx F Y ( x ) = f X ( x ) 1 2 x -f X (- x ) -1 2 x . 32...
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Ask a homework question - tutors are online | 575 | 1,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-50 | longest | en | 0.857318 |
https://stats.stackexchange.com/questions/408150/bayesian-methods-are-about-averaging-over-uncertainty-rather-than-optimization | 1,561,014,931,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00075.warc.gz | 592,773,899 | 36,012 | # Bayesian methods are about averaging over uncertainty rather than optimization. Explain?
I came across the statement "The key ingredient in Bayesian methods is to average over your uncertain variables and parameters, rather than to optimize". Can someone explain why this is?
My confusion is that in Bayesian model selection, if you compute the model evidence, surely that involves some element of optimization... you then compute the posterior probability to get the Bayes factor, for example, but it seems there is some element of optimization still. Is this incorrect?
• @Xi'an This is from a lecture by Gharamani but I've seen related statements by Gelman, MacKay etc – questionmark May 14 at 10:45
When doing model selection, each model $$\mathfrak M$$ is given an evidence $$\mathfrak e(\mathfrak M)$$ that writes as the corresponding integrated likelihood $$\mathfrak e(\mathfrak M) = \int f_{\mathfrak M}(\mathbf x|\theta_{\mathfrak M})\,\text{d}\theta_{\mathfrak M}$$Each model is then given a posterior probability $$\pi(\mathfrak M|\mathbf x)$$. The decision to select a model, if need be, is based on the maximisation of a utility function $$\arg\max_{\mathfrak M} \mathbb E[U(\mathfrak M,\theta_{\mathfrak M})|\mathbf x]$$Hence, it is correct that in the decision, optimisation occurs. | 327 | 1,304 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-26 | longest | en | 0.874521 |
http://www.aimsuccess.in/2017/09/questions-asked-in-ibps-rrb-office_17.html | 1,516,453,877,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889617.56/warc/CC-MAIN-20180120122736-20180120142736-00246.warc.gz | 404,760,854 | 72,844 | # Questions Asked in IBPS RRB Office Assistant Prelims - 17 September 2017 (All Slots)
## Questions Asked in IBPS RRB Office Assistant Prelims - 17 September 2017 (All Slots)
Dear Readers, The IBPS RRB Office Assistant exam is started from today. The exam will took place on 16th, 17th. 23rd & 24th September 2017 in 4 different slots across various cities. In this article, we are sharing the Questions Asked in IBPS RRB Office Assistant Prelims 16th Sep 2017 (All Slots) which is important to analyse your exam. This will be beneficial for the aspirants who are going to appear for IBPS RRB Office Assistant Prelims Exam on further dates Moreover, you can also know the number of good attempts that will keep you at safer side. So go through the Questions Asked in IBPS RRB Office Assistant Prelims - 16 September 2017 (All Slots)
## QUESTIONS ASKED IN SHIFT 01/SLOT 01: 17 SEP 2017
### Questions Asked from Quantitative Aptitude
1. Floor Based Puzzle :
2. Linear Sitting Arrangement :
• Seven people H, I, J, K, L, M and N are standing in a straight line facing north, with equal distance between them, not necessarily in the same order. L is second to the left of H. H sits at one of the ends J is third to the right of K. M is not an immediate neighbour of l. K is second to the left of L. I is second left.
### Questions Asked from Reasoning Ability
1. Data Interpretation : Tabular Graph , based on the day-wise calculation of 5 entities (A, B, C, D, & E).
2. Simple Ratio, Percentage and average based questions were asked
3. Number Series :
1. 6,8,13,23,?,56 - Logic: Difference of Difference 3, 5, 7 ANSWER : 40
2. 7,8,18,57,232,? - Logic: *1+1 ,*2+ 2, *3+3 ANSWER : 1165
3. 8,5,6,10,21,? - Logic:*0.5 +1, *1+1, *1.5 +1 ANSWER : 53.5
4. 4,18,46,102,?,438 - Logic: difference 14,28,56 ANSWER : 204
5. 109,110,102,129,165,? - Logic: Difference of 13, 23, 33, 43, 53 ANSWER : 290
4. Sum of money invested in simple interest for 4 years at 7.5% per annum, interest earned 720 Rs, if the same sum invested for 2 years at 5%, the total amount will be?
5. The ratio between X% of 240 and 75% of Y is 8:5, and X% of Y=98. Then find the value of Y?
## QUESTIONS ASKED IN SHIFT 02/SLOT 02: 17 SEP 2017
### Questions Asked from Reasoning Ability
1. Puzzle :
• Seven people I, J, K, L, M, N, O lives in 7 different floors. First floor numbered as 1 and so on top floor numbered as 7. J lives on odd floor above floor number 3. Two people live between J and L. Four people live between H and N. H lives above N. M lives on odd numbered floor above I.
2. Linear Sitting Arrangement – All people facing north & sitting in a single row.
### Questions Asked from Quantitative Aptitude
1. Number Series :
• 10, 4, 3, 3.5, 6, ? ANSWER - 14
• 6, 5, 8, 21, 80, ? ANSWER - 395
• 98, 111, 85, 137, ?, 241 ANSWER - 215
• 7, 8, 16, 43, 107, ? ANSWER - 232
• 7, 7, 10, 18, ?, 57 ANSWER - 33
2. A man spends 25% his salary for house hold expenses 15 % for mutual fund and 10% of remaining for miscellaneous expenses, and the remaining he saves. What is his savings? If his salary is Rs.1200.00
## QUESTIONS ASKED IN SHIFT 03/SLOT 03: 17 SEP 2017
### Questions Asked from Reasoning Ability
1. Puzzle
• H,I,J,K,L,M and N are living in a seven store building. The lowermost floor is numbered one and the topmost floor is numbered seven. K lives in a even numbered floor below 6th floor. There are only tw floors between M and K. M lives in one of the floor above K. H lives immediately above J, who lives in an odd numbered floor. Number of floors above M is same as below Y. L lives immediately above N.
2. Sitting Arrangement
• S,T,U,V,W,X & Y are sitting in a straight line and all are facing north but not in same order. T sits third to the right of X. More than one person sit to the right of T. W sits second to the left of Y. Y is not an immediate neighbor of T. More than one person sit between Y and S. Only Two person sit between S and U. V is not an immediate neighbor of X.
3. Syllogism questions were straight and aspirants could have solved it in within the respective time frame had by their strategies. Overall the reasoning questions were easier and many would have solved more than 35.
### Questions Asked from Quantitative Aptitude
1. Number Series :
• 7 6 10 36 280 ? ANSWER - 4464
• 13 7 8 13.5 29 ? ANSWER - 75
• 4 7 16 43 124 ? ANSWER - 367
• 69 67 62 52 ? 9 ANSWER - 35
• 614 78 294 806 ? ANSWER -
## QUESTIONS ASKED IN SHIFT 04/SLOT 04: 17 SEP 2017
### Questions Asked from Reasoning Ability
1. Linear Arrangement - 7 people facing North
2. Puzzle - (7 people lives in different floor)
3. Syllogism - 3 statement & 2 conclusion
4. Inequality - Direct Question -with 2 conclusion
5. Numeric Series - 7 8 5 6 4 3 2 1 8 7 5 4 - number based
6. Coding Decoding - Alphabet/ Symbol coding with three condition
### Questions Asked from Quantitative Aptitude
1. Number Series -
• 1. 266,267,240,365,?
• 2. 2,6,22,86,342,?
• 3. 17,8,7,9,16,40,?
• 4. 78,78,75,67,?
2. Data Interpretation - Tabular based - 6 bakery sales on different days
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3.7 Set Types -- set, frozenset
# 3.7 Set Types -- set, frozenset
A set object is an unordered collection of immutable values. Common uses include membership testing, removing duplicates from a sequence, and computing mathematical operations such as intersection, union, difference, and symmetric difference. New in version 2.4.
Like other collections, sets support `x in set`, `len(set)`, and `for x in set`. Being an unordered collection, sets do not record element position or order of insertion. Accordingly, sets do not support indexing, slicing, or other sequence-like behavior.
There are currently two builtin set types, set and frozenset. The set type is mutable -- the contents can be changed using methods like add() and remove(). Since it is mutable, it has no hash value and cannot be used as either a dictionary key or as an element of another set. The frozenset type is immutable and hashable -- its contents cannot be altered after is created; however, it can be used as a dictionary key or as an element of another set.
Instances of set and frozenset provide the following operations:
Operation Equivalent Result
`len(s)` cardinality of set s
`x in s` test x for membership in s
`x not in s` test x for non-membership in s
`s.issubset(t)` `s <= t` test whether every element in s is in t
`s.issuperset(t)` `s >= t` test whether every element in t is in s
`s.union(t)` s | t new set with elements from both s and t
`s.intersection(t)` s & t new set with elements common to s and t
`s.difference(t)` s - t new set with elements in s but not in t
`s.symmetric_difference(t)` s ^ t new set with elements in either s or t but not both
`s.copy()` new set with a shallow copy of s
Note, the non-operator versions of union(), intersection(), difference(), and symmetric_difference(), issubset(), and issuperset() methods will accept any iterable as an argument. In contrast, their operator based counterparts require their arguments to be sets. This precludes error-prone constructions like `set('abc') & 'cbs'` in favor of the more readable `set('abc').intersection('cbs')`.
Both set and frozenset support set to set comparisons. Two sets are equal if and only if every element of each set is contained in the other (each is a subset of the other). A set is less than another set if and only if the first set is a proper subset of the second set (is a subset, but is not equal). A set is greater than another set if and only if the first set is a proper superset of the second set (is a superset, but is not equal).
Instances of set are compared to instances of frozenset based on their members. For example, "set('abc') == frozenset('abc')" returns `True`.
The subset and equality comparisons do not generalize to a complete ordering function. For example, any two disjoint sets are not equal and are not subsets of each other, so all of the following return `False`: `a<b`, `a==b`, or `a>b`. Accordingly, sets do not implement the __cmp__ method.
Since sets only define partial ordering (subset relationships), the output of the list.sort() method is undefined for lists of sets.
Set elements are like dictionary keys; they need to define both __hash__ and __eq__ methods.
Binary operations that mix set instances with frozenset return the type of the first operand. For example: "frozenset('ab') | set('bc')" returns an instance of frozenset.
The following table lists operations available for set that do not apply to immutable instances of frozenset:
Operation Equivalent Result
`s.update(t)` s |= t update set s, adding elements from t
`s.intersection_update(t)` s &= t update set s, keeping only elements found in both s and t
`s.difference_update(t)` s -= t update set s, removing elements found in t
`s.symmetric_difference_update(t)` s ^= t update set s, keeping only elements found in either s or t but not in both
`s.add(x)` add element x to set s
`s.remove(x)` remove x from set s; raises KeyError if not present
`s.discard(x)` removes x from set s if present
`s.pop()` remove and return an arbitrary element from s; raises KeyError if empty
`s.clear()` remove all elements from set s
Note, the non-operator versions of the update(), intersection_update(), difference_update(), and symmetric_difference_update() methods will accept any iterable as an argument.
The design of the set types was based on lessons learned from the sets module. | 1,100 | 4,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-40 | longest | en | 0.682443 |
https://plainmath.net/17700/related-functions-described-chapter-describe-sequence-transformations | 1,642,817,489,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00476.warc.gz | 466,872,067 | 7,768 | # h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. h(x) = (x - 2)^3 + 2
Joni Kenny 2021-06-12 Answered
h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. $$h(x) = (x - 2)^3 + 2$$
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## Expert Answer
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Answered 2021-06-13 Author has 21292 answers
(b)We are starting with the parent function $$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}$$
STEP 1: Shift the graph hy 2 units to the right, to get $$\displaystyle{y}={\left({x}-{2}\right)}^{{3}}$$
STEP 2: Shift the graph hy 2 units upwards, to get $$\displaystyle{y}={\left({\left({x}-{2}\right)}^{{3}}\right)}+{2}$$, which is the required function h(x) Shift the graph by 2 units the right and 2 units upwards
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... | 402 | 1,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-05 | latest | en | 0.84296 |
https://mcqquestions.guru/rs-aggarwal-class-6-solutions-chapter-3-whole-numbers-ex-3e/ | 1,660,641,318,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00088.warc.gz | 357,010,389 | 13,516 | # RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E
## RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E
These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E.
Other Exercises
Question 1.
Solution:
(i) By actual division, we have
Question 2.
Solution:
(i) By actual division, we have
Question 3.
Solution:
(i) We know that any number (non-zero) divided by 1 gives the number itself
65007 ÷ 1 = 65007
(ii) We know that 0 divided by any natural number gives 0
0 ÷ 879 = 0
(iii) 981 + 5720 ÷ 10 = 981 + (5720 ÷ 10)
= 981 + 572
= 1553
(iv) 1507 – 625 ÷ 25 = 1507 – (625 ÷ 25)
= 1507 – 25
= 1482
(v) 32277 ÷ (648 – 39)
= 32277 ÷ 609
32277 ÷ (648 – 39) = 53
(vi) 1573 ÷ 1573 – 1573 ÷ 1573
= (1573 ÷ 1573) – (1573 ÷ 1573)
= 1 – 1
= 0
Question 4.
Solution:
We have n ÷ n = n
let n = 1, 1 ÷ 1 = 1
1 = 1
which is true
∴ Hence 1 is the required whole number.
Question 5.
Solution:
Product of two numbers = 504347
One number = 317
Other number = 504347 ÷ 317
∴ Other number = 1591
Question 6.
Solution:
Here Dividend = 59761, Quotient = 189
∴ Remainder = 37
We know that Dividend = Divisor x Quotient + Remainder
59761 = Divisor x 189 + 37
59761 – 37 = Divisor x 189
59724 = Divisor x 189
Divisor x 189 = 59724
∴ Divisor = 59724 ÷ 189
∴ Divisor = 59724 ÷ 189 = 316
Question 7.
Solution:
Here dividend = 55390,
Divisor = 299 and Remainder = 75
By division algorithm, we have
Dividend = Quotient x Divisor + Remainder
55390 = Quotient x 299 + 75
55390 – 75 = Quotient x 299
55315 = Quotient x 299
Quotient x 299 = 55315
Quotient = 55315 ÷ 299
∴ Required quotient = 185
Question 8.
Solution:
On dividing 13601 by 87, we have
It is clear that if we subtract 29 from 13601, the resulting number will be exactly divisible by 87.
∴ The required least number = 29.
Question 9.
Solution:
Here dividend = 1056, Divisor = 23
By actual division, we have
It is clear that if we add 2 to 21, it will become 23 which is divisible by 23.
∴ Required least number = 2.
Question 10.
Solution:
Greatest 4-digit number = 9999
On, dividing by 16, we get remainder as 15
∴ The required largest 4-digit number = 9999 – 15
= 9984
Question 11.
Solution:
Largest number of 5-digits = 99999
On dividing 99999 by 653, we have
∴ Quotient = 153, Remainder = 90
Check : By division algorithm Dividend = Divisor x Quotient + Remainder
= 653 x 153 + 90
= 99909 + 90
= 99999
Question 12.
Solution:
The least 6-digit number = 100000
On dividing 100000 by 83, we have
It is clear that if we add 15 to 68, it will become 83 which is divisible by 83.
∴ Required least 6-digit number = 100000 + 15
= 100015
Question 13.
Solution:
Cost of 1 dozen bananas = Rs. 29
Bananas can be purchase in Rs. 1392
1392 ÷ 29
= 48 dozens
Question 14.
Solution:
Total number of trees = 19625
Total number of rows = 157
Number of trees in each row = 19625 ÷ 157
∴ Number of trees in each row = 125
Question 15.
Solution:
Total population of the town = 517530
Since there is one educated person out of 15
Total number of educated persons in the town = 517530 ÷ 15
∴ Total number of educated persons in the town = 34502.
Question 16.
Solution:
Cost of 23 colour TV sets = Rs. 570055
Cost of 1 colour TV set
∴ Cost of 1 color TV set
= Rs. 570055 ÷ 23
= Rs. 24785
Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | 1,241 | 3,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2022-33 | latest | en | 0.771196 |
http://isabelle.in.tum.de/repos/isabelle/diff/c94eefffc3a5/src/ZF/int_arith.ML | 1,575,968,209,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527010.70/warc/CC-MAIN-20191210070602-20191210094602-00335.warc.gz | 71,804,875 | 4,762 | src/ZF/int_arith.ML
changeset 27237 c94eefffc3a5 parent 27154 026f3db3f5c6 child 29269 5c25a2012975
``` 1.1 --- a/src/ZF/int_arith.ML Mon Jun 16 17:54:50 2008 +0200
1.2 +++ b/src/ZF/int_arith.ML Mon Jun 16 17:54:51 2008 +0200
1.3 @@ -6,101 +6,15 @@
1.4 Simprocs for linear arithmetic.
1.5 *)
1.6
1.7 -
1.8 -(** To simplify inequalities involving integer negation and literals,
1.9 - such as -x = #3
1.10 -**)
1.11 -
1.12 -Addsimps [OldGoals.inst "y" "integ_of(?w)" @{thm zminus_equation},
1.13 - OldGoals.inst "x" "integ_of(?w)" @{thm equation_zminus}];
1.14 -
1.15 -AddIffs [OldGoals.inst "y" "integ_of(?w)" @{thm zminus_zless},
1.16 - OldGoals.inst "x" "integ_of(?w)" @{thm zless_zminus}];
1.17 -
1.18 -AddIffs [OldGoals.inst "y" "integ_of(?w)" @{thm zminus_zle},
1.19 - OldGoals.inst "x" "integ_of(?w)" @{thm zle_zminus}];
1.20 -
1.21 -Addsimps [OldGoals.inst "s" "integ_of(?w)" @{thm Let_def}];
1.22 -
1.23 -(*** Simprocs for numeric literals ***)
1.24 -
1.25 -(** Combining of literal coefficients in sums of products **)
1.26 -
1.27 -Goal "(x \$< y) <-> (x\$-y \$< #0)";
1.28 -by (simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.29 -qed "zless_iff_zdiff_zless_0";
1.30 -
1.31 -Goal "[| x: int; y: int |] ==> (x = y) <-> (x\$-y = #0)";
1.32 -by (asm_simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.33 -qed "eq_iff_zdiff_eq_0";
1.34 -
1.35 -Goal "(x \$<= y) <-> (x\$-y \$<= #0)";
1.36 -by (asm_simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.37 -qed "zle_iff_zdiff_zle_0";
1.38 -
1.39 -
1.40 -(** For combine_numerals **)
1.41 -
1.42 -Goal "i\$*u \$+ (j\$*u \$+ k) = (i\$+j)\$*u \$+ k";
1.43 -by (simp_tac (simpset() addsimps [@{thm zadd_zmult_distrib}]@ @{thms zadd_ac}) 1);
1.44 -qed "left_zadd_zmult_distrib";
1.45 -
1.46 -
1.47 -(** For cancel_numerals **)
1.48 -
1.49 -val rel_iff_rel_0_rls = map (OldGoals.inst "y" "?u\$+?v")
1.50 - [zless_iff_zdiff_zless_0, eq_iff_zdiff_eq_0,
1.51 - zle_iff_zdiff_zle_0] @
1.52 - map (OldGoals.inst "y" "n")
1.53 - [zless_iff_zdiff_zless_0, eq_iff_zdiff_eq_0,
1.54 - zle_iff_zdiff_zle_0];
1.55 -
1.56 -Goal "(i\$*u \$+ m = j\$*u \$+ n) <-> ((i\$-j)\$*u \$+ m = intify(n))";
1.57 -by (simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]) 1);
1.58 -by (simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.59 -by (simp_tac (simpset() addsimps @{thms zadd_ac}) 1);
1.60 -qed "eq_add_iff1";
1.61 -
1.62 -Goal "(i\$*u \$+ m = j\$*u \$+ n) <-> (intify(m) = (j\$-i)\$*u \$+ n)";
1.63 -by (simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]) 1);
1.64 -by (simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.65 -by (simp_tac (simpset() addsimps @{thms zadd_ac}) 1);
1.66 -qed "eq_add_iff2";
1.67 -
1.68 -Goal "(i\$*u \$+ m \$< j\$*u \$+ n) <-> ((i\$-j)\$*u \$+ m \$< n)";
1.69 -by (asm_simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]@
1.70 - @{thms zadd_ac} @ rel_iff_rel_0_rls) 1);
1.71 -qed "less_add_iff1";
1.72 -
1.73 -Goal "(i\$*u \$+ m \$< j\$*u \$+ n) <-> (m \$< (j\$-i)\$*u \$+ n)";
1.74 -by (asm_simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]@
1.75 - @{thms zadd_ac} @ rel_iff_rel_0_rls) 1);
1.76 -qed "less_add_iff2";
1.77 -
1.78 -Goal "(i\$*u \$+ m \$<= j\$*u \$+ n) <-> ((i\$-j)\$*u \$+ m \$<= n)";
1.79 -by (simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]) 1);
1.80 -by (simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.81 -by (simp_tac (simpset() addsimps @{thms zadd_ac}) 1);
1.82 -qed "le_add_iff1";
1.83 -
1.84 -Goal "(i\$*u \$+ m \$<= j\$*u \$+ n) <-> (m \$<= (j\$-i)\$*u \$+ n)";
1.85 -by (simp_tac (simpset() addsimps [@{thm zdiff_def}, @{thm zadd_zmult_distrib}]) 1);
1.86 -by (simp_tac (simpset() addsimps @{thms zcompare_rls}) 1);
1.87 -by (simp_tac (simpset() addsimps @{thms zadd_ac}) 1);
1.88 -qed "le_add_iff2";
1.89 -
1.90 -
1.91 structure Int_Numeral_Simprocs =
1.92 struct
1.93
1.94 (*Utilities*)
1.95
1.96 -val integ_of_const = Const (@{const_name "Bin.integ_of"}, @{typ "i => i"});
1.97 -
1.98 -fun mk_numeral n = integ_of_const \$ NumeralSyntax.mk_bin n;
1.99 +fun mk_numeral n = @{const integ_of} \$ NumeralSyntax.mk_bin n;
1.100
1.101 (*Decodes a binary INTEGER*)
1.102 -fun dest_numeral (Const(@{const_name "Bin.integ_of"}, _) \$ w) =
1.103 +fun dest_numeral (Const(@{const_name integ_of}, _) \$ w) =
1.104 (NumeralSyntax.dest_bin w
1.105 handle Match => raise TERM("Int_Numeral_Simprocs.dest_numeral:1", [w]))
1.106 | dest_numeral t = raise TERM("Int_Numeral_Simprocs.dest_numeral:2", [t]);
1.107 @@ -113,8 +27,6 @@
1.108 val zero = mk_numeral 0;
1.109 val mk_plus = FOLogic.mk_binop @{const_name "zadd"};
1.110
1.111 -val zminus_const = Const (@{const_name "zminus"}, @{typ "i => i"});
1.112 -
1.113 (*Thus mk_sum[t] yields t+#0; longer sums don't have a trailing zero*)
1.114 fun mk_sum [] = zero
1.115 | mk_sum [t,u] = mk_plus (t, u)
1.116 @@ -132,7 +44,7 @@
1.117 | dest_summing (pos, Const (@{const_name "zdiff"}, _) \$ t \$ u, ts) =
1.118 dest_summing (pos, t, dest_summing (not pos, u, ts))
1.119 | dest_summing (pos, t, ts) =
1.120 - if pos then t::ts else zminus_const\$t :: ts;
1.121 + if pos then t::ts else @{const zminus} \$ t :: ts;
1.122
1.123 fun dest_sum t = dest_summing (true, t, []);
1.124
1.125 @@ -245,8 +157,8 @@
1.126 val prove_conv = ArithData.prove_conv "inteq_cancel_numerals"
1.127 val mk_bal = FOLogic.mk_eq
1.128 val dest_bal = FOLogic.dest_eq
1.129 - val bal_add1 = eq_add_iff1 RS iff_trans
1.130 - val bal_add2 = eq_add_iff2 RS iff_trans
1.131 + val bal_add1 = @{thm eq_add_iff1} RS iff_trans
1.132 + val bal_add2 = @{thm eq_add_iff2} RS iff_trans
1.133 );
1.134
1.135 structure LessCancelNumerals = CancelNumeralsFun
1.136 @@ -254,8 +166,8 @@
1.137 val prove_conv = ArithData.prove_conv "intless_cancel_numerals"
1.138 val mk_bal = FOLogic.mk_binrel @{const_name "zless"}
1.139 val dest_bal = FOLogic.dest_bin @{const_name "zless"} @{typ i}
1.140 - val bal_add1 = less_add_iff1 RS iff_trans
1.141 - val bal_add2 = less_add_iff2 RS iff_trans
1.142 + val bal_add1 = @{thm less_add_iff1} RS iff_trans
1.143 + val bal_add2 = @{thm less_add_iff2} RS iff_trans
1.144 );
1.145
1.146 structure LeCancelNumerals = CancelNumeralsFun
1.147 @@ -263,8 +175,8 @@
1.148 val prove_conv = ArithData.prove_conv "intle_cancel_numerals"
1.149 val mk_bal = FOLogic.mk_binrel @{const_name "zle"}
1.150 val dest_bal = FOLogic.dest_bin @{const_name "zle"} @{typ i}
1.151 - val bal_add1 = le_add_iff1 RS iff_trans
1.152 - val bal_add2 = le_add_iff2 RS iff_trans
1.153 + val bal_add1 = @{thm le_add_iff1} RS iff_trans
1.154 + val bal_add2 = @{thm le_add_iff2} RS iff_trans
1.155 );
1.156
1.157 val cancel_numerals =
1.158 @@ -298,7 +210,7 @@
1.159 val dest_sum = dest_sum
1.160 val mk_coeff = mk_coeff
1.161 val dest_coeff = dest_coeff 1
1.162 - val left_distrib = left_zadd_zmult_distrib RS trans
1.163 + val left_distrib = @{thm left_zadd_zmult_distrib} RS trans
1.164 val prove_conv = prove_conv_nohyps "int_combine_numerals"
1.165 fun trans_tac _ = ArithData.gen_trans_tac trans
1.166
``` | 2,896 | 7,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.383982 |
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A197183 Number of partitions of n^2 into distinct factorials. 4
1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 0 COMMENTS a(n) = A115944(n^2); a(A014597(n)) > 0. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 PROG (Haskell) a197183 = a115944 . a000290 CROSSREFS Cf. A197182, A000290. Sequence in context: A322586 A147612 A323509 * A295405 A267001 A141735 Adjacent sequences: A197180 A197181 A197182 * A197184 A197185 A197186 KEYWORD nonn AUTHOR Reinhard Zumkeller, Dec 04 2011 STATUS approved
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Last modified October 20 20:24 EDT 2019. Contains 328273 sequences. (Running on oeis4.) | 552 | 1,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.627708 |
https://www.honingjs.com/challenges/leetcode/1758-minimum-changes-to-make-alternating-binary-string | 1,679,601,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00377.warc.gz | 920,107,033 | 196,317 | # LeetCode: Minimum Changes To Make Alternating Binary String Solution
## Approach
Create 2 possible output
s1
,
s2
Calculate the number of steps from
s
to each of those
Find min between 2 of them
## Implementation
`.css-ds3kc{display:table-row;}.css-1t8atru{display:table-cell;opacity:0.5;padding-right:var(--chakra-space-6);-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;text-align:right;}1.css-2qghsv{display:table-cell;}/**2 * @param {string} s3 * @return {number}4 */5var minOperations = function (s) {6 const N = s.length7 const chars = s.split("").map(Number)8 const s1 = Array.from({ length: N }, (_, i) => (i % 2 === 0 ? 1 : 0))9 const s2 = Array.from({ length: N }, (_, i) => (i % 2 === 0 ? 0 : 1))10 return Math.min.apply(11 null,12 chars.reduce(13 (acc, el, i) => [14 acc[0] + (el !== s1[i] ? 1 : 0),15 acc[1] + (el !== s2[i] ? 1 : 0),16 ],17 [0, 0]18 )19 )20}`
leetcode
array
greedy
## Next Post
LeetCode: Count Number Of Homogenous Substrings
Feb 14, 2021
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#### Resources tagged with Working systematically similar to Natural Born Mathematicians:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Encouraging Primary Children to Work Systematically
##### Stage: Early years, 1 and 2
This article for primary teachers suggests ways in which to help children become better at working systematically.
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##### Stage: 1, 2, 3, 4 and 5
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
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##### Stage: 1 and 2
This article for teachers suggests activities based on pegboards, from pattern generation to finding all possible triangles, for example.
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
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##### Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
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##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
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Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods?
### More on Mazes
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There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
### Jumping Squares
##### Stage: 1 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
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##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
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##### Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
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##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
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##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
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##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
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##### Stage: 2 Challenge Level:
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##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Mrs Beeswax
##### Stage: 1 Challenge Level:
In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins?
### Diagonal Sums Sudoku
##### Stage: 2, 3 and 4 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
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##### Stage: 1 Challenge Level:
Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land.
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How many trains can you make which are the same length as Matt's, using rods that are identical?
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##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
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##### Stage: 1 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
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##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
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##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
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##### Stage: 2 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
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##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
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##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
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##### Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
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##### Stage: 1 Challenge Level:
How many different triangles can you draw on the dotty grid which each have one dot in the middle?
### One to Fifteen
##### Stage: 2 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
### Winning the Lottery
##### Stage: 2 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### 1 to 8
##### Stage: 2 Challenge Level:
Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.
### Making Trains
##### Stage: 1 Challenge Level:
Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it?
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##### Stage: 2 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Creating Cubes
##### Stage: 2 and 3 Challenge Level:
Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour.
### Inky Cube
##### Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Triangle Animals
##### Stage: 1 Challenge Level:
How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Cover the Camel
##### Stage: 1 Challenge Level:
Can you cover the camel with these pieces?
### How Long Does it Take?
##### Stage: 2 Challenge Level:
In this matching game, you have to decide how long different events take. | 2,102 | 9,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2017-39 | latest | en | 0.912993 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Registered: 26.06.2002
From:
Posted: Monday 20th of Aug 08:09 Hello friends, I misplaced my math textbook yesterday. It’s out of stock and so I can’t find it in any of the shops near my place. I have an option of hiring a private instructor but then I live in a very far off place so any tutor would charge high rates to come over. Now the problem is that I have my assessment next week and I can’t study since I lost my textbook. I couldn’t read the chapters on teaching 4th grade exponents and teaching 4th grade exponents. A few more topics such as sum of cubes, scientific notation, subtracting fractions and powers are still confusing me. I need some help guys!
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Tuesday 21st of Aug 16:19 Believe me, it’s sometimes quite difficult to learn a topic by your own because of its complexity just like teaching 4th grade exponents. It’s sometimes better to request someone to teach you the details rather than knowing the topic on your own. In that way, you can understand it very well because the topic can be explained clearly. Fortunately, I encountered this new software that could help in solving problems in math . It’s a cheap quick hassle-free way of understanding algebra concepts. Try making use of Algebrator and I guarantee you that you’ll have no trouble solving algebra problems anymore. It shows all the useful solutions for a problem. You’ll have a good time learning math because it’s user-friendly. Try it .
Jrahan
Registered: 19.03.2002
From: UK
Posted: Wednesday 22nd of Aug 19:54 Hello there , Thanks for the instantaneous reply. But could you let me know the details of genuine websites from where I can make the purchase? Can I get the Algebrator cd from a local book mart available in my area?
Anonog2112
Registered: 22.04.2006
From:
Posted: Friday 24th of Aug 14:20 I have tried a number of math related software. I would not name them here, but they were useless. I hope this one is not like the one’s I’ve used in the past .
MichMoxon
Registered: 21.08.2001
From:
Posted: Sunday 26th of Aug 08:42 Here http://www.mathpoint.net/rational-expressions.html. Happy problem solving! | 790 | 3,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | longest | en | 0.923666 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/4/34/a/a/ | 1,603,710,453,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891203.69/warc/CC-MAIN-20201026090458-20201026120458-00442.warc.gz | 805,143,646 | 81,853 | # Properties
Label 4.34.a.a Level $4$ Weight $34$ Character orbit 4.a Self dual yes Analytic conductor $27.593$ Analytic rank $0$ Dimension $3$ CM no Inner twists $1$
# Learn more about
## Newspace parameters
Level: $$N$$ $$=$$ $$4 = 2^{2}$$ Weight: $$k$$ $$=$$ $$34$$ Character orbit: $$[\chi]$$ $$=$$ 4.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$27.5931315524$$ Analytic rank: $$0$$ Dimension: $$3$$ Coefficient field: $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ Defining polynomial: $$x^{3} - x^{2} - 65185566 x - 173679864984$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$2^{22}\cdot 3^{3}\cdot 7\cdot 11\cdot 29$$ Twist minimal: yes Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( 30830596 + \beta_{1} ) q^{3} + ( -17960227962 - 1535 \beta_{1} + \beta_{2} ) q^{5} + ( 1513669971464 - 122346 \beta_{1} + 180 \beta_{2} ) q^{7} + ( 2010788144563341 + 75248670 \beta_{1} - 10530 \beta_{2} ) q^{9} +O(q^{10})$$ $$q +(30830596 + \beta_{1}) q^{3} +(-17960227962 - 1535 \beta_{1} + \beta_{2}) q^{5} +(1513669971464 - 122346 \beta_{1} + 180 \beta_{2}) q^{7} +(2010788144563341 + 75248670 \beta_{1} - 10530 \beta_{2}) q^{9} +(75872552434149900 + 2215839115 \beta_{1} + 256840 \beta_{2}) q^{11} +(90990147405786254 + 6383188809 \beta_{1} - 3668535 \beta_{2}) q^{13} +(-10714157827080368616 - 382917087630 \beta_{1} + 33406668 \beta_{2}) q^{15} +(31012396103938905618 + 1069301430158 \beta_{1} - 184208050 \beta_{2}) q^{17} +($$$$44\!\cdots\!76$$$$+ 11850406238205 \beta_{1} + 320516280 \beta_{2}) q^{19} +(-$$$$76\!\cdots\!84$$$$- 57340224604180 \beta_{1} + 4392064620 \beta_{2}) q^{21} +($$$$38\!\cdots\!32$$$$- 94999163347438 \beta_{1} - 49295676820 \beta_{2}) q^{23} +($$$$86\!\cdots\!07$$$$+ 615284668657260 \beta_{1} + 277757168364 \beta_{2}) q^{25} +($$$$38\!\cdots\!08$$$$+ 2919170542770738 \beta_{1} - 973938527640 \beta_{2}) q^{27} +($$$$96\!\cdots\!26$$$$- 17291424904695515 \beta_{1} + 1814914414885 \beta_{2}) q^{29} +($$$$52\!\cdots\!76$$$$+ 3377924000980200 \beta_{1} + 1180388959200 \beta_{2}) q^{31} +($$$$17\!\cdots\!60$$$$+ 98072137402840890 \beta_{1} - 18904063453830 \beta_{2}) q^{33} +($$$$34\!\cdots\!16$$$$+ 51799596297276380 \beta_{1} + 55139397898832 \beta_{2}) q^{35} +($$$$11\!\cdots\!14$$$$- 1038005378158598451 \beta_{1} - 52265853646515 \beta_{2}) q^{37} +($$$$45\!\cdots\!56$$$$+ 1463249000741125850 \beta_{1} - 130471960050900 \beta_{2}) q^{39} +(-$$$$22\!\cdots\!54$$$$- 641632766910683060 \beta_{1} + 448924689742540 \beta_{2}) q^{41} +(-$$$$51\!\cdots\!40$$$$+ 9557870528390708475 \beta_{1} - 210366843981840 \beta_{2}) q^{43} +(-$$$$27\!\cdots\!62$$$$- 29103707165110673535 \beta_{1} - 950908515500799 \beta_{2}) q^{45} +(-$$$$24\!\cdots\!92$$$$+ 13537930853481011508 \beta_{1} - 576631108849560 \beta_{2}) q^{47} +(-$$$$15\!\cdots\!03$$$$+ 496168262546838840 \beta_{1} + 10601270675344440 \beta_{2}) q^{49} +($$$$80\!\cdots\!12$$$$+$$$$13\!\cdots\!10$$$$\beta_{1} - 14436065202263640 \beta_{2}) q^{51} +($$$$18\!\cdots\!78$$$$- 27337706038779017987 \beta_{1} - 40829066390340995 \beta_{2}) q^{53} +($$$$24\!\cdots\!60$$$$-$$$$84\!\cdots\!50$$$$\beta_{1} + 158548384447532820 \beta_{2}) q^{55} +($$$$92\!\cdots\!16$$$$+$$$$87\!\cdots\!06$$$$\beta_{1} - 119258077651337610 \beta_{2}) q^{57} +($$$$67\!\cdots\!92$$$$+$$$$66\!\cdots\!35$$$$\beta_{1} - 355142845670518240 \beta_{2}) q^{59} +($$$$19\!\cdots\!22$$$$+$$$$11\!\cdots\!45$$$$\beta_{1} + 864805591092629745 \beta_{2}) q^{61} +(-$$$$41\!\cdots\!56$$$$-$$$$39\!\cdots\!06$$$$\beta_{1} - 321105448391693580 \beta_{2}) q^{63} +(-$$$$75\!\cdots\!04$$$$-$$$$25\!\cdots\!20$$$$\beta_{1} - 991922296965526108 \beta_{2}) q^{65} +(-$$$$96\!\cdots\!76$$$$+$$$$61\!\cdots\!09$$$$\beta_{1} + 745104731828763960 \beta_{2}) q^{67} +(-$$$$50\!\cdots\!72$$$$+$$$$14\!\cdots\!80$$$$\beta_{1} + 150330017751397380 \beta_{2}) q^{69} +($$$$56\!\cdots\!48$$$$+$$$$17\!\cdots\!70$$$$\beta_{1} + 4210979720475567620 \beta_{2}) q^{71} +($$$$43\!\cdots\!34$$$$-$$$$54\!\cdots\!86$$$$\beta_{1} - 9892260507865032810 \beta_{2}) q^{73} +($$$$67\!\cdots\!76$$$$+$$$$30\!\cdots\!55$$$$\beta_{1} - 1689547931514628848 \beta_{2}) q^{75} +($$$$69\!\cdots\!40$$$$-$$$$13\!\cdots\!40$$$$\beta_{1} + 25628357491802757380 \beta_{2}) q^{77} +($$$$14\!\cdots\!68$$$$+$$$$14\!\cdots\!60$$$$\beta_{1} - 24410580389385198840 \beta_{2}) q^{79} +($$$$20\!\cdots\!09$$$$+$$$$38\!\cdots\!30$$$$\beta_{1} + 11004304993611004530 \beta_{2}) q^{81} +(-$$$$23\!\cdots\!88$$$$-$$$$34\!\cdots\!63$$$$\beta_{1} - 30337551274260013440 \beta_{2}) q^{83} +(-$$$$45\!\cdots\!28$$$$-$$$$41\!\cdots\!90$$$$\beta_{1} - 21434939812222425306 \beta_{2}) q^{85} +(-$$$$84\!\cdots\!64$$$$-$$$$34\!\cdots\!14$$$$\beta_{1} +$$$$21\!\cdots\!80$$$$\beta_{2}) q^{87} +(-$$$$64\!\cdots\!66$$$$+$$$$68\!\cdots\!30$$$$\beta_{1} -$$$$17\!\cdots\!70$$$$\beta_{2}) q^{89} +(-$$$$12\!\cdots\!36$$$$-$$$$23\!\cdots\!40$$$$\beta_{1} -$$$$19\!\cdots\!40$$$$\beta_{2}) q^{91} +($$$$18\!\cdots\!96$$$$+$$$$50\!\cdots\!76$$$$\beta_{1} - 15215953620938720400 \beta_{2}) q^{93} +(-$$$$68\!\cdots\!92$$$$-$$$$46\!\cdots\!10$$$$\beta_{1} +$$$$56\!\cdots\!16$$$$\beta_{2}) q^{95} +($$$$18\!\cdots\!74$$$$-$$$$10\!\cdots\!46$$$$\beta_{1} +$$$$87\!\cdots\!90$$$$\beta_{2}) q^{97} +($$$$75\!\cdots\!00$$$$+$$$$14\!\cdots\!15$$$$\beta_{1} -$$$$27\!\cdots\!60$$$$\beta_{2}) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$3q + 92491788q^{3} - 53880683886q^{5} + 4541009914392q^{7} + 6032364433690023q^{9} + O(q^{10})$$ $$3q + 92491788q^{3} - 53880683886q^{5} + 4541009914392q^{7} + 6032364433690023q^{9} + 227617657302449700q^{11} + 272970442217358762q^{13} - 32142473481241105848q^{15} + 93037188311816716854q^{17} +$$$$13\!\cdots\!28$$$$q^{19} -$$$$22\!\cdots\!52$$$$q^{21} +$$$$11\!\cdots\!96$$$$q^{23} +$$$$25\!\cdots\!21$$$$q^{25} +$$$$11\!\cdots\!24$$$$q^{27} +$$$$28\!\cdots\!78$$$$q^{29} +$$$$15\!\cdots\!28$$$$q^{31} +$$$$51\!\cdots\!80$$$$q^{33} +$$$$10\!\cdots\!48$$$$q^{35} +$$$$34\!\cdots\!42$$$$q^{37} +$$$$13\!\cdots\!68$$$$q^{39} -$$$$67\!\cdots\!62$$$$q^{41} -$$$$15\!\cdots\!20$$$$q^{43} -$$$$82\!\cdots\!86$$$$q^{45} -$$$$72\!\cdots\!76$$$$q^{47} -$$$$47\!\cdots\!09$$$$q^{49} +$$$$24\!\cdots\!36$$$$q^{51} +$$$$54\!\cdots\!34$$$$q^{53} +$$$$72\!\cdots\!80$$$$q^{55} +$$$$27\!\cdots\!48$$$$q^{57} +$$$$20\!\cdots\!76$$$$q^{59} +$$$$59\!\cdots\!66$$$$q^{61} -$$$$12\!\cdots\!68$$$$q^{63} -$$$$22\!\cdots\!12$$$$q^{65} -$$$$29\!\cdots\!28$$$$q^{67} -$$$$15\!\cdots\!16$$$$q^{69} +$$$$16\!\cdots\!44$$$$q^{71} +$$$$13\!\cdots\!02$$$$q^{73} +$$$$20\!\cdots\!28$$$$q^{75} +$$$$20\!\cdots\!20$$$$q^{77} +$$$$43\!\cdots\!04$$$$q^{79} +$$$$60\!\cdots\!27$$$$q^{81} -$$$$71\!\cdots\!64$$$$q^{83} -$$$$13\!\cdots\!84$$$$q^{85} -$$$$25\!\cdots\!92$$$$q^{87} -$$$$19\!\cdots\!98$$$$q^{89} -$$$$38\!\cdots\!08$$$$q^{91} +$$$$55\!\cdots\!88$$$$q^{93} -$$$$20\!\cdots\!76$$$$q^{95} +$$$$56\!\cdots\!22$$$$q^{97} +$$$$22\!\cdots\!00$$$$q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{3} - x^{2} - 65185566 x - 173679864984$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$-128 \nu^{2} + 542720 \nu + 5562320768$$$$)/25$$ $$\beta_{2}$$ $$=$$ $$($$$$343168 \nu^{2} + 6776698880 \nu - 14915325889408$$$$)/125$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$5 \beta_{2} + 2681 \beta_{1} + 109756416$$$$)/ 329269248$$ $$\nu^{2}$$ $$=$$ $$($$$$1325 \beta_{2} - 3308935 \beta_{1} + 894316769246208$$$$)/20579328$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−6032.20 9172.25 −3139.05
0 −6.39317e7 0 −2.18954e11 0 −4.92542e13 0 −1.47179e15 0
1.2 0 2.16957e7 0 6.04966e11 0 1.12234e14 0 −5.08836e15 0
1.3 0 1.34728e8 0 −4.39893e11 0 −5.84388e13 0 1.25925e16 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 4.34.a.a 3
4.b odd 2 1 16.34.a.d 3
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
4.34.a.a 3 1.a even 1 1 trivial
16.34.a.d 3 4.b odd 2 1
## Hecke kernels
This newform subspace is the entire newspace $$S_{34}^{\mathrm{new}}(\Gamma_0(4))$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ 1
$3$ $$1 - 92491788 T + 9599774056706745 T^{2} -$$$$84\!\cdots\!36$$$$T^{3} +$$$$53\!\cdots\!35$$$$T^{4} -$$$$28\!\cdots\!52$$$$T^{5} +$$$$17\!\cdots\!67$$$$T^{6}$$
$5$ $$1 + 53880683886 T +$$$$46\!\cdots\!75$$$$T^{2} -$$$$45\!\cdots\!00$$$$T^{3} +$$$$54\!\cdots\!75$$$$T^{4} +$$$$73\!\cdots\!50$$$$T^{5} +$$$$15\!\cdots\!25$$$$T^{6}$$
$7$ $$1 - 4541009914392 T +$$$$13\!\cdots\!97$$$$T^{2} -$$$$39\!\cdots\!80$$$$T^{3} +$$$$10\!\cdots\!79$$$$T^{4} -$$$$27\!\cdots\!08$$$$T^{5} +$$$$46\!\cdots\!43$$$$T^{6}$$
$11$ $$1 - 227617657302449700 T +$$$$19\!\cdots\!93$$$$T^{2} +$$$$71\!\cdots\!00$$$$T^{3} +$$$$45\!\cdots\!83$$$$T^{4} -$$$$12\!\cdots\!00$$$$T^{5} +$$$$12\!\cdots\!91$$$$T^{6}$$
$13$ $$1 - 272970442217358762 T +$$$$13\!\cdots\!95$$$$T^{2} -$$$$28\!\cdots\!04$$$$T^{3} +$$$$75\!\cdots\!35$$$$T^{4} -$$$$90\!\cdots\!58$$$$T^{5} +$$$$19\!\cdots\!77$$$$T^{6}$$
$17$ $$1 - 93037188311816716854 T +$$$$10\!\cdots\!75$$$$T^{2} -$$$$76\!\cdots\!52$$$$T^{3} +$$$$41\!\cdots\!75$$$$T^{4} -$$$$15\!\cdots\!26$$$$T^{5} +$$$$65\!\cdots\!53$$$$T^{6}$$
$19$ $$1 -$$$$13\!\cdots\!28$$$$T +$$$$39\!\cdots\!05$$$$T^{2} -$$$$35\!\cdots\!80$$$$T^{3} +$$$$61\!\cdots\!95$$$$T^{4} -$$$$33\!\cdots\!68$$$$T^{5} +$$$$39\!\cdots\!79$$$$T^{6}$$
$23$ $$1 -$$$$11\!\cdots\!96$$$$T +$$$$18\!\cdots\!73$$$$T^{2} -$$$$14\!\cdots\!40$$$$T^{3} +$$$$16\!\cdots\!59$$$$T^{4} -$$$$87\!\cdots\!44$$$$T^{5} +$$$$64\!\cdots\!87$$$$T^{6}$$
$29$ $$1 -$$$$28\!\cdots\!78$$$$T +$$$$43\!\cdots\!95$$$$T^{2} -$$$$47\!\cdots\!60$$$$T^{3} +$$$$79\!\cdots\!55$$$$T^{4} -$$$$95\!\cdots\!38$$$$T^{5} +$$$$59\!\cdots\!69$$$$T^{6}$$
$31$ $$1 -$$$$15\!\cdots\!28$$$$T +$$$$13\!\cdots\!01$$$$T^{2} -$$$$66\!\cdots\!72$$$$T^{3} +$$$$21\!\cdots\!91$$$$T^{4} -$$$$42\!\cdots\!68$$$$T^{5} +$$$$44\!\cdots\!71$$$$T^{6}$$
$37$ $$1 -$$$$34\!\cdots\!42$$$$T +$$$$58\!\cdots\!87$$$$T^{2} -$$$$50\!\cdots\!80$$$$T^{3} +$$$$32\!\cdots\!39$$$$T^{4} -$$$$11\!\cdots\!78$$$$T^{5} +$$$$17\!\cdots\!73$$$$T^{6}$$
$41$ $$1 +$$$$67\!\cdots\!62$$$$T +$$$$59\!\cdots\!11$$$$T^{2} +$$$$21\!\cdots\!68$$$$T^{3} +$$$$98\!\cdots\!31$$$$T^{4} +$$$$18\!\cdots\!42$$$$T^{5} +$$$$46\!\cdots\!61$$$$T^{6}$$
$43$ $$1 +$$$$15\!\cdots\!20$$$$T +$$$$22\!\cdots\!29$$$$T^{2} +$$$$19\!\cdots\!20$$$$T^{3} +$$$$18\!\cdots\!47$$$$T^{4} +$$$$10\!\cdots\!80$$$$T^{5} +$$$$51\!\cdots\!07$$$$T^{6}$$
$47$ $$1 +$$$$72\!\cdots\!76$$$$T +$$$$60\!\cdots\!05$$$$T^{2} +$$$$22\!\cdots\!28$$$$T^{3} +$$$$91\!\cdots\!35$$$$T^{4} +$$$$16\!\cdots\!04$$$$T^{5} +$$$$34\!\cdots\!83$$$$T^{6}$$
$53$ $$1 -$$$$54\!\cdots\!34$$$$T +$$$$28\!\cdots\!83$$$$T^{2} -$$$$80\!\cdots\!40$$$$T^{3} +$$$$23\!\cdots\!59$$$$T^{4} -$$$$34\!\cdots\!86$$$$T^{5} +$$$$50\!\cdots\!17$$$$T^{6}$$
$59$ $$1 -$$$$20\!\cdots\!76$$$$T +$$$$56\!\cdots\!29$$$$T^{2} -$$$$65\!\cdots\!96$$$$T^{3} +$$$$15\!\cdots\!91$$$$T^{4} -$$$$15\!\cdots\!16$$$$T^{5} +$$$$20\!\cdots\!39$$$$T^{6}$$
$61$ $$1 -$$$$59\!\cdots\!66$$$$T +$$$$26\!\cdots\!95$$$$T^{2} -$$$$28\!\cdots\!40$$$$T^{3} +$$$$21\!\cdots\!95$$$$T^{4} -$$$$40\!\cdots\!26$$$$T^{5} +$$$$55\!\cdots\!41$$$$T^{6}$$
$67$ $$1 +$$$$29\!\cdots\!28$$$$T +$$$$81\!\cdots\!37$$$$T^{2} +$$$$11\!\cdots\!00$$$$T^{3} +$$$$14\!\cdots\!19$$$$T^{4} +$$$$96\!\cdots\!32$$$$T^{5} +$$$$60\!\cdots\!03$$$$T^{6}$$
$71$ $$1 -$$$$16\!\cdots\!44$$$$T +$$$$30\!\cdots\!45$$$$T^{2} -$$$$32\!\cdots\!60$$$$T^{3} +$$$$37\!\cdots\!95$$$$T^{4} -$$$$25\!\cdots\!24$$$$T^{5} +$$$$18\!\cdots\!31$$$$T^{6}$$
$73$ $$1 -$$$$13\!\cdots\!02$$$$T +$$$$36\!\cdots\!75$$$$T^{2} -$$$$19\!\cdots\!84$$$$T^{3} +$$$$11\!\cdots\!75$$$$T^{4} -$$$$12\!\cdots\!78$$$$T^{5} +$$$$29\!\cdots\!37$$$$T^{6}$$
$79$ $$1 -$$$$43\!\cdots\!04$$$$T +$$$$15\!\cdots\!89$$$$T^{2} -$$$$35\!\cdots\!44$$$$T^{3} +$$$$62\!\cdots\!71$$$$T^{4} -$$$$75\!\cdots\!84$$$$T^{5} +$$$$73\!\cdots\!19$$$$T^{6}$$
$83$ $$1 +$$$$71\!\cdots\!64$$$$T +$$$$66\!\cdots\!93$$$$T^{2} +$$$$26\!\cdots\!20$$$$T^{3} +$$$$14\!\cdots\!59$$$$T^{4} +$$$$32\!\cdots\!16$$$$T^{5} +$$$$97\!\cdots\!47$$$$T^{6}$$
$89$ $$1 +$$$$19\!\cdots\!98$$$$T +$$$$63\!\cdots\!75$$$$T^{2} +$$$$76\!\cdots\!20$$$$T^{3} +$$$$13\!\cdots\!75$$$$T^{4} +$$$$88\!\cdots\!78$$$$T^{5} +$$$$97\!\cdots\!09$$$$T^{6}$$
$97$ $$1 -$$$$56\!\cdots\!22$$$$T -$$$$50\!\cdots\!73$$$$T^{2} +$$$$35\!\cdots\!00$$$$T^{3} -$$$$18\!\cdots\!21$$$$T^{4} -$$$$75\!\cdots\!38$$$$T^{5} +$$$$49\!\cdots\!33$$$$T^{6}$$
show more
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# 4. Ir is an oversimplified view of cattle raising to say
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4. Ir is an oversimplified view of cattle raising to say [#permalink]
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25 Aug 2003, 04:50
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4.
Ir is an oversimplified view of cattle raising to say that [all one has to do with cattle is leave them alone while they feed themselves, corral them and to] drive them to market when the time is ripe.
a] all one has to do with cattle is leave them alone while they feed themselves, corral them and to
b] all one has to do with cattle is to leave them alone to feed themselves, to corral them and
c] all one has to do with cattle is leave them alone while they feed themselves and then corral them and
d] the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them and
e] the only thing that has to be doen with cattle is leave them alone while they feed themselves, to corral them and
If you have any questions
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26 Aug 2003, 02:29
aps_can wrote:
4.
Ir is an oversimplified view of cattle raising to say that [all one has to do with cattle is leave them alone while they feed themselves, corral them and to] drive them to market when the time is ripe.
a] all one has to do with cattle is leave them alone while they feed themselves, corral them and to
b] all one has to do with cattle is to leave them alone to feed themselves, to corral them and
c] all one has to do with cattle is leave them alone while they feed themselves and then corral them and
d] the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them and
e] the only thing that has to be doen with cattle is leave them alone while they feed themselves, to corral them and
Parallelism !
