url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://socratic.org/questions/how-do-you-use-the-quotient-rule-to-differentiate-x-6-x-6 | 1,611,506,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00233.warc.gz | 568,241,290 | 5,727 | # How do you use the quotient rule to differentiate x^6/(x-6)?
Apr 14, 2016
$f ' \left(x\right) = \frac{5 {x}^{6} - 36 {x}^{5}}{x - 6} ^ 2$
#### Explanation:
Let $f = {x}^{6} \mathmr{and} g = x - 6$
Then$f ' = 6 {x}^{5} \mathmr{and} g ' = 1$
Quotient Rule:
$f ' \left(x\right) = \frac{g f ' - f g '}{g} ^ 2$
$f ' \left(x\right) = \frac{\left(x - 6\right) \left(6 {x}^{5}\right) - \left({x}^{6}\right)}{x - 6} ^ 2$
$= \frac{6 {x}^{6} - 36 {x}^{5} - {x}^{6}}{x - 6} ^ 2$
$= \frac{5 {x}^{6} - 36 {x}^{5}}{x - 6} ^ 2$ | 268 | 522 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 4.40625 | 4 | CC-MAIN-2021-04 | longest | en | 0.397644 |
https://www.physicsforums.com/threads/what-happends-when-we-oscillate-three-pendulums-next-to-each-other.362855/ | 1,720,804,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00531.warc.gz | 733,228,408 | 17,116 | # What happends when we oscillate three pendulums next to each other?
• Idrees
In summary, when three pendulums with the same mass but different lengths are oscillated together, the first pendulum will transmit its energy to the other two. The second pendulum, with the same length as the first, will resonate and oscillate with a larger amplitude, while the third pendulum, with a slightly longer length, will oscillate with a smaller amplitude. This energy will then be transmitted back to the first pendulum and so on, resulting in a continuous transformation of energy from gravitational potential to kinetic energy and back again.
Idrees
what happends when we oscillate three pendulums next to each other??
## Homework Statement
if we have three pendula hanging from one string...and each of them separated by 5 cm, and each pendulum has same amount of mass hanging from it,, but the first two pendulums are 2.0 m long and the third one is 2.5 m long... if i set the first 2.0 m long pendulum to oscillate what will happened to the other two? please give me some hints or full answer
## Homework Equations
we don't use equations but we can think about the different heights of each pendulum
## The Attempt at a Solution
i would think once you start oscillating the first pendulum, it will moeve and strt oscillating the 2nd pendulum which are both of same height, but nothing will happened to the third pendulum, it will have no motion
you don't use equations? well the pendulum will simply start oscillating at a specific angular frequency. It won't necessarily oscillate like a simple pendulum however. There's a different moment of inertia and center of mass. Without using equations, I'd say average out the distances apart to find your center of mass. As a rule of thumb, I'd imagine that the moment of inertia of this physical pendulum would probably be similar to a point mass at that center of mass.
Thanks for youre reply,, its very useful.. but also i wanted to know what will simply happen to the motion, would one pendulum moving oscillate the second one ,, please eloborate on what will simply happpen to the motion and how will the energy be transformed when this happends? please help out
They might start out at a different oscillation periods, but a forced oscillation will quickly take over if that would happen. In which the system will resonate and oscillate all together about the center of mass.
thank you for the second reply.. and i think by energy transformation they mean what will happen to kinetic and gravitational energy.. which i can answer easily
well more correctly it's gravitational potential to kinetic energy, but you have the right idea.
I think you have a problem related to forced oscillations and resonance.
The first pendulum will "transmit" some of its energy to the other two. The one with the same length is at resonance with the first one so it will oscillate with a relatively large amplitude. The third one will oscillate with a smaller amplitude, as it's out of resonance (but not by much).
After a while the energy will be transmitted back to the first pendulum and so on.
thank you nasu heaps.. i completely understood your answer properly and thank you to godtripp aswell.. the question asked "DESCRIBE THE TRANSFORMATION OF THE ENERGY THAT OCCURS WHEN THESE PENDULAS are oscillated?? should i wirte about how the gravitational energy transforms to kinetic energy... can please help me on how to write this part aswell? thanks heaps.
## 1. What is the relationship between the oscillation of three pendulums next to each other?
The oscillation of three pendulums next to each other is known as a coupled pendulum system. This means that the motion of each pendulum affects the motion of the other pendulums in the system. The pendulums are connected through a shared support point, which allows for energy to be transferred between them.
## 2. How does the length of each pendulum affect the oscillation?
The length of each pendulum plays a crucial role in the oscillation of the system. The longer the pendulum, the slower the oscillation period, and the shorter the pendulum, the faster the oscillation period. This is because the length of the pendulum determines the distance traveled by the pendulum in one cycle, also known as the amplitude of oscillation.
## 3. What happens when one pendulum in the system is set in motion?
When one pendulum is set in motion, it transfers energy to the other pendulums through the shared support point. This leads to a phenomenon known as sympathetic vibration, where the other pendulums start oscillating at the same frequency as the first pendulum. The motion of the pendulums will continue until the energy is dissipated due to friction and air resistance.
## 4. How does the amplitude of one pendulum affect the other pendulums in the system?
The amplitude of one pendulum has a direct effect on the amplitudes of the other pendulums in the system. If the amplitude of one pendulum is increased, the amplitude of the other pendulums will also increase. This is because the energy transfer between pendulums is dependent on the amplitude of oscillation.
## 5. What factors can affect the oscillation period of the pendulums in the system?
The oscillation period of the pendulums can be affected by several factors, including the length of the pendulums, the mass of the pendulums, and the angle at which they are released. Additionally, external factors such as air resistance and friction can also affect the period of oscillation. These factors can be manipulated to create different patterns and behaviors in the coupled pendulum system.
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### Why Sparse Priors Like Total Variation Opts to Concentrate Derivatives at a Small Number of Pixels?
When performing image deconvolution (deblurring), people often make use of priors to get rid of the illness of the problem. One very common prior is total variation, a sparse prior. Compared to ... | 758 | 3,366 | {"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} | 2.96875 | 3 | CC-MAIN-2019-43 | latest | en | 0.87955 |
https://brilliant.org/discussions/thread/is-01-then-find-the-mistake/?sort=new | 1,603,316,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00167.warc.gz | 257,990,355 | 17,014 | # Is 0=1? Then find the mistake
Consider the integral, $\displaystyle I =\int \dfrac{1}{x}dx$
We know that $I= \log x$
But, $\displaystyle I =\int \dfrac{1}{x}(1)dx =(x)\dfrac{1}{x}-\int \bigg(-\dfrac{1}{x^2} \bigg) xdx$
$1+I=I \Rightarrow 1=0$
What is the mistake?
I do not know the mistake
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You cannot cancel the integrals on both sides because each function has infinitely many anti-derivatives off by a constant. That's why, for good reason, you should always include "+C" in your answers.
- 2 years, 10 months ago
And what happens without limits? $\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}$
- 3 years, 9 months ago
It's still correct.. Integration constants need not be same
- 3 years, 9 months ago
Oh, right, thank you
- 3 years, 9 months ago
Welcome.. :)
- 3 years, 9 months ago
you actually got $\int \frac{1}{x} -\int \frac{1}{x}=1$. There is absolutely nothing wrong with that. Those are indefinite integrals.
- 3 years, 9 months ago
How is LHS not equal to 0
- 3 years, 9 months ago
Integration constants need not be equal.
- 3 years, 9 months ago
K.. I agree on that
- 3 years, 9 months ago
- 3 years, 9 months ago
It is $\frac{x}{x} |_{a}^{b}$. That yeilds zero. 1 there has limits.
- 3 years, 9 months ago
Exactly... Then he should get correctly.
- 3 years, 9 months ago
I didn't quite get u, u meant $\dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1$
- 3 years, 9 months ago
I can't believe you just stated that...
- 3 years, 9 months ago
Correct me if I am wrong, isn't $\dfrac{2-1}{2-1}=\dfrac{1}{1}=1$ ?
- 3 years, 9 months ago
its $\frac{b}{b}-\frac{a}{a}$ and still $1|_a^b = 1_a-1_b=1-1=0$.
- 3 years, 9 months ago
Oops! I can't believe I made such a mistake... Anyways thx..
- 3 years, 9 months ago
Can you post this as a problem so others can learn from this misconception?
Staff - 3 years, 9 months ago
Ok, surely
- 3 years, 9 months ago
- 3 years, 9 months ago
What if I add limits from 1 to 2..you still get 1=0
- 3 years, 9 months ago | 1,215 | 3,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 19, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-45 | longest | en | 0.828594 |
https://www.coursehero.com/file/5667938/Mgmt-200-Fall-2007-Exam-Final417/ | 1,496,078,730,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612502.45/warc/CC-MAIN-20170529165246-20170529185246-00148.warc.gz | 1,098,393,546 | 139,337 | Mgmt_200_Fall_2007_Exam_Final(4,17)
# Mgmt_200_Fall_2007_Exam_Final(4,17) - Name PUID Purdue...
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Name: ________________________ PUID: ________________________ Purdue University Krannert School of Management MGMT 200 – Introductory Financial Accounting Fall 2007 Final Exam – December 11, 2007 This exam consists of 4 questions on 11 pages (excluding this cover page and the present value table page) for a total of 100 points. Time allowed: 90 minutes. Answer all questions. To ensure full credit and to maximize partial credit, clearly show all supporting calculations. The exam is closed book. A calculator is permitted. Present value tables are provided on the last page. GOOD LUCK . Question 1 (25 points) ________ Question 2 (25 points) ________ Question 3 (25 points) ________ Question 4 (25 points) ________ TOTAL (100 points) ________
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Question 1. Bonds Payable (25 points) Jim Corporation borrowed the majority of the funds needed for the purchase of new factory equipment that cost \$440,000. Jim Corporation raised the funds with the issuance of the following loan on January 1, 2006: \$600,000 face value, 4% coupon (interest payable annually on December 31), twenty-year bonds. The market rate of interest on January 1, 2006 was 7%. Required: a. At what price (where par = 100) would these bonds be issued? b. Prepare the journal entry to record the sale of the 4% coupon bonds on January 1, 2006. c. Prepare the journal entry to record interest expense and the interest payment on the 4% coupon bonds on December 31, 2006, assuming the straight-line amortization method is used. Question 1 continued over . . . Mgmt 200 – Final Exam – Fall 2007 – page 1
Question 1 continued. d.
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## This note was uploaded on 11/22/2009 for the course MGMT 200 taught by Professor Greigg during the Spring '08 term at Purdue.
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Ask a homework question - tutors are online | 592 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-22 | longest | en | 0.867038 |
http://linuxcertif.com/man/3/clarzt/ | 1,627,586,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00098.warc.gz | 24,499,615 | 3,823 | clarzt
Langue: en
Version: 322648 (ubuntu - 08/07/09)
Section: 3 (Bibliothèques de fonctions)
Sommaire
NAME
CLARZT - form the triangular factor T of a complex block reflector H of order > n, which is defined as a product of k elementary reflectors
SYNOPSIS
SUBROUTINE CLARZT(
DIRECT, STOREV, N, K, V, LDV, TAU, T, LDT )
CHARACTER DIRECT, STOREV
INTEGER K, LDT, LDV, N
COMPLEX T( LDT, * ), TAU( * ), V( LDV, * )
PURPOSE
CLARZT forms the triangular factor T of a complex block reflector H of order > n, which is defined as a product of k elementary reflectors. If DIRECT = 'F', H = H(1) H(2) . . . H(k) and T is upper triangular;
If DIRECT = 'B', H = H(k) . . . H(2) H(1) and T is lower triangular.
If STOREV = 'C', the vector which defines the elementary reflector H(i) is stored in the i-th column of the array V, and
H = I - V * T * V'
If STOREV = 'R', the vector which defines the elementary reflector H(i) is stored in the i-th row of the array V, and
H = I - V' * T * V
Currently, only STOREV = 'R' and DIRECT = 'B' are supported.
ARGUMENTS
DIRECT (input) CHARACTER*1
Specifies the order in which the elementary reflectors are multiplied to form the block reflector:
= 'F': H = H(1) H(2) . . . H(k) (Forward, not supported yet)
= 'B': H = H(k) . . . H(2) H(1) (Backward)
STOREV (input) CHARACTER*1
Specifies how the vectors which define the elementary reflectors are stored (see also Further Details):
= 'R': rowwise
N (input) INTEGER
The order of the block reflector H. N >= 0.
K (input) INTEGER
The order of the triangular factor T (= the number of elementary reflectors). K >= 1.
V (input/output) COMPLEX array, dimension
(LDV,K) if STOREV = 'C' (LDV,N) if STOREV = 'R' The matrix V. See further details.
LDV (input) INTEGER
The leading dimension of the array V. If STOREV = 'C', LDV >= max(1,N); if STOREV = 'R', LDV >= K.
TAU (input) COMPLEX array, dimension (K)
TAU(i) must contain the scalar factor of the elementary reflector H(i).
T (output) COMPLEX array, dimension (LDT,K)
The k by k triangular factor T of the block reflector. If DIRECT = 'F', T is upper triangular; if DIRECT = 'B', T is lower triangular. The rest of the array is not used.
LDT (input) INTEGER
The leading dimension of the array T. LDT >= K.
FURTHER DETAILS
Based on contributions by
A. Petitet, Computer Science Dept., Univ. of Tenn., Knoxville, USA
The shape of the matrix V and the storage of the vectors which define the H(i) is best illustrated by the following example with n = 5 and k = 3. The elements equal to 1 are not stored; the corresponding array elements are modified but restored on exit. The rest of the array is not used.
DIRECT = 'F' and STOREV = 'C': DIRECT = 'F' and STOREV = 'R':
______V_____
( v1 v2 v3 ) / ( v1 v2 v3 ) ( v1 v1 v1 v1 v1 . . . . 1 )
V = ( v1 v2 v3 ) ( v2 v2 v2 v2 v2 . . . 1 )
( v1 v2 v3 ) ( v3 v3 v3 v3 v3 . . 1 )
( v1 v2 v3 )
. . .
. . .
1 . .
1 .
1
DIRECT = 'B' and STOREV = 'C': DIRECT = 'B' and STOREV = 'R':
______V_____
1 / . 1 ( 1 . . . . v1 v1 v1 v1 v1 )
. . 1 ( . 1 . . . v2 v2 v2 v2 v2 )
. . . ( . . 1 . . v3 v3 v3 v3 v3 )
. . .
( v1 v2 v3 )
( v1 v2 v3 )
V = ( v1 v2 v3 )
( v1 v2 v3 )
( v1 v2 v3 ) | 1,194 | 3,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.680092 |
https://www.roseindia.net/answers/viewqa/Java-Beginners/20243-length-in-array.html | 1,719,324,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00715.warc.gz | 833,847,897 | 15,129 | # length in array
hai wat s length in array and where it s used
July 18, 2011 at 1:52 PM
The length determines the number of elements stored in the array.
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Array's I have to finish an "order page". that checks 2 dropdown boxes and a set of radio buttons and put the results into a set of textboxes, 1 line at a time, then calculate a total. I know I need to create an array, but I'm | 2,264 | 9,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.532033 |
https://math.stackexchange.com/questions/2360315/give-me-such-a-3n%C2%B1q-problem-that-we-do-not-know-a-counter-example | 1,563,644,649,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526560.40/warc/CC-MAIN-20190720173623-20190720195623-00188.warc.gz | 471,778,415 | 33,268 | # Give me such a $“3n±Q”$ problem that we do not know a counter-example [closed]
Maybe, This question is stupid.But, I want to ask.Because, really I dont Know answer.This problem may be similar to others. My Question is:
$$f(n) = \begin{cases} Pn±Q & \text {if n is odd} \\ \frac{n}{2} & \text {if n is even} \end{cases} ,$$ and we can find such $k$ $$f^{k}(n)=1$$ Here $P,Q\in \mathbb{N}$
For Example: We know counter examples, for $"3n-5","3n-1",3n+5"$ problems. So that $f^{k}(n)≠1$
Give me such a problem that we do not have a counter-example,(In shortly $f^{k}(n)=1$)
If the example you give is $"3n ± Q"$, and "$Q≠1"$ it was very good.
(Please, edit or improve question for me,because Unfortunately, I'm not as knowledgeable as you.) Thanks so much!
## closed as unclear what you're asking by Did, kingW3, Sahiba Arora, Xam, user251257Jul 17 '17 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• I think it's not so interesting. Collatz took already his place with his $3n+1$ conjecture. – Michael Rozenberg Jul 16 '17 at 7:59
• It is not my question. I want to know , is there a problem like a collatz conjecture ("$3n+K"$) ,which that we dont know counter-examples.$"K≠1"$ – Elvin Jul 16 '17 at 8:05
• You wrote that “We all know that,There is no counter-examples for Collatz Conjecture”. Actually, if we all knew that, then it wouldn't be a conjecture anymore, right? – José Carlos Santos Jul 16 '17 at 8:09
• @Idontknow There are several things that I do not understand, even after your latest edition. – José Carlos Santos Jul 16 '17 at 9:01
• @Idontknow I am sorry, but I do not have time for this right now. – José Carlos Santos Jul 16 '17 at 9:07
Surely the article "On the '3x+1'-Problem" of R.E.Crandall of 1978 fits your question well (it is online you can find it). Here is a screenshot of a part of that article dealing with the general $qx+r$-problem-variant:
As an interesting sidenote: I've found a second cycle with the $q=181$ - problem. And also the $q=3511$ (having a similar property as $q=1093$ being a wieferich prime) should have been looked at.
• You say that, there are not $P$ and $Q$?? Always $f^{k}(n)=1$?? ($P≠3$ and $Q≠1$) – Elvin Jul 16 '17 at 15:10 | 761 | 2,484 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-30 | latest | en | 0.935933 |
https://barisalcity.org/common-multiples-of-2-and-6/ | 1,660,283,843,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00554.warc.gz | 135,776,463 | 5,001 | LCM the 2 and also 6 is the the smallest number amongst all usual multiples that 2 and also 6. The first couple of multiples that 2 and also 6 space (2, 4, 6, 8, . . . ) and also (6, 12, 18, 24, 30, 36, 42, . . . ) respectively. There are 3 frequently used techniques to discover LCM that 2 and 6 - by listing multiples, by division method, and also by prime factorization.
You are watching: Common multiples of 2 and 6
1 LCM that 2 and also 6 2 List the Methods 3 Solved Examples 4 FAQs
Answer: LCM that 2 and 6 is 6.
Explanation:
The LCM of two non-zero integers, x(2) and also y(6), is the smallest optimistic integer m(6) that is divisible through both x(2) and y(6) without any type of remainder.
The approaches to find the LCM of 2 and 6 are defined below.
By Listing MultiplesBy department MethodBy prime Factorization Method
### LCM the 2 and also 6 by Listing Multiples
To calculation the LCM the 2 and 6 by listing the end the usual multiples, we can follow the given listed below steps:
Step 1: list a couple of multiples that 2 (2, 4, 6, 8, . . . ) and 6 (6, 12, 18, 24, 30, 36, 42, . . . . )Step 2: The common multiples indigenous the multiples that 2 and 6 room 6, 12, . . .Step 3: The smallest common multiple that 2 and also 6 is 6.
∴ The least typical multiple that 2 and also 6 = 6.
### LCM that 2 and 6 by division Method
To calculate the LCM the 2 and 6 through the department method, we will divide the numbers(2, 6) by their prime factors (preferably common). The product of this divisors offers the LCM the 2 and 6.
Step 3: continue the procedures until just 1s space left in the critical row.
The LCM of 2 and 6 is the product of every prime numbers on the left, i.e. LCM(2, 6) by division method = 2 × 3 = 6.
### LCM the 2 and 6 by prime Factorization
Prime administer of 2 and 6 is (2) = 21 and also (2 × 3) = 21 × 31 respectively. LCM the 2 and 6 have the right to be acquired by multiplying prime determinants raised to their respective highest possible power, i.e. 21 × 31 = 6.Hence, the LCM that 2 and also 6 by element factorization is 6.
## FAQs top top LCM that 2 and also 6
### What is the LCM the 2 and also 6?
The LCM that 2 and also 6 is 6. To find the least common multiple the 2 and 6, we require to find the multiples of 2 and 6 (multiples of 2 = 2, 4, 6, 8; multiples the 6 = 6, 12, 18, 24) and also choose the smallest multiple that is specifically divisible by 2 and 6, i.e., 6.
### If the LCM that 6 and also 2 is 6, find its GCF.
LCM(6, 2) × GCF(6, 2) = 6 × 2Since the LCM of 6 and also 2 = 6⇒ 6 × GCF(6, 2) = 12Therefore, the greatest common factor (GCF) = 12/6 = 2.
### How to uncover the LCM that 2 and also 6 by prime Factorization?
To uncover the LCM that 2 and also 6 utilizing prime factorization, we will uncover the prime factors, (2 = 2) and also (6 = 2 × 3). LCM that 2 and 6 is the product of prime determinants raised to your respective greatest exponent among the number 2 and 6.⇒ LCM that 2, 6 = 21 × 31 = 6.
### What is the Relation between GCF and LCM that 2, 6?
The adhering to equation can be provided to refer the relation between GCF and LCM of 2 and also 6, i.e. GCF × LCM = 2 × 6.
See more: What Type Of Oil To Use In Ez Go Golf Cart Use? What Kind Of Oil Do I Put In My Ez Go Golf Cart
### What is the least Perfect Square Divisible through 2 and also 6?
The least number divisible by 2 and 6 = LCM(2, 6)LCM that 2 and also 6 = 2 × 3 ⇒ least perfect square divisible by every 2 and also 6 = LCM(2, 6) × 2 × 3 = 36 Therefore, 36 is the required number. | 1,154 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-33 | latest | en | 0.943758 |
http://www.numbersaplenty.com/21096 | 1,582,455,453,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00427.warc.gz | 219,868,054 | 3,441 | Search a number
21096 = 2332293
BaseRepresentation
bin101001001101000
31001221100
411021220
51133341
6241400
7115335
oct51150
931840
1021096
1114939
1210260
1397aa
14798c
1563b6
hex5268
21096 has 24 divisors (see below), whose sum is σ = 57330. Its totient is φ = 7008.
The previous prime is 21089. The next prime is 21101. The reversal of 21096 is 69012.
It can be written as a sum of positive squares in only one way, i.e., 12996 + 8100 = 114^2 + 90^2 .
It is a tau number, because it is divible by the number of its divisors (24).
It is a Harshad number since it is a multiple of its sum of digits (18).
It is an alternating number because its digits alternate between even and odd.
It is an unprimeable number.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 75 + ... + 218.
221096 is an apocalyptic number.
It is an amenable number.
21096 is an abundant number, since it is smaller than the sum of its proper divisors (36234).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
21096 is a wasteful number, since it uses less digits than its factorization.
21096 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 305 (or 298 counting only the distinct ones).
The product of its (nonzero) digits is 108, while the sum is 18.
The square root of 21096 is about 145.2446212429. The cubic root of 21096 is about 27.6312186133.
The spelling of 21096 in words is "twenty-one thousand, ninety-six". | 455 | 1,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-10 | latest | en | 0.909667 |
https://gmatclub.com/forum/out-of-seven-models-all-of-different-heights-five-models-56445.html?fl=similar | 1,498,574,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321426.45/warc/CC-MAIN-20170627134151-20170627154151-00606.warc.gz | 749,200,459 | 69,030 | It is currently 27 Jun 2017, 07:48
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# Out of seven models, all of different heights, five models
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Out of seven models, all of different heights, five models [#permalink]
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30 Nov 2007, 08:54
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Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
A. 6
B. 11
C. 17
D. 72
E. 210
[Reveal] Spoiler: OA
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30 Nov 2007, 11:01
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I got 17, is it correct?
Ways in which you can sit all models in oder: 7!/5!2! = 21
Then we figure out in how many of those 21 options model 4 and 6 are sitting together.
I usually draw something like this:
Option 1: _ _ _ _ _ Model 4, 6 could take this spaces, leaving last spot for model 7. You can have 3 in which you can arrange the first 3 models in the first 2 spots = 3
Option 2: _ _ _ _ _ Model 4, 6 could also be on the last 2 spots, but there's only one option on this one, because there are only 3 models to fill the first 3 spots = 1
21 - (3+1) = 17
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Re: PS: Permutations & Combinations [#permalink]
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22 Dec 2007, 10:45
tarek99 wrote:
If the five models are to stand in a line from shortest to tallest
I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?
CEO
Joined: 29 Mar 2007
Posts: 2559
Re: PS: Permutations & Combinations [#permalink]
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22 Dec 2007, 17:45
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CaspAreaGuy wrote:
tarek99 wrote:
If the five models are to stand in a line from shortest to tallest
I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?
I agree this messed part up as well.
7!/5!*2! = 21 ways
12346 No
12467 No
23467 No
13467 No
so 17 ways.
Director
Joined: 03 Sep 2006
Posts: 871
Re: PS: Permutations & Combinations [#permalink]
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22 Dec 2007, 18:09
tarek99 wrote:
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
a) 6
b) 11
c) 17
d) 72
e) 210
Say we have the models numbered 1~7 ( assume, 1 is shortest and 7 is tallest in that order )
1, 2, 3, 4, 5, 6, 7.
Question says, 4th and 6th can't be adjacent, which implies question "implies" 4th and 6th are always selected amongst the 5 models but never sit/stand together.
(4,6)XXX
The remaining three can be selected in 5C3 ways, and (4 and 6 ) can arrange amongst themselves in 2! ways.
Therefore number of ways choosing,5 models so that 4 and 6 are akways included are
5C3*2! = 20
This also includes the number of ways in which(4,6) are together.
Then, the number of ways in which (4,6) will be always together can be computed
If (4,6) occupy any of the two adjacent places, then remaining three places can be occupied by (xxx) in 3! ways.
Or
46xxx
x46xx
xx46x
(2!*5C3)-3! = 17
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Joined: 01 Sep 2007
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Re: PS: Permutations & Combinations [#permalink]
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22 Dec 2007, 23:37
GMATBLACKBELT wrote:
CaspAreaGuy wrote:
tarek99 wrote:
If the five models are to stand in a line from shortest to tallest
I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?
I agree this messed part up as well.
7!/5!*2! = 21 ways
12346 No
12467 No
23467 No
13467 No
so 17 ways.
Ok, agreed. They do stand in ascending order. Therefore, 7!/(5!2!)-4 = 17
Director
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Re: PS: Permutations & Combinations [#permalink]
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14 Aug 2008, 13:15
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Here is how I think...
7 models say 1,2,3,4,5,6,7
Now find out the arrangements where 4 and 6 are NOT adjacent or seat together = total arrangements - Seat together
total arrangements = 7C5 = 21
Seat together = If assume that 4 and 6 are selected in the pool already, need to find out other three members. Therefore we have 7-3 = 4 members left. I deduct 3 because we have to neglect 5 also otherwise 4 and 6 can't be adjacent (the five models are to stand in a line from shortest to tallest - here is the significance). Now select 3 people from 4 members in 4C3 ways.
Lets form the equation:
the arrangements where 4 and 6 are NOT adjacent or stand together = 7C5 - 4C3 = 21 - 4 = 17.
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Re: PS: Permutations & Combinations [#permalink]
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05 Sep 2009, 10:32
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lets say 1 2 3 4 5 6 7 are models with 1 to 7 from smallest to tallest.
no of ways to slect any 5 models from 7 models =7C5 =21
no of arrangement for any selection would be =1
there for no of arrangement for 21 selection =21
now 4 and 6 should not be adjusant ,so lets subtract the cases in which 4 and 6 are adjusant.
4 and 6 can be adjusant only when 5 is not selected . so we have decided about selecting two models 4 & 6
and not selecting 5 .
no of models left (1 2 3 4 5 6 7)- (4,6 )-5 =1 2 3 7 only four models are left for selection
so to make a group of 5 models we have to select 3 out of 4 (or drop 1 out of 4) =4C3 (or 4C1)=4
21-4=17
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Re: PS: Permutations & Combinations [#permalink]
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27 Sep 2009, 22:31
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Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
a) 6
b) 11
c) 17
d) 72
e) 210
Soln:
Total number of ways of choosing 5 out of 7 models is = 7C5
Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3
Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2
= 7C5 - 5C3 + 4C2
= 21 - 10 + 6
= 17
Ans is B
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Re: PS: Permutations & Combinations [#permalink]
### Show Tags
14 Nov 2009, 23:24
kudos srivas, u made it sound simple.......
I was actually planning to post this problem, thank god i searched for it before......
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Posts: 359
Re: PS: Permutations & Combinations [#permalink]
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16 Feb 2010, 12:51
srivas wrote:
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
a) 6
b) 11
c) 17
d) 72
e) 210
Soln:
Total number of ways of choosing 5 out of 7 models is = 7C5
Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3
Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2
= 7C5 - 5C3 + 4C2
= 21 - 10 + 6
= 17
Ans is B
Could someone explain me the part highlighted in red? Thanks
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Posts: 39719
Re: PS: Permutations & Combinations [#permalink]
### Show Tags
16 Feb 2010, 13:22
2
KUDOS
Expert's post
4
This post was
BOOKMARKED
jeeteshsingh wrote:
srivas wrote:
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
a) 6
b) 11
c) 17
d) 72
e) 210
Soln:
Total number of ways of choosing 5 out of 7 models is = 7C5
Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3
Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2
= 7C5 - 5C3 + 4C2
= 21 - 10 + 6
= 17
Ans is B
Could someone explain me the part highlighted in red? Thanks
When we choose 4 and 6 and they are not adjacent means that we must choose 5 too (to stand between them). So for this case we must choose 4, 5, and 6 (3C3) and 2 other from 4 left (4C2) = 3C3*4C2=4C2.
This can be solved in another way:
If we choose 4 and 6, we must also choose 5 (to stand between them) =3C3*4C2=4C2=6
We can choose either 4 or 6 = 2*1C1*5C4=10
We can choose neither 4 nor 6 = 5C5=1
6+10+1=17.
Hope it helps.
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Re: PS: Permutations & Combinations [#permalink]
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16 Feb 2010, 14:11
Bunuel wrote:
jeeteshsingh wrote:
srivas wrote:
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
a) 6
b) 11
c) 17
d) 72
e) 210
Soln:
Total number of ways of choosing 5 out of 7 models is = 7C5
Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3
Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2
= 7C5 - 5C3 + 4C2
= 21 - 10 + 6
= 17
Ans is B
Could someone explain me the part highlighted in red? Thanks
When we choose 4 and 6 and they are not adjacent means that we must choose 5 too (to stand between them). So for this case we must choose 4, 5, and 6 (3C3) and 2 other from 4 left (4C2) = 3C3*4C2=4C2.
This can be solved in another way:
If we choose 4 and 6, we must also choose 5 (to stand between them) =3C3*4C2=4C2=6
We can choose either 4 or 6 = 2*1C1*5C4=10
We can choose neither 4 nor 6 = 5C5=1
6+10+1=17.
Hope it helps.
Thanks Bunuel... I missed "are to stand in a line from shortest to tallest"... and hence I was expecting arrangements like x4xx6... too!... My bad
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Re: Out of seven models, all of different heights, five models [#permalink]
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06 Jan 2014, 20:46
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Re: Out of seven models, all of different heights, five models [#permalink]
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07 Jan 2014, 12:48
2
KUDOS
So 7C5 gives us total number of combinations =21
We need to subtract the number of ways that the 4th and 6th can be next to each other.
The 4th and 6th tallest will only be next to each other when the 5th is NOT selected. The number of combinations where the 5th is not selected is 4 - (two not selected from 7, one of them is the 5th, the others can be 1st, 2nd, 3rd and 7th.)
So we have 21-4 = 17 ways.
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Re: Out of seven models, all of different heights, five models [#permalink]
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10 Jul 2015, 19:28
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Re: Out of seven models, all of different heights, five models [#permalink]
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08 Feb 2016, 09:04
Seven models in order: ABCDEFG. D and F should not be adjacent to each other.
Total arrangements: 7C5 = 21.
When D and F are next to each other:
ABCDF
ABDFG
BCDFG
ACDFG
so 21-4 = 17 ways.
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Re: Out of seven models, all of different heights, five models [#permalink]
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09 Feb 2017, 23:25
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Re: Out of seven models, all of different heights, five models [#permalink]
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16 Jun 2017, 23:41
tarek99 wrote:
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?
A. 6
B. 11
C. 17
D. 72
E. 210
Okay..here's my take..
Let the models be(in ascending order of height)
A B C D E F G
Now..out of these 7..we have to select 5 and make them stand in ascending order of height.
According to the question, B and D cannot be standing together.
Imagine this, if we select any 5 of these, there will only be one way to make them stand.
Considering the situation, if both B and D were selected, and let's say that B and D were actually selected in the group; then they will stand together every time C is not the part of the group(as there will be no one to stand in the middle). So, our complement event is that B and D are selected, and C is not considered at all for selection. In this way, they will always stand together.
Number of ways to do this..
B and D are already in the group, so we have to select the remaining 3 out of 5. But wait, we also have to never consider C for selection. Finally then, we have to select the remaining 3 out of 4(where C is excluded)
$$4C3 = 4$$
This has to be subtracted from the total possible combinations.
$$7C5 - 4 = 17$$
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Re: Out of seven models, all of different heights, five models [#permalink] 16 Jun 2017, 23:41
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# Out of seven models, all of different heights, five models
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,244 | 17,899 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-26 | latest | en | 0.917514 |
https://www.physicsforums.com/threads/simple-math-question-will-help-me-at-work.651482/ | 1,524,532,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946314.70/warc/CC-MAIN-20180424002843-20180424022843-00281.warc.gz | 852,393,222 | 15,617 | # Simple math question, will help me at work
1. Nov 11, 2012
### SimpleSoCal
So my white collar job demands lots of number crunching as of lately. Now this is most likely simple, and I guess I should know this but.... here it goes.
I just need to know how to calculate a percentage difference on the calculator.
For example. I take 1.46 * .28 which equals .4088. I then take 2.47 * .5 which obviously halves it to 1.235.
I know how to find percentages, like a simple X * whatever percent I want. What I'm asking is, what simple thing would I do on the calculator to get the percentage difference between .4088 and 1.235?
Thanks, hope I made sense!
2. Nov 11, 2012
### chiro
Hey SimpleSoCal.
If you want to find how much larger one thing is in terms of another then you need to look at the ratio of the two objects.
For example with your numbers, the ratio of the two numbers 0.4088/1.235 = 0.3310121 and the difference between the two is found by subtracting 1 from the answer which gives -0.6689879 or in other words 0.4088 is -66.89879% larger than 1.235.
Now if you reverse the order you get 1.235/0.4088 = 3.021037 and minus 1 gives 2.021037 or 1.235 is 202.1037% larger than 0.4088.
You can interpret negative numbers like -66.89879% to be 66.89879% smaller than the larger value (so if you talk about the positive % then you are talking about much smaller something is).
If two things are equal then the ratio of both is equal to 1 and 1-1 = 0 which is a 0% difference which is what we expect.
3. Nov 11, 2012
### LeonhardEuler
It depends what you want the difference as a percentage of. 1.235 - 0.4088 = 0.8262. If you want the difference as a percentage of 1.235, it is
$$100*\frac{0.8262}{1.235} = 66.9\%$$
If you want the difference as a percentage of 0.4088, it is
$$100*\frac{0.8262}{0.4088} = 202.1\%$$
And similarly to get the difference as a percentage of any other number.
4. Nov 11, 2012
### SimpleSoCal
Thanks, guys! That helps tons. I hope I made sense on my end. What I really meant to say was percentage increase I guess. Like how much more percent bigger is 1.235 than .4088
5. Nov 11, 2012
### chiro
You'll want to use the formula I posted in that case.
Remember that an increase or decrease means you have to subtract 100% from the answer which just means subtracting 1.
6. Nov 11, 2012
### SimpleSoCal
Gotcha. Man, this place is great. More than I need but it's all useful. Thank God for the internet.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
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Very simple maths question to determine my fate Sep 3, 2011 | 806 | 2,808 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-17 | longest | en | 0.943522 |
https://studylib.net/doc/18750241/standard-7 | 1,606,488,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00311.warc.gz | 501,652,039 | 12,868 | # Standard 7
advertisement
```Domain: Number and Operations in Base Ten
Grade: 5
Core Content
Cluster Title: Perform operations with multi-digit whole numbers and with decimals to hundredths.
Standard 7: Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies
based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the
strategy to a written method and explain the reasoning used.
MASTERY Patterns of Reasoning:
Conceptual:
Students will understand the relationship between addition and subtraction when adding, subtracting, multiplying
and dividing decimals.
Students will understand the properties of operations in relationship to adding, subtracting, multiplying, and dividing
decimals to the hundredths place.
Students will understand the significance of place value when adding, subtracting, multiplying and dividing as it
applies to decimals.
Procedural:
Students can add, subtract, multiply, and divide decimals to hundredths. Include decimal dividends and divisors.
Students can divide whole numbers by 0.1 and 0.01 to build understanding of the place value significance in
division of decimal numbers.
Representational:
Students can use concrete models or drawings and strategies based on place value, properties of operations,
and/or the relationship between addition and subtraction.
Students can relate the strategy to a written method and explain the reasoning used.
Code: 5.NBT.7
Domain: Number and Operations in Base Ten
Grade: 5
Supports for Teachers
Critical Background Knowledge
Conceptual:
Students will understand the inverse relationship of multiplication and division.
Students will understand the relationship between all operations (+ - x ÷) when working with whole numbers.
Students will understand place value up to the thousandths place.
Students will understand the properties of operations in relationship to adding, subtracting, multiplying, and dividing
multi-digit whole numbers.
Procedural:
Students can fluently add, subtract, multiply, and divide multi-digit whole numbers using the standard algorithms.
Representational:
Students can model whole number division.
Students can model multi-digit whole number multiplication.
Students can accurately represent multi-digit whole numbers with concrete models and drawings.
Academic Vocabulary and Notation
properties of operations, operation notations, multiplication symbols [3 x a, 3 . a, 3 * a, 3(a), 3a]
𝑎
division notation (𝑏 , 𝑎�𝑏, 𝑎 ÷ 𝑏, a)b )
Instructional Strategies Used
Build understanding of place value by dividing by 0.1 and
0.01 before moving to other tenths and hundredths.
Explore how the result is related to using the powers of
ten to multiply or divide.
Use a variety of measurement contexts for addition and
subtraction of decimal numbers.
Code: 5.NBT.7
Resources Used
National Library of Virtual Manipulatives:
http://nlvm.usu.edu/en/nav/frames_asid_264_g_2_t_1.html?fr
om=category_g_2_t_1.html
Domain: Number and Operations in Base Ten
Grade: 5
Explain the significance of decimal places when working
with measurement (time, weight, length, money, area,
volume).
Have students make up a story problem that involves
22.8 ÷ 6 . Ask them to share their number stories and
explain how they got the answer to their problem.
Assessment Tasks Used
Skill-Based Task:
Calculate the following and show your work.
1.
2.
3.
4.
3.4 + 6.2
7.7 – 4.1
5.6 × 2.4
8.4 ÷ 2.1
Problem Task:
A tabletop has the measurements 3.5 meters by 1.2 meters.
what is the area in square meters? If your brother cut 0.3
meters off of one side, how would that affect the area of the
table top. Does it matter which side is cut? Show your work,
including a diagram. You can use graph paper if needed.
I divided 6.12 by 3 and got the quotient 2.4. What did I do
wrong? Give a similar problem where I might make the same
error.
In this calculation some numbers are missing. What might
they be? How do you know?
3.??
- ?.7?
1.?3
I added 3 decimal numbers together and got exactly 4. What
might those 3 decimal numbers be?
How many different ways can you make your calculator show
a number with a particular decimal, such as 12.34, without
pressing the decimal point button?
Code: 5.NBT.7
``` | 998 | 4,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-50 | latest | en | 0.895476 |
https://learnshareit.com/how-to-get-the-number-of-minutes-between-two-dates-in-javascript/ | 1,695,417,407,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00798.warc.gz | 411,289,313 | 51,932 | # How To Get The Number Of Minutes Between Two Dates In JavaScript
## Get The Number Of Minutes Between Two Dates In JavaScript
These are some methods that can help you get the number of minutes between two Dates in JavaScript when working with the Date object in JavaScript. You can learn more about the Date object in my previous tutorial.
### Calculate the duration indirectly
You can get the number of minutes between two Dates indirectly by calculating the duration between two Dates and converting it to minutes.
First, you calculate the durations between two dates and January 1, 1970 by the getTime() method.
Second, subtracting them. The result is the duration between two dates by milliseconds.
Finally, your remaining task is to convert the result from milliseconds to minutes. It will be the result of the calculation.
The getTime() method returns the number of milliseconds since January 1, 1970.
Syntax
getTime()
Parameters: None.
Return Value: The number of the milliseconds since January 1, 1970.
<!DOCTYPE html>
<html>
<title>
How To Get The Number Of Minutes Between Two Dates In JavaScript
</title>
<body style="text-align: left;">
<h1 style="color: rebeccapurple;">
How To Get The Number Of Minutes Between Two Dates In JavaScript
</h1>
<h2 style="color: black;">
Get The Number Of Minutes Between Two Dates In JavaScript
</h2>
<h3 style="color: black;">
Calculate the duration indirectly
</h3>
<script>
const firstDate = new Date("10/10/2022");
document.write("<br>The first date: " + firstDate);
const secondDate = new Date();
document.write("<br>The second date: " + secondDate);
// The function that gets the number of minutes between two Dates.
function getDuration(firstDate, secondDate) {
const seconds = (secondDate.getTime() - firstDate.getTime()) / 1000;
const minutes = Math.floor(seconds / 60);
return minutes
}
document.write("<br>The Number Of Minutes Between Two Dates: " + getDuration(firstDate, secondDate));
</script>
</body>
</html
Output
### Calculate the duration directly
You can get the number of minutes between two Dates directly by calculating the duration between two Dates and converting it to minutes.
First, you get the different days between two Dates by getDate() .
Second, you get the different times of two Dates and convert them to minutes by getHours() and getMinutes() methods.
Finally, you can get the result by adding the result of the step one to the result of the step 2.
<!DOCTYPE html>
<html>
<title>
How To Get The Number Of Minutes Between Two Dates In JavaScript
</title>
<body style="text-align: left;">
<h1 style="color: rebeccapurple;">
How To Get The Number Of Minutes Between Two Dates In JavaScript
</h1>
<h2 style="color: black;">
Get The Number Of Minutes Between Two Dates In JavaScript
</h2>
<h3 style="color: black;">
Calculate the duration directly
</h3>
<script>
const firstDate = new Date("10/10/2022");
document.write("<br>The first date: " + firstDate);
const secondDate = new Date();
document.write("<br>The second date: " + secondDate);
// The function that gets the number of minutes between two Dates.
function getDuration(firstDate, secondDate) {
const diffDate = secondDate.getDate() - firstDate.getDate();
const result = diffDate * 24 * 60 + (secondDate.getHours() * 60 + secondDate.getMinutes()) - (firstDate.getHours() * 60 + firstDate.getMinutes());
return result;
}
document.write("<br>The Number Of Minutes Between Two Dates: " + getDuration(firstDate, secondDate));
</script>
</body>
</html
Output
## Summary
You can calculate the duration of two Dates directly or indirectly and convert it to minutes to get the number of minutes between two Dates in JavaScript. Choose the way that is best for you. We hope this tutorial is helpful to you. Thanks!
Maybe you are interested: | 853 | 3,800 | {"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} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.701522 |
https://codereview.stackexchange.com/questions/131835/circular-primes-below-101025 | 1,718,459,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00460.warc.gz | 147,618,760 | 42,582 | # Circular primes below 10^1025
## The problem
Project Euler: 35 is stated in the following way
Circular primes
Problem 35
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
## My attempt
I solved this question some time ago, and have now come back to it because I saw it pop up a few times here on CR. I wanted to over-engineer the problem and instead of finding the circular primes under a million I wanted to find them under some crazy big number.
It is not terribly quick, but uses that all circular primes above a million are repunit primes with a prime number of digits.
I included two different ways of checking whether a number (or in this case a list of numbers) is a circular prime. I use the following
def is_circular_2(p):
for i in range(len(p)):
if not isprime(int(''.join(map(str, p[i:]+p[:i])))):
return False
return True
However after some quick googling, the following seem to be the "preferred" way
def lst_2_int(lst):
return int(''.join(map(str, lst)))
def is_circular(p):
return all(isprime(lst_2_int(p[i:]+p[:i])) for i in range(len(p)))
However after some light speedtests using timeit, this version is a tad slower. As many other answers do I cycle through the products of the odd integers [1, 3, 7, 9] since no prime can end in an even integer, or 5.
I do not bother saving the primes or non primes already found. Is it "worth" it? isprime from the primefac libary already seems quite fast. To find the really big circular primes I iterate over the repunit primes. Eg primes only consisting of 1s. No need to check if it is circular. I also used the small increase that a repunit number is prime if and only if it contains a prime number of digits.
Looking for any sort of general comments on my code. Not sure if my code can be improved speedwise, but comments there are appreciated as well.
## The code
from primefac import isprime, primes
from itertools import product # cartesian product
'''
Circular primes
Project Euler: Problem 35
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
'''
def lst_2_int(lst):
return int(''.join(map(str, lst)))
def is_circular(p):
return all(isprime(lst_2_int(p[i:]+p[:i])) for i in range(len(p)))
def is_circular_2(p):
for i in range(len(p)):
if not isprime(lst_2_int(p[i:]+p[:i])):
return False
return True
def repunit(k):
return (10**k-1)//9
def find_circular_primes_under(digits):
if type(digits) != int or digits <= 0:
raise ValueError(
"Error: power needs to be a positive integer, 10^input")
circular_lst = []
# Circular primes under 10
total = 0
for k in xrange(10):
if isprime(k):
circular_lst.append(k)
# Circular primes under 10^6
for k in xrange(2, min(digits+1, 7)):
for combo in product([1, 3, 7, 9], repeat=k):
if is_circular_2(list(combo)):
circular_lst.append(lst_2_int(combo))
# All circular primes over 10^6 are repunit primes (1...1 where ... =
# index)
if digits > 6:
prime_lst = primes(digits+1)
# All repunit primes have a prime number of digits
for prim in prime_lst[3::]:
if isprime(repunit(prim)):
circular_lst.append('1 ... 1 ['+str(prim)+']')
return circular_lst
if __name__ == '__main__':
circular_primes = find_circular_primes_under(400)
for prim in circular_primes:
print prim
print "Found a total of", len(circular_primes), 'circular primes.'
• I have no idea what repunit primes are, did you mean repeating primes instead?
– Mast
Commented Jun 13, 2016 at 8:15
• A repunit is a number like 11, 111, or 1111 that contains only the digit 1. A repunit prime is a repunit that is also a prime number. Commented Jun 13, 2016 at 8:24
• That's just 1 and 11 though. All others are not prime AFAIK.
– Mast
Commented Jun 13, 2016 at 9:07
• Out of curiosity, why are candidates over 10^6 all repunits ? Commented Jun 13, 2016 at 9:25
• A circular prime is less strict than a permutable prime, e.g. 197 is circular but not permutable. Are you certain only repunits can be circular primes above one million? Commented Jun 13, 2016 at 20:36
An optimisation for values under 10^6 (or if you want to ignore the conjecture)
Primality tests can be expensive, especially for big values. At the moment, you often perform the same primality check on the same value multiple times. For instance, with the same "base number" 199933, you'll:
• consider 199933 and check the primality of its rotations (199933, 319993, 331999, 933199, 993319, 999331)
• consider 319993 and check the primality of its rotations (319993, 331999, 933199, 993319, 999331, 199933).
• etc.
It may be a better option when considering a new permutation to handle all the rotations in one go. For instance, you could decide that a given permutation is interesting only if it corresponds to the smallest of its roations.
Then you'd have something like (a set is used to get rid of duplicated rotations):
# Circular primes
for k in xrange(2, digits+1):
for p in product(['1', '3', '7', '9'], repeat=k):
p_int = int(''.join(p))
perm = {int(''.join(p[i:] + p[:i])) for i in range(len(p))}
if p_int == min(perm) and all(isprime(n) for n in perm):
circular_sets.append(perm)
Now the primality is never checked more than once for a given number.
It's hard to review this kind of code unless there's a clear idea of what your self-imposed constraints are. What I mean by this is that the code makes use of an assumption (all permutable primes over $10^6$ are repunits) that it does not check. But if it is legitimate to make use of assumptions that the program itself does not check, then it would be much faster to avoid all those calls to isprime and write:
# https://oeis.org/A004023
A004023 = 19, 23, 317, 1031, 49081, 86453, 109297, 270343
circular_lst.extend((10**p - 1) // 9 for p in A004023 if p <= digits)
` | 1,721 | 6,139 | {"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} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.880975 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=2365&CurriculumID=40&NQ=8&Num=4.16 | 1,560,897,553,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998844.16/warc/CC-MAIN-20190618223541-20190619005541-00448.warc.gz | 263,322,991 | 4,322 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation4.16 Integers - Positive and Negative Integers
Natural Numbers: Natural Numbers are 1,2,3,4,5,................ Whole Numbers: Whole Numbers are 0,1,2,3,4,5,............... Positive Numbers: Positive numbers are, 1,2 ,3 ,4 ,5................. Negative Numbers: Negative numbers are, ............-3, -2, -1. They are read as negative three, negative two, negative one. Integers: A set containing the positive numbers, 1, 2, 3, 4, ...., and negative numbers,............-3, -2, -1, together with zero is called a set of integers. In other words, Integers are defined as set of whole numbers and their opposites. Z = {..., -3, -2, -1, 0, 1, 2, 3, ...} {1, 2, 3, . . .} is the set of positive integers { . . . -3, -2, -1} is the set of negative integers Zero is not positive nor negative, but is both. A positive integer has a (+) or addition sign in front of the number and looks like ( 5 or (+ 5). Also note that a positive integer doesn't need a sign. A negative integer has a (-) or negative sign in front of the number and looks like (- 7). A negative integer can be thought of as owing someone something. Think of it as (-8) means owing someone 8 items or things. Zero in neither positive or negative and is always larger than a negative integer. Positive integers are always larger than zero and any negative number. The larger a positive integer is, the bigger it is. Therefore a +7 is greater than a +5, written another way +7 > +5. The larger a negative integer the smaller is its value. Therefore, (-6) < (-5). Directions: Answer the following questions. Write your own examples of positive and negative integer problems.