E has tense problems and ellipsis
The only thing that has to be done with cattle is "to" leave them alone
C Too many "and's" !!
B. parallelism
to leave them, to corral them and " " drive them
A . again parallelism
leave them, corral them and "to" drive them
if anyone has a better explanation, please do share with us.
Thanks
Praetorian[/b]
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26 Aug 2003, 06:37
I think that E and D are both wrong because they start with "the only thing" and then they mention more than one thing.
From A,B and C the best one is C.
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26 Aug 2003, 14:56
MartinMag wrote:
I think that E and D are both wrong because they start with "the only thing" and then they mention more than one thing.
From A,B and C the best one is C.
I dont like so many ands though.
thanks
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28 Aug 2003, 20:54
I pick B.
because A, C, D, E are putting "is" and "leave" together.
But B is not perfect. It would be better if there is an "to" after "and." But GMAT is not looking for perfect answer but the best answer.
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26 Oct 2003, 22:19
aps_can wrote:
4.
Ir is an oversimplified view of cattle raising to say that [all one has to do with cattle is leave them alone while they feed themselves, corral them and to] drive them to market when the time is ripe.
a] all one has to do with cattle is leave them alone while they feed themselves, corral them and to
b] all one has to do with cattle is to leave them alone to feed themselves, to corral them and
c] all one has to do with cattle is leave them alone while they feed themselves and then corral them and
d] the only thing that has to be done with cattle is leave them alone while they feed themselves, corral them and
e] the only thing that has to be doen with cattle is leave them alone while they feed themselves, to corral them and
We can narrow to C and D.
D clearly change the meaning and worng. The only thing is A, B, and C
C has many end but they are in perfect correct. "... leave them and THEN (corral and drive)..." = do one thign and Then (do 2 more things)
_________________
I have 2 month for gmat.
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# The Tax System Explained in Beer
## Suppose that every day, ten men go out for beer and the bill for all ten comes to \$100.
If they paid their bill the way we pay our taxes, it would go something like this:
The first four men (the poorest) would pay nothing.
The fifth would pay \$1.
The sixth would pay \$3.
The seventh would pay \$7.
The eighth would pay \$12.
The ninth would pay \$18.
The tenth man (the richest) would pay \$59.
So, that's what they decided to do. The ten men drank in the bar every day and seemed quite happy with the arrangement, until one day, the owner threw them a curve ball. "Since you are all such good customers," he said, "I'm going to reduce the cost of your daily beer by \$20." Drinks for the ten men would now cost just \$80.
The group still wanted to pay their bill the way we pay our taxes. So the first four men were unaffected. They would still drink for free. But what about the other six men? How could they divide the \$20 windfall so that everyone would get his fair share? They realized that \$20 divided by six is \$3.33. But if they subtracted that from everybody's share, then the fifth man and the sixth man would each end up being paid to drink his beer.
So, the bar owner suggested that it would be fair to reduce each man's bill by a higher percentage the poorer he was, to follow the principle of the tax system they had been using, and he proceeded to work out the amounts he suggested that each should now pay.
The fifth man, like the first four, now paid nothing (100% saving).
The sixth now paid \$2 instead of \$3 (33% saving).
The seventh now paid \$5 instead of \$7 (28% saving).
The eighth now paid \$9 instead of \$12 (25% saving).
The ninth now paid \$14 instead of \$18 (22% saving).
The tenth now paid \$49 instead of \$59 (16% saving).
Each of the six was better off than before. And the first four continued to drink for free. But, once outside the bar, the men began to compare their savings.
"I only got a dollar out of the \$20 saving," declared the sixth man. He pointed to the tenth man, "but he got \$10!"
"Yeah, that's right," exclaimed the fifth man. "I only saved a dollar, too. It's unfair that he got ten times more benefit than me!"
"That's true!" shouted the seventh man. "Why should he get \$10 back, when I got only \$2? The wealthy get all the breaks!"
"Wait a minute," yelled the first four men in unison, "we didn't get anything at all. This new tax system exploits the poor!"
The nine men surrounded the tenth and beat him up.
The next night the tenth man didn't show up for drinks, so the nine sat down and had their beers without him. But when it came time to pay the bill, they discovered something important – they didn't have enough money between all of them for even half of the bill!
And that, boys and girls, journalists and government ministers, is how our tax system works. The people who already pay the highest taxes will naturally get the most benefit from a tax reduction. Tax them too much, attack them for being wealthy, and they just may not show up anymore. In fact, they might start drinking overseas, where the atmosphere is somewhat friendlier.
David R. Kamerschen, Ph.D.
Professor of Economics
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https://codereview.stackexchange.com/questions/28453/fast-integer-handling | 1,702,162,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00310.warc.gz | 198,545,806 | 44,488 | Fast integer handling
Each data structure has its own time complexities. The biggest one that jumps out at you is the hash table. The average times for Insertion, Deletion and Searching are all O(1). But it's really just constant time, since there can be multiple reps through each of these to find the right spot. The real question is, can we do each of those operations in exactly one operation, if space wasn't an issue?
Recently I was working on the radix sort, and figuring out how to speed it up as much as possible. The one thing I settled on was doing a 2-byte counting sort, thus you would only have to loop once through. Then I got to thinking about the trie I used before, with integers where I traversed through by modding the number by 10 each time through.
This seemed to be faster than any other data structure that I've tried, but wanted to make it faster. What I tried today was treating the trie as a guaranteed 2-level tree: each node with $2^{16}$ (65536) children. That way I can reference each number like this: root->link[twoByte1]->link[twoByte2]. Where twoByte1 is the first 16 bits, and twoByte2 is the second 16-bit chunk. Problem is, this is a lot of memory. Instead, I decided to make an array of structures, with $2^{16}$ bool (uint8_t) values instead, and ran across the same issue. To fix this, I made the following structure.
struct trie {
};
I decided to use 2048 unsigned integers (or signed, wasn't sure if there would be a difference) instead. That way I can treat each bit as the true/false value during look up. In order to reference each position, I would take the first 16 bytes, and reference the element he same way as before, but the 2nd I would divide by 32 to find out which element of the integer array to use, and mod the number again by 32 to find out which bit to look at. The following is the code I used to reference it:
uint8_t getData (int data, uint16_t& dir, uint16_t& pos, uint8_t& x) {
dir = (data >> 0) & 0xffff;
pos = (data >> 16) & 0xffff;
x = pos / 32;
return pos % 32;
}
The insert function was simply finding the spot and turning the bit on. To do this, I would get the data from the function above, then OR the specific bit with 1. Similarly with deletion, only I would invert the bit string, then AND it, to turn it off. Then to search for that element, I would just return the specific bit. Help with how to clear the bit was found in How do you set, clear, and toggle a single bit? on SO.
Here is the code for those three functions:
void insert(trie *root, int data) {
uint16_t dir, pos;
uint8_t x, y = getData(data, dir, pos, x);
root[dir].link[x] |= 1 << y;
}
bool find(trie *root, int data) {
uint16_t dir, pos;
uint8_t x, y = getData(data, dir, pos, x);
return root[dir].link[x] & (1 << y);
}
void remove(trie *root, int data) {
uint16_t dir, pos;
uint8_t x, y = getData(data, dir, pos, x);
root[dir].link[x] &= ~(1 << y);
}
From here it was as simple as initializing the array of structs. The great thing is that with the number of elements coming in, the data structure will never grow any farther out. Also, in the case of threading, since I'm only turning the bits on during insertion, I don't think I would need to synchronize any of the threads. Therefore it could be speed up with threading. What are your thoughts on this? I made a test case on Ideone. I ran a test on a Linux machine running on an i7, and was able to process 250 million integers every 2.5 seconds.
First there is no C++ here.
You are simply doing C (use class and member functions that way you will not be passing pointers around).
So what you have implemented is a chunked set. Each chunk trie uses 2048 integers to hold existence information and somewhere outside the scope of code review you are creating arrays of theses to define your set.
Looking at the code:
uint16_t dir, pos;
uint8_t x, y = getData(data, dir, pos, x);
This is awful. The first think I am thinking is that x is undefined because you think that functions can return multiple values. BUT I read on and find that x is passed in and is assigned inside the function. This is just horrible in terms of style and readability.
Rues of thumb:
1. Declare one variable per line (and make their names meaningful).
2. Don't mix input/output parameters and return values all in the same function.
When you use output parameters the result is usually a success value.
What is trie.
bool find(trie *root, int data)
It looks like you have to pass an array of them through for this to work. From inside this function I can not tell how legal it is to access multiple elements. I am assuming the memory management is done somewhere else but who know.
You are making the assumption that an unsigned int is 32 bits.
return pos % 32; // assigned to y
// All your code then calculates bit offsets like this
(1 << y)
The standard only requires 16. https://stackoverflow.com/a/271132/14065
Overall I think this is marginal for C code as you can not do much more.
But for C++ code this absolutely terrible. There is no encapsulation, there is no protection from using the code wrong. There is no initialization to make sure it is all set to zero.
int data; data >> 16 is actually Undefined Behavior. My suggestion, in addition to most of what's already been said, is to stick to unsigned ints as much as possible in C.
Edit: uppercased Undefined Behaviour again (after someone unhelpfully lowercased it) because it’s a technical term from the standard.
It’s Undefined Behaviour if data is negative, that is.
• Technically, it's UB iff int is 16 bits or less. Or am I missing something? Jul 30, 2013 at 18:20
• @Listor: int is a signed type. Jul 11, 2018 at 14:47 | 1,405 | 5,699 | {"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} | 2.96875 | 3 | CC-MAIN-2023-50 | longest | en | 0.957398 |
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##### Wk10 Q6 Calculating Cost of Debt Jiminy's Cric...
Description
Solution
Question
Wk10 Q6 Calculating Cost of Debt Jiminy's Cricket Farm issued a 30-year, 8 percent semiannual bond 7 years ago. The bond currently sells for 95 percent of its face value. The book value of the debt issue is \$80 million. The company's tax rate is 35 percent. In addition, the company has a second debt issue on the market, a zero coupon bond with seven years left to maturity; the book value of this issue is \$35 million, and the bonds sell for 61 percent of par. The company's total book value of debt is \$___. Its total market value of debt is \$___. (Enter your answers in dollars, not millions of dollars, e.g, \$1,234,567.) Your best estimate of the aftertax cost of debt is ___ percent. (Do not include the percent sign (%). Round your answer to 2 decimal places.(e.g., 32.16)),Please return answer in excel. Thank you.
Paper#5385 | Written in 18-Jul-2015
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# Composite Functions
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## Mr Mattock
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#### Transcript of Composite Functions
Composite Functions
Starter
f(x) = 3x + 4 g(x) = x - 5
(a) Find f(5)
(c) Find g(5)
Main Activity
Complete the composite function worksheet
Composite functions
A composite function is a function built out of two or more functions.
The function fg(x) is a composite that means "first do g to x, then do f to the result".
L.O. - Understand how to form the
composite of 2 functions (Grade 6+)
Starter
f(x) = 3x + 4 g(x) = x - 5
(a) Find f(5)
= 19
g(19) = 356
(c) Find g(5)
= 20
f(20) = 64
It makes a big difference which order you use the
functions.
Writing Composites
f(x) = 3x + 4 g(x) = x - 5
(a) Find the composite fg(x)
(b) Find the composite gf(x)
(c) Find the composite ff(x)
2
2
2
Writing Composites
f(x) = 3x + 4 g(x) = x - 5
(a) Find the composite fg(x)
(b) Find the composite gf(x)
(c) Find the composite ff(x)
2
x
-5
2
x - 5
2
x 3 + 4
3(x - 5) + 4 = 3x - 11
2
2
x
x 3 + 4
3x + 4
2
-5
(3x + 4) - 5 =
9x + 24x + 11
2
2
x
x 3 + 4
3x + 4
x 3 + 4
3(3x + 4) + 4 = 9x + 16
Dice of Ultimate Power
1) Give a key word for the lesson.
2) Summarise one key point for the
lesson.
3) Explain one thing you have learned this
lesson.
4) State one thing you still don't understand following this lesson.
5) Suggest one thing you would like to find out
based on this lesson.
6) Free choice from 1 to 5, or another you make
up.
Main Activity
(a) -3 (b) -4 (c) 17 (d) 86 (e) 5 (f) -7 (g) 41 (h) 26
(a) 2(1 - 3x) + 1 = 3 - 6x
(b) 2(2x + 1) + 1 = 4x + 3
(c) 2(x + 4) + 1 = 2x + 9
(d) (2x + 1) + 4 = 4x + 4x + 5
(e) 1 - 3(x + 4) = -3x - 11
(f) 1 - 3(1 - 3x) = 9x - 2
(g) (1 - 3x) + 4 = 9x - 6x + 5
(h) 1 - 3(2x + 1) = -6x - 2
2
2
2
2
2
2
2
2
(a) 3 + 2(4 - x) = 11 - 2x
(b) 2(3 + 2x) + 7 = 8x + 24x + 25
(c) 4 - (2x + 7) = -2x - 3
(d) 3 + 2(3 + 2x) = 9 + 4x
(e) 3 + 2(2x + 7) = 4x + 17
(f) 4 - (4 - x) = x
(g) 4 - (3 + 2x) = 1 - 2x
(h) 2(4 - x) + 7= 2x - 16x + 39
2
2
2
2
2
2
2
2
(a) fg(x)
(b) gf(x)
(c) gg(x)
(d) ff(x)
(e) gff(x)
(f) gfg(x)
Key
Examples
Activities
Activity
Worked
Example
Extra example for 10b2
f(x) = 7 - 3x g(x) =
3
x
Find gf(x)
Find fg(x)
Find ff(x)
Find gg(x)
Extra example for 10b2
f(x) = 7 - 3x g(x) =
3
x
Find gf(x)
Find fg(x)
Find ff(x)
Find gg(x)
x-3
+ 7
reciprocal
x3
gf(x) = 3
7-3x
x-3
+ 7
reciprocal
x3
fg(x) = + 7
-9
x
x-3
+ 7
x-3
+ 7
ff(x) = -3(7 - 3x) + 7 = 9x -14
reciprocal
x3
reciprocal
x3
gg(x) = x
Worked
Example
Extra starter for 10b2
f(x) = 7 - 3x g(x) =
3
x
Find f(2)
Find g(2)
Find fg(2)
Find gf(2)
Extra starter for 10b2
f(x) = 7 - 3x g(x) =
3
x
Find f(2)
= 1
Find g(2)
=
1
Find fg(2)
= -2
Find gf(2)
= 3
1
2
1
2
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Basic algebra questions Solving multi step equations Mathematics apps for students App to check math problems Math homework | 776 | 3,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-49 | latest | en | 0.960943 |
http://cn.metamath.org/mpeuni/predel.html | 1,656,639,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00330.warc.gz | 13,595,151 | 3,315 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > predel Structured version Visualization version GIF version
Theorem predel 5858
Description: Membership in the predecessor class implies membership in the base class. (Contributed by Scott Fenton, 11-Feb-2011.)
Assertion
Ref Expression
predel (𝑌 ∈ Pred(𝑅, 𝐴, 𝑋) → 𝑌𝐴)
Proof of Theorem predel
StepHypRef Expression
1 elinel1 3942 . 2 (𝑌 ∈ (𝐴 ∩ (𝑅 “ {𝑋})) → 𝑌𝐴)
2 df-pred 5841 . 2 Pred(𝑅, 𝐴, 𝑋) = (𝐴 ∩ (𝑅 “ {𝑋}))
31, 2eleq2s 2857 1 (𝑌 ∈ Pred(𝑅, 𝐴, 𝑋) → 𝑌𝐴)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∈ wcel 2139 ∩ cin 3714 {csn 4321 ◡ccnv 5265 “ cima 5269 Predcpred 5840 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-v 3342 df-in 3722 df-pred 5841 This theorem is referenced by: predpo 5859 predpoirr 5869 predfrirr 5870 dftrpred3g 32038
Copyright terms: Public domain W3C validator | 609 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-27 | latest | en | 0.204903 |
https://kiiky.com/articles/how-long-is-60-days/ | 1,716,400,062,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00483.warc.gz | 298,987,263 | 51,184 | How Long is 60 Days? (Explaining the Calculation Process)
Time, an intangible yet omnipresent force, governs our lives in ways both subtle and profound. We measure our days, weeks, and months, but what does it truly mean when we say something will happen in “60 days”? Is it a blink of an eye, an eternity, or something in between? In this article, we embark on a journey through the concept of time, exploring its various dimensions and how we perceive the passing of 60 days. By delving into the psychological, practical, and even philosophical aspects of this seemingly straightforward span, we hope to uncover a deeper understanding of the intricacies that lie within the ticking clock. So, join us as we unravel the mystery of just how long 60 days can be.
How long is 60 Days?
When we mention that something will occur in “60 days,” it appears to be a straightforward statement regarding time. However, the actual experience of those 60 days can vary significantly depending on the context, individual perception, and the way we measure time. Let’s dive into the calculation process and explore the factors that influence our understanding of this seemingly fixed unit of time.
#1. Basic Calculation
The simplest way to calculate how long 60 days is involves basic arithmetic. There are 24 hours in a day, so:
60 days × 24 hours/day = 1,440 hours
This gives us a total of 1,440 hours in 60 days. It’s important to note that this calculation is based on the conventional Gregorian calendar, where each day is divided into 24 hours.
#2. Daily Routine
The perception of time can vary from person to person based on their daily routines and activities. For someone with a hectic schedule, 60 days might pass quickly, filled with appointments, work, and responsibilities. Conversely, for someone on vacation or experiencing a period of leisure, these same 60 days could feel luxuriously long.
#3. Psychological Time
Time often seems to pass more slowly when we eagerly anticipate an event or during moments of stress or boredom. Conversely, when we’re engrossed in an enjoyable activity or “in the flow,” time may feel like it’s flying by. This psychological aspect of time perception can significantly influence how we experience 60 days.
#4. Cultural and Contextual Differences
Different cultures and regions may have unique ways of measuring or perceiving time. Some cultures place a greater emphasis on punctuality and efficiency, while others prioritize a more relaxed and leisurely approach. These cultural variations can affect how people perceive the passage of 60 days.
#5. Seasonal and Environmental Factors
Seasonal changes and environmental conditions can also play a role in our perception of time. For example, the long, sunny days of summer might make 60 days feel shorter, while the dark and cold days of winter can make it seem longer.
In the age of instant communication and global connectivity, our perception of time can be distorted. We’re accustomed to receiving information and updates within seconds, which can affect how we gauge the duration of 60 days.
#7. Personal Events and Milestones
Significant personal events, such as weddings, vacations, or life-changing experiences, can greatly influence how we remember and perceive the passage of 60 days. These events create lasting memories that can make the time feel either brief or substantial.
While a simple calculation can determine that 60 days equals 1,440 hours, the actual experience of this time span is far from uniform. It is a dynamic concept shaped by a multitude of factors, including our daily routines, psychological state, cultural influences, and the events that fill our lives. Understanding how we perceive and calculate the passage of 60 days sheds light on the intricate relationship between time and human experience.
People also search for; How can I become an Aircraft mechanic in 2023?
What is the equivalent of 60 days in hours?
60 days is equal to 1,440 hours.
Why does time feel different in 60 days?
Time perception varies due to factors like daily routines, psychological state, and personal events.
Can 60 days ever feel shorter than it is?
Yes, when we are busy or engaged in enjoyable activities, 60 days can feel shorter.
What cultural influences affect our perception of 60 days?
Cultural attitudes towards time and punctuality can impact how we perceive the duration of 60 days.
How do technological advances affect our sense of 60 days?
Instant communication and global connectivity can distort our perception of time, making 60 days seem shorter.
Conclusion
The duration of 60 days is not merely a fixed interval on the calendar but a dynamic concept influenced by a myriad of factors, including our routines, psychology, culture, and personal experiences. Understanding the multifaceted nature of time perception within these 60 days unveils the intriguing interplay between objective measurement and the subjective human experience of time.
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How to Decline Admission to a College
Are you facing the dilemma of declining admission to a college? It’s a significant decision that requires careful… | 1,149 | 5,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.93973 |
http://tex.stackexchange.com/questions/73981/align-multiple-equations-under-longer-equation | 1,466,922,207,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783394987.40/warc/CC-MAIN-20160624154954-00082-ip-10-164-35-72.ec2.internal.warc.gz | 311,434,762 | 17,001 | # Align multiple equations under longer equation
I want to do what this MWE would do if the hspace{-1in} actually moved that second alignment column to the left, underneath the big quadratic formula.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{alignat*}{2}
x &= \frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(0)}}{2(7)}\\
&= \frac{13 \pm \sqrt{169}}{14}\\
&= \frac{13 + 13}{14}& \hspace{-1in} &\phantom{=} \frac{13-13}{14}\\
&= \frac{26}{14}& &= \frac{0}{14}\\
&= \frac{13}{7}& &= 0
\end{alignat*}
\end{document}
I'm trying to get rid of the unnecessary white space, and both equations in the third line are solutions of the equation in the first line, so it makes sense to have them underneath. But I still want the left one aligned vertically with the first two equal signs and the second one aligned vertically.
Any ideas?
-
You could enclose the quadratic in an rlap:
you could also add a horizontal space to separate out the two equations:
## Code:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{alignat*}{2}
x &= \rlap{$\displaystyle \frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(0)}}{2(7)}$}\\
&= \frac{13 \pm \sqrt{169}}{14}\\
&= \frac{13 + 13}{14} &&\phantom{=} \frac{13-13}{14}\\
&= \frac{26}{14} &&= \frac{0}{14}\\
&= \frac{13}{7} &&= 0
\end{alignat*}
\begin{alignat*}{2}
x &= \rlap{$\displaystyle \frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(0)}}{2(7)}$}\\
&= \frac{13 \pm \sqrt{169}}{14}\\
&= \frac{13 + 13}{14} &\qquad&\phantom{=} \frac{13-13}{14}\\
&= \frac{26}{14} &&= \frac{0}{14}\\
&= \frac{13}{7} &&= 0
\end{alignat*}
\end{document}
## Code: (with mathtools):
\documentclass{article}
\usepackage{mathtools}% includes amsmath
\begin{document}
\begin{alignat*}{2}
x &= \mathrlap{\frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(0)}}{2(7)}}\\
&= \frac{13 \pm \sqrt{169}}{14}\\
&= \frac{13 + 13}{14} &\qquad&\phantom{=} \frac{13-13}{14}\\
&= \frac{26}{14} &&= \frac{0}{14}\\
&= \frac{13}{7} &&= 0
\end{alignat*}
\end{document}
-
Perfect! I hadn't seen \rlap before. That's just what I needed. – Keila Sep 25 '12 at 3:23 | 797 | 2,099 | {"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} | 3.125 | 3 | CC-MAIN-2016-26 | longest | en | 0.551023 |
https://edufixers.com/values-brought-to-the-philosophical-approach-to-research-2/ | 1,702,334,007,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00798.warc.gz | 248,093,063 | 44,446 | # Values Brought to the Philosophical Approach to Research
Subject: Tech & Engineering 3 592 2 min
An important class of cryptographic algorithms is Message Authentication Codes (MAC) algorithms. Such algorithms are designed to ensure message integrity and authentication of their source. To do this, the MAC produced by such an algorithm is added to the message – a binary set depending on the message and on the secret key. Algorithms for the development of MAC are devoted to a large number of different sources. Increasing demands on the throughput and security of communication systems and networks lead to the need to develop new cryptographic algorithms with high performance.
In this regard, the use of generalized cellular automata is of great interest. Crypto algorithms based on generalized cellular automata show very high performance in hardware implementation. A Message Authentication Codes (MAC) is a sequence of data of a fixed length, which is generated according to a certain rule from open data and a key, either before encrypting the message or in parallel with encryption.
The MAC are transmitted via the communication channel or to the computer memory after the encrypted data. The development of an imitation MAC provides protection against the imposition of false data, the probability of non-detection of which is dependent on the duration of the imitation. The received encrypted data is decrypted, and a new imitation insert is generated from the received data blocks, which is then compared with the imitation MAC received from the communication channel or from the computer memory. In the event of a mismatch between the insets, all decrypted data is considered false.
A hash function is a mathematical or other function that, for a string of arbitrary length, calculates some integer value or some other string of fixed length. The main purpose of the hash function is to calculate the characteristic feature of the inverse image, which can be considered as a value of the hash function. This value usually has a certain fixed size, for example, 64 or 128 bits. The hash code can be further analyzed to solve a problem. Thus, for example, hashing can be used to compare data: if two data arrays have different hash codes, the arrays are guaranteed to differ. If they are the same, the arrays are most likely the same.
In general, there is a lack of instances of unambiguous correspondence among the source data and the hash code, because the hash function number and values is always smaller than the data options of the input directory. Therefore, there are many input messages that give the same hash codes. The probability of collisions plays an important role in evaluating the quality of hash functions. It is important to note that it is highly difficult to determine the message M, because the latter does not possess a similar hash value as the first message.
Therefore, it challenging to define the pair mechanism of two random messages, which have the same x -function. A block cipher is a type of symmetric cipher that operates with groups of bits of a fixed-length — blocks whose characteristic size varies between 64–256 bits. If the source text or its remainder is smaller than the size of a block, it is supplemented before encryption. In fact, a block cipher is a substitution in the alphabet of blocks, which, as a result, can be mono- or polyalphabetic. Block cipher is an important component of many cryptographic protocols and is widely used to protect data transmitted over the network. | 672 | 3,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.922407 |
http://www.iancgbell.clara.net/maths/bezier.htm | 1,553,613,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00264.warc.gz | 273,168,621 | 4,980 | Bezier Curves
An N-dimensional Bezier curve of order M is a curve of the form p(s) = åi=0M MCi si(1-s)M-ipi where the control points p0,...,pM are N-dimensional vectors. Of greatest interest to programmers are Beziers of order 3. These are given by
p(s) = (1-s)3p0 + 3s(1-s)2p1 + 3s2(1-s)p2 + s3p3 = s3a3 + s2a2 + sa1 + a0
where a0 = p0 ; a1 = -3p0+3p1 ; a2 = 3p0-6p1+3p2 ; a3 = -p0+3p1-3p2+p3 are the coefficient points of the bezier.
Subdividing Beziers
A Bezier curve with control points p0,p1,p2,p3 can be split (at s=½) into two Beziers having control points p00,p01,p02,p03 and p10,p11,p12,p13 where
p00 = p(0)=p0
p01 = (p0+p1)/2
p02 = (p0+2p1+p2)/4
p03 = p10 = p(½) = (p0+3(p1+p2)+p3)/8
p11 = (p3+2p2+p1)/4
p12 = (p2+p3)/2
p13 = p(1)=p3.
To split at a general s=t, it is easier to manipulate coefficient rather than control points. We can construct a bezier for the [0,t] portion by substituting s=ut to obtain coefficients (a0, a1t, a2t2, a3t3). The [t,1] portion is given by the coefficients of u in a(t + u(1-t)).
Bezier Approximations
To Cubic
The cubic a3s3 + a2s2 + a1s + a0 is precisely equivalent to a Bezier with control points
p3 = (a0+a1+a2+a3)/4 ; p2 = (3a0+2a1+a2)/3 ; p1 = (3a0+a1)/3 ; p0 = a1 .
To Circular Arc
With acknowledgment to the work of David Seal
Suppose we wish to construct a 2D bezier curve which approximates the unit circular arc from ( cosa,- sina) to ( cosa, sina).
If we impose a second order fit at the arc endpoints then our bezier must have control points
p0 = ( cosa,- sina) ; p1 = p0 + b( sina, cosa) ; p2 = p3 + b( sina,- cosa) ; p3 = ( cosa, sina) ;
whence the curve is given by
p1(s) = 3b sinas(1-s) + cosa p2(s) = (-4 sina +6b cosa)s3 + (6 sina - 9b cosa)s2 + 3b cosas - sina = (2s-1)((2 sina-3b cosa)s(1-s) + sina) = (2s-1)((1+2s(1-s)) sina -3b cosas(1-s)) Writing z = s(1-s) p1(s) = 3b sinaz + cosa p2(s) = (2s-1)((1+2z) sina -3b cosaz)
Consider the error terms g2(s) = p(s)2 - 1 and g(s) = |p(s)| - 1.
g2(s)= p(s)2-1 = (|p(s)|+1)g(s) = (2+g(s))g(s) whence g(s) » g2(s)/2.
g2(s) = (3b sinaz + cosa)2 + (2s-1)2((1+2z) sina -3b cosaz)2 -1 = (9b2sin2az2 + cos2a) -1 + (1-4z)((1+2z)2sin2a -6b cosaz(1+2z) sina +9b2cos2az2) = (9b2sin2az2 + cos2a) -1 + (1-4z)(z2(4sin2a - 12b cosa sina + 9b2cos2a) +z(4sin2a -6b cosa sina) + sin2a) = z3(48b cosa sina - 16sin2a - 36b2cos2a) +z2( 9b2sin2a + (4sin2a - 12b cosa sina + 9b2cos2a) - 4 ( 4 sin2a -6b cosa sina )) = -4z3(3b cosa-2 sina)2 + z2(9b2 + 12b cosa sina - 12sin2a)
dg2/ds = (1-2s)(-12z2(3b cosa-2 sina)2 +2z(9b2 + 12b cosa sina - 12sin2a)) = (1-2s)z(-12z(3b cosa-2 sina)2 +2(9b2 + 12b cosa sina - 12sin2a)) = 6(1-2s)z(-2z(3b cosa-2 sina)2 + (3b2 + 4b cosa sina - 4sin2a))
Hence dg2/ds = 0 when s=0,1, ½, or when z = (3b2 + 4b cosa sina - 4sin2a))/(-2z(3b cosa-2 sina)2. These give g2(s) = 0, 0, (-(3b cosa-2 sina)2+(9b2+12b cosa sina-12sin2a))/16 = (9b2sin2a+24b cosa sina-16sin2a)/16, and
(3b2+4b cosa sina-4sin2a)3/(4(3b cosa-2 sina)4) repsectively.
We can ensure g2(½)=0 by choosing b0 such that p1(½)=1 Þ 3b0 sina½(1-½) + cosa = 1 Þ b0 = 4(1- cosa)/(3 sina).
The maximal error is then
g2max = (3b02+4b0 cosa sina-4sin2a)3/ (4(3b0 cosa-2 sina)4) = (16(1- cosa)2/(3sin2a)+16(1- cosa) cosa/3-4sin2a)3 / (4(4(1- cosa) cosa/ sina-2 sina)4) = ... = (1- cosa)3/(27(1+ cosa)) Þ gmax = (1- cosa)3/(54(1+ cosa)) » (1- cosa)3/108 when a small
a gmax Pixel radius for > ½ pixel error p/2 0.019 27 p/3 .0015 324 p/4 2.8 x 10-4 1831 p/8 4.24 x 10-6 117770 p/16 6.6 x 10-8 7538744 d d6/864 432 d-6
To General 1-D Curve
Suppose we wish to construct a 1D bezier curve q(s)=s3q3 + s2q2 + sq1 + q0 with given values q(0), q(1) and gradients q'(0), q'(1) .
æ 1 1 1 1 ö æ q0 ö = æ q(1) ö ç 0 1 2 3 ÷ ç q1 ÷ ç q'(1) ÷ ç 0 1 0 0 ÷ ç q2 ÷ ç q'(0) ÷ è 1 0 0 0 ø è q3 ø è q(0) ø
is easily solved by q0 = q(0) ; q1 = q'(0) ; q2 = 1/3(q'(1)-2q1-q0) = 1/3(q'(1)-2q'(0)-q(0)) = ; q3 = q(0)-q0-q1-q2 = -(q1+q2) = -1/3(q'(1)+q'(0)-q(0)) .
To General N-D Curve
Suppose we wish to construct a 2D bezier curve q(s)=s3q3 + s2q2 + sq1 + q0 through given points q(0), q(t) and q(1).
Suppose we also require q'(0) = m0 g0 , q'(1)= m1g1 where g0 and g1 are given. If g0 and g1 are non parallel this may be achieved. If g0 and g1 are (close to) parallel then a (viable) fit is only possible if q(0), q(t) and q(1) are colinear and in accordance with the desired gradient.
æ 1 t t2 t3 ö æ q0 ö = æ q(t) ö ç 1 1 1 1 ÷ ç q1 ÷ ç q(1) ÷ ç 0 1 2 3 ÷ ç q2 ÷ ç m1g1 ÷ ç 0 1 0 0 ÷ è q3 ø ç m0g0 ÷ è 1 0 0 0 ø è q(0) ø
Substituting q0 = q(0), then q1 = m0g0, and finally q2 = q(1)-q0-m0g0 - q3 we obtain:
æ t t2 t3 ö æ q1 ö = æ q(t) - q0 ö ç 1 1 1 ÷ ç q2 ÷ ç q(1)-q0 ÷ ç 1 2 3 ÷ è q3 ø ç m1g1 ÷ è 1 0 0 ø è m0g0 ø
Substituting q1 = m0g0 :
æ t2 t3 ö æ q2 ö = æ q(t) - q0 - tm0g0 ö ç 1 1 ÷ è q3 ø ç q(1)-q0-m0g0 ÷ è 2 3 ø è m1g1-m0g0 ø
Substituting q2 = q(1)-q0-m0g0 - q3 and then eliminating q3:
t2(q(1)-q0-m0g0 - q3) + t3q3 = q(t) - q0 - tm0g0 2(q(1)-q0-m0g0 - q3) + 3q3 = m1g1-m0g0 Þ m0g0(1-t)/t - m1g1 = q(t)/t2(1-t) + q(1)(2t-3)/(1-t) - q0(2t3 - 3t2 +1)/t2(1-t))
If t is known and g0 and g1 are not parallel this may be solved for m0,m1 whereupon we can set
• q0 = q(0)
• q1 = m0g0
• q3 = m1g1 + q1 -2q(1) +2q0
• q2 = q(1) - q3 - q1 - q0
What is a sound choice for t ?
t =½ is an obvious candidate, the equations for m0 and m1 becoming
m0g0 - m1g1 = 8q(½) - 4q(1) - 4q0
The point of least curvature is given by minimising q"(t)2 = (6q3t + 2q2)2 which is minimised by t = -q2.q3/3 so if this lies comfortably inside (0,1) it may be a good choice. Please let me know if progress is made in this direction.
Glossary Contents Author
Copyright (c) Ian C G Bell 1998
Web Source: www.iancgbell.clara.net/maths or www.bigfoot.com/~iancgbell/maths
Latest Edit: 09 Jun 2007. | 2,749 | 5,768 | {"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} | 3.828125 | 4 | CC-MAIN-2019-13 | latest | en | 0.705212 |
http://compaland.com/margin-of/what-does-within-the-margin-of-error-mean.html | 1,519,438,379,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00782.warc.gz | 74,345,039 | 6,264 | ## Repair What Does Within The Margin Of Error Mean Tutorial
Home > Margin Of > What Does Within The Margin Of Error Mean
# What Does Within The Margin Of Error Mean
## Contents
Because the results of most survey questions can be reported in terms of percentages, the margin of error most often appears as a percentage, as well. Interpretation: If one obtains many unbiased samples of the same size from a defined population, the difference between the sample percent and the true population percent will be within the margin Because it is impractical to poll everyone who will vote, pollsters take smaller samples that are intended to be representative, that is, a random sample of the population.[3] It is possible If an approximate confidence interval is used (for example, by assuming the distribution is normal and then modeling the confidence interval accordingly), then the margin of error may only take random http://compaland.com/margin-of/what-does-a-5-margin-of-error-mean.html
On the other hand, if those percentages go from 50 percent to 54 percent, the conclusion is that there is an increase in those who say service is "very good" albeit Census Bureau. Si vous les avez manquées, mes project... Tags: confidence intervals, population Before posting, create an account!Stop this in-your-face noticeReserve your usernameFollow people you like, learn fromExtend your profileGain reputation for your contributionsNo annoying captchas across siteAnd much more!
## Acceptable Margin Of Error
Cette course 2016 fût bien différente de celle de 20... 73% of Canadians want a referendum on electoral reform according to Ipsos poll I already wrote in details why I was Stokes, Lynne; Tom Belin (2004). "What is a Margin of Error?" (PDF). Un front PQ-QS (et vert) dans Verdun? For election surveys in particular, estimates that look at “likely voters” rely on models and predictions about who will turn out to vote that may also introduce error.
Sur l'île de Montréal, c'est NPD vs PLC alors que ... ► August (6) ► July (3) ► June (6) ► May (4) ► April (4) ► March (2) ► 2014 It is a subsidiary of The Pew Charitable Trusts. Normally researchers do not worry about this 5 percent because they are not repeating the same question over and over so the odds are that they will obtain results among the Margin Of Error Confidence Interval Calculator You need to do different calculations.
and R.J. The tick marks include 45 twice. Harry Contact iSixSigma Get Six Sigma Certified Ask a Question Connect on Twitter Follow @iSixSigma Find us around the web Back to Top © Copyright iSixSigma 2000-2016. Analysts should be mindful that the samples remain truly random as the sampling fraction grows, lest sampling bias be introduced.
But a series of polls showing a gradual increase in a candidate’s lead can often be taken as evidence for a real trend, even if the difference between individual surveys is Margin Of Error Calculator If p moves away from 50%, the confidence interval for p will be shorter. The level of observed change from one poll to the next would need to be quite large in order for us to say with confidence that a change in the horse-race Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
## Margin Of Error Synonym
In particular they got lucky enough not to poll BC 2013 or Alberta 2012. http://www.pewresearch.org/fact-tank/2016/09/08/understanding-the-margin-of-error-in-election-polls/ About Fact Tank Real-time analysis and news about data from Pew Research writers and social scientists. Acceptable Margin Of Error More » Login Form Stay signed in Forgot your password? Margin Of Error In Polls In other words, Company X surveys customers and finds that 50 percent of the respondents say its customer service is "very good." The confidence level is cited as 95 percent plus
It's probably because the margins of error don't correctly represent all the uncertainty that exist (people can lie, change their mind, etc). this contact form Notice that each pollster misses at least one party (i.e: doesn't have this party within the margins of error). The margin of error for the difference between two percentages is larger than the margins of error for each of these percentages, and may even be larger than the maximum margin But polls often report on subgroups, such as young people, white men or Hispanics. Margin Of Error Sample Size
And with responses rates of 10% sometimes, there is just as much of a selection process (after all, you need to answer your phone and accept to answer the questions). Projections finales pour la course au PQ 2016 Nous y voici enfin! For example, in a random survey of 1,000 eligible voters, a result of 50% has a margin of error of +/- 3.1 percentage points, but a result of 2% has a have a peek here With this comes the caveat that we'll see variations.
Quite often, they only get half of them! Margin Of Error Vs Standard Error The twice the margin is only valid if there are only two parties. Pollsters report the margin of error for an estimate of 50% because it is the most conservative, and for most elections featuring two candidates, the levels of support tend to be
## I mean if I took a sample of 1000 from a population of 2000 I would think the results would have a smaller margin of error than if I took a
Blackwell Publishing. 81 (1): 75–81. Journal of the Royal Statistical Society. The definition is that 95 percent of the time, the sampled result should fall within that margin of the result you’d get by sampling everybody. What Does Margin Of Error Mean In Confidence Intervals The Gallup poll reported a margin of error of plus or minus 2 percent, while the UConn/Hartford Courant poll reported a 3 percent margin of error — so even if you
Being super accurate shouldn't always be expected. Many poll watchers know that the margin of error for a survey is driven primarily by the sample size. So, statisticians use the laws of probability to ensure that at least 95% of the time, the difference between the sample percent and the population percent will be within the margin Check This Out Sampling: Design and Analysis.
The Liberals were at 32.3% and the Tories at 28.9%. For example, a survey may have a margin of error of plus or minus 3 percent at a 95 percent level of confidence. Therefore, if 100 surveys are conducted using the same customer service question, five of them will provide results that are somewhat wacky. Reply TPRJones I don't understand how the margin of error calculation doesn't take the population size into consideration.
Reply New JobCentura HealthManager Value Optimization Main Menu New to Six Sigma Consultants Community Implementation Methodology Tools & Templates Training Featured Resources What is Six Sigma? For a subgroup such as Hispanics, who make up about 15% of the U.S. Being "within the margin of error" simply means that if the two parties were tied for real and you were to sample a 100 times, there would be more than 5% This information means that if the survey were conducted 100 times, the percentage who say service is "very good" will range between 47 and 53 percent most (95 percent) of the
It's not that simple. Plus, a collection of individual polls showing insignificant leads can actually give us an overall good picture of the race. 3. Election Canada has relatively strict guidelines on how to report polls, but somehow nothing about this. But how can we distinguish real change from statistical noise?
Thanks f Reply James Jones Great explanation, clearly written and well appreciated. In Poll B, which also has a 3-point margin of error for each individual candidate and a 6-point margin for the difference, the Republican lead of 8 percentage points is large People need to realize that in 2015, phone samples are also anything but random. | 1,688 | 7,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-09 | latest | en | 0.817932 |
https://mathoverflow.net/questions/351584/theory-of-integration-for-functions-from-mathbbz-p-to-mathbbz-q-fo | 1,660,536,151,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00217.warc.gz | 349,224,507 | 24,874 | # Theory of integration for functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{q}$ for distinct primes $p,q$
Let $$p$$ and $$q$$ be prime numbers. When $$p=q$$, Mahler's Theorem gives a complete description of $$C\left(\mathbb{Z}_{p};\mathbb{Z}_{p}\right)$$, the space of continuous functions from $$\mathbb{Z}_{p}$$ to $$\mathbb{Z}_{p}$$. I'm wondering (possibly in vain) if there might be a comparable classification of $$C\left(\mathbb{Z}_{p};\mathbb{Z}_{q}\right)$$ when $$p$$ and $$q$$ are distinct.
I ask only because I've been doing $$p$$-adic harmonic analysis, but have found myself having to brave the wilds of $$L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$$, the space of all $$f:\mathbb{Z}_{p}\rightarrow\mathbb{C}_{q}$$ so that:$$\sup_{\mathfrak{z}\in\mathbb{Z}_{p}}\left|f\left(\mathfrak{z}\right)\right|_{q}<\infty$$
Pontryagin duality lets me do Fourier analysis on $$L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}\right)$$; for $$p=q$$, on the other hand, I can use things like the volkenborn integral, or the amice transform / mazur-mellin transform—$$p$$-adic distributions, in general. The problem is, without a structure theorem like Mahler's for the $$p\neq q$$ case, though I can define “integration” on $$L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$$ by elements of its dual space (continuous functionals $$\varphi:L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)\rightarrow\mathbb{C}_{q})$$, I don't see a way to do useful computations for the specific, non-abstract functions that I'm trying to fourier analyze.
So, I guess what I'm really asking is: how do you take the "integral" or "fourier transform" of such a function?
Any thoughts? Reference recommendations? Etc.?
$${\mathbb Z}_p$$ and $${\mathbb Z}_q$$ are homeomorphic; hence so are $$C({\mathbb Z}_p,{\mathbb Z}_p)$$ and $$C({\mathbb Z}_p,{\mathbb Z}_q)$$. | 610 | 1,863 | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.70383 |
https://web2.0calc.com/questions/help_13792 | 1,566,446,049,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00233.warc.gz | 689,894,151 | 5,714 | +0
# HELP
+1
276
2
+736
What is the smallest five-digit palindrome that is divisible by 11?
Jun 5, 2018
#1
+1
I think it is 11 0 11 = 1,001 x 11
Jun 5, 2018
edited by Guest Jun 5, 2018
#2
+736
+1
I got it and its 10901 but thanks!!!
MIRB16 Jun 5, 2018 | 118 | 262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.944702 |
https://www.spe.org/en/jpt/jpt-article-detail/?art=1356 | 1,566,226,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00067.warc.gz | 980,748,521 | 14,132 | ## Modeling Transient Wellbore Temperature During Diagnostic Fracture Injection Tests
Fig. 1—BHP, bottomhole-temperature (BHT), and WHP data gathered during a falloff test.
Diagnostic fracture injection tests (DFITs) have gained widespread usage in the evaluation of unconventional reservoirs. In typical field operations, pressure is measured at the wellhead, not at the bottom of the hole. The bottomhole pressure (BHP) is obtained by adding a constant hydrostatic head of the water column to the wellhead pressure (WHP) at each timestep. One can question the soundness of this practice because of significant changes in temperature that occur in the wellbore, leading to changes in density and compressibility throughout the fluid column. This paper offers an analytical model for estimating the transient temperature at a given depth and timestep, for computing the BHP.
## Temperature Model During Pressure-Falloff Test
After a well is shut in at the surface, afterflow at the sandface is negligible because of low formation permeability. Upon cessation of injection, the cold injection water begins to gain heat from the surroundings. Heat transfer during flowing and shut-in conditions has been modeled with an energy-balance equation, accounting for various resistances to heat transfer. The resulting differential equations are solved numerically. However, with robust assumptions, an analytical expression can be developed for transient fluid temperature during falloff. The heat flow from the formation into the wellbore raises the internal energy of the fluid and of the composite tubing/casing/cement material. Please see the complete paper for the quantitative expression.
This article, written by Special Publications Editor Adam Wilson, contains highlights of paper SPE 166120, “Modeling Wellbore Transient Fluid Temperature and Pressure During Diagnostic Fracture Injection Testing in Unconventional Reservoirs,” by B. Nojabaei, SPE, The Pennsylvania State University, A.R. Hasan, SPE, Texas A&M University, and C.S. Kabir, SPE, Hess, prepared for the 2013 SPE Annual Technical Conference and Exhibition, New Orleans, 30 September–2 October. The paper has not been peer reviewed.
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Modeling Transient Wellbore Temperature During Diagnostic Fracture Injection Tests
01 February 2015
Volume: 67 | Issue: 2
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https://ntv.ifmo.ru/en/article/13702/odnomernye_zadachi_gazovoy_dinamiki_i_ih_resheniepri_pomoschi_raznostnyh_shem.htm | 1,701,628,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00474.warc.gz | 494,757,150 | 9,391 | doi: 10.17586/2226-1494-2015-15-4-731-740
# ONE-DIMENSIONAL GAS DYNAMICS PROBLEMS AND THEIR SOLUTION BASED ON HIGH-RESOLUTION FINITE DIFFERENCE SCHEMES
P. V. Bulat, K. N. Volkov
Article in Russian
For citation: Bulat P.V., Volkov K.N. Оne-dimensional gas dynamics problems and their solution based on high-resolution finite difference schemes. Scientific and Technical Journal of Information Technologies, Mechanics and Optics, 2015, vol.15, no. 4, pp. 731–740.
Abstract
One-dimensional unsteady gas dynamics problems are revealing tests for the accuracy estimation of numerical solution with respect to simulation of supersonic flows of inviscid compressible gas. Numerical solution of Euler equations describing flows of inviscid compressible gas and conceding continuous and discontinuous solutions is considered. Discretization of Euler equations is based on finite volume method and WENO finite difference schemes. The numerical solutions computed are compared with the exact solution of Riemann problem. Monotonic correction of derivatives makes possible avoiding new extremes and ensures monotonicity of the numerical solution near the discontinuity, but it leads to the smoothness of the existing minimums and maximums and to the accuracy loss. Calculations with the use of WENO schemes give the possibility for obtaining accurate and monotonic solution with the presence of weak and strong gas dynamical discontinuities.
Keywords: gas dynamics, finite difference scheme, shock wave, rarefaction wave, contact discontinuity, Riemann problem, Sod problem, Lax problem.
References
1. Kulikovskii A.G., Pogorelov N.V., Semenov A.Yu. Matematicheskie Voprosy Chislennogo Resheniya Giperbolicheskikh Sistem Uravnenii [Mathematical Problems in the Numerical Solution of Hyperbolic Systems]. Moscow, Fizmatlit Publ., 2001, 608 p.
2. Wesseling P. Principles of Computational Fluid Dynamics. Springer, 2000, 664 p. doi: 10.1007/978-3-642- 05146-3
3. Volkov K.N. Raznostnye skhemy rascheta potokov povyshennoi razreshayushchei sposobnosti i ikh primenenie dlya resheniya zadach gazovoi dinamiki [High-resolution difference schemes of flux calculation and their application to solving gas dynamics problems]. Vychislitel'nye Metody i Programmirovanie, 2005, vol. 6, no. 1, pp. 146–167.
4. Volkov K.N., Deryugin Yu.N., Emel'yanov V.N., Kozelkov A.S., Teterina I.V. Raznostnye Skhemy v Zadachakh Gazovoi Dinamiki na Nestrukturirovannykh Setkakh [Difference Schemes in Gas Dynamic Problems on Unstructured Grids]. Moscow, Fizmatlit Publ., 2014, 412 p.
5. Wolf W.R., Azevedo J.L.F. High-order ENO and WENO schemes for unstructured grids. International Journal for Numerical Methods in Fluids, 2007, vol. 55, no. 10, pp. 917–943. doi: 10.1002/fld.1469
6. Castro M., Costa B., Don W.S. High order weighted essentially non-oscillatory WENO-Z schemes for hyperbolic conservation laws. Journal of Computational Physics, 2011, vol. 230, no. 5, pp. 1766–1792. doi: 10.1016/j.jcp.2010.11.028
7. Clain S., Diot S., Loubere R. A high-order finite volume method for systems of conservation laws-multidimensional optimal order detection (MOOD). Journal of Computational Physics, 2011, vol. 230, no. 10, pp. 4028–4050. doi: 10.1016/j.jcp.2011.02.026
8. Hu G.H., Li R., Tang T. A robust WENO type finite volume solver for steady Euler equations on unstructured grids. Communications in Computational Physics, 2011, vol. 9, no. 3, pp. 627–648. doi: 10.4208/cicp.031109.080410s
9. Su X., Sasaki D., Kazuhiro N. Efficient implementation of WENO scheme on structured meshes. Proc. 25th
Computational Fluid Dynamics Symposium. Osaka, Japan, 2011, no. C01-3, 9 p.
10. Tsoutsanis P., Titarev V.A., Drikakis D. WENO schemes on arbitrary mixed-element unstructured meshes in three space dimensions. Journal of Computational Physics, 2011, vol. 230, no. 4, pp. 1585–1601. doi: 10.1016/j.jcp.2010.11.023
11. Vincent P.E., Castonguay P., Jameson A. A new class of high-order energy stable flux reconstruction schemes. Journal of Scientific Computing, 2011, vol. 47, no. 1, pp. 50–72. doi: 10.1007/s10915-010-9420-z
12. Rozhdestvenskii B.L., Yanenko N.N. Sistemy Kvazilineinykh Uravnenii i ikh Prilozheniya k Gazovoi Dinamike [Systems of Quasilinear Equations and their Applications to Gas Dynamics]. Moscow, Nauka Publ., 1978, 688 p.
13. van der Heul D.R., Vuik C., Wesseling P. A conservative pressure-correction method for flow at all speeds. Computers and Fluids, 2003, vol. 32, no. 8, pp. 1113–1132. doi: 10.1016/S0045-7930(02)00086-5
14. Xiao F. Unified formulation for compressible and incompressible flows by using multi-integrated moments. I. One-dimensional inviscid compressible flow. Journal of Computational Physics, 2004, vol. 195, no. 2, pp. 629–654. doi: 10.1016/j.jcp.2003.10.014
15. Sod G.A. A survey of several finite difference methods for systems of nonlinear hyperbolic conservation
laws. Journal of Computational Physics, 1978, vol. 27, no. 1, pp. 1–31. doi: 10.1016/0021-9991(78)90023-2
16. Lax P.D. Weak solutions of nonlinear hyperbolic equations and their numerical computation. Communications on Pure and Applied Mathematics, 1954, vol. 7, no. 1, pp. 159–193.
17. Arora M., Roe P.L. A well-behaved TVD limiter for high-resolution calculations of unsteady flow. Journal of Computational Physics, 1997, vol. 132, no. 1, pp. 3–11. doi: 10.1006/jcph.1996.5514
18. Woodward P.R., Colella P. The numerical simulation of two-dimensional fluid flow with strong shocks. Journal of Computational Physics, 1984, vol. 54, no. 1, pp. 115–173. doi: 10.1016/0021-9991(84)90142-6
19. Einfeldt B., Munz C.D., Roe P.L., Sjogren B. On Godunov-type methods near low densities. Journal of Computational Physics, 1991, vol. 92, no. 2, pp. 273–295. doi: 10.1016/0021-9991(91)90211-3
20. Sjogreen B., Yee H.C. Variable high order multiblock overlapping grid methods for mixed steady and unsteady multiscale viscous flow. Communications in Computational Physics, 2009, vol. 5, no. 2–4, pp. 730–744. | 1,817 | 5,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.734261 |
http://betterlesson.com/lesson/443421/practice-with-measures-of-central-tendency | 1,488,056,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00064-ip-10-171-10-108.ec2.internal.warc.gz | 30,832,539 | 20,188 | # Practice with Measures of Central Tendency
Unit 5: Modeling With Statistics
Lesson 3 of 19
## Big Idea: This lesson gives students the opportunity to apply what they have learned about the measures of central tendency.
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Subject(s):
Math, measures of central tendency, Statistics, modeling, mean, median, mode, range, 9th grade
41 minutes
### James Bialasik
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Nothing to upload details close | 319 | 1,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-09 | latest | en | 0.889885 |
https://sonorarural.com/qa/question-how-can-you-tell-how-old-a-forest-is.html | 1,618,853,879,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00316.warc.gz | 639,494,138 | 7,969 | # Question: How Can You Tell How Old A Forest Is?
## How do you determine the age of a forest?
Here is the formula: Diameter X Growth Factor = Approximate Tree Age.
Let’s use a red maple to calculate age.
A red maple’s growth factor has been determined to be 4.5 and you have determined that its diameter is 10 inches: 10 inch diameter X 4.5 growth factor = 45 years..
## How can you determine the age of a tree without cutting it?
You can get a rough estimate of the age of a tree without cutting it down and counting the rings. The girth of a tree can be used to estimate its age, as roughly a tree will increase it’s girth by 2.5cm in a year. So, simply measure around the trunk of the tree (the girth) at about 1m from the ground.
## What is considered old growth wood?
Old-growth wood refers to wood from trees that belonged to forests that grew up over hundreds of years. A majority of today’s lumber is harvested from trees that have been cultivated to grow rapidly, so the wood is not as dense.
## What does an old growth forest look like?
Typical characteristics of old-growth forest include presence of older trees, minimal signs of human disturbance, mixed-age stands, presence of canopy openings due to tree falls, pit-and-mound topography, down wood in various stages of decay, standing snags (dead trees), multilayered canopies, intact soils, a healthy …
## How tall is a 10 year old oak tree?
Under optimal conditions, northern red oak is fast growing and a 10-year-old tree can be 15–20 feet tall. In many forests, it grows straight and tall, to 90 ft, exceptionally to 140 ft tall, with a trunk of up to 20–40 inches diameter. Trees may live up to 500 years.
## How can you tell how old a poplar tree is?
How to Tell the Age of a Tree Without Cutting it DownWrap the tape measure around the tree at about four and a half feet above the ground. This measurement is the tree’s circumference. … Use the circumference to find the diameter of the tree. … Determine the age of the tree by multiplying the diameter by the growth factor.
## How old is a tree?
If you know when the tree was planted and the age of the tree at the time of planting, obviously, you can easily and accurately determine its age. Most trees are between 5 – 10 years when they come out of the nursery. The second most accurate way to estimate tree age is to count the annual rings of wood growth.
## How can you work out the age of a tree?
This is the girth or circumference of the tree. Roughly, every 2.5cm of girth represents about one year’s growth. So to estimate the age of a living tree, divide the girth by 2.5. For example a tree with a girth of 40cm will be sixteen years old.
## Does Pine get harder with age?
Because wood does gain strength as it loses moisture content. At around 12 percent moisture content, it might be as much as 50% stronger than as rough-cut green lumber! Aged lumber, unlike fine wine or whiskey, generally does not get better with age.
## Can you tell the age of a tree by the rings?
If you know when the tree was planted, you can easily and accurately determine its age. The second most accurate way to estimate tree age is to count the annual rings of wood growth. … For trees that are dead and have been cut down, you can count the rings on the stump.
## What do dark tree rings mean?
The light and dark rings of a tree. … These rings can tell us how old the tree is, and what the weather was like during each year of the tree’s life. The light-colored rings represent wood that grew in the spring and early summer, while the dark rings represent wood that grew in the late summer and fall. | 842 | 3,638 | {"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} | 3.015625 | 3 | CC-MAIN-2021-17 | latest | en | 0.942339 |
https://flatearthsoc.com/flat-earth-proofs-flat-earth-news-blog-games.html | 1,571,454,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688674.52/warc/CC-MAIN-20191019013909-20191019041409-00419.warc.gz | 496,384,356 | 8,431 | If the Earth were a globe, rolling and dashing through "space" at the rate of "a hundred miles in five seconds of time," the waters of seas and oceans could not, by any known law, be kept on its surface - the assertion that they could be retained under these circumstances being an outrage upon human understanding and credulity! But as the Earth - that is, the habitable world of dry land - is found to be "standing out of the wafer and in the water" of the "mighty deep," whose circumferential boundary is ice, we may throw the statement back into the teeth of those who make it and flaunt before their faces the flag of reason and common sense, inscribed with a proof that the Earth is not a globe.
35) If the Earth were truly a globe, then every line of latitude south of the equator would have to measure a gradually smaller and smaller circumference the farther South travelled. If, however, the Earth is an extended plane, then every line of latitude south of the equator should measure a gradually larger and larger circumference the farther South travelled. The fact that many captains navigating south of the equator assuming the globular theory have found themselves drastically out of reckoning, moreso the farther South travelled, testifies to the fact that the Earth is not a ball.
Another MAJOR problem with the Ball earth model is that the path of the total eclipse shadow that is coming on August 21 in North America (and all paths of the solar eclipses) is only 73 miles across! How can a shadow be SMALLER than the object casting the shadow? This is physically impossible! We know from experience that shadows can be the same size or larger than the object casting the shadow, but it can never be smaller. We are told that the moon is 2,159 miles in diameter. So shouldn’t the moon’s shadow on earth be at LEAST 2,159 miles wide? But instead we are given the path of the next eclipse across the United States and it is only 73 miles wide. You have to be in a very specific location to even see the total eclipse.
34) Ship captains in navigating great distances at sea never need to factor the supposed curvature of the Earth into their calculations. Both Plane Sailing and Great Circle Sailing, the most popular navigation methods, use plane, not spherical trigonometry, making all mathematical calculations on the assumption that the Earth is perfectly flat. If the Earth were in fact a sphere, such an errant assumption would lead to constant glaring inaccuracies. Plane Sailing has worked perfectly fine in both theory and practice for thousands of years, however, and plane trigonometry has time and again proven more accurate than spherical trigonometry in determining distances across the oceans.
Look up Admiral Byrd's tape from the 1920's (you-tube) he told us of the many continents beyond ours, all on the same plane. He said it would take years to explore them all. Then he retired from the navy and they started NASA to hide all that stuff. Yep thousands of NASA employees know the truth and are trained to keep us stupid. shows like coast to coast will never talk about the flat earth, wont even discus it, only to say NASA is correct case closed. Sounds like brain washing to me.
72) October 16, 1854 the Times newspaper reported the Queen’s visit to Great Grimsby from Hull recording they were able to see the 300 foot tall dock tower from 70 miles away. On a ball-Earth 25,000 miles in circumference, factoring their 10 foot elevation above the water and the tower’s 300 foot height, at 70 miles away the dock tower should have remained an entire 2,600 feet below the horizon.
135) Not only is the Moon clearly self-luminescent, shining its own unique light, but it is also largely transparent. When the waxing or waning Moon is visible during the day it is possible to see the blue sky right through the Moon. And on a clear night, during a waxing or waning cycle, it is even possible to occasionally see stars and “planets” directly through the surface of the Moon! The Royal Astronomical Society has on record many such occurrences throughout history which all defy the heliocentric model.
#### If the Earth were a globe, people - except those on the top - would, certainly, have to be "fastened" to its surface by some means or other, whether by the "attraction" of astronomers or by some other undiscovered and undiscoverable process! But, as we know that we simply walk on its surface without any other aid than that which is necessary for locomotion on a plane, it follows that we have, herein, a conclusive proof that Earth is not a globe.
If we stand on the sands of the sea-shore and watch a ship approach us, we shall find that she will apparently "rise" - to the extent, of her own height, nothing more. If we stand upon an eminence, the same law operates still; and it is but the law of perspective, which causes objects, as they approach us, to appear to increase in size until we see them, close to us, the size they are in fact. That there is no other "rise" than the one spoken of is plain from the fact that, no matter how high we ascend above the level of the sea, the horizon rises on and still on as we rise, so that it is always on a level with the eye, though it be two-hundred miles away, as seen by Mr. J. Glaisher, of England, from Mr. Coxwell's balloon. So that a ship five miles away may be imagined to be "coming up" the imaginary downward curve of the Earth's surface, but if we merely ascend a hill such as Federal Hill, Baltimore, we may see twenty-!five miles away, on a level with the eye - that is, twenty miles level distance beyond the ship that we vainly imagined to be " rounding the curve," and "coming up!" This is a plain proof that the Earth is not a globe.
22.) God's Truth never – no, never – requires a falsehood to help it along. Mr. Proctor, in his " Lessons," says: Men " have been able to go round and round the Earth in several directions." Now, in this case, the word " several will imply more than two, unquestionably: whereas, it is utterly impossible to circumnavigate the Earth in any other than an easterly or a westerly direction; and the fact is perfectly consistent and clear in its relation to Earth as a Plane.. Now, since astronomers would not be so foolish as to damage a good cause by misrepresentation, it is presumptive evidence that their cause is a bad one, and – a proof that Earth is not a globe.
81) The distance from which various lighthouse lights around the world are visible at sea far exceeds what could be found on a ball-Earth 25,000 miles in circumference. For example, the Dunkerque Light in southern France at an altitude of 194 feet is visible from a boat (10 feet above sea-level) 28 miles away. Spherical trigonometry dictates that if the Earth was a globe with the given curvature of 8 inches per mile squared, this light should be hidden 190 feet below the horizon.
With increasing distance from the object, the earth’s curvature causes the surface of the water to fall away from the beam of light. Over one mile, the amount of drop is eight inches, but the drop increases quadratically with distance. Consequently, after three miles the drop is six feet, and after six miles the drop is 24 feet. This is the point of the Bedford level experiment—the curvature of the earth ought to intervene to prevent the mast of the boat being visible from much more than three miles, let alone six miles. However, for the light from the distant object not to be visible, it would have to travel in a straight line. But with a temperature inversion, straight-line motion would carry the light from a cooler layer of air into a warmer layer of air at nearly a grazing angle. The light cannot do this, so it continually is internally reflected, causing the light to bend around the edge of the earth. Therefore, with a temperature inversion, one can see objects that lie well beyond the edge of the earth’s curvature when viewing close to the surface of water.
### Consider a flat plane. The center of mass of a flat plane is in its center, so the force of gravity will pull anything on the surface toward the middle of the plane. That means that if you stand on the edge of the plane, gravity will be pulling you sideways toward the plane's middle, not straight down like you usually experience when you stand on Earth.
As for flight paths and what Appears to be the silly way for a ball earth but makes sense for a flat earth, it reminds me of the child quiz. There is a spider in the corner of the room on the floor and he wants to get to the opp corner on the ceiling. Which is the quickest path? We instantly say, across the floor and up the wall join. BUT, if we flatten the room we then draw a straight line, we find the quickest path is diagonally up one wall and then diagonally across the ceiling, which Looks longer but is best.
Considerably more than a million Earths would be required to make up a body like the Sun -the astronomers tell us: and more than 53,000 suns would be wanted to equal the cubic contents of the star Vega. And Vega is a "small star!" And there are countless millions of these stars! And it takes 30,000,000 years for the light of some of those stars to reach us at 12,000,000 miles in a minute! And, says Mr. Proctor, "I think a moderate estimate of the age of the Earth would be 500,000,000 years! "Its weight," says the same individual, "is 6,000,000,000,000,000,000,060 tons!" Now, since no human being is able to comprehend these things, the giving of them to the world is an insult - an outrage. And though they have all risen from the one assumption that Earth is a planet, instead of upholding the assumption, they drag it down by the weight of their own absurdity, and leave it lying in the dust - a proof that Earth is not a globe.
# 3) The natural physics of water is to find and maintain its level. If Earth were a giant sphere tilted, wobbling and hurdling through infinite space then truly flat, consistently level surfaces would not exist here. But since Earth is in fact an extended flat plane, this fundamental physical property of fluids finding and remaining level is consistent with experience and common sense.
85.) There are rivers which flow east, west, north, an south – that is, rivers are flowing in all directions over the Earth's surface, and at the same time. Now, if the Earth were a globe, some of these rivers would be flowing up-hill and others down, taking it for a fact that there really is an "up" and a "down" in nature, whatever form she assumes. But, since rivers do not flow up-hill, and the globular theory requires that they should, it is a proof that the Earth is not a globe.
135) Not only is the Moon clearly self-luminescent, shining its own unique light, but it is also largely transparent. When the waxing or waning Moon is visible during the day it is possible to see the blue sky right through the Moon. And on a clear night, during a waxing or waning cycle, it is even possible to occasionally see stars and “planets” directly through the surface of the Moon! The Royal Astronomical Society has on record many such occurrences throughout history which all defy the heliocentric model.
Mr. Lockyer says: "The appearances connected with the rising and setting of the Sun and stars may be due either to our earth being at rest and the Sun and stars traveling round it, or the earth itself turning round, while the Sun and stars are at rest." Now, since true science does not allow of any such beggarly alternatives as these, it is plain that modern theoretical astronomy is not true science, and that its leading dogma is a fallacy. We have, then, a plain proof that the Earth is not a globe.
38.) When the Sun crosses the equator, in March, and begins to circle round the heavens in north latitude, the inhabitants of high northern latitudes see him slimming round their horizon and forming the break of their long day, in a horizontal course, not disappearing again for six months, as he rises higher and higher in the heavens whilst he makes his twenty-four hour circle until June, when he begins to descend and goes on until he disappears beyond the horizon in September. Thus, in the northern regions, they have that which the traveler calls the "midnight Sun," as he sees that luminary at a time when, in his more southern latitude, it is always midnight. If,
The ancient Hebrew view of the earth was of a flat earth with a glass firmament that separated the “waters below” from the “waters above” (Gen. 1:6-7; Job 37:28), and in which the sun and moon were placed and had their daily circuit around and above the Flat earth (Gen. 1:14-18; Joshua 10:13; Job 22:14). The earth did not move and was set on a firm foundation, on pillars in the waters of the “deep”, a great ocean underneath the flat earth (Gen. 1:2; 1 Samuel 2:8; 1 Chronicles 16:30; Psalm 104:5; Proverbs 8:28-29). They did not believe in a spinning globe. If the false globe ideas were taught in ancient times, they were taught by the Babylonian Pagans, not God’s people.
```When astronomers assert that it is "necessary" to make "allowance for curvature" in canal construction, it is, of course, in order that, in their idea, a level cutting may be had, for the water. How flagrantly, then, do they contradict themselves when the curved surface of the Earth is a "true level!" What more can they want for a canal than a true level? Since they contradict themselves in such an elementary point as this, it is an evidence that the whole thing is a delusion, and we have a proof that the Earth is not a globe.
```
146) The ball-Earth model claims the Moon orbits around the Earth once every 28 days, yet it is plain for anyone to see that the Moon orbits around the Earth every single day! The Moon’s orbit is slightly slower than the Sun’s, but follows the Sun’s same path from Tropic to Tropic, solstice to solstice, making a full circle over the Earth in just under 25 hours.
9) Engineer, W. Winckler was published in the Earth Review regarding the Earth’s supposed curvature, stating, “As an engineer of many years standing, I saw that this absurd allowance is only permitted in school books. No engineer would dream of allowing anything of the kind. I have projected many miles of railways and many more of canals and the allowance has not even been thought of, much less allowed for. This allowance for curvature means this - that it is 8” for the first mile of a canal, and increasing at the ratio by the square of the distance in miles; thus a small navigable canal for boats, say 30 miles long, will have, by the above rule an allowance for curvature of 600 feet. Think of that and then please credit engineers as not being quite such fools. Nothing of the sort is allowed. We no more think of allowing 600 feet for a line of 30 miles of railway or canal, than of wasting our time trying to square the circle”
Consider a flat plane. The center of mass of a flat plane is in its center, so the force of gravity will pull anything on the surface toward the middle of the plane. That means that if you stand on the edge of the plane, gravity will be pulling you sideways toward the plane's middle, not straight down like you usually experience when you stand on Earth.
23) Ball-believers often claim “gravity” magically and inexplicably drags the entire lower-atmosphere of the Earth in perfect synchronization up to some undetermined height where this progressively faster spinning atmosphere gives way to the non-spinning, non-gravitized, non-atmosphere of infinite vacuum space. Such non-sensical theories are debunked, however, by rain, fireworks, birds, bugs, clouds, smoke, planes and projectiles all of which would behave very differently if both the ball-Earth and its atmosphere were constantly spinning Eastwards at 1000mph. | 3,532 | 15,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-43 | latest | en | 0.948152 |
https://www.jiskha.com/questions/1116025/suppose-a-ball-is-thrown-straight-up-into-the-air-and-the-height-of-the-ball-above-the | 1,604,013,459,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00576.warc.gz | 780,394,921 | 5,007 | # Calculus
Suppose a ball is thrown straight up into the air, and the height of the ball above the ground is given by the function h(t) = 6 + 37t – 16t2, where h is in feet and t is in seconds. At what time t does the ball stop going up and start returning to earth?
1. 👍 0
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3. 👁 517
1. you are given h(t), to you need to take the derivatave of h'(t) and set it to zero.
The derivative of height is velocity.
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👨🏫
bobpursley
2. Ohhhhhhhhhhhhhhhhhhhhh, Thanks!
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A ball is thrown straight upward with a speed of +12 m/s. A) what's the ball's acceleration just after it is thrown? B) how much time does it take for the ball to rise to its maximum height? C) what's the approximate maximum | 980 | 3,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-45 | latest | en | 0.92617 |
https://numbermatics.com/n/91176902501001413556088/ | 1,610,929,918,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00488.warc.gz | 484,065,736 | 8,471 | # 91176902501001413556088
## 91,176,902,501,001,413,556,088 is an even composite number composed of three prime numbers multiplied together.
What does the number 91176902501001413556088 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 112 divisors.
91176902501001413556088 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of one hundred twelve divisors.
## Prime factorization of 91176902501001413556088:
### 23 × 2696 × 3113
(2 × 2 × 2 × 269 × 269 × 269 × 269 × 269 × 269 × 311 × 311 × 311)
See below for interesting mathematical facts about the number 91176902501001413556088 from the Numbermatics database.
### Names of 91176902501001413556088
• Cardinal: 91176902501001413556088 can be written as Ninety-one sextillion, one hundred seventy-six quintillion, nine hundred two quadrillion, five hundred one trillion, one billion, four hundred thirteen million, five hundred fifty-six thousand and eighty-eight.
### Scientific notation
• Scientific notation: 9.1176902501001413556088 × 1022
### Factors of 91176902501001413556088
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 12
• Sum of prime factors: 582
### Divisors of 91176902501001413556088
• Number of divisors d(n): 112
• Complete list of divisors:
• Sum of all divisors σ(n): 172148121212594146317360
• Sum of proper divisors (its aliquot sum) s(n): 80971218711592732761272
• 91176902501001413556088 is a deficient number, because the sum of its proper divisors (80971218711592732761272) is less than itself. Its deficiency is 10205683789408680794816
### Bases of 91176902501001413556088
• Binary: 100110100111010110101101101001011010010111010010001100101000001110111011110002
• Base-36: EUI7CZ31MA80KFS
### Squares and roots of 91176902501001413556088
• 91176902501001413556088 squared (911769025010014135560882) is 8313227549677117822332230543950316813921863744
• 91176902501001413556088 cubed (911769025010014135560883) is 757974337765549456930889000147944282557858750540796671757651637673472
• The square root of 91176902501001413556088 is 301955133258.2398963381
• 91176902501001413556088 is a perfect cube number. Its cube root is 45008542
### Scales and comparisons
How big is 91176902501001413556088?
• 91,176,902,501,001,413,556,088 seconds is equal to 2,899,143,470,855,000 years, 3 weeks, 3 days, 15 hours, 30 minutes, 7 seconds.
• To count from 1 to 91,176,902,501,001,413,556,088 would take you about eleven quadrillion, five hundred ninety-six trillion, five hundred seventy-three billion, eight hundred eighty-three million, four hundred twenty thousand years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 91176902501001413556088 cubic inches would be around 3750711.8 feet tall.
### Recreational maths with 91176902501001413556088
• 91176902501001413556088 backwards is 88065531410010520967119
• The number of decimal digits it has is: 23
• The sum of 91176902501001413556088's digits is 82
• More coming soon! | 1,008 | 3,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.722283 |
https://www.jobilize.com/online/course/12-6-convection-heat-and-heat-transfer-methods-by-openstax?qcr=www.quizover.com&page=2 | 1,568,761,524,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00301.warc.gz | 911,106,845 | 20,497 | # 12.6 Convection (Page 3/4)
Page 3 / 4
Some interesting phenomena happen when convection is accompanied by a phase change . It allows us to cool off by sweating, even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.
## Calculate the flow of mass during convection: sweat-heat transfer away from the body
The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (This evaporation might occur when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)
Strategy
Energy is needed for a phase change ( $Q={\text{mL}}_{\text{v}}$ ). Thus, the energy loss per unit time is
We divide both sides of the equation by ${L}_{\text{v}}$ to find that the mass evaporated per unit time is
Solution
(1) Insert the value of the latent heat from [link] , . This yields
Discussion
Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.
Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere. Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise, where it is colder. More condensation occurs in these colder regions, which in turn drives the cloud even higher. Such a mechanism is called positive feedback, since the process reinforces and accelerates itself. These systems sometimes produce violent storms, with lightning and hail, and constitute the mechanism driving hurricanes.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Got questions? Join the online conversation and get instant answers! | 1,212 | 5,453 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.93033 |
https://leanprover-community.github.io/mathlib4_docs/Mathlib/Topology/Algebra/Module/LinearPMap.html | 1,713,251,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00750.warc.gz | 339,153,086 | 6,376 | # Partially defined linear operators over topological vector spaces #
We define basic notions of partially defined linear operators, which we call unbounded operators for short. In this file we prove all elementary properties of unbounded operators that do not assume that the underlying spaces are normed.
## Main definitions #
• LinearPMap.IsClosed: An unbounded operator is closed iff its graph is closed.
• LinearPMap.IsClosable: An unbounded operator is closable iff the closure of its graph is a graph.
• LinearPMap.closure: For a closable unbounded operator f : LinearPMap R E F the closure is the smallest closed extension of f. If f is not closable, then f.closure is defined as f.
• LinearPMap.HasCore: a submodule contained in the domain is a core if restricting to the core does not lose information about the unbounded operator.
## Main statements #
• LinearPMap.closable_iff_exists_closed_extension: an unbounded operator is closable iff it has a closed extension.
• LinearPMap.closable.exists_unique: there exists a unique closure
• LinearPMap.closureHasCore: the domain of f is a core of its closure
## References #
• [J. Weidmann, Linear Operators in Hilbert Spaces][weidmann_linear]
## Tags #
Unbounded operators, closed operators
### Closed and closable operators #
def LinearPMap.IsClosed {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] (f : E →ₗ.[R] F) :
An unbounded operator is closed iff its graph is closed.
Equations
Instances For
def LinearPMap.IsClosable {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] (f : E →ₗ.[R] F) :
An unbounded operator is closable iff the closure of its graph is a graph.
Equations
Instances For
theorem LinearPMap.IsClosed.isClosable {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
A closed operator is trivially closable.
theorem LinearPMap.IsClosable.leIsClosable {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} {g : E →ₗ.[R] F} (hf : ) (hfg : g f) :
If g has a closable extension f, then g itself is closable.
theorem LinearPMap.IsClosable.existsUnique {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
∃! (f' : E →ₗ.[R] F),
The closure is unique.
noncomputable def LinearPMap.closure {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] (f : E →ₗ.[R] F) :
If f is closable, then f.closure is the closure. Otherwise it is defined as f.closure = f.
Equations
• = if hf : then else f
Instances For
theorem LinearPMap.closure_def {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
theorem LinearPMap.closure_def' {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
theorem LinearPMap.IsClosable.graph_closure_eq_closure_graph {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
The closure (as a submodule) of the graph is equal to the graph of the closure (as a LinearPMap).
theorem LinearPMap.le_closure {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] (f : E →ₗ.[R] F) :
A LinearPMap is contained in its closure.
theorem LinearPMap.IsClosable.closure_mono {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} {g : E →ₗ.[R] F} (hg : ) (h : f g) :
theorem LinearPMap.IsClosable.closure_isClosed {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
If f is closable, then the closure is closed.
theorem LinearPMap.IsClosable.closureIsClosable {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : ) :
If f is closable, then the closure is closable.
theorem LinearPMap.isClosable_iff_exists_closed_extension {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} :
∃ (g : E →ₗ.[R] F), f g
### The core of a linear operator #
structure LinearPMap.HasCore {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] (f : E →ₗ.[R] F) (S : ) :
A submodule S is a core of f if the closure of the restriction of f to S is f.
• le_domain : S f.domain
• closure_eq :
Instances For
theorem LinearPMap.hasCore_def {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} {S : } (h : ) :
theorem LinearPMap.closureHasCore {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] (f : E →ₗ.[R] F) :
For every unbounded operator f the submodule f.domain is a core of its closure.
Note that we don't require that f is closable, due to the definition of the closure.
### Topological properties of the inverse #
theorem LinearPMap.closure_inverse_graph {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : LinearMap.ker f.toFun = ) (hf' : ) (hcf : LinearMap.ker .toFun = ) :
If f is invertible and closable as well as its closure being invertible, then the graph of the inverse of the closure is given by the closure of the graph of the inverse.
theorem LinearPMap.inverse_isClosable_iff {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : LinearMap.ker f.toFun = ) (hf' : ) :
Assuming that f is invertible and closable, then the closure is invertible if and only if the inverse of f is closable.
theorem LinearPMap.inverse_closure {R : Type u_1} {E : Type u_2} {F : Type u_3} [] [] [] [Module R E] [Module R F] [] [] [] [] [] [] [] {f : E →ₗ.[R] F} (hf : LinearMap.ker f.toFun = ) (hf' : ) (hcf : LinearMap.ker .toFun = ) :
If f is invertible and closable, then taking the closure and the inverse commute. | 2,149 | 6,297 | {"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} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.879382 |
http://spotidoc.com/doc/161266/activity-1--whole-numbers-times-fractions-purpose--materi.. | 1,540,192,990,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00291.warc.gz | 334,647,501 | 8,001 | Activity 1 Whole numbers times fractions Purpose: Materials:
```MULTIPLICATION 4.NF.4
Whole Numbers Times Fractions
Purpose:
To multiply whole numbers times fractions
Materials:
Fraction Bars and "Whole Numbers Times Fractions on Number Lines" (attached)
TEACHER MODELING/STUDENT COMMUNICATION
Activity 1 Whole numbers times fractions
Fraction
Bars
pencils and
paper
1. Find the Fraction Bar for the fraction 2/3.
How many shaded parts would there be if there were three of these bars? (3 × 2 = 6)
How many whole bars would this equal? (2, because there are 3 shaded parts in one
of these whole bars.)
How can this be indicated by an addition equation? (2/3 + 2/3 + 2/3 = 6/3 = 2)
How can this be indicated by a multiplication equation? (3 × 2/3 = 6/3 = 2. Discuss
that 3 × 2 represents three times the number of shaded parts in a 2/3 bar.)
2. Find the Fraction Bar for the fraction 5/6.
How many shaded parts would there be if there were four of these bars? (4 × 5 = 20)
How many whole bars would this equal? (3 and 2/6, because there are 6 shaded parts
in one of these whole bars.)
Write both the addition equation and the multiplication equation for four of the 5/6 bars.
5
6
+
5
6
+
5
5
20
2
+
=
=3
6
6
6
6
and 4 ×
5
4!5
20
=
=
6
6
6
Discuss the convenience of using multiplication, and that (4 × 5) in the numerator of
the multiplication equation represents four times the number of shaded parts in a 5/6 bar.
This provides a connection between multiplication involving whole numbers and fractions.
3. Find any whole Fraction Bar and write the
multiplication equation for 5 times the fraction for the bar.
(Look at some examples. Point out that the fraction for a
whole bar can be replaced by 1, that is, 5 × 4/4 = 5 × 1 = 5.
4. Find any zero Fraction Bar and write the
multiplication equation for 5 times the fraction for the
bar. (Point out that the fraction for a zero bar can be
replaced by 0, that is, 5 × 0/4 = 5 × 0 = 0.
5. List a few examples from above. State a rule for multiplying a whole number times a fraction.
(Multiply the whole number times the numerator and keep the denominator.)
6. Discuss the similarity between multiplying a whole number times a whole number and
multiplying a whole number times a fraction, and illustrate with examples. In both of the
following products, the number "5" indicates how many times the whole number 7 will occur
and how many times the fraction 1/3 will occur.
5 × 7 = 7 + 7 + 7 + 7 + 7 and
5×
1
3
=
1
3
+
1
3
+
1
3
+
1
3
+
1
3
That is, multiplication by a whole number can be computed as repeated addition whether a
whole number is multiplied times a whole number, or a whole number is multiplied times a
fraction. This will help students make sense of products when multiplying a whole number
times a fraction.
activity
sheets,
pencils
and
paper
Activity 2 Products of fractions on number lines
1. Distribute the activity sheet "Whole Numbers Times Fractions on Number Lines" to
students. The first two examples of number lines from his activity sheet are completed here.
Once the activity sheet is completed, discuss how jumps on the number line illustrate repeated
addition of fractions, and repeated addition can be written as a whole number times a fraction.
For 6 jumps of size 3/10, the distance on the number line is
3
3
3
3
3
3
18
+
+
+
+
+
=
10
10
10
10
10
10
10
Also discuss that any number of jumps of the same size can be represented as a whole number
times the numerator of the fraction to obtain the total distance on the number line. Cite examples.
For 6 jumps of size 3/10, the total distance on the number line is
6×
Fraction
Bars and
Die
3
6!3
18
8
=
=
=1
10
10
10
10
Game: Each player in turn selects a Fraction Bar and rolls a die. The
player's score is the number from the die times the fraction from the
bar rounded to the nearest whole number. For the roll of the die and the
bar shown here, the player scores 4 points: 5 × ¾ = 15/4 = 3 ¾ which
rounds to 4. The first player to score 11 points wins the game.
INDEPENDENT PRACTICE and ASSESSMENT
Worksheets 4.NF.4 #4 and #5
4.NF.4
Name:
Date
.
Whole Numbers Times Fractions on Number Lines
1. This number line shows 6 jumps of size
3
, starting at the 0 point.
10
a. Write the mixed number under the line for the point after 6 jumps.
b. Write the improper fraction and mixed number to complete the following equations:
3
3
3
3
3
3
+ + + + +
= ____ =
10 10 10 10 10 10
____ ,
and 6 ×
3
= ____ =
10
2. Starting at the 0 point on the following number line draw 7 jumps of size
____
3
.
12
a. Write the mixed number under the line for the point after 7 jumps.
b. Write improper fraction and mixed number to complete the following equations:
3
3
3
3
3
3
3
+ + + + + +
= ____ =
12 12 12 12 12 12 12
____ ,
and 7 ×
3. Starting at the 0 point on the following number line draw 3 jumps of size
3
= ____ =
12
____
3
.
6
a. Write the mixed number under the line for the point after 3 jumps.
b. Write the improper fraction and mixed number:
3 3 3
+ + = ____ =
6 6 6
____ , and 3 ×
4. Starting at the 0 point on the following number line draw 3 jumps of size
3
= ____ =
6
____
3
= ____ =
5
____
3
.
5
a. Write the mixed number under the line for the point after 3 jumps.
b. Write the improper fraction and mixed number:
3 3 3
+ + = ____ =
5 5 5
____ , and 3 ×
``` | 1,549 | 5,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2018-43 | latest | en | 0.914626 |
https://forum.arduino.cc/t/generate-random-number-display-on-screen-and-require-number-input-for-lock/621945 | 1,628,179,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00485.warc.gz | 259,644,435 | 5,564 | # Generate random number, display on screen and require number input for lock
Hello, I am wondering if it is possible to program an Arduino to generate a random, 4 digit number, display that number on a screen of some sort, and also have that number be necessary to open a lock. If at all possible, I would like to have it do this on a button press.
I have done the research to find the generate random number code, which is awesome. I think that the code I would need is this:
``````long randNumber;
void setup() {
Serial.begin(9600);
// noise will cause the call to randomSeed() to generate
// different seed numbers each time the sketch runs.
// randomSeed() will then shuffle the random function.
}
void loop() {
// print a random number from 0 to 9999
randNumber = random(9999);
Serial.println(randNumber);
delay(50);
}
``````
Is that correct? I am pretty new to this stuff, I have been working through a kit I bought and I am really excited to learn more.
Dylan Fleming
Is that correct?
Does it do what you want/expect when you run it ?
Do you want 9999 ever to be generated ?
UKHeliBob:
Does it do what you want/expect when you run it ?
Do you want 9999 ever to be generated ?
Unfortunately, I am not near my arduino in order to test it out, at current. I will be able to test it later today. I am mostly reaching out in order to get a starting point. I am relatively new to most of these things.
As for 9999 being generated, if it is, fine. If it is not, that is also fine. I take it by your question that I need to change the parameters of my coding,
randNumber = random(10000);
Is that more what I should be doing?
Thanks for the response!
``````randNumber = random(10000);
``````
That should give you a number between 0 and 9999 which is what you are aiming for as I understand it.
Keep in mind, as I have found out, random() is not very random.
Your code will generate a new number every 50ms or so. To do what you describe you need to generate one number, save it then compare it to what is entered.
Ditch the delay as soon as you can. Delay is bad.
Something that might be of interest, if you let your code keep generating random numbers, but don't use them until the user enters something that means a new random number is required, then you are adding an new source of randomness that will make the numbers more random. You are making the random number dependent on when the user does whatever it is they do to get the number, rather than just getting the next number in the sequence. When the user does something is pretty much a random event that you cannot predict.
[EDIT]
I was typing this while Danger was typing his response. I didn't set out to reply to him; I didn't know he was going to say that. However, you might view the above as my reply.
PerryBebbington:
Your code will generate a new number every 50ms or so. To do what you describe you need to generate one number, save it then compare it to what is entered.
Ditch the delay as soon as you can. Delay is bad.
Something that might be of interest, if you let your code keep generating random numbers, but don't use them until the user enters something that means a new random number is required, then you are adding an new source of randomness that will make the numbers more random. You are making the random number dependent on when the user does whatever it is they do to get the number, rather than just getting the next number in the sequence. When the user does something is pretty much a random event that you cannot predict.
[EDIT]
I was typing this while Danger was typing his response. I didn't set out to reply to him; I didn't know he was going to say that. However, you might view the above as my reply.
If I'm following your method correctly, that's pretty much how I wrote my random function in a "for fun" sketch. | 889 | 3,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.929656 |
http://fuelzilla.com/solar/PV-into-caps-27418-1.htm | 1,631,969,105,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00230.warc.gz | 28,525,129 | 6,800 | # PV into caps? - Page 2
register :: Login Password :: Lost Password?
Posted by Roderick on December 5, 2009, 2:07 am
Assuming an ideal capacitor, a real world solar panel will give
slightly more current when short circuited, so the rate at which
charge is accumulating is initially high. But this does not mean that
the rate that energy is being stored (i.e., the power) is high. It
takes more power to push a fixed amount of charge against a high
voltage than a low voltage.
If you play with the equations V = C [integral of] i dt, and E = 1/2 C
V^2, you'll see that for constant current, the energy in the cap is
proportional to t^2. So if you store 1 unit of energy between 0 and 1
second, you'll store 10^2 - 9^2 = 19 units of energy between the 9th
and 10th seconds.
The power flowing into the capacitor will continue to increase until
physical constraints (the panel is not an ideal current source) take
over.
This Thread
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Re: PV into caps? Mauried 11-20-2009
Re: PV into caps? Josepi 11-20-2009
Re: PV into caps? CG Audio Labora... 11-20-2009
Re: PV into caps? Ken and Jane Be... 11-24-2009
Re: PV into caps? Roderick 12-05-2009
please rate this thread | 371 | 1,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.898521 |
http://at.metamath.org/ilegif/elpr2.html | 1,722,643,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00283.warc.gz | 1,989,970 | 4,309 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > elpr2 Unicode version
Theorem elpr2 3397
Description: A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.)
Hypotheses
Ref Expression
elpr2.1
elpr2.2
Assertion
Ref Expression
elpr2
Proof of Theorem elpr2
StepHypRef Expression
1 elprg 3395 . . 3
21ibi 165 . 2
3 elpr2.1 . . . . . 6
4 eleq1 2100 . . . . . 6
53, 4mpbiri 157 . . . . 5
6 elpr2.2 . . . . . 6
7 eleq1 2100 . . . . . 6
86, 7mpbiri 157 . . . . 5
95, 8jaoi 636 . . . 4
10 elprg 3395 . . . 4
119, 10syl 14 . . 3
1211ibir 166 . 2
132, 12impbii 117 1
Colors of variables: wff set class Syntax hints: wb 98 wo 629 wceq 1243 wcel 1393 cvv 2557 cpr 3376 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 This theorem depends on definitions: df-bi 110 df-tru 1246 df-nf 1350 df-sb 1646 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-v 2559 df-un 2922 df-sn 3381 df-pr 3382 This theorem is referenced by: elxr 8696
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# Ohio vtracs..Root sum 9.. 12/3-12/6
Topic closed. 17 replies. Last post 3 years ago by Mo-butta216.
Page 1 of 2
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
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Posted: December 3, 2013, 11:52 am - IP Logged
Suppose I told you Digit 7 or 9 will fall in the first position or second position within 3 days...would you believe me?
__________________
Suppose I told you Digit 4 or 2 will fall in the first position or second position within 3 days...would you believe me?
__________________
And Suppose I told you also Root sum 1or 6 will fall within the same 3 days...would you believe me?
And Suppose I told you also Root sum 7or 9 will fall within the same 3 days...would you believe me?
_________________
And Suppose I told you within those same 3 days
Short sum 7 or 9 falls..would you believe me?
then brace yourself
(((((((((((((((((((((((((((((((((((((((((
The Power of Vtrac 3(7s or 2s)
71x-78x-76x-73x
26x-23x-21x-28x
Winning facts:
Yes a combo having
Digit 0 or 1 hits
You will see Root sum 3,9 or 6 hit..fact
********************
I would rather see
The group below hit
129 135 138 156 159 168 189
024 027 045 048 057 069 078 147
012 015 018 036 039 123 126
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
OPTIMAL SUMS..9,13,16, or 18 DUE
100% chance...
034 079 124 169 178 259 349 367 457 007 448 466 277 889 799
027 045 189 234 279 369 378 459 468 567 009 117 144 477 099 999
049 067 139 148 157 247 346 589 679 004 229 337 445 778 499
Wyncote,Pa
United States
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January 3, 2004
61367 Posts
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Posted: December 3, 2013, 12:07 pm - IP Logged
GET SHORTY
Extracting can be tough
time to check
the longest out short sums
2 039 048 057 129 147 237 246 345 589 679 002 228 255 778 499 444
8 017 026 125 134 189 279 369 378 459 468 567 224 233 044 477 288 099
9 027 045 126 234 289 379 469 478 009 117 225 144 559 667 577 199
2 039 048 129 138 237 246 589 679 002 228 255 778 688 499
8 026 125 189 279 369 378 459 468 008 224 233 558 288 099
9 018 027 126 234 289 379 469 478 568 009 225 559 388 199
youngstown, ohio
United States
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September 30, 2009
4291 Posts
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Posted: December 3, 2013, 12:15 pm - IP Logged
Thanks Blackapple!
Wyncote,Pa
United States
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January 3, 2004
61367 Posts
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Posted: December 3, 2013, 12:26 pm - IP Logged
The Future
520 523 528 570 573 578 620 623
628 670 673 678 920 923 928
970 973 978 208 256 276 278
306 308 356 358 378 906 908 956 958 976
]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[
The 5s and 0s....The Past..Sometimes what recently fell comes back
054 059 064 065 069 104 105 109 154 159
164 165 169 204 205 209 254 259 264 265
269 021 094 521 594 620 621 654 694
004 005 009 055 155 255 020 554
Wyncote,Pa
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Posted: December 3, 2013, 12:36 pm - IP Logged
RICK G's workout
Digit 1 is all you need to know
102 104 107 152 154 157 301 312 314 317
351 712 714 901 912
914 917 951 101 112 114
117 151 311 717 911 111
This post is a battle of workouts
Wyncote,Pa
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Posted: December 3, 2013, 1:06 pm - IP Logged
Tue, Dec 3, 2013 3-6-0
This post is a battle of workouts Winning facts:Yes a combo havingDigit 0 or 1 hitsYou will see Root sum 3,9 or 6 hit..fact******************** I would rather seeThe group below hit129 135 138 156 159 168 189024 027 045 048 057 069 078 147012 015 018 036BX 039 123 126 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Posted: Today, 12:26 pm - IP Logged
The Future
520 523 528 570 573 578 620 623
628 670 673 678 920 923 928
970 973 978 208 256 276 278
306BX 308 356 358 378 906 908 956 958 976
]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
Wyncote,Pa
United States
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January 3, 2004
61367 Posts
Online
Posted: December 3, 2013, 8:04 pm - IP Logged
Tue, Dec 3, 2013 3-2-9
Posted: Today, 12:26 pm - IP Logged
The Future
520 523 528 570 573 578 620 623
628 670 673 678 920 923BX 928
970 973 978 208 256 276 278
306BX 308 356 358 378 906 908 956 958 976
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 3, 2013, 8:05 pm - IP Logged
I need to see Digit 4 or 7...be ready
ohio
United States
Member #125198
March 26, 2012
1905 Posts
Offline
Posted: December 3, 2013, 8:11 pm - IP Logged
I need to see Digit 4 or 7...be ready
wa u see
.
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 4, 2013, 1:50 pm - IP Logged
Suppose I told you Digit 4 or 2 will fall in the first position or second position within 3 days...would you believe me?
Wed, Dec 4, 2013 2-2-9
__________________
OPTIMAL SUMS..9,13,16, or 18 DUE
100% chance...
679 004 229ST 337 445 778 499
STILL need to see Digit 4 or 7 ....
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 5, 2013, 1:49 pm - IP Logged
Thu, Dec 5, 2013 1-6-5
Posted: December 3, 2013, 12:26 pm - IP Logged
Winning facts:
Yes a combo having
Digit 0 or 1 hits
You will see Root sum 3,9 or 6 hit..fact
********************
I would rather see
The group below hit
129 135 138 156BX 159 168 189
024 027 045 048 057 069 078 147
012 015 018 036 039 123 126
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[
The 5s and 0s....The Past..Sometimes what recently fell comes back
054 059 064 065 069 104 105 109 154 159
164 165ST 169 204 205 209 254 259 264 265
269 021 094 521 594 620 621 654 694
004 005 009 055 155 255 020 554
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 5, 2013, 1:50 pm - IP Logged
STILL need to see Digit 4 or 7 ....
ohio
United States
Member #125198
March 26, 2012
1905 Posts
Offline
Posted: December 5, 2013, 1:58 pm - IP Logged
STILL need to see Digit 4 or 7 ....
u got the 7 last night 973
.
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 5, 2013, 7:03 pm - IP Logged
THANKS smooth11484
FOR THE UPDATE...Did'nt check last nite's results
Posted: December 3, 2013, 11:52 am - IP Logged
Suppose I told you Digit 7 or 9 will fall in the first position or second position within 3 days...would you believe me?
Wed, Dec 4, 2013 9-7-3
_________________
__________________
And Suppose I told you also Root sum 1or 6 will fall within the same 3 days...would you believe me?
Wed, Dec 4, 2013 9-7-3
And Suppose I told you within those same 3 days
Short sum 7 or 9 falls..would you believe me?
then brace yourself
Wed, Dec 4, 2013 9-7-3
(((((((((((((((((((((((((((((((((((((((((
Posted: December 3, 2013, 8:04 pm - IP Logged
Tue, Dec 4, 2013 3-2-9
Wed, Dec 4, 2013 9-7-3
Posted: Today, 12:26 pm - IP Logged
The Future
520 523 528 570 573 578 620 623
628 670 673 678 920 923BX 928
970 973BX 978 208 256 276 278
306BX 308 356 358 378 906 908 956 958 976
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Posted: December 3, 2013, 12:07 pm - IP Logged
GET SHORTY
Extracting can be tough
time to check
the longest out short sums
2 039 048 057 129 147 237 246 345 589 679 002 228 255 778 499 444
8 017 026 125 134 189 279 369 378 459 468 567 224 233 044 477 288 099
9 027 045 126 234 289 379BX 469 478 009 117 225 144 559 667 577 199
2 039 048 129 138 237 246 589 679 002 228 255 778 688 499
8 026 125 189 279 369 378 459 468 008 224 233 558 288 099
9 018 027 126 234 289 379BX 469 478 568 009 225 559 388 199
Wyncote,Pa
United States
Member #3206
January 3, 2004
61367 Posts
Online
Posted: December 5, 2013, 7:07 pm - IP Logged
Posted: December 3, 2013, 11:52 am - IP Logged
The Power of Vtrac 3(7s or 2s)
71x-78x-76x-73x
26x-23x-21x-28x
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Results......boy those vtracs
329
973
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# Chapter 9
1. Because n = 10, r = 7, p = 0.5
2. Because n = 6, r = 3, p = 0.75
3. The probability of making at least 6 of 8 is P[6,8] + P[7,8] + P[8,8]. Because n = 8, p = 0.8
Therefore, the probability of making at least 6 out of 8 is 0.2936 + 0.3355 + 0.1678 = 0.7969.
4. Because n = 12, r = 6, p = 0.2
5. The probability of no more ...
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No credit card required | 234 | 700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-26 | latest | en | 0.845255 |
https://www.math-only-math.com/calculation-of-sales-tax.html | 1,723,693,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00884.warc.gz | 662,929,216 | 14,035 | # Calculation of Sales Tax
The calculation of sales tax is very easy as it involves very simple concept of percentage.
The government of every country needs money the following:
(i) to meet their administrative expenses,
(ii) to execute social welfare and development schemes,
(ii) to meet the expenses on salaries of its employees, etc.
One of the many sources of collecting money (revenue) by the government from the citizen on the sale of goods within their respective territories. This purpose is known as sales tax.
It is levied by a government on the sale of different commodity. The sales tax is the sum of money a buyer pays over and above the price of a commodity to buy it.
The rates of tax on purchase of different commodities within a country are different. Also, the rates of tax on the same commodity in different country are different. Some commodities may be be exempted from sales tax by a government. Sales tax is one of the many forms of indirect taxes that a government imposes on its citizens.
If P be the printed price or marked price of a commodity, the rate of sales tax be r% and S be the selling price (i.e., the price a customer has to pay) then
S = P(1 + $$\frac{r}{100}$$) and sales tax = S - P = $$\frac{Pr}{100}$$
Solved examples on relation between printed price, rate of sales tax and selling price:
1. The printed price of a bi-cycle is $4200. The rate of sales tax on it is 10%. Find the price at which the cycle can be purchased. Solution: Here, the printed price P =$ 4200, the rate of sales tax = 10%, i.e. r = 10
Therefore, the selling price S = P(1 + $$\frac{r}{100}$$)
= $4200 × (1 + $$\frac{10}{100}$$) =$ 4200 × $$\frac{11}{10}$$
= $4620 Therefore, the cycle can be purchased for$ 4620.
3. Mason purchased a pair of shoes costing $850. Calculate the total amount to be paid by him, if the rate of Sales Tax is 6%. Solution: The sale price of shoes =$ 850
and, the sales tax = 6 % of $850 =$ 51
Therefore, the total amount to be paid by Rohit = $850 +$ 51 = $901 4. John bought a printing machine for$5136, which includes sales tax. If the listed price of the printing machine is $4800, what was the rate of sales tax? Solution: Here, the printed price (or listed price) P =$ 4800, the selling price S = $5136. If the rate of sales tax be r% then, S = P(1 + $$\frac{r}{100}$$) ⟹$ 5136 = $4800(1 + $$\frac{r}{100}$$) ⟹ 1 + $$\frac{r}{100}$$ = $$\frac{5136}{4800}$$ Therefore, $$\frac{r}{100}$$ = $$\frac{5136}{4800}$$ - 1 $$\frac{r}{100}$$ = $$\frac{336}{4800}$$ ⟹ r = $$\frac{336}{48}$$ ⟹ r = 7. Therefore, the rate of sales tax was 7%. 5. Jacob purchased an article for$ 702 including Sales Tax. If the rate of Sales Tax is 8%, find the sale price of the article.
Solution:
Let the sale price of the article be $x Therefore, x + 8% of x =$ 702
⟹ x + $$\frac{8x}{100}$$ = $702 ⟹ $$\frac{108x}{100}$$ =$ 702
⟹ x = $702 × $$\frac{100}{108}$$ ⟹ x =$ 650
Therefore, sale price of the article = $650 6. Find the marked price of a motorbike which is bought at$ 36300 after paying a sales tax at the rate of 10%.
Solution:
Here, the marked price or listed price P, the selling price S = $36300 and the rate of sales tax = 10%, i.e. r = 10 S = P(1 + $$\frac{r}{100}$$) ⟹$ 36300 = P(1 + $$\frac{10}{100}$$)
⟹ P × $$\frac{11}{10}$$ = $36300 ⟹ P =$ 36300 × $$\frac{10}{11}$$
⟹ P = $3300 × 10 ⟹ P =$ 33000
Therefore, the marked price is $33000. 4. A refrigerator is marked for sale at$ 17600 inclusive of sales tax at the rate of 10%. Calculate the sales tax on the refrigerator.
Solution:
Let the price of the refrigerator without sales tax = P.
Here, the selling price S = $17600 and the rate of sales tax i.e., r = 10 S = P(1 + $$\frac{r}{100}$$) ⟹$ 17600 = P(1 + $$\frac{10}{100}$$)
⟹ P ∙ $$\frac{11}{10}$$ = $17,600 Therefore, P =$17,600 × $$\frac{11}{10}$$ = $16,00 × 10 =$ 16,000
Therefore, the sale tax = Selling price - Printed price
= $17,600 -$ 16,000
= $1,600 5. If the rate of sale tax increases by 55, the selling price of an article goes up by$ 40. Find the marked price of the article.
Solution:
Let the marked price = P(1 + $$\frac{r}{100}$$)
When the rate of sales tax increases by 5%,
the selling price = p(1 + $$\frac{r + 5}{100}$$)
From the equation, P(1 + $$\frac{r + 5}{100}$$) – P(1 + $$\frac{r}{100}$$) = $40 ⟹ $$\frac{5P}{100}$$ =$ 40
⟹ P = $40 × 20 =$ 800
Therefore, the marked price of the article is \$ 800.
● Sales Tax and Value Added Tax
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• Questions (17 to 20) :
A family consists of six members P, Q, R, S, T, and U.
T is the daughter of Q.
Q and S are married couple.
P is the brother of Q.
R is the son of S but S is not the mother of R.
U is the brother of S.
1.
Find the paternal uncle of T?
A.) U
B.) R
C.) S
D.) P
"U" is the paternal-uncle of T.
• Questions (6 to 10) :
Blood Relations for a family consisting of L, M, N, R, P, Q is given below.
Answer the following questions based on the relation given.
R is the daughter of L. Q is the brother of L.
N and L are married couples.
P is the brother of N. M is the son of N but N is not the mother of M.
2.
Find the maternal uncle of R?
A.) Q
B.) N
C.) M
D.) P
Q is the Maternal-uncle of R.
3.
What is the relationship of P to A?
A.) Grandson
B.) Son
C.) Daughter
D.) Father
P is Grandson of A.
4.
Find the maternal uncle of R?
A.) Q
B.) M
C.) P
D.) N
"Q" is the maternal uncle of R.
5.
Find how is G related to D?
A.) Maternal Uncle
B.) Nephew
C.) Granddaughter
D.) Paternal Uncle
Hence, G isGranddaughter of D.
• Questions (14 to 15) :
A is the sister of B. B is married to C. C is the son of D. A is the mother of E. F is the father of G. F has only 1 son and 1 daughter. G is the daughter of A. H is the son of B.
6.
How is H related to C?
A.) Son
B.) Daughter
C.) Sister
D.) Brother
Hence, H is "Son" of C.
7.
What is the relationship of T to R?
A.) Brother
B.) Sister
C.) Father
D.) Mother
T is the sister of R.
8.
What is the relationship of P to B?
A.) Nephew
B.) Neice
C.) Paternal Uncle
D.) Paternal Aunt
P is "Nephew" (i.e., sister's son) of B.
• Questions (16) : (i) In a family of six persons A, B, C, D, E and F, there are two married couples. (ii) D is grandmother of A and mother of B. (iii) C is wife of B and mother of F. (iv) F is the grand daughter of E
9.
What is the relationship of B with E?
A.) Son
B.) Mother
C.) GrandSon
D.) GrandDaughter
Son
10.
Find the maternal uncle of T?
A.) Q
B.) R
C.) P
D.) U
The maternal Uncle of T is "P"
• Questions (11 to 13) :
P and Q are brothers of R. Q is son of S and T. S is the daughter of U. A is the father-in-law of T. B is son of U.
11.
What is the relationship of Q to B?
A.) Nephew
B.) Neice
C.) Maternal Uncle
D.) Paternal Aunt
Q is theNephew (i.e., sister's son) of B.
12.
How many female members are there in the family?
A.) One
B.) Two
C.) Three
D.) Four
There are two female members in the family.
13.
Who is the brother of R?
A.) L
B.) M
C.) N
D.) P
M is the brother of R.
14.
How many female members are there in the family?
A.) One
B.) Two
C.) Three
D.) Four
There are two female members in the family (i.e., L and R)
15.
In what relation M is related to R?
A.) Sister
B.) Father
C.) Mother
D.) Uncle
M is the sister of R.
16.
How many male members are there in the family?
A.) Four
B.) Three
C.) Two
D.) One
There are four male members in the family. (i.e., P, N, Q and M)
• Questions (1 to 5) : Answer the following questions based on the given data.
A family consists of six people L, M, N, P, Q and R.
R is the son of L but L is not the mother of R.
P and L are married couples.
N is the brother of L.
M is the daughter of P.
Q is the brother of P
17.
Find the maternal uncle of M?
A.) P
B.) Q
C.) R
D.) L
Q is the maternal-Uncle of M.
18.
Which person in the following option do not belong to the same generation of Q?
A.) P
B.) N
C.) M
D.) N
M and R are next Generation of Q.
19.
How many male members are there in the family?
A.) One
B.) Two
C.) Three
D.) Four
There are 4 male members in the family i.e., N, L, Q and R.
20.
How many female members are there in the family?
A.) One
B.) Two
C.) Three
D.) Four | 1,231 | 3,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-24 | longest | en | 0.894757 |
https://thekidsworksheet.com/linear-equations-word-problems-worksheet-with-answers-pdf/ | 1,669,539,064,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00821.warc.gz | 614,143,876 | 22,110 | # Linear Equations Word Problems Worksheet With Answers Pdf
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Solve Systems Of Equations Given Word Problems Exam Mrs Math Systems Of Equations Word Problems Equations | 1,025 | 5,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.931066 |
http://hablemosdepeces.com/pizza/what-size-dough-ball-for-detroit-style-pizza.html | 1,685,820,794,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00708.warc.gz | 20,925,008 | 48,935 | # What Size Dough Ball For Detroit Style Pizza?