Q 1: _________ numbers are used for counting.NaturalNegativeReal Q 2: If an integer is less than zero, we say that its sign is negative.truefalse Q 3: If an integer is greater than zero, we say that its sign is negative.falsetrue Q 4: Positive integers are all the whole numbers greater than zero: 1, 2, 3, 4, 5, ... .falsetrue Q 5: Negative integers are all the opposites of these whole numbers: -1, -2, -3, -4, -5, …truefalse Q 6: Zero is neither positive nor negative.falsetrue Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 663 | 2,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-26 | latest | en | 0.865525 |
https://liangminblog.wordpress.com/2017/02/23/feature-engineering/ | 1,660,173,315,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00085.warc.gz | 356,372,416 | 28,984 | machine learning
# Feature Engineering
### Data Pre-processing(Transformation)
Normalization
sigmoid normalization
0-1 normalization ((bla – min(bla)) / ( max(bla) – min(bla) ))
z-score
Gaussian normalization (Gaussian kernel)
Box-cox transformation
log transformation
Tukey’s Ladder of Powers
### Feature Engineering
image
speech
text
time series: entropy, approximate entropy, sample entropy
### Data Visualization
1. Statistics
2. Histogram
3. Density estimation: kernal density estimation (Parzen–Rosenblatt window)
### Feature Selection
How to choose a proper feature selection method for your data? Go from easier ones to complicated ones, go from linear ones to non-linear ones.
The combinations of individually good features do not necessarily lead to good classification performance. “The m best features are not the best m features”
Similarity Measure
Euclidean distance
Cosine distance
Gaussian distance
is a multi-dimensional generalization of the idea of measuring how many standard deviations away a point P is from the mean of the distribution D. The Mahalnobis distance transforms the random vector into a zero mean vector with an identity matrix for covariance. In that space, the Euclidean distance is safely applied. It can be used to identify outliers, which are data points away from the distribution of the data. We can consider each feature a time (multivariate reduced to univariate), then the covariance matrix reduces to a diagonal matrix. Then, we can rank the features by the distance, and delete one feature a time to identify the best combination of features by investigating the changes of the metric.
Statistical Tests
Hypothesis testing to test whether the difference of one feature is significant among classes: t-test
Filter Methods
• Correlation
• F-statistics
• Mutual information
MI is more general and determines how similar the joint distribution p(X,Y) is to the products of factored marginal distribution p(X)p(Y). I (i) is a measure of dependency between the density of variable xi and the density of the target y. Intuitively, mutual information measures the information that X and Y share: it measures how much knowing one of these variables reduces uncertainty about the other. I(X; Y) = 0 if and only if X and Y are independent random variables. Moreover, mutual information is nonnegative (i.e. I(X;Y) ≥ 0; see below) and symmetric (i.e. I(X;Y) = I(Y;X)).
The area contained by both circles is the joint entropy H(X,Y). The circle on the left (red and violet) is the individual entropy H(X), with the red being the conditional entropy H(X|Y). The circle on the right (blue and violet) is H(Y), with the blue being H(Y|X). The violet is the mutual information I(X;Y), which is equivalent to the amount of uncertainty in Y which is removed by knowing X
How to estimate mi of continuous variables: data discretization; density estimation method (e.g., Parze windows with Gaussian window)
• MRMR(Minimum Redundancy Maximum Relevance)
Maximal relevance: selecting the features with the highest relevance to
the target class c. Relevance is usually characterized in terms of correlation of mutual information.
Minimal redundancy:
combine MRMR:
We can get a ranking list of all the features, and apply wrapper method with the rank.
Linear Methods
FDA (Fisher’s discriminant analysis)
Tree Based Methods(embedded)
Adaboost with a tree stump (Variable importance is measured by how much error the variable reduced each time it was used in a tree’s split/branch), CART, BART (tree model), random forest
univariate methods,linear models and regularization and random forests for feature selection, stability selection, Recursive feature elimination
Greedy Selection(wrapper)
Greedily select, use the performance of a reliable classifier. Combined with data partition (subsampling)
Dimension Reduction Methods(covered in representation)
Dimension reduction method, which will change the feature space
Regularization/sparsity(embedded)
Ensemble Feature Selection
• Harmony search
• Data Reliability Based Feature Selection: a feature is considered reliable (or relevant) if its values are tightly grouped together.
• Stability selection: apply a feature selection algorithm on different subsets of data and with different subsets of features (bootstrap)
• Boosting
!Categorical features
Random Permutation
## Imbalanced Data
Under-Sampling
• Condensed Nearest-Neighbor
• One-sided Selection
Over-Sampling
• SMOTE: First it finds the n-nearest neighbours in the minority class for each of the samples in the class . Then it draws a line between the the neighbours an generates random points on the lines.
• ADASYN: After creating those sample it adds a random small values to the points thus making it more realistic. In other words instead of all the sample being linearly correlated to the parent they have a little more variance in them i.e they are bit scattered.
Cross Validation for imbalanced data
Ensemble different resampled datasets: When deal with imbalanced data set, create multiple balanced data sets from the original imbalanced data set via sampling, and subsequently evaluate feature subsets using an ensemble of base classifiers each trained on a balanced data set. And add ratio.
Cluster the majority class: Instead of relying on random samples to cover the variety of the training samples, he suggests clustering the abundant class in r groups, with r being the number of cases in r. For each group, only the medoid (centre of cluster) is kept. The model is then trained with the rare class and the medoids only.
under-sampling, over-sampling, increasing minority samples and decreasing majority samples simultaneously, synthesize “new” samples from the minority class, bootstrap
GAN for data augmentation
## Automated Feature Engineering
DFS (Deep Feature Synthesis) | 1,244 | 5,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.834198 |
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=368.msg12054 | 1,618,967,133,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00399.warc.gz | 160,186,938 | 13,089 | NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
April 21, 2021, 08:17:27 am
Wisdom is to teach our students how to teach themselves. ...Wisdom
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Author Topic: RLC AC circuit (Read 39000 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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« Embed this message on: June 06, 2006, 05:02:33 pm »
-*-
This applet simulatie an RLC circuit.
You can adjust angular frequency for the AC source , R, L, C .
2007/08/05:
The code has been modified so that you can remove one or two elements.
Now you can study pure resistor circuit, RC, RL, LC or RLC circuit with this simulation.
I also add small circle along the circuit loop to simulate the average motion of charge particle in the loop.
The current flow I in the loop in proportional to (in phase) the voltage of the resistor VR.
However, there is a 90 degree out of phase between voltage of inductor/capacitor and the current I.
The slider bar on the left is used to adjust the voltage of the source. V
Embed a running copy of this simulation
Embed a running copy link(show simulation in a popuped window)
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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fredcfreitas
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Posts: 2
« Embed this message Reply #1 on: September 22, 2007, 01:32:27 am »
Hi,
Where is the java applet file? I am a undergraduate student and I need this applet for a work in learning of physics. If you have this file already, send-me please.
Thank's
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Fu-Kwun Hwang
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« Embed this message Reply #2 on: September 23, 2007, 12:42:43 am »
Click "get files for offline use" button, and you will get related files.
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fredcfreitas
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« Embed this message Reply #3 on: March 07, 2008, 06:48:49 pm »
Thanks for attention!
I'm sorry for my donkey question. Anyway, I do my work and your help was very important.
Thanks again.
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DaZyS
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Posts: 14
« Embed this message Reply #4 on: October 10, 2010, 09:48:45 pm »
Hello sir Fu-Kwun Hwang, I have a circuit with one current source and one memristor.
I want to make something similar to his simulation,
but I don't understand it.
How do you move the balls around the circuit?
Thanks!
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Fu-Kwun Hwang
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« Embed this message Reply #5 on: October 10, 2010, 10:08:02 pm »
You can always download EJS jar file. Double click to run it. then Click right mouse button, select "open EJS model"
to view how the EJS simulation was created.
I use particleSet to represent movement of charge particles. You need to set coordinate and velocity for those particles.
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DaZyS
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« Embed this message Reply #6 on: October 12, 2010, 06:27:10 pm »
but only one ball move and the square and triangular wave have changed!
I'm sending you my EJS's file, could you look it? Thank you.
*** There are 1 more attached files. You need to login to acces it! Logged
Fu-Kwun Hwang
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« Embed this message Reply #7 on: October 12, 2010, 09:20:31 pm »
The simulation presented above is a circuit, and I add particles to re-present movement of charged particle in the circuit.
I do not know why you are talking about square wave or triangle wave?
You should not change your model if you just want to add view element.
There is no need to open your file because I do not really understand what you want to do/show?
And I do not know what to look for?
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lookang
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http://weelookang.blogspot.com
« Embed this message Reply #8 on: April 18, 2011, 09:50:57 pm »
this is an excellent applet. can see a lot of effort in making it more useful to teachers and students.
below is a slightly remixed version
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smith88
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« Embed this message Reply #9 on: August 18, 2011, 04:07:10 am »
I also add small circle along the circuit loop to simulate the average motion of charge particle in the loop.
The current flow I in the loop in proportional to (in phase) the voltage of the resistor VR.
However, there is a 90 degree out of phase between voltage of inductor/capacitor and the current I.
The slider bar on the left is used to adjust the voltage of the source. V
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bun
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« Embed this message Reply #10 on: October 03, 2019, 10:25:28 am »
Thank you. I need to study the series RLC.
It is possible to change the range of the R, L and C? Thank you
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Fu-Kwun Hwang
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« Embed this message Reply #11 on: October 21, 2019, 08:42:22 am »
Please give me an example what kind of range of the R,L,C values you are interested.
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Wisdom is to teach our students how to teach themselves. ...Wisdom | 1,427 | 5,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.869182 |
https://alan-turing-institute.github.io/rse-course/html/module01_introduction_to_python/01_09_iteration.html | 1,660,418,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00029.warc.gz | 120,065,665 | 8,982 | # Iteration¶
Our other aspect of control is looping back on ourselves.
We use forin to “iterate” over lists:
mylist = [3, 7, 15, 2]
for whatever in mylist:
print(whatever ** 2)
9
49
225
4
Each time through the loop, the variable in the value slot is updated to the next element of the sequence.
## Iterables¶
Any sequence type is iterable:
vowels = "aeiou"
sarcasm = []
for letter in "Okay":
if letter.lower() in vowels:
repetition = 3
else:
repetition = 1
sarcasm.append(letter * repetition)
"".join(sarcasm)
'OOOkaaay'
The above is a little puzzle, work through it to understand why it does what it does.
### Dictionaries are Iterables¶
All sequences are iterables. Some iterables (things you can for loop over) are not sequences (things with you can do x[5] to), for example sets and dictionaries.
import datetime
now = datetime.datetime.now()
founded = {"James": 1976, "UCL": 1826, "Cambridge": 1209}
current_year = now.year
for thing in founded:
print(thing, "is", current_year - founded[thing], "years old.")
James is 46 years old.
UCL is 196 years old.
Cambridge is 813 years old.
## Unpacking and Iteration¶
Unpacking can be useful with iteration:
triples = [[4, 11, 15], [39, 4, 18]]
for whatever in triples:
print(whatever)
[4, 11, 15]
[39, 4, 18]
for first, middle, last in triples:
print(middle)
11
4
# A reminder that the words you use for variable names are arbitrary:
for hedgehog, badger, fox in triples:
11
4
for example, to iterate over the items in a dictionary as pairs:
things = {
"James": [1976, "Kendal"],
"UCL": [1826, "Bloomsbury"],
"Cambridge": [1209, "Cambridge"],
}
print(things.items())
dict_items([('James', [1976, 'Kendal']), ('UCL', [1826, 'Bloomsbury']), ('Cambridge', [1209, 'Cambridge'])])
for name, year in founded.items():
print(name, "is", current_year - year, "years old.")
James is 46 years old.
UCL is 196 years old.
Cambridge is 813 years old.
## Break, Continue¶
• Continue skips to the next turn of a loop
• Break stops the loop early
for n in range(50):
if n == 20:
break
if n % 2 == 0:
continue
print(n)
1
3
5
7
9
11
13
15
17
19
These aren’t useful that often, but are worth knowing about. There’s also an optional else clause on loops, executed only if you don’t break, but I’ve never found that useful.
## Classroom exercise: the Maze Population¶
Take your maze data structure. Write a program to count the total number of people in the maze, and also determine the total possible occupants. | 714 | 2,490 | {"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} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.807246 |
http://www.jiskha.com/display.cgi?id=1224032947 | 1,493,614,073,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917127681.50/warc/CC-MAIN-20170423031207-00267-ip-10-145-167-34.ec2.internal.warc.gz | 557,130,864 | 3,856 | # physics
posted by on .
A 45 kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.0 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her).
• physics - ,
Impulse = changeinMomentum
force*time=mass*changeinvelocity
Now note: it is the wall which pushes her, so the wall force is in the direction of her motion. The force she exerts is opposite to the wall.
• physics - ,
thank you | 162 | 674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-17 | longest | en | 0.927694 |
https://essayzoo.org/essay/other/mathematics-and-economics/math-quiz.php | 1,604,143,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00569.warc.gz | 300,873,822 | 6,521 | Not register? Register Now!
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# Math Quiz (Essay Sample)
Instructions:
Mathematics and Economics Essay: Math Quiz source..
Content:
Running head: Math Quiz
TEST #3 FORM C
Customer Inserts His/Her Name
Customer Inserts Grade Course
Customer Inserts Tutor`s Name
Writer Inserts Date Here (Day, Month, Year)
Solve the following equations. Give Exact Solutions.
8x-3= 143x
(2)3(x-3)= ( 2)(-2)(3x)
3(x-3) = -6x
3x - 9 = -6x
3x + 6x = 9
9x = 9
x = 1
log5125 =x
5x= 125
5x= ( 53 )12
x=32
logx164 = -3
x- 3= 164
x- 3= 4- 3
x = 4
log2x+ log2x+3= log218
Using log property, sum of logs is equal to their products,
log2x(x +3) = log218
log2x2+3x = log218
x2+3x=18
x2+3x-18=0
Factorizing the above quadratic equation, we get:
x2+6x-3x-18=0
x(x + 6) - 3( x + 6) = 0
(x + 6) (x - 3) = 0
x = - 6, x= 3.
log103x+17- log102x=1
Using log property, subtraction of logs is equal to their division
log103x+172x=1
Using log property, log of same base is equal to 1.
log103x+172x = log10 10
3x+172x=10
3x + 17 = 20x
3x - 20x = - 17
-17x = - 17
x = 1
logbx+ logb4- 3logby
Using log property, addition of log is equal to their product
= logb4)(x- 3logby
= logb4x - logby3
Using log property, subtraction of logs is equal to their division
= logb4xy3
Graph each function. List at least three points
f(x) = 3x
x
f(x)
10
f(10) = 310 =0.059 x 106
16
f(16) = 316 =4.3 x 107
20
f(20) = 320 =3.48 x 109
g(x) = log3x
x
f(x)
3
g(3) = log33 = 1
8
g(8) = log38 = 1.89
10
g(10) = log310 = 2.09
Solve the following inequalities and graph the solution set: | 705 | 1,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-45 | latest | en | 0.766287 |
http://kitada.com/List/time/199909/0085.html | 1,558,738,658,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257781.73/warc/CC-MAIN-20190524224619-20190525010619-00418.warc.gz | 117,768,485 | 5,393 | # [time 733] Re: [time 730] Entanglement defines the fundamental bi-simulation?
Stephen P. King (stephenk1@home.com)
Thu, 09 Sep 1999 17:04:02 -0400
Dar Matti,
Matti Pitkanen wrote:
snip
[SPK]
> > Frieden has pointed out that the Shannon entropy is a "global measure",
> > should we not be conserned with the local measure, since the particular
> > observer's perceptions are restricted to local measures? Or, is this how
> > you define the NMP so that it choses from a global set?
[MP]
> Shannon entropy defined here is local in the sense that it applies
> inside each self, be it DNA triplet or human. Locality
> is always relative to some scale.
This is an example of what Hitoshi calls <glocal>. The localization is
relative to the scale set by the LSs propagator... I was thinking more
of the difference between Shannon and Fisher information as discussed in
Frieden's paper "Lagrangians of physics and the game of
Fisher-information transfer" By Frieden and Soffer. I'll include it in
my package of papers that I am putting together for you...
snip
[MP]
> > > In p-adic context one must defined logarithm appropriately
> > > and this leads to some exotic effects (entanglement without
> > > entanglement entropy).
> > > ***************
[SPK]
> > Umm, that is interesting but I wish you could give an example of
> > "entanglement without entanglement entropy". :-) I am still building my
> [MP] p-Adic logarith is counter part of p-based
> real logarithm is defined in such a manner that it satisfies
> the usual sum rule
>
> Log_p[SUM(n=n0) x_n p^n ]= n_0.
>
> It is easy to verify that Log_p(xy)= Log_p/x) + Log_p(y).
> This guarantees the addivity of entropy.
How do we consider the case where x and y interact?
> When x is of form
>
> x_0 +x_1p+...with x_=1,2...., or p-1
>
> one has n_0=0 and p-adic logarith vanishes!
>
> This means that p-adic entanglement entropy vanishes if
> For instance, p_n could be of form
>
> p_k = n_k/N
>
> N= SUM(k) n_k
>
> such that n_k and N are not divisible by p.
This is facinating! The p-adicity makes the entanglement entropy
selective to the value of the prime! Thank you for explaining this to
me. I would like to have a concrete example to test my intuitions
against... :-)
[SPK]
> > > I think that the determination of which "subsystem-complement"
> > > pair has
> > > the minimal quantum entropy is given by a tournament of games "played"
> > > between the pairs. The winner of the tournament is the quantum state
> > > that is the most informative. I see the "tournament" as modelable by a
> > > periodic gossiping on graphs formalism.
> > > http://www.cs.wvu.edu/~chif/cs418/1.html
> > >
> > > [MP] This would represent attempt to reduce quantum jump to
> > > classical computation. What makes me sceptic are Bell inequalities
> > > plus my belief that genuine (not completely) free will resides in quantum
> > > jump. Quantum jump is not reducible to process, quantum jump
> > > is the Spirit, the Godly.
[SPK]
> > I agree, tenatively! I am not sure how to derive the Bell Inequalities
> > from the statistics of tournaments, but I am certain that they can be
> > given since the distinction between classical and quantum computation is
> > that the former does not consider ensembles of systems while the later
> > does.
>
> [MP] I think that the unitary time development defining
> large number of parallel computations (N computations for N-dimensional
> system) is the basic difference. Quantum jump halting the computation
> selects one computation. This is what makes quantum computation so
> effective that it could make finding of prime decompositions of integeres
> child's play some day.
Ok, I think that we need to look long and hard at this issue! I have
found several papers on quantum computation; I'll pass them along to you
if you like.
> > > [SPK] The main ideas presupposes that "subsystem-complement" pairs can
> > > communicate with each other. I suspect that this follows some thing like
> > > this: Subsystem A <-> Complement B, Subsystem B <-> Complement A. If the
> > > complement of subsystem A is subsystem B and the complement of subsystem
> > > B the subsystem A, then subsystems A and B have identical entanglement
> > > entropy or information.
> > >
> > > [MP] I have for a long time pondered the problem whether this is
> > > indeed suggested by quantum measurement theory.
> > >
> > > The notion of self seems to resolve the question finally:
> > > communication is *not* in question in the sense one might
> > > think. The self containing *both* the subsystem *and* its complement is
> > > the basic experiencer. Not the subsystem or/and its complement.
[SPK]
> > Yes, I believe you are right about that. I needed to test a
> > hypothesis... The key point is that we can model the interaction of
> > observers as resulting in an equilibration in both thermodynamic and
> > information theoretical terms!
> [MP] Average entanglement entropy probably corresponds to thermodynamic
> entropy but I am somewhat cautious here. The point is that
> thermodynamic entropy is concept characterizing ensemble.
> Entanglement entropy characterizes single subsystem: purely
> quantum mechanical concept is in question. No idealizations
> brought in by thermodynamics.
Yes, but the equilibration of temperatures, etc. is a real phenomenon!
The same ensemble approach that props up out thinking of probability
waves and information theory so we must be consistent here! Perhaps we
need to look at this concept more closely!
[MP]
> > > Note: the map m-->M(m) defined by the diagonalized
> > > density matrix maps the states of the subsystem
> > > of self to the states of its complement in self and is
> > > perhaps analogous to *'bi-simulation map'* that Stephen has
> > > been talking.
> > >
> > > Entanglement would define the fundamental bisimulation.
> > > Subsystems of self would simulate each other just at the
> > > moment when they wake-up and reduce quantum entanglement
> > > to zero. When they are selves they do not anymore bisimulate.
> > > This would be sub-conscious bisimulation. Note that
> > > any entangled subsystem of self would unconsciously bisimulate its
> > > complement.
[SPK]
> > Bisimulation, by Peter's definition, captures the notion "underwhat
> > conditions do two systems have the same behaviour". The difficulty that
> > I see the "atomisity" of systems is not necessarily an absolute. We can
> > consider it to be such if we only are considering possible systems that
> > have similar enough subjective measures, e.g. clocking and gauging
> > standards.
> >
> [MP] Entanglement does not define bisimulation in this sense.
> It characterizes only measurement: map of the states of system
> A to those of B.
Round and round we go... What is a measurement?
[SPK]
> > > Now, what is a given pair of subsystems do not have complete
> > > agreements, but do share some information? (I see "information sharing"
> > > as the existence of identical configurations in the configuration space
> > > of each subsystem, following the logic that "identical configurations
> > > encode identical information".) Can we model how, given an initial
> > > common information, a pair of subsystems can evolve such that they
> > > become equivalent? This is what happens in the periodic gossiping
> > > situation, so I suspect that it may be useful.
> > > [MP] This would require precise specification of a model for
> > > interaction. As I mentioned: quantum entanglement defines
> > > a map between states of subsystem and its complement
> > > resembling bisimulation: M(m) simulates m and vice versa.
> > > Schrodinger cat bravely simulates atomic nucleus whose
> > > transition leads to the opening of the bottle of poison.
[SPK]
> > Can we think of this as a process, like Fitini Markopoulou's idea of
> > evolving sets? I see a loose analogy in her thinking and your q-jumps,
> > Matti, in that the set of past events is "updated", but there are a lot
> > of differences. I think that you sould take a hard look at the Category
> > theory approach!
> [MP] Category theory might provide stimulate some ideas. If I only had the
> time to do all those things I want to do! In any case, m-->M(m) does not
> mean that two systems have same behaviour: this is unfortunately the case.
Ok, but can there behaviour be similar enough so that the information
content can be related?
Later,
Stephen
This archive was generated by hypermail 2.0b3 on Sat Oct 16 1999 - 00:36:40 JST | 2,142 | 8,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.886583 |
http://observations.rene-grothmann.de/technical-details-in-proofs/ | 1,680,303,038,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00178.warc.gz | 36,877,696 | 9,225 | # Technical Details in Proofs
Einstein is reported to have said that you should explain things as easy as possible, but not easier. In a talk, you may disobey this in favor of communicating the main ideas. But you should leave behind the impression that you know that there are some nasty details left for the reader.
An example is the following proof by Cantor that the reals have the same cardinality as the power set of the natural numbers. I heard it in a talk recently. The idea is to map any subset M of the natural numbers to a real number
$$0.i_1i_2i_3\ldots_2 = \sum\limits_{k=1}^\infty \dfrac{i_k}{2^k}$$
in dual representation by the rule
$$i_k = \begin{cases} 1 & k \in M \\ 0 & k \notin M \end{cases}$$
The idea is that all numbers in [0,1] have a dual representation of this kind, and that we produce all subsets of the natural numbers with these representations and vice versa. It is an easy (?) geometrical task to show that [0,1] has the cardinality of the real numbers.
The problem of this idea is the small technical detail that the mapping constructed above is not really a bijection. It is merely surjective. The problem is that e.g. the following representations produce identical reals.
$$0.0100110011111\ldots_2 = 0.01001101_2$$
So how can we fix the bijection? The reason we can do it is that the number of exceptions is countable. In general, any infinite set retains its cardinality, if we remove a countable set from it, but leave at least another countable set. This is a general result, and it is not too difficult to prove.
In our case, we are facing the problem that two subsets may map to the same real number, one of it finite and the other containing all elements k from some K on. The number of pairs of these subsets is countable however. So we can omit one subset of each pair. E.g., we can delete all finite sets from the power set first.
Set theory was never easy for me. There are various theorems, where I would not know if complicate proofs involving the axiom of choice are necessary or not. An example is the theorem that if you have a surjection from A to B and one from B to A, then there is a bijection between A and B. Tough!
Diese Website verwendet Akismet, um Spam zu reduzieren. Erfahre mehr darüber, wie deine Kommentardaten verarbeitet werden. | 571 | 2,310 | {"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} | 3.6875 | 4 | CC-MAIN-2023-14 | latest | en | 0.94255 |
https://www.fmaths.com/interesting-about-math/often-asked-what-is-a-percentage-in-math.html | 1,632,387,312,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00312.warc.gz | 805,654,496 | 10,497 | # Often asked: What Is A Percentage In Math?
## How do you explain percentages?
Percentages as Decimals and Fractions It can therefore be written as both a decimal and a fraction. To write a percentage as a decimal, simply divide it by 100. For example, 50% becomes 0.5, 20% becomes 0.2, 1% becomes 0.01 and so on. We can calculate percentages using this knowledge.
## What is 10 as a percentage of 100?
How much is 10 out of 100 written as a percent value? Convert fraction (ratio) 10 / 100 Answer: 10 %
## How do you find a percentage in math?
In mathematics, a percentage is a number or ratio that can be expressed as a fraction of 100. If we have to calculate percent of a number, divide the number by whole and multiply by 100.