Each dough ball should weigh 17.5 to 18 ounces. Coat the dough balls with oil and refrigerate, covered, for 24 hours. (Doing this provides more flavor and better texture. If you prefer, you can use the fresh dough right away.)
## How big should a pizza dough ball be?
For Neapolitan pizza, the weight of each ball should be between 180g – 250g. Personally, I tend to go for around 250g or just under. With this you should be able to achieve a 10 inch pizza with quite puffy crusts. The weight that you choose will depend on how you shape the pizza.
## How thick is Detroit pizza?
Randazzo says that the crust should be about 1.5 inches thick for true Detroit-style pizza. The buttery flavor of the crust results from a small quantity of oil and the melting properties of the mozzarella and Wisconsin brick cheeses.
## What size is a Detroit-style pizza pan?
The classic Detroit pizza pan is a deep, 10- by 14-inch rectangle of metal with black surfaces (for better conduction) and sides that flare gently away from the bottom. While the original pans were made from blue steel, most modern pans are made from anodized aluminum and come with a nonstick coating.
## How many ounces should a pizza dough ball be?
If you want a good starting point, go with 1-ounce of dough per inch of diameter for any size up to 16 inches. Add or subtract dough weight until you are satisfied with the finished pizza.
## How much dough do I need for a 10 inch thin crust pizza?
As an example, if we want to make a 10-inch pizza in addition to the 12-inch pizza, the correct dough weight for the 10-inch would be calculated as 3.14 X 25 = 78.5 (square inches) X 0.08849 (ounces per square inch) = 6.946 (7-ounces).
## How many ounces of dough do I need for a 16 inch pizza?
Round that off to 19.5 ounces of dough needed to make the 16-inch pizza crust.
## Is Detroit-style pizza thick or thin?
Detroit-style pizza is a thick, square-cut pizza with a crunchy, fried bottom layer of crust overflowing with delicious melted cheese. Like most rectangular pan pizzas served in America, Detroit-style pizza is a variation of the Sicilian pizza.
## What kind of cheese is used on Detroit-style pizza?
The best kind of cheese to use on a traditional Detroit style pizza is Wisconsin Brick cheese. Brick cheese is a kind of cheese native to the Midwestern USA that melts well due to its high fat content and has a somewhat buttery flavor, making it a perfect fit for Detroit style pizza.
## What’s the difference between deep dish and Detroit-style pizza?
Detroit-style pizza is fired at roughly 500°F and takes less about 12-15 minutes to bake, but deep dish takes it low and slow at 350-425°F for 30 minutes or more. The sauce of a Detroit-style pizza is as smooth and sweet with lots of aromatics.
## Which is the best Detroit style pizza pan?
Top 8 Best Deep Dish Pizza Pan Reviews
1. Chicago Metallic Deep Dish Pizza Pan.
2. LloydPans Kitchenware Detroit Style Pizza Pan.
3. Chicago Metallic Perforated Mini-Deep Dish Pizza Pan Set.
4. Nordic Ware Commercial Deep Dish Pizza Pan.
5. LloydPans 12 x 2.25, Pre-Seasoned Cast Aluminum Deep Dish Pan for Pizza Making.
## How much dough do I need for a 8×10 pan?
You can make your own pizza dough if you like, or use store-bought—either way, for an 8 x 10-inch pan, you’ll want a 10-ounce portion of dough (a little more for a slightly larger pan, or you can stretch it for a thinner pizza if absolutely necessary).
## How much dough do I need for a 13 inch pizza?
I use a thickness coefficient of 3.8. All this means is that for every 1 square inch of pizza there will be 3.8 grams of dough. So if your pan is 13 inches by 8 inches, 13×8= 104 square inches of pizza multiplied by the thickness coeffecient of 3.8, 104×3. 8 = 395 grams of dough.
## How much dough do I need for an 8 inch pizza?
Yield: Yields four balls of dough for four individual 8-inch pizzas; 1-3/4 pounds total.
## How much dough do I need for a 7 inch pizza?
For your 7-inch pan, you may want to start with a dough ball weight of 10.95 ounces. The weight of all-purpose flour for this weight of dough ball is 10.95 oz. divided by 2.05 (the sum of all baker’s percents for the recipe divided by 100), or 5.34 ounces.
## Detroit-Style Pizza: Shawn Randazzo Shares His Recipe for the Detroit Twist
• Latest news: Shawn Randazzo, one of the country’s most beloved advocates for Detroit-style pizza, died in December 2020 at the age of 44 after a long battle with cancer.
• The following story appeared in the December 2018 issue of PMQ Pizza Magazine and is reprinted with permission.
• Detroit is a thriving industrial metropolis that is well-known for its music and automobiles.
Some people, on the other hand, are still unaware that it is also famed for its pizza.In the United States, Detroit-style pizza is distinguished by its thick and airy crust, trademark rectangular form, and caramelized cheese around the edges.It is prepared in a pan that was originally used in automobile manufacturing to capture oil and trash parts.
Despite the fact that this variety of pizza has been available since 1946, it has only lately begun to gain popularity throughout the country.In light of the recent increase in the number of independents serving a form of Detroit-style pizza, I wanted to speak with someone I knew could shed some light on this delectable take on pizza.It was with great interest that I reached out to Detroit’s favorite son—and, no, I am not referring to Eminem or Robocop; rather, I am referring to Shawn Randazzo of Detroit Style Pizza Co.
In the wake of multiple competition victories with his pizza, Randazzo decided to take it upon himself to begin spreading the word about this Detroit institution.Because he learnt from the masters of the art, he has made it his duty to share his expertise of this pillowy pie with any other pizzaiolo who is ready to learn from him.And he was gracious enough to take a seat in the Chef’s Corner and provide a dish that can be easily replicated at home or in a restaurant setting.To produce an authentic Detroit-style pizza, I believe there are five fundamental guidelines to follow: The pie must have a rectangular form and be baked in a deep pan.
The hydration level should be at least 70% in order to add to the airiness and crispiness of the dish.The sauce is applied on top.Some people refer to it as a ‘upside-down pizza.’ The cheese—Brick cheese is essential for achieving the desired taste and caramelization.Thickness—approximately 112″ for authentic Detroit style.″ —Shawn Randazzo, author
### The Detroit Twist
• Ingredients for the dough: 135 c.
• 85 °F water 12 c.
• semolina flour (optional) 2 teaspoons of sugar 2 teaspoons active dry yeast 312 cup bread flour or baker’s flour 2 teaspoons of salt Instructions for making the dough: In a large mixing basin, combine the water, semolina flour, sugar, and yeast.
Allow for around 15 minutes of resting time after mixing.In a large mixing basin, combine the bread/bakers flour and salt.For one minute, on the lowest speed, combine all of the ingredients.
Scrape the edges of the basin with a plastic dough scraper/divider to remove any excess batter.Three minutes on a low/medium pace will enough.Take the dough out of the mixing bowl and split it in half, as shown.
Each dough ball should be between 17.5 and 18 ounces in weight.Refrigerate the dough balls for 24 hours after coating them with oil and keeping them covered.(Doing so enhances the flavor and improves the texture of the dish.) It is possible to utilize the freshly made dough straight immediately if you like.Using 8 to 10 pushes, flatten the dough on a baking pan and set aside.
Allow for 15 to 20 minutes of resting time in the pan with the lid on.Fully push out the dough in the pan, all the way to the sides and corners of the pan.Allow the dough to rest at room temperature, covered, for 112 to 212 hours.You want it to rise about a third of the way up the pan, at the very least.Once the dough has doubled in size, you’re ready to start making pizza!
This recipe makes two huge 10″ by 14″ pizzas.Ingredients for a pizza: 1 teaspoon dried oregano A 24 oz.shredded cheese combination is preferred, especially a 50/50 mix of brick cheese and low-moisture mozzarella.
1 red onion, peeled and chopped 2 roma tomatoes, peeled and diced 4 ounces of freshly chopped arugula 1 teaspoon freshly grated lemon zest (optional) Dressing with Lemon Oil (also known as Lemon Vinaigrette) (see recipe below) Vinaigrette with Lemon 1 cup extra-virgin olive oil 14 cup red wine 8 tablespoons freshly squeezed lemon juice (and more to taste) 8 tsp.Dijon mustard (optional) 4 pinches of rock salt freshly ground black pepper (four teaspoons) 4 teaspoons of sugar Shake all of the ingredients together in a spray bottle until they get emulsified.This recipe makes enough to make multiple pizzas or salads.Instructions for making pizza: Dry oregano can be sprinkled straight on the dough.
12 ounces of cheese should be spread on the pan from edge to edge, making sure the cheese is equally distributed.Cut the tomatoes into slices and arrange them on top of the cheese.Save a few slices and cut them up to combine with arugula for a refreshing salad (to be added after baking).Place the red onion on top of the pizza once it has been sliced.Preheat the oven to 450 degrees Fahrenheit.
Once the oven is prepared, bake the pizzas for around 15 minutes, or until the tops are browned and the bottoms are golden brown, depending on your preference.Toss the arugula with a drizzle of lemon oil dressing in a large mixing dish.Place the pizza on a cutting board when it has been removed from the oven.Using a sharp knife, cut into 8 pieces.Place the arugula-tossed dressing on top of the pizza and serve immediately.
Garnish with lemon zest for color and taste, and then serve immediately.Finally, pizza lovers: the award-winning Detroit Twist is at your disposal.Enjoy!
## Balling pizza dough
• Balling pizza dough is an important stage in the pizza-making process, yet it is also a simple one.
• Making dough balls may be done in a variety of methods, just like kneading dough is done.
• Once we have completed the preparation of our pizza dough balls, we may allow them to proof one more time.
Once our dough balls have proven themselves, we will be able to mold and cook some delicious pizzas.If you haven’t already, be sure to read parts 1 and 2 of the series, which cover combining the dough, kneading the dough, and testing the dough.
## Pizza dough balls video
A handful of alternative simple approaches for forming pizza dough balls are demonstrated in the fourth installment of this series. I also go through the best sorts of containers to use for proving your dough balls in detail. Take a look at the video below: Making Neapolitan Pizza Dough from Scratch
## Why ball pizza dough?
• Several factors influence our decision to ball pizza dough: For the purpose of creating a spherical shape for when we stretch the pizza
• In order to increase the strength of the dough before the final proof
• In order to make certain that we have the proper amount of dough for each pizza
• To put it simply, we’re attempting to create a circular dough ball that is rather tight.
• We may also verify that each dough ball has the appropriate weight for each pizza by weighing the dough before rolling it into balls.
• As the dough ball proves, the strain that we create in it will offer strength to the dough ball.
This will assist the dough in maintaining its round shape, which will make shaping the dough much easier later on.There is no need to overtighten the dough, however, as we do not want the dough to rip when baking.
## Neapolitan pizza dough ball weight
• The weight of each ball should be between 180g and 250g for a Neapolitan pizza, according to the recipe.
• Personally, I want to stick to a weight of approximately 250g or little less.
• If you follow these instructions, you should be able to produce a 10 inch pizza with puffy crusts.
Depending on how you shape the pizza, you will need to pick a different weight than the last one.Various forms of pizza shape may be found even within the Neapolitan pizza tradition.Some individuals prefer huge, soft crusts on their pizza, while others prefer smaller, less raised crusts on their pizza.
I prefer pizzas with somewhat bigger crusts that are somewhere in the middle of the spectrum.For a 10 inch pizza with a fairly thick crust, I think that 240g-250g is the right amount of cheese.This will allow you to make the centre of the pizza extremely thin while yet maintaining a substantial thickness on the crusts.
A ball weight of 200g to 210g is recommended for a smaller crustm size preference.This will allow you to make a pizza that is around 10 inches in diameter with a very thin crust.
## When to ball pizza dough
Once the pizza dough has proofed, it should be rolled into balls.In most cases, pizza dough is proofed in a single large dough ball before being balled up.The bulk ferment, also known as the bulk prove, is the stage at which the bulk ferment is completed.
1. After the dough has been allowed to proof, it is formed into dough balls.
2. The dough balls must be allowed to prove again after being balled before being formed into pizzas, as this is critical to their success.
3. This is due to the fact that a significant amount of the air in the dough (which has been built up during the bulk prove) is lost during the balling process.
4. It is necessary to prove the dough balls a second time in order for them to expand and become airy again.
5. It is the air that has accumulated throughout the proving process that gives the crust its lovely texture.
Additionally, when the dough balls loosen throughout the proofing process.It is important that the pizza dough has time to rest after we have worked with it for a long period of time.It is critical that you do not neglect this step, even if you are only providing a brief demonstration of your product.When it comes time to make the pizzas, this will guarantee that the dough is extremely soft and simple to stretch when we get started.
## Do you knead pizza dough after it rises?
In general, once the pizza dough has risen, it should not be kneaded any further.Kneading the dough at this point will remove all of the air that has accumulated in the dough throughout the proving process.If you believe your pizza dough is lacking in strength, it can be kneaded again, but this should be done before proving it in the oven.
1. If you need to refer back to Part 2 of this series on kneading, you may do so by visiting this link.
2. True, the pizza would prove itself once again, but we should restrict the number of times this occurs.
3. The more times we have to proof our dough, the denser the dough grows as a result of this.
## Proofing pizza dough balls
Especially if you are doing a long prove, as I recommend, it is critical to shape your dough balls after a bulk prove has been completed.If we form our dough balls as soon as we have done kneading them, they will lose all of their power by the time the lengthy prove is completed (24 hours).In order to get a 24 hour prove, I recommend proving your dough for around 18 hours (for a 48 hour prove).
1. After that, we can shape our dough balls and set them aside to proof for the remaining 6 hours.
2. This will ensure that the dough balls are properly proofed without any loss of strength as a result of the lengthy proofing period.
3. If you are only proving your dough for a short period of time (up to 6 hours), you can shape your dough balls immediately after kneading.
## Proofing dough balls at room temperature
It is critical that your dough balls be proofed at room temperature before using them.This is necessary because when it comes time to stretch the pizza, we want the dough to be as soft as possible.It is important to remove the dough balls for their final proofing, even if you are using a cold proofing method.
1. I actually recommend removing them a couple of hours before you plan to go ballistic.
2. This will make the process of forming the dough balls as simple as it possibly can be.
## Container for proofing pizza dough
• If you’re making pizza dough, there are a plethora of various containers you may use to prove it. No worries if you don’t have an appropriate lid. Simply cover the container with cling film to keep the food fresh (plastic wrap). Here’s a list of containers that you may use to store your items: Cling film for the bowl
• Cling film for the large dish or roasting tray
• Cling film for the chopping board
• Tupperware containers
• Pizza proving box
### Proving pizza in a bowl
A bowl is a popular choice for many individuals. Everyone has one, and it is used to keep the dough balls separated from one another. The drawback is that the dough is difficult to remove from the bowl without degassing it, which is undesirable. We want to keep as much air as possible in the dough when baking.
### Proving pizza in a large dish
A big dish works well for proving pizza dough since it has plenty of space. Cling film is a simple solution for this problem. In order to remove the balls for shape, we may simply insert our hands into the dough or use a dough scraper/wall scraper to assist us. This guarantees that when it comes time to shape the pizza, we can maintain as much air in the dough as we possibly can.
### Proofing on a chopping board
A chopping board is another wonderful tool for making your point. The dough may be removed off a chopping board with relative ease; however, the cling film may adhere to the tops of the dough balls, requiring a little extra time to set up. It is critical to ensure that the balls are sealed tightly to prevent them from drying out.
### Proofing pizza dough in a tupperware box
Tupperware containers in small sizes are a fantastic choice. 2 dough balls may be accommodated in each container, and the lid can be used to keep them airtight. The dough scraper/wall scraper should be sufficient to remove them at the beginning of the process.
### Using a pizza proving box
Last but not least, you may get a pizza proving box.They are reasonably priced, and they are excellent quality.I’ve had mine for quite some time, and I’ve used it to prove 12 dough balls at the same time!
1. The dough balls may be removed with relative ease, and the cover prevents them from drying out.
2. They are also a fantastic space saver and make it simple to move dough while working with huge quantities of ingredients.
3. Whatever method you use, make certain that the dough does not dry out too much.
4. Don’t be concerned about it; I’ve successfully created excellent pizza crust using all of the containers listed above.
## Dough Ball Weights
If you’re just getting started in the pizza industry, you might be curious about how to establish the proper dough weight for each of the pizza sizes you’ll be serving.Choose a size (any size will do).Working with a 12-inch pizza or something similar is my preferred method of expressing myself creatively.
1. Then, using Pi X R squared as our method for calculating surface area, we can figure out how much surface area there is.
2. Let’s assume our pizzas are available in three different sizes: 10-inch, 12-inch, and 16-inch.
3. Here’s how the math works: 3.14 x 25 = 78.5 square inches for a ten-inch square.
4. Inches 12 inches: 3.14 x 36 = 113.44 square inches 3.14 x 64 = 200.96 square inches for a 16-inch screen.
5. Test different dough weights to see which one produces the pizza you desire.
In order to have a suitable beginning point, use one ounce of dough per inch of diameter for any size up to and including sixteen inches.Continue to increase or decrease the dough weight until you are pleased with the completed product.We’re ready to get the calculator out of the drawer once more.Calculate the weight of the dough by multiplying it by the surface area of the pan, disk, or screen that you used to make your pizza(s).
1. This will provide you with the amount of dough to be loaded per square inch of pan surface.
2. Consider the following scenario: you were constructing a thin crust pizza and discovered that 10 ounces of dough yielded the 12-inch pie you desired.
3. Here’s how the math works: In this case, 0.0884642 ounce of dough per square inch of pan surface area is 10 ounces divided by 113.04 ounces.
4. All you have to do to figure out how much dough you’ll need for each of your other pan sizes is multiply this quantity (0.0884642) by the surface area of each of your other pan sizes.
5. Here’s how the math works: 10-inch: 78.5 square inches multiplied by 0.0884642 equals 6.9444-ounces (7-ounces) 160 square inches times 0.0884642 Equals 17.777 ounces for the 16-inch (17.75-ounces).
You may use this to figure out the size and type of pizza you want to make.The biggest advantage of following this technique is that all of your pizzas will now have a comparable quantity of dough under them; the only difference will be in the size (diameter) of the pizzas themselves.Using an air impingement oven or any other sort of conveyor oven, this implies that all of your pizzas with comparable toppings will bake at around the same time, regardless of their size (within reason).This will make the process of setting up your conveyor oven(s) much simpler.
## Dough Math – Delco Foods
Tom Lehmann has written a piece for Bellissimo Foods.Trying to figure out how much dough to use while making different size pizzas has always been a bit of a mystery to me.This riddle, on the other hand, can be simply answered with a little elementary mathematics.
1. Try out different sizes of pizza until you find one that suits your tastes and preferences.
2. Make many batches of dough and experiment with different weights to find which weight produces the crust you desire.
3. Take, for example, a 12-inch diameter pizza that you’ve been experimenting with, and you’ve discovered that 10-ounces of dough produces the desired crust thickness.
4. To find out how much space is on the surface of the 12-inch pizza (Pi X R squared or 3.14 X R squared), multiply 3.14 by 36 to get 3.14 x 36 = 113 square inches.
5. We get 0.08849-ounces of dough per square inch of surface area when we divide the weight of the dough by the size of the pan.
Then, to figure out how much dough you’ll need for different sizes of pizza, all you have to do is figure out how much surface area you’ll need for each size of pizza you’re making and multiply that number by your dough loading per square inch figure (0.08849) to come up with the correct dough weight you’ll need for that size of pizza.In the case of making a 10-inch pizza in addition to a 12-inch pizza, the right dough weight for the 10-inch would be calculated as 3.14 X 25 = 78.5 (square inches) X 0.08849 (ounces per square inch) = 6.946 (ounces per square inch) (7-ounces).Additionally, if you bake your pizza in one of the conveyor ovens, you will have a greater chance of having both sizes bake at the same time and temperature, which will result in a more consistent product.Return to the Tips and Articles page.
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## How to calculate how much dough you need for ANY size of pizza
Do you want to know how to calculate the amount of dough required to produce any size pizza?It’s not an issue!Simply use the mathematical constant ″pi″ to compute the surface area of a circle, and then utilize that amount to produce a dough density value.
1. It may appear to be difficult to understand, but it isn’t.
2. Here’s how it’s done in the proper manner.
3. Tom Lehmann is a well-known author.
4. Pizza Today’s Dough Expert on the Spot Consider the following scenario: you want to manufacture pizzas with diameters of 12, 14, and 16 inches, and you want to know what the appropriate dough weight will be for each size.
5. Choose a size that you are comfortable working with as a starting point (any size at all will work).
We’ll presume that we choose the 12-inch size for our project.The first step is to prepare our dough, after which we will scale and ball a number of dough balls using a variety of scaling weights.The aim here is to construct pizzas out of various dough ball weights and then, depending on the features of the final pizza, choose the dough ball weight that produces the pizza that we want in terms of crust look, texture, and thickness, among other things.Make a mental note of how much weight you have.
1. To illustrate, let us suppose that 11 ounces of dough provides us with the result we were aiming for.
2. Next, we’ll figure out how to calculate the dough density, which is critical in establishing the dough weights for the other sizes.
3. To begin, determine the surface area of the size of pizza for which you want to determine the dough weight by calculating the surface area of the pizza.
4. In this instance, the pizza is 12 inches in diameter.
5. To get the surface area of a circle, multiply pi x R squared by the radius of the circle.
Pi is equal to 3.14, and R is half the circumference of the circle.To square it, we just multiply it by itself many times.Here’s how the math works out in practice: 3.14 x 6 x 6 (or 36) Equals 113.04 square inches (in metric units).It will be necessary to divide the dough weight by the number of square inches in order to arrive at the density of the dough number.We have 11 ounces of dough per square inch of surface area on our 12-inch pizza, which is 0.0973106 ounces of dough per square inch of surface area.The ″dough density number″ is the number that represents the density of the dough.
• Following that, we’ll need to figure out how many square inches of surface area we’ll need for each of the other sizes we’d like to construct.
• Along with the 12-inch pizza, we’d want to bake two additional 14- and 16-inch pizzas to serve as appetizers.
• It has a surface area of 3.14 x 49 (seven times seven equals fifty-seven) = 153.86 square inches for a pizza that is 14 inches in diameter.
All that remains is to multiply the surface area of the 14-inch pizza by the dough density number (0.0973106) in order to determine the dough scaling weight for the 14-inch pizza — 153.86 x 0.0973106 = 14.972208 ounces of dough — to obtain the dough scaling weight.To produce a 14-inch pizza crust, you’ll need 15 ounces of dough, or 15 ounces of dough total.For the 16-inch pizza, multiply 3.14 times 64 (8 x 64 = 200.96 square inches) to get a total surface area of 200.96 square inches.
1. To calculate the dough weight necessary to manufacture our 16-inch crusts, multiply the above figure by the dough density factor.
2. — 200.96 multiplied by 0.0973106 is 19.555538 ounces of dough This comes out to 19.5 ounces of dough, which is the amount needed to produce the 16-inch pizza crust.
3. Summary: For our 12-, 14-, and 16-inch pizza crusts, the following dough weights will be required: 12-, 14-, and 16-inch dough weights The following sizes are available: 12-inch (11 ounces), 14-inch (15 ounces), and 16-inch (19.5 ounces).
1. Apart from being used to calculate the weights of dough for various pizza sizes, this technique may also be used to determine the weights of sauce and cheese, depending on the type of sauce and cheese used.
2. Simply substituting the dough weight with the sauce or cheese weight that you have determined would result in the finest pizza for you is all that is required in these situations.
3. This will supply you with a specific sauce or cheese weight, which can then be used in precisely the same way to calculate the amount of sauce or cheese necessary for every other size pizza you choose to build using the same method as previously described.
4. Let’s assume we really like the pizza when it has five ounces of sauce on it, so let’s use the 12-inch pizza as an example.
Assuming we already know that a 12-inch pizza has a surface area of 113.04 square inches, we can divide five ounces by 113.04 to get 0.0442321 ounces of sauce per square inch of surface area on the pizza.Our sauce density is 0.0442321, which is a very small value.Knowing that the 14-inch pizza has a surface area of 153.86 square inches, we may estimate its weight.
• To discover the exact quantity of sauce to use on our 14-inch pizza, we just multiply the sauce density figure by 153.86 in order to obtain the correct amount of sauce to use on our 14-inch pizza — 153.86 x 0.0442321 = 6.80 ounces of sauce to be used on our 14-inch pizza We know that the 16-inch pizza has a surface area of 200.96 square inches since it is 16 inches in diameter.
• In order to determine how much sauce to put on our 16-inch pizza, we just multiply the number of slices by the sauce density factor (200.96 divided by 0.0442321 = 8.88 ounces of sauce).
• Again, we will use the 12-inch pizza to determine the quantity of cheese to use, and we will experiment with different quantities of cheese until we discover the amount that works best for our needs.
To get the surface area of our test pizza, multiply this number by its height (a 12-inch, which has 113.04 inches of surface area).Take, for example, the case where we discovered that six ounces of cheese worked effectively in our application.A six-ounce portion of cheese divided by 113.04 is 0.0530785 ounce of cheese per square inch of surface area of the cheese.Our cheese density is 0.0530785 grams per cubic meter of cheese.
• The total surface area of a 14-inch pizza is 153.86 square inches.
• To calculate the amount of cheese to use on our 14-inch pizza, multiply the amount of cheese by the cheese density figure.
• For example, 153.86 x 0.0530785 = 8.16 ounces of cheese to be used on our 14-inch pizza.
• The total surface area of a 16-inch pizza is 200.96 square inches.
• To determine the amount of cheese to use on our 16-inch pizza, multiply this figure by the cheese density number.
For example, 200.96 x 0.0530785 = 10.66 ounces of cheese should be used on our 16-inch pizza if the cheese density value is 0.Calculating the weights of your dough, sauce, and cheese for each of your pizza sizes will help to ensure that your pizzas bake in a consistent manner, regardless of size.This is especially important if you are baking in one of the conveyor ovens, where the baking time is fixed and you want to be able to bake all of your pizza sizes at the same baking time.Most of the time, this enables us to bake pizzas with one to three toppings on one conveyor, regardless of their size, and pizzas with four or more toppings on another conveyor, also regardless of their size.Tom Lehmann is a former director of the American Institute of Baking in Manhattan, Kansas, and Pizza Today’s resident dough expert.
He formerly served as director of the American Institute of Baking.
• Is there anything that distinguishes Detroit-style pizza from other types of pizza?
• What is the process of making Detroit-style pizza?
• What Is the Difference Between Detroit-Style Pizza and Other Types of Pizza?
Detroit is a powerful city that is well-known for a variety of things, including automobiles, music, and pizza.Detroit’s pizza, which is known as the ″Detroit-style square,″ has garnered a reputation as one of the greatest slices in the country.In fact, Detroit-style pizza was made for savoring, thanks to its unusual rectangular form, light and airy dough, crispy caramelized cheese edge, and wonderful crunch that distinguishes it from other styles of pizza.
1. But what precisely distinguishes a Detroit-style pan pizza from a regular one, and why is a substantial Detroit slice so widely sought after?
2. In this post, we’ll cover all you need to know about this delectable Detroit cuisine, from the characteristics that distinguish a Detroit-style pizza to the process of making a Detroit-style pizza to the differences between a Detroit-style pie and other forms of deep-dish pizza.
## What Are the Defining Characteristics of Detroit-Style Pizza?
Detroit-style pizza is a thick, square-cut pizza with a crispy, fried bottom layer of dough that is oozing with luscious melted cheese on the top and bottom.Detroit-style pizza, like the majority of rectangular pan pizzas offered in the United States, is a derivation of the Sicilian pizza.Sicilian pizza, formerly known as ″sfincione,″ which translates as ″small sponge,″ has a fluffier bread foundation than most other types of pizza in the United States.
1. When compared to traditional Sicilian pizza, Detroit-style pizza is distinguished by its cheese-covered crust, which is generally so crispy that the cheese is somewhat scorched, and its long, vertical streaks of sauce running through it on top of the cheese.
2. Because of this unusual pizza technique, the core of the pizza has a gooey, doughy center with a crispy outer crust and caramelized cheese wrapping around the edges.
3. Even though pepperoni is a common topping for a Detroit-style pizza, not just any pepperoni is used in the creation of this form of pizza.
4. The pepperoni on Detroit-style pizza is smaller and thicker, and as they cook, they curl up into little cups, as opposed to the wide, flat circular pepperoni that appears on most other pizzas.
5. During the baking process, these adorable miniature pepperoni cups retain their fat and improve the flavor of the pizza.
## How Is Detroit-Style Pizza Made?
The dough is the first step in making Detroit-style pizza.Given that Detroit-style pizza is noted for its lighter center and crispier crust, it is essential that the dough is perfectly hydrated.This requires finding the ideal ratio of water to flour.
1. Detroit-style pizza is made using a drier dough in order to get the crust’s famed open, fluffy, and chewy interior with a crisp outside crust.
2. The pan is the second most critical consideration when attempting to make a faultless Detroit-style pizza.
3. Detroit-style pizza, which is a deep-dish rectangular pizza, is best cooked in a steel industrial pan with edges that are slanted outwards so that the area at the top of the pan is bigger than the area at the bottom of the pan.
4. In order to ensure that the dough fills the whole pan and that the sides are nicely crisped, a 10-by-14-inch baking dish with black surfaces is excellent.
5. Detroit-style pizza, in contrast to most other pizza varieties, stacks the dough with cheese and toppings underneath the sauce before adding the sauce.
In contrast to the more common mozzarella, traditional Detroit-style pizza makes use of Wisconsin brick cheese, a softly flavored semi-soft cheese with a high fat content that is minimally flavored.This pie’s crust is flavored with buttery taste thanks to the fat from Wisconsin brick cheese used in its preparation, but the cheese itself remains gooey in the middle while spreading outward to form a golden cheesy crust on its borders and edges of the pie.Although the cheese is occasionally put first and the toppings are sometimes placed directly on the dough, the sauce is usually ladled onto the pizza last as a last layer of taste to complete the dish.It is traditional to serve a Detroit-style pizza with a tomato-based sauce that is a perfect compliment to the milder brick cheese, which has a flavor comparable to cheddar.
1. The sauce can be spread in blotchy dollops or in thick lines, known as ″racing stripes,″ to create a unique look.
2. It is customary in certain recipes to add the sauce after the pizza has been baked, which is referred to as ″red topping″ because the sauce is the last topping on a baked pizza.
3. By adding the sauce at the very end of the cooking process, the crust is kept from ever becoming soggy.
4. Regardless of whether the sauce is poured before or after the pizza is placed in the oven, a Detroit-style pizza should be baked at around 440 degrees Fahrenheit for approximately 13 minutes to achieve a perfectly cooked pie.
5. When the pizza is finished cooking, it is chopped into squares to give it a distinct Detroit flavor and appearance.
Because these square slices of pizza are served sizzling hot, many customers prefer to eat Detroit-style pizza with a fork and knife, however some diehard Detroit pizza aficionados are willing to withstand the heat and eat their pizza without utensils if the opportunity presents itself.
## How Is Detroit-Style Pizza Different From Other Types of Pizza?
In most cases, when someone mentions ″pizza,″ the image that springs to mind is a circular, thin-crust pie with stringy melted mozzarella cheese dripping off of thin triangular pieces.Detroit-style pizza, with its rectangular form, sauce-topped cheese, and a dough that is so thick that it would be difficult to fold, deviates completely from these traditions of pizza making.In spite of the fact that Detroit-style pizza is radically different from traditional thin-crust pizzas such as a New York slice, it does have some characteristics in common with other pan pizzas.
1. In other words, what distinguishes a piece of Detroit-style pizza from any other Sicilian pizza descendant?
2. Detroit-style pizza’s biggest distinguishing characteristic as compared to other pan pizzas is its use of mozzarella cheese.
3. A Detroit-style pie has cheese strewn all over the top, which results in a characteristic crispy coating of burnt cheese around the edges that gives it its distinctive flavor.
4. During the cooking of the pizza, the fat drippings from the Wisconsin brick cheese that is used to top the pie also aid to sear the sides and bottom of the dough.
5. While other styles of pizza may be classified as deep dish, Detroit-style pizza distinguishes itself by having a soft center and a crispy crust on the exterior.
While Chicago-style pizza has a flakier, thinner deep crust, similar to that of a regular pie, Detroit-style pizza has a thick crust that is more similar to focaccia in appearance.Due to its airier texture, fluffy chew, and crispy underbelly, the Detroit-style crust distinguishes itself from the more traditional Sicilian crust.A Detroit-style pizza is similar to a Chicago deep-dish pizza in that the sauce is scooped over the top of the cheese.A Detroit-style pizza, on the other hand, does not use nearly as much sauce as a Chicago-style pizza and is distinguished by its spotted dollops of sauce and sleek red racing stripes.
1. In fact, the distinctive look of Detroit-style pizza, with its tomato sauce beautifully contrasted with its scorched cheese edges, has helped to make the cuisine a social media sensation.
You’ve found the spot to go if you’re seeking for the best Detroit-style pizza in the Metro Detroit region; Green Lantern Pizza has the pizza you’ve been looking for.Since the early 1950s, Green Lantern Pizza has been meticulously hand-cutting each of its ingredients and creating its delectable dough and sauce from scratch each and every day.Every single slice of Green Lantern Pizza is a testament to the impact that high-quality ingredients can make in the final product.
1. When it comes to pepperoni pizza, Green Lantern Pizza is recognized as the ″King of Pepperoni″ for its excellent pepperoni-topped deep-dish pies.
2. Check out the full menu to see all of the different toppings available.
3. In order to sample the greatest Detroit-style pizza in the Metro-Detroit region, locate the Green Lantern Pizza restaurant nearest you or place an online order for a Green Lantern pie.
4. Green Lantern Pizza is owned and operated by the Green Lantern Pizza Company.
5. Please get in touch with Green Lantern Pizza right away if you have any inquiries regarding the menu or Detroit-style pizza in general.
## Best Cheese Blends For Detroit Style Pizza
The cheese on Detroit style pizza is one of its distinguishing characteristics.Detroit style pizza, in contrast to most other types of pizza, does not rely on standard low-moisture or fresh mozzarella cheese to generate its distinctive flavor.Wisconsin Brick cheese is the greatest type of cheese to use on a typical Detroit style pizza, according to experts.
1. Due to its high fat content and buttery flavor, brick cheese is a type of cheese specific to the Midwestern United States that melts nicely and is a perfect match for the style of pizza popular in the city of Detroit.
2. When it comes to making their pizza, many Detroit-style pizzerias employ a combination of Brick cheese and low-moisture mozzarella.
3. However, this does not imply that you must always use Brick cheese while making a Detroit style pizza, especially when you consider that Brick cheese is quite difficult to come by outside of the Midwestern United States.
4. Let’s clear up some misunderstandings and discuss why Brick cheese is so delicious on Detroit style pizza, as well as several common options if you don’t have access to Brick cheese.
## What Kind Of Cheese Is Used On Detroit Style Pizza?
Wisconsin In the United States, brick cheese is a sort of cheese that is particularly popular in the Midwestern region of the country.It is often created in the shape of a giant brick, hence the name.A medium-soft cheese with a rich, buttery taste that develops as it matures, this cheese is a favorite among cheese lovers.
1. The increased fat level of Brick cheese is the secret to its success on Detroit-style pizza, and it is also the reason for its popularity.
2. This additional fat is a result of the higher temperatures at which the cheese is cultured, and it is this fat that gives the cheese its distinctive buttery flavor and ideal melting characteristics.
## Brick Cheese & Low-Moisture Mozzarella Are A Great Combo
Brick cheese is ideal for Detroit style pizza for all of the reasons we’ve discussed thus far, but it’s even better when combined with whole-milk low-moisture mozzarella, as we’ve demonstrated.The use of a combination of low-moisture mozzarella and Wisconsin Brick cheese results in the best of both worlds when it comes to Detroit style pizza.The mozzarella cheese provides it the classic pizza flavor that we all know and love, while the fatty Brick cheese adds a delicious, buttery flavor and excellent melting properties.
1. Many well-known Detroit-style pizzerias build their pizza by combining brick with whole-milk low-moisture mozzarella cheese and baking it at a high temperature.
## Brick Cheese Substitutes For Detroit Style Pizza
If you reside outside of the Midwestern United States, or even just outside of Wisconsin, you’ve undoubtedly noticed that Brick cheese isn’t commonly accessible on the shelves of your local grocery store or market.If you live in the United States, you may always purchase some from Amazon, although this option may not be suitable for everyone.Fortunately, there are several really acceptable substitutes to classic Brick cheese that will provide you with many of the same characteristics on your pizza as the original.
1. Shawn Randazzo, owner of the highly popular Detroit Style Pizza Company in Detroit, suggests using Muenster or Monterrey Jack cheese for the Brick cheese in this recipe.
2. Both of these cheeses are readily accessible in practically any store and have fat content and melting properties that are comparable to Brick cheese.
3. As with mozzarella, you may use nearly any type of cheese and combine it with mozzarella, so long as it meets the requirements of having a little acidic flavor and a larger fat content than regular mozzarella.
4. A list of common cheeses that may be used as a Brick cheese alternative on Detroit style pizza can be found below the recipe (in no particular order).
5. All of these cheeses have a greater fat content than others and a pleasant taste.
Please keep in mind that these cheese substitutes will not necessarily have the same flavor as Brick cheese.For the sake of this article, I’m just presenting them as common cheeses that have many of the same characteristics as Brick cheese, particularly their melting ability and greater fat content.Finally, the finest Brick cheese replacement will be determined by which cheese you prefer the flavor of the most and which is most affordable.Are you looking for a recipe for Detroit-style pizza?
1. Take a look at my detailed instructions here.
2. Are you looking for a decent Detroit style pizza sauce?
3. You’re going to enjoy my recipe right here.
4. Look no farther than this recipe for no-knead Detroit style pizza dough if you’re looking for one.
5. Looking for a high-quality pan to bake your Detroit-style pizza in?
## Chicago Deep Dish vs. Detroit-Style Pizza – Slice Pizza Blog
All right, let’s get this party started.Choose to sit on the sofa, or any other position that is more comfortable for you.Soldier Field will host a matchup between the Lions and the Bears on Sunday in a matchup of two long-time divisional rivals.
1. Both clubs have endured a difficult season thus far.
2. As a result, we’ve decided to devote our attention to more vital matters.
3. Specifically, the extremely wonderful pizza served in Chicago and Detroit.
4. Remember, it’s not about whether or not they win; it’s about how you consume the game.
5. Deep dish pizza has been the preferred form of pizza in the Midwest for many years.
Mountains of sauce, cheese, and ″toppings″ on the inside create a mouthwatering gut rush of delight.As a surprise, it has gained popularity outside of Da Bears’ home zone as well as within.However, in recent years, Detroit-style pizza has risen to the top of the pizza rankings.The gospel of its burned cheesy edges, which was once exclusive for Motown, has now traveled from coast to coast.
1. Both pies are out of this world.
2. While both are deserving of being included on your ordering list, they are not the same thing at the same time.
3. In this week’s Styles Spotlight: Pizza, we’ll tell you everything you need to know about these thick, cheesy, and wonderful pies.
4. The Crust is a kind of crust.
5. A thick and solid castle of dough might seem like the most logical way to keep deep dish pizza contained at first appearance.
However, this is not the case.According to the structural integrity of the deep dish, the crust is remarkably shallow, indicating a weakening of the structure.Although it is a stark contrast to the abundance of kindness contained within, it is a magnificent contrast nonetheless.In my opinion, the extra effort required to use a fork and knife is more than worthwhile.When it comes to deep dish pizza, attempting to palm a sizzling hot slice is a sure-fire way to end up on the injured reserve list.Please make use of the utensils provided.
• After all, you’ll need those hands to pick up the delicious squares of Detroit-style pizza that you’ll be eating.
• When the Detroit crust is finished, it looks similar to a foccacia or a Sicilian slice, but the end result is somewhere in the middle – it has smaller air bubbles than the foccaccia and a bite that is crispier than a typical Sicilian slice.
• Because they are the offensive linemen of pizza, no regular pan would suffice in this situation.
See also: How To Make Pizza On The Grill With A Stone?
Deep dish and Detroit-style pizzas, on the other hand, are baked in heavy steel pans that are more akin to dessert cookware than pizza pans in their appearance.The Spectator’s Cheese A lot of cheese is used in Detroit-style pizza.The deep dish in Chicago weighs a ****ton.
1. Michigan’s Brick cheese, a cross between white cheddar and limburger, is used to make Detroit-style pizza.
2. While well-aged brick cheeses may carry a strong fragrance and taste that can quickly clear a room, don’t worry: the Brick cheese that tops Detroit-style pizza is on the younger side, making it nearly as mild as mozzarella in flavor and aroma.
3. At the same time, it retains a strong taste, giving it an excellent complement to the buttery foundation underneath.
1. Meanwhile, cheese is seeping out of each and every piece of deep dish pizza.
2. Typically, this refers to mozzarella made using whole milk.
3. What’s in the Sauce We can see some similarities between Detroit-style and Chicago deep dish pies here, since both pies have a sauce layer on top of the cheese.
4. It is common for melty cheese to blend delightfully with the dough on Detroit-style pizza, which gives the dish its distinctive texture and flavor.
When making a deep dish pie, the order of operations is reversed for a different reason: the sauce is placed on top of the pie to protect the rest of the pie from burning during the extended cooking phase.Deep dish pizza is baked at 350-425°F for 30 minutes or longer, but Detroit-style pizza is baked at 500°F for 12-15 minutes and is fired at around 500°F for fewer than 12-15 minutes.An authentic Detroit-style pizza has a sauce that is silky and sweet with plenty of aromatics.
• When it comes to deep dish, thick and chunky tomatoes are used, which explode with every mouthful.
• A Brief Overview of the Past According to Chicago-based culinary writer Jeff Ruby, the history of deep dish pizza is a ″enigma wrapped in a pie crust.″ Ike Sewell, the founder of Pizzeria Uno, is credited with being the originator of deep dish pizza, according to some.
• His goal, according to that version of the story, was for him to develop a more filling type of pizza that was thicker, denser, and cheesier than the typical pie.
Others believe that Rudy Malnati, Lou Malnati’s father, was the true creator of the deep dish pizza.The stories of Sewell and Malnati are intertwined in the same way that the layers of a pizza are hard to separate.The Malnati family claims that Rudy worked as an employee of Ike’s and was an active participant in the operation of the restaurant.Rudy died in a car accident on the way to work.
• Rudy’s descendants claim that he invented the pizza, despite the fact that he never sought credit or recognition for it.
• The question of who developed deep dish pizza will likely never be resolved (″Every day, it feels a bit more lost to history,″ Ruby says), but things turned out rather nicely for both families in the end.
• When Sewell sold the franchise rights to Pizzeria Uno, it was the latter that was responsible for introducing the Monster Of The Midway to people all over the world.
• Lou Malnati’s pizza, on the other hand, has more than 50 locations around the country and is still acknowledged to be one of Chicagoland’s top suppliers of deep dish.
• In Detroit, there is no such disagreement.
After borrowing a set of blue steel industrial pan from a buddy working at an automobile plant in 1946, Gus Guerra was credited for inventing Detroit-style pizza.Those with a lip on them were the ideal vessel for his Sicilian-style crust, which was made even better by the heavy metal, which made it soft on the inside and crispy on the outside.After that, Gus’ restaurant, Buddy’s Rendezvous, went from 0 to 60 in a matter of minutes.Can you tell me where I can get Deep Dish and Detroit-Style Pizza near me?You can now buy true Chicago deep dish and Detroit-style pizza from fantastic local shops just about everywhere these days, which is a welcome development.
With the Slice app, you can place an order for football Sunday or any other day of the week.
## Detroit-Style Pizza: Description, Tips, and Recipe
We feature goods that we believe will be of interest to our readers.If you make a purchase after clicking on one of the links on this page, we may receive a small commission.Here’s how we went about it.
1. Pizza made in the way of Detroit is a lovely thing.
2. Fortunately, you are no longer need to be in the Motor City in order to sample it (nor do you have to pay a premium to order it online).
3. A Detroit-style pizza recipe from Matt Hyland of Emmy Squared was provided, along with professional suggestions and information on the gear you’ll need to make the crispy crust to your liking.
4. The good news is that neither a wood-fired oven nor a convection oven is required.
5. It’s all going to waste.
According to Hyland, Detroit-style pizza may be distinguished by its rectangular form and intensely crispy, caramelized edges (which he compares to ″when a mozzarella stick bursts in a deep fryer″).Fortunately, the toppings cover the dough from one lovely edge to another, which means there are no exposed crusts to be eaten or (depending upon your preferences) to become pizza bones dumped on the dish.You’ll also notice that the sauce is normally placed on top of the cheese, however depending on the other elements in the dish, it may be difficult to distinguish the inverted order of ingredients.While the sauce and dough are created in the traditional way, the mild and creamy Wisconsin brick cheese that is used as a topping is believed to be the right dairy topper for this dish.
1. However, because it might be difficult to get in some areas, a low-moisture mozzarella can suffice in this situation.
2. Some individuals recommend a combination of shredded mozzarella and muenster or Monterey Jack cheese to more precisely resemble the flavor of the original.
3. The pan, on the other hand, is non-negotiable.
4. An auto worker is said to have carried home a tray frequently used to house spare parts at the plant and repurposed it as a pizza pan, thereby giving birth to the Detroit-style pizza tradition.
5. While the narrative is frustratingly lacking in facts, it is a truth that using the proper pan is essential to obtaining those crispy edges.
A deep, rectangular, blue steel pan (8 x 10 or 10 x 14 inches) produced in Michigan is suitable for this recipe.You may prepare your own pizza dough if you like, or you can use store-bought; in any case, you’ll need a 10-ounce amount of dough for an 8 × 10-inch pan of pizza (a little more for a slightly larger pan, or you can stretch it for a thinner pizza if absolutely necessary).Lightly oil a baking pan and lay the dough in it to rest at room temperature until it begins to soften; use your hands to nudge the dough into the borders of the pan until it is evenly distributed (and appreciate the lack of pressure of trying to stretch the perfect circle).Once the dough has rested for approximately half an hour in the pan, spread the entire thing with shredded cheese, being careful to go all the way to the edges so that they crisp up on the hot metal of the baking pan.Spread two racing stripes of sauce down the length of the pizza using the bottom of a ladle to make them a bit more even in appearance.After that, top with anything you want—Emmy Squared’s Colony2 pie (as seen in the video above) includes pickled jalapenos, thick pepperoni* that cooks up into crispy cups full of fatty delight, and a sprinkling of pecorino cheese to finish it off.
• As soon as it comes out of the oven, it is smothered with a sweet honey glaze, which pairs perfectly with the fiery ingredients of the pie.
• On Detroit-style pizza, pepperoni is frequently placed beneath the cheese and sauce, but for the crispiest edges, it should be placed on top.
• Preheat the oven to 500 degrees Fahrenheit and allow plenty of time for it to reach that temperature (at least 20 minutes).
To assist conduct the heat, you may place the pan on top of a pizza stone that has been preheated as well.The pizza should be baked for 8-11 minutes, or until the edges are well caramelized, before removing it from the oven and lifting it onto a cooling rack to prevent the bottom from becoming too hot and mushy.You may transfer it to a chopping board to slice it up, but you should return it to the rack to maintain its crispness afterward.
1. Pro tip: Using clamps to remove a pan from the oven is not only more entertaining than using kitchen towels or pot holders, but it also makes clean-up easier in the future (no greasy silicone or oily fabric to deal with).
2. Furthermore, you have a lower risk of burning yourself.
3. More recipes and professional insights may be found in the cookbook Matt and his wife Emily authored, which is available here: Also, check out this video to understand more about the exact technologies that Emmy Squared employs:
## Easy Pizza Dough – Recipe
The recipe makes four balls of dough that may be used to make four individual 8-inch pizzas, for a total weight of 1-3/4 pounds.You may get a head start on supper by making homemade pizza dough the day before or a couple of weeks in advance.Place the individual balls in zip-top bags and place them in the refrigerator overnight or freeze them for extended periods of time.
1. Browse our slideshow for pizza topping ideas, or use our Recipe Maker to build your own pizza recipe (thick- or thin-crusted) using your favorite ingredients.
## Ingredients
• Active-dry yeast (1 package, 2-1/4 tsp. )
• 1-1/2 cups very warm water (110°F)
• 18 oz. (4 cups) all-purpose flour
• More flour for dusting
• 1-1/2 tsp. salt
• 2 tbsp. olive oil
• 530 calories (kcal)
• 70 calories (kcal) from fat
• 8 grams of fat
• 1 gram of saturated fat
• 1 gram of polyunsaturated fat
• 5 grams of monounsaturated fat
• 0 mg cholesterol
• 880 mg sodium
## Preparation
### Making and dividing the dough
• Using a Pyrex 2-cup measure, dissolve the yeast in the warm water and set it aside (make sure the cup isn’t too cold or it will be difficult to pour). In the meantime, combine the flour and salt in a food processor equipped with a steel blade and pulse quickly to incorporate. Continue to feed the machine the water-yeast combination in a constant stream while the machine is operating. Remove the processor from the machine and pour in the oil. Pulse a couple of times to incorporate the oil
• To make it easier to work with, scrape the soft dough out of the machine and onto a lightly floured work surface. Knead the dough rapidly using lightly dusted hands until it becomes a smooth mass, being sure to include any flour or dough from the processor bowl that didn’t get incorporated in earlier. Using a knife or a dough scraper, divide the dough into four equal pieces and set aside. Make a tight, smooth ball out of each piece, kneading it to get all of the air out.
### Rising and storing the dough
• Which method you choose to use will depend on whether you want to cook pizza the traditional manner or at a later time.
• For quickest results, place the dough balls on a gently floured board, cover them with a clean dishtowel, and allow them to rise until they have about doubled in size, about 45 minutes. In the meantime, preheat your oven, with the baking stone inside, to ensure that the stone is completely heated. The dough can be proofed in as little as 45 minutes. These dough balls are ready to be formed into various shapes.
• Prepare a baking sheet with a floured dishtowel and place the dough balls on it. Cover the dough balls with plastic wrap, allowing them to expand (they’ll practically double in size), and set aside in the refrigerated overnight if you want to make the pizzas tomorrow.
• In order to use dough that has been refrigerated overnight, simply remove it from the refrigerator 15 minutes before forming the dough into a pizza.
• As soon as you finish making the dough balls, dust each one thoroughly with flour and place each one in a separate zip-top bag until you are ready to use them. For optimal results, freeze dough overnight (or at least 10-12 hours before you plan to use it). Transfer frozen dough to the refrigerator the night before (or at least 10-12 hours before you plan to use it). However, I’ve discovered that dough balls that are withdrawn directly from the freezer and allowed to warm up on the counter will be totally defrosted in around 1-1/2 hours. Unlike other doughs, this one is virtually unbreakable. | 13,283 | 59,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.915898 |
https://e2e.ti.com/support/amplifiers-group/amplifiers/f/amplifiers-forum/947138/tle2022am-input-bias-current?tisearch=e2e-sitesearch&keymatch=TLE2022AM | 1,716,298,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00067.warc.gz | 186,605,181 | 26,248 | If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
# TLE2022AM: Input Bias current
Part Number: TLE2022AM
Hi,
Per the datasheet of TLE2022AMD the input bias current is specified as +90nA max over temperature (page 24), but as per the graph in figure 9 the input bias current is given as negative current.
As the input stage of the TLE2022AMD consists of PNP transistors is it advisable to consider -VE current for input bias current or we should consider bias current can be either +VE or -VE.
Thanks and Regards,
Midhun
• Hello Midhun,
There have been different practices used for specifying the input bias current polarity sign among the various divisions, times and products. That can lead to some confusion in terms what that actual current flow direction is for an op amp and what gets stated in the Electrical Characteristics table. As you mention the TLE2022AMD lists a maximum input bias current over temperature of 90 nA. Since it doesn't have sign in front of the number one would presume the maximum input current is positive (+90 nA).
The current convention applied to many integrated circuits is that current coming out of a pin is termed negative, and current flowing into the pin positive. The TLE2022AMD has PNP input transistors as you noted. The PNP transistor emitter arrow points into the transistor and the emitter current is equal to IE = IC + IB. The base current, which is the majority of op amp input bias current is IB and would flow out of each op amp input. Therefore, the bias current flow in reality would be negative and thus a maximum -90 nA. Since the input consists of PNP transistors and a some low leakage ESD cells the input polarity would be considered negative for the TLE2022ADM inputs.
In many circuit applications it doesn't matter if the input bias current in reality has one polarity, or the other. However, the sign of the input bias current can affect the output referred voltage offset that includes the effect of the input bias currents flowing through their circuit paths. The bias current can result in a somewhat higher, or lower output referred voltage offset than would be had by the contribution from input voltage offset alone.
Regards, Thomas
Precision Amplifiers Applications Engineering
• Thanks for the confirmation Thomas. | 513 | 2,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.906192 |
https://www.topperlearning.com/doubts-solutions/find-the-sum-up-to-n-terms-11-22-33-nn-lz3rni11/ | 1,540,007,856,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512504.64/warc/CC-MAIN-20181020034552-20181020060052-00348.warc.gz | 1,093,100,860 | 47,275 | 1800-212-7858 (Toll Free)
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# Find the sum up to n terms: 1.1!+2.2!+3.3!+........+n.n!
Asked by 11th August 2012, 9:30 PM
We have:
a*a! = (a+1 - 1)*a! = (a+1)a! - a! = (a+1)! - a!
Hence, 1*1! + 2*2! + .... +n*n! = (2!-1!) + (3! - 2!) + .....(n + 1)! - n!.
= -1! + (n+1)! or (n+1)! - 1
Answered by Expert 11th August 2012, 9:39 PM
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You have rated this answer /10 | 279 | 568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-43 | longest | en | 0.812481 |
http://betterlesson.com/lesson/reflection/5250/more-bang-for-your-buck | 1,477,390,754,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720026.81/warc/CC-MAIN-20161020183840-00173-ip-10-171-6-4.ec2.internal.warc.gz | 27,976,058 | 21,694 | ## Reflection: Developing a Conceptual Understanding Naming Arrays - Section 1: Vocabulary Rule Building
I have used this strategy of giving the students the answer to a problem and then having them define what it is or why it works for several years now, knowing that it helps build individual understanding. However, now that I am paying more attention to the Common Core and the 8 Mathematical Practices, I recognize this teaching strategy also points students to practice attending to precision and making sense of a problem and persevering to solve them. A lot of bang for your buck
Developing a Conceptual Understanding: More Bang for Your Buck
# Naming Arrays
Unit 2: Understanding Multiplication
Lesson 2 of 15
## Big Idea: After partners have worked with each other and have some practice creating arrays, it is time to challenge them to define the concept demonstrated by an array and how it can be used to solve problems.
Print Lesson
12 teachers like this lesson
Standards:
Subject(s):
65 minutes
### Michelle Marcus
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https://people.maths.bris.ac.uk/~matyd/GroupNames/481/C3xC3%5E2sD9.html | 1,708,627,494,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00761.warc.gz | 452,347,371 | 5,728 | Copied to
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## G = C3×C32⋊D9order 486 = 2·35
### Direct product of C3 and C32⋊D9
Series: Derived Chief Lower central Upper central
Derived series C1 — C3×C9 — C3×C32⋊D9
Chief series C1 — C3 — C32 — C3×C9 — C32×C9 — C3×C32⋊C9 — C3×C32⋊D9
Lower central C3×C9 — C3×C32⋊D9
Upper central C1 — C3
Generators and relations for C3×C32⋊D9
G = < a,b,c,d,e | a3=b3=c3=d9=e2=1, ab=ba, ac=ca, ad=da, ae=ea, dbd-1=ebe=bc=cb, cd=dc, ece=c-1, ede=d-1 >
Subgroups: 708 in 138 conjugacy classes, 32 normal (16 characteristic)
C1, C2, C3, C3, C3, S3, C6, C9, C32, C32, C32, D9, C3×S3, C3⋊S3, C3×C6, C3×C9, C3×C9, C33, C33, C33, C3×D9, C9⋊S3, S3×C32, C3×C3⋊S3, C32⋊C9, C32⋊C9, C32×C9, C32×C9, C34, C32⋊D9, C3×C9⋊S3, C32×C3⋊S3, C3×C32⋊C9, C3×C32⋊D9
Quotients: C1, C2, C3, S3, C6, C32, D9, C3×S3, C3×C6, C3×D9, C32⋊C6, C9⋊C6, S3×C32, C32⋊D9, C32×D9, C3×C32⋊C6, C3×C9⋊C6, C3×C32⋊D9
Smallest permutation representation of C3×C32⋊D9
On 54 points
Generators in S54
(1 33 41)(2 34 42)(3 35 43)(4 36 44)(5 28 45)(6 29 37)(7 30 38)(8 31 39)(9 32 40)(10 21 54)(11 22 46)(12 23 47)(13 24 48)(14 25 49)(15 26 50)(16 27 51)(17 19 52)(18 20 53)
(1 36 38)(2 8 5)(3 43 35)(4 30 41)(6 37 29)(7 33 44)(9 40 32)(10 13 16)(11 19 49)(12 47 23)(14 22 52)(15 50 26)(17 25 46)(18 53 20)(21 24 27)(28 34 31)(39 45 42)(48 51 54)
(1 44 30)(2 45 31)(3 37 32)(4 38 33)(5 39 34)(6 40 35)(7 41 36)(8 42 28)(9 43 29)(10 24 51)(11 25 52)(12 26 53)(13 27 54)(14 19 46)(15 20 47)(16 21 48)(17 22 49)(18 23 50)
(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)
(1 24)(2 23)(3 22)(4 21)(5 20)(6 19)(7 27)(8 26)(9 25)(10 44)(11 43)(12 42)(13 41)(14 40)(15 39)(16 38)(17 37)(18 45)(28 53)(29 52)(30 51)(31 50)(32 49)(33 48)(34 47)(35 46)(36 54)
G:=sub<Sym(54)| (1,33,41)(2,34,42)(3,35,43)(4,36,44)(5,28,45)(6,29,37)(7,30,38)(8,31,39)(9,32,40)(10,21,54)(11,22,46)(12,23,47)(13,24,48)(14,25,49)(15,26,50)(16,27,51)(17,19,52)(18,20,53), (1,36,38)(2,8,5)(3,43,35)(4,30,41)(6,37,29)(7,33,44)(9,40,32)(10,13,16)(11,19,49)(12,47,23)(14,22,52)(15,50,26)(17,25,46)(18,53,20)(21,24,27)(28,34,31)(39,45,42)(48,51,54), (1,44,30)(2,45,31)(3,37,32)(4,38,33)(5,39,34)(6,40,35)(7,41,36)(8,42,28)(9,43,29)(10,24,51)(11,25,52)(12,26,53)(13,27,54)(14,19,46)(15,20,47)(16,21,48)(17,22,49)(18,23,50), (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), (1,24)(2,23)(3,22)(4,21)(5,20)(6,19)(7,27)(8,26)(9,25)(10,44)(11,43)(12,42)(13,41)(14,40)(15,39)(16,38)(17,37)(18,45)(28,53)(29,52)(30,51)(31,50)(32,49)(33,48)(34,47)(35,46)(36,54)>;
G:=Group( (1,33,41)(2,34,42)(3,35,43)(4,36,44)(5,28,45)(6,29,37)(7,30,38)(8,31,39)(9,32,40)(10,21,54)(11,22,46)(12,23,47)(13,24,48)(14,25,49)(15,26,50)(16,27,51)(17,19,52)(18,20,53), (1,36,38)(2,8,5)(3,43,35)(4,30,41)(6,37,29)(7,33,44)(9,40,32)(10,13,16)(11,19,49)(12,47,23)(14,22,52)(15,50,26)(17,25,46)(18,53,20)(21,24,27)(28,34,31)(39,45,42)(48,51,54), (1,44,30)(2,45,31)(3,37,32)(4,38,33)(5,39,34)(6,40,35)(7,41,36)(8,42,28)(9,43,29)(10,24,51)(11,25,52)(12,26,53)(13,27,54)(14,19,46)(15,20,47)(16,21,48)(17,22,49)(18,23,50), (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), (1,24)(2,23)(3,22)(4,21)(5,20)(6,19)(7,27)(8,26)(9,25)(10,44)(11,43)(12,42)(13,41)(14,40)(15,39)(16,38)(17,37)(18,45)(28,53)(29,52)(30,51)(31,50)(32,49)(33,48)(34,47)(35,46)(36,54) );
G=PermutationGroup([[(1,33,41),(2,34,42),(3,35,43),(4,36,44),(5,28,45),(6,29,37),(7,30,38),(8,31,39),(9,32,40),(10,21,54),(11,22,46),(12,23,47),(13,24,48),(14,25,49),(15,26,50),(16,27,51),(17,19,52),(18,20,53)], [(1,36,38),(2,8,5),(3,43,35),(4,30,41),(6,37,29),(7,33,44),(9,40,32),(10,13,16),(11,19,49),(12,47,23),(14,22,52),(15,50,26),(17,25,46),(18,53,20),(21,24,27),(28,34,31),(39,45,42),(48,51,54)], [(1,44,30),(2,45,31),(3,37,32),(4,38,33),(5,39,34),(6,40,35),(7,41,36),(8,42,28),(9,43,29),(10,24,51),(11,25,52),(12,26,53),(13,27,54),(14,19,46),(15,20,47),(16,21,48),(17,22,49),(18,23,50)], [(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)], [(1,24),(2,23),(3,22),(4,21),(5,20),(6,19),(7,27),(8,26),(9,25),(10,44),(11,43),(12,42),(13,41),(14,40),(15,39),(16,38),(17,37),(18,45),(28,53),(29,52),(30,51),(31,50),(32,49),(33,48),(34,47),(35,46),(36,54)]])
63 conjugacy classes
class 1 2 3A 3B 3C ··· 3N 3O ··· 3T 3U ··· 3Z 6A ··· 6H 9A ··· 9AA order 1 2 3 3 3 ··· 3 3 ··· 3 3 ··· 3 6 ··· 6 9 ··· 9 size 1 27 1 1 2 ··· 2 3 ··· 3 6 ··· 6 27 ··· 27 6 ··· 6
63 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 6 6 6 6 type + + + + + + image C1 C2 C3 C3 C6 C6 S3 D9 C3×S3 C3×D9 C32⋊C6 C9⋊C6 C3×C32⋊C6 C3×C9⋊C6 kernel C3×C32⋊D9 C3×C32⋊C9 C32⋊D9 C3×C9⋊S3 C32⋊C9 C32×C9 C34 C33 C33 C32 C32 C32 C3 C3 # reps 1 1 6 2 6 2 1 3 8 24 1 2 2 4
Matrix representation of C3×C32⋊D9 in GL8(𝔽19)
7 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11
,
1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 5 11 0 0 0 0 0 0 15 0 7 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 5 11 0 0 0 0 0 0 15 0 7
,
1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 7
,
9 0 0 0 0 0 0 0 0 17 0 0 0 0 0 0 0 0 9 0 15 0 0 0 0 0 0 0 2 0 0 0 0 0 0 7 10 0 0 0 0 0 0 0 0 17 15 0 0 0 0 0 0 0 2 11 0 0 0 0 0 0 10 0
,
0 2 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 17 15 0 0 0 0 0 0 0 2 11 0 0 0 0 0 0 10 0 0 0 9 0 15 0 0 0 0 0 0 0 2 0 0 0 0 0 0 7 10 0 0 0
G:=sub<GL(8,GF(19))| [7,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11],[1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,5,15,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,0,1,5,15,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,7],[1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,0,7,0,0,0,0,0,0,0,0,7],[9,0,0,0,0,0,0,0,0,17,0,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,7,0,0,0,0,0,15,2,10,0,0,0,0,0,0,0,0,17,0,0,0,0,0,0,0,15,2,10,0,0,0,0,0,0,11,0],[0,10,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,7,0,0,0,0,0,15,2,10,0,0,17,0,0,0,0,0,0,0,15,2,10,0,0,0,0,0,0,11,0,0,0,0] >;
C3×C32⋊D9 in GAP, Magma, Sage, TeX
C_3\times C_3^2\rtimes D_9
% in TeX
G:=Group("C3xC3^2:D9");
// GroupNames label
G:=SmallGroup(486,94);
// by ID
G=gap.SmallGroup(486,94);
# by ID
G:=PCGroup([6,-2,-3,-3,-3,-3,-3,4755,873,453,3244,11669]);
// Polycyclic
G:=Group<a,b,c,d,e|a^3=b^3=c^3=d^9=e^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,d*b*d^-1=e*b*e=b*c=c*b,c*d=d*c,e*c*e=c^-1,e*d*e=d^-1>;
// generators/relations
×
𝔽 | 4,649 | 6,877 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.365537 |
https://www.reference.com/web?q=mean+mode+median+and+range+answer+key&qo=relatedSearchBing&o=600605&l=dir | 1,558,370,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256082.54/warc/CC-MAIN-20190520162024-20190520184024-00329.warc.gz | 930,536,530 | 14,481 | http://www.math-aids.com/Mean_Mode_Median/
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Calculate average (mean), median, mode, and range of number sets. ... Use the data on the line plot to answer the range, median, mean, and mode questions.
https://www.math-salamanders.com/mean-median-mode-range.html
Here you will find our selection of Mean Median Mode Range Worksheets for kids by the ... Mean Help; Median Help; Mode Help; Range Help; Mean Median Mode Range Worksheets .... Median Mean Mode and Range Problems 1 · Answers. | 500 | 2,014 | {"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} | 3.953125 | 4 | CC-MAIN-2019-22 | latest | en | 0.797199 |
https://buziospousadas.com/a-basic-guide-to-the-game-of-poker/ | 1,702,042,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00119.warc.gz | 177,939,824 | 13,171 | A basic guide to the game of poker. In this article, we will discuss the Rules, Variations, Lingo, and Betting intervals. As an added bonus, we’ll explore how players make their decisions and use statistics to make better decisions. For further details, visit our Poker Rules page. In addition, we’ll explain the lingo and psychology of poker. In addition, you’ll discover the most important aspects of the game.
## Rules
The “Rules of Poker” are the rules of the game of poker. They are widely adopted and freely copied. As long as they are not sold for profit, people are free to use the rules portions for their own purposes. In general, however, you must give proper credit to the original source. The primary goal of any rulebook is to create the best set of rules for poker that exists. It is the goal of any idn play author to make his work the best it can be.
## Variations
While all forms of poker are similar, variations vary. In addition to the rules of the game, variations can differ in betting structures, betting limits, and other aspects. Understanding these differences can help you learn the rules of each game and impress other players. Some examples of these differences include the Omaha and Lowball variants, as well as the Dr. Pepper variant. To learn more about the differences between the two types of poker, keep reading! Listed below are some of the most common variations.
## Lingo
When it comes to poker, the lingo used to describe the hands of players is extremely important. For example, the term “wild card” is often used to describe the lowest card in a game. The term “wild card” is also used to describe the unpredictable person in a game. Learn to use poker lingo to better understand the game and maximize your winnings. The following are some examples of poker lingo you should learn.
## Betting intervals
Betting intervals for poker games vary, depending on the number of players and the game being played. Generally, the first player to act places a bet and all players to his or her left must raise their bets proportionally to their previous contributions. This cycle is repeated until one player remains with the most chips in the pot. If the betting intervals are not followed, the winner is determined by the next player’s move.
## Best possible hand in poker
A royal flush is the best possible hand in poker. This hand contains all five cards of the same suit: ace, king, queen, and jack. Although four aces can be a winning hand in poker, they never beat a royal flush. The royal flush is the highest possible hand in poker, and it’s hard to beat it. Here’s how to get a royal flush in Texas Hold’em. | 566 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-50 | longest | en | 0.961555 |
https://www.prepbharat.com/EntranceExams/JEEMain/MotionPlane/motionplane-10.html | 1,709,346,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00535.warc.gz | 942,638,324 | 5,188 | ## Motion in a Plane Questions and Answers Part-10
1. A car is moving with high velocity when it has a turn. A force acts on it outwardly because of
a) Centripetal force
b) Centrifugal force
c) Gravitational force
d) All the above
Explanation: Centrifugal force
2. A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
a) Double
b) Half
c) 4 times
d) $\frac{1}{4} times$
Explanation:
3. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is $\left(g =10ms^{-2}\right)$
a) $40ms^{-1}$
b) $20ms^{-1}$
c) $15ms^{-1}$
d) $10ms^{-1}$
Explanation:
4. An athlete completes one round of a circular track of radius 10 m in 40 sec. The distance covered by him in 2 min 20 sec is
a) 70 m
b) 140 m
c) 110 m
d) 220 m
Explanation:
5. A proton of mass $1.6\times10^{-27}kg$ goes round in a circular orbit of radius 0.10 m under a centripetal force of $4\times10^{-13}N$ . then the frequency of revolution of the proton is about
a) $0.08\times10^{8}$ cycles per sec
b) $4\times10^{8}$ cycles per sec
c) $8\times10^{8}$ cycles per sec
d) $12\times10^{8}$ cycles per sec
Explanation:
6. A particle is moving in a circle with uniform speed v. In moving from a point to another diametrically opposite point
a) The momentum changes by mv
b) The momentum changes by 2mv
c) The kinetic energy changes by $\left(1\diagup 2\right)mv^{2}$
d) The kinetic energy changes by $mv^{2}$
Explanation: Momentum changes by 2mv but kinetic energy remains same
7. In uniform circular motion
a) Both the angular velocity and the angular momentum vary
b) The angular velocity varies but the angular momentum remains constant
c) Both the angular velocity and the angular momentum stay constant
d) The angular momentum varies but the angular velocity remains constant
Explanation: L = I $\omega$
In U.C.M. $\omega$ = constant, L = constant
8. When a body moves in a circular path, no work is done by the force since
a) There is no displacement
b) There is no net force
c) Force and displacement are perpendicular to each other
d) The force is always away from the centre
Explanation: W = FS cos $\theta$
$\theta$ = 90°
9. Which of the following statements is false for a particle moving in a circle with a constant angular speed
a) The velocity vector is tangent to the circle
b) The acceleration vector is tangent to the circle
c) The acceleration vector points to the centre of the circle
d) The velocity and acceleration vectors are perpendicular to each other
10. If $a_{r}$ and $a_{t}$ represent radial and tangential accelerations, the motion of a particle will be uniformly circular if
a) $a_{r}=0$ and $a_{t}=0$
b) $a_{r}=0$ but $a_{t}\neq0$
c) $a_{r}\neq0$ but $a_{t}=0$
d) $a_{r}\neq0$ and $a_{t}\neq0$ | 835 | 2,869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-10 | latest | en | 0.799948 |
http://mattdickenson.com/tag/research/ | 1,419,024,982,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768977.107/warc/CC-MAIN-20141217075248-00090-ip-10-231-17-201.ec2.internal.warc.gz | 174,661,681 | 30,248 | # Classifying Olympic Athletes by Sport and Event (Part 2)
This is the second post in a three-part series. The first post, giving some background and describing the data, is here. In that post I pointed out David Epstein’s claim that he could identify an Olympian’s event knowing only her height and weight. The sheer number of Olympians–about 10,000–makes me skeptical, but I decided to see whether machine learning could accurately produce the predictions Mr. Epstein claims he could.
To do this, I tried four different machine learning methods. These are all well-documented methods implemented in existing R packages. Code and data for is here (for sports) and here (for events).
The first two methods, conditional inference trees (using the party package) and evolutionary trees (using evtree), are both decision tree-based approaches. That means that they sequentially split the data based on binary decisions. If the data falls on one side of the split (say, height above 1.8 meters) you continue down one fork of the tree, and if not you go down the other fork. The difference between these two methods is how the tree is formed: the first recursively partitions the data based on conditional probability, while the second method (as the name suggests) uses an evolutionary algorithm. To get a feel for how this actually divides the data, see the figure below and this post.
If a single tree is good, a whole forest must be better–or at least that’s the thinking behind random forests, the third method I used. This method generates a large number of trees (500 in this case), each of which has access to only some of the features in the data. Once we have a whole forest of trees, we combine their predictions (usually through a voting process). The combination looks a little bit like the figure below, and a good explanation is here.
The fourth and final method used–artificial neural networks–is a bit harder to visualize. Neural networks are sort of a black box, making them difficult to interpret and explain. At a coarse level they are intended to work like neurons in the brain: take some input, and produce output based on whether the input crosses a certain threshold. The neural networks I used have a single hidden layer with 30 (for sports classification) or 50 hidden nodes (for event classification). To get a better feel for how neural networks work, see this three part series.
That’s a very quick overview of the four machine learning methods that I applied to classifying Olympians by sport and event. The data and R code are available at the link above. In the next post, scheduled for Friday, I’ll share the results.
# Classifying Olympic Athletes by Sport and Event (Part 1)
Note: This post is the first in a three-part series. It describes the motivation for this project and the data used. When parts two and three are posted I will link to them here.
Can you predict which sport or event an Olympian competes in based solely on her height, weight, age and sex? If so, that would suggest that physical features strongly drive athletes’ relative abilities across sports, and that they pick sports that best leverage their physical predisposition. If not, we might infer that athleticism is a latent trait (like “grit“) that can be applied to the sport of one’s choice.
David Epstein argues that sporting success is largely based on heredity in his book, The Sports Gene. To support his argument, he describes how elite athletes’ physical features have become more specialized to their sport over time (think Michael Phelps). At a basic level Epstein is correct: males and females differ at both a genetic level and in their physical features, generally speaking.
However, Epstein advanced a stronger claim in an interview (at 29:46) with Russ Roberts:
Roberts: [You argue that] if you simply had the height and weight of an Olympic roster, you could do a pretty good job of guessing what their events are. Is that correct?
Epstein: That’s definitely correct. I don’t think you would get every person accurately, but… I think you would get the vast majority of them correctly. And frankly, you could definitely do it easily if you had them charted on a height-and-weight graph, and I think you could do it for most positions in something like football as well.
I chose to assess Epstein’s claim in a project for a machine learning course at Duke this semester. The data was collected by The Guardian, and includes all participants for the 2012 London Summer Olympics. There was complete data on age, sex, height, and weight for 8,856 participants, excluding dressage (an oddity of the data is that every horse-rider pair was treated as the sole participant in a unique event described by the horse’s name). Olympians participate in one or more events (fairly specific competitions, like a 100m race), which are nested in sports (broader categories such as “Swimming” or “Athletics”).
Athletics is by far the largest sport category (around 20 percent of athletes), so when it was included it dominated the predictions. To get more accurate classifications, I excluded Athletics participants from the sport classification task. This left 6,956 participants in 27 sports, split into a training set of size 3,520 and a test set of size 3,436. The 1,900 Athletics participants were classified into 48 different events, and also split into training (907 observations) and test sets (993 observations). For athletes participating in more than one event, only their first event was used.
What does an initial look at the data tell us? The features of athletes in some sports (Basketball, Rowing, Weightlifting, and Wrestling) and events (100m hurdles, Hammer throw, High jump, and Javelin) exhibit strong clustering patters. This makes it relatively easy to guess a participant’s sport or event based on her features. In other sports (Archery, Swimming, Handball, Triathlon) and events (100m race, 400m hurdles, 400m race, and Marathon) there are many overlapping clusters making classification more difficult.
Well-defined (left) and poorly-defined clusters of height and weight by sport.
Well-defined (left) and poorly-defined clusters of height and weight by event.
The next post, scheduled for Wednesday, will describe the machine learning methods I applied to this problem. The results will be presented on Friday.
# What Really Happened to Nigeria’s Economy?
You may have heard the news that the size Nigeria’s economy now stands at nearly \$500 billion. Taken at face value (as many commenters have seemed all to happy to do) this means that the West African state “overtook” South Africa’s economy, which was roughly \$384 billion in 2012. Nigeria’s reported GDP for that year was \$262 billion, meaning it roughly doubled in a year.
How did this “growth” happen? As Bloomberg reported:
On paper, the size of the economy expanded by more than three-quarters to an estimated 80 trillion naira (\$488 billion) for 2013, Yemi Kale, head of the National Bureau of Statistics, said at a news conference yesterday to release the data in the capital, Abuja….
The NBS recalculated the value of GDP based on production patterns in 2010, increasing the number of industries it measures to 46 from 33 and giving greater weighting to sectors such as telecommunications and financial services.
The actual change appears to be due almost entirely to Nigeria including figures in GDP calculation that had been excluded previously. There is nothing wrong with this, per se, but it makes comparisons completely unrealistic. This would be like measuring your height in bare feet for years, then doing it while wearing platform shoes. Your reported height would look quite different, without any real growth taking place. Similar complications arise when comparing Nigeria’s new figures to other countries’, when the others have not changed their methodology.
Nigeria’s recalculation adds another layer of complexity to the problems plaguing African development statistics. Lack of transparency (not to mention accuracy) in reporting economic activity makes decisions about foreign aid and favorable loans more difficult. For more information on these problems, see this post discussing Morten Jerven’s book Poor NumbersIf you would like to know more about GDP and other economic summaries, and how they shape our world, I would recommend Macroeconomic Patterns and Stories (somewhat technical), The Leading Indicators, and GDP: A Brief but Affectionate History.
# “The Impact of Leadership Removal on Mexican Drug Trafficking Organizations”
That’s the title of a new article, now online at the Journal of Quantitative Criminology. Thanks to fellow grad students Cassy Dorff and Shahryar Minhas for their feedback. Thanks also to mentors at the University of Houston (Jim Granato, Ryan Kennedy) and Duke University (Michael D. Ward, Scott de Marchi, Guillermo Trejo) for thoughtful comments. The anonymous reviewers at JQC and elsewhere were also a big help.
Here is the abstract:
### Objectives
Has the Mexican government’s policy of removing drug-trafficking organization (DTO) leaders reduced or increased violence? In the first 4 years of the Calderón administration, over 34,000 drug-related murders were committed. In response, the Mexican government captured or killed 25 DTO leaders. This study analyzes changes in violence (drug-related murders) that followed those leadership removals.
### Methods
The analysis consists of cross-sectional time-series negative binomial modeling of 49 months of murder counts in 32 Mexican states (including the federal district).
### Results
Leadership removals are generally followed by increases in drug-related murders. A DTO’s home state experiences more subsequent violence than the state where the leader was removed. Killing leaders is associated with more violence than capturing them. However, removing leaders for whom a \$30m peso bounty was offered is associated with a smaller increase than other removals.
### Conclusion
DTO leadership removals in Mexico were associated with an estimated 415 additional deaths during the first 4 years of the Calderón administration. Reforming Mexican law enforcement and improving career prospects for young men are more promising counter-narcotics strategies. Further research is needed to analyze how the rank of leaders mediates the effect of their removal.
I didn’t shell out \$3,000 for open access, so the article is behind a paywall. If you’d like a draft of the manuscript just email me.
# Mexico Update Following Joaquin Guzmán’s Capture
As you probably know by now, the Sinaloa cartel’s leader Joaquin Guzmán was captured in Mexico last Saturday. How will violence in Mexico shift following Guzman’s removal?
(Alfredo Estrella/AFP/Getty Images)
I take up this question in an article forthcoming in the Journal of Quantitative Criminology. According to that research (which used negative binomial modeling on a cross-sectional time series of Mexican states from 2006 to 2010), DTO leadership removals in Mexico are generally followed by increased violence. However, capturing leaders is associated with less violence than killing them. The removal of leaders for whom a 30 million peso bounty (the highest in my dataset, which generally identified high-level leaders) been offered is also associated with less violence. The reward for Guzmán’s capture was higher than any other contemporary DTO leader: 87 million pesos. Given that Guzmán was a top-level leader and was arrested rather than killed, I would not expect a significant uptick in violence (in the next 6 months) due to his removal. This follows President Pena Nieto’s goal of reducing DTO violence.
My paper was in progress for a while, so the data is a few years old. Fortunately Brian Phillips has also taken up this question using additional data and similar methods, and his results largely corroborate mine:
Many governments kill or capture leaders of violent groups, but research on consequences of this strategy shows mixed results. Additionally, most studies have focused on political groups such as terrorists, ignoring criminal organizations – even though they can represent serious threats to security. This paper presents an argument for how criminal groups differ from political groups, and uses the framework to explain how decapitation should affect criminal groups in particular. Decapitation should weaken organizations, producing a short-term decrease in violence in the target’s territory. However, as groups fragment and newer groups emerge to address market demands, violence is likely to increase in the longer term. Hypotheses are tested with original data on Mexican drug-trafficking organizations (DTOs), 2006-2012, and results generally support the argument. The kingpin strategy is associated with a reduction of violence in the short term, but an increase in violence in the longer term. The reduction in violence is only associated with leaders arrested, not those killed.
A draft of the full paper is here.
# Visualizing the Indian Buffet Process with Shiny
(This is a somewhat more technical post than usual. If you just want the gist, skip to the visualization.)
N customers enter an Indian buffet restaurant, one after another. It has a seemingly endless array of dishes. The first customer fills her plate with a Poisson(α) number of dishes. Each successive customer i tastes the previously sampled dishes in proportion to their popularity (the number of previous customers who have sampled the kth dish, m_k, divided by i). The ith customer then samples a Poisson(α) number of new dishes.
That’s the basic idea behind the Indian Buffet Process (IBP). On Monday Eli Bingham and I gave a presentation on the IBP in our machine learning seminar at Duke, taught by Katherine Heller. The IBP is used in Bayesian non-parametrics to put a prior on (exchangeability classes of) binary matrices. The matrices usually represent the presence of features (“dishes” above, or the columns of the matrix) in objects (“customers,” or the rows of the matrix). The culinary metaphor is used by analogy to the Chinese Restaurant Process.
Although the visualizations in the main paper summarizing the IBP are good, I thought it would be helpful to have an interactive visualization where you could change α and N to see how what a random matrix with those parameters looks like. For this I used Shiny, although it would also be fun to do in d3.
One realization of the IBP, with α=10.
In the example above, the first customer (top row) sampled seven dishes. The second customer sampled four of those seven dishes, and then four more dishes that the first customer did not try. The process continues for all 10 customers. (Note that this matrix is not sorted into its left-ordered-form. It also sometimes gives an error if α << N, but I wanted users to be able to choose arbitrary values of N so I have not changed this yet.) You can play with the visualization yourself here.
Interactive online visualizations like this can be a helpful teaching tool, and the process of making them can also improve your own understanding of the process. If you would like to make another visualization of the IBP (or another machine learning tool that lends itself to graphical representation) I would be happy to share it here. I plan to add the Chinese restaurant process and a Dirichlet process mixture of Gaussians soon. You can find more about creating Shiny apps here.
# Constitutional Forks Revisited
Around this time last year, we discussed the idea of a constitutional “fork” that occurred with the founding of the Confederate States of America. That post briefly explains how forks work in open source software and how the Confederates used the US Constitution as the basis for their own, with deliberate and meaningful differences. Putting the two documents on Github allowed us to compare their differences visually and confirm our suspicions that many of them were related to issues of states’ rights and slavery.
Caleb McDaniel, a historian at Rice who undoubtedly has a much deeper and more thorough knowledge of the period, conducted a similar exercise and also posted his results on Github. He was faced with similar decisions of where to obtain the source text and which differences to retain as meaningful (for example, he left in section numbers where I did not). My method identifies 130 additions and 119 deletions when transitioning between the USA and CSA constitutions, whereas the stats for Caleb’s repo show 382 additions and 370 deletions.
What should we draw from these projects? In Caleb’s words:
My decisions make this project an interpretive act. You are welcome to inspect the changes more closely by looking at the commit histories for the individual Constitution files, which show the initial text as I got it from Avalon as well as the changes that I made.
You can take a look at both projects and conduct a difference-in-differences exploration of your own. More generally, these projects show the need for tools to visualize textual analyses, as well as the power of technology to enhance understanding of historical and political acts. Caleb’s readme file has great resources for learning more about this topic including the conversation that led him to this project, a New York Times interactive feature on the topic, and more.
# The Economy That Is Stanford
Five of the six most-visited websites in the world are here, in ranked order: Facebook, Google, YouTube (which Google owns), Yahoo! and Wikipedia. (Number five is a Chinese-language site.) If corporations founded by Stanford alumni were to form an independent nation, it would be the tenth largest economy in the world, with an annual revenue of \$2.7 trillion, as some professors at that university recently calculated. Another new report says: ‘If the internet was a country, its gross domestic product would eclipse all others but four within four years.’
That’s from this London Review of Books piece by Rebecca Solnit. The October, 2012, research report on which the claim is based is here, based on survey data. Solnit’s piece is interesting throughout, including a discussion of parallels and differences between the tech boom and the Gold Rush.
# Political Forecasting and the Use of Baseline Rates
As Joe Blitzstein likes to say, “Thinking conditionally is a condition for thinking.” Humans are not naturally good at this skill. Consider the following example: Kelly is interested in books and keeping things organized. She loves telling stories and attending book clubs. Is it more likely that Kelly is a bestselling novelist or an accountant?
Many of the “facts” about Kelly in that story might lead you to answer that she is a novelist. Only one–her sense of organization–might have pointed you toward an accountant. But think about the overall probability of each career. Very few bookworms become successful novelists, and there are many more accountants than (successful) authors in the modern workforce. Conditioning on the baseline rate helps make a more accurate decision.
I make a similar point–this time applied to political forecasting–in a recent blog post for the blog of Mike Ward’s lab (of which I am a member):
One piece of advice that Good Judgment forecasters are often reminded of is to use the baseline rate of an event as a starting point for their forecast. For example, insurgencies are a very rare event on the whole. For the period January, 2001 to August, 2013, insurgencies occurred in less than 10 percent of country-months in the ICEWS data set.
From this baseline, we can then incorporate information about the specific countries at hand and their recent history… Mozambique has not experienced an insurgency for the entire period of the ICEWS dataset. On the other hand, Chad had an insurgency that ended in December, 2003, and another that extended from November, 2005, to April, 2010. For the duration of the ICEWS data set, Chad has experienced an insurgency 59 percent of the time. This suggests that our predicted probability of insurgency in Chad should be higher than for Mozambique.
I started writing that post before rebels in Mozambique broke their treaty with the government. Maybe I spoke too soon, but the larger point is that baselines are the starting point–not the final product–of any successful forecast.
Having more data is useful, as long as it contributes more signal than noise. That’s what ICEWS aims to do, and I consider it a useful addition to the toolbox of forecasters participating in the Good Judgment Project. For more on this collaboration, as well as a map of insurgency rates around the globe as measured by ICEWS, see the aforementioned post here.
# Visualizing Political Unrest in Egypt, Syria, and Turkey
The lab of Michael D. Ward et al now has a blog. The inaugural post describes some of the lab’s ongoing projects that may come up in future entries including modeling of protests, insurgencies, and rebellions, event prediction (such as IED explosions), and machine learning techniques.
The second post compares two event data sets–GDELT and ICEWS–using recent political unrest in the Middle East as a focal point (more here):
We looked at protest events in Egypt and Turkey in 2011 and 2012 for both data sets, and we also looked at fighting in Syria over the same period…. What did we learn from these, limited comparisons? First, we found out first hand what the GDELT community has been saying: the GDELT data are in BETA and currently have a lot of false positives. This is not optimal for a decision making aid such as ICEWS, in which drill-down to the specific events resulting in new predictions is a requirement. Second, no one has a good ground truth for event data — though we have some ideas on this and are working on a study to implement them. Third, geolocation is a boon. GDELT seems especially good a this, even with a lot of false positives.
The visualization, which I worked on as part of the lab, can be found here. It relies on CartoDB to serve data from GDELT and ICEWS, with some preprocessing done using MySQL and R. The front-end is Javascript using a combination of d3 for timelines and Torque for maps.
GDELT (green) and ICEWS (blue) records of protests in Egypt and Turkey and conflict in Syria
If you have questions about the visualizations or the technology behind them, feel free to mention them here or on the lab blog. | 4,649 | 22,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.93661 |
https://kr.mathworks.com/matlabcentral/cody/problems/42876-rotate-matrix-180-degree/solutions/1588365 | 1,585,613,862,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497309.31/warc/CC-MAIN-20200330212722-20200331002722-00021.warc.gz | 460,615,781 | 15,774 | Cody
# Problem 42876. Rotate Matrix @180 degree
Solution 1588365
Submitted on 20 Jul 2018 by Tom Holz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(rot180(x),y_correct))
2 Pass
x=[8 1 6; 3 5 7; 4 9 2], y_correct= [2 9 4; 7 5 3; 6 1 8] assert(isequal(rot180(x),y_correct))
x = 8 1 6 3 5 7 4 9 2 y_correct = 2 9 4 7 5 3 6 1 8
3 Pass
A=[2 4;8 6]; y_correct =[ 6 8; 4 2] assert(isequal(rot180(A),y_correct))
y_correct = 6 8 4 2
4 Pass
A=[1 2 3; 4 5 6]; y_correct =[ 6 5 4;3 2 1] assert(isequal(rot180(A),y_correct))
y_correct = 6 5 4 3 2 1 | 309 | 709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.518968 |
https://www.assignmenthelp.net/forum/topic/the-circumference-of-a-circle-is-equal-to-72-pi-find-the-radius-of-this-circle/ | 1,670,374,536,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00805.warc.gz | 687,396,214 | 8,846 | Forums Assignment Help The circumference of a circle is equal to 72 pi. Find the radius of this circle.
Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #1934
john Smith
Participant
The circumference of a circle is equal to 72 pi. Find the radius of this circle.
#10519
ahprofessional
Participant
C = 2 pi r, where r is the radius of the circle.
Substitute C by 72 pi to obtain the equation
72 pi = 2 pi r
Simplify and solve for r to obtain
r = 36
Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic. | 163 | 558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.858921 |
https://www.convertunits.com/from/pertica/to/pace | 1,591,233,367,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00236.warc.gz | 678,767,023 | 7,706 | ## ››Convert pertica to pace [great]
pertica pace
Did you mean to convert pertica to pace [great] pace [Roman]
How many pertica in 1 pace? The answer is 0.51486486486486.
We assume you are converting between pertica and pace [great].
You can view more details on each measurement unit:
pertica or pace
The SI base unit for length is the metre.
1 metre is equal to 0.33783783783784 pertica, or 0.65616797900262 pace.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pertica and pace [great].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pertica to pace
1 pertica to pace = 1.94226 pace
5 pertica to pace = 9.71129 pace
10 pertica to pace = 19.42257 pace
15 pertica to pace = 29.13386 pace
20 pertica to pace = 38.84514 pace
25 pertica to pace = 48.55643 pace
30 pertica to pace = 58.26772 pace
40 pertica to pace = 77.69029 pace
50 pertica to pace = 97.11286 pace
## ››Want other units?
You can do the reverse unit conversion from pace to pertica, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 442 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-24 | latest | en | 0.754441 |
https://bankersway.com/arun-sharma-quantitative-aptitude/ | 1,721,634,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00446.warc.gz | 105,940,986 | 22,494 | # No #1 Platform For Job Updates
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## Content Of Arun Sharma Quantitative Aptitude PDF
1: Time & Work
2: Percentage
3: Mixture & Allegation
4: Simple Interest
5: Compound Interest
6: Trigonometry
7: Power And Indices
8: Simplification
9: Ratio And Proportion
10: Mensuration
11: Pipe And Cistern
12: Co-Ordinate Geometry
13: Discount
14: Profit & Loss
15: Number System
16: LCM & HCF
17: Average Speed
18: Time & Distance Train
19: Algebra
20: Boat & Stream
21: Partnership
22: Height & Distance
23: Data Interpretation
24: Simplification
### Some important questions about Quantitative Aptitude
Q1. If 3/4 of a number is 7 more than 1/6 of the number then 5/3 of the number is?
(a) 12
(b) 18
(c) 15
(d) 20
S1. Ans.(d)
Sol. Let the number be x
According to the question,
3x/4-x/6 = 7
(9x β 2x)/12 = 7
7x = 7 Γ 12
x = 12
Then 5/3 of the number will be
(x Γ5)/3 β (12 Γ 5)/3 = 20
Q2. If x = 1/β(2 + 1) then (x + 1) equal to?
(a) β2+1
(b) β2-1
(c) β2
(d) 2
S2. Ans.(c)
Sol. Given
x=1/(β2+1)
Find (x + 1) = ?
β x + 1 = 1/(β2Β + 1)+1
β(β2+1+1)/(β2+1)
β(β2+2)/(β2+1)
β(β2 (β2+1))/((β2+1) )
β β2
Q3.Β If a * b = a + b + a/b, then the value of 12 * 4 is?
(a) 20
(b) 21
(c) 48
(d) 19
S3. Ans.(d)
Sol. 12 * 4 = 12 + 4 + 12/4
= 12 + 4 +3 = 19
Q4. Find the maximum number of trees which can be planted, 20 meters apart, on two sides of a straight road 1760 meters long?
(a) 180
(b) 178
(c) 174
(d) 176
S4. Ans.(b)
Sol. No of trees planted on one side of Road
= (Length of road)/(Distance between trees) + 1 β first tree
=1760/20+1
= 88 + 1 = 89
Total trees (both sides) = 89 Γ 2 = 178
### Quantitative Aptitude By Arun Sharma PDF
Q5. Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is?
(a) 15
(b) 20
(c) 25
(d) 10
S5. Ans.(b)
Sol. Mohan does βxβ correct sums
According to question
β 3x β 60 + 2x = 40
5x = 100
x = 20
Q6. Two trains 108 m and 112 m in length are running towards each other on the parallel lines at a speed of 45 km/hr and 54 km/hr respectively. To cross each other after they meet, it will take?
(a) 12 sec
(b) 9 sec
(c) 8 sec
(d) 10 sec
S6. Ans.(c)
Sol. TimeΒ takenΒ by them to cross each other
=(l1+l2)/(Relative speed in opposite direction)
Time = ((108 + 112))/((45 + 54)Γ5/18)=(220 Γ 18)/(99 Γ 5)
Time = 8 second
Q7. Two trains 150 m and 120 m long respectively moving from opposite directions cross each other in 10 seconds. If the speed of the second train is 43.2 km/hr, then the speed of the first train is?
(a) 54 km/hr
(b) 50 km/hr
(c) 52 km/hr
(d) 51 km/hr
## Why should you follow Arun Sharma Quantitative Aptitude for CAT book pdf?
• This pdf covers all essential topics in great depth that are scoring and carries high marks in the competitive examinations.
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• Arun Sharma Quantitative Aptitude Book covers all important topics that are scoring and highly weightage in competitive examination.
• Arun Sharma Quantitative Aptitude Book also provided previous year question bank that is very useful for every aspirant.
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This book Arun Sharma Quantitative Aptitude Book contains the latest sets of questions as per the new syllabus from all the competitive examinations such as CAT (Common Admission Test), SSC, and for the Banking sector.
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https://mcqslearn.com/cost-accounting/capital-budgeting-and-inflation-multiple-choice-questions.php | 1,723,619,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00348.warc.gz | 307,448,967 | 16,898 | BBA Finance Degree Courses
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## Capital Budgeting and Inflation MCQs: Questions and Answers
MCQ 1:
If the nominal rate is 26% and the inflation rate is 12%, then the real rate can be
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2. 11.65%
3. 12.50%
4. 13.50%
MCQ 2:
If the real rate is 16% and an inflation rate is 8%, then the nominal rate of return will be
1. 27.28%
2. 25.28%
3. 22.28
4. 21.28
MCQ 3:
The rate of return to cover a risk of investment and decrease in purchasing power, as a result of inflation is known as
1. nominal rate of return
2. accrual accounting rate of return
3. real rate of return
4. required rate of return
MCQ 4:
The rate of return, which is made up of risk free and business risk element is known as
1. nominal rate of return
2. accrual accounting rate of return
3. real rate of return
4. required rate of return
MCQ 5:
The rate of required return to cover the risk of investment, in absence of inflation is classified as
1. real rate of return
2. required rate of return
3. nominal rate of return
4. none of above
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Marketing Principles App (iOS & Android) | 658 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.852928 |
http://mathhelpforum.com/math-topics/127130-suvat-time-print.html | 1,498,755,857,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00380.warc.gz | 258,881,198 | 3,322 | # SUVAT(time)
• Feb 4th 2010, 04:49 AM
Mist
SUVAT(time)
A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.
S=20 V=5.5 A=2 T=?
I used the formula $s=vt-1/2at^2$ but I got the wrong answer.
Any help?
• Feb 4th 2010, 05:06 AM
e^(i*pi)
Quote:
Originally Posted by Mist
A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.
S=20 V=5.5 A=2 T=?
I used the formula $s=vt-1/2at^2$ but I got the wrong answer.
Any help?
I can't see any other method to solve this. In this case I am using A to denote acceleration as to not confuse with a in the quadratic formula
$s = vt + \frac{1}{2}At^2$
$\frac{1}{2}At^2 + vt -s = 0$
$t = \frac{-v\pm \sqrt{v^2+2As}}{A}$
Sub in what you know and I get t = 8
• Feb 4th 2010, 07:31 AM
Mist
Quote:
Originally Posted by e^(i*pi)
Sub in what you know and I get t = 8
The formula has been rearranged wrong
• Feb 4th 2010, 08:06 AM
e^(i*pi)
Quote:
Originally Posted by Mist
The formula has been rearranged wrong
From what I can gather you've took the initial speed, u as the final speed, v. If we define O as the start point and O to A is the positive direction then at point O the speed given is the initial speed.
My equation is fine, I didn't evaluate both roots in my previous post - the second root is 2.5
$s = ut+\frac{1}{2}at^2$ (this is one of the suvat equations)
s = 20, u = 5.5, t = t, a = 2
$20 = 5.5t+\frac{1}{2}(2)t = 5.5t+t^2$
As this is a quadratic rearrange into the form $ax^2+bx+c=0$
$t^2+5.5t-20 = 0$
Solving that using the quadratic formula gives $t = 8$ and $t= 2.5$. | 636 | 1,862 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-26 | longest | en | 0.934266 |
http://tex.stackexchange.com/questions/57812/bounding-box-for-each-letter?answertab=votes | 1,462,286,399,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121561.0/warc/CC-MAIN-20160428161521-00078-ip-10-239-7-51.ec2.internal.warc.gz | 289,640,471 | 21,543 | # Bounding box for each letter
How can I take input text and replace each character with a solid (and/or hollow) rectangle representing the bounding box for that character? In the case where two characters are closer together from kerning (e.g. microtype) the boxes would overlap. The addition of the boxes should not change the spacing of the text - I'm looking for a "draft mode" for the letters.
# Solutions and Problems
Both solutions presented by Yiannis Lazarides and egreg seem to do a reasonable job for single words, though as mentioned, it seems that kerning was not completely respected. Below are the results (without markup/egreg/Yiannis):
Both solutions fail however when multiple words are involved. One of the answers completely eats a space while the other overcompensates. Both of them seem to choke on a line break as well.
-
this is quite similar to Exercise 11.5 from The TeXbook. Knuth provides a solution- not sure about posting it here though because of copyright... – cmhughes May 29 '12 at 23:17
Am pretty sure there is a question here doing exactly this but can't seem to locate it... – Peter Grill May 30 '12 at 0:18
@cmhughes The code of the TeXbook is available for study and quoting a passage from a book is always allowed (with proper attribution). – egreg May 30 '12 at 7:57
fwiw, vafa khalighi has created a package showcharinbox which answers this (nearly 2-months-old) question. see ctan.org/pkg/showcharinbox – wasteofspace Jul 24 '12 at 10:09
A LuaTeX solution. Should work in all situations that I am aware of:
\documentclass{article}
\usepackage{luacode,luatexbase}
\begin{document}
\begin{luacode*}
local number_sp_in_a_pdf_point = 65782
function math.round(num)
return math.floor(num * 1000 + 0.5) / 1000
end
-- width/height/depth of a glyph and the whatsit node
local wd,ht,dp,w
-- head is a linked list (next/prev entries pointing to the next node)
-- a hbox/vbox
-- Create a pdf_literal node to draw a box with the dimensions
-- of the glyph
w = node.new("whatsit","pdf_literal")
-- draw a dashed line if depth not zero
if dp ~= 0 then
w.data = string.format("q 0.2 G 0.1 w 0 %g %g %g re S f [0.2] 0 d 0 0 m %g 0 l S Q",-dp,-wd,dp + ht,-wd)
else
w.data = string.format("q 0.2 G 0.1 w 0 %g %g %g re S f Q",-dp,-wd,dp + ht,-wd)
end
-- insert this new node after the current glyph and move pointer (head) to
-- the new whatsit node
end
end
return true
end
\end{luacode*}
A \emph{wonderful} serenity has taken {\large possession} of my entire soul, like these
\textsl{sweet}
\textbf{mornings} of spring which I enjoy with my whole heart. I am alone, and feel the
charm of existence in this spot, \textbf{which} was created for the bliss of souls like
mine. I am so happy, my dear friend, so absorbed in the exquisite sense of
mere tranquil existence, that I neglect my talents. I should be incapable of
drawing a single stroke at the present moment; and yet I feel that I never was
a greater artist than now.
\end{document}
which yields:
(detail)
Bonus: it draws the base line if the depth of the glyph is not 0.
Here is a solution that replaces the glyphs by black rectangles (rules):
\documentclass{article}
\usepackage{luacode,luatexbase,microtype}
\begin{document}
\begin{luacode*}
-- head is a linked list (next/prev entries pointing to the next node)
-- parent it the surrounding h/vbox
-- a hbox/vbox
r = node.new("rule")
-- replace the glyph by
-- the rule by changing the
-- pointers of the next/prev
-- entries of the rule node
-- first glyph in a list
parent.list = r
else
end
-- now the glyph points to
-- nowhere and we should remove
-- it from the memory
end
end
return true
end
\end{luacode*}
\hsize6cm
A wonderful serenity has taken possession of my entire soul, like these sweet
mornings of spring which I enjoy with my whole heart. I am alone, and feel the
charm of existence in this spot, which was created for the bliss of souls like
mine. I am so happy, my dear friend, so absorbed in the exquisite sense of
mere tranquil existence, that I neglect my talents. I should be incapable of
drawing a single stroke at the present moment; and yet I feel that I never was
a greater artist than now.
\end{document}
-
Really nice solution! – Gonzalo Medina Jun 2 '12 at 20:56
Glad to see you back! Thanks for the great solution - this works great! How would you modify the code to draw a filled rectangle instead of a hollow one? – Hooked Jun 2 '12 at 21:10
I've just fixed an embarrassing rounding error in my code – topskip Jun 2 '12 at 21:17
@Hooked Just change the re S f in the w.data lines above to re b. Please note that I have updated my code in the mean time. – topskip Jun 2 '12 at 21:25
@Sigur well, LuaTeX makes it easy to draw these lines. PDFTeX and other engines are completely lacking this kind of support. So there is no way I could implement this behaviour for other engines. – topskip Sep 29 '14 at 9:12
Here is a solution that respects kerning (but not ligatures):
\documentclass{article}
\usepackage{xcolor}
\makeatletter
\def\showboxes#1{%
\begingroup\fboxrule=.1pt \fboxsep=-\fboxrule
\@showboxes#1\@showboxes\@empty
\endgroup}
\def\@showboxes#1#2{%
\ifx#2\@showboxes
\fbox{\color{gray}#1}\expandafter\@gobble
\else
\setbox0=\hbox{#1\kern0pt#2}\setbox2=\hbox{#1#2}%
\dimen0=\wd0 \advance\dimen0 -\wd2 % \dimen0 contains the kern between the two chars
\fbox{\color{gray}#1}\kern-\dimen0
\expandafter\@showboxes
\fi#2}
\begin{document}
\showboxes{AVov}
\end{document}
For doing more words just add
\usepackage{xparse}
\def\showboxesaux#1{\showboxes{#1} }
\NewDocumentCommand\Showboxes{>{\SplitList{ }}m}
{\ProcessList{#1}\showboxesaux\unskip}
and use
\Showboxes{The quick brown fox jumped over the lazy dog.}
but hyphenation won't be taken care of.
-
Great solution! Can this be modified to accept multiple words? At the moment the spaces are swallowed. (See question update). – Hooked May 30 '12 at 19:43
@Hooked It can, but I don't think it can be a substitute for the "real thing", that is, doing this with LuaTeX. – egreg May 30 '12 at 19:52
Excuse my Lua-ignorance, are you saying that a LuaTeX solution would be able to handle the bounding box problem in a more robust way? – Hooked May 30 '12 at 19:55
@Hooked I'm sure it is (but I can't give a solution). You were unlucky that our main LuaTeX expert, Patrick Gundlach, is on his well deserved holidays. – egreg May 30 '12 at 19:58
This solution doesn't work with xetex as is. Wouldn't it be better to replace \setbox0=\hbox{{#1}{#2}} by \setbox0=\hbox{#1\kern0pt#2}, because xetex doesn't parse the grouping as expected for unknown reason? – Henri Menke Jul 1 '14 at 15:44
Probably the best is the code from the TeXbook and this I will leave up to you. Another alternative is to study the code from the soul package and use the scanners provided.
\documentclass{article}
\usepackage{soul,graphicx,xcolor}
\fboxrule=0.1pt
\fboxsep=-\fboxrule
\begin{document}
\makeatletter
\def\SOUL@soeverytoken{%
\fbox{\color{gray}\the\SOUL@token}}
\makeatother
\scalebox{7}{\color{purple}\so{ailbcdefgh}}
\end{document}
Change the gray color to white to have the letters disappear.
You will need to change the \fbox to basic TeX primitives, if you need to compensate for the 0.1pt rule or use the suggestion of egreg in the comments, which I have incorporated.
-
\fboxsep=-\fboxrule would solve the issue. But you lose kerning information in this way. – egreg May 30 '12 at 10:09
@egreg Thanks I added your code. Best is to go to a Knuth-box (I had it as a macro somewhere ...) – Yiannis Lazarides May 30 '12 at 10:16
@YiannisLazarides Thank you for the solution, I'll have to play around with soul a bit more. I've updated my question based off your solution, it seems that in addition your interword spacing is too large as well. – Hooked May 30 '12 at 19:42
I remember I saw something like this in fontchart.tex – morbusg Jul 24 '12 at 17:21
I like this solution, but it doesn't work well for math mode. \so{$(f+g)'(x) = f'(x)+g'(x)$} just puts everything in one box. – Joe Corneli Dec 9 '12 at 21:49 | 2,362 | 8,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-18 | latest | en | 0.902899 |
https://study.com/academy/topic/overivew-of-inequalities-homework-help.html | 1,590,722,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401260.16/warc/CC-MAIN-20200529023731-20200529053731-00497.warc.gz | 553,903,023 | 17,740 | # Ch 3: Overview of Inequalities: Homework Help
The Overview of Inequalities chapter of this College Algebra Homework Help course helps students complete their inequalities homework and earn better grades. This homework help resource uses simple and fun videos that are about five minutes long.
## How it works:
• Identify which concepts are covered on your inequalities homework.
• Find videos on those topics within this chapter.
• Watch fun videos, pausing and reviewing as needed.
• Complete sample problems and get instant feedback.
• Finish your inequalities homework with ease!
## Topics from your homework you'll be able to complete:
• Graphing 1- and 2-variable inequalities
• Using set and interval notation
• Working with compound inequalities
• Graphing systems of inequalities
• Solving and graphing absolute value inequalities
6 Lessons in Chapter 3: Overview of Inequalities: Homework Help
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
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Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
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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. | 309 | 1,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-24 | latest | en | 0.92229 |
http://interpoint-power.com/w352J90Ol/mG35313aH/ | 1,571,690,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00536.warc.gz | 101,530,131 | 13,197 | Coordinate Plane Calculator Computer Worksheet Intermediate Algebra And Trigonometry Is Over Of Math Formula Best Graphing For Factoring Word Problems Evaluate Equation About Polynomials
Published at Wednesday, 18 September 2019. algebra. By .