## What is 60% of a number?
You have learned that to find 1% of a number means finding 1/100 of it. Similarly, finding 60 % of a number means finding 60 /100 (or 6/10) of it.
## Why do we need percentages?
Percentages are important for understanding the financial aspects of everyday life. The symbol for percent is % and its similarity to /100 reminds us of its meaning. Percentage is another way to write fractions with a denominator of 100. For example, 8% means.
You might be interested: How To Pass A Math Test Without Studying?
## How do I figure out the percentage of a number?
To find the percentage of a number when it is in decimal form, you just need to multiply the decimal number by 100. For example, to convert 0.5 to a percentage, 0.5 x 100 = 25% The second case involves a fraction. If the given number is in fractional form, first convert it to a decimal value and multiply by 100.
## What is 3 as a percentage of 100?
How much is 3 out of 100 written as a percent value? Convert fraction (ratio) 3 / 100 Answer: 3 %
## What number is 10% of 80?
Percentage Calculator: What is 10 percent of 80? = 8.
## What is 20 out of 100 as a percentage?
Therefore the fraction 20 / 100 as a percentage is 20 %.
## How do I calculate a percentage between two numbers?
First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100.
## How do I figure out a percentage of two numbers?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage! To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage.
## What is 60 out of 100 as a percentage?
Therefore the fraction 60 / 100 as a percentage is 60 %.
## What number is 70% of 60?
Percentage Calculator: What is 70 percent of 60? = 42.
9 is 15 % of 60. | 695 | 2,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-39 | latest | en | 0.936012 |
https://www.gamedev.net/forums/topic/467730-changing-degrees-into-directions/ | 1,531,877,481,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00415.warc.gz | 906,367,854 | 23,187 | # OpenGL changing degrees into directions
This topic is 3935 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi all, Can anyone show me how to change degrees into directions, for example I have 4 objects which move off at 0,90,180,270 degrees, how do I change these into x and y opengl movement. It could be any number of objects it's haow to do it mathmatically rather than the above example. Thanks
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You want to convert polar coordinates to Cartesian? The transformation is straight forward:
x = r • cos(a)
y = r • sin(a)
Where r>0 is the distance from origin to your point and a is the angle (0≤a<2π)
For normalized directions you have r=1, so you get
x = cos(a)
y = sin(a)
Hope this helps
EDIT: And make sure you don't mix up radians and degrees. Most implementations of sin and cos expect input in radians, whilst OpenGL I believe uses degrees.
You can convert degrees to radians by a simple transformation
y = T(x) = x • π / 180;
Where the input x in degrees and the output y will be in radians. And the inverse converts radians to degrees
x = T-1(y) = y • 180 / π
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The right triangle is one such that the hypotenuse (which represents your direction of movement in this case) forms the angle of movement with the adjacent, while the opposite completes the triangle. Assuming sector 1, the adjacent represents movement in x and the opposite represents movement in y.
The following relationships then hold:
1. the sine of the angle is equivalent to the length of the opposite divided by the length of the hypotenuse;
2. the cosine of the angle is equivalent to the length of the adjacent divided by the length of the hypotenuse; and
3. the tangent of the angle is equivalent to the length of the opposite divided by the length of the adjacent
You may remember them from trigonometry classes. Assuming a unit movement in the hypotenuse, then:
x-displacement = cos(angle)
y-displacement = sin(angle)
Scale these displacement by the number of units of real movement you want. Test for values of 90, 180, 270 and 0/360 degrees.
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× | 590 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.885744 |
https://www.physicsforums.com/threads/the-width-of-a-finite-potential-well.892447/ | 1,566,633,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00231.warc.gz | 916,293,132 | 20,108 | # The width of a finite potential well
#### TheSodesa
1. Homework Statement
An electron is enclosed in a potential well, whose walls are $V_0 = 8.0eV$ high. If the energy of the ground state is $E = 0.50eV$, approximate the width of the well.
Answer: $0.72nm$
2. Homework Equations
For an electron in a potential well, whose energy is less than the height of the potential walls, outside the well the time-independent Schrödinger equation becomes:
\Psi ''(x) = \frac{2m}{\hbar^2}(V_0 - E) \Psi (x) = \alpha^2 \Psi (x),
where
\alpha^2 = \frac{2m}{\hbar} (V_0 - E) > 0
Inside the well, where $V(x) = 0$, it is the familiar:
\Psi ''(x) = -k^2 \Psi (x), k^2 = \frac{2mE}{\hbar^2}
By doing a whole bunch of math (by requiring that $\Psi$ be continuously differentiable at the potential walls and solving the resulting linear system with the assumption that it has more than one solution, meaning $det() = 0$), we and up with the result
\frac{\sin (kL)}{\cos (kL)} = \tan (kL) = \frac{\alpha}{k}.
Substituting $\alpha$ from $(2)$ and $k$ from $(3)$, we get
\tan (kL) = \sqrt{\frac{V_0 - E}{E}}
3. The Attempt at a Solution
Now my idea was to use $(5)$ to solve for $L$ as follows:
$$\tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\ \iff\\ kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\ \iff\\ L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}}) = \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\ = \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\ \arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\ = 2.084 751 \cdot 10^{-8}m,$$
which is a bit off. Any idea what I'm doing wrong?
Last edited:
Related Introductory Physics Homework Help News on Phys.org
#### TSny
Homework Helper
Gold Member
I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.
Also, make sure your calculator is in the radian mode for taking the arctan.
#### TheSodesa
I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.
Also, make sure your calculator is in the radian mode for taking the arctan.
Yeah, it was the radian thing, and the fact that the well goes from -L to L. The online material for our book (Tipler, Modern Physics, itself glosses over this part, the mathematics in particular) actually uses the letter $a$ in place of $L$, but I figured I could just change the place of the origin without doing anything to make up for it, so to speak.
Thanks again.
Last edited:
#### Simon Bridge
Homework Helper
Apparently there are even and odd solutions. The online material for our book states, that the $-\cot(ka)$ actually corresponds to the odd case.
Yes?
#### TheSodesa
I as just noting that the link (the part of the page) you provided used $-\cot$ instead of $\tan$. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
#### Simon Bridge
Homework Helper
I as just noting that the link (the part of the page) you provided used $-\cot$ instead of $\tan$. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
Yes? Where are you up to in your solution?
#### TheSodesa
Yes? Where are you up to in your solution?
Gah, sorry. I just fell of the map for a week, didn't I? I basically forgot to use radians in my calculation and also forgot, that the formula applies to a potential well from $-a$ to $a$, not $0$ to $L$.
Should I have posted my final numerical answer in this thread? I guess that would have been in good housekeeping...
The answer I got was $\approx 0.73nm$.
"The width of a finite potential well"
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• Solo and co-op problem solving | 1,260 | 4,211 | {"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": 5, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-35 | latest | en | 0.830129 |
https://gateoverflow.in/47349/permutation-of-string | 1,575,580,954,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00429.warc.gz | 391,769,822 | 18,221 | 248 views
How to generate permutation of a string? What will be the complexity?
retagged | 248 views
0
Can it be done by backtracking?
0
Try recursion. Permute for length l-1, add the last char at each position of these permuted strings.
0
+1
s is the starting index of the string,e is the ending index of the string.
void permute(char *x, int s, int e)
{
int i;
if (s== e)
printf("%s\n", x);
else
{
for (i = s; i <= e; i++)
{
swap((x+s), (x+i));
permute(x, s+1, e);
swap((x+s), (x+i));
}
}
}
There are man methods and best case complexity is $O(n!)$ as there are $n!$ permutations for a string. I'm using Java code as it is recommended for most top tier interviews. (C is not suitable for many interviews as in Google).
Here permute and permute1 are two different permutation methods.
import java.io.*;
import java.util.ArrayList;
public class permuteString{
ArrayList<String> permute(String str){
ArrayList<String> al = new ArrayList<String>();
if(str.length() == 0)
{
return al;
}
ArrayList<String> words = permute(str.substring(1, str.length()));
char c = str.charAt(0);
for(String word: words)
{
for(int i = 0; i<= word.length(); i++)
}
return al ;
}
void permute1(String prefix, String s) {
if(s.length() == 0)
{
System.out.println(prefix);
}
for(int i =0; i < s.length(); i++)
{
permute1(prefix+s.charAt(i), s.substring(0,i)+s.substring(i+1, s.length()));
}
}
public static void main(String args[]) {
permuteString mystring = new permuteString(); //Initialize class
mystring.permute1("", args[0]);
System.out.println("____________________________");
ArrayList<String> words = mystring.permute(args[0]);
for(String word:words)
System.out.println(word);
}
}
by Veteran (424k points)
0
Sir, Is not this can be solved by using Tower of Hanoi recursion.?
0
How?
0
void permute(char *x, int s, int e)
{
int i;
if (s== e)
printf("%s\n", x);
else
{
for (i = s; i <= e; i++)
{
swap((x+s), (x+i));
permute(x, s+1, e);
swap((x+s), (x+i));
}
}
} | 564 | 1,969 | {"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} | 3.078125 | 3 | CC-MAIN-2019-51 | longest | en | 0.498679 |
https://www.acmicpc.net/problem/21847 | 1,624,513,131,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00381.warc.gz | 535,578,525 | 10,130 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
4 초 512 MB 0 0 0 0.000%
## 문제
As one can’t make a living from the prizes awarded at computer science competitions, you have decided to get into the art business—or, more precisely, into a certain museum you have chosen for your debut as an art thief. Unfortunately, this museum is guarded quite well: there are $K$ watchmen patrolling the building, each touring on his own simple closed path through the museum.
To coordinate your operation, you use a map of the area which describes the museum as a set of $M$ corridors connecting $N$ corners, numbered from 1 to $N$. You start at corner 1, whereas your target—a valuable exhibit—is located at corner $N$. Of course, one can reach any corner of the museum from any other corner, but you are not sure whether this is possible without getting noticed, that is, without ever being at the same corner as a watchman and without passing a watchman in a corridor.
Fortunately, you got your hands on the guard roster. So you know for each watchman where he is located at the beginning and which route he is taking. Each minute, a watchman moves from his current position to the next corner on his route, and you can either stay at your current position or move to an adjacent corner. You observed that no two of the watchmen’s routes intersect and that neither your starting position nor your target is contained in any of them.
Write a program that uses this information to either calculate the minimum amount of time in minutes you need to safely reach your target* without being noticed or to decide that this is impossible.
* Once you reach the exhibit, you will open a window and leave the museum by means of the fancy wingsuit you won in your national computer science competition, so there is no need to plan a safe route back.
Of course, you are a gentleman thief who would never lay a finger on the watchmen!
## 입력
The first line of input contains the integers $N$ and $M$ described above. The following $M$ lines each contain two integers $u$ and $v$ ($1 \le u, v \le N$, $u \ne v$), meaning that there is a corridor directly connecting corners $u$ and $v$. It is guaranteed that at most one corridor directly connects any two given corners.
The next line contains the single integer $K$ described above. Then $K$ lines follow; the $i$th of these lines contains a sequence of integers, describing the route of the $i$th watchman as follows: The first integer $\ell_i$ gives the number of distinct corners on the route. Then $\ell_i$ pairwise distinct integers $v_1, \cdots , v_{\ell_i}$specify the sequence of corners the watchman passes on his route. More precisely, the watchman starts at corner $v_1$, in one minute he will be at $v_2$, and so on; after $\ell_i$ minutes he will be at $v_1$ again.
## 출력
Your program should output a single line. This should consist of an integer, the minimum amount of time in minutes you need to safely reach your target, or the string impossible if there is no way to achieve this.
## 제한
• $1 \le N \le 250 000$
• $1 \le M \le 3 000 000$
• $3 \le \ell_i \le 1 500$
• $\ell_1 + \cdots + \ell_K \le 2 750$
## 서브태스크
번호 배점 제한
1 5
$N, M \le 100~000$, $K = 1$, $\ell_1 \le 125$
2 10
$N, M \le 100~000$, $\ell_1 + \cdots + \ell_K \le 125$ and no corridor connects the routes of two distinct watchmen.
3 10
$\ell_i \le 200$, $\ell_1 + \cdots + \ell_K \le 350$ and no corridor connects the routes of two distinct watchmen.
4 10
No corridor connects the routes of two distinct watchmen.
5 25
$\ell_1 + \cdots + \ell_K \le 125$
6 20
$\ell_i \le 200$, $\ell_1 + \cdots + \ell_K \le 350$
7 20
No further constraints.
## 예제 입력 1
6 6
1 2
2 3
3 4
4 5
5 2
5 6
1
4 3 2 5 4
## 예제 출력 1
4
## 예제 입력 2
6 6
1 2
2 3
3 4
4 5
5 2
5 6
1
4 4 5 2 3
## 예제 출력 2
5
## 예제 입력 3
11 13
1 2
2 3
3 4
4 2
3 5
5 6
6 7
7 5
6 8
8 9
9 10
10 8
9 11
3
3 4 2 3
3 7 6 5
3 10 8 9
## 예제 출력 3
impossible
## 힌트
The following picture corresponds to the situation of the first sample case above:
Here, the corner where the watchman is located at the beginning is shown in bold and his route is indicated by shoe prints. An optimal route for you would be the following: you wait at your starting location (corner 1) for one minute, and then go to corners 2, 5, and finally 6 without further delay.
The second sample has the same museum layout, but the starting location and direction of the watchman differ. A possible optimal route: go from 1 to 2, 3, 4, 5, and finally 6.
## 채점 및 기타 정보
• 예제는 채점하지 않는다. | 1,340 | 4,524 | {"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} | 2.859375 | 3 | CC-MAIN-2021-25 | latest | en | 0.962872 |
https://www.physicsforums.com/threads/transconductance-of-a-differential-pair.332834/ | 1,516,653,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00054.warc.gz | 957,806,510 | 15,892 | # Transconductance of a differential pair
1. Aug 26, 2009
### bitrex
I'm trying to figure out what the transconductance of a bipolar differential pair with a certain tail current would be, but I'm getting tangled up in thinking like "well the pair sees half the input swing, but the transconductance might be double if there are two transistors.."etc. Could someone explain to me the proper way to calculate the transconductance of such a circuit with both a resistive load and a current mirror load? Thanks!
2. Aug 26, 2009
### Bob S
Maybe this will help. If the first (assume npn) transistor has a zero resistance collector load, then it is an emitter follower. If the second transistor has the base resistor(s) bypassed with a capacitor, then the second transistor is a common base circuit. So the first transistor is driving a common base transistor in parallel with the "tail" resistor. These are nice low signal amplifiers.
3. Aug 27, 2009
### bitrex
By looking through some references I managed to find a really elegant derivation of the gain of a differential amplifier...I never thought the hyperbolic tangent function would show up here! I'll write it out since I need practice with LaTeX and it might help someone who is working through the same material.
$$I_e = e^\frac{Vbe}{V_t}$$
So
$$V_{be} = V_t ln I_e$$
$$V_{dif} = V_{be1} - V_{be2} = V_t ln I_{e1} - V_t ln I_{e2} = Vt ln \frac{I_e1}{I_e2}$$
So
$$e^\frac{V_{dif}}{V_t} = \frac{I_e1}{I_e2}$$
$$\frac{I_{c1}}{I_{c2}} = \frac{\alpha I_{e1}}{\alpha I_{e2}} = \frac{I_{e1}}{I_{e2}}$$
The differential current ratio is:
$$\frac{I_{c1} - I_{c2}}{I_{c1} + I_{c2}} = \frac{e^\frac{V_{dif}}{V_t} - 1}{e^\frac{V_{dif}}{V_t} + 1} = tanh(\frac{V_{diff}}{2V_t})$$
So by multiplying the differential current ratio top and bottom by the collector load resistance we can do this:
$$\frac{R_c(I_{c1} - I_{c2})}{Rc(I_{c1} + I_{c2})} = \frac{V_{o1} - V_{o2}}{Rc(I_{c1} + I_{c2})} = \frac{V_{od}}{R_c \alpha I_o}$$
Where Io is the current through both emitters of the differential pair and alpha is the common base current gain.
Setting the two equations equal we finally have:
$$\frac{V_{od}}{R_c \alpha I_o} = tanh(\frac{V_{dif}}{2V_t})$$
$$V_{od} = R_c \alpha I_o tanh(\frac{V_{dif}}{2V_t})$$.
For small signals, $$tanh(\frac{V_{dif}}{2V_t}) \approx \frac{V_{dif}}{2V_t}$$ and alpha can be taken to be 1, so we get for small signal gain:
$$V_{od} = R_c\frac{I_o}{2V_t}$$.
For large signals, the transfer function of the amplifier behaves just like the hyperbolic tangent function: it's linear in a small region around the quiescent point, but goes asymptotic as the voltage increases or decreases beyond this linear region and the input transistor saturates or goes into cutoff.
4. Aug 28, 2009
### Bob S
hello bitrex-
This a nice derivation. I do have some comments. You are using Vt which is about 26 millivolts at room temperature. It is actually kBT/q (Boltzmanns constant, temperature, electron charge). Some of your subscripts are not subscripts. This derivation I think applies to differential inputs and outputs, which is not always the case. I usually use only one (base) input and one (collector) output, for example. The large common emitter tail resistance is more important in this case. It does not appear at all in your derivation because you assume that the emitter tail current is constant (current sink). I have sometimes actually used an npn transistor in the tail to make a constant current sink.
Bob S
5. Aug 28, 2009
### uart
It's quite common to use it like that Bob. BTW it can be considered as CC CB cascade in that case. | 1,063 | 3,631 | {"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} | 3.75 | 4 | CC-MAIN-2018-05 | longest | en | 0.846168 |
http://www.slideserve.com/zeal/physics-2 | 1,492,987,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118831.16/warc/CC-MAIN-20170423031158-00602-ip-10-145-167-34.ec2.internal.warc.gz | 694,321,321 | 17,807 | # Physics 2 - PowerPoint PPT Presentation
1 / 6
Physics 2. Chapter 12 problems. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
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## Physics 2
Chapter 12 problems
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
12.8 An 8kg point mass and a 15kg point mass are held in place 50cm apart. A particle of mass m is released from a point between the two masses 20cm from the 8kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
• 12.50 A uniform sphere with mass 60kg is held with its center at the origin, and a second uniform sphere with mass 80kg is held with its center at the point x=0, y=3m.
• What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.5kg placed at the point x=4m, y=0?
• Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
12.53 An experiment is performed in deep space with two uniform spheres, one with mass 25kg, and the other with mass 100kg. They have equal radii, r=0.2m. The spheres are released from rest with their centers 40m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres.
(a) Explain why linear momentum is conserved.
(b) When their centers are 20m apart, find
(i) the speed of each sphere and
(ii) the magnitude of the relative velocity with which one sphere is approaching the other.
(c) How far from the initial position of the center of the 25kg sphere do the surfaces of the two spheres collide?
Actually, their surfaces will touch before they get there. Subtract the radius of one of the spheres to get 31.9m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
12.56 A landing craft with mass 12,500kg is in a circular orbit 5.75x105m above the surface of a planet. The period of the orbit is 5800s. The astronauts in the lander measure the diameter of the planet to be 9.6x106m. The lander sets down at the north pole of the planet. What is the weight of a 85.6kg astronaut as he steps out onto the planet’s surface?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
12.76 As Mars orbits the sun in its elliptical orbit, its distance of closest approach to the center of the sun (at perihelion) is 2.067x1011m, and its maximum distance from the center of the sun (at aphelion) is 2.492x1011m. If the orbital speed of Mars at aphelion is 2.198x104 m/s, what is the orbital speed at perihelion? (You can ignore the influence of the other planets.)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB | 887 | 3,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.92863 |
https://programming-idioms.org/impl-edit/11/828 | 1,563,920,215,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529737.79/warc/CC-MAIN-20190723215340-20190724001340-00405.warc.gz | 508,340,144 | 6,108 | Implementation
Haskell
Be concise.
Be useful.
All contributions dictatorially edited by webmasters to match personal tastes.
Please do not paste any copyright violating resource.
Please try to avoid dependencies to third-party libraries and frameworks.
Other implementations
import java.util.Random;
x.get(new Random().nextInt(x.size()))
x[Math.floor(Math.random() * x.length)]
import random
random.choice(x)
import "math/rand"
x[rand.Intn(len(x))]
#include <stdlib.h>
x[rand() % x_length];
\$x[ array_rand(\$x) ]
my @x = ('a', 'list', 'of', 'random', 'items');
print \$list[ rand(@x) ], "\n";
import std.random;
x.randomSample(1);
x[new Random().nextInt(x.length)];
element := x[random(length(x))];
uses classes;
element := x.Items[random(x.count)];
Enum.random(x)
https://github.com/inaka/erlang-katana
ktn_random:pick(X)
lists:nth(rand:uniform(length(X)), X).
x[math.random(#x)]
x.sample
import scala.util.Random
val x = List(1, 2, 3, 4)
x.apply(Random.nextInt(x.size))
using System;
using System.Collections.Generic;
x[new Random().Next(x.Count)];
import "math/rand"
func pickT(x []T) T {
return x[rand.Intn(len(x))]
}
use rand::{self, Rng};
x[rand::thread_rng().gen_range(0, x.len())]
(rand-nth x)
import System.Random (randomRIO)
(l !!) <\$> randomRIO (0, length l - 1)
#include <random>
std::mt19937 gen;
std::uniform_int_distribution<size_t> uid (0, x.size () - 1);
x[uid (gen)];
extern crate rand;
use rand::{thread_rng, Rng};
let choice = thread_rng().choose(&x).unwrap();
import java.util.concurrent.ThreadLocalRandom;
x.get(ThreadLocalRandom.current().nextInt(0, x.size())) | 428 | 1,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.29255 |
https://stackoverflow.com/questions/39131995/longest-increasing-subsequence-with-maximum-sum-plus-constraint-allowed-number | 1,540,005,549,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00281.warc.gz | 808,748,047 | 27,326 | # Longest increasing subsequence with maximum sum plus constraint - allowed number of elements can be skipped
The longest increasing sub-sequence with maximum sum (http://www.geeksforgeeks.org/dynamic-programming-set-14-maximum-sum-increasing-subsequence/) is a classic algorithm problem and there exist a lot of solutions on the web. However, I just encountered a variation of this problem and have no idea how to solve it.
Compared with the original question, now you are also given a number m which indicates the number of elements you can skip at most from a continuous sub-range in order to find the LIS with maximum sum. For example, with the following array,
[1, 200, 300, 3, 4, 5, 6]
The LIS is 1,3,4,5,6 and the maximum sum is 19. However, if m is 1, it means that at most one element can be skipped in a continuous sub-range in order to find the LIS. Hence the above solution is not right because between 1 and 3, two elements are skipped (200, 300 in this case). The new solution should be 3,4,5,6 since no elements are skipped in a continuous sub-range. The question is to find the LIS with the maximum sum and return the sub-sequence (not the sum or the length of the sub-sequence) when the array and the number m is given. I have been stuck with this problem for several days so any help is appreciated.
Edit: O(n^2) solution is good enough for now since I have complete no idea where to start.
Edit: m is the cumulative steps can be skipped for the entire array, not the steps can be skipped between two separate increasing sub-sequence.
• BTW, the original LIS problem can be solved in O(n^2) or O(nlgn). For this question, even a O(n^2) solution is good for me. – Optimus Prime Aug 24 '16 at 20:02
• Is O(n^2 log n) ok? – Pham Trung Aug 25 '16 at 5:55
This problem can be solved by using Dynamic programming technique.
Call the input array `data` length `n`.
Assume we have an array `dp[n][n + 1]` which entry `dp[i][j]` store the nearest index, which from `i` to `dp[i][j]`, the length of increasing sub-sequence start at `i` is `j`. If we have this `dp`, the result for your question is straight forwards.
Now, how to calculate `dp[i][j]` for a specific `j`? moving `i` backward from index `n - 1` to `0`, assume that, we maintain another array `list[n + 1]`, with `list[i]` storing all index `k`, which has a increasing sub-sequence start at `k` and length `i`. We need to maintain the property of `list[j]`: `list[j]` is decreasing list, with element at index `x` and `y` in `list[j]`, then `data[x] > data[y]` if and only if `x < y`. If we have `list[j]` for each length `j`, for `dp[i][j + 1]`, we only need to binary search inside `list[j]` to find the smallest element in list[j] which is greater than data[i].
``````int[][]dp = new int[n][n + 1];
fill(dp, -1);
List<Integer>[]lists = new List[n + 1];
for(int i = n - 1; i >= 0; i--){
for(int j = 1; j <= n; j++){
if(j == 1){
dp[i][j] = i;
}else if(!list[j - 1].isEmpty()){
int index = binary search in list[j - 1] to get the nearest index that greater than data[i];
dp[i][j] = dp[index][j - 1];
}
if(dp[i][j] == -1)
continue;
while(data[list[j].peekLast()] <= data[i]){
//Remove all entries which is smaller than i in list, we can easily see that all entries which is smaller than i can only end at point at least as near as end point of i.
list[j].pollLast();
}
if(list[j].isEmpty() || dp[list[j].peekLast()][j] > dp[i][j]){
//Only add entry to list if result of new entry is nearer.