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Algebra is a challenge which is worth facing, Let’s face it – algebra can be hard and there will be a point for everyone when they find using algebra difficult. However algebra can also give a great sense of achievement and for those who become good at it in school, it can give a real feeling of satisfaction every time a problem is solved. In fact algebra can easily become the favorite area of mathematics for some pupils! Even it is a real challenge to you at school, try and talk to someone who struggled to get a grade C but finally managed it, or someone who has gone back to study maths later in life. Overcoming a difficult hurdle in life can feel really worthwhile and says a lot about you as a person.
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67 of 100 by 596 users | 424 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.932143 |
https://piziadas.com/en/2015/02/geometria-proyectiva-aplicacion-de-los-haces-superpuestos-de-segundo-orden.html | 1,685,742,893,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00618.warc.gz | 502,904,629 | 22,527 | # Geometría proyectiva: Application of second-order overlapping beams
Projective concepts that we have developed in studying the overlapping beams of second order, whose base is a conical, They allow to solve problems of determination of points of contact in the tangent of a Conic defined by five tangent or five restrictions through the combination of tangent and with their respective tangent points.
To solve this type of problems we will remember that given two beams of second order, When sectioning them from two homologous elements are obtained perspectives series that projected from the projective center beams (Brianchon point). In the following figure, the homologous points a-a.’ They determine the double point of the perspectives series, While the AB'-to 'B-AC' to ' C projected the straight lines 1 and 2 that contain your prospective Center respectively (“V..” It is the projective Center of bundles of second order mentioned above)
#### General model for Brianchon point
Homologous rays that serve as bases for these perspectives series can be any of the three pairs that define the is between the beams of the second order. We can see that if we cut from all of them we get three straight (1,2 and 3) containing it to the Brianchon point, from which the double lines be drawn (tangent if any) beams (It will be imaginary if this point is the conical interior).
#### Center of Brianchon point of tangency
The exposed projective model allows to relate the tangent of the cone with their points of tangency, thinking that a point of tangency is the intersection of two tangent infinitely close.
For example, If we move the tangent line “c” from the figure above to match the straight line “b'” keeping the geometric constraints of this figure, We will have to b-c’ It has become a point of tangency which will belong to the straight “3” passing through the Centre of the projective “V..”.
Brianchon point with a point of tangency
#### Brianchon point with two points of tangency
Matching a second pair of tangents as b-c’ (It could also be a-c’ or ’-c) We will obtain a variant of the previous model, but in this case with two points of tangency.
Brianchon point with two points of tangency
#### Brianchon point with three points of contact
If we do agree to two tangent three, for example a-c ', b-a’ and c-b ’, We will have three points of tangents in this variant of the general model. You can use other combinations of the tangents, but will have to wear each pair one of each of the beams and in any case two counterparts (as a-a. ’, b-b’ or c-c ’).
Brianchon point with three points of contact
### Statement of issues
These figures allow us to pose problems of determination of points of contact in the tangents that determine the Conic as shown in an example, the reader leaving the resolution of the remaining.
The problems that can arise, understanding the Conic as envelope of the tangent, its:
1. Given a Conic tangent five, determine the point of tangency on one of them.
2. Given a tangent with its point of contact and three additional tangent of a conical, determine the point of contact in another of the tangents.
3. Given two tangent with their respective points of contact and an additional tangent, determine the contact point is tangent.
### Application to problem solving
We will resolve the first of the problems raised as an example:
Given the straight lines p, q, r, s and t tangent to a conical, determine the point “T” contact the straight “t“.
#### 1.-Determination of the figure of analysis of application
We will use as a figure of analysis to solve the problem that we have tagged as “Brianchon point with a point of tangency”, as in this variant of the “General model” We have a point of contact in one of the tangents.
#### 2.- Allocation of the corresponding labels
We shall first proceed to identify the straight lines of the wording of the problem with the tangent to the Conic Figure analysis, taking into account that, in this case, We assign each second-order beam straight to the straight “t” in which we want to find the point of contact.
#### 3.- Determination of the is
Once determined the elements of the beams, We obtain the projective Center them (Brianchon point).
#### 4.- Resolving the problem
Finally determine the tangency point knowing that this, point B'C, It will be screened from the projective Center with its homologous point BC’
Similarly, we solve the two remaining cases.
Can you solve them? | 998 | 4,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-23 | longest | en | 0.92675 |
https://www.simplilearn.com/tutorials/c-sharp-tutorial/what-is-graphs-in-c-sharp | 1,726,033,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00329.warc.gz | 938,480,618 | 85,376 | Graphs are are an integral part of communication networks, maps, data models and much more. Graphs are used to represent information with appealing visuals. For example, organization hierarchy is represented using graphs. Graph transformation systems use rules to manipulate graphs in memory and much more.
This article will help us unravel the detailed technicalities of graphs in C#, their terminologies, representations, and the operations performed in graphs, such as graph traversal along with important graph applications.
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## What Is Graph Data Structure in C#?
A graph is a non-linear data structure trumped-up of nodes and edges. Edges are lines or arcs that link any two nodes in a graph, and nodes are also called vertices. A graph can be more explicitly defined as,
A graph comprises a finite number of vertices (or nodes) and a set of Edges that connect them.
The set of vertices V = {A,B,C,D,E} and the set of edges E ={AB, AC, BD, BE} are shown in the graph above.
In this post, we'll look at some different forms of graphs after learning what a graph data structure is.
## Types of Graphs in C#
There are various sorts of graphs based on the position of these nodes and vertices, such as:
• ### Undirected Graph
A directed graph is defined as an ordered pair of vertices. The Graph's edges represent a precise path from one vertex to the next. The direction is from V1 to V2 when an edge is represented as (V1, V2). The starting node, or start vertex, is the first element, V1. The terminal node, also known as the end vertex, is the second part V2.
• ### Directed Graph
A directed graph (digraph) is a graph with edges that have different orientations, i.e., an edge (x, y) is not the same as an edge (x, y) (y, x).
• ### Cyclic Graph
An acyclic Graph is defined as one that has at least one cycle.
• ### Acyclic Graph
An acyclic graph does not have any cycles in it.
• ### Directed Acyclic Graph(DAG)
A DAG (Directed Acyclic Graph) is a directed graph with no cycles.
• ### Multi Graph
A multigraph is an undirected graph that allows numerous edges (and occasionally loops). Two or more edges that connect the same two vertices are multiple edges. A loop is a directed or undirected edge that connects a vertex to itself, and it may or may not be permitted.
• ### Simple Graph
In contrast to a multigraph, a simple graph is an undirected graph in which multiple edges and loops are forbidden. Every vertex in a simple graph with n vertices has a degree of n-1.
• ### Weighted Graph
Every edge in a weighted graph has a value (weight) associated with it. Instead of weight, we can use the phrases cost or length.
• ### Unweighted Graph
Every edge in an unweighted graph does not have a value (weight) associated with it. To put it another way, an unweighted graph is a weighted graph with all edge weights equal to one. By default, all graphs are presumed to be unweighted unless otherwise stated.
• ### Complete Graph
Every two vertices in a complete graph are adjacent, and all edges that could exist are present.
• ### Connected Graph
Every pair of vertices has a path linking them in a connected graph. To put it another way, there are no inaccessible vertices. A graph that is not connected is referred to as a disconnected graph.
• ### Disconnected
If at least two of the Graph's vertices are not connected by a path, the Graph is disconnected. When a graph G is disconnected, every maximally connected subgraph of G is referred to as a connected component of G.
Following our graphs discussion in C#, we'll look at certain graph terminologies.
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## Graph Terminologies
The terms we utilized in the graph data structure are listed below.
• Edges are two basic pieces from which graphs are built (together with vertices). Each edge has two ends, called vertices, to which it is connected.
• If two vertices are endpoints of the same edge, they are neighboring.
• A vertex's outgoing edges are directed edges in which the vertex is the origin.
• A vertex's incoming edges are directed edges with the vertex as the destination.
• The number of edges occurring to a vertex in a graph is its degree.
• The total number of outgoing edges is the out-degree of a vertex in a directed graph, while the total number of receiving edges is the in-degree.
• A source vertex is an in-degree zero, while a sink vertex is a vertex without-degree zero.
• A vertex of degree zero that is not an edge's endpoint is called an isolated vertex.
• Each edge has two ends, called vertices, to which it is connected.
• A cycle is defined as a journey that begins and finishes simultaneously.
• A path with distinct vertices is known as a simple path.
• For any pair of vertices u, v, a graph is Strongly Connected if it accommodates a directed path from u to v and a directed path from v to u.
• When undirected edges replace the directed edges in a directed graph, the Graph becomes connected (undirected). A weakly linked graph's vertices have at least one out-degree or in-degree.
• The largest connected subgraph of an unconnected graph is called a connected component.
• A bridge is an edge that, if removed, would cause the Graph to be disconnected.
• Forest is a graph that does not have any cycles.
• A tree is a linked graph that does not have any cycles. When all the cycles in a DAG (Directed Acyclic Graph) are removed, it becomes a tree, and when an edge in a tree is removed, it becomes a forest.
• The spanning tree of an undirected graph is a subgraph that contains all of the Graph's vertices.
After learning about graph terminologies, we'll look at different graph representations in further depth.
## Representation Of Graph in C#
In graph theory, a graph representation stores a graph in a computer's memory.
The collection of vertices and the neighbors of each vertex are required to represent a graph (vertices that are directly connected to it by an edge). The weight will be assigned to each edge if the Graph is weighted.
Depending on the density of the Graph's edges, the type of operations to be performed, and the simplicity of usage, there are several ways to describe it optimally.
A square matrix represents a finite graph as an adjacency matrix. The matrix's elements show whether two vertices in the Graph are adjacent or not. The adjacency matrix for a basic unweighted graph with vertex set V is a square |V| |V| matrix A with the element:
When an edge is formed between vertex i to vertex j, Aij = 1, and there isn't Aij = 0.
Each column in the matrix represents destination vertices, while each row represents source vertices. Because edges from a vertex to itself, i.e., loops, are not allowed in simple graphs, the diagonal members of the matrix are all 0. The adjacency matrix will be symmetric if the Graph is undirected. Aij can also indicate edge weights in a weighted graph.
An adjacency matrix uses n*n space since it stores a value (1/0/edge-weight) for every pair of vertices, whether or not the edge exists. They can only be used effectively when the Graph is dense.
Each vertex in the Graph is associated with a collection of its neighboring vertices or edges in an adjacency list form, i.e., each vertex keeps a list of adjacent vertices. There are many different ways to express an adjacency list depending on the implementation. This data format allows additional data to be stored on the vertices. However, it is only viable when the Graph has a few edges. The Graph is sparse, in other words.
Following that, we'll see various graph operations in C#.
## Operations on Graphs in C#
The succeeding is the operations you can perform on graphs in data structures:
• Creating graphs
• Insert vertex
• Delete vertex
• Insert edge
• Delete edge
You'll go over each operation one by one in detail:
### Creating Graphs
A graph can be created using one of two methods:
The connection matrix, or adjacency matrix, of a basic labeled graph, is a matrix with rows and columns labeled by graph vertices and a 1 or 0 in position depending on whether they are adjacent or not.
An adjacency list, made up of unordered lists, represents a finite graph. Each unordered list within an adjacency list describes the set of neighbors of a specific vertex.
### Insert Vertex
When you add or insert a vertex to a graph after adding one or more vertices or nodes, the Graph increases by one, increasing the row and column sizes of the matrix.
### Delete Vertex
The term "delete a vertex" refers to the act of removing a specific node or vertex from a previously saved graph. The matrix returns a removed node if it appears in the Graph. The matrix returns the node not available if a deleted node does not appear in the Graph.
### Insert Edge
An edge can be added to a graph by connecting two supplied vertices.
### Delete Edge
An edge can be deleted by removing the connection between the vertices or nodes.
After that, we'll look at two distinct sorts of graph traversal.
## Graph Traversal in C#
A search approach for identifying a vertex in a graph is graph traversal. Graph traversal is also used in the search process to decide the order in which vertices are visited. A graph traversal finds the edges used in the search process without creating loops. That is, we can visit all of the Graph's vertices without walking through a looping path via graph traversal.
There are two techniques for traversing graphs, which are as follows.
• DFS (Depth First Search)
### DFS (Depth First Search)
A spanning tree is the result of a DFS traversal of a graph. A graph with no loops is known as a spanning tree. We use a Stack data structure with a maximum size equal to the number of vertices in the Graph to implement DFS traversal.
To implement DFS traversal, we take the following stages.
Step 1: Create a Stack with the number of vertices in the Graph as the size.
Step 2: Select any vertex as the traversal's beginning point. Please pay a visit to that vertex and add it to the Stack.
Step 3: Push any of the non-visited adjacent vertices of a vertex at the top of the Stack to the top of the Stack.
Step 4: Repeat steps 3 and 4 up to there are no additional vertices to visit from the vertex at the top of the Stack.
Step 5: If there are no new vertices to visit, go back and pop one from the Stack using backtracking.
Step 6: Continue using steps 3, 4, and 5 until the Stack is empty.
Step 7: When the Stack is empty, create the final spanning tree by deleting the Graph's unused edges.
#### Example
Let's see an example of how the Depth First Search algorithm works. An undirected graph with five vertices is used.
Starting with vertex 1, the DFS method places it in the Visited list and adds its neighboring vertices to the Stack.
Next, we visit the top of the Stack, element 2, and nearby nodes. We go to 3 instead of 1 because one has already been visited.
We add vertex 3 to the top of the Stack and visit it because it has an unvisited adjacent vertex in 5.
We have finished the Depth First Search Traversal of the Graph after visiting the last element 4, which has no unvisited nearby nodes.
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The outcome of a BFS traversal of a graph is a spanning tree. A spanning tree is a graph that is devoid of loops. To implement BFS traversal, we utilize a Queue data structure with a maximum size equal to the number of vertices in the Graph.
To build BFS traversal, we apply the steps below.
Step 1: Create a Queue with a size equal to the number of vertices in the Graph.
Step 2: Choose any vertex as the traversal's beginning point. Make a trip to that vertex and add it to the Queue.
Step 3: Insert all of the non-visited adjacent vertices of the vertex at the front of the Queue into the Queue.
Step 4: Delete the vertex in front of the Queue if no new vertex has to be visited.
Step 5: Continue using steps 3 and 4 until the Queue is empty.
Step 6: When the Queue is empty, create the final spanning tree by eliminating the Graph's unused edges.
Let's look at an illustration of how the Breadth-First Search algorithm works. An undirected graph with five vertices is used.
Starting with vertex 1, the BFS algorithm places it in the Visited list and adds its neighboring vertices to the Stack.
Following that, we go to the element at the front of the Queue, i.e., 2, and its nearby nodes. We go to 3 instead of 1 because one has already been visited.
We move vertex 3 to the back of the Queue and visit 4, which is at the front because it has an unvisited adjacent vertex in 5.
Since the sole adjacent node of 4, namely 1, has already been accessed, only five remain in the Queue. We go there.
#### Output
Last but not least, we'll look at various graph applications in the graphs in the c# article.
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## Applications of Graphs in C#
The applications of graphs in real life are listed below.
The Graph API on Facebook is likely the best example of graphs used to solve real-world problems. The Graph API is a game-changer for large-scale data delivery.
Everything is a vertice or node in the Graph API. Users, Pages, Places, Groups, Comments, Photos, Photo Albums, Stories, Videos, Notes, Events, and so on are examples of this. A vertice is anything that has data-storing properties.
And every link or interaction is a vantage point. This may be something like a user uploading a photo, video, or comment, or a user updating their profile with their birthplace, relationship status, or a friend's photo, among other things.
A knowledge graph is defined as a graph-based representation of knowledge that connects data and graphs. It is still unclear what it can and cannot achieve.
### 3. Yelp's Local Graph
The Local Graph API claims to make it easy for developers to incorporate Yelp's data into their products and share amazing local companies.
By describing the business problem as a graph within its schema, GraphQL takes advantage of the capabilities of graph data structures.
GraphQL is most commonly used to manipulate graph data structures. Both the Apollo Client and Relay use GraphQL data in a normalized graph.
### 4. Path Optimizing Algorithm
Path optimizations are generally concerned with determining the optimum link that meets a set of predetermined criteria, such as speed, safety, and fuel consumption, or a combination of criteria, such as procedures and routes.
The Shortest Path of a graph in an unweighted graph is the path with the fewest edges. Breadth-First Search (BFS) is used to locate the shortest paths in graphs—with breadth graph traversals; we always reach a node from another node with the fewest number of edges.
Using either BFS or Depth First Search, any Spanning Tree is a Minimum Spanning Tree unweighted graph.
Shortest Path APIs like Google Maps and Routes are classics. Using edge-weighted directed graphs (digraphs), this is an easy graph problem to answer using edge-weighted directed graphs (digraphs).
The goal of a Map API is to discover the shortest way from your current location to any other destination on the map, as in a single source shortest path variant.
### 6. Flight Networks
In the case of aircraft networks, efficient route optimizations suit graph data structures perfectly. Airport procedures can be efficiently modeled and optimized using graph models.
Algorithm engineering is used to determine the optimal connections in aviation networks.
Graph data structures are employed in-flight networks to compute shortest pathways and fuel usage in route planning, commonly done in a multi-modal setting.
Now, in the graphs in c#, we'll summarize all of the subjects we've studied thus far in this post.
## Next Steps
We grasped everything there is to know about graphs in this c# tutorial, as well as certain sorts of graphs and graph terminologies. Later on, we observed two types of graph representation: adjacency matrix and list and a variety of graph operations. Then there are two types of graph traversal: depth-first and breadth-first, and finally, there are some graph applications in the real world.
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https://classroom.synonym.com/easy-projects-use-scientific-methods-8422788.html | 1,531,773,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00323.warc.gz | 626,440,464 | 21,118 | Assigning easy science projects is a way to introduce the scientific method to your kids or students. There are six main steps in a scientific method, most of which are included in every experiment your students will complete, regardless of the project's simplicity. The steps include asking a question, researching the issue, forming a hypothesis, testing the hypothesis with an experiment, analyzing data to arrive at a conclusion and stating results. Lead your students through potentially dangerous elements of an experiment, and ensure that both you and they wear the appropriate safety clothing at all times.
Plant Population Density
Challenge your students to write a prediction -- known as a hypothesis -- on what they think will happen during this plant-population-density experiment. Separate the class into two halves based on students' hypotheses; instruct students to work as a group to develop a two-minute speech based on why they have made their prediction. Lead the classroom through the experiment as you set up two identical parts of an experiment design next to one another. Take two identical glass jars and fill them with the same amount of damp soil. Mark one glass "few" and plant two runner bean seedlings in it and mark the other glass "many" and plant 20 runner bean seedlings in it. Leave the jars on the same windowsill, so they receive the same environmental conditions, and water them every other day. Have students observe their growth and compare how the overcrowded and roomy growing conditions affect plant growth.
Acids vs. Bases
To conduct this experiment -- with the common scientific method of using controls and variables -- students should empty a medicine dropper of water, white vinegar (the acid) and ammonia (the base) into three separate, identical plastic cups. Students should then douse three single sheets of paper towel in one of the liquids before wrapping each around a mild steel nail. Have students leave their nails wrapped in paper towels on a side where they will be undisturbed overnight. Students should return and note their observations as well as take photographs that can be used in a presentation or at a science fair. Have students check on their nails once a day for a school week. Students should draw comparisons between the three different conditions and consider how the conditions affected the rate of the nail's rusting.
Magnets and Charge
This simple physics project gets students to think about magnetism and electric charge. Have your students take an iron nail that is roughly 6 inches long and approximately 20 inches of copper wire. Students should strip an inch of insulation from each end of the copper wire and wrap the middle section of the wire around the nail into a coil; ensure students leave at least 2 inches of copper wire at each end of the coil. Students should take both parts of the wire that are not coiled on the nail and attach them to the positive and negative terminals of a 9-volt battery. Get students to experiment with their electromagnet by passing it near iron filings or metal paper clips. Have students unravel half of the coils on the nail, pass the electromagnet near these items once more and see if they notice any difference in the magnet's strength.
Hawaiian Islands Geology
Easy science projects do not have to be based on students carrying out their own experiments; instead, they can focus on research from library books, journals and articles as well as Internet resources. One idea for a science project that centers on the background research phase of the scientific method gets your students to research the formation of the Hawaiian Islands. Students can look at historic ideas behind how the Hawaiian Islands were formed before thinking specifically about Hawaii's different volcanoes and mountain ranges. Educate your students about the appropriate ways of using resources and referencing the work they included during their research; this will prove valuable, particularly for students looking to continue into college. Suggest a specific style of referencing, such as the Harvard system, where your students must write the author name, then open brackets, and enclose the publication's year before closing brackets. | 800 | 4,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-30 | latest | en | 0.954787 |
https://www.wiseowl.co.uk/c-sharp/exercises/standard/modular-code/2805/ | 1,653,303,694,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00360.warc.gz | 1,255,410,441 | 9,991 | # Visual C# | Modular code exercise | Create a function to calculate your BMI
This exercise is provided to allow potential course delegates to choose the correct Wise Owl Microsoft training course, and may not be reproduced in whole or in part in any format without the prior written consent of Wise Owl.
Software ==> Visual C# (79 exercises) Version ==> Any version of C# Topic ==> Modular code (3 exercises) Level ==> Average difficulty Subject ==> C# training
Before you can do this exercise, you'll need to download and unzip this file (if you have any problems doing this, click here for help).
You need a minimum screen resolution of about 700 pixels width to see our exercises. This is because they contain diagrams and tables which would not be viewable easily on a mobile phone or small laptop. Please use a larger tablet, notebook or desktop computer, or change your screen resolution settings.
To avoid having to draw the form needed for this exercise, right-click on the name of your project in Solution Explorer then choose Add --> Existing Item... (you can also press SHIFT + ALT + A to do the same thing).
Choose only the file called frmBMI.cs in the above folder to import it into your project, then edit Program.cs to make this the default form. When you run your application, you should now see this:
You can type in your stats in imperial units, but the button doesn't work yet!
Uncomment the call to show the message box in the code-behind file. Write the BMI function that this calls, using the constants included in the code. Some notes to help you:
• A person's body mass index is their weight in kilograms divided by their height in metres squared.
• There are 12 inches in a foot, and 14 pounds in a stone.
• In an ideal world your BMI should be between 20 and 25, although there are lots of good reasons why this might vary.
Test out your system, then go and have a biscuit!
You can unzip this file to see the answers to this exercise, although please remember this is for your personal use only. | 440 | 2,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.898714 |
http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects%5B%5D=Physical+sciences%3AElectricity+and+magnetism&topicsSubjects%5B%5D=Mathematics%3AMeasurement&instructionalStrategies=Nonlinguistic+representations | 1,548,246,079,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00353.warc.gz | 172,867,706 | 15,653 | NASA Wavelength is transitioning to a new location on Jan. 30, 2019, read notice »
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# Solar Week Monday: Do the Activity - Measuring Solar Activity
This is an online lesson associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. Outside of Solar Week, information, activities, and... (View More)
# Solar Week Wednesday: Do the Activity: Activity 1: Measuring the Motion of a Coronal Mass Ejection
This is an activity associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. Outside of Solar Week, information, activities, and resources... (View More)
# Soda Bottle Magnetometer
This is an activity about Earth's magnetic field. Learners will construct a soda bottle magnetometer, collect data, and analyze the results to detect magnetic storm events. Ideally, learners should collect data for at least a month. If several... (View More)
Audience: Middle school
# Design Challenge - Stacking the Satellites
This is an activity about area and volume. Learners will use fabrication software to determine the optimal size of a satellite which can fit within a given rocket cylinder. To complete this activity, fabrication software is required (an example is... (View More)
Audience: Middle school
# Model of the MMS Satellite
In this lesson, learners will construct a 3D scale model of one of the MMS satellites. After, they will calculate the octagonal area of the top and bottom of the satellites, given the measurements of the satellite. Then, learners will compare the... (View More)
# Design Challenge - Deploying the Satellites' Antennae
This is an activity about using models to solve a problem. Learners will use a previously constructed model of the MMS satellite to determine if the centrifugal force of the rotating MMS model is sufficient to push the satellite's antennae outward,... (View More)
Audience: Middle school
# Launch of the Satellites
In this lesson, learners will research facts about Atlas V rockets, which launched the MMS satellites. After, they will compute the speed of the launch rocket, given a data chart of time vs. distance from lift-off. Then, they will write a report... (View More)
# MRC: Overview of the Solar System (Grades 3-5)
This is a lesson about the size and scale of planets in the solar system. Learners will kinesthetically model the order of the planets outward from the sun. Then they will use a string and beads to create a model to represent the relative distances... (View More)
# MRC: Overview of the Solar System (Grades 6-8)
Learners will review the structure, content and size of the Solar System. This lesson is designed using the 5E instructional model and includes: teacher training, unit pacing guides, essential questions, a black-line master science notebook, a... (View More)
Audience: Middle school
Materials Cost: \$1 - \$5 per group of students
# Dusty Dilemma
Learners will be introduced to the concepts of error analysis, including standard deviation. They will apply the knowledge of averages (means), standard deviation from the mean, and error analysis to their own classroom distribution of heights. They... (View More)
«Previous Page1234 Next Page» | 812 | 3,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-04 | latest | en | 0.905378 |
https://hextobinary.com/unit/acceleration/from/mis2/to/inmin2 | 1,621,266,863,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00113.warc.gz | 332,914,910 | 14,245 | Mile/Second Squared
Inch/Minute Squared
###### How many Inch/Minute Squared are in a Mile/Second Squared?
The answer is one Mile/Second Squared is equal to 228096000 Inch/Minute Squared. Feel free to use our online unit conversion calculator to convert the unit from Mile/Second Squared to Inch/Minute Squared. Just simply enter value 1 in Agate Line and see the result in Inch/Minute Squared.
###### How to Convert Mile/Second Squared to Inch/Minute Squared (mi/s2 to in/min2)
By using our Mile/Second Squared to Inch/Minute Squared conversion tool, you know that one Mile/Second Squared is equivalent to 228096000 Inch/Minute Squared. Hence, to convert Mile/Second Squared to Inch/Minute Squared, we just need to multiply the number by 228096000. We are going to use very simple Mile/Second Squared to Inch/Minute Squared conversion formula for that. Pleas see the calculation example given below.
Convert 1 Mile/Second Squared to Inch/Minute Squared1 Mile/Second Squared = 1 × 228096000 = 228096000 Inch/Minute Squared
###### What is Mile/Second Squared Unit of Measure?
Mile/Second Squared or Mile per Second Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 mile/second squared, that means its speed is increased by 1 mile per second every second.
###### What is the symbol of Mile/Second Squared?
The symbol of Mile/Second Squared is mi/s2. This means you can also write one Mile/Second Squared as 1 mi/s2.
###### What is Inch/Minute Squared Unit of Measure?
Inch/Minute Squared or Inch per Minute Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 inch/minute squared, that means its speed is increased by 1 inch per minute every minute.
###### What is the symbol of Inch/Minute Squared?
The symbol of Inch/Minute Squared is in/min2. This means you can also write one Inch/Minute Squared as 1 in/min2.
###### Mile/Second Squared to Inch/Minute Squared Conversion Table
Mile/Second Squared [mi/s2]Inch/Minute Squared [in/min2]
1228096000
2456192000
3684288000
4912384000
51140480000
61368576000
71596672000
81824768000
92052864000
102280960000
10022809600000
1000228096000000
###### Mile/Second Squared to Other Units Conversion Chart
Mile/Second Squared [mi/s2]Output
1 Mile/Second Squared in Meter/Second Squared is Equal to1609.34
1 Mile/Second Squared in Attometer/Second Squared is Equal to1.609344e+21
1 Mile/Second Squared in Centimeter/Second Squared is Equal to160934.4
1 Mile/Second Squared in Decimeter/Second Squared is Equal to16093.44
1 Mile/Second Squared in Dekameter/Second Squared is Equal to160.93
1 Mile/Second Squared in Femtometer/Second Squared is Equal to1609344000000000000
1 Mile/Second Squared in Hectometer/Second Squared is Equal to16.09
1 Mile/Second Squared in Kilometer/Second Squared is Equal to1.61
1 Mile/Second Squared in Micrometer/Second Squared is Equal to1609344000
1 Mile/Second Squared in Millimeter/Second Squared is Equal to1609344
1 Mile/Second Squared in Nanometer/Second Squared is Equal to1609344000000
1 Mile/Second Squared in Picometer/Second Squared is Equal to1609344000000000
1 Mile/Second Squared in Meter/Hour Squared is Equal to20857098240
1 Mile/Second Squared in Millimeter/Hour Squared is Equal to20857098240000
1 Mile/Second Squared in Centimeter/Hour Squared is Equal to2085709824000
1 Mile/Second Squared in Kilometer/Hour Squared is Equal to20857098.24
1 Mile/Second Squared in Meter/Minute Squared is Equal to5793638.4
1 Mile/Second Squared in Millimeter/Minute Squared is Equal to5793638400
1 Mile/Second Squared in Centimeter/Minute Squared is Equal to579363840
1 Mile/Second Squared in Kilometer/Minute Squared is Equal to5793.64
1 Mile/Second Squared in Kilometer/Hour/Second is Equal to5793.64
1 Mile/Second Squared in Inch/Hour/Minute is Equal to13685760000
1 Mile/Second Squared in Inch/Hour/Second is Equal to228096000
1 Mile/Second Squared in Inch/Minute/Second is Equal to3801600
1 Mile/Second Squared in Inch/Hour Squared is Equal to821145600000
1 Mile/Second Squared in Inch/Minute Squared is Equal to228096000
1 Mile/Second Squared in Inch/Second Squared is Equal to63360
1 Mile/Second Squared in Feet/Hour/Minute is Equal to1140480000
1 Mile/Second Squared in Feet/Hour/Second is Equal to19008000
1 Mile/Second Squared in Feet/Minute/Second is Equal to316800
1 Mile/Second Squared in Feet/Hour Squared is Equal to68428800000
1 Mile/Second Squared in Feet/Minute Squared is Equal to19008000
1 Mile/Second Squared in Feet/Second Squared is Equal to5280
1 Mile/Second Squared in Knot/Hour is Equal to11261932.14
1 Mile/Second Squared in Knot/Minute is Equal to187698.87
1 Mile/Second Squared in Knot/Second is Equal to3128.31
1 Mile/Second Squared in Knot/Millisecond is Equal to3.13
1 Mile/Second Squared in Mile/Hour/Minute is Equal to216000
1 Mile/Second Squared in Mile/Hour/Second is Equal to3600
1 Mile/Second Squared in Mile/Hour Squared is Equal to12960000
1 Mile/Second Squared in Mile/Minute Squared is Equal to3600
1 Mile/Second Squared in Yard/Second Squared is Equal to1760
1 Mile/Second Squared in Gal is Equal to160934.4
1 Mile/Second Squared in Galileo is Equal to160934.4
1 Mile/Second Squared in Centigal is Equal to16093440
1 Mile/Second Squared in Decigal is Equal to1609344
1 Mile/Second Squared in G-unit is Equal to164.11
1 Mile/Second Squared in Gn is Equal to164.11
1 Mile/Second Squared in Gravity is Equal to164.11
1 Mile/Second Squared in Milligal is Equal to160934400
1 Mile/Second Squared in Kilogal is Equal to160.93 | 1,590 | 5,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-21 | latest | en | 0.911174 |
https://visualfractions.com/unit-converter/convert-500-cm-to-nmi/ | 1,652,996,435,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00515.warc.gz | 661,454,578 | 6,324 | # Convert 500 cm to nmi
So you want to convert 500 centimeters into nautical miles? If you're in a rush and just need the answer, the calculator below is all you need. The answer is 0.0026997840172786 nautical miles.
## How to convert centimeters to nautical miles
We all use different units of measurement every day. Whether you're in a foreign country and need to convert the local imperial units to metric, or you're baking a cake and need to convert to a unit you are more familiar with.
Luckily, converting most units is very, very simple. In this case, all you need to know is that 1 cm is equal to 5.3995680345572E-6 nmi.
Once you know what 1 cm is in nautical miles, you can simply multiply 5.3995680345572E-6 by the total centimeters you want to calculate.
So for our example here we have 500 centimeters. So all we do is multiply 500 by 5.3995680345572E-6:
500 x 5.3995680345572E-6 = 0.0026997840172786
## What is the best conversion unit for 500 cm?
As an added little bonus conversion for you, we can also calculate the best unit of measurement for 500 cm.
What is the "best" unit of measurement? To keep it simple, let's say that the best unit of measure is the one that is the lowest possible without going below 1. The reason for this is that the lowest number generally makes it easier to understand the measurement.
For 500 cm the best unit of measurement is fathoms, and the amount is 2.7340332458443 fm. | 374 | 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} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.91813 |
https://mmsanotherstage2019.com/what-is-1-over-4-as-a-decimal/ | 1,638,669,737,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00417.warc.gz | 473,039,127 | 5,169 | Four friends, Rocky, Rooney, David, and also Sam separated a donut equally among themselves. Sam asked, "What part of the donut do I have?" Rocky replied, " One fourth." Sam asked again, "Is it possible to write it in a form that has actually denominator 1?".
You are watching: What is 1 over 4 as a decimal
Let us aid Sam solve his questions in mere few minutes.
This section focuses on representing ( frac14 ) as a decimal together with some interactive examples and questions on the depiction of ( frac38 ) together a decimal, ( frac23 ) as a decimal, and also decimal to fraction conversion. Perform not forget to shot practice inquiries at the finish of the web page for a rapid fun revision.
## Lesson Plan
1 What is 1/4 as a Decimal? 2 Important notes on 1/4 as a Decimal? 3 Think the end of Box! 4 Solved examples on 1/4 together a Decimal 5 Interactive inquiries on 1/4 as a Decimal
## What is 1/4 as a Decimal?
When we convert a portion into decimal form, we convert it into a number that has actually denominator 1. So the numerator takes a form that has a decimal in it.
The decimalform of(dfrac14,) is 0.25
Let's see exactly how to acquire that.
## How to change ¼ as a decimal
The fraction(dfrac14,)can be convert to a form with denomiator 1 by recognize its decimal form.
To gain the decimal form of (dfrac14,), various methods can be used.
### Method 1
In this method, we use long division.
Here, molecule = 1 and also Denominator = 4. Let's division 1 by 4
So,
(dfrac14=0.25..... extDecimal form)
We deserve to use this technique forconverting any fraction to decimal form.
### Method 2
The other technique is to transform the portion into its equivalent portion with denominator as a power of 10
The very first power that 10 is 10 i m sorry is nota multiple of 4. So, let's look in ~ the 2nd multiple, 100, i m sorry is a multiple of 4
In thefraction(dfrac14), denominator = 4
eginalign dfrac14&=dfrac1 imes 254 imes 25\&=dfrac25100endalign
Now, observe that there are 2 zeros in the denominator.
So, there will be two digits after ~ decimal in the numerator.
(dfrac14=0.25)
Similarly, if us havethefraction(dfrac38), we can convert it right into decimal by convert it into its equivalent portion with denominator 1000
eginalign dfrac38&=dfrac3 imes 1258 imes 125\&=dfrac3751000\&=0.375endalign
We deserve to even convert a decimal come fraction. Because that this, we take into consideration the variety of digits after ~ the decimal.
Example: take it 0.65. There space two number after the decimal.
eginalign 0.65&=dfrac65100\&=dfrac1320endalign
Example 1
Harry is struggling come express(dfrac23,) together decimal. Deserve to you aid him by utilizing long department method to transform a fraction to decimal?
Solution
As denominator is 3, i m sorry is no a variable of 100, we will convert(dfrac23,) as a decimal by long division.
The quotient is 0.6666.....= (0.overline6cdotcdot )Digit 6 is recurring right here as the remainder is walking on repeating.
So, we round turn off 0.6666..... To 0.67 (rounding off upto 2 decimal places).
See more: What Does Ntc Mean In Spanish, (You'Re A Whore!) Person2:
( hereforedfrac23=0.67)
Example 2
Mia desires to express(dfrac38,) as a decimal number using long division. Aid her to with the correct answer. | 880 | 3,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-49 | latest | en | 0.916954 |
http://www.justintools.com/unit-conversion/data-storage.php?k1=gigabytes&k2=megabits | 1,568,617,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572491.38/warc/CC-MAIN-20190916060046-20190916082046-00483.warc.gz | 278,730,283 | 18,238 | 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 :)
# Convert [Gigabytes] to [Megabits], (GB to Mbit)
## DATA STORAGE
1 Gigabytes
= 8000 Megabits
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: data storage
Conversion: Gigabytes to Megabits
The base unit for data storage is bytes (Non-SI/Derived Unit)
[Gigabytes] symbol/abbrevation: (GB)
[Megabits] symbol/abbrevation: (Mbit)
How to convert Gigabytes to Megabits (GB to Mbit)?
1 GB = 8000 Mbit.
1 x 8000 Mbit = 8000 Megabits.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [data storage] => (bytes), 1 Gigabytes (GB) is equal to 1000000000 bytes, while 1 Megabits (Mbit) = 125000 bytes.
1 Gigabytes to common data-storage units
1 GB =1000000000 bytes (B)
1 GB =1000000 kilobytes (KB)
1 GB =1000 megabytes (MB)
1 GB =1 gigabytes (GB)
1 GB =0.001 terabytes (TB)
1 GB =8000000000 bits (bit)
1 GB =8000000 kilobits (kbit)
1 GB =8000 megabits (Mbit)
1 GB =8 gigabits (Gbit)
1 GB =0.008 terabits (Tbit)
Gigabytes to Megabits (table conversion)
1 GB =8000 Mbit
2 GB =16000 Mbit
3 GB =24000 Mbit
4 GB =32000 Mbit
5 GB =40000 Mbit
6 GB =48000 Mbit
7 GB =56000 Mbit
8 GB =64000 Mbit
9 GB =72000 Mbit
10 GB =80000 Mbit
20 GB =160000 Mbit
30 GB =240000 Mbit
40 GB =320000 Mbit
50 GB =400000 Mbit
60 GB =480000 Mbit
70 GB =560000 Mbit
80 GB =640000 Mbit
90 GB =720000 Mbit
100 GB =800000 Mbit
200 GB =1600000 Mbit
300 GB =2400000 Mbit
400 GB =3200000 Mbit
500 GB =4000000 Mbit
600 GB =4800000 Mbit
700 GB =5600000 Mbit
800 GB =6400000 Mbit
900 GB =7200000 Mbit
1000 GB =8000000 Mbit
2000 GB =16000000 Mbit
4000 GB =32000000 Mbit
5000 GB =40000000 Mbit
7500 GB =60000000 Mbit
10000 GB =80000000 Mbit
25000 GB =200000000 Mbit
50000 GB =400000000 Mbit
100000 GB =800000000 Mbit
1000000 GB =8000000000 Mbit
1000000000 GB =8.0E+12 Mbit
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Vol.07 No.01(2015), Article ID:53435,12 pages
10.4236/ns.2015.71005
The Classical Binary and Triplet Distribution Functions for Dilute Relativistic Plasma
N. A. Hussein1, D. A. Eisa2, E. G. Sayed1
1Mathematics Department, Faculty of Science, Assiut University, Assiut, Egypt
2Mathematics Department, Faculty of Science, Assiut University, New Valley, Egypt
Email: aragamal@yahoo.com
Received 1 January 2015; accepted 19 January 2015; published 22 January 2015
ABSTRACT
The aim of this paper is to calculate the binary and triplet distribution functions for dilute relativistic plasma in terms of the thermal parameter where, is the mass of charge; is the speed of light; is the Boltzmann’s constant; and is the absolute temperature. Our calculations are based on the relativistic Bogoliubov-Born-Green-Kirkwood-Yvon (BBGKY) hierarchy. We obtain classical binary and triplet distribution functions for one- and two-component plasmas.
Keywords:
Relativistic Plasma, Binary and Triplet Distribution Functions, BBGKY Hierarchy
1. Introduction
Studying the properties of plasma has received a great interest in both astrophysical and laboratory plasma application. The relativistic binary distribution function is one of the most important functions of statistical mechanics. The importance of the distribution function in statistical mechanics is due to the fact that all the thermodynamic quantities, such as the pressure, the internal energy and the free energies, can be calculated from it.
Relativistic statistical mechanics has a long story, but we may notice that, whereas the theory of relativistic ideal gases has received deep and detailed developments, little has been achieved in order to account for mutual interactions between particles [1] . In statistical physics, the BBGKY hierarchy (Bogoliubov-Born-Green- Kirkwood-Yvon hierarchy, sometimes called Bogoliubov hierarchy) is a set of equations describing the dynamics of a system of a large number of interacting particles. The equation for an s-particle distribution function (probability density function) in the BBGKY hierarchy includes the -particle distribution function thus forming a coupled chain of equations [2] . Their history of 70 years has brought enormous progress in the investigation of the transition from the microscopic to the macroscopic world, and they are still an attractive starting point for new developments. In particular, great advance has come by clever methods of truncation, approxi- mation and scaling limits of the hierarchy, providing in various cases a justification of the kinetic equations describing particle systems on mesoscopic (intermediate) scales [3] . Many authors studied the BBGKY hierarchy [4] and [5] ; Hussein and Eisa [6] [7] calculated the binary and triplet distribution function for one- and two- component plasmas for quantum and classical non-relativistic plasma. In this paper, we will calculate these distributions in the relativistic case in terms of plasma thermal parameter. Arendt and Eilek [8] showed that the
pair-plasma distribution functions could be described by a thermal parameter that was moderately
relativistic. The thermal parameter value plays a significant role for the stability of our system. Barcons and Lapiedra (1984) [9] gave explicit expressions for the thermodynamic functions of a high-temperature electron- positron plasma and gave expression for distribution functions for a classical dilute arbitrarily hot plasma in equilibrium which we compared our results with it.
Special relativity, however, does not permit velocities greater than the speed of light and is thus incompatible with a Maxwellian distribution that predicts a non-zero probability for every velocity [10] . It is possible to obtain the relativistic Maxwellian distribution (Jüttner distribution) of a moving gas when the so-called Planck-Einstein case is analyzed. It should be reiterated that the Planck-Einstein theory suffered a modification, principally in the transformation law of energy (the Planck-Einstein case) [11] . Treumann et al. [12] have also shown that the isotropic thermal equilibrium distribution function in relativistic plasma analytically has the form of a modified Jüttner distribution.
A little is known about relativistic distribution functions involving more than two particles, and in particular about the three-particle (or triplet) distribution function. This is of course due to the greater mathematical complexity of higher order correlation functions, and to a lack of a direct link with experiment. Although one can consider experimental determination of triplet distribution function from triple elastic scattering similar to that of the binary distribution function, but to our knowledge, the measurement of three-body correlation function requires the knowledge of the positions of three particles at the same time which is technically very demanding to obtain in 3D samples [13] ; this means that the experiment to measure the triplet distribution function directly requires high precision so that the multiple scattering can be differentiated from single scattering. Essentially, one follows the phase-space trajectory of the system as it evolves in time, and has thus the same amount of information as one obtains in a simulation. Similarly the particle positions have measured in dusty plasmas [14] . Lapiedra et al. [15] have undertaken an application of predictive methods to relativistic statistical mechanics. Coulomb forces between point charges are purely repulsive and charges approach very close to each other. Coulomb systems, such as plasma or electrolytes, are made of charged particles interacting through Coulomb’s law. The simplest model of a Coulomb system is the one-component plasma (OCP), also called jellium: an assembly of identical point charges, embedded in a neutralizing uniform background of the opposite sign. Here we consider the classical (i.e. non-quantum) equilibrium statistical mechanics of the OCP. It is rather straightforward to calculate higher-order correlation functions from the measured configurations. Moreover we study the model of two-component plasma (TCP) i.e. neutral system of point like particles of positive and negative charges such as electrons and ions. For the numerical calculation we restrict ourselves to the case of (TCP) which anti-symmetric with respect to the charges and therefore symmetrical with respect to the densities. To simplify the numeric investigations, we simulated so far only mass symmetrical electron-ion plasma with.
Homogeneous plasma is characterized by two parameters: the density of particles and the temperature; if there are several kinds of particles, it is also necessary to state their concentrations. There are different energies associated with the plasma, namely:
1) Energy of the rest mass per particle;
2) Kinetic energy per particle of order;
3) Coulomb energy of order per particle.
The ratios of these energies give us the two main dimensionless parameters of the plasma: the thermal para-
meter of plasma and the dilution parameter. The value of is very important. There
are four different regimes characterized by it: plasma with is relativistic, weakly relativistic plasma, low plasma temperature and ultra-relativistic high temperatures plasma. In our study
we note that the system is safely classical if where, , is the Planck constant, and
is the typical linear momentum of particles. A relativistic plasma with a thermal distribution function has temperatures greater than around 260 keV, or 3.0 GK (5.5 billion degrees Fahrenheit), where approximately 10% of
the electrons have where is the Lorentz factor. It is occurs in many environments in
astrophysics, including gamma-ray bursts, AGN jets, and pulsar winds. In physics, a particle is called ultrarelativistic when its speed is very close to the speed of light. Or, similarly, in the limit where the Lorentz factor is very large and. Our study is only valid for dilute plasmas. The plasma temperatures and velocities also plays a significant role for the stability of our system Figure 1 show qualitative sketch of the four different regimes characterized by the thermal parameter and Lorentz factor. Our model lies in the region of high speeds, and.
In the astrophysical environment, the fraction of ionized particles varies widely from nearly no ionization in cold regions to fully ionized in regions of high temperature. This leads to a wide range of parameters where astrophysical plasmas can exist. While the astrophysical environment is frequently dominated by the presence of the plasma, this plasma is often strongly influenced by and coupled to the presence of embedded particulates (i.e., dust). These dust grains which range in size from a few nanometers to micron-sized objects can become either positively or negatively charged due to interactions with the background plasma environment and ionizing radiation sources in the astrophysical environment [16] . Understanding the processes that govern these plasma particle interactions is critical to the study of astrophysics. The agglomeration and growth of larger particles from single atoms and dust grains leads to the eventual formation of objects so large that gravity becomes the dominant force controlling their subsequent evolution [17] .
Another important point is that of the walls. We shall study plasma which is homogeneous and isotropic but the plasma must be confined or it will expand. We may assume that the plasma is confined by some kind of walls which prevent the escape of particles, but that the container is so large that the effects of the walls are negligible. Our calculations are based on the phase-space distribution function; this is defined as the number of particles per unit volume of space per unit volume of velocity space: At time t, number of particles in elementary volume of space, with velocities in range.
2. The Basic Equations and Hierarchy
The statistical state of a macroscopic system of N particles is in a complete―though in an intractably complex―
Figure 1. Qualitative sketch of the four different regimes characterized by the thermal parameter μ and Lorentz factor γ: our model lies in the region of high speeds, γ > 2 and.
way described by the distribution function in dimensional phase space, which is spanned by the coordinates and velocities of all individual particles, , being the three posi-
tion of particle, and the spatial components of its four velocity in a given frame.
The s-particle reduced distribution is giving by:
(1)
The relativistic BBGKY hierarchy [15] is given by
(2)
where is the acceleration of the charge in the presence of the charge.
The Einstein summation convention is only valid for Greek labels. Therefore, for each value of s we have s equations, since. As in the non-relativistic case, the determination of the reduced generalized distribution function can only be made when the hierarchy is cut off somewhere, that is, when for some value s we give as a function of the other functions with.
According to what we done in non-relativistic case [7] we set, whatever particles 1, 2, 3 are,
(3)
(4)
Let us consider the case of homogeneous plasma in equilibrium. Then is the one-particle distribution function of an ideal gas see Figure 2 which shows one particle relativistic distribution function in the particle velocity interval (0, 0.9c) for different values of the thermal parameter μ. That is, in a frame relative to which the system is macroscopically at rest, we must set for the relativistic Maxwellian distribution [18]
(5)
where denotes the modified Bessel function.
Figure 2. The one particle relativistic distribution function in the particle velocity interval (0, 0.9c) for different values of the thermal parameter μ.
3. The Binary Relativistic Distribution Function
At first sight, the calculation of the relativistic interaction between two charged particles seems rather involved, because the force on particle 1 at time would depend on the position and velocity of particle 2 at a retarded time. Also, the position and velocity of 1 at that time depends on the position and velocity of 1 at an earlier time, and so on. If one chooses a frame of reference, the acceleration on particle 1 due to the presence of 2 can be calculated from the position and velocity of 2 at that time. This does not mean that actions propagate instantaneously, but rather that there is a precise scheme to take retardation into account automatically, through the equations of the theory. All we shall need in that chapter is the time component of the acceleration of particle 2 due to particle 1 [19] .
To calculate the binary relativistic distribution function substituting from Equations (3) and (5) into (2) for we obtain
(6)
(7)
Then, by solving the integro-differential Equation (7) and by using the Fourier transform of
i.e.,. We get the relativistic binary
generalized distribution function at in the following form
(8)
Let us now study the two-particle relativistic distribution for the model of two-component plasma (TCP) i.e. neutral system of point like particles of positive and negative charges such as electrons and ions. For the numerical calculation we restrict ourselves to the case of two-component plasma which anti-symmetric with respect to the charges and therefore symmetrical with respect to the densities.
For two-component plasma we can use the two-particle correlation function which is given by
(9)
(10)
where is the Debye-Hiickel solution [20] for two-component plasma, by substituting Equation (9) into Equation (3) we get the binary distribution function for two-component plasma in the following form:
(11)
with given by Equation (5) (in fact only terms of up to must be retained). It can be verified that given by Equation (9), with satisfies identically the condition (6) for, to first order in.
The standard two-body distribution function for dilute slightly relativistic plasma has been calculated previously by Kosachev and Trubnikov [21] starting from the Darwin Lagrangian. Lapiedra and Santos [19] result agrees with theirs to order, but not to higher-order terms. We think that these terms are meaningless unless one goes beyond the Darwin Lagrangian, which is only correct to order. The Lagrangian approach introduces some modification of the expression for the statistical sum and what is more important the use of the restricted Breit-Darwin Hamiltonian leads to the wrong behavior of the pair correlation function and therefore to the incorrect expressions for various thermodynamical quantities [22] .
According to our knowledge few scientists studied the binary distribution function for relativistic dilute plasma [15] ; the new in our article is using effects of thermal parameter on the values of binary relativistic distribution function for one and two-component plasma.
4. The Triplet Relativistic Distribution Function
The triplet distribution function is defined in such a way that, the calculation of the relativistic interaction between three charged particles seems rather involved, because the force on particle 1 at time t would depend on the position and velocity of particles 2 and 3 at a retarded time. Also, the position and velocity of 1 at that time depends on the position and velocity of 1 at an earlier time.
Substituting Equation (4) and (5) into (2) for we obtain
(12)
If we used the Kirkwood superposition approximation (KSA) [23] ; which is consisting of the assumption that the potential in a set of three particles is the sum of the three pair potentials, this is equivalent to assuming that the triplet distribution function is the product of the three radial distribution functions
(13)
It has been mentioned that before 2003 there is no direct measurement of three-body correlation function and this is the first study for the triplet distribution function in the case of dilute relativistic plasma. Such measurement requires the knowledge of the positions of three particles at the same time which is technically very demanding to obtain in 3D samples [13] ; this means that the experiment to measure the triplet distribution function directly requires high precision so that the multiple scattering can be differentiated from single scattering. Now because of the existence of video-microscopy, a modern experimental technique applied to colloidal systems to directly measure all particles’ positions at all times. Essentially, one follows the phase-space trajectory of the system as it evolves in time, and has thus the same amount of information as one obtains in a simulation. Similarly the particle positions have measured in dusty plasmas [24] .
From Equation (12) we can obtain the classical TDF for two-component plasma in the following form:
(14)
And we also can used (KSA) which is given in Equation (13) to get it in the form
(15)
The triplet and quadruple distribution functions as well as binary distribution function must be incorporated for a more accurate and complete discussion of macroscopic equilibrium properties. A little is known about distribution functions involving more than two particles, and in particular about the three-particle (or triplet) distribution functions. This is of course due to the greater mathematical complexity of higher order correlation functions, and to a lack of a direct link with experiment [25] . And this is considered the first study for the triplet distribution functions for dilute relativistic plasma.
5. Conclusions
In many physical systems, the description of a plasma as a Coulomb system is sufficient to reproduce most of the properties of interest. If the system is cold enough, the mean velocities of the particles are much smaller than the speed of light, and the charges may be assumed to interact via the instantaneous Coulomb potential. However, at sufficiently high temperatures, this approximation is no longer valid, and the contributions of the relativistic effects (which include, apart from the trivial kinetic corrections and of course all the retardation effects) must be incorporated when studying the equilibrium properties of the system [26] .
In the classical (non-quantum) case, the systematic approach adopted here follows the traditional route of the relativistic Bogoliubov-Born-Green-Kirkwood-Yvon (BBGKY) hierarchy for the reduced distribution functions by formal density expansion. Interestingly, this study is the first to display the effect of thermal parameter of plasma in the classical binary and triplet distribution function. Also the triplet relativistic distribution function for dilute plasma was calculated from the relativistic BBGKY hierarchy. We used the results to obtain the analytical forms of the classical triplet distribution functions for one- and two-component plasmas.
In Figure 3 and Figure 4 we noticed that the value of both binary and triplet relativistic distribution function increased when the value decreased at very high temperature. Physically these results seem acceptable because from the definition of the phase-space distribution function which means the number of particles per unit volume of space per unit volume of velocity space, and at high temperature, the velocity of the particles increases and as a result the number of particles increases per unit volume.
Our calculations are grounded in the classical relativistic statistical mechanics. Plasma is non-degenerate. The system is not dense, so one may neglect the contributions of higher order particle interactions. Figure 5 and Figure 6 show that the two- and three-particle distribution functions have become concentrated to ever-smaller region of speed, dramatically increasing the thermal parameter. When we made a comparison between the two-particle relativistic distribution function from Equation (11) and the result of Barcons and Lapiedra as shown in Figure 7, the results were nearly similar at very high velocities.
Figure 3. The two-particle relativistic distribution function in the particle velocity interval (0, 0.9c) for one-compo- nent plasma.
Figure 4. The three-particle relativistic distribution function in the particle velocity interval (0, 0.9c) for one-compo- nent plasma for different values of μ, v1 = v2 = v3 = v.
Figure 5. The two-particle relativistic distribution function for two-component plasma in the particle velocity interval (−c, c) at μ = 1; μ = 0.3 and v1 ≠ v2.
Figure 8 and Figure 9 show that the two- and three-particle distribution functions in the thermal parameter of plasma interval for one-component plasma for different values of speed, , and. We note that the curves are very close to each other, dramatically increasing the thermal parameter
Figure 6. The three-particle relativistic distribution function in the particle velocity interval (0, 0.9c) for one-compo- nent plasma for different values of μ; v1 = v2 = v3 = v.
Figure 7. The comparison between F(2) from our result and from Barcons and Lapiedra [9] for v = 0.86c and v = 0.9c.
Figure 8. The two-particle relativistic distribution function in the thermal parameter of plasma μ interval (0, 1) for one-component plasma for different values of speed, v = 0.8c, v = 0.86c, v = 0.9c.
and for three-particle distribution function more than two-particle distribution function.
Figure 10 shows the comparison between the three-particle relativistic distribution function from Equations (14) and (15). One of them is based on the Kirkwood superposition approximation (KSA) which is consisting of the assumption that the potential in a set of three particles is the sum of the three pair potentials. This is equivalent
Figure 9. The three-particle relativistic distribution function in the thermal parameter of plasma μ interval (0, 1) for one-component plasma for different values of speed, v = 0.8c, v = 0.86c, v = 0.9c.
Figure 10. The comparison between F(3) from Equation (14) and F(3) from (KSA) for two-component plasma for μ = 1; at v1 = v2 = v3 = v.
Figure 11. The two-particle relativistic distribution function in the particle velocity interval (−c, c) for μ > 1 (weakly relativistic case).
to assume that the triplet distribution function is the product of the three radial distribution functions, and the other form is calculated by using the relativistic BBGKY hierarchy.
Figure 12. The two-particle relativistic distribution function in the particle velocity interval (−c, c) for μ = 1 (relativistic case).
Figure 13. The two-particle relativistic distribution function in the particle velocity interval (−c, c) for μ > 1 (weakly relativistic case).
Figure 14. The three-particle relativistic distribution function in the particle velocity interval (−c, c) for μ = 1 (relativistic case).
Finally, we can note that from Figures 11-14, the distribution function has become more concentrated to ever- smaller region of speed, dramatically increasing the thermal parameter and for three-particle distribution function more than two-particle distribution function.
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24. Bonitz, M., Henning, C. and Block, D. (2010) Complex Plasmas: A Laboratory for Strong Correlations. Reports on Progress in Physics, 73, Article ID: 066501. http://dx.doi.org/10.1088/0034-4885/73/6/066501
25. Barrat, J.L., Hansen, J.P. and Pastore, G. (1988) On the Equilibrium Structure of Dense Fluids: Triplet Correlations, Integral Equations and Freezing. Molecular Physics, 63, 747-767. http://dx.doi.org/10.1080/00268978800100541
26. Kalman, G.J., Rommel, J.M., Blagoev, K. and Blagoev, K. (1998) Strongly Coupled Coulomb Systems. Springer, Berlin. http://www.springer.com/physics/particle+and+nuclear+physics/book/978-0-306-46031-9 | 6,505 | 28,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.888818 |
https://www.glassdoor.com/Interview/Brain-Teaser-Given-a-room-with-3-light-bulbs-which-have-their-respective-three-switches-On-Off-outside-the-room-You-do-QTN_367137.htm | 1,516,722,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00597.warc.gz | 935,256,539 | 59,128 | Halliburton Interview Question: Brain Teaser: Given a room wi... | Glassdoor
## Interview Question
Intern Interview(Student Candidate) Houston, TX
# Brain Teaser: Given a room with 3 light bulbs which have
their respective three switches (On/Off) outside the room. You don't know which switch corresponds to which light bulb. You can do whatever you want with the switches but you can enter the room only once. How will you find which light switch corresponds to which light bulb?
6
Hint 1: Think of what happens when a light bulb glows?
Hint 2: It gets heated
Answer: Switch on one bulb for some time. Then switch on another bulb and go into the room. The bulb which is on and hot is the corresponding to the switch you switched on first. The other glowing bulb corresponds to the switch you switched on second.
Interview Candidate on Oct 21, 2012
1
Hint 1: Imagine that the room has a door.
Hint 2: Open it.
Answer: flip the switches and observe which one corresponds to which bulb. You don't even need to enter it.
Anonymous on Apr 29, 2013 | 246 | 1,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-05 | latest | en | 0.942004 |
https://nathanbrixius.wordpress.com/2012/09/28/detecting-the-sources-of-model-infeasibility-using-gurobi/ | 1,500,625,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00190.warc.gz | 679,269,142 | 33,979 | # Detecting the Sources of Model Infeasibility using Gurobi
Today I come to sing the praises of Irreducible Infeasible Sets (IIS). An IIS is a minimal subset of the constraints of an optimization problem that are self-contradictory. In other words, if you remove any single constraint in an IIS, then the remaining constraints are feasible. Most modern linear and mixed-integer solvers have programming APIs to return IIS information for an infeasible model.
The reason IIS is so cool is that it helps diagnose why a model is not able to be solved. This is especially useful when the optimization problem to be solved is formed with the help of input from a user who may not know anything about operations research, or when the model is complicated. IIS is like the “spell check” of linear programming. Building optimization models without IIS is like debugging using “printf”.
Here is a simple example from the lp_solve documentation:
```min x + y
x >= 6
y >= 6
x + y <= 11```
6 plus 6 is more than 11, people. In this case, all three constraints are part of the IIS.
Here is some C# code using the Gurobi API that obtains the IIS for a very simple model. First, here is the code that forms and solves the model. This is easy to figure this out by reading the Gurobi docs. The only twist is that I have expressed the first two constraints as lower bounds on the variables x and y.
```GRBEnv env = new GRBEnv();
GRBModel model = new GRBModel(env);
GRBVar x = model.AddVar(6, Double.MaxValue, 1, GRB.CONTINUOUS, "x");
GRBVar y = model.AddVar(6, Double.MaxValue, 1, GRB.CONTINUOUS, "y");
model.Update();
GRBConstr c3 = model.AddConstr(x + y, GRB.LESS_EQUAL, 11, "c3");
model.Optimize();
model.ComputeIIS();
```
To find the IIS you need to call another method called ComputeIIS. Then you can query the constraints and variables to see if they are part of the IIS. This is accomplished using the Get method. I simply print out the constraint and variable names that are in the IIS.
```// Print the names of all of the constraints in the IIS set.
foreach (var c in model.GetConstrs())
{
if (c.Get(GRB.IntAttr.IISConstr) > 0)
{
Console.WriteLine(c.Get(GRB.StringAttr.ConstrName));
}
}
// Print the names of all of the variables in the IIS set.
foreach (var v in model.GetVars())
{
if (v.Get(GRB.IntAttr.IISLB) > 0 || v.Get(GRB.IntAttr.IISUB) > 0)
{
Console.WriteLine(v.Get(GRB.StringAttr.VarName));
}
}```
If you want to write the IIS information to a text file for review, simply use the write method with the “ilp” extension on the file name:
`model.Write("model.ilp");`
One last thing: using LINQ (a feature of the C# language), you can simplify how we query for the IIS constraints and variables. For example, I can get all of the constraint names in one statement as follows:
```var allConstraintNames =
from c in model.GetConstrs()
where c.Get(GRB.IntAttr.IISConstr) > 0
select c.Get(GRB.StringAttr.ConstrName);```
LINQ is very handy for creating models and querying results. I use it often! | 779 | 3,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.896682 |
http://www.chegg.com/homework-help/questions-and-answers/motorist-drives-along-straight-road-constant-speedof-225-m-s-shepasses-parked-motorcycle-p-q654533 | 1,448,872,349,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461132.22/warc/CC-MAIN-20151124205421-00096-ip-10-71-132-137.ec2.internal.warc.gz | 355,565,792 | 13,135 | A motorist drives along a straight road at a constant speedof 22.5 m/s. Just as shepasses a parked motorcycle police officer, the officer starts toaccelerate at4.00 m/s2 to overtake her.Assume that the officer maintains this acceleration.
(a) Determine the time it takes the police officer to reach themotorist.
(b) Find the speed of the officer as he overtakes themotorist.
(c) Find the total displacement of the officer as he overtakes themotorist. | 106 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2015-48 | latest | en | 0.954349 |
https://docs.google.com/forms/d/e/1FAIpQLSe1rwVd1ltKDFVllLKbdenb4o7aGYFvev4UFftj345nkSxKIA/viewform?usp=send_form | 1,606,635,015,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197278.54/warc/CC-MAIN-20201129063812-20201129093812-00542.warc.gz | 255,285,940 | 36,275 | Physics Quiz 1 Chapter 7
Alternating Current
NAME OF STUDENT *
CLASS *
MOBILE NO. *
Q .1 An electrical element X when connected to an alternating voltage source has current through it leading the voltage by π/2 radian. Identify X *
1 point
Q. 2 Average power dissipated in an ideal inductor in a.c. circuit is: *
1 point
Q. 3 A Capacitor blocks *
1 point
Q.4 A transformer steps up 220 V to2200 V. What is the transformation ratio? *
1 point
Q.5 In which of the following circuits the maximum power dissipation is observed? *
1 point
Q.6 The instantaneous current in an ac circuit is i=2.0 sin314t, what is its frequency? *
1 point
Q.7 The instantaneous current in an ac circuit is i=2.0 sin314t, what is rms value of the current *
1 point
Q.8 An L,C, R series circuit is connected to a.c. source. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of *
1 point
Q.9 In an a.c. circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is 0.5henry and of capacitance is 8μF. The angular frequency of the input A.C. Voltage must be equal to *
1 point
Q.10 In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is 4A, then that in the secondary is *
1 point
Q.11 In a series LCR circuit, resistance R = 10 and the impedance Z = 20. The phase difference between the current and the voltage is *
1 point
Q.12 In the case of an inductor : *
1 point
Q.13 In an AC circuit, V and I are given by V=150 sin (150t) volts and I = 150 sin (150t) amperes. The power dissipated in the circuit is: *
1 point
Q. 14 An A.C source is connected to a resistive circuit. Which of the following is true ? *
1 point
Q.15 The output voltage of an ideal transformer, connected to a 240V a.c. mains is 24V. When this transformer is used to light a bulb with rating (24V, 24W), calculate the current in the primary coil of the circuit. *
1 point
Q. 16 In an LCR-series ac circuit, the voltage across each of the component L, C and R is 50 V. The voltage across the LC-combination will be *
1 point
Q. 17 The core of any transformer is laminated so as to. *
1 point
Q. 18 For an ideal-step-down transformer, the quantity which is constant for both the coils is. *
1 point
Q.19 To reduce the resonant frequency in an LCR series circuit with a generator . *
1 point
Q. 20 When an AC voltage of 220 V is applied to the capacitor C . *
1 point
A copy of your responses will be emailed to the address you provided. | 700 | 2,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-50 | latest | en | 0.882805 |
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# All intelligent people are nearsighted. I am very
Author Message
Director
Joined: 20 Apr 2005
Posts: 585
All intelligent people are nearsighted. I am very [#permalink]
### Show Tags
09 May 2005, 16:59
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22. All intelligent people are nearsighted. I am very nearsighted. So I must be a genius.
Which one of the following exhibits both of the logical flaws exhibited in the argument above?
(A) I must be stupid because all intelligent people are nearsighted and I have perfect eyesight.
(B) All chickens have beaks. This bird has a beak. So this bird must be a chicken.
(C) All pigs have four legs, but this spider has eight legs. So this spider must be twice as big as any pig.
(D) John is extremely happy, so he must be extremely tall because all tall people are happy.
(E) All geniuses are very nearsighted. I must be very nearsighted since I am a genius.
Manager
Joined: 28 Jan 2005
Posts: 98
### Show Tags
09 May 2005, 18:00
Intelligent= near sight ASSUMPTION very near sight = very intelligent, genius
Tall = happy Very Tall = Very happy
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore
### Show Tags
09 May 2005, 19:09
All I -> N
Very N -> Very I (Genius)
The flaw here is to give you a general relationship, and then reverse
the relationship illogically under an extreme circumstance.
Here, the general relationship is Intelligent -> Nearsighted (All I->N)
The author then reverse the relationship illogically (N->I, when it should be Not N -> Not I)
The extreme circumstance given is Genius (very intelligent) because he is Very nearsighted.
So now look at the answer choices:
(A) All I->N, S->P. Different logic.
(B) All C->B. B->C. Different logic.
(C) All P->4L. S->8L. S->2*P. Different logic.
(D) All T->H. John: Extemely happy -> Extremely tall. Same logic as passage.
(E) All G->V.N. I: V.N->G. Different logic
D for me.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore
### Show Tags
09 May 2005, 19:36
shalinikhatri wrote:
Ans B.
B only commits 1 of the 2 flaws in the original passage. There is no extreme condition given in choice (B).
Manager
Joined: 07 Apr 2005
Posts: 81
Location: Lontano da dove
### Show Tags
09 May 2005, 22:10
D
Wilfred is right. We would need to have bigger beak for the bird and a "super" chicken, so to speak, to make B correct.
Intern
Joined: 04 May 2005
Posts: 32
### Show Tags
10 May 2005, 04:48
Dudes it has to be (D)
Manager
Joined: 28 Jan 2005
Posts: 98
### Show Tags
10 May 2005, 06:19
Those who chose B pls explain why
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany
### Show Tags
10 May 2005, 08:49
B)...
argument: all A are B ! i am B ! so i am A !
B) all A are B ! it is B ! so it is A ! => same reasoning
D) all A are B ! he is C ! so he is D ! => not the same reasoning, because we cannot deduce that happy = extremely happy and tall = extremely tall.
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
VP
Joined: 18 Nov 2004
Posts: 1433
### Show Tags
10 May 2005, 09:05
Christoph, it shud be "D"....argument is actually twisted it actually says
all A are B ! I am C (extreme B) ! so I must be D (extreme A)! ......"D" has the same logic. I missed this nuance in the first go as well.
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany
### Show Tags
10 May 2005, 09:11
banerjeea_98 wrote:
Christoph, it shud be "D"....argument is actually twisted it actually says
all A are B ! I am C (extreme B) ! so I must be D (extreme A)! ......"D" has the same logic. I missed this nuance in the first go as well.
baner, you are right. thx
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
Director
Joined: 20 Apr 2005
Posts: 585
### Show Tags
11 May 2005, 16:26
The OA is D.
Current Student
Joined: 29 Jan 2005
Posts: 5218
### Show Tags
12 May 2005, 21:51
Late, but D.
D can be rewritten as:
All happy people are tall. John is extremely happy, so he must be extremely tall.
which is very similiar to...
All intelligent people are nearsighted. I am very nearsighted. So I must be a genius.
Eureka!
12 May 2005, 21:51
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Display posts from previous: Sort by | 1,580 | 5,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-26 | latest | en | 0.923803 |
https://www.physicsforums.com/threads/rings-units-nilpotents-idempotents.463330/ | 1,590,751,773,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00363.warc.gz | 883,666,411 | 16,151 | # Rings- units, nilpotents, idempotents
## Homework Statement
Find the units, nilpotents and idempotents for the ring R =
[$$\Re$$ $$\Re$$]
[0 $$\Re$$]
(Those fancy R's are suppose to be the set of Reals by the way.. not good with this typing math stuff)
## The Attempt at a Solution
I'm not actually sure I understand the ring itself. So it is a matrix with entries $$\Re$$ in the 1-1, 1-2, 2-2 positions and 0 in the 2-1 position..
So is it the entire set of the real numbers?? :S
Anyways for the units, I said all elements are units except for when $$\Re$$=0.
But then again I wasn't sure if I'm suppose to use the entire set as those positions, or is it just random numbers from the reals? like a,b,c. OR is it just any number from the reals, but each position has the same #, say x belonging to the reals?
Eughh.. just clarification on the actual question I guess is what I need some help with.
Thanks :)
Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
It's got to be just all matrices [[a,b],[0,c]] where a, b, and c are real numbers. It's your 'random numbers from the reals' theory. Where can you go from there?
ok well for the units i'm attempting to solve some systems of equations..
[[a,b],[0,c]]*[[a',b'],[d',c']] = [[1,0],[0,1]]
and
[[a',b'],[d',c']]*[[a,b],[0,c]] = [[1,0],[0,1]]
but for matrices, isn't everything invertible whose det is not 0?
Dick | 410 | 1,409 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-24 | longest | en | 0.913974 |
https://www.physicsforums.com/threads/sample-space.184952/ | 1,539,888,850,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511897.56/warc/CC-MAIN-20181018173140-20181018194640-00363.warc.gz | 1,024,872,041 | 12,587 | # Homework Help: Sample Space
1. Sep 16, 2007
### EugP
1. The problem statement, all variables and given/known data
I'm having some trouble understanding how to write a sample space in a problem.
Here's an example:
Shuffle a deck of cards and turn over the first card. What is the sample space of this experiment? How many outcomes are in the event that the first card is a heart?
2. Relevant equations
$$C_k^n = {n \choose k} = \frac{n!}{k!(n - k)!}$$
3. The attempt at a solution
From what I was explained, sample space is the mutually exclusive and collectively exhaustive set of all possible outcomes. So in my case, wouldn't it be {2-A of hearts, 2-A of spades, 2-A of clubs, 2-A of diamonds} ? Those together create all the possiblities in the deck.
For the second part, isn't it simply 52 choose 13? If it is, it will just be
$${52 \choose 13} = \frac{52!}{13!(52 - 13)!} = 635,013,559,600 \approx 11 6.350135596 \cdot 10^{11}$$
But I'm not sure this is right. There are no answers in my book. If someone could help me on this it would be greatly appreciated.
2. Sep 16, 2007
### matness
yes. You are right.
Sample space consists of all cards in the deck
and the answer of the second part is in mattmns 's post
Last edited: Sep 16, 2007
3. Sep 16, 2007
### EugP
Thanks for verifying!
4. Sep 16, 2007
### mattmns
I disagree with your second answer, I think it should be 13. The possible outcomes of a heart being the first card flipper over are A hearts, K hearts, ... , 2 hearts. | 438 | 1,502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-43 | latest | en | 0.923793 |
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# Integrals Maths 12 Science CBSE Solutions for MCQ in English
Integrals Maths 12 Science CBSE Solutions for MCQ in English to enable students to get Solutions in a narrative video format for the specific question.
Expert Teacher provides Integrals Maths 12 Science CBSE Solutions for MCQ through Video Solutions in English language. This video solution will be useful for students to understand how to write an answer in exam in order to score more marks. This teacher uses a narrative style for a question from Integrals not only to explain the proper method of answering question, but deriving right answer too.
Please find the question below and view the Solution in a narrative video format.
Question:
Solution Video in English:
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Prolog Notes
# Prolog Notes
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Prolog Notes
Prolog Notes
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Given a complex term of which you don’t know what it looks like, what kind of infor-
mation would be interesting to get? Probably, what’s the functor, what’s the arity and
what do the arguments look like. Prolog provides built-in predicates that answer these
questions. The first two are answered by the predicate functor/3. Given a complex
term functor/3 will tell us what the functor and the arity of this term are.
?- functor(f(a,b),F,A).
A = 2
F = f
yes
?- functor(a,F,A).
A = 0
F = a
yes
?- functor([a,b,c],X,Y).
X = ’.’
Y = 2
yes
So, we can use the predicate functor to find out the functor and the arity of a term,
but we can also use it to construct terms, by specifying the second and third argument
and leaving the first undetermined. The query
?- functor(T,f,8).
for example, returns the following answer:
T = f(_G286, _G287, _G288, _G289, _G290, _G291, _G292, _G293)
yes
Note, that either the first argument or the second and third argument have to be instanti-
ated. So, Prolog would answer with an error message to the query functor(T,f,N). If
you think about what the query means, Prolog is reacting in a sensible way. The query
134
Chapter 9. A Closer Look at Terms
is asking Prolog to construct a complex term without telling it how many arguments to
provide and that is something Prolog can just not do.
In the previous section, we saw built-in predicates for testing whether something is
an atom, a number, a constant, or a variable. So, to make the list complete, we were
actually missing a predicate for testing whether something is a complex term. Now,
we can define such a predicate by making use of the predicate functor. All we have
to do is to check that the term is instantiated and that it has arguments, i.e. that its arity
is greater than zero. Here is the predicate definition.
complexterm(X) :-
nonvar(X),
functor(X,_,A),
A > 0.
In addition to the predicate functor there is the predicate arg/3 which tells us about
arguments of complex terms. It takes a number N and a complex term T and returns
the Nth argument of T in its third argument. It can be used to access the value of an
argument
?- arg(2,loves(vincent,mia),X).
X = mia
yes
or to instantiate an argument.
?- arg(2,loves(vincent,X),mia).
X = mia
yes
Trying to access an argument which doesn’t exist, of course fails.
?- arg(2,happy(yolanda),X).
no
The third useful built-in predicate for analyzing term structure is ’=..’/2. It takes a
complex term and returns a list that contains the functor as first element and then all
the arguments. So, when asked the query ’=..’(loves(vincent,mia),X) Prolog
will answer X = [loves,vincent,mia]. This predicate is also called univ and can
be used as an infix operator. Here are a couple of examples.
Yes
?- X =.. [a,b(c),d].
X = a(b(c), d)
Yes
?- footmassage(Y,mia) =.. X.
Y = _G303
X = [footmassage, _G303, mia]
Yes
9.3. Examining Terms
135
Univ (’=..’) is always useful when something has to be done to all arguments
of a complex term. Since it returns the arguments as a list, normal list processing
strategies can be used to traverse the arguments. As an example, let’s define a pred-
icate called copy_term which makes a copy of a term replacing variables that occur
in the original term by new variables in the copy. The copy of dead(zed) should
be dead(zed), for instance. And the copy of jeallou(marcellus,X) should be
jeallous(marcellus,_G235); i.e. the variable X in the original term has been re-
places by some new variable.
So, the predicate copy_term has two arguments. It takes any Prolog term in the first
argument and returns a copy of this Prolog term in the second argument. In case the
input argument is an atom or a number, the copying is simple: the same term should
be returned.
copy_term(X,X) :- atomic(X).
In case the input term is a variable, the copy should be a new variable.
copy_term(X,_) :- var(X).
With these two clauses we have defined how to copy simple terms. What about com-
plex terms? Well, copy_term should return a complex term with the same functor and
arity and all arguments of this new complex term should be copies of the correspond-
ing arguments in the input term. That means, we have to look at all arguments of the
input term and copy them with recursive calls to copy_term. Here is the Prolog code
for this third clause:
copy_term(X,Y) :-
nonvar(X),
functor(X,F,A),
A > 0,
functor(Y,F,A),
X =.. [F|ArgsX],
Y =.. [F|ArgsY],
copy_terms_in_list(ArgsX,ArgsY).
copy_terms_in_list([],[]).
copy_terms_in_list([HIn|TIn],[HOut|TOut]) :-
copy_term(HIn,Hout),
copy_terms_in_list(TIn,TOut).
So, we first check whether the input term is a complex term: it is not a variable and
its arity is greater than 0. We then request that the copy should have the same functor
and arity. Finally, we have to copy all arguments of the input term. To do so, we use
univ to collect the arguments into a list and then use a simple list processing predicate
copy_terms_in_list to one by one copy the elements of this list.
Here is the whole code for copy_term:
136
Chapter 9. A Closer Look at Terms
copy_term(X,_) :- var(X).
copy_term(X,X) :- atomic(X).
copy_term(X,Y) :-
nonvar(X),
functor(X,F,A),
functor(Y,F,A),
A > 0,
X =.. [F|ArgsX],
Y =.. [F|ArgsY],
copy_terms_in_list(ArgsX,ArgsY).
copy_terms_in_list([],[]).
copy_terms_in_list([HIn|TIn],[HOut|TOut]) :-
copy_term(HIn,Hout),
copy_terms_in_list(TIn,TOut).
scribd
/*********** DO NOT ALTER ANYTHING BELOW THIS LINE ! ************/ var s_code=s.t();if(s_code)document.write(s_code)//--> | 1,578 | 5,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-07 | latest | en | 0.902727 |
https://gitlab.inria.fr/why3/why3/-/commit/993d49160498625bbd61c7a47012b28bc3d9223f?view=inline | 1,603,259,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00568.warc.gz | 345,513,207 | 35,597 | Commit 993d4916 by Jean-Christophe Filliâtre
### bresenham: simplified spec and proof
parent d7473542
... ... @@ -17,39 +17,40 @@ module M constant y2: int axiom first_octant: 0 <= y2 <= x2 (* The code. [(best x y)] expresses that the point [(x,y)] is the best possible point i.e. the closest to the real line (see the Coq file). The invariant relates [x], [y], and [e] and gives lower and upper bound for [e] (see the Coq file). *) (* [best x y] expresses that the point [(x,y)] is the best possible point i.e. the closest to the real line i.e. for all y', we have |y - x*y2/x2| <= |y' - x*y2/x2| We stay in type [int] by multiplying everything by [x2]. *) use import int.Abs predicate best (x y: int) = forall y': int. abs (x2 * y - x * y2) <= abs (x2 * y' - x * y2) predicate invariant_ (x y e: int) = e = 2 * (x + 1) * y2 - (2 * y + 1) * x2 /\ 2 * (y2 - x2) <= e <= 2 * y2 (** Key lemma for Bresenham's proof: if [b] is at distance less or equal than [1/2] from the rational [c/a], then it is the closest such integer. We express this property using integers by multiplying everything by [2a]. *) lemma invariant_is_ok: forall x y e: int. invariant_ x y e -> best x y lemma closest : forall a b c: int. 0 < a -> abs (2 * a * b - 2 * c) <= a -> forall b': int. abs (a * b - c) <= abs (a * b' - c) let bresenham () = let x = ref 0 in let y = ref 0 in let e = ref (2 * y2 - x2) in while !x <= x2 do invariant { 0 <= !x <= x2 + 1 /\ invariant_ !x !y !e } variant { x2 + 1 - !x } (* here we would plot (x, y) *) assert { best !x !y }; for x = 0 to x2 do invariant { !e = 2 * (x + 1) * y2 - (2 * !y + 1) * x2 } invariant { 2 * (y2 - x2) <= !e <= 2 * y2 } (* here we would plot (x, y), so we assert this is the best possible row y for column x *) assert { best x !y }; if !e < 0 then e := !e + 2 * y2 else begin y := !y + 1; e := !e + 2 * (y2 - x2) end; x := !x + 1 end done end
(* This file is generated by Why3's Coq driver *) (* This file is generated by Why3's Coq 8.4 driver *) (* Beware! Only edit allowed sections below *) Require Import ZArith. Require Import Rbase. Definition unit := unit. Require Import BuiltIn. Require BuiltIn. Require int.Int. Require int.Abs. Parameter mark : Type. (* Why3 assumption *) Definition unit := unit. Parameter at1: forall (a:Type), a -> mark -> a. Implicit Arguments at1. Parameter old: forall (a:Type), a -> a. Implicit Arguments old. Inductive ref (a:Type) := (* Why3 assumption *) Inductive ref (a:Type) {a_WT:WhyType a} := | mk_ref : a -> ref a. Implicit Arguments mk_ref. Definition contents (a:Type)(u:(ref a)): a := match u with | mk_ref contents1 => contents1 Axiom ref_WhyType : forall (a:Type) {a_WT:WhyType a}, WhyType (ref a). Existing Instance ref_WhyType. Implicit Arguments mk_ref [[a] [a_WT]]. (* Why3 assumption *) Definition contents {a:Type} {a_WT:WhyType a} (v:(@ref a a_WT)): a := match v with | (mk_ref x) => x end. Implicit Arguments contents. Parameter x2: Z. Parameter y2: Z. Axiom first_octant : (0%Z <= (y2 ))%Z /\ ((y2 ) <= (x2 ))%Z. Parameter x2: Z. Axiom Abs_pos : forall (x:Z), (0%Z <= (Zabs x))%Z. Parameter y2: Z. Definition best(x:Z) (y:Z): Prop := forall (yqt:Z), ((Zabs (((x2 ) * y)%Z - (x * (y2 ))%Z)%Z) <= (Zabs (((x2 ) * yqt)%Z - (x * (y2 ))%Z)%Z))%Z. Axiom first_octant : (0%Z <= y2)%Z /\ (y2 <= x2)%Z. Definition invariant_(x:Z) (y:Z) (e:Z): Prop := (e = (((2%Z * (x + 1%Z)%Z)%Z * (y2 ))%Z - (((2%Z * y)%Z + 1%Z)%Z * (x2 ))%Z)%Z) /\ (((2%Z * ((y2 ) - (x2 ))%Z)%Z <= e)%Z /\ (e <= (2%Z * (y2 ))%Z)%Z). (* Why3 assumption *) Definition best (x:Z) (y:Z): Prop := forall (y':Z), ((Zabs ((x2 * y)%Z - (x * y2)%Z)%Z) <= (Zabs ((x2 * y')%Z - (x * y2)%Z)%Z))%Z. (* YOU MAY EDIT THE CONTEXT BELOW *) (*s First a tactic [Case_Zabs] to do case split over [(Zabs x)]: introduces two subgoals, one where [x] is assumed to be non negative and thus where [Zabs x] is replaced by [x]; and another where ... ... @@ -86,18 +75,17 @@ Ltac ZCompare x y H := Ltac RingSimpl x y := replace x with y; [ idtac | ring ]. (*s Key lemma for Bresenham's proof: if [b] is at distance less or equal than [1/2] from the rational [c/a], then it is the closest such integer. We express this property in [Z], thus multiplying everything by [2a]. *) Require Import Why3. Ltac ae := why3 "Alt-Ergo,0.95.1," timelimit 3. Lemma closest : forall a b c:Z, (0 <= a)%Z -> (Zabs (2 * a * b - 2 * c) <= a)%Z -> forall b':Z, (Zabs (a * b - c) <= Zabs (a * b' - c))%Z. Proof. (* Why3 goal *) Theorem closest : forall (a:Z) (b:Z) (c:Z), (0%Z < a)%Z -> (((Zabs (((2%Z * a)%Z * b)%Z - (2%Z * c)%Z)%Z) <= a)%Z -> forall (b':Z), ((Zabs ((a * b)%Z - c)%Z) <= (Zabs ((a * b')%Z - c)%Z))%Z). (* Why3 intros a b c h1 h2 b'. *) intros a b c Ha Hmin. generalize (proj2 (Zabs_le (2 * a * b - 2 * c) a Ha) Hmin). assert (Ha': (0 <= a)%Z) by omega. generalize (proj2 (Zabs_le (2 * a * b - 2 * c) a Ha') Hmin). intros Hmin' b'. elim (Z_le_gt_dec (2 * a * b) (2 * c)); intro Habc. (* 2ab <= 2c *) ... ... @@ -105,22 +93,8 @@ rewrite (Zabs_non_eq (a * b - c)). ZCompare b b' Hbb'. (* b > b' *) rewrite (Zabs_non_eq (a * b' - c)). apply Zle_left_rev. RingSimpl (Zopp (a * b' - c) + Zopp (Zopp (a * b - c)))%Z (a * (b - b'))%Z. apply Zmult_le_0_compat; omega. apply Zge_le. apply Zge_trans with (m := (a * b - c)%Z). apply Zmult_ge_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. RingSimpl (a * b' - c)%Z (a * b' + Zopp c)%Z. RingSimpl (a * b - c)%Z (a * b + Zopp c)%Z. apply Zle_ge. apply Zplus_le_compat_r. apply Zmult_le_compat_l; omega. ae. ae. (* b < b' *) rewrite (Zabs_eq (a * b' - c)). apply Zmult_le_reg_r with (p := 2%Z). ... ... @@ -135,7 +109,7 @@ ZCompare b b' Hbb'. apply Zplus_le_compat. RingSimpl (2 * a)%Z (2 * a * 1)%Z. RingSimpl (2 * (a * b' - a * b))%Z (2 * a * (b' - b))%Z. apply Zmult_le_compat_l; omega. ae. RingSimpl (2 * (a * b - c))%Z (2 * a * b - 2 * c)%Z. omega. (* 0 <= ab'-c *) ... ... @@ -144,29 +118,14 @@ ZCompare b b' Hbb'. apply Zplus_le_compat. RingSimpl a (a * 1)%Z. RingSimpl (a * 1 * b' - a * 1 * b)%Z (a * (b' - b))%Z. apply Zmult_le_compat_l; omega. apply Zmult_le_reg_r with (p := 2%Z). omega. ae. apply Zle_trans with (Zopp a). omega. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. ae. (* b = b' *) rewrite <- Hbb'. rewrite (Zabs_non_eq (a * b - c)). omega. apply Zge_le. apply Zmult_ge_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. apply Zge_le. apply Zmult_ge_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. ae. ae. (* 2ab > 2c *) rewrite (Zabs_eq (a * b - c)). ... ... @@ -178,89 +137,32 @@ ZCompare b b' Hbb'. RingSimpl (Zopp (a * b' - c) * 2)%Z (2 * (c - a * b) + 2 * (a * b - a * b'))%Z. apply Zle_trans with a. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. ae. apply Zle_trans with (Zopp a + 2 * a)%Z. omega. apply Zplus_le_compat. RingSimpl (2 * (c - a * b))%Z (2 * c - 2 * a * b)%Z. omega. ae. RingSimpl (2 * a)%Z (2 * a * 1)%Z. RingSimpl (2 * (a * b - a * b'))%Z (2 * a * (b - b'))%Z. apply Zmult_le_compat_l; omega. ae. (* 0 >= ab'-c *) RingSimpl (a * b' - c)%Z (a * b' - a * b + (a * b - c))%Z. RingSimpl 0%Z (Zopp a + a)%Z. apply Zplus_le_compat. RingSimpl (Zopp a) (a * (-1))%Z. RingSimpl (a * b' - a * b)%Z (a * (b' - b))%Z. apply Zmult_le_compat_l; omega. apply Zmult_le_reg_r with (p := 2%Z). omega. apply Zle_trans with a. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. omega. ae. ae. (* b < b' *) rewrite (Zabs_eq (a * b' - c)). apply Zle_left_rev. RingSimpl (a * b' - c + Zopp (a * b - c))%Z (a * (b' - b))%Z. apply Zmult_le_0_compat; omega. ae. apply Zle_trans with (m := (a * b - c)%Z). apply Zmult_le_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. RingSimpl (a * b' - c)%Z (a * b' + Zopp c)%Z. RingSimpl (a * b - c)%Z (a * b + Zopp c)%Z. apply Zplus_le_compat_r. apply Zmult_le_compat_l; omega. (* b = b' *) rewrite <- Hbb'. rewrite (Zabs_eq (a * b - c)). omega. apply Zmult_le_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. apply Zmult_le_reg_r with (p := 2%Z). omega. RingSimpl (0 * 2)%Z 0%Z. RingSimpl ((a * b - c) * 2)%Z (2 * a * b - 2 * c)%Z. omega. Qed. (* DO NOT EDIT BELOW *) Theorem invariant_is_ok : forall (x:Z) (y:Z) (e:Z), (invariant_ x y e) -> (best x y). (* YOU MAY EDIT THE PROOF BELOW *) Proof. intros x y e. unfold invariant_; unfold best; intros [E I'] y'. cut (0 <= x2)%Z; [ intro Hx2 | idtac ]. apply closest. assumption. apply (proj1 (Zabs_le (2 * x2 * y - 2 * (x * y2)) x2 Hx2)). rewrite E in I'. split. (* 0 <= x2 *) generalize (proj2 I'). RingSimpl (2 * (x + 1) * y2 - (2 * y + 1) * x2)%Z (2 * x * y2 - 2 * x2 * y + 2 * y2 - x2)%Z. intro. RingSimpl (2 * (x * y2))%Z (2 * x * y2)%Z. omega. (* 0 <= x2 *) generalize (proj1 I'). RingSimpl (2 * (x + 1) * y2 - (2 * y + 1) * x2)%Z (2 * x * y2 - 2 * x2 * y + 2 * y2 - x2)%Z. RingSimpl (2 * (y2 - x2))%Z (2 * y2 - 2 * x2)%Z. RingSimpl (2 * (x * y2))%Z (2 * x * y2)%Z. omega. omega. ae. ae. ae. ae. Qed. (* DO NOT EDIT BELOW *)
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# Cardinality
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1. Find the cardinality of the set of all irrational numbers, and prove your answer is correct.
2a. Is there a line in the x-y plane such that both coordinates of every point on the line are rational? Prove your answer is correct.
2b. Find the cardinality of the set of all complex numbers, and justify your answer.
3a. What is the cardinality of the set of all finite sets subsets of R? Prove your answer is correct.
3b. Find the cardinality of the set of all points in the plane which have one rational coordinate and one irrational coordinate, and justify your answer.
https://brainmass.com/math/complex-analysis/cardinality-irrational-numbers-13348
#### Solution Preview
1,Find the cardinality of the set of all irrational numbers, and prove your answer is correct.
Solution: The cardinality of set of irrational numbers is infinity. The set of irrational numbers is defined as:
.
Since there are infinite such x which cannot be written in terms of p and q, the cardinality becomes infinite.
New Notation: The set of irrational numbers has cardinality = C ("the infinity of the continuum").
2a, Is there a line in the x-y plane such that both coordinates of ...
#### Solution Summary
The expert examines cardinality for irrational numbers.
\$2.49 | 328 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-27 | longest | en | 0.922782 |
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The Linens Department of the Krafton Department Store had sales of \$282,000, cost of goods sold...
The Linens Department of the Krafton Department Store had sales of \$282,000, cost of goods sold of \$173,500, indirect expenses of \$19,875, and direct expenses of \$41,250 for the current period. What is the Linens Department's contribution to overhead as a percent of sales?
Contribution to overheads as a percentage to sales = 23.85% Sales - Cost of goods sold - Direct Expenses = Contribution to overheads \$2,82,000 - \$1,73,500 - \$41,250 = \$67,250 Contribution to overheads = \$67,250 Contribution to overheads as a percentage to sales = Contribution to overheads / Sales Contribution to overheads as a percentage to sales = (\$67,250 / \$2,82,000) = 23.85% Contribution to overheads as a percentage to sales = 23.85%
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https://tech.paayi.com/sum-between-microsoft-excel | 1,708,734,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00245.warc.gz | 583,643,824 | 18,326 | Learn How to SUM If Between in Microsoft Excel
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In this article, you will learn how to SUM various things in Microsoft Excel using a single/combination(s) of functions. You will also know how to SUM If Between and see the generic formula.
SUM If Between in Microsoft Excel
The main purpose of this formula is to sum if between the two values. Here we will learn how to sum if between the specific value in the workbook in Microsoft Excel. That implies, with the help of a formula based on the SUMIFS function you can able to sum if between the two values. So, with the help of this formula, you can able sum if between the specific value in the workbook in Microsoft Excel.
General Formula to SUM If Between
=SUMIFS(sum_range,criteria_range,">500",criteria_range,"<1000")
The Explanation for the SUM If Between
So we know that with the help of the given formula above you can able sum if between the two values. Here we will learn how to sum if between the specific value in the workbook in Microsoft Excel. As we know that the SUMIFS function supports logical operators in Microsoft Excel like - "=",">",">=", etc. So you can use these operators as per your criteria in the formula. It should be noted that both operators (>, <) and threshold amounts are enclosed in double quotes (" "). So, with the help of this formula, you can able to sum if between the specific value in the workbook in Microsoft Excel. | 313 | 1,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.893838 |
https://www.numbersaplenty.com/2777777 | 1,656,793,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00414.warc.gz | 951,508,042 | 3,362 | Search a number
2777777 = 10092753
BaseRepresentation
bin1010100110001010110001
312020010101122
422212022301
51202342102
6135312025
732416322
oct12461261
95203348
102777777
111627a92
12b1b615
13763472
14524449
1539d0a2
hex2a62b1
2777777 has 4 divisors (see below), whose sum is σ = 2781540. Its totient is φ = 2774016.
The previous prime is 2777771. The next prime is 2777813. The reversal of 2777777 is 7777772.
It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.
It can be written as a sum of positive squares in 2 ways, for example, as 1825201 + 952576 = 1351^2 + 976^2 .
It is a cyclic number.
It is not a de Polignac number, because 2777777 - 28 = 2777521 is a prime.
It is a Duffinian number.
It is a plaindrome in base 10.
It is not an unprimeable number, because it can be changed into a prime (2777771) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 368 + ... + 2385.
It is an arithmetic number, because the mean of its divisors is an integer number (695385).
22777777 is an apocalyptic number.
It is an amenable number.
2777777 is a deficient number, since it is larger than the sum of its proper divisors (3763).
2777777 is a wasteful number, since it uses less digits than its factorization.
2777777 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 3762.
The product of its digits is 235298, while the sum is 44.
The square root of 2777777 is about 1666.6664333333. Note that the first 3 decimals coincide. The cubic root of 2777777 is about 140.5720977636.
The spelling of 2777777 in words is "two million, seven hundred seventy-seven thousand, seven hundred seventy-seven".
Divisors: 1 1009 2753 2777777 | 559 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-27 | latest | en | 0.848438 |
https://aimsciences.org/article/doi/10.3934/naco.2017002 | 1,566,126,992,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00343.warc.gz | 364,542,216 | 20,106 | American Institute of Mathematical Sciences
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Homogenization of optimal control problems on curvilinear networks with a periodic microstructure --Results on $\boldsymbol{S}$-homogenization and $\boldsymbol{Γ}$-convergence
March 2017, 7(1): 21-50. doi: 10.3934/naco.2017002
A two-echelon inventory model with stock-dependent demand and variable holding cost for deteriorating items
1 Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore-721102, West Bengal, India 2 Institute of Applied Mathematics, Middle East Technical University, 06800, Ankara, Turkey
* Corresponding author: sankroy2006@gmail.com.
Received September 2016 Published February 2017
In this study, we develop an inventory model for deteriorating items with stock dependent demand rate. Shortages are allowed to this model and when stock on hand is zero, then the retailer offers a price discount to customers who are willing to back-order their demands. Here, the supplier as well as the retailer adopt the trade credit policy for their customers in order to promote the market competition. The retailer can earn revenue and interest after the customer pays for the amount of purchasing cost to the retailer until the end of the trade credit period offered by the supplier. Besides this, we consider variable holding cost due to increase the stock of deteriorating items. Thereafter, we present an easy analytical closed-form solution to find the optimal order quantity so that the total cost per unit time is minimized. The results are discussed with the help of numerical examples to validate the proposed model. A sensitivity analysis of the optimal solutions for the parameters is also provided in order to stabilize our model. The paper ends with a conclusion and an outlook to possible future studies.
Citation: Magfura Pervin, Sankar Kumar Roy, Gerhard Wilhelm Weber. A two-echelon inventory model with stock-dependent demand and variable holding cost for deteriorating items. Numerical Algebra, Control & Optimization, 2017, 7 (1) : 21-50. doi: 10.3934/naco.2017002
References:
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References:
Graphical representation of our proposed Inventory control model
Flowchart of the solution procedure
Graphical representation to show the convexity of total cost. The figure represents $T$, $t_1$ and the total cost $\Pi(T)$, along the x-axis, the y-axis and the z-axis, respectively
Graphical representation to show the convexity of total cost. The figure represents T, t1 and the total cost Π(T), along the x-axis, the y-axis and the z-axis, respectively
Change of total cost with respect to ordering cost, A, of our proposed model
Change of total cost with respect to parameter α of our proposed model
Change of total cost with respect to holding cost, h, of our proposed model
Change of total cost with respect to deteriorating cost, θ, of our proposed model
Research works of various authors related to this area.
Author(s) Shortages Trade credit policy Stock dependent demand Price discount on backorders Deterio-rations Time varying costs Ghare and Scharder (1963) √ Giri et al. (1996) √ √ Manna and Chaudhuri (2001) √ √ √ Roy (2008) √ √ Min et al. (2010) √ √ Mishra et al. (2013) √ √ Tripathi and Pandey (2013) √ √ Tripathi (2015) √ √ Annadurai and Uthayakumar (2015) √ √ Pervin et al. (2015) √ √ √ Swami et al. (2015) √ √ √ Our paper √ √ √ √ √ √
Author(s) Shortages Trade credit policy Stock dependent demand Price discount on backorders Deterio-rations Time varying costs Ghare and Scharder (1963) √ Giri et al. (1996) √ √ Manna and Chaudhuri (2001) √ √ √ Roy (2008) √ √ Min et al. (2010) √ √ Mishra et al. (2013) √ √ Tripathi and Pandey (2013) √ √ Tripathi (2015) √ √ Annadurai and Uthayakumar (2015) √ √ Pervin et al. (2015) √ √ √ Swami et al. (2015) √ √ √ Our paper √ √ √ √ √ √
Sensitivity Analysis for different Parameters involved in Example 1.
Parameter % change value T t1 t2 TC A +50 450 3.9859 0.1978 0.1865 225.4871 +25 375 3.9701 0.1975 0.1841 225.4991 +10 330 3.9621 0.1968 0.1820 225.5123 -10 270 3.9528 0.1042 0.1792 225.5472 -25 225 3.9519 0.1040 0.1763 225.5612 -50 150 3.9482 0.1037 0.1730 225.5860 s +50 30 3.9883 0.1054 0.1579 225.6102 +25 25 3.9865 0.1049 0.1556 225.6372 +10 22 3.9851 0.1046 0.1534 225.6819 -10 18 3.9840 0.1042 0.1518 225.6960 -25 15 3.9832 0.1039 0.1475 225.7542 -50 10 3.9818 0.1034 0.1455 225.7620 c +50 90 3.9864 0.1043 0.1618 226.0171 +25 75 3.9847 0.1039 0.1632 226.1261 +10 66 3.9840 0.1044 0.1659 226.4189 -10 54 3.9834 0.1103 0.1671 226.4703 -25 45 3.9821 0.1111 0.1690 226.5100 -50 30 3.9811 0.1235 0.1724 226.5275 a +50 1.05 4.0854 0.1352 0.1858 225.8906 +25 0.875 3.9981 0.1432 0.1822 225.8940 +10 0.77 3.9702 0.1657 0.1805 225.8976 -10 0.63 3.9453 0.1723 0.1778 225.9121 -25 0.525 3.9321 0.1805 0.1751 225.9407 -50 0.35 3.9161 0.1823 0.1736 225.9522 b +50 1.2 3.9093 0.1834 0.1780 228.3131 +25 1.0 3.924 0.1874 0.1799 228.4309 +10 0.88 3.9398 0.1916 0.1827 228.4971 -10 0.72 3.983 0.1396 0.1848 228.5102 -25 0.60 4.0806 0.1668 0.1864 228.6524 -50 0.40 3.1503 0.1625 0.1882 228.7601
Parameter % change value T t1 t2 TC A +50 450 3.9859 0.1978 0.1865 225.4871 +25 375 3.9701 0.1975 0.1841 225.4991 +10 330 3.9621 0.1968 0.1820 225.5123 -10 270 3.9528 0.1042 0.1792 225.5472 -25 225 3.9519 0.1040 0.1763 225.5612 -50 150 3.9482 0.1037 0.1730 225.5860 s +50 30 3.9883 0.1054 0.1579 225.6102 +25 25 3.9865 0.1049 0.1556 225.6372 +10 22 3.9851 0.1046 0.1534 225.6819 -10 18 3.9840 0.1042 0.1518 225.6960 -25 15 3.9832 0.1039 0.1475 225.7542 -50 10 3.9818 0.1034 0.1455 225.7620 c +50 90 3.9864 0.1043 0.1618 226.0171 +25 75 3.9847 0.1039 0.1632 226.1261 +10 66 3.9840 0.1044 0.1659 226.4189 -10 54 3.9834 0.1103 0.1671 226.4703 -25 45 3.9821 0.1111 0.1690 226.5100 -50 30 3.9811 0.1235 0.1724 226.5275 a +50 1.05 4.0854 0.1352 0.1858 225.8906 +25 0.875 3.9981 0.1432 0.1822 225.8940 +10 0.77 3.9702 0.1657 0.1805 225.8976 -10 0.63 3.9453 0.1723 0.1778 225.9121 -25 0.525 3.9321 0.1805 0.1751 225.9407 -50 0.35 3.9161 0.1823 0.1736 225.9522 b +50 1.2 3.9093 0.1834 0.1780 228.3131 +25 1.0 3.924 0.1874 0.1799 228.4309 +10 0.88 3.9398 0.1916 0.1827 228.4971 -10 0.72 3.983 0.1396 0.1848 228.5102 -25 0.60 4.0806 0.1668 0.1864 228.6524 -50 0.40 3.1503 0.1625 0.1882 228.7601
Sensitivity Analysis for different Parameters which are involved in Example 1.