• @OptimusPrime with that input array, so, for index 0, dp[0][1] = 0, dp[0][2] = 1, dp[0][3] = 2, dp[0][4] = 7, dp[0][5] = 8. For sequence starts from 0 and have length 3, so it will start from 0 and end at dp[0][3]. So the number of characters that need to be removed is 0. If we need a sequence start from 0 and has length 4, the number of characters need to be removed is (dp[0][4] - 0 + 1) - 4 = 3. Basically, with `dp`, we know the length of the segment and the length of increasing sequence in that segment, to calculate how many character need to be removed is trivial. – Pham Trung Aug 26 '16 at 3:04 | 1,176 | 4,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-43 | latest | en | 0.924448 |
https://oeis.org/A124694 | 1,542,485,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743732.41/warc/CC-MAIN-20181117185331-20181117211331-00438.warc.gz | 703,640,016 | 3,803 | This site is supported by donations to The OEIS Foundation.
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A124694 Sets of digits such that the product of the digits is 10 times the sum of the digits. Each set is arranged as a number with nondecreasing digits. 2
459, 1566, 2259, 2355, 11558, 12445, 111567, 112356, 122245, 1113345, 1222225, 11111568, 11112357, 11112455, 11122335, 111122255, 1111111569, 1111112358, 11111111578, 11111112456, 111111112359, 111111112555, 111111113445 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS 4*5*9 = 180 and 4 + 5 + 9 = 18. Each term must include the digit 5, so it is a subsequence of A011535. - Chai Wah Wu, Dec 08 2015 LINKS Chai Wah Wu, Table of n, a(n) for n = 1..931 MATHEMATICA FromDigits /@ DeleteDuplicates@ Map[Sort, IntegerDigits@ Select[Range[10^7], Times @@ # == 10 Total@ # &@ IntegerDigits@ # &]] (* Michael De Vlieger, Dec 09 2015 *) CROSSREFS A062043 (Numbers for which the product of the digits is 10 times their sum) is created by permuting digits in every number of this sequence. Sequence in context: A121970 A055162 A055159 * A227754 A253527 A183964 Adjacent sequences: A124691 A124692 A124693 * A124695 A124696 A124697 KEYWORD base,nonn AUTHOR Tanya Khovanova, Dec 25 2006 EXTENSIONS Extended by D. S. McNeil, Dec 16 2010 STATUS approved
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Last modified November 17 14:50 EST 2018. Contains 317276 sequences. (Running on oeis4.) | 545 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-47 | latest | en | 0.679437 |
https://moretubeviews.com/winning-number-tips-in-powerball-game/ | 1,624,545,166,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00081.warc.gz | 368,694,526 | 9,166 | ## Winning Number Tips in Powerball Game
Because of the recognition of Powerball, you’ll find several sites that talk about the game, including how it operates, what advantages it offers, and what individuals who have really gained in the game may state about their luck. This informative article lets you know ways to develop into a Powerball winner by showing you a couple of items that Powerball winners reveal as their key weapons.
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Powerball doesn’t have uncertainty achieved the top of its popularity these past months. Internet searches for websites on the internet that host Powerballs have doubled in numbers. Furthermore, more and more people are recording online to discover how to choose winning Powerball figures in the hopes of earning another Powerball pull and using home the jackpot or any Powerball prize. But what exactly is Powerball in the initial position? How do you play it?
Powerball, the first game ever to utilize two drums in their key sport, can be an all-American lottery sport that is being sold through United Claims lotteries as one of many provided jackpot share games. The Multi-State Lottery Association or MUSL, which is a non-profit company produced by the official agreement recognized by all member lotteries, coordinates the processes of Powerball, including the growth of new games, management of finances, conduction of pulls, and different services. Therefore, there are several claims – thirty-one all in all – offering Powerball. Since Powerball employs not merely one but two drums, it allows more flexibility as far as game design it concerned, attracting more players weighed against different lotto games. It features a larger possibility of reduced chances for other rewards but large chances for the jackpot. That two-drum function has become being employed by the United Kingdom’s EuroMillions and Thunderball, Australia’s Powerball, and the United States’Super Millions. | 889 | 4,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-25 | latest | en | 0.967706 |
https://math.answers.com/other-math/What_number_between_49_and_95_that_are_multiples_of_2_8_10 | 1,701,838,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00504.warc.gz | 427,146,030 | 45,835 | 0
# What number between 49 and 95 that are multiples of 2 8 10?
Updated: 4/28/2022
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Q: What number between 49 and 95 that are multiples of 2 8 10?
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Related questions
60 and 90.
60, 90.
### What multiples of 359 are between 49 and 95?
A multiple has to be equal to or larger than the number itself, so there are no multiples for 359 that are between 49 and 95. 359 is a prime number, so it has no factors between 49 and 95 either.
### What number between 67 and 113 is a multiple of 5 6 and 10?
What number between 49 and 95 are multiples of 4, 6, and 9
It is 80
It is: 72
### What number between 49 and 95 that is a multiple of 69 and 18?
No number between 49 and 95 is a multiple of 69 and 18. 69 is a multiple of itself. 54, 72 and 90 are multiples of 18. So no number within that range are multiples of both.
It is: 72
60 and 90
### Is 49 a multiple of 10?
Nope. The multiples of 10 always ends with 0 (zero). 49 does not end in 0 so it is NOT a multiple of 10. Multiples of 10 is 10,20,30,40,50,60,70,80,90,100,110 etc.
### What are multiples of 4 between 23-49?
The multiples of 4 between 23 and 49 are: 24, 28, 32, 36, 40, 44, 48.
### What are the first ten multiples of 49?
The first 10 multiples of 49: 49, 98, 147, 196, 245, 294, 343, 392, 441, 490 | 521 | 1,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-50 | latest | en | 0.914205 |
https://math.libretexts.org/TextMaps/Calculus/Book%3A_Calculus_(Guichard)/04._Transcendental_Functions | 1,542,227,166,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00119.warc.gz | 689,661,308 | 15,378 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4: Transcendental Functions
[ "article:topic-guide", "Transcendental Functions", "authorname:guichard" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
So far we have used only algebraic functions as examples when finding derivatives, that is, functions that can be built up by the usual algebraic operations of addition, subtraction, multiplication, division, and raising to constant powers. Both in theory and practice there are other functions, called transcendental, that are very useful. Most important among these are the trigonometric functions, the inverse trigonometric functions, exponential functions, and logarithms. | 343 | 1,162 | {"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} | 3.984375 | 4 | CC-MAIN-2018-47 | latest | en | 0.637122 |
https://communities.sas.com/t5/SAS-Programming/prevalence-rate-and-95-CI-for-BMI-across-age-groups-and-gender/m-p/902907 | 1,702,068,112,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00577.warc.gz | 211,701,945 | 27,906 | Calcite | Level 5
## prevalence rate and 95%CI for BMI across age groups and gender
Hi Apologies this is my first post to this group.
I wish to calculate the prevalence rate and 95% confidence intervals for people by BMI category and age group by sex to produce the table below. Could you advise? Thanks for your help!
*********************************************************************************************************************************************
| Sex | age group | underweight | healthy | overweight | obese | Total
*********************************************************************************************************************************************
| M | 20-<30years | Prev (95%CI) | Prev (95%CI) |Prev (95%CI) |Prev (95%CI) |Prev (95%CI) | etc
*********************************************************************************************************************************************
PROC FORMAT;
VALUE agefmt
20-<30 = '20-<30 years'
30-<40 = '30-<40 years'
40-<50 = '40-<50 years'
;
value bmifmt
9-<18.5 = 'Underweight (9-<18.5)'
18.5-<25 = 'Healthy (18.5-<25)'
25-<30 = 'Overweight (25-<30)'
30-51 = 'Obese (30+)';
VALUE sexfmt
0 = 'Female'
1 = 'Male'
;
RUN;
DATA have;
INPUT age bmi sex @@;
DATALINES;
20 18.5 1 25 23.5 1 26 24.5 1 20 15.5 1 20 16.2 1 21 18.0 1 20 30.2 1 21 26.5 1 22 29.0 1 22 29.5 1
20 22.5 0 21 23.5 0 20 24.5 0 23 25.5 0 23 25.2 0 25 19.0 0 26 38.2 0 23 36.5 0 24 29.0 0 25 16.5 0
30 22.5 0 31 23.5 0 30 24.5 0 33 25.5 0 33 25.2 0 35 19.0 0 36 38.2 0 33 36.5 0 34 29.0 0 35 16.5 0
30 18.5 1 35 20.5 1 36 24.5 1 30 15.5 1 30 16.2 1 31 18.0 1 30 30.2 1 31 26.1 1 32 29.1 1 32 29.5 1
40 22.5 0 40 23.5 0 40 24.5 0 42 25.5 0 42 20.2 0 45 39.0 0 46 38.2 0 43 36.5 0 44 29.0 0 45 30.5 0
40 12.5 1 41 28.5 1 40 24.5 1 43 25.5 1 43 25.2 1 45 19.0 1 46 38.2 1 43 36.5 1 44 29.0 1 45 16.5 1
;
DATA want;
SET have;
FORMAT age agefmt.;
FORMAT bmi bmifmt. ;
FORMAT sex sexfmt.;
RUN;
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# Limits
On question #3 the last two mutliple choice options didn't scan they are:
c.Yes, the limit exists, and is equal to -1.
d. Yes, the limit exists, and is equal to 0.
#### Solution Preview
4. (b) The limit of a constant is the constant itself.
5. f(x->1)=2
6.By using ...
#### Solution Summary
Problems involving limits are solved.
\$2.19 | 104 | 378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.907079 |
https://www.wyzant.com/resources/answers/16029/p_x_p_5_if_p_r_4r_2_5 | 1,524,680,563,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00367.warc.gz | 936,504,244 | 16,910 | 0
p(x)+p(5), if p(r)=4r^2+5
p(x)+p(5), if p(r)=4r^2+5 I do not understand how to do this question I have read over the chapter and I still don't understand how to do this function.
1 Answer by Expert Tutors
Branden A. | Math, Sci, & Computer Help All Ages - Especially Senior Citizens!Math, Sci, & Computer Help All Ages - Es...
2
Marked as Best Answer
This is a question about how to add two functions together. The way you solve these types of problems, is to first substitute the given variables into the function, and then perform the addition.
Given that p(r) = 4r^2 + 5,
you can start by substituting the variables given in your problem in to the function.
First, substitute x for r in p(r)
p(x) = 4x^2 + 5,
Second, substitute 5 for r in p(r)
p(5) = 4*5^2 + 5 = 4*25 + 5 = 100 + 5 = 105.
Next, add p(x) to p(r), and simplify
[4x^2 + 5] + [4*5^2 + 5]
[4x^2 + 5] + [105] as shown above
[4x^2 + 5 + 105 = 4x^2 + 110.
The answer is 4x^2 + 110.
I hope you understand. If you need any more clarification, or to solve any more problems, feel free to contact me. | 355 | 1,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.877208 |
https://mrmanojpandey.com/resonance-in-l-c-r-circuit-power-in-ac-circuit/ | 1,675,535,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00531.warc.gz | 411,174,466 | 59,847 | Maths And Physics With Pandey Sir Ncert 12th Physics Chapter-7 Resonance in L-C-R Circuit || Power in AC Circuit
# Resonance in L-C-R Circuit || Power in AC Circuit
Chapter-7|Alternating Current |NCERT 12th Physics:
Resonance in L-C-R Circuit:
In LCR circuit, when phase(Ø) between current and voltage is zero, the circuit is said to be resonant circuit.
Power in AC Circuit:
We know that, V= V₀Sinωt applied to a series LCR circuit drives a current in the circuit. | 130 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.859891 |
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# Visual Logic Help
### #1 mduran7599 Posted 15 June 2008 - 01:57 PM
mduran7599
New Member
• Member
• 1 posts
I am at the end of my term in CIS115 in devry and i am having some problems with my last 2 questions of homework before the final if i could get any help at all it would be greatly appreciated.
here they are
High-Low: Write a program that allows the computer to randomly generate a number between 1 and 1000. The user will then enter a guessed value. If the guessed value is lower than the secret number, the computer should tell user to “enter a higher value”. If the guessed value is greater than the secret number then the computer should tell the user to “enter a lower value”. This process should continue till the user guesses the secret value. Then the final output should say: ” CORRECT! You guessed the secret number which is ….” “It took you # of guesses” (Hint: you need to use while loop for the guessed value. Also you need to create a counter to count the number of guesses)
• 0
### #2 Metallica Posted 18 June 2008 - 04:20 AM
Metallica
Spyware Veteran
• GeekU Moderator
• 33,054 posts
Hi mduran7599,
Your teacher should be happier with a failing attempt made by yourself then with a working program that you hardly understand.
Hint:
Random(5) A random number between 0 and 4
Random(100) + 1 A random number between 1 and 100
Regards,
Pieter
• 0
### #3 DeepPurple Posted 08 June 2009 - 01:54 PM
DeepPurple
New Member
• Member
• 1 posts
how about a hint as to how to set up the two arrays. Someone gave me this program and I am not doing this for a school but my self and the password problems looks interesting.
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0 members, 0 guests, 0 anonymous users | 558 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.919817 |
http://forum.enjoysudoku.com/puzzle-54-mike-barker-ur-3x-1sl-t32526-15.html | 1,537,846,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00273.warc.gz | 92,939,883 | 9,069 | ## Puzzle #54: Mike Barker UR+3X/1SL
Post the puzzle or solving technique that's causing you trouble and someone will help
### Re: Puzzle #54: Mike Barker UR+3X/1SL
sultan vinegar wrote:I can't find any instances of the pattern in your two examples, ...
That seems to be the common consensus from what I've read. However, everyone has done a lot of work on these two puzzles ... AND ... I'm going to carefully review all of the responses to learn what I can about the URs in these two puzzles.
Out of the 62 example puzzles posted by Mike Barker, there are a few others that I couldn't understand/resolve. I'll give everyone a break before presenting the next troublesome UR pattern.
Thanks to everyone for your effort !!!
Regards, Danny
_
daj95376
2014 Supporter
Posts: 2624
Joined: 15 May 2006
### Re: Puzzle #55: Mike Barker UR+3X/1SL
daj95376 wrote:Now, it's time to see how much of the above logic holds together on Mike Barker's second puzzle for this pattern. Soggestions?
Code: Select all
` Puzzle #55: Mike Barker UR+3X/1SL +-----------------------+ | 6 9 5 | . 2 . | . . . | | . . 3 | 4 . . | . . . | | . . . | 9 . . | . 1 . | |-------+-------+-------| | . . 9 | . . 2 | 4 . . | | . . . | . 3 . | 1 . . | | 4 . . | . . . | 3 5 . | |-------+-------+-------| | . 6 . | . . . | . . . | | . 1 . | 8 . . | . 6 . | | 9 8 4 | . 1 . | . . . | +-----------------------+ +--------------------------------------------------------------+ | 6 9 5 | 3 2 1 | 78 478 478 | | 1 27 3 | 4 678 578 | 2569 *29 *29+5 | | 27 4 8 | 9 67 57 | 256 1 3 | |--------------------+--------------------+--------------------| | 78 3 9 | 1 5 2 | 4 78 6 | | 28 5 6 | 7 3 4 | 1 *29+8 *29+8 | <- SL on <9> | 4 27 1 | 6 89 89 | 3 5 27 | |--------------------+--------------------+--------------------| | 35 6 27 | 25 479 379 | 2589 2489 1 | | 35 1 27 | 8 479 379 | 259 6 2459 | | 9 8 4 | 25 1 6 | 257 3 257 | +--------------------------------------------------------------+ # 61 eliminations remain`
I've marked a potential UR(29) at r25c89. For the UR+3X\1SL to crack the puzzle, we need to eliminate <2> in both of r5c89
Eliminating 2r5c9, leaves skyscraper <2>r35\c1, that eliminates 2r2c8 and cracks the puzzle.
Code: Select all
`+------------+--------------+---------------------+| 6 9 5 | 3 2 1 | 78 478 478 || 1 27 3 | 4 678 578 | 2569 (29) (259) || 27 4 8 | 9 67 57 | 256 1 3 |+------------+--------------+---------------------+| 78 3 9 | 1 5 2 | 4 78 6 || 28 5 6 | 7 3 4 | 1 28(9) (89-2) || 4 27 1 | 6 89 89 | 3 5 (27) |+------------+--------------+---------------------+| 35 6 27 | 25 479 379 | 2589 2489 1 || 35 1 27 | 8 479 379 | 259 6 2459 || 9 8 4 | 25 1 6 | 257 3 (257) |+------------+--------------+---------------------+(2=57)r69c9 - 5r2c9 [ U = 57r69c9, Y = 5r2c9 ]From that, we get: 2r5c9 => 9r2c9.Then with the SL on <9> and the bivalue in r2c8, we have: 2r5c9 => 9r5c8,2r2c8,9r2c9 ... UR contradiction => -2r5c9`
blue
Posts: 683
Joined: 11 March 2013
### Re: Puzzle #54: Mike Barker UR+3X/1SL
Code: Select all
`--- UR+3X/1SL: (reformatted) Constraint #1: includes the extra cell(s) "(ab)U..." such that "U" is a locked set which includes "Y", "abY" is seen by all of the cells of "(ab)U..." which contain elements of "Y", Constraint #2a: "(ab)U..." can contain "a", and "(ab)U..." can contain "b" if all of its cells which contain "b" are seen by "abX" => "b" can be removed from "abX". Similarly, Constraint #2b: "(ab)U..." can contain "b", and "(ab)U..." can contain "a" if all of its cells which contain "b" are seen by "abX" => "b" can be removed from "ab(Z)". ab abX | |a |abY ab(Z) (ab)U...`
[Solution Withdrawn: matches case #2 in a posting earlier by blue.]
_
Last edited by daj95376 on Sat Jun 27, 2015 2:46 am, edited 1 time in total.
daj95376
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Posts: 2624
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### Re: Puzzle #54: Mike Barker UR+3X/1SL
To cover both of Mike's examples, I would suggest this alteration:
Code: Select all
`--- UR+3X/1SL: (re-interpreted) Includes the extra cell(s) "(ab)U" such that "U" is a locked set which includes "Y". The cells in "(ab)U" needn't be in the same line as "abY" and "ab(Z)" (as the diagram below might seem to suggest). Constraint #1: "abY" is seen by all of the cells of "(ab)U" which contain elements of "Y", Constraint #2a: Cells in "(ab)U" that contain "a", can see "ab(Z)" or "ab" Cells in "(ab)U" that contain "b", can see "abX" => "b" can be removed from "abX". Similarly, Constraint #2b: Cells in "(ab)U" that contain "b", can see "ab(Z)" or "ab" Cells in "(ab)U" that contain "a", can see "abX" => "b" can be removed from "ab(Z)". ab abX | |a |abY ab(Z) ... (ab)U`
blue
Posts: 683
Joined: 11 March 2013
### Re: Puzzle #54: Mike Barker UR+3X/1SL
blue wrote:To cover both of Mike's examples, I would suggest this alteration:
Code: Select all
`--- UR+3X/1SL: (re-interpreted) Includes the extra cell(s) "(ab)U" such that "U" is a locked set which includes "Y". The cells in "(ab)U" needn't be in the same line as "abY" and "ab(Z)" (as the diagram below might seem to suggest). Constraint #1: "abY" is seen by all of the cells of "(ab)U" which contain elements of "Y", Constraint #2a: Cells in "(ab)U" that contain "a", can see "ab(Z)" or "ab" Cells in "(ab)U" that contain "b", can see "abX" => "b" can be removed from "abX". Similarly, Constraint #2b: Cells in "(ab)U" that contain "b", can see "ab(Z)" or "ab" Cells in "(ab)U" that contain "a", can see "abX" => "b" can be removed from "ab(Z)". ab abX | |a |abY ab(Z) ... (ab)U`
Mike Barker doesn't fully qualify several of his descriptions. For example, he never indications as to whether a pattern is in a band or a stack. He leaves it to the reader to consider both possibilities. As to (ab)U, I believe that he was only indicating that there must be at least one cell of (ab)U that sees abY ... and that Y must be limited to (ab)U cells that see abY.
Note: I an concerned that he marked Z as optional in ab(Z). This may be a typo. If so, maybe there are others.
_
daj95376
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### Re: Puzzle #54: Mike Barker UR+3X/1SL
sultan vinegar wrote:I can't find any instances of the pattern in your two examples, and I can't follow DPBs second chain, but after I search of the forum I found an example of a puzzle where the method works <here>.
It's just an AIC with a derived strong inference through the UR - not a separate technique.
Code: Select all
`+--------------------------+--------------------------+--------------------------+| 2 46 1 | 69 456 59 | 3 7 8 || 8 67abU 459 |267abZ 267abY 3 | 459 1 459 || 3 479 459 |78 478 1 | 459 6 2 |+--------------------------+--------------------------+--------------------------+| 6 489 3 | 1 258 7 | 4589 289 459 || 1 5 2 | 68 9 4 | 7 38 36 || 7 489 49 | 3 2568 25 | 1 289 4569 |+--------------------------+--------------------------+--------------------------+| 4 3 8 |279abX 27ab 6 | 29 5 1 || 9 1 6 | 5 3 28 | 28 4 7 || 5 2 7 | 4 1 89 | 6 389 39 |+--------------------------+--------------------------+--------------------------+`
a=2, b=7, U=6, X=9, Y=6, Z=6.
(7=6)r2c2 - (6)r2c45 = (9)r7c4[AUR27r27c45] - (2)r7c4 = (2)r2c4 => r2c4 <> 7.
[Withdrawn comment: I (yet again) incorrectly assigned the wrong cells to the SL.]
_
Last edited by daj95376 on Sat Jun 27, 2015 6:41 am, edited 1 time in total.
daj95376
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Posts: 2624
Joined: 15 May 2006
### Re: Puzzle #54: Mike Barker UR+3X/1SL
daj95376 wrote:Nice try, but it doesn't match the pattern. Your referenced puzzle has three SLs in the UR: <2> in [r2], <2> in [c4], and <7> in [r7]. Your identify abY and abZ using the SL in [r2], and you used the SL in [c4] to derive the elimination in your chain.
_
Have another look. There is a SL on 2 in c4 between the abX and abZ cells (a is 2 in this example), just as in Mike's pattern. Yes, there happen to be other strong links in this particular UR but these are not used by the pattern. There is no SL between abY and abZ in Mike's pattern.
More generally,
5 Truths = {1V1 1C4 2N567}
6 Links = {789r2 12n4 2b2}
2 Eliminations --> r12c4<>2
As AICs:
(6)r1c4 = QNT(789)r2c3467[AUR12:r12c34] - (789=2)r2c5 => r1c4 <> 2.
(1)r2c4 = (1-6)r1c4 = QNT(789)r2c3467[AUR12:r12c34] - (789=2)r2c5 => r2c4 <> 2.
Like I said, nothing more than AICs with a strong link through the AUR. If you can do AICs with AURs, then you can do this. There are no new techniques to memorise.
sultan vinegar
Posts: 81
Joined: 27 August 2013
### Re: Puzzle #55: Mike Barker UR+3X/1SL
Sultan: I've updated my post above. Sorry over the confusion.
eleven wrote:
daj95376 wrote:For the UR+3X\1SL to crack the puzzle, we need to eliminate <2> in both of r5c89.
My alternative is
*9r5c8->*2r2c8->2r3c1->*2r5c9->75r69c9->*9r2c9 => r5c8<>9
Here's the grid with the UR cells identified.
Code: Select all
` ab=r2c8; abY=r2c9; abX=r5c8; ab(Z)=r5c9 +--------------------------------------------------------------+ | 6 9 5 | 3 2 1 | 78 478 478 | | 1 27 3 | 4 678 578 | 2569 *29 *29+5 | | 27 4 8 | 9 67 57 | 256 1 3 | |--------------------+--------------------+--------------------| | 78 3 9 | 1 5 2 | 4 78 6 | | 28 5 6 | 7 3 4 | 1 *29+8 *29+8 | <- SL on a=<9> | 4 27 1 | 6 89 89 | 3 5 U27 | |--------------------+--------------------+--------------------| | 35 6 27 | 25 479 379 | 2589 2489 1 | | 35 1 27 | 8 479 379 | 259 6 2459 | | 9 8 4 | 25 1 6 | 257 3 U257 | +--------------------------------------------------------------+ # 61 eliminations remain`
The UR+3X\1SL pattern only identifies possible eliminations for b=2 in cells abX(r5c8) and/or abZ(r5c9). Both eliminations are needed to crack the puzzle.
I've identified an (ab)U in r69c9 and derived:
Code: Select all
` 2r5c9 -> r5c8=9, r6c9=7 -> r9c9=5 -> r2c9=9 -> r2c8=2 -> DP => -2 r5c9`
daj95376
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Previous | 4,278 | 11,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-39 | longest | en | 0.427997 |
https://www.mathblog.dk/project-euler-79-secret-numeric-passcode/ | 1,656,759,709,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00796.warc.gz | 918,662,019 | 23,553 | # Project Euler 79: By analysing a user’s login attempts, can you determine the secret numeric passcode?
Before even starting this problem I would like to say that I have solved the major part of Problem 79 of Project Euler by hand. The problem reads
A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.
The text file, keylog.txt, contains fifty successful login attempts.
Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.
I actually just wanted to take a look at the data before figuring out a solution, and as I saw there were some multiple entries I wanted to remove them. So I imported the input file into a spread sheet and sorted the entries to remove the duplicate entries.
After doing that I was down to 31 entries to analyse – a tad easier to get an overview over. I noticed one thing. 0 never comes first or second. So I could safely place 0 at the end of the string. I also noted that 4 and 5 was not present at all and that 7 always comes first.