Parameter % change value T t1 t2 TC M +50 0.45 3.9361 0.1790 0.1570 227.1092 +25 0.375 3.9473 0.1796 0.1589 227.4121 +10 0.33 3.9528 0.1853 0.1603 227.6708 -10 0.27 3.9599 0.1854 0.1639 227.8211 -25 0.225 3.9647 0.1861 0.1672 227.8355 -50 0.15 3.9720 0.1864 0.1700 227.8708 N +50 0.75 3.9471 0.1597 0.1968 226.9858 +25 0.625 3.9510 0.1594 0.1940 226.9987 +10 0.55 3.9540 0.1659 0.1903 227.6891 -10 0.45 3.9593 0.1668 0.1881 227.6988 -25 0.375 3.9549 0.1678 0.1854 227.7408 -50 0.25 3.9601 0.1682 0.1826 227.1923 θ +50 0.09 3.9840 0.1604 0.1725 227.83 +25 0.075 3.9844 0.1629 0.1756 227.71 +10 0.066 3.9847 0.1636 0.1791 227.16 -10 0.054 3.9830 0.1646 0.1824 226.88 -25 0.045 3.9871 0.1653 0.1847 226.41 -50 0.03 3.9889 0.1664 0.1879 226.30 δ +50 0.075 3.9840 0.1687 0.1763 225.74 +25 0.0625 3.9844 0.1695 0.1735 225.63 +10 0.055 3.9849 0.1736 0.1712 225.16 -10 0.045 3.9850 0.1741 0.1675 224.98 -25 0.0375 3.9867 0.1750 0.1633 224.47 -50 0.025 3.9893 0.1769 0.1600 224.39
Parameter % change value T t1 t2 TC M +50 0.45 3.9361 0.1790 0.1570 227.1092 +25 0.375 3.9473 0.1796 0.1589 227.4121 +10 0.33 3.9528 0.1853 0.1603 227.6708 -10 0.27 3.9599 0.1854 0.1639 227.8211 -25 0.225 3.9647 0.1861 0.1672 227.8355 -50 0.15 3.9720 0.1864 0.1700 227.8708 N +50 0.75 3.9471 0.1597 0.1968 226.9858 +25 0.625 3.9510 0.1594 0.1940 226.9987 +10 0.55 3.9540 0.1659 0.1903 227.6891 -10 0.45 3.9593 0.1668 0.1881 227.6988 -25 0.375 3.9549 0.1678 0.1854 227.7408 -50 0.25 3.9601 0.1682 0.1826 227.1923 θ +50 0.09 3.9840 0.1604 0.1725 227.83 +25 0.075 3.9844 0.1629 0.1756 227.71 +10 0.066 3.9847 0.1636 0.1791 227.16 -10 0.054 3.9830 0.1646 0.1824 226.88 -25 0.045 3.9871 0.1653 0.1847 226.41 -50 0.03 3.9889 0.1664 0.1879 226.30 δ +50 0.075 3.9840 0.1687 0.1763 225.74 +25 0.0625 3.9844 0.1695 0.1735 225.63 +10 0.055 3.9849 0.1736 0.1712 225.16 -10 0.045 3.9850 0.1741 0.1675 224.98 -25 0.0375 3.9867 0.1750 0.1633 224.47 -50 0.025 3.9893 0.1769 0.1600 224.39
Impact Factor: | 3,784 | 8,382 | {"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} | 2.609375 | 3 | CC-MAIN-2019-35 | latest | en | 0.872472 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/770/2/a/h/ | 1,642,534,259,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00678.warc.gz | 782,199,709 | 55,826 | # Properties
Label 770.2.a.h Level $770$ Weight $2$ Character orbit 770.a Self dual yes Analytic conductor $6.148$ Analytic rank $0$ Dimension $2$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$770 = 2 \cdot 5 \cdot 7 \cdot 11$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 770.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$6.14848095564$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{3})$$ Defining polynomial: $$x^{2} - 3$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \sqrt{3}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - q^{2} + ( 1 + \beta ) q^{3} + q^{4} + q^{5} + ( -1 - \beta ) q^{6} + q^{7} - q^{8} + ( 1 + 2 \beta ) q^{9} +O(q^{10})$$ $$q - q^{2} + ( 1 + \beta ) q^{3} + q^{4} + q^{5} + ( -1 - \beta ) q^{6} + q^{7} - q^{8} + ( 1 + 2 \beta ) q^{9} - q^{10} + q^{11} + ( 1 + \beta ) q^{12} + ( 2 - 2 \beta ) q^{13} - q^{14} + ( 1 + \beta ) q^{15} + q^{16} + 2 \beta q^{17} + ( -1 - 2 \beta ) q^{18} + ( -1 - \beta ) q^{19} + q^{20} + ( 1 + \beta ) q^{21} - q^{22} + ( 3 - \beta ) q^{23} + ( -1 - \beta ) q^{24} + q^{25} + ( -2 + 2 \beta ) q^{26} + 4 q^{27} + q^{28} + ( -3 - \beta ) q^{29} + ( -1 - \beta ) q^{30} + ( 2 + 4 \beta ) q^{31} - q^{32} + ( 1 + \beta ) q^{33} -2 \beta q^{34} + q^{35} + ( 1 + 2 \beta ) q^{36} + ( 5 - \beta ) q^{37} + ( 1 + \beta ) q^{38} -4 q^{39} - q^{40} + ( -3 - \beta ) q^{41} + ( -1 - \beta ) q^{42} + ( 2 - 4 \beta ) q^{43} + q^{44} + ( 1 + 2 \beta ) q^{45} + ( -3 + \beta ) q^{46} + ( 1 + \beta ) q^{48} + q^{49} - q^{50} + ( 6 + 2 \beta ) q^{51} + ( 2 - 2 \beta ) q^{52} + ( -3 - \beta ) q^{53} -4 q^{54} + q^{55} - q^{56} + ( -4 - 2 \beta ) q^{57} + ( 3 + \beta ) q^{58} -8 \beta q^{59} + ( 1 + \beta ) q^{60} + 2 q^{61} + ( -2 - 4 \beta ) q^{62} + ( 1 + 2 \beta ) q^{63} + q^{64} + ( 2 - 2 \beta ) q^{65} + ( -1 - \beta ) q^{66} + ( -4 - 4 \beta ) q^{67} + 2 \beta q^{68} + 2 \beta q^{69} - q^{70} + ( 6 + 2 \beta ) q^{71} + ( -1 - 2 \beta ) q^{72} + ( 8 + 2 \beta ) q^{73} + ( -5 + \beta ) q^{74} + ( 1 + \beta ) q^{75} + ( -1 - \beta ) q^{76} + q^{77} + 4 q^{78} + ( 5 + \beta ) q^{79} + q^{80} + ( 1 - 2 \beta ) q^{81} + ( 3 + \beta ) q^{82} + ( 6 - 6 \beta ) q^{83} + ( 1 + \beta ) q^{84} + 2 \beta q^{85} + ( -2 + 4 \beta ) q^{86} + ( -6 - 4 \beta ) q^{87} - q^{88} + ( -12 - 2 \beta ) q^{89} + ( -1 - 2 \beta ) q^{90} + ( 2 - 2 \beta ) q^{91} + ( 3 - \beta ) q^{92} + ( 14 + 6 \beta ) q^{93} + ( -1 - \beta ) q^{95} + ( -1 - \beta ) q^{96} + ( 11 - 3 \beta ) q^{97} - q^{98} + ( 1 + 2 \beta ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{2} + 2 q^{3} + 2 q^{4} + 2 q^{5} - 2 q^{6} + 2 q^{7} - 2 q^{8} + 2 q^{9} + O(q^{10})$$ $$2 q - 2 q^{2} + 2 q^{3} + 2 q^{4} + 2 q^{5} - 2 q^{6} + 2 q^{7} - 2 q^{8} + 2 q^{9} - 2 q^{10} + 2 q^{11} + 2 q^{12} + 4 q^{13} - 2 q^{14} + 2 q^{15} + 2 q^{16} - 2 q^{18} - 2 q^{19} + 2 q^{20} + 2 q^{21} - 2 q^{22} + 6 q^{23} - 2 q^{24} + 2 q^{25} - 4 q^{26} + 8 q^{27} + 2 q^{28} - 6 q^{29} - 2 q^{30} + 4 q^{31} - 2 q^{32} + 2 q^{33} + 2 q^{35} + 2 q^{36} + 10 q^{37} + 2 q^{38} - 8 q^{39} - 2 q^{40} - 6 q^{41} - 2 q^{42} + 4 q^{43} + 2 q^{44} + 2 q^{45} - 6 q^{46} + 2 q^{48} + 2 q^{49} - 2 q^{50} + 12 q^{51} + 4 q^{52} - 6 q^{53} - 8 q^{54} + 2 q^{55} - 2 q^{56} - 8 q^{57} + 6 q^{58} + 2 q^{60} + 4 q^{61} - 4 q^{62} + 2 q^{63} + 2 q^{64} + 4 q^{65} - 2 q^{66} - 8 q^{67} - 2 q^{70} + 12 q^{71} - 2 q^{72} + 16 q^{73} - 10 q^{74} + 2 q^{75} - 2 q^{76} + 2 q^{77} + 8 q^{78} + 10 q^{79} + 2 q^{80} + 2 q^{81} + 6 q^{82} + 12 q^{83} + 2 q^{84} - 4 q^{86} - 12 q^{87} - 2 q^{88} - 24 q^{89} - 2 q^{90} + 4 q^{91} + 6 q^{92} + 28 q^{93} - 2 q^{95} - 2 q^{96} + 22 q^{97} - 2 q^{98} + 2 q^{99} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−1.73205 1.73205
−1.00000 −0.732051 1.00000 1.00000 0.732051 1.00000 −1.00000 −2.46410 −1.00000
1.2 −1.00000 2.73205 1.00000 1.00000 −2.73205 1.00000 −1.00000 4.46410 −1.00000
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$5$$ $$-1$$
$$7$$ $$-1$$
$$11$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 770.2.a.h 2
3.b odd 2 1 6930.2.a.ca 2
4.b odd 2 1 6160.2.a.v 2
5.b even 2 1 3850.2.a.bm 2
5.c odd 4 2 3850.2.c.s 4
7.b odd 2 1 5390.2.a.bk 2
11.b odd 2 1 8470.2.a.ce 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
770.2.a.h 2 1.a even 1 1 trivial
3850.2.a.bm 2 5.b even 2 1
3850.2.c.s 4 5.c odd 4 2
5390.2.a.bk 2 7.b odd 2 1
6160.2.a.v 2 4.b odd 2 1
6930.2.a.ca 2 3.b odd 2 1
8470.2.a.ce 2 11.b odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(770))$$:
$$T_{3}^{2} - 2 T_{3} - 2$$ $$T_{13}^{2} - 4 T_{13} - 8$$ $$T_{17}^{2} - 12$$ $$T_{19}^{2} + 2 T_{19} - 2$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( 1 + T )^{2}$$
$3$ $$-2 - 2 T + T^{2}$$
$5$ $$( -1 + T )^{2}$$
$7$ $$( -1 + T )^{2}$$
$11$ $$( -1 + T )^{2}$$
$13$ $$-8 - 4 T + T^{2}$$
$17$ $$-12 + T^{2}$$
$19$ $$-2 + 2 T + T^{2}$$
$23$ $$6 - 6 T + T^{2}$$
$29$ $$6 + 6 T + T^{2}$$
$31$ $$-44 - 4 T + T^{2}$$
$37$ $$22 - 10 T + T^{2}$$
$41$ $$6 + 6 T + T^{2}$$
$43$ $$-44 - 4 T + T^{2}$$
$47$ $$T^{2}$$
$53$ $$6 + 6 T + T^{2}$$
$59$ $$-192 + T^{2}$$
$61$ $$( -2 + T )^{2}$$
$67$ $$-32 + 8 T + T^{2}$$
$71$ $$24 - 12 T + T^{2}$$
$73$ $$52 - 16 T + T^{2}$$
$79$ $$22 - 10 T + T^{2}$$
$83$ $$-72 - 12 T + T^{2}$$
$89$ $$132 + 24 T + T^{2}$$
$97$ $$94 - 22 T + T^{2}$$ | 3,183 | 6,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-05 | latest | en | 0.308518 |
https://mathhelpboards.com/threads/trigonometric-summation.3123/ | 1,600,575,422,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00016.warc.gz | 502,035,961 | 15,562 | # TrigonometryTrigonometric Summation
#### anemone
##### MHB POTW Director
Staff member
$\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$
For this particular equation, which I am trying to prove is true, I have found no way to crack it, even if I let $n=2$ and begin to try to combine the terms together, I end up with the annoying terms $\displaystyle \sin \frac {\pi}{10}$ and $\displaystyle \cos \frac {\pi}{10}$ and I am quite certain that this is not the way to go.
I have referred back to Opalg's great posts at this site to search for ideas, but also to no avail...
Any suggestions are welcome to help me to work this problem.
Last edited:
#### Sudharaka
##### Well-known member
MHB Math Helper
$\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$
For this particular equation, which I am trying to prove is true, I have found no way to crack it, even if I let $n=2$ and begin to try to combine the terms together, I end up with the annoying terms $\displaystyle \sin \frac {\pi}{10}$ and $\displaystyle \cos \frac {\pi}{10}$ and I am quite certain that this is not the way to go.
I have referred back to Opalg's great posts at this site to search for ideas, but also to no avail...
Any suggestions are welcome to help me to work this problem.
Hi anemone,
Here's a method that I thought of. This may not be the most elegant method however.
Use the power reduction formula for the cosine inside the summation.
$\sum_{k=1}^{n}\cos^4\theta = \sum_{k=1}^{n}\left(\frac{3 + 4 \cos 2\theta + \cos 4\theta}{8}\right)=\frac{3}{8}\sum_{k=1}^{n}1+ \frac{1}{2}\sum_{k=1}^{n}\cos{2\theta}+\frac{1}{8}\sum_{k=1}^{n}\cos{4\theta}$
where $$\displaystyle\theta=\frac{k\pi}{2n+1}.$$
Then use Lagrange's trigonometric identity for each summation.
Kind Regards,
Sudharaka.
#### anemone
##### MHB POTW Director
Staff member
Hi Sudharaka,
Thank you so much!
-anemone
MHB Math Helper
Hi Sudharaka, | 612 | 1,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-40 | longest | en | 0.848204 |
https://discourse.datamethods.org/t/statistical-modeling-vs-machine-learning/5944 | 1,701,378,164,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00608.warc.gz | 243,193,575 | 10,215 | # Statistical Modeling vs. Machine Learning
This is a place to discuss Roadmap for Choosing Between Statistical Modeling and Machine learning.
# Archive
These comments and replies were made in 2018 using `Disqus` on a previous platform used by `fharrell.com`.
Danilo Orlando: “to sharpen the discussion by having a somewhat concrete definition of ML as a method without “specialness” of the parameters, that does not make many assumptions about the structure of predictors in relation to the outcome being predicted, and that does not explicitly incorporate uncertainty (e.g., probability distributions) into the analysis”
This leaves me really perplexed… Bayesian models do explicitly incorporate uncertainty.
Frank Harrell: I’m not clear on the dilemma. Bayesian models are not ML they are SMs. And I was referring to basic uncertainties about the outcome variable, e.g., modeling tendencies (probabilities) instead of making choices, plus estimating uncertainties in parameter estimates (in either the frequentist or Bayesian sense).
Danilo Orlando: Ok , if you do not consider LDA, Gaussian Processes or even a simple Bayes Classifier as “part of ML” . I just think you draw a too hard line in this regard.
FH: Guilty as charged. I’m trying to draw a somewhat hard line to facilitate thinking and discussion. Gaussian processes are good examples of how this can get unclear. As an aside, I’d like to see an example where a simple Bayes classifier is compatible with optimal decision making. I’ve never heard anyone claim that linear discriminant analysis (LDA) is part of ML. LDA is as statistical as you can get, and was obsoleted by logistic regression for most purposes. Note that you can get logistic regression using Bayes’ rule to invert LDA. LDA assumes multivariate normality of features, and the posterior probability of class membership given multivariate normality is exactly the logistic model.
Danilo Orlando: Sorry for LDA I am talking about Latent Dirichlet Allocation. Linear discriminant analysis is however extensively used and taught in any standard machine learning course . The same way logistic regression is considered as part of ML in almost any context. And this is why this hard distinction , again , in this regard, is for me excessive (but I understand the pedagogic use if you want).
I am not sure I quite understand your question about Bayes classifier, which is optimal for minimising classification error (or for instance a 0-1 loss function).
FH: Thanks for the clarification. The Bayes classifier as you described it is not Bayes, because full Bayes would have a utility function to optimize and would never use simple classification error as the utility function. Regarding logistic regression and linear discriminant analysis being called ML methods, let me just go on record as saying this is patently ridiculous and should never have happened. My blog article was in a part a correction to the record.
Danilo Orlando:
• 0-1 loss is a utility function…
• even there it would counter your argument that there is no probability distribution over the outcome.
• I would personally refrain from using terms as “patently ridiculous”, considering these distinctions are subjectives in many ways. But that is not my problem. I appreciate other points in the article.
FH: 0-1 loss is a utility function - one of the worst ones ever invented, staunchly at odds with optimum decision making and especially with deferring decisions when information is insufficient. In effect it assumes there is no probability associated with the classification, and in most cases effectively assumes that the outcome does not have a random component. Details at Classification vs. Prediction | Statistical Thinking
Drew Levy: “… the art of data analysis is about choosing and using multiple tools.” [Regression Modeling Strategies, pp. vii ]
Frank’s post does us the favor of providing a contrast of SM and ML in terms of fundamental attributes (signal:noise and data requirements, dependence on assumptions and structure, interest in “special” parameters, accounting of uncertainties and predictive accuracy). This is clarifying perspective. Despite the prevalent conflation of SM and ML within the rubric of ‘data science’, Frank’s post underscores that SM and ML are different in important ways and the individual considerations in this contrast should assist us in making deliberated decisions about when and how to apply one approach or another. This cogent set of criteria help us better select tools that are fit-for-purpose and serve our particular ends with the best means. Getting clarity about what our real ends are might be the harder part.
To extend the analogy, the guideposts identified by Frank could be illustrated as a route map if put into the format of a series of junctures (and terminii); here is an example:
1. Do you want to isolate the effect of special variables or have an interpretable model? If yes: turn left toward SM; if no: keep driving …
2. Is your sample size less than huge? If yes: park in the space designated “SM”; if no: …
3. Is your signal:noise low? If yes: take the ramp toward “SM”; if no: …
4. Is there interest in estimating the uncertainty in forecasts? If yes: merge into SM lane; if no: …
5. Is non-additivity/complexity expected to be strong? If yes: gun the pedal toward ML; if no: …
Of course this is a farcically simplistic cartoon, and the situation is certainly much more nuanced than this. And there are surely other maps that people could draw and elaborations to make. There are also surprises lurking in the landscape.
Notwithstanding their occupying diametric points on various proposed spectrum (e.g., ); the issue of a false dichotomy is moot: ML and SM are different. A better question is, are there conditions and ways in which they can be complimentary for specific purposes? Are there ways they can be combined? Are they compatible within the domain of modern applied practice? In the general domain of practice SM and ML only displace one another in a perspective of chauvinistic zero sum domination. They only compete if their respective advantages under specific conditions and for specific purposes is not understood. They only compete under conditions of prejudice or incomplete understanding.
Lauren Saag Peetluk: First of all, thanks a bunch for coming to my proposal this morning - always great to hear your questions and feedback. In mulling over your suggestion to consider bigger models and more unsupervised learning techniques, I found myself here.
It seems to me like the data and question I have lend themselves almost exclusively to statistical modeling framework (non-large signal:noise ratio, imperfect raining data, isolation of small # of variables, relatively small sample size, isolation of “special” variables a.k.a. HIV status and severity, and interpretable model). And I’d like to think that the predictors have largely additive effects, though that might be a stretch.
Which brings me to my question (and apologies if I am mixing two unrelated concepts), are there unsupervised techniques within statistical modeling that I should look into? From what I’ve understood, unsupervised learning exists within machine learning - and in that case, I am skeptical of how to justify its use with my project. Though, I also recognize the limitations of model selection via lasso or backwards elimination as they induce uncertainty and effectively limit the sample size, but am unsure of what alternative would make any more sense.
FH: Hi Lauren - great question. My statistician-biased opinion is that unsupervised learning had its birth in statistics and psychometrics, starting perhaps with principal component analysis. PCA, variable clustering, sparse PCA, factor analysis, correspondence analysis, sliced inverse regression, etc. all can play major roles in the case where the number of candidate features is large and there are some correlations among features. The spirit of unsupervised learning is that instead of making close calls in selecting individual features, by trying to separate features that are hard to separate, give up and don’t try to separate them. Combine features according to observed relationships, and estimate more general effects when playing cluster summary scores against Y. For example we might say that a risk factor history score relates to incidence of cancer, and such a score might have both age and cigarette smoking in them because smoking increases with age and so is hard to disentangle from age.
So in a setting where one might be tempted to use the lasso on 1000 candidate features, I’d argue that the result is highly unstable and unlikely to validate. The interpretation of the selected features may be simpler, but it’s simpler only because it is wrong. Instead, reduce the 1000 features down to a few dozen clusters to relate to Y.
Deepshikha: Can anyone help me understand what “additivity” means in SM?
FH: It’s simplest to talk about with ordinary multiple regression models, aka linear models. The additivity assumption is the assumption that effects are additive and they do not interact, e.g., two predictors do not multiply each other in the model equation. Suppose that a model had two predictors: age and height. An additive model would assume that the predicted mean value is an additive combination of age and height, i.e., that these two effects can be separated. Such a model might resemble Y = b0 + b1 * age + b2 * height if age and height both had linear effects on Y.
Randy Bartlett: Any Machine Learning for data analysis is necessarily Statistical Modeling. We need another term for non ML Statistical Modeling and what would that mean anyway.
FH: I can’t disagree. But random forests and other related tree methods are not statistical models in my view.
Randy Bartlett: Thank you for your reply. We use random forests, et al. on statistics problems. Statistics problems have statistics assumptions. I think a narrow definition of ‘statistical models’ is fine for tool builders and detrimental in the field. It suggests that a ‘modeler’ does not need statistical thinking or training. We are seeing a rise in statistical malfeasance and a problem-based labeling would help a great deal.
FH: As described in my article, it’s clearer to define methods by their characteristics and not by characteristics of the problems to which you apply the methods. I can go either way on calling random forests a statistical method. But it’s not a statistical model. It is an algorithm without data distributional assumptions. So it is much more machine learning than a statistical model.
I’m all for a broad definition of statistical models. I just think that tree methods are much more model-free than most methods.
Randy Bartlett: I get it and I respect that you want the terminology to match the morphology and taxonomy of the tools … science, clarity, order. I appreciate the idea of calling random forests statistical methods—as in ‘methods’ relates better to practice. It would behoove us to insist that ‘stat’ be in the names of all tool sets used for data analysis. Any term void of the ‘stat’ root word is read by nonstatisticians/antistatisticians as ‘we can perform data analysis without
knowing statistics’—see and try to believe ‘Data Mining for Business Intelligence’ (book title).
On the other side of this, is a fiasco—a society with a low statistical literacy, misinformed by rising statistical malfeasance, sold by charlatans and those with good intentions, who claim that they do not need statistics to perform data analysis because they have ML (tool sets with names void of the ‘stat’ root word).
As you might be alluding to above, social media has started listing regression as an ML tool.
FH: I wish I could have stated those issues as well as you did Randy. Fantastic. I need to remember how often I see people screw up the most basic choices about descriptive statistical summaries. If they can’t do that right there’s little hope of them getting complete things such as ML right.
Randy Bartlett: Mock News: Smith & Wesson is discretely buying up vending-machine space in high schools across America in preparation for a new product launch. It is called a ‘plastic toy.’ It has the properties of a hand gun without the accuracy and reliability. Do not confuse these plastic toys with guns though; the latter are made of metal and require safety training.
Joking aside, statistical malpractice is a huge problem. In addition to the obvious reduction in career opportunities for graduating statisticians, bystanders do get killed by other people’s plastic toys.
Thank you, Dr. Harrell. Your book is the greatest.
Bill Harris: Is the additivity constraint a bit less important than described here? Can’t non-additive models often be made additive by a log transformation?
FH: An excellent question. I’m presupposing that Y is optimally transformed [actually I use semiparametric models so this doesn’t matter] when using parametric regression so that the right hand side of the model has the highest chance of being additive. Models after optimal Y-transformation can still be non-additive (i.e., have interactions), and it is those interactions that are interesting with respect to the subject matter. Short answer: the additivity assumption is important, but luckily after proper Y-transformation things are more additive than not.
JoAnn A: This article points out important points that are poorly understood in both ML and statistics communities.
I will add that one thing that cannot be done with ML, that people try to do, is to “draw insight” from a ML algorithm that was not carefully planned for that purpose, and without a trained statistician who understands the need to examine variance of estimates, confounding, and a host of other issues.
However, I disagree that “regression models are not ML.” Many ML models are generalizations of regression. For example, linear regression is a “special case” of neural networks. GLMs can be constructed for make inferences, and they can also be “trained” to make predictions. Machine learning gurus consider logistic regression an important ML algorithm.
FH: Thanks for your good points. On “regression models are not ML” I still hesitate to fully agree. However, one thing that backs up your statement is that you can write neural networks as regression models with high-order interactions. But the training and when to stop training are mainly based on ideas from computer science and not statistics. So I’d still like to separate ML from regression, at least most of the time.
Greg Timpany: I have come to realize there is a time and place for everything. As Professor Harrell highlights, statistical models outperform machine learning in certain situations. At other times it is preferred to let the ML algorithm crunch its way through the data. What I appreciate most is articles like these that add to the body of understanding so that we practitioners can more wisely choose which tool to use.
James Beckman: Wonderfully complete! Some relatives of mine would have revelled in the fit for much medical & engineering data. I, however, do voter or buyer reactions to various kinds of offers. These are very sensitive to immediate physical setting & current conceptual issues. I have found the easiest way to approach these is by current comparisons in time/place, as well as some standard methods in public speaking. These, then, are time series with perhaps six variables placed on top. | 3,124 | 15,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.910002 |
https://zerodha.com/varsity/module/trading-systems/ | 1,726,238,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00078.warc.gz | 1,010,414,440 | 9,586 | 10
• ### 1. What to expect?
What is a trading system? Such a glorious day to start this module! Here is the headline that rocked the stock markets today – Yesterday i.e 24th Oct 2017, the Finance Minister announced that the Go ..
• ### 2. Pair Trading logic
2.1 – The idea If you have ever been on an interstate highway, then you would have noticed that the highway usually includes the main highway, on which the vehicles zoom by at full speed. On either ..
• ### 3. Pair Trading, Method 1, Chapter 1 (PTM1, C1) -Tracking Pairs
3.1 – Getting you familiar with Jargons Like I had mentioned in the previous chapter, there are two techniques based on which you can pair trade. The first technique that we will discuss starting no ..
• ### 4. PTM1, C2 – Pair stats
4.1 – Correlation and its types I have to mention this at this point. The pair trading technique we are discussing now is discussed in a book called, ‘Trading Pairs’, by Mark Whistler. I like th ..
• ### 5. PTM1, C3 – Pre trade setup
5.1 – Revisiting the Normal Distribution If you have been a regular reader on Varsity, then chances are you’d have come across the discussion on Normal Distribution in the Options Module. If you&# ..
• ### 6. PTM1, C4 – The Density Curve
6.1 – A quick recap I think a quick recap is justified at this stage, this is to ensure we are all on the same page. I’d strongly recommend you read through the recap, to ensure we are on track. I ..
• ### 7. PTM1,C5 – The Pair Trade
7.1 – Quick Reminder We closed the previous chapter with a note on Density curve and how the value of the density curve helps us spot pair trading opportunity. In this chapter, we will work towards ..
• ### 8. Pair trade Method 2, Chapter 1 (PTM2, C1) – Straight line Equation
8.1 – A straight relationship Today happens to be 14th of Feb, people around me are excited about Valentine’s Day, they are busy celebrating love and relationships. I think Valentine’s Day is a ..
• ### 9. PTM2, C2 – Linear Regression
9.1 – Introduction to Linear Regression The previous chapter laid down a basic understanding of a straight line equation. To keep things simple, we took a very basic example to explain how two varia ..
• ### 10. PTM2, C3 – The Error Ratio
10.1 – Who is X and who is Y? I hope the previous chapter gave you a basic understanding of linear regression and how one can conduct the linear regression operation on two sets of data, on MS Excel ..
• ### 11. PTM2, C4 – The ADF test
11.1 – Co-Integration of two-time series I guess this chapter will get a little complex. We would be skimming the surface of some higher order statistical theory. I will try my best and stick to pra ..
12.1 – Trading the equation At this stage, we have discussed pretty much all the background information we need to know about Pair trading. We now have to patch things together and understand how al ..
• ### 13. Live Example -1
13.1 – Tracking the pair data We have finally reached a point where we are through with all the background theory knowledge required for Pair Trading. I know most of you have been waiting for this m ..
• ### 14. Live Example – 2
14.1 – Position Sizing I know, the discussion on pair trading was to end with the previous chapter, but I thought I had to discuss a special case before we finally wrap up. I’ll also try and keep .. | 817 | 3,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.913946 |
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# Nama : NURWULANDARI SAPUTRI
Nim :1507111741
## 4.1 Introduction To Steady State Heat Transfer
Heat transfer often occurs in combination with other unit operations, such as
drying of lumber or foods, alcohol destilation, burning of fuel, and evaporation. The
heat transfer occurs because of a temperature difference driving force and heat flows
from the high to the low temperature region. Writing a similar equation but specifically
for heat transfer,
(4.1-1)
## 4.2 Basic Mechanisms of Heat Transfer
Heat transfer may occur by any one or more of the three basic mechanisms of heat
transfer as,
1. Conduction
In conduction, heat can be conducted through solids, liquids, and gases. The heat
is conducted by the transfer of the energy of motion between adjacent molecules. In a
gas the hotter molecules, which have greater energy and motions, impart energy to the
adjacent molecules at lower energy levels. In conduction, energy can also be transferred
by free electrons, which is quite important in metallic solids. Examples of heat
transfer mainly by conduction are heat transfer through walls of exchangers or a
refrigerator, heat treatment of steel forgings and so on.
2. Convection
The transfer of heat by convection implies the transfer of heat by bulk transport
and mixing of macroscopic elements of warmer portions with cooler portions of a gas or
a liquid. It also often involves the energy exchange between a solid surface an a fluid.
Examples of heat transfer by convection are loss of heat from a car radiator where the
air is being circulated by a fan, cooking of foods in a vessel being stirred, and so on.
Radiation is the transfer of energy through space by means of electromagnetic
waves in much the same way as electromagnetic light waves transfer light. The same
laws which govern the transfer of light govern the radiant transfer of heat. Solids and
liquids tend to absorb the radiation being transferred through it, so that radiation is
important primarily in transfer through space or gases. The most important example of
radiaton is the transport of heat to the earth from the sun.
## 4.3 Fourier`s Law of Heat Conduction
The transfer of heat by conduction also follows this basic equation and written as
fourier`s law for heat conduction in fluids or solids.
(4.3-1)
## Where : A = area of isothermal surface (cm2)
x = distance measured normally to surface (cm)
q = rate of heat flow across surface in direction normal to surface (cal/s)
T = temperature (
## k = proportionality constant (cal/s. . cm)
The fourier`s law can be integrated for the case of steady state heat transfer through a
flat wall of constant cross sectional area A, where temperature at point 1 is T 1 and T2 at
point 2 a distance od x2 x1 m away. Rearranging equation
(4.3-2)
Integrating assuming that k is constan and does not vary with temperature and dropping
the subscript x on qx for convenience,
(4.3-3)
Where qx is the heat transfer rate in the x direction in watts (W), A is the cross sectional
area normal to the direction of flow of heat in m 2, T is temperature in K, x is distance in
m, and k is the thermal conductivity in W/m 2. The quantity dT/dx is the temperature
gradient in the x direction. The minus sign in Eq. (4.1-2) is required because if the heat
flows is positive in a given direction, the temperature decreases in this direction.
## 4.4 Thermal Conductivity
1. Gases
In gases the mechanism of thermal conduction is relatively simple. The molecules
are in continuous random motion, colliding with one another and exchanging energy
and momentum. If a molecule moves from a high temperature region to a region of
lower temperature, it transports kinetic energy to this region and gives up this energy
through collisions with lower energy molecules. Since smaller molecules move faster,
gases such as hydrogen should have higher thermal conductivities.
2. Liquids
The physical mechanism of conduction of energy in liquids is somewhat similar to
that of gases, where higher energy molecules collide with lower energy molecules,
however, the molecules are packed so closely together that molecular force fields exert
a strong effect on the energy exchange. The thermal conductivity of liquids varies
moderately with temperature and often can be expressed as a linear variation.
(4.4-1)
Where a and b are empirical constants. Thermal conductivities of liquids are essentially
independent of pressure.
4. Solids
Heat or energy is conducted through solids by two mechanisms. In the first, which
applies primarily to metallic solid, heat, like electricity, is conducted by free electrons
which move through the metallic lattice. In the second mechanism, present in all solids,
heat is conducted by transmission of energy of vibration between adjacent atoms.
## Example 4.1.1 heat loss through an insulating wall
Calculate the heat loss per m2 of surface area for an insulating wall composed of
25,4 mm thick fiber insulating board, where the inside temperature is 352,7 K and
the outside temperature is 297,1 K.
Solution
Form appendix A.3, the thermal conductivity of fiber insulating board is 0,048
W/m.K. the thickness x2 x1 = 0,0254 m. substituting into equation :
105,1 W/m2
(105,1 W/m2)
## = 33,30 btu/h . ft2
Simple examples of steady state conduction are shown in figure 4.1 in figure
4.1(a) a flat walled insulated tank contains a refrigerant at perhaps -10 , while the air
outside the tank is at 28 . The temperature falls linearly with distance across the layer
of insulation as heat flows from the air to the refrigerant. As we will see in a later
section, there may actually be a temperature drop between the bulk of the air and the
outside surface of the insulation, but it is assumed to be negligible in fig 4.1 (a), fig
4.1(b) shows a similar tank containing boiling water at 100 , losing heat to air at 20
.
The rate of heat flows is found as follows, assuming that k is independent of
temperature, q is constant along the path of heat flows. If x is the distance from the hot
side.
(4.5-1)
(4.5-2)
(4.5-3)
## When the thermal conductivity k varies linearly with temperature, in accordance
still can be used rigorously by taking an average value k for k, which may be found
either by using the arithmetic average of the individual values of k for the two surface
temperatures, T1 and T2 or by calculating the arithmetic average of the temperatures and
using the value of k at that temperature.
(4.5-4)
4.6 Compound Resistances in Series
Consider a flat wall constructed of a series of layers, as shown in figure 4.2 let the
thicknesses of the layers be BA, BB, and Bc and the average conductivities of the
(4.6-1)
## Equation 10.5 can be written for each layer, using in place of k
(4.6-2)
(4.6 -3)
Since in steady heat flow, all the heat that passes through the first resistance must pass
through the second and in turn pass through the third, qA, qB, and qC are equal and all
can be denoted by q. using this fact and solving for q/A gives
(4.6-4)
## Where RA, RB, RC = resistance of individual layers
R = overall resistance
## 4.7 Heat Flow Through a Cylinder
Consider a hollow cylinder of length L with an inside radius r i and an outside ro.
The cylinder is made of material with a thermal conductivity k. The temperature of the
outside surface is To that of the inside surface is Ti with Ti To. At radius r from the
center the heat flow rate is q and the area through which it flows is A.
## Figure 4.3 Flow of Heat Through Thick Walled Cylinder
Consider a very thin cylinder, concentric with the main cylinder, of radius r, where r is
between ri and ro. the thickness of the wall of this cylinder is dr and if dr is small enough
with respect to r for the lines of heat to be considered parallel
(4.7-1)
(4.7-2)
(4.7-3)
(4.7-4)
## That AL is the area of a cylinder of length L and radius rL where
(4.7-5)
The logarithmic mean is less convenient than the arithmetic mean and the latter can be
used without appreciable error for thin walled tubes, where r 0/ri is nearly 1. The ratio of
figure 4.5. | 1,938 | 8,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.945776 |
https://betterlesson.com/lesson/reflection/19267/we-sing | 1,511,171,649,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805977.0/warc/CC-MAIN-20171120090419-20171120110419-00793.warc.gz | 569,467,727 | 27,481 | ## Reflection: Shared Expectations Making M&M Treats Part I - Section 1: Opening
I use songs in the classroom on a regular basis. Sometimes I write content-related lyrics to the tune of a well known nursery rhyme, such as London Bridge is Falling Down, or I find another teacher's song on You Tube.
I use songs for many reasons. First, songs help reinforce high-level vocabulary as repeated exposure is one of the number one ways to develop vocabulary. Second, songs bring learning alive. What is often dry, uninteresting content can easily become fun and engaging. Finally, songs help students learn procedural steps. For example, when learning long division, there are several steps that can be challenging for students to learn. Instead of telling students the steps when they forget, I'll often begin humming or singing the song. Students will excitedly say, "Oh, now I get it!"
WE SING...
Shared Expectations: WE SING...
# Making M&M Treats Part I
Unit 7: Multi-Digit Division
Lesson 2 of 9
## Big Idea: Students solve division problems with 3-digit and 4-digit dividends by dividing M&Ms and using the standard algorithm.
Print Lesson
49 teachers like this lesson
Standards:
Subject(s):
Math, long division, standard aglorithm
100 minutes
### Kara Nelson
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Nothing to upload details close | 448 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-47 | latest | en | 0.885138 |
https://rebelsky.cs.grinnell.edu/Courses/CSC161/2010F/Labs/multi-arrays-lab.html | 1,713,103,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00390.warc.gz | 449,463,307 | 3,632 | # Lab: Multi-Dimensional Arrays
Summary: In today's laboratory, we explore issues pertaining to mutli-dimensional arrays.
Contents:
## Preparation
a. Create a directory for this lab.
b. Add the standard Makefile to that directory. (Note that the `-Wall` flag will give you warnings for some of the code we provide. You can safely ignore some of the warnings.)
## Exercises
### Exercise 1: Initializing Multi-Dimensional Arrays
Here is a sample initialization of a one-dimensional array of integers.
```int ant[5] = { 5, 2, 7, 3, 4 };
```
Figure out how to initialize the two-dimensional array `bat` so that the row zero contains 8, 16, 32, and 64; row one contains 5, 7, 9, and 11; and row two contains 0, 1, 2, 3.
```int bat[3][4] = figure-this-out;
```
### Exercise 2: Multi-Dimensional Arrays as Single-Dimensional Arrays
Suppose we've declared `bat` as above and `cow` and `i` as follows:
```int *cow;
int i;
```
a. What do you expect the effect of the following code to be?
``` cow = (int *) bat;
for (i = 0; i < 12; i++) {
printf("cow[%d]: %d\n", i, cow[i]);
} // for
```
### Exercise 3: Bounds Violations in Multi-Dimensional Arrays
Consider again the declaration of `bat` above.
a. What values do you expect to get for the following?
``` printf ("bat[0][4] = %d\n", bat[0][4]);
printf ("bat[0][7] = %d\n", bat[0][7]);
printf ("bat[1][7] = %d\n", bat[1][7]);
printf ("bat[2][4] = %d\n", bat[2][4]);
```
### Exercise 4: Declarations, Revisited
Consider the following code.
``` int rabbit[2][3] = { 1, 2, 3, 4, 5, 6 };
int r, c;
for (r = 0; r < 2; r++)
for (c = 0; c < 3; c++)
printf ("rabbit[%d,%d] = %d\n", r, c, rabbit[r][c]);
```
a. What do you expect to happen when you try to compile this code?
c. What do you expect to happen when you try to run this code?
### Exercise 5: Printing Arrays
a. Suppose `a` is a NxM array. Write instructions for printing `a` as a grid. For example, for `rabbit` above, we might print
``` 1 2 3
4 5 6
```
b. Rewrite your code to work with `rabbit`.
c. Rewrite your code to work with `bat`.
### Exercise 6: Three-Dimensional Arrays
Consider the following declaration of a three-dimensional array:
```int chinchilla[2][3][4];
```
a. How many elements does `chinchilla` have?
b. Check your answer with `sizeof`.
c. Can one initialize `chincilla` while declaring it?
e. Where in memory is `chinchilla[i][j][k]`?
## For Those with Extra Time
Write a function, `printIntMatrix (int rows, int cols, int matrix[rows][cols])`, that prints a matrix as in problem 5.
## History
Monday, 17 February 2003 [Samuel A. Rebelsky]
Tuesday, 2 November 2010 [Samuel A. Rebelsky]
Disclaimer: I usually create these pages on the fly, which means that I rarely proofread them and they may contain bad grammar and incorrect details. It also means that I tend to update them regularly (see the history for more details). Feel free to contact me with any suggestions for changes.
This document was generated by Siteweaver on Tue Nov 2 10:48:01 2010.
The source to the document was last modified on Tue Nov 2 10:48:00 2010.
This document may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/CSC161/2010F/Labs/multi-arrays-lab.html`.
Samuel A. Rebelsky, rebelsky@grinnell.edu | 972 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.771498 |
https://www.nagwa.com/en/videos/639196596109/ | 1,718,735,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00675.warc.gz | 818,421,187 | 37,207 | Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa
# Question Video: Finding an Unknown Matrix Using Operations on Matrices Mathematics • First Year of Secondary School
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Consider the matrices π΄ = [0, β4 and 2, 1], π΅ = [β3, β7, 6 and 6, β4, 3], and πΆ = [7, β7 and β7, 2 and β7, 7]. Determine the matrix π that satisfies βπ^(π) = π΄Β² + (π΅πΆ)^(π).
06:28
### Video Transcript
Consider the matrices π΄ is the two-by-two matrix zero, negative four, two, one; π΅ is the three-by-two matrix negative three, negative seven, six, six, negative four, three; and πΆ is the two-by-three matrix seven, seven, negative seven, two, negative seven, seven. Determine the matrix π that satisfies negative π transpose is equal to π΄ squared plus π΅πΆ all transposed.
In this question, weβre given a matrix equation involving three known matrices π΄, π΅, and πΆ. We need to use this to determine the matrix π. To do this, letβs start by evaluating the right-hand side of this equation, and we can do this term by term. First, we need to evaluate π΄ squared. And we recall squaring a matrix means we multiply it by itself. π΄ squared is π΄ times π΄. Therefore, π΄ squared is given by the following expression.
And we recall to multiply two matrices together, we need to find the sum of the products of the corresponding entries in the rows of the first matrix and the columns of the second matrix. For example, to find the entry in row one, column one of matrix π΄ squared, we need to find the product of zero and zero and add this to the product of negative four and two. And we can evaluate this expression. Itβs equal to negative eight. This is the entry in row one, column one of matrix π΄ squared.
We can follow the same process to find the entry in row one, column two of matrix π΄ squared. Itβs equal to zero multiplied by negative four plus negative four multiplied by one, which we can calculate is equal to negative four. And we can follow the same procedure to find the final two elements of π΄ squared. We see that matrix π΄ squared is equal to the two-by-two matrix negative eight, negative four, two, negative seven. And now that we found an expression for matrix π΄ squared, letβs find an expression for the second term of the right-hand side of our equation. Thatβs the transpose of π΅πΆ. And to find the transpose of matrix π΅πΆ, letβs start by multiplying matrix π΅ by matrix πΆ. And although itβs not necessary, letβs check that we can multiply matrix π΅ by matrix πΆ.
Remember, to multiply two matrices together, we need the number of columns of the first matrix to be equal to the number of rows of the second matrix. We can see that these are both equal to three. Then the order of the products of these two matrices will be the number of rows of the first matrix by the number of columns of the second matrix. The product of these two matrices is a two-by-two matrix. And we find the products of these two matrices in the same way we did for the matrix π΄ squared.
We need to find the sum of the products of the corresponding elements of the rows of matrix π΅ with the columns of matrix πΆ. For example, the entry in row one, column one of matrix π΅πΆ will be negative three times seven plus negative seven times negative seven plus six multiplied by negative seven, which if we evaluate we see is equal to negative 14. So the entry in row one, column one of π΅πΆ is negative 14.
We can follow the same process to find the entry in row one, column two. Itβs negative three times seven plus negative seven times two plus six multiplied by seven, which if we calculate is equal to seven. We can do the same to find the remaining two entries of this matrix. π΅ times πΆ is the two-by-two matrix negative 14, seven, 49, 55. Now weβre ready to find an expression for the second term of the right-hand side of our equation. We need to take the transpose of matrix π΅πΆ. And to do this, letβs recall how we find the transpose of a matrix. This means we write the rows of the matrix as the columns of the new matrix. So the first column of the transpose of this matrix is found by using the first row of this matrix.
The first column of the transpose is negative 14, seven. And we can do the same to find the second column of this matrix. We use the second row of π΅πΆ. The second column is 49, 55. Therefore, weβve shown the transpose of π΅πΆ is the two-by-two matrix negative 14, 49, seven, 55. And now that we found π΄ squared and the transpose of π΅πΆ, we can evaluate the right-hand side of the equation. We just need to add π΄ squared to the transpose of matrix π΅πΆ.
Remember, to add two matrices of the same order together, we just need to add the corresponding entries together. For example, the entry in row one, column one of the sum of these two matrices will be negative eight plus negative 14, which we can calculate is negative 22. Similarly, the element in row one, column two of our matrix will be negative four plus 49, which we can calculate is 45. And we can continue this process to find the final two elements of the matrix. We have two plus seven is nine and negative seven plus 55 is 48. So our matrix is the two-by-two matrix negative 22, 45, nine, 48.
Now, we can substitute the matrix we found for the entire right-hand side of the equation weβre given. This gives us that negative π transpose is the two-by-two matrix negative 22, 45, nine, 48. And remember, we want to solve this equation for the matrix π, and thereβs a few different ways of doing this. For example, we could notice multiplying by the scalar and taking the transpose of a matrix does not change its order. Therefore, we can conclude that matrix π is also a two-by-two matrix. So we could write π as a two-by-two matrix and find all of the elements of matrix π. And this would work. However, we can solve this matrix equation directly.
First, weβll multiply both sides of our equation through by negative one. When we do this, on the left-hand side of our equation, we get negative one times negative one which is one multiplied by the transpose of π. And on the right-hand side of our equation, we get negative one multiplied by our matrix. And remember, to multiply a matrix by a scalar, we just multiply all of the entries of the matrix by our scalar. And since our scalar is negative one, this just means we switch the sign of all of the entries of the matrix. This gives us the two-by-two matrix 22, negative 45, negative nine, negative 48.
Now, we can solve the matrix π by taking the transpose of both sides of the equation. When we do this, on the left-hand side of the equation, we get the transpose of the transpose of π. And on the right-hand side of our equation, we get the transpose of the given two-by-two matrix. And we can then simplify this. First, for any matrix π, the transpose of the transpose of π is just equal to π. So the left-hand side of our equation is equal to π. π is just the transpose of the given two-by-two matrix. And therefore, we can find matrix π by taking the transpose of this matrix. π is the two-by-two matrix 22, negative nine, negative 45, negative 48.
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• Realistic Exam Questions | 2,018 | 7,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.850565 |
http://root.cern.ch/root/html520/TGraphQQ.html | 1,581,985,726,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00512.warc.gz | 121,962,148 | 10,791 | # class TGraphQQ: public TGraph
```
This class allows to draw quantile-quantile plots
Plots can be drawn for 2 datasets or for a dataset and a theoretical
distribution function
2 datasets:
Quantile-quantile plots are used to determine whether 2 samples come from
the same distribution.
A qq-plot draws the quantiles of one dataset against the quantile of the
the other. The quantiles of the dataset with fewer entries are on Y axis,
with more entries - on X axis.
A straight line, going through 0.25 and 0.75 quantiles is also plotted
for reference. It represents a robust linear fit, not sensitive to the
extremes of the datasets.
If the datasets come from the same distribution, points of the plot should
fall approximately on the 45 degrees line. If they have the same
distribution function, but location or scale different parameters,
they should still fall on the straight line, but not the 45 degrees one.
The greater their departure from the straight line, the more evidence there
is, that the datasets come from different distributions.
The advantage of qq-plot is that it not only shows that the underlying
distributions are different, but, unlike the analytical methods, it also
gives information on the nature of this difference: heavier tails,
different location/scale, different shape, etc.
Some examples of qqplots of 2 datasets:
```
/* */
```
1 dataset:
Quantile-quantile plots are used to determine if the dataset comes from the
specified theoretical distribution, such as normal.
A qq-plot draws quantiles of the dataset against quantiles of the specified
theoretical distribution.
(NOTE, that density, not CDF should be specified)
A straight line, going through 0.25 and 0.75 quantiles can also be plotted
for reference. It represents a robust linear fit, not sensitive to the
extremes of the dataset.
As in the 2 datasets case, departures from straight line indicate departures
from the specified distribution.
" The correlation coefficient associated with the linear fit to the data
in the probability plot (qq plot in our case) is a measure of the
goodness of the fit.
Estimates of the location and scale parameters of the distribution
are given by the intercept and slope. Probability plots can be generated
for several competing distributions to see which provides the best fit,
and the probability plot generating the highest correlation coefficient
is the best choice since it generates the straightest probability plot."
From "Engineering statistic handbook",
http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm
Example of a qq-plot of a dataset from N(3, 2) distribution and
TMath::Gaus(0, 1) theoretical function. Fitting parameters
are estimates of the distribution mean and sigma.
```
/* */
```//
References:
http://www.itl.nist.gov/div898/handbook/eda/section3/qqplot.htm
http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm
```
## Function Members (Methods)
public:
protected:
virtual Double_t** TGraph::Allocate(Int_t newsize) Double_t** TGraph::AllocateArrays(Int_t Narrays, Int_t arraySize) virtual void TGraph::CopyAndRelease(Double_t** newarrays, Int_t ibegin, Int_t iend, Int_t obegin) virtual Bool_t TGraph::CopyPoints(Double_t** newarrays, Int_t ibegin, Int_t iend, Int_t obegin) Bool_t TGraph::CtorAllocate() virtual void TObject::DoError(int level, const char* location, const char* fmt, va_list va) const Double_t** TGraph::ExpandAndCopy(Int_t size, Int_t iend) virtual void TGraph::FillZero(Int_t begin, Int_t end, Bool_t from_ctor = kTRUE) void MakeFunctionQuantiles() void MakeQuantiles() void TObject::MakeZombie() void Quartiles() Double_t** TGraph::ShrinkAndCopy(Int_t size, Int_t iend) virtual void TGraph::SwapPoints(Int_t pos1, Int_t pos2) static void TGraph::SwapValues(Double_t* arr, Int_t pos1, Int_t pos2)
## Data Members
public:
enum TGraph::[unnamed] { kClipFrame kNotEditable }; enum TObject::EStatusBits { kCanDelete kMustCleanup kObjInCanvas kIsReferenced kHasUUID kCannotPick kNoContextMenu kInvalidObject }; enum TObject::[unnamed] { kIsOnHeap kNotDeleted kZombie kBitMask kSingleKey kOverwrite kWriteDelete };
protected:
TF1* fF theoretical density function, if specified Color_t TAttFill::fFillColor fill area color Style_t TAttFill::fFillStyle fill area style TList* TGraph::fFunctions Pointer to list of functions (fits and user) TH1F* TGraph::fHistogram Pointer to histogram used for drawing axis Color_t TAttLine::fLineColor line color Style_t TAttLine::fLineStyle line style Width_t TAttLine::fLineWidth line width Color_t TAttMarker::fMarkerColor Marker color index Size_t TAttMarker::fMarkerSize Marker size Style_t TAttMarker::fMarkerStyle Marker style Int_t TGraph::fMaxSize !Current dimension of arrays fX and fY Double_t TGraph::fMaximum Maximum value for plotting along y Double_t TGraph::fMinimum Minimum value for plotting along y TString TNamed::fName object identifier Int_t TGraph::fNpoints Number of points <= fMaxSize Int_t fNy0 size of the fY0 dataset TString TNamed::fTitle object title Double_t* TGraph::fX [fNpoints] array of X points Double_t fXq1 x1 coordinate of the interquartile line Double_t fXq2 x2 coordinate of the interquartile line Double_t* TGraph::fY [fNpoints] array of Y points Double_t* fY0 !second dataset, if specified Double_t fYq1 y1 coordinate of the interquartile line Double_t fYq2 y2 coordinate of the interquartile line
## Function documentation
TGraphQQ(const TGraphQQ& )
```default constructor
```
TGraphQQ(Int_t n, Double_t* x)
```Creates a quantile-quantile plot of dataset x.
Theoretical distribution function can be defined later by SetFunction method
```
TGraphQQ(Int_t n, Double_t* x, TF1* f)
```Creates a quantile-quantile plot of dataset x against function f
```
TGraphQQ(Int_t nx, Double_t* x, Int_t ny, Double_t* y)
```Creates a quantile-quantile plot of dataset x against dataset y
Parameters nx and ny are respective array sizes
```
```Destroys a TGraphQQ
```
void MakeFunctionQuantiles()
```Computes quantiles of theoretical distribution function
```
void MakeQuantiles()
```When sample sizes are not equal, computes quantiles of the bigger sample
by linear interpolation
```
void Quartiles()
``` compute quartiles
a quartile is a 25 per cent or 75 per cent quantile
```
void Paint(Option_t* opt = "")
``` paint this graphQQ. No options for the time being
```
void SetFunction(TF1* f)
```Sets the theoretical distribution function (density!)
and computes its quantiles
```
TGraphQQ(const TGraphQQ& )
Author: Anna Kreshuk 18/11/2005
Last change: root/graf:\$Id: TGraphQQ.h 20882 2007-11-19 11:31:26Z rdm \$
Last generated: 2008-06-25 08:46 | 1,683 | 6,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-10 | latest | en | 0.876772 |
https://www.freelancer.pl/projects/engineering/mathematica-aircraft-guidance-project/ | 1,547,713,478,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00063.warc.gz | 799,413,536 | 36,312 | # Mathematica Aircraft guidance project
An aircraft flying under the guidance of a nondirectional beacon (a fixed radio transmitter, abbreviated
NDB) moves so that its longitudinal axis always points toward the beacon (see Figure 3.19).
A pilot sets out toward an NDB from a point at which the wind is at right angles to the initial
direction of the aircraft; the wind maintains this direction. Assume that the wind speed and the
speed of the aircraft through the air (its “airspeed”) remain constant. (Keep in mind that the latter
is different from the aircraft’s speed with respect to the ground.)
(a) Locate the flight in the xy-plane, placing the start of the trip at and the destination
at . Set up the differential equation describing the aircraft’s path over the ground.
Hint:
(b) Make an appropriate substitution and solve this equation.
(c) Use the fact that x 2 and y 0 at t 0 to determine the appropriate value of the
arbitrary constant in the solution set.
(d) Solve to get y explicitly in terms of x. Write your solution in terms of a hyperbolic function.
(e) Let be the ratio of windspeed to airspeed. Using a software package, graph the
solutions for the cases 0.1, 0.3, 0.5, and 0.7 all on the same set of axes. Interpret
these graphs.
(
Umiejętności: Inżynieria, Matematyka, Matlab i Mathematica, Inżynieria mechaniczna, Fizyka
O pracodawcy:
( 3 ocen ) Sannicolau-Mare, Romania
Numer ID Projektu: #15733691
## 5 freelancerów złożyło ofertę na kwotę €31 do tego projektu
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tunmack
As a mathematics masters degree holder, currently onto my PhD, I will like to solve this maths questions for you
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I am always good in mathematics when it comes to vectors. I am also good in kinematics and visualizing mathematical equations. I will try my very best to help.
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0.0 | 565 | 2,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-04 | latest | en | 0.702712 |
https://astarmathsandphysics.com/a-level-maths-notes/c4/5450-differrentiating-and-integrating-odd-and-even-functions.html | 1,720,790,913,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00274.warc.gz | 86,172,213 | 10,044 | ## Differrentiating and Integrating Odd and Even Functions
The condition for functions to be odd or even are
$f(x)=-f(-x)$
$f(x)=f(-x)$
respectively.
When you differentiate these equations you get
$f'(x)=--f'(-x)=f'(-x)$
(1)
$f'(x)=-f'(-x)$
(2)
Differentiating an odd function gives an even function and vice versa. The same is not true when integrating in general, because of the role of the arbitrary constant .Integrating(1)and(2)gives
$f(x)=-f(-x)+c$
$f(x)=f(-x)+c$
so the odd and even conditions fail without the further condition that
$c=0$
.
You have no rights to post comments | 175 | 590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.761575 |
http://mathhelpforum.com/trigonometry/7748-tangent-sum.html | 1,480,765,132,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00500-ip-10-31-129-80.ec2.internal.warc.gz | 183,323,181 | 9,875 | 1. ## tangent sum
Evaluate $\sum_{n=0}^{\infty}\cot^{-1}(n^2+n+1)$
and this one is similar.
Evaluate $\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{n^2+n+1}\right)$
2. Originally Posted by putnam120
Evaluate $\sum_{n=0}^{\infty}\cot^{-1}(n^2+n+1)$
and this one is similar.
Evaluate $\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{n^2+n+1}\right)$
I presume that these are not problems that you do not know the answer to.
If this is the case please put them in the "Maths Games and Puzzles" forum
where we will talk about them at our leisure rather than reduce the time
we spend helping people with problems that they can't do.
Thank you
RonL | 213 | 642 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-50 | longest | en | 0.832576 |
http://nrich.maths.org/public/leg.php?code=71&cl=1&cldcmpid=8081 | 1,477,254,811,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719416.57/warc/CC-MAIN-20161020183839-00453-ip-10-171-6-4.ec2.internal.warc.gz | 189,240,380 | 7,210 | # Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Take One Example:
Filter by: Content type:
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Challenge level:
### There are 24 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Primary Proof?
##### Stage: 1
Proof does have a place in Primary mathematics classrooms, we just need to be clear about what we mean by proof at this level.
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Odd Times Even
##### Stage: 1 Challenge Level:
This problem looks at how one example of your choice can show something about the general structure of multiplication.
### Two Numbers Under the Microscope
##### Stage: 1 Challenge Level:
This investigates one particular property of number by looking closely at an example of adding two odd numbers together.
### Walking Round a Triangle
##### Stage: 1 Challenge Level:
This ladybird is taking a walk round a triangle. Can you see how much he has turned when he gets back to where he started?
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Three Neighbours
##### Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### A Bag of Marbles
##### Stage: 1 Challenge Level:
Use the information to describe these marbles. What colours must be on marbles that sparkle when rolling but are dark inside?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Always, Sometimes or Never? KS1
##### Stage: 1 Challenge Level:
Are these statements relating to calculation and properties of shapes always true, sometimes true or never true?
### Air Nets
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you visualise whether these nets fold up into 3D shapes? Watch the videos each time to see if you were correct.
##### Stage: 1 and 2 Challenge Level:
Who said that adding couldn't be fun?
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Always, Sometimes or Never? Shape
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Logic
##### Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Breaking the Equation ' Empirical Argument = Proof '
##### Stage: 2, 3, 4 and 5
This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field. | 1,005 | 4,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2016-44 | longest | en | 0.854737 |
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