This lead me to write a small piece of code to pin out what numbers were before and after what other numbers. I wrote this
```int[,] input = ReadInput(filename);
bool[,] edge = new bool[10, 10];
bool[] found = new bool[10];
for (int i = 0; i < input.GetLength(0); i++) {
edge[input[i, 1], input[i, 0]] = true;
edge[input[i, 2], input[i, 0]] = true;
edge[input[i, 2], input[i, 1]] = true;
}
for (int i = 0; i < 10; i++) {
Console.Write("Comes before " + i + ": ");
for (int j = 0; j < 10; j++) {
if (edge[i, j]) Console.Write(j + ", ");
}
Console.WriteLine();
}
for (int i = 0; i < 10; i++) {
Console.Write("Comes after " + i + ": ");
for (int j = 0; j < 10; j++) {
if (edge[j, i]) Console.Write(j + ", ");
}
Console.WriteLine();
}
```
```private int[,] ReadInput(string filename) {
.Select(x => x.Trim()).ToArray();
int[,] retVal = new int[message.Length, 3];
for(int i= 0; i < message.Length; i++){
for (int j = 0; j < message[i].Length; j++) {
retVal[i, j] = int.Parse(message[i].Substring(j,1));
}
}
return retVal;
}
```
Executing that gave me
```Comes before 0: 1, 2, 3, 6, 7, 8, 9,
Comes before 1: 3, 7,
Comes before 2: 1, 3, 6, 7,
Comes before 3: 7,
Comes before 4:
Comes before 5:
Comes before 6: 1, 3, 7,
Comes before 7:
Comes before 8: 1, 2, 3, 6, 7,
Comes before 9: 1, 2, 3, 6, 7, 8,
Comes after 0:
Comes after 1: 0, 2, 6, 8, 9,
Comes after 2: 0, 8, 9,
Comes after 3: 0, 1, 2, 6, 8, 9,
Comes after 4:
Comes after 5:
Comes after 6: 0, 2, 8, 9,
Comes after 7: 0, 1, 2, 3, 6, 8, 9,
Comes after 8: 0, 9,
Comes after 9: 0,
```
And from that we can see that 4,5 is indeed not present in the input.
Further more by inspection we can see that there are no repeated digits so we should be able to produce a sequence without repeated digits.
Using what comes before the others we can see that 7 comes first, and the only thing that is before 3 is 7, so we can write 73 as the first digits. This can be continued by hand and yields the sequence 73162890. We can then check that it also fits the other way around.
## Wrapping up
I don’t have the full solution through an algorithm, but the piece of source code I have can be downloaded here..
Later on I have found out that since we don’t have repeated digits in the solution we could have used topological sorting where the nodes of the input graph is the digits 0-9 and there is a directed edge between two nodes if the number comes before the other. This can then be fed to this algorithm and the answer should be provided. However, a combination of laziness and … well just laziness means that I haven’t implemented the algorithm. If you have I would love to see it.
Today’s blog image is provided by Marc Falardeau who shared it under the creative commons license.
### Posted by Kristian
SuprDewd
I remember this being a pretty weird problem. Me and my friend had found the solution by hand (which wasn’t too hard), but I had trouble creating a program. So I wrote down the steps that I was using to solve the problem by hand, and created a flowchart of sorts. After I did that, it was trivial to convert that into working code. I guess that was the first “algorithm” that I created.
Kristian
Nice way of solving it SuprDewd. I guess that is a good first solution for almost any algorithm. It probably wont be the best algorithm, but it is a good first attempt.
SuprDewd
This is actually my original code for the problem:
```List<string> codes = new List<string> {
/* ... Here were the codes ... */
};
codes = codes.Distinct().OrderBy(i => i).ToList();
string code = "";
while (codes.Count > 0)
{
List<char> possibleNext = new List<char>();
for (int i = 0; i < codes.Count; i++)
{
if (!possibleNext.Contains(codes[i][0]))
{
}
}
for (int i = 0; i < codes.Count; i++)
{
for (int j = 1; j < codes[i].Length; j++)
{
for (int c = 0; c < possibleNext.Count; c++)
{
if (possibleNext[c] == codes[i][j])
{
possibleNext.RemoveAt(c);
}
}
}
}
char next = possibleNext.Single();
code += next;
List<string> buffer = new List<string>();
for (int i = 0; i < codes.Count; i++)
{
codes[i] = codes[i].Replace(next, ' ').Trim();
if (codes[i].Length > 0)
{
}
}
codes = buffer;
}
Console.WriteLine(code);```
This is another solution I created later. I think this one is faster.
```string[] codes = File.ReadAllLines("Problem79.txt").Distinct().ToArray();
char nextChar = codes[0][0];
List<char> triedChars = new List<char> { nextChar };
List<char> code = new List<char>();
while (true)
{
if (codes.All(c => !c.Skip(1).Contains(nextChar))) { code.Add(nextChar); codes = codes.Select(l => l.TrimStart(nextChar)).ToArray(); triedChars.Clear(); }
codes = codes.Where(c => c != "").Distinct().ToArray();
if (codes.Length > 0)
{
nextChar = codes.Select(i => i.First()).Where(i => !triedChars.Contains(i)).First();
}
else break;
}
Console.WriteLine(new String(code.ToArray()));```
[…] I was researching the web for approaches to solving this problem, I found Kristian’s MathBlog article quite useful (as usual), and so I took up his suggestion to try implementing a topological sorting […]
Robert Tsai
Thanks Kristian for another great blog post. I took up your advice, and tried to implement a topological sort algorithm.
Kristian
Cool. I guess I have to make a test implementation of that as well. Did it work well?
Robert Tsai
Yes – it worked beautifully!!!
Khaur
Nice one spotting the relation with topological sorting!
I just solved this using gedit and the replace function: find the digit that only occurs at the beginning, write it down, delete it and repeat. I think this instance of the problem is not very interesting because the greedy solution is too easy.
It would be interesting to try and find an algorithm that works even when there are repeated digits, and what are the conditions on the sampling of the password for the algorithm to successfully recover it. Put that way, it sounds a lot like grammatical inference (and it actually is, can you find the class of grammar that we are trying to learn?).
S.Ged
Why all of numbers coming at most 1 time ?
Ethereal_shore
SuprDewd,I faced the same problem while trying to create the program for the problem I easily solved by observation. Now will try to implement a algorithm.
Martin Kirk
1: 319
2: 680 -> 319680
3: 180 tells us 1 < 8 319680 = ok
4: 690 -> swap 9 & 6 -> 316980
5: 129 -> inject 2 after 1 -> 3162980
6: 620 = ok
7: 762 -> inject 7 before 6 -> 31762980
8: 689 -> swap 8 & 9 -> 31762890
9: 762 (repeat of 7)
10: 318 = ok
11: 368 = ok
12: 710 -> swap 1 & 7 -> 37162890
.. ok
20: 736 -> swap 7 & 3 -> 73162890 (solved)
..
50: all ok
should be pretty easy to build an algorithm on that.
Martin Kirk
Solution:
var solution = new List();
var lines = from txt in File.ReadAllLines(Path.GetDirectoryName (Util.CurrentQueryPath) + “\\p079_keylog.txt”)
let num = Convert.ToInt32(txt).ToArr()
select new
{
A = num[0],
B = num[1],
C = num[2]
};
foreach (var line in lines)
{
if(solution.IndexOf(line.B) < solution.IndexOf(line.A))
{
var tmp = solution.IndexOf(line.A);
solution[solution.IndexOf(line.B)] = line.A;
solution[tmp]= line.B;
}
if(solution.IndexOf(line.C) < solution.IndexOf(line.B))
{
var tmp = solution.IndexOf(line.B);
solution[solution.IndexOf(line.C)] = line.B;
solution[tmp]= line.C;
}
}
String.Join("", solution).Dump();
Martin Kirk
solution
```var solution = new List<int>();
var lines = from txt in File.ReadAllLines(Path.GetDirectoryName (Util.CurrentQueryPath) + "\\p079_keylog.txt")
let num = Convert.ToInt32(txt).ToArr()
select new
{
A = num[0],
B = num[1],
C = num[2]
};
foreach (var line in lines)
{
if(solution.IndexOf(line.B) < solution.IndexOf(line.A))
{
var tmp = solution.IndexOf(line.A);
solution[solution.IndexOf(line.B)] = line.A;
solution[tmp]= line.B;
}
if(solution.IndexOf(line.C) < solution.IndexOf(line.B))
{
var tmp = solution.IndexOf(line.B);
solution[solution.IndexOf(line.C)] = line.B;
solution[tmp]= line.C;
}
}
String.Join("", solution).Dump();
```
Jeremy Miller
There is a much simpler algorithm.
1. Determine the digits in the passcode by making a list of digits in the keys, without duplicates.
2. Make a list of pairs of numbers from the order in the keys.
3. Find the unique digit which never appears at the beginning of any of the pairs. This is the last digit in the passcode, because nothing comes after it.
4. Remove pairs which include the last digit in the second entry of the pair, and repeat step 3 to find the second-last digit.
5. Loop this process n times, where n is the length of the passcode until you have constructed the whole passcode.
fileName = raw_input(“Enter the file name: “)
entries = []
with open(fileName) as inputKeys:
#print “entries \t\t= \n\n”, entries, “\n”
L = len(entries[0])
N = len(entries)
#print “key length \t\t= “, L, “\n”
#print “number of entries \t= “, N, “\n”
digitsInPasscode = set()
for attempt in entries:
#print attempt
#attemptDigits = map(int,str(attempt))
attemptDigits = attempt
#print attemptDigits
for digit in attemptDigits:
passcodeLength = len(digitsInPasscode)
#print ‘Digits in the passcode are’, digitsInPasscode, ‘\n’
#print ‘Number of digits in the passcode =’, passcodeLength, ‘\n’
#print ‘range(passcodeLength) =’, range(passcodeLength), ‘\n’
###################################################################
pairOrder = []
for n in range(N):
entryDigits = entries[n]
for i in range(L-1):
for j in range(i+1,L):
pairOrder = pairOrder + [entryDigits[i],entryDigits[j]]
pairOrder = [pairOrder[i:i+2] for i in range(0, len(pairOrder),2)]
def uniq(input):
output = []
for x in input:
if x not in output:
output.append(x)
return output
pairOrder = uniq(pairOrder)
#print “The order of pairs of numbers \t= \n\n”, pairOrder, “\n”
###################################################################
solution = []
digitsInPasscodeList = list(digitsInPasscode)
possibleDigits = digitsInPasscodeList
for d in range(passcodeLength):
for pair in pairOrder:
if pair[0] in possibleDigits:
possibleDigits.remove(pair[0])
for i in range(len(pairOrder)):
for pair in pairOrder:
if (pair[1] == possibleDigits[0]):
pairOrder.remove(pair)
solution.insert(0,possibleDigits[0])
del digitsInPasscodeList
del possibleDigits
digitsInPasscodeList = list(digitsInPasscode)
possibleDigits = digitsInPasscodeList
for n in solution:
possibleDigits.remove(n)
#print “The solution so far starting from the end = “, solution, “\n”
#print “The new pair order = \n”, pairOrder, “\n”
#print “The new possibleDigits = “, possibleDigits, “\n”
###################################################################
Solution = map(int, solution)
print “\nSolution: “, solution, “\n”
###################################################################
Aravind
What I really did was first solving it by hand. And it seemed that, it was somewhat easy enough to implemented.
public class checker {
public static void main(String args[]) {
String temp[] = { “319”, “680”, “180”, “690”, “129”, “620”, “762”, “689”, “762”, “318”, “368”, “710”, “720”,
“710”, “629”, “168”, “160”, “689”, “716”, “731”, “736”, “729”, “316”, “729”, “729”, “710”, “769”, “290”,
“719”, “680”, “318”, “389”, “162”, “289”, “162”, “718”, “729”, “319”, “790”, “680”, “890”, “362”, “319”,
“760”, “316”, “729”, “380”, “319”, “728”, “716” };
for (int countLogs = 0; countLogs -1; count–) {
if (!(tempPassWord + “”).contains(temp[countLogs].charAt(count) + “”) && count == 2) {
} else if (!(tempPassWord + “”).contains(temp[countLogs].charAt(count) + “”)) {
temp[countLogs].charAt(count));
}
}
for (int count = 0; count tempPassWord
.indexOf(String.valueOf(temp[countLogs].charAt(count + 1)))) {
int index2 = tempPassWord.indexOf(String.valueOf(temp[countLogs].charAt(count + 1)));
tempChar = temp[countLogs].charAt(count);
tempPassWord.replace(index1, index1 + 1, String.valueOf(temp[countLogs].charAt(count + 1)));
tempPassWord.replace(index2, index2 + 1, tempChar + “”);
}
}
}
}
}
Ranjith
I tried using java for the problem it checks using the same way manual work does.
public class checker {
public static void main(String args[]) {
String temp[] = { “319”, “680”, “180”, “690”, “129”, “620”, “762”, “689”, “762”, “318”, “368”, “710”, “720”,
“710”, “629”, “168”, “160”, “689”, “716”, “731”, “736”, “729”, “316”, “729”, “729”, “710”, “769”, “290”,
“719”, “680”, “318”, “389”, “162”, “289”, “162”, “718”, “729”, “319”, “790”, “680”, “890”, “362”, “319”,
“760”, “316”, “729”, “380”, “319”, “728”, “716” };
for (int countLogs = 0; countLogs -1; count–) {
if (!(tempPassWord + “”).contains(temp[countLogs].charAt(count) + “”) && count == 2) {
} else if (!(tempPassWord + “”).contains(temp[countLogs].charAt(count) + “”)) {
temp[countLogs].charAt(count));
}
}
for (int count = 0; count tempPassWord
.indexOf(String.valueOf(temp[countLogs].charAt(count + 1)))) {
int index2 = tempPassWord.indexOf(String.valueOf(temp[countLogs].charAt(count + 1)));
tempChar = temp[countLogs].charAt(count);
tempPassWord.replace(index1, index1 + 1, String.valueOf(temp[countLogs].charAt(count + 1)));
tempPassWord.replace(index2, index2 + 1, tempChar + “”);
}
}
}
}
}
prakhar
what if the success full login attempts are 012 and 210, passcode need not have any digit only once, it is a wrong assumption.
Mateen
Made this in python, seemed simple and I didn’t see any other python solutions:
[code language="python"][def problem_79():
from operator import itemgetter
f = open('p079_keylog.txt', 'r')
f.close()
a = [x.replace('''\n''', '') for x in a]
a = list(set(a))
b = [x[0] for x in a]
c = [x[1] for x in a]
d = [x[2] for x in a]
b = [x for x in b if x not in c+d]
d = [x for x in d if x not in c+b]
c = [x for x in c if x not in b+d]
dic = {str(k):0 for k in range(10) if str(k) in b+c+d}
for x in a:
if dic[x[0]] >= dic[x[1]]:
for s in dic:
if dic[s] > dic[x[1]]:
dic[s] += 1
dic[x[1]] = dic[x[0]] + 1
if dic[x[1]] >= dic[x[2]]:
for s in dic:
if dic[s] > dic[x[2]]:
dic[s] += 1
dic[x[2]] = dic[x[1]] + 1
final = sorted(dic.items(), key=itemgetter(1))
print(''.join([x[0] for x in final]))]
Mateen
Made this in python, seemed simple and I didn’t see any other python solutions:
```def problem_79():
from operator import itemgetter
f = open('p079_keylog.txt', 'r')
f.close()
a = [x.replace('''\n''', '') for x in a]
a = list(set(a))
b = [x[0] for x in a]
c = [x[1] for x in a]
d = [x[2] for x in a]
b = [x for x in b if x not in c+d]
d = [x for x in d if x not in c+b]
c = [x for x in c if x not in b+d]
dic = {str(k):0 for k in range(10) if str(k) in b+c+d}
for x in a:
if dic[x[0]] >= dic[x[1]]:
for s in dic:
if dic[s] > dic[x[1]]:
dic[s] += 1
dic[x[1]] = dic[x[0]] + 1
if dic[x[1]] >= dic[x[2]]:
for s in dic:
if dic[s] > dic[x[2]]:
dic[s] += 1
dic[x[2]] = dic[x[1]] + 1
final = sorted(dic.items(), key=itemgetter(1))
print(''.join([x[0] for x in final]))```
Well, I spent a long time thinking about some generic solution to this problem to the point I gave up continuing. I thought about several things, from linked lists to topological sorting. The linked list solution turned out to be wrong, and it was not obvious that each digit must appear only once in the secret passcode to use topological sorting. After some time, I decided to check solutions from other people and I got disappointed when most of them solved it by hand. I know being aware of the conditions of the problem and of what is possible to do to solve it is part of problem-solving. But still, I wanted to somehow see the beauty in this problem. | 4,785 | 16,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | longest | en | 0.895784 |
https://www.convertunits.com/from/cubic+metre/day/to/ounce/minute+%5BUK%5D | 1,601,136,412,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00650.warc.gz | 800,732,495 | 8,461 | ## ››Convert cubic metre/day to ounce/minute [UK]
cubic metre/day ounce/minute [UK]
Did you mean to convert cubic metre/day to ounce/minute [US] ounce/minute [UK]
How many cubic metre/day in 1 ounce/minute [UK]? The answer is 0.04091481.
We assume you are converting between cubic metre/day and ounce/minute [UK].
You can view more details on each measurement unit:
cubic metre/day or ounce/minute [UK]
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 86400 cubic metre/day, or 2111704.7836712 ounce/minute [UK].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic meters/day and ounces/minute.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cubic metre/day to ounce/minute [UK]
1 cubic metre/day to ounce/minute [UK] = 24.44103 ounce/minute [UK]
2 cubic metre/day to ounce/minute [UK] = 48.88206 ounce/minute [UK]
3 cubic metre/day to ounce/minute [UK] = 73.32308 ounce/minute [UK]
4 cubic metre/day to ounce/minute [UK] = 97.76411 ounce/minute [UK]
5 cubic metre/day to ounce/minute [UK] = 122.20514 ounce/minute [UK]
6 cubic metre/day to ounce/minute [UK] = 146.64617 ounce/minute [UK]
7 cubic metre/day to ounce/minute [UK] = 171.08719 ounce/minute [UK]
8 cubic metre/day to ounce/minute [UK] = 195.52822 ounce/minute [UK]
9 cubic metre/day to ounce/minute [UK] = 219.96925 ounce/minute [UK]
10 cubic metre/day to ounce/minute [UK] = 244.41028 ounce/minute [UK]
## ››Want other units?
You can do the reverse unit conversion from ounce/minute [UK] to cubic metre/day, 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! | 600 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.812495 |
http://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-8-dynamics-ii-motion-in-a-plane-exercises-and-problems-page-211/8 | 1,524,346,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945448.38/warc/CC-MAIN-20180421203546-20180421223546-00241.warc.gz | 440,363,571 | 12,656 | Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
The correct banking angle of the road is $7.3^{\circ}$
We can convert the speed to units of m/s: $v = (90~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$ $v = 25~m/s$ We can use the equation for the speed around a banked curve to find the banking angle. $v = \sqrt{rg~tan(\theta)}$ $v^2 = rg~tan(\theta)$ $tan(\theta) = \frac{v^2}{rg}$ $\theta = arctan(\frac{v^2}{rg})$ $\theta = arctan(\frac{(25~m/s)^2}{(500~m)(9.80~m/s^2)})$ $\theta = 7.3^{\circ}$ The correct banking angle of the road is $7.3^{\circ}$ | 221 | 600 | {"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} | 3.984375 | 4 | CC-MAIN-2018-17 | latest | en | 0.751503 |
https://answers.yahoo.com/question/index?qid=20140301000051KK00003 | 1,579,987,510,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00320.warc.gz | 322,924,112 | 25,259 | Anonymous
Anonymous asked in 科學及數學其他 - 科學 · 6 years ago
# heat transfer
(1) 如果一個輻射儀量測到某地區海表面輻射出的長波輻射為370 W/m^2, 試估算海表面的溫度為多少? ( 請自行查詢相關所需參數，並明確說明 )
(2)假設某個物體的長波輻射率為0.9，溫度為攝氏25 度，試比較下列兩個因子變化對於長波輻射的影響 (i) 長波輻射率減少20% (ii)該物體絕對溫度下降 5%
You can answer either in Chinese or English.
### 1 Answer
Rating
• 天同
Lv 7
6 years ago
Favorite Answer
(1) Using Stefan's Law,
370 = (5.7x10^-8)T^4
where (5.7x10^-8 W/m^2.T^4 is the Stefan's Constant), and T is the absolute temperature of the sea surface.
Hence, T = 284 K = 10.8 deg.C
(2)(i) New emissivity (輻射率) = 0.9 x 0.8 = 0.72
hence, intensity of radiation
= 0.72 x (5.7x10^-8) x (273+25)^4 W/m^2 = 324 W/m^2
Original intensity = 0.9 x (5.7x10^-8) x (273+25)^4 W/m^2 = 405 W/m^2
Hence, percentage decrease in intensity
= [(405 - 324)/405] x 100% = 20%
(ii) New temperature = (273+25) x 0.95 K = 283.1 K
hence, intensity of radiation = 0.9 x (5.7x10^-8) x 283.1^4 W/m^2
= 330 W/m^2
Percentage decrease in intensity
= [(405 - 330)/405] x 100%
= 18.5%
• Login to reply the answers
Still have questions? Get your answers by asking now. | 507 | 1,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-05 | latest | en | 0.654879 |
https://socratic.org/questions/how-do-you-factor-completely-2u-3-w-4-2u-3 | 1,575,603,997,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484477.5/warc/CC-MAIN-20191206023204-20191206051204-00051.warc.gz | 562,214,625 | 6,007 | # How do you factor completely: 2u^3 w^4 -2u^3?
Jul 24, 2015
#### Answer:
$2 {u}^{3} {w}^{4} - 2 {u}^{3} = 2 {u}^{3} \left(w - 1\right) \left(w + 1\right) \left({w}^{2} + 1\right)$
#### Explanation:
First separate out the common factor $2 {u}^{3}$ to get:
$2 {u}^{3} {w}^{4} - 2 {u}^{3} = 2 {u}^{3} \left({w}^{4} - 1\right)$
Then use the difference of squares identity (twice):
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
$2 {u}^{3} \left({w}^{4} - 1\right)$
$= 2 {u}^{3} \left({\left({w}^{2}\right)}^{2} - {1}^{2}\right)$
$= 2 {u}^{3} \left({w}^{2} - 1\right) \left({w}^{2} + 1\right)$
$= 2 {u}^{3} \left({w}^{2} - {1}^{2}\right) \left({w}^{2} + 1\right)$
$= 2 {u}^{3} \left(w - 1\right) \left(w + 1\right) \left({w}^{2} + 1\right)$
${w}^{2} + 1$ has no linear factors with real coefficients since ${w}^{2} + 1 \ge 1 > 0$ for all $w \in \mathbb{R}$ | 437 | 879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 4.5625 | 5 | CC-MAIN-2019-51 | latest | en | 0.559136 |
https://cracku.in/blog/discount-questions-for-ibps-clerk/ | 1,716,941,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00033.warc.gz | 158,970,463 | 37,526 | 0
1632
# Discount Questions For IBPS Clerk
Download important Discount Questions PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Discount Question and Answers for IBPS Clerk Exam.
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Question 1: A trader sells an item to a retailer at 20% discount, but charges 10% on the discounted price, for delivery and packaging. The retailer sells it for Rs. 2046 more, thereby earning a profit of 25%. At what price had the trader marked the item?
a) Rs. 9400
b) Rs. 9000
c) Rs. 8000
d) Rs. 12000
e) Rs. 9300
Question 2: A profit of 25% is earned on goods when a discount of 20% is allowed on the marked price. What profit percentage will be earned when a discount of 10% is allowed on the marked price?
a) $45\frac{9}{11}$
b) $42\frac{3}{4}$
c) $40\frac{5}{8}$
d) $37\frac{2}{3}$
e) None of these
Question 3: A shopkeeper sold first unit of an article to Aria at 20% discount and the second unit of the same article at 40% discount. If the shopkeeper earned an overall profit of 5%, the marked price of the article was what percent of the cost price of the article ?
a) 200
b) 150
c) 125
d) 160
e) 120
Question 4: Deepa bought a calculator with 30% discount on the listed price. Had she not got the discount she would have paid Rs 82.50. extra At what price did she buy the calculator ?
a) Rs 192.50
b) Rs 275
c) Rs 117.85
d) Cannot be determined
e) None of these
Question 5: A profit of 8% is made by selling a shirt after offering a discount of 12%. If the marked price of the shirt is Rs.1,080/-, find its cost price.
a) Rs.890/-
b) Rs. 780/-
c) Rs.880/-
d) Rs.900/-
e) None of these
Question 6: A shopkeeper labelled the price of his articles so as to earn a profit of 30% on the cost price. He then sold the articles by offering a discount of 10% on the labelled price. What is the actual percent profit earned in the deal?
a) 18%
b) 15%
c) 20%
d) Cannot be determined
e) None of these
Question 7: A person bought an article on 40% discount and sold it at 50% more than the marked price. What profit did he get?
a) 250 %
b) 150 %
c) 350 %
d) 200 %
e) None of these
Question 8: An article was purchased for Rs.78,350. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price?
a) 4
b) 7
c) 5
d) 3
e) 6
Question 9: A shopkeeper sold an article at 20% discount and earned d profit of 4%. By what percent the marked price of the article more than the cost price of the article ?
a) 20%
b) 15%
c) 40%
d) 25%
e) 30%
Question 10: A shopkeeper sold a T.V. set for Rs. 17,940/-, with a discount of 8% and gained 19.6% If no discount is allowed, what will be his gain per cent?
a) 25%
b) 26.4%
c) 24.8%
d) Cannot be determined
e) None of these
Question 11: A seller marks the price 50% above the cost price and gives 10% discount on an item.While selling, he cheats customer by giving 20% less in weight. Find his overall profit percent (approximate) ?
a) 26%
b) 65%
c) 68%
d) 72%
e) 76%
Question 12: A shopkeeper bought 84 identical shirts priced at Rs. 240 each. He spent a total of Rs. 3200 on transportation and packaging. He put the label of marked price of Rs.420 on each shirt. He offered a discount of 15% on each shirt at the marked price. What is the total profit of the shopkeeper in the whole transaction ?
a) Rs. 6258
b) Rs. 6528
c) Rs. 6268
d) Rs. 6628
e) None of these
Question 13: Vishal sold an article for Rs 1,840 and made 15 per cent profit on the discounted price he bought.If the discount was 20 per cent.What was the original price ?
a) Rs 1,900
b) Rs 1,600
c) Rs 2,400
d) Can’t be determined
e) None of these
Question 14: A merchant bought some goods worth Rs. 6000 and sold half of them at 12% profit. At what profit per cent should he sell the remaining goods so as to make an overall profit of 18%?
a) 24
b) 28
c) 18
d) 20
e) 26
Question 15: Cost price of two beds are equal. One bed is sold at a profit of 30% and the other one for Rs. 5,504/less than the first one. If the overall profit earned after selling both the beds is 14%, what is the cost price of each bed ?
a) Rs. 17,000
b) Rs. 16,800
c) Rs. 17,600
d) Rs. 17,800
e) Rs. 17,200
Let Marked price of item = $Rs. 100x$
=> Selling price of trader = Cost price of retailer = $100x \times \frac{80}{100} \times \frac{110}{100}$
= $Rs. 88x$
Selling price of retailer = $Rs. (88x + 2046)$
Profit % = $\frac{(88x + 2046) – 88x}{88x} \times 100 = 25$
=> $\frac{2046}{88x} = \frac{25}{100} = \frac{1}{4}$
=> $x = \frac{2046 \times 4}{88} = 93$
$\therefore$ Marked price = $100 \times 93 = Rs. 9,300$
Let the Marked Price = Rs. 100
=> Selling price = $\frac{80}{100} \times 100 = 80$
$\therefore$ Cost price = $\frac{100}{125} \times 80 = 64$
When discount of 10% is allowed
=> S.P. = $\frac{90}{100} \times 100 = 90$
$\therefore$ Required Profit % = $\frac{90 – 64}{64} \times 100$
= $\frac{325}{8} = 40\frac{5}{8} \%$
Let the cost price of each article = Rs. $100x$
=> C.P. of both articles = $2 \times 100x = 200x$
=> Selling price = $\frac{105}{100} \times 200x = 210x$ ——————Eqn(1)
Let the marked price = Rs. $100y$
Selling price of 1st article after 20% discount = $\frac{80}{100} \times 100y = 80y$
Selling price of 2nd article after 40% discount = $\frac{60}{100} \times 100y = 60y$
=> Total selling price = $80y + 60y = 140y$ ——————–Eqn(2)
Comparing eqns(1) & (2), we get :
=> $210x = 140y$
=> $\frac{y}{x} = \frac{210}{140} = \frac{3}{2}$
$\therefore$ Required % = $\frac{3}{2} \times 100 = 150 \%$
Let the listed price of the calculator be L.
Hence, the price at which Deepa bought the calculator is 0.7L.
The difference between the two is 0.3L = 82.50
Hence, L = 82.50/0.3 = 275
Hence, the price at which Deepa bought the calculator is 275 – 82.50 = 192.50
Let cost price = Rs. $100x$
=> Selling price = $100x + \frac{8}{100} \times 100x$
= $108x$
Now, after offering 12% discount, Marked price = $\frac{88}{100} \times 1080 = 108x$
=> $\frac{88}{10} = \frac{108x}{108}$
=> $x = \frac{88}{10} = 8.8$
$\therefore$ C.P. = 100 * 8.8 = Rs. 880
Let the cost price of the article is 100
In order to earn 30% profit, the shopkeeper will sell it for 130
Discount = 10% of 130 = 13
SP = 130-13 = 117
Profit % =$\frac{117-100}{100}*100$ = 17%
Let the marked price be Rs. 100.
Then cost price = 100 – 40 = Rs. 60
Selling price = 100+ 50 = Rs. 150
Profit = 150 – 60 = 90
Profit%=90/60×100=150%
Marked price = 78,350 + 30% of 78,350 = Rs 1,01,855
Price after discount = 101855 -20% of 101855 = Rs 81,484
Profit = Rs 3,134
Let the cost price of each article = Rs. $100x$
=> Selling price = $\frac{104}{100} \times 100x = 104x$ ——————Eqn(1)
Let the marked price = Rs. $100y$
Selling price of article after 20% discount = $\frac{80}{100} \times 100y = 80y$ ————Eqn(2)
Comparing eqns(1) & (2), we get :
=> $104x = 80y$
=> $\frac{y}{x} = \frac{104}{80} = \frac{13}{10}$
$\therefore$ Required % by which marked price is more than the cost price
= $\frac{13 – 10}{10} \times 100 = 30 \%$
Let C.P. = Rs. $100x$
=> Selling price after profit of 19.6 %
= $100x + \frac{19.6}{100} \times 100x = 119.6x$
Acc. to ques,
=> $119.6x = 17,940$
=> $x = \frac{17,940}{119.6} = 150$
=> C.P. = Rs. $15,000$
Marked price after discount of 8%
= $\frac{100}{92} \times 17,940$
= $19,500$
$\therefore$ Profit when no discount is given
= $\frac{19,500 – 15,000}{15,000} \times 100$
= $\frac{45}{15} \times 10 = 30 \%$
Let the cost price be Rs 100 and amount of product be 100 gm
Now there is 50% markup so the marked price becomes Rs 150
Now after 10% discount on Marked price the selling price = Rs (150 – 15) = Rs 135
Now it is said that he only sale 80 % of the quantity So,
he is selling 80 gm in 135 Rs which means SP of 1 gm is = Rs 1.6875
Cost price of 1 gm = Rs 1
So profit percentage is 68.75 % ~ 68%
total cost of buying 84 shirts at Rs 240 each = Rs(84×240) = Rs 20160
Transportation cost = Rs 3200
Total cost = 3200 + 20160 = Rs 23360
As 15 % discount is given on marked price of 420 so selling price = 0.85 x 420 = Rs 357
total selling price of 84 shirts = 84 x 357 = Rs 29988
Total profit = 29988 – 23360 = Rs 6628
let the original price be Rs y
So after 20% discount it becomes = 0.8 y
Now as the article is sold at 15 % profit so Selling Price = 1.15 × 0.8 y = 1840
y = Rs 2000
Total profit to be made = 18%
Profit already made = 6% (12% on half of total value means 6% on total value)
Profit remaining = 12%
Since 12% profit is to be earned on 100% of the quantity, 24% of profit is to be made on half of the qunatity.
Let the cost price of each bed Rs C per bed
Selling price of one bed (S1) = 1.3 C
Selling Price of another friend (S2) = 1.3C – 5504
it is given that overall profit is 14 % on selling both beds so ,
S1 + S2 = 1.14(2C)
1.3C + 1.3C – 5504 = 2.28C
0.32C = 5504
C = Rs 17200
We hope this Discount questions and answers for IBPS Clerk preparation will be helpful to you. | 3,170 | 9,128 | {"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} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.867841 |
https://www.physicsforums.com/threads/transfer-function-needed-help-please.862646/ | 1,511,128,752,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00542.warc.gz | 842,074,962 | 17,822 | # Transfer function needed -- help please
1. Mar 18, 2016
### Suiluas
Hi guys, I'm new to this engineering problem solving and I just wanted to ask for your help getting a transfer function of the input voltage Uin(t) and the output voltage Uout(t).
1. Uin(t) = U_R(t) + u_out(t)
2. i(t) = i_L(t) + i_C(t)
3. u_r(t) = R_i(t)
4. u_out(t) = L d_iL / dt
5. i_C = C d_uout / dt
3. The attempt:
Dv_c(t) / dt = 1/C i_c(t) = [ 1/C (i(t) - i_out(t) ] = - 1/RC_vc(t) + 1/C i (t)
Di(t) / dt = 1/L v_L(t) = 1/L...
Last edited by a moderator: Mar 18, 2016
2. Mar 18, 2016
### Staff: Mentor
Welcome to the PF.
Is there a circuit diagram that you could UPLOAD for us to look at. For me, it's hard to visualize the problem. Thanks
3. Mar 18, 2016
### Suiluas
Sure, attached is the circuit. :) http://postimg.org/image/7a4mj5c6b/
<< Link removed by Mentor >>
4. Mar 18, 2016
### Staff: Mentor
5. Mar 18, 2016
### Suiluas
Sorry for that, here is the picture of the system.
#### Attached Files:
• ###### Capture2.PNG
File size:
8.1 KB
Views:
31
6. Mar 18, 2016
### Staff: Mentor
No worries. Thanks for the diagram.
I'm not able to follow what you are doing with your equations, but it seems like you are working with each component individually (I could be wrong). Instead, I would write the one KCL equation for the node between the input resistor and the RLC load. Write that differential equation and solve for the currents and voltages. Can you give that approach a try?
7. Mar 18, 2016
### Suiluas
I will try, but it's not as easy for me. anyways, thanks for advice.
8. Mar 19, 2016
### LvW
Suiluas - are you aware that a system`s "transfer function" requires to find voltage-current relations in the frequency domain?
You have started in the time domain - this is not necessary and requires application of the Laplace transformation. Instead, you can start directly with impedances in the frequency domain
Examples: inductice impedance: ZL=jωL, capcitive impedance: ZC=1/jωC.
Last edited: Mar 19, 2016 | 622 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-47 | longest | en | 0.892731 |
https://kalkicode.com/nelson-number-program | 1,701,730,933,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00105.warc.gz | 403,883,932 | 9,255 | Posted on by Kalkicode
Code Number
# Nelson Number Program
The Nelson number problem is an interesting mathematical concept. A Nelson number is a number that is divisible by 111. It is named after the mathematician Edward Nelson, who worked in the field of logic and mathematical physics. In this problem, we are given a number, and our task is to determine whether it is a Nelson number or not.
## Problem Statement
Given an integer input, we need to check whether it is a Nelson number or not. If the number is divisible by 111, then it is a Nelson number; otherwise, it is not.
## Explanation with Suitable Example
Let's take an example to understand the Nelson number problem better. Consider the number 666. To check if it is a Nelson number, we divide it by 111:
666 ÷ 111 = 6
Since the remainder is 0, 666 is a Nelson number.
## Standard Pseudocode
``````function isNelsonNumber(number):
if number % 111 == 0:
return True
else:
return False
``````
## Algorithm with Explanation
1. Start the program.
2. Define the function `isNelsonNumber(number)` that takes an integer `number` as input.
3. Check if the `number` is divisible by 111 using the modulo operator (`%`).
4. If the result of `number % 111` is equal to 0, return `True`, indicating that the number is a Nelson number.
5. If the result is not 0, return `False`, indicating that the number is not a Nelson number.
6. End the function.
7. In the `main()` function, call the `isNelsonNumber()` function for different test cases to check if they are Nelson numbers or not.
8. Print the result for each test case.
9. End the program.
## Code Solution
Here given code implementation process.
``````// C program
// Nelson Number Program
#include <stdio.h>
// Determine that given number is Nelson Number or not
void isNelsonNo(int number)
{
if ((number % 111) == 0)
{
// When number is nelson
printf(" %d is nelson number \n", number);
}
else
{
printf(" %d is not nelson number \n", number);
}
}
int main()
{
// Test Case
isNelsonNo(444);
isNelsonNo(113);
isNelsonNo(333);
return 0;
}``````
#### input
`````` 444 is nelson number
113 is not nelson number
333 is nelson number``````
``````/*
Java Program for
Nelson Number
*/
class NelsonNo
{
// Determine that given number is Nelson Number or not
public void isNelsonNo(int number)
{
if ((number % 111) == 0)
{
// When number is nelson
System.out.println(" [" + number + "] is nelson number");
}
else
{
System.out.println(" [" + number + "] is not nelson number");
}
}
public static void main(String[] args)
{
// Test Case
}
}``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````// Include header file
#include <iostream>
using namespace std;
/*
C++ Program for
Nelson Number
*/
class NelsonNo
{
public:
// Determine that given number is Nelson Number or not
void isNelsonNo(int number)
{
if ((number % 111) == 0)
{
// When number is nelson
cout << " [" << number << "] is nelson number" << endl;
}
else
{
cout << " [" << number << "] is not nelson number" << endl;
}
}
};
int main()
{
// Test Case
return 0;
}``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````// Include namespace system
using System;
/*
Csharp Program for
Nelson Number
*/
public class NelsonNo
{
// Determine that given number is Nelson Number or not
public void isNelsonNo(int number)
{
if ((number % 111) == 0)
{
// When number is nelson
Console.WriteLine(" [" + number + "] is nelson number");
}
else
{
Console.WriteLine(" [" + number + "] is not nelson number");
}
}
public static void Main(String[] args)
{
// Test Case
}
}``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````<?php
/*
Php Program for
Nelson Number
*/
class NelsonNo
{
// Determine that given number is Nelson Number or not
public function isNelsonNo(\$number)
{
if ((\$number % 111) == 0)
{
// When number is nelson
echo " [".\$number.
"] is nelson number".
"\n";
}
else
{
echo " [".\$number.
"] is not nelson number".
"\n";
}
}
}
function main()
{
// Test Case
}
main();``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````/*
Node JS Program for
Nelson Number
*/
class NelsonNo
{
// Determine that given number is Nelson Number or not
isNelsonNo(number)
{
if ((number % 111) == 0)
{
// When number is nelson
console.log(" [" + number + "] is nelson number");
}
else
{
console.log(" [" + number + "] is not nelson number");
}
}
}
function main()
{
// Test Case
}
main();``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````# Python 3 Program for
# Nelson Number
class NelsonNo :
# Determine that given number is Nelson Number or not
def isNelsonNo(self, number) :
if ((number % 111) == 0) :
# When number is nelson
print(" [", number ,"] is nelson number")
else :
print(" [", number ,"] is not nelson number")
def main() :
# Test Case
if __name__ == "__main__": main()``````
#### input
`````` [ 444 ] is nelson number
[ 113 ] is not nelson number
[ 333 ] is nelson number``````
``````# Ruby Program for
# Nelson Number
class NelsonNo
# Determine that given number is Nelson Number or not
def isNelsonNo(number)
if ((number % 111) == 0)
# When number is nelson
print(" [", number ,"] is nelson number", "\n")
else
print(" [", number ,"] is not nelson number", "\n")
end
end
end
def main()
# Test Case
end
main()``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number
``````
``````/*
Scala Program for
Nelson Number
*/
class NelsonNo()
{
// Determine that given number is Nelson Number or not
def isNelsonNo(number: Int): Unit = {
if ((number % 111) == 0)
{
// When number is nelson
println(" [" + number + "] is nelson number");
}
else
{
println(" [" + number + "] is not nelson number");
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: NelsonNo = new NelsonNo();
// Test Case
}
}``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
``````/*
Swift 4 Program for
Nelson Number
*/
class NelsonNo
{
// Determine that given number is Nelson Number or not
func isNelsonNo(_ number: Int)
{
if ((number % 111) == 0)
{
// When number is nelson
print(" [", number ,"] is nelson number");
}
else
{
print(" [", number ,"] is not nelson number");
}
}
}
func main()
{
// Test Case
}
main();``````
#### input
`````` [ 444 ] is nelson number
[ 113 ] is not nelson number
[ 333 ] is nelson number``````
``````/*
Kotlin Program for
Nelson Number
*/
class NelsonNo
{
// Determine that given number is Nelson Number or not
fun isNelsonNo(number: Int): Unit
{
if ((number % 111) == 0)
{
// When number is nelson
println(" [" + number + "] is nelson number");
}
else
{
println(" [" + number + "] is not nelson number");
}
}
}
fun main(args: Array < String > ): Unit
{
// Test Case
}``````
#### input
`````` [444] is nelson number
[113] is not nelson number
[333] is nelson number``````
## Resultant Output Explanation
Here's the explanation of each test case:
1. Test Case 1 (444): 444 is divisible by 111, so it is a Nelson number.
2. Test Case 2 (113): 113 is not divisible by 111, so it is not a Nelson number.
3. Test Case 3 (333): 333 is divisible by 111, so it is a Nelson number.
## Time Complexity of the Code
The time complexity of the provided code is O(1) because the function `isNelsonNumber()` performs a constant number of operations, regardless of the magnitude of the input number. The operations involved are simple arithmetic operations and a single modulo operation. Thus, the time complexity remains constant for any input size.
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
Categories
Relative Post | 2,249 | 7,973 | {"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} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.729881 |
https://www.mysciencework.com/publication/show/connected-resolvability-graphs-170801a5?search=1 | 1,603,412,877,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00373.warc.gz | 813,138,695 | 12,829 | # Connected Resolvability of Graphs
Authors
• 1 Western Michigan University, Department of Mathematics and Statistics, Kalamazoo, MI, 49008, USA , Kalamazoo
Type
Published Article
Journal
Czechoslovak Mathematical Journal
Publisher
Publication Date
Dec 01, 2003
Volume
53
Issue
4
Pages
827–840
Identifiers
DOI: 10.1023/B:CMAJ.0000024524.43125.cd
Source
Springer Nature
Keywords
For an ordered set W = {w1, w2,..., wk} of vertices and a vertex v in a connected graph G, the representation of v with respect to W is the k-vector r(v|W) = (d(v, w1), d(v, w2),... d(v, wk)), where d(x, y) represents the distance between the vertices x and y. The set W is a resolving set for G if distinct vertices of G have distinct representations with respect to W. A resolving set for G containing a minimum number of vertices is a basis for G. The dimension dim(G) is the number of vertices in a basis for G. A resolving set W of G is connected if the subgraph 〈W〉 induced by W is a nontrivial connected subgraph of G. The minimum cardinality of a connected resolving set in a graph G is its connected resolving number cr(G). Thus 1 ≤ dim(G) ≤ cr(G) ≤ n−1 for every connected graph G of order n ≥ 3. The connected resolving numbers of some well-known graphs are determined. It is shown that if G is a connected graph of order n ≥ 3, then cr(G) = n−1 if and only if G = Kn or G = K1,n−1. It is also shown that for positive integers a, b with a ≤ b, there exists a connected graph G with dim(G) = a and cr(G) = b if and only if \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\left( {a,b} \right) \notin \left\{ {\left( {1,k} \right):k = 1\;{\text{or}}\;k \geqslant 3} \right\}$$ \end{document} Several other realization results are present. The connected resolving numbers of the Cartesian products G × K2 for connected graphs G are studied. | 592 | 2,009 | {"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} | 3.171875 | 3 | CC-MAIN-2020-45 | latest | en | 0.856302 |
http://web2.0calc.com/questions/point-y-is-on-a-circle-and-point-p-lies-outside-the | 1,512,970,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00340.warc.gz | 323,550,069 | 5,550 | +0
# Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the
+2
318
2
Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?
Guest Mar 20, 2017
Sort:
#1
+79654
+2
We can use the secant-tangent theorem, here....specifically........
PA * PB = (PY)^2
15* PB = (9)^2
15 * PB = 81 divide both sides by 15
PB = 81 / 15
PB = 27 / 5
So...AB = PA - PB = 15 - 27/5 = 45/5 - 27/5 = 18/5
CPhill Mar 20, 2017
#2
+26357
+1
Typo in the last line Chris. 15 = 75/5 so: PB = 75/5 - 27/5 = 48/5
.
Alan Mar 21, 2017
### 9 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 360 | 1,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-51 | latest | en | 0.843505 |
https://tumblebearmobile.com/useful-for-parents/question-what-number-is-in-the-tens-and-ones-activity-kindergarten.html | 1,660,166,522,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00177.warc.gz | 535,829,640 | 11,187 | # Question: What Number Is In The Tens And Ones Activity Kindergarten?
## What number is tens and ones?
10 can be thought of as a bundle of ten ones — called a “ten.” The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
## How do you teach kids about tens?
An activity to help children understand tens and ones Use beans or rocks. Place a pile of them on a table and show that it is easier to count them in groups of ten. First make groups of ten, then count the ten-groups and the individual beans separately. Say, “I have here five ten-groups, and four individual beans.”
## What number can be written as four tens two ones?
Sal shows 42 as 4 tens and 2 ones.
## How do you teach place value in Year 1?
In our experience the answer to teaching place value in Year 1 is to keep things as simple as possible. Year 1 pupils may start to recognise place value when working with coins, but a good way to explain the topic in more explicit terms is by using beads. One idea is to use strings of beads in sets of ten.
You might be interested: How To Teach Writing In Kindergarten?
## How do you teach number value to kindergarten?
How to Teach Numbers to Preschoolers
1. Teach Counting with Number Rhymes.
2. Incorporate Numbers into Daily Tasks.
3. Play Number Games with a Group of Kids.
4. Write Down Numbers and Make the Child Draw that Quantity.
5. Point Out Numbers on Ad Boards and Vehicles.
6. Teach the Order of Numbers with Connect the Dots.
7. Count Fingers and Toes.
## How do you teach one’s tens and hundreds?
Create simple place value charts that are reusable by including a place for hundreds, tens, and ones. This layout mimics the way the number is written from left to right. In the ones section, ensure that there are two ten-frames to promote the concept of a group of ten and eliminate the need for one-by-one counting.
## What is the number equal to 30 ones?
S: 30 ones = 2 tens 10 ones. Repeat the process for 32, 38, 40, 41, 46, 50, 63, and 88.
## Is there only one way to break up a 2 digit number into tens and ones?
Subtract two-digit numbers by breaking the second number down into its tens and ones. Subtract the tens from the original first number, find the answer and then subtract the ones from that answer for the final result.
## What is the value in math?
In math, value is a number signifying the result of a calculation or function. So, in the example above, you could tell your teacher that the value of 5 x 6 is 30 or the value of x + y if x = 6 and y = 3 is 9. Value can also refer to a variable or constant. A constant is a fixed or well-defined number.
You might be interested: FAQ: How To Draw For Kindergarten?
## What manipulatives can be used to teach place value?
Base-10 blocks can be used to help students understand the concepts behind place value. Base-10 blocks also can be used to explain decimals. Other place-value manipulatives are Unifix cubes, snap cubes, plastic clips, and bean sticks/beans.
## What is the value of 9 in 90?
Place value of 9 is 9 × 10 = 90 and the place is tens. Place value of 5 is 5 × 1 = 5 and the place is ones. Note: The place value of digit 0 at any place is 0. | 851 | 3,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-33 | latest | en | 0.930854 |
http://ask.sagemath.org/question/73/multivariate-polynomials-over-rational-function | 1,386,350,212,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052216/warc/CC-MAIN-20131204131732-00033-ip-10-33-133-15.ec2.internal.warc.gz | 10,866,524 | 8,003 | # Multivariate Polynomials over Rational Function Fields
2 Is it possible to define multivariate polynomials where the coefficients lie in a rational function field and do Groebner basis computations on them? Maple, Reduce and Axiom support this. For example I would like to be able to compute the Groebner basis of the polynomials {v * x^2 + y, u* x * y + y^2} where the polynomials belong to the ring Q(u,v)[x,y]. I tried the following B. = PolynomialRing(QQ, 'u', 'v') R. = PolynomialRing(B, 'x', 'y') I = R.ideal(v * x^2 + y, u* x * y + y^2) g = I.groebner_basis() This fails with the error TypeError: Can only reduce polynomials over fields. asked Aug 26 '10 Sameer Agarwal 21 ● 2 ● 3 Kelvin Li 443 ● 10 ● 17
4 Make u and v be in the Fraction field: sage: B. = Frac(QQ['u,v']) sage: R. = PolynomialRing(B, 'x', 'y') sage: I = R.ideal(v * x^2 + y, u* x * y + y^2) sage: g = I.groebner_basis() sage: g [y^3 + u^2/v*y^2, x^2 + 1/v*y, x*y + 1/u*y^2] posted Aug 26 '10 William Stein 1369 ● 5 ● 19 ● 44 http://wstein.org/ Thanks, that works perfectly. Now my quest for the Elimination ideal I2 = I.elimination_ideal([x]) fails with TypeError: Cannot call Singular function 'eliminate' with ring parameter of type '' Sameer Agarwal (Aug 26 '10)
1 When I try this, the error I get is a little more informative: sage: I2 = I.elimination_ideal([x]) ... TypeError: Cannot call Singular function 'eliminate' with ring parameter of type '' The function eliminte comes from sage: eliminate = sage.libs.singular.ff.eliminate which is a wrapper for the Singular function and is supposed to convert Sage's rings to rings that Singular understands . . . however it seems that this wrapper does not understand the _polydict_domain rings. Perhaps one could convert R to a ring that the wrapper does understand (or file this as a bug in the wrapper and fix it). Checking, I see that there is a ring type MPolynomialRing_libsingular, but it only allows base rings from a very small list (like ZZ and finite fields). Maybe someone who knows more about the Singular interface can help here? posted Aug 26 '10 niles 3605 ● 7 ● 45 ● 101 http://nilesjohnson.net/
[hide preview] | 624 | 2,167 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2013-48 | longest | en | 0.778688 |
https://www.hackmath.net/en/math-problem/22463?tag_id=144 | 1,603,245,671,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00465.warc.gz | 752,860,310 | 15,400 | # Annulus from triangle
Calculate the content of the area bounded by a circle circumscribed and a circle inscribed by a triangle with sides a = 25mm, b = 29mm, c = 36mm
Correct result:
S = 831.0003 mm2
#### Solution:
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# Quadratic
-1
46
1
The equation y = -16t^2 - 60t + 54 describes the height (in feet) of a ball thrown downward at 60 feet per second from a height of 54 feet from the ground, where t represents time, measured in seconds. What is the maximum height of the ball? Express your answer as a decimal rounded to the nearest hundredth.
Jan 1, 2022
### 1+0 Answers
#1
+13015
+1
What is the maximum height of the ball?
Hello Guest!
Since the ball is thrown down, its maximum height is that of the starting point: 54.00 feet.
!
Jan 1, 2022 | 159 | 543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-05 | longest | en | 0.921413 |
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## 42 Listing Results Quizizz Quadratic Formula
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3 hours ago 30 seconds. Report an issue. Q. What is the first step to solving THIS equation by completing the square? a2 + 10a + 21 = 0. answer choices. Set the equation equal to zero. Divide 10 by 2 and add the result to both sides. Add a 2 and 10a together.
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5 hours ago Q. If the discriminant equals 0, then the quadratic has: Q. If the discriminant is negative, then the quadratic has: Q. If the discriminant is positive, then the quadratic has: Q. If the graph of a quadratic does not intercept the x axis at any point, then it has: Q.
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7 hours ago The quadratic formula is derived from a quadratic equation in standard form when solving for x by completing the square. The steps involve creating a perfect square trinomial, isolating the trinomial, and taking the square root of both sides. Solving Quadratics: Quadratic Formula Quiz - Quizizz. Preview this quiz on Quizizz. If the
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1 hours ago Quizizz.com. Each multiple choice question is worth 5 points. 1. Fill in the blank to make a perfect square trinomial (i.e. complete the square): m2 – 12m + _____ A. -144 B. 144 C. -36 D. 36 2. Fill in the blank to complete the next step in solving the quadratic equation using the complete the square method: c2 + 10c + 8 = 0
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3 hours ago Radical Equations and Inequalities Quiz - Quizizz Solving Radical Equations and Inequalities A radical equationcontains a variable within a radical. She began her career as a student teacher in middle school and has taught students from 7th to 12th grade. Solve your math problems using our free math solver with step-by-step solutions.
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8 hours ago The three main ways to solve quadratic equations are: to factor, to use the quadratic formula, or to complete the square. For the following problems, practice choosing the best method by solving for x in the quadratic equation. For example: 2x 2 - 3x - 5 = 0 By quadratic formula: 6. Preview. Download.
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8 hours ago 4.7 The Quadratic Formula 1. 4.7 THE QUADRATIC FORMULA 2. SOLVING QUADRATIC EQUATIONS Remember: when we solve quadratic equations, we are finding the values of the zeros (also called the x – intercepts and the roots) So far, we have learned to solve quadratic equations by factoring, graphing, and completing the square
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8 hours ago Quadratic equations by grouping method for real solutions for quick quadratic equation has no real roots made changes have a pantry stocked with. With art a few clicks, fast, completing the youth and using the quadratic formula. Begin by factoring the left side of the equation. Quizizz is free and we rely on users to spread the word. Later
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1 hours ago The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, which means it contains at least one term that is squared. Only one person can edit a quiz at a time. Your account has been deleted. Factoring to Solve Quadratic Equations. The Quizizz creator is not fully compatible with touch devices.
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3 hours ago Calculator Use. This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula. The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots.
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3 hours ago Quadratic functions together can be called a family, and this particular function the parent, because this is the most basic quadratic function (i.e., not transformed in any way).We can use this function to begin generalizing domains and ranges of quadratic functions. To determine the domain and range of any function on a graph, the general idea is to assume …
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2 hours ago Lesson 9-1 Chapter 9 5 Glencoe Algebra 1 Characteristics of Quadratic Functions Quadratic Function a function described by an equation of the form f(x) = ax 2 + bx + c, where a {≠ 0 Example: y which is the maximum. = 2x 2 O + 3 x + 8 The parent graph of the family of quadratic fuctions is y = x 2.
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1 hours ago Quizizz with quadratic formula. If there may want students. When does the ball reach the maximum height? In general form with flashcards because our chief goal here, practice links do you were created by hand, then experiment with a game or more. There are inside different types of quadratic equations, as these examples show.
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4 hours ago A: In this question, we need to find the number of equations and variables from the augmented matrix an question_answer Q: Diane determined that the \$25.50 raise she received in her last paycheck represented a 3% increase o
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8 hours ago slideshare.netImage: slideshare.netStudents learn the sum and product of roots formula, which states that if the roots of a quadratic equation are given, the quadratic equation can be written as 0 = x^2 – (sum of roots)x + (product of roots).For example, to write a quadratic equation that has the given roots –9 and 4, the first step is to find the sum and product of the roots.
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7 hours ago Solving Quadratic Equations By Completing The Square. Quadratic Equations using Quadratic Formula. Activity # 2. Activity # 3. Activity # 4. Activity # 5. Activity # 6. Activity # 7. Videos and links are sourced from YouTube and various websites and are meant to be used as supplemental material to the learning activities for the purpose of
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5 hours ago Download Free 41 Name Solving Quadratic And Other Equations 3 6 Solving Quadratic Equations 7. I can solve by factoring. 8. I can solve by taking the square root. 9. I can perform operations with imaginary numbers. 10. I can solve by completing the square. 11. I can solve equations using the quadratic formula (with rationalized denominators). 12.
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Just Now M-Review distribute and combine like terms then quizizz W - Double/triple distribute and then practice. Like radicals have the same root and radicand. 3) Complete the "Radicals" Worksheet. Example of the quadratic formula to solve an equation. Not only that, this app also gives you a step by step explanation on how to reach the answer!.
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8 hours ago Using the Quadratic Formula Date_____ Period____ Solve each equation with the quadratic formula. 1) v2 + 2v − 8 = 0 2) k2 + 5k − 6 = 0 3) 2v2 − 5v + 3 = 0 4) 2a2 − a − 13 = 2 5) 2n2 − n − 4 = 2 6) b2 − 4b − 14 = −2 7) 8n2 − 4n = 18 8) 8a2 + 6a = −5 9) 10 x2 + 9 = x 10) n2 = 9n − 20 11) 3a2 = 6a − 3 12) x2 = −3x + 40
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3 hours ago Mathway widget below is practice solving quadratic equations by factoring. Quizizz PRO for teachers! Solving a valid date of squares, tag standards to break this is one variable that can invite link copied to your changes have as a fun! Find the vertex of the quadratic equation. Understanding quadratic
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## Catalogs Updated
### How do you calculate quadratic formula?
The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by: For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root.
### Why should I know the quadratic formula?
The quadratic equation is used in the design of almost every product in stores today. The equation is used to determine how safe products are and the life expectancy of products, such as when they can expect to quit working. Designers can then see what needs to be changed in the product to make it last longer.
### How can I derive the quadratic formula?
Steps on How to Derive the Quadratic Formula Then simplify it further. Add the output of step #5 to both sides of the equation. Simplify the right side of the equation. Be careful when you add fractions with different denominators. Express the trinomial on the left side of the equation as the square of a binomial. Simplify and we are done! ...
### What careers use the quadratic formula?
There are a wide variety of jobs that use the quadratic equation. Actuaries, mathematicians, statisticians and computer engineers are a few of the directly related jobs that use the quadratic equation. Others include engineers, chemists, physicists and even nurses. | 3,505 | 14,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-05 | longest | en | 0.807812 |
https://responsivedesigntool.com/qa/quick-answer-how-much-does-2-liters-weigh-in-lbs.html | 1,618,579,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00083.warc.gz | 586,734,910 | 8,761 | # Quick Answer: How Much Does 2 Liters Weigh In Lbs?
## Is milk heavier than water?
A cubic meter of milk (1000 liters) weigh between 27 and 33 kg heavier than water.
That’s because milk is about 87% water and all the other substances, excluding fat, are heavier than water..
## How many Litres is 1 kg of milk?
l to kg conversion table:0.1 liter = 0.1 kg2.1 liters = 2.1 kg7 liters = 7 kg0.8 liter = 0.8 kg2.8 liters = 2.8 kg14 liters = 14 kg0.9 liter = 0.9 kg2.9 liters = 2.9 kg15 liters = 15 kg1 liter = 1 kg3 liters = 3 kg16 liters = 16 kg1.1 liter = 1.1 kg3.1 liters = 3.1 kg17 liters = 17 kg16 more rows
## How much does 2 liters of fluid weigh?
1 liter of water weighs 2.20 pounds, so we multiply that times 2 and get 4.40 pounds. The answer is 2 liters of water weighs 4.40 pounds.
## How many pounds is 1.75 liters?
Convert 1.75 Pounds to Liters1.75 Pounds (lbs)0.793787 Liters (L)1 lbs = 0.453592 L1 L = 2.204623 lbs
## What does 4 liters weigh?
1 litre of water weighs 1 kg or 2.2lbs. Hence, 4 litres weigh; (4/1)* (2.2) = 8.80 lbs.
## How many liters is 50 lbs?
22.68 LitersConvert 50 Pounds to Liters50 Pounds (lbs)22.68 Liters (L)1 lbs = 0.453592 L1 L = 2.204623 lbs
## How much does 4 liters of sand weigh?
The answer is: The change of 1 L ( liter ) volume unit of beach sand measure equals = to weight 3.37 lb ( pound ) as the equivalent measure within the same beach sand substance type.
## Is sand heavier than water?
A pound of sand and a pound of water weigh exactly the same. If you’re talking about density, rather than weight, an individual grain of sand is almost always more dense than water. You can tell this because sand sinks in water. … A pound of sand and a pound of water weigh exactly the same.
## How much does 2 Litres of water weigh in lbs?
Specifically and roughly speaking, the density of water is 1000kg/m3. In practical terms, this means that one liter of water weighs about one kilogram, so two liters weigh two kilograms or a little under 4.5 pounds.
## How much does a 2 Litre of milk weigh?
Imran, The density of milk is approximately 1.03 kilograms per litre so a litre of milk weighs very close to 1 kilogram.
## How much does 2 Litres of sand weigh?
One liter of beach sand converted to kilogram equals to 1.53 kg – kilo. How many kilograms of beach sand are in 1 liter? The answer is: The change of 1 L ( liter ) volume unit of beach sand measure equals = to weight 1.53 kg – kilo ( kilogram ) as the equivalent measure within the same beach sand substance type.
## How many liters is 40 lbs?
18.14 LitersConvert 40 Pounds to Liters40 Pounds (lbs)18.14 Liters (L)1 lbs = 0.453592 L1 L = 2.204623 lbs
## How much cups are in a pound?
3.6 cupsCups in a pound of all purpose flourPoundsCups (US)1 lb3.6 cups2 lbs7.2 cups3 lbs10.8 cups4 lbs14.4 cups
## How much does 1 liter of cement weigh?
3.32 lbOne liter of Portland cement converted to pound equals to 3.32 lb. How many pounds of Portland cement are in 1 liter? The answer is: The change of 1 L ( liter ) unit of Portland cement measure equals = to 3.32 lb ( pound ) as the equivalent measure for the same Portland cement type.
## How many pounds does 3 liters of water weigh?
3 liters of H2O at STP (Standard temperature and pressure) weigh 3 kilograms – approximately 6.6 pounds.
## How many liters is 25 lbs?
11.34 LitersConvert 25 Pounds to Liters25 Pounds (lbs)11.34 Liters (L)1 lbs = 0.453592 L1 L = 2.204623 lbs
## What does 4 liters of water weigh?
1 liter of water (l) = 2.20 pounds of water (lb wt.)
## What’s bigger 1kg or 1 Litre?
A kilogram is a unit of mass, a litre of volume. One litre of water has a mass of approximately one kilogram (depending on temperature, isotope mix etc), but one litre of molten lead has significantly more mass!!
## How many liters are in a pound?
0.45359237 liter1 pound (lb) = 0.45359237 liter (l). Pound (lb) is a unit of Weight used in Standard system. Liter (l) is a unit of Volume used in Metric system.
## How many lbs is 4 liters?
Conversion Tableliters to poundsllb24.409236.613948.818517 more rows
## How many liters is 10 pounds of water?
lbs to l conversion table:0.1 lbs = 0.0454 l2.1 lbs = 0.953 l7 lbs = 3.18 l0.4 lbs = 0.181 l2.4 lbs = 1.09 l10 lbs = 4.54 l0.5 lbs = 0.227 l2.5 lbs = 1.13 l11 lbs = 4.99 l0.6 lbs = 0.272 l2.6 lbs = 1.18 l12 lbs = 5.44 l0.7 lbs = 0.318 l2.7 lbs = 1.22 l13 lbs = 5.9 l13 more rows | 1,375 | 4,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-17 | latest | en | 0.917698 |
https://bettersolutions.com/excel/functions/isoweeknum-function.htm | 1,718,266,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00433.warc.gz | 114,031,227 | 4,617 | # ISOWEEKNUM
ISOWEEKNUM(date)
Returns the number of the ISO week of the year for a given date.
date The date.
REMARKS
* This function returns a number between 1 and 54.* The week number returned is according to the European ISO standard. All weeks start on a Monday and week number 1 is assigned to the first week that contains a Thursday.* If "date" is not a valid date, then #VALUE is returned.* If "date" is not a valid number, then #NUM is returned.* You can use the NOW function to return the serial number of the current system date and time.* You can use the TODAY function to return the serial number of the current system date.* You can use the WEEKNUM function to return the week number in the year for a given date.* This function was added in Excel 2013.* For the Microsoft documentation refer to support.microsoft.com* For the Google documentation refer to support.google.com
A 1 =TODAY() = Saturday, June 01, 2024 2 =NOW() = Saturday, June 01, 2024, 06:36 3 =WEEKNUM(A1, 21) = 22 4 =ISOWEEKNUM(A1) = 22 5 =ISOWEEKNUM(A1+7) = 23 6 =ISOWEEKNUM(A2+14) = 24
1 - What is the current system date. This cell has been formatted with the number format "dddd, mmmm dd, yyyy".2 - What is the current system date and time. This cell has been formatted with the number format "dddd, mmmm dd, yyyy, hh:mm". | 366 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.847928 |
https://communities.sas.com/t5/General-SAS-Programming/Many-to-Many-Merge-using-Hashing/td-p/112428?nobounce | 1,532,319,116,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00580.warc.gz | 633,818,785 | 41,118 | Contributor
Posts: 34
# Many to Many Merge using Hashing..?
How to perform many to many megrge using Hashing.....?
We need to pefomr merge on two tables A and B...
Table B needs to be loaded into memory....I need to implement three scenarios:
If A=1 and B=1 then
...........
.........
We can implement this using RC=0
If A=1 and B=0 then
.......
We can implement this using RC<>0
If A=0 and B=1 then
..............
What about this scenario...How to implement this scenario(Basically Right Join) using Hashing...
Please suggest..
Contributor
Posts: 34
## Re: Many to Many Merge using Hashing..?
Please find the existing data step too... I need to perfom this using Hashing...
Ideas are welcomed...
data matching;
merge abc(in=a) def (in = b);
by x;
if a and b then aa = 1;
if b and not a then bb = 2;
if a and not b then cc = 3;
run;
Contributor
Posts: 34
## Re: Many to Many Merge using Hashing..?
Any updates by any Experts..?
Respected Advisor
Posts: 3,167
## Re: Many to Many Merge using Hashing..?
@bhoopesh:
I am no expert, but Any updates on sample data sets, both input and output? Your question is too general to be answered efficiently, so please facilitate it a little by give people something to work on.
Haikuo
Contributor
Posts: 34
## Re: Many to Many Merge using Hashing..?
Hi Kuo,
Basically I need to implement Right Join using Hash....
I know Left join can be easily done by loading smaller table into memory and then math it one by one using RC....
But what about RIGHT JOIN...Is it possible using Hash...?
I think this is the most difficult question as of now...
PROC Star
Posts: 8,169
## Re: Many to Many Merge using Hashing..?
You've posted this a couple of times but, thus far, still have not described the problem sufficiently for at least some of us to understand. It would help if you provided to example datasets (provided as datasteps), the code you have tried, and a third dataset showing the results you want to obtain.
Why do you have to do this using a hash? How many records will exist in the real two datasets?
Contributor
Posts: 34
## Re: Many to Many Merge using Hashing..?
Hi Arthur and All,
I was not able to post the code, yesterday as i was busy... Sorry for inconvinence..
Please find the code as attached:
data load;
infile datalines dsd dlm='';
input numid:8. name: \$8.;
datalines;
22 "john"
11 "peter"
33 "terry"
44 "dony"
55 "randy"
66 "tony"
77 "mrna"
88 "joe"
99 "jack"
100 "sam"
;
run;
data main;
infile datalines dsd dlm='';
input numid:8. address: \$8.;
datalines;
12 "california"
22 "texas"
13 "indiana"
11 "nyc"
33 "xyz"
14 "def"
17 "dff"
18 "deee"
99 "jjjj"
19 "kkkk"
;
run;
proc sort data=load; by numid; quit;
proc sort data=main; by numid; quit;
data final;
merge
main(in=a)
load(in=b);
by numid;
if a and b then output;
if a and not b then output;
if b and not a then output;
run;
Merge works fine...but it takes lot of time...Thats why i need to implement Hashing....As i want to load table 'load' in memory and will read each record from Main table...
Main table is huge...So that why need to implement Hashing on it... Please suggest the Hashing code for above scenario...Thanks in advance....This task is very important to me...So pls help asap...
Respected Advisor
Posts: 3,167
## Re: Many to Many Merge using Hashing..?
I can't comment on your real life problem since you have not posted any sample data. However, as far as methodology goes, I don't see any fundamental difference in between two joins, left and right in term of hash processing. We need to know what you have, and what you are expecting to be the outcome, something tangible to work on.
Haikuo
Respected Advisor
Posts: 3,167
## Re: Many to Many Merge using Hashing..?
Hi,
Here is a Hash() approach that can mimic your outcome (except the order):
data final_hash;
if _n_=1 then do;
if 0 then do; set load; set main; end;
declare hash load(dataset:'load');
load.definekey('numid');
load.definedata(all:'y');
load.definedone();
declare hash main(dataset:'main');
main.definekey('numid');
main.definedata(all:'y');
main.definedone();
declare hiter hl('load');
declare hiter hm('main');
end;
rc=hl.first();
do rc=0 by 0 while (rc=0);
rc=main.find();
if rc ne 0 then call missing(address);
output;
rc=hl.next();
end;
rc=hm.first();
do rc=0 by 0 while (rc=0);
rc=load.find();
if rc ne 0 then do; call missing(name); output;end;
rc=hm.next();
end;
stop;
drop rc;
run;
Now some comments:
1. I don't know why you call it "many to many" merge, as your data shows otherwise. Your data addressed an "one to one" merge, since for any obs in one table, there is only ONE match at most from another table. Maybe you have "many to many" scenarios in your real data, then you need to lay out your rule of what to do if they happen. For now, I can only treat it as is: "one to one".
2. I have commented out part of your merge code, cause it does not make any difference in term of final outcome. In your code, you have already exhausted all of the merging possibilities without outputting to different target, so it might not reflecting your intention?
data final;
merge
main(in=a)
load(in=b);
by numid;
/*if a and b then output;
if a and not b then output;
if b and not a then output;*/
run;
3. Efficiency wise, I am not sure if Hash() holds significant advantage over 'data step merge', unless we are talking about millions of records with thousands of variables, while Hash() may save some time for direct access, but then the incoming table may have the risk for being too large to be accommodated by your computer RAM. In general, Hash() will be more efficient if comparing to SQL join when Cartesian product happens.
Haikuo
Contributor
Posts: 34
## Re: Many to Many Merge using Hashing..?
Hi Kuo,
I appreciate your work on creating this Hash code to me... Thanks very much...
But the problem is..
Actual Dataset Load is having few hundreds of rows and that can be loaded to memory(RAM) succesfully.
But other Dataset Main is having 30 millions of records and that cannot be loaded into the memory.
So i doubt that above code will work...Actual Main dataset was giving memory fill error to me...when i try to load into Hashbuckets using HASEXP:20...
Pls suggest
Respected Advisor
Posts: 3,167
## Re: Many to Many Merge using Hashing..?
Like I said, why not using Merge? Hash(), if forced to use to serve your purpose, may not have an edge comparing to "data step merge". Also, you need to make sure that you only process those variables you need, variable numbers do make a big difference when sorting and merging.
Haikuo
PROC Star
Posts: 8,169
## Re: Many to Many Merge using Hashing..?
Bhoopesh: A couple of comments. If your data are already sorted, then a hash or array approach probably won't save you much time. However, if you data have to be sorted with each run, then another approach probably will be quicker.
First, though, if you continue to use a merge approach, and want all of the matched and non-match records, why do you include the ina and inb statements and the three if then conditions at the end? Are there any other possibilities I'm not thinking of? I think you could get what you want/need with just:
data final;
merge main load;
by numid;
run;
And, if you do use a hash approach, I would think you only want to use it to build an index of the data in your load file. However, I'll leave that for our more hash inclined colleagues.
A similar array approach, that also doesn't require the data to be sorted, might be:
data final (drop=i loc check: numids: names;
array numids(&nrecs.);
array names(&nrecs.) \$8.;
array check(&nrecs.);
do i=1 to &nrecs.;
set load;
numids(i)=numid;
names(i)=name;
end;
do until (eofmain);
set main end=eofmain;
loc=whichn(numid,of numids(*));
if loc then do;
name=names(loc);
check(loc)=1;
end;
else call missing(name);
output;
end;
call missing(address);
do i=1 to &nrecs.;
if not check(i) then do;
name=names(i);
numid=numids(i);
output;
end;
end;
run;
However, given sorted data, it runs at about the same speed as a merge approach.
Contributor
Posts: 41
## Re: Many to Many Merge using Hashing..?
Dear Hai.Kuo,
could you please explain this part of your code?
if 0 then do;
set main;
set load;
end;
What's the meaning of it if conditional expression is always false (zero).
Thank you beforehand.
upd. I think I've got it. Is it necessary to define this datasets in order to tell SAS in compilation step about this datasets? So then it's possible to use dataset: tag, right?
Respected Advisor
Posts: 3,167
## Re: Many to Many Merge using Hashing..?
Posted in reply to ghastly_kitten
Hi,
To my understanding (which could be off), Before using rc=hiter.first() method (or any method that brings in Hash() data into PDV), Hash() requires there are already variables defined in PDV. So you need to load ONE record into PDV ahead of time.
Haikuo
Contributor
Posts: 41
## Re: Many to Many Merge using Hashing..?
Well, it's still seem to be a little more tricky than I could realize, but this part of code:
if 0 then do;
set ...;
set ...;
end;
does not load any variables to PDV.
However, at the moment of executing of data step, SAS perfoms "compilation" and during the "forward run" calculates the length of PDV and initializes some variables (like one you place in nobs= option).
As SAS does not check if some condition is always FALSE/TRUE, he prepare the structure of PDV in order to be ready to load something from datasets which were used in set statement (as well as for variables used in datastep code which weren't defined in input datasets).
In the nutshell, the PDV is really empty (except automatic variables), but you can use variable definitions. The key idea, which I still want to check, is that SAS also opens and locks (for r/w) that datasets, so you can use links to them in dataset: tag in declare hash statement without reading any data.
Anyway, that's a good idea to make an empty PDV, in case you need to have a foolproof from "accidential" ouptut. However, if the main reason is to read the first record again somewhere after hash definition (I have no example for that), I think the alternative could be the use of (firstobs=) option.
Thanks, anyway!
Discussion stats
• 15 replies
• 5898 views
• 3 likes
• 4 in conversation | 2,674 | 10,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-30 | latest | en | 0.886881 |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-68e52840-b25a-488c-a775-8f1d0bdf0669/chapter-10-review-exercises-page-485/4 | 1,718,220,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00420.warc.gz | 730,911,805 | 11,856 | ## Elementary Geometry for College Students (6th Edition)
a.)(2,-3) and (2,5) The mid point of the given line segment that joins the given points is($\frac{x1+x2}{2}$, $\frac{y1+y2}{2}$ ($\frac{2+2}{2}$, $\frac{-3+5}{2}$ (2,1) b.)(3,-2) and (-7,-2) The mid point of the given line segment that joins the given points is($\frac{x1+x2}{2}$, $\frac{y1+y2}{2}$ ($\frac{3-7}{2}$, $\frac{-2-2}{2}$ (-2,-2) c.)(-4,1) and (4,5) The mid point of the given line segment that joins the given points is($\frac{x1+x2}{2}$, $\frac{y1+y2}{2}$ ($\frac{-4+4}{2}$, $\frac{1+5}{2}$ (0,3) d.)(x-2,y-3) and (x+4,y+5) The mid point of the given line segment that joins the given points is($\frac{x1+x2}{2}$, $\frac{y1+y2}{2}$ ($\frac{x-2+x+4}{2}$, $\frac{y-3+y+5}{2}$ (x+1,y+1) | 319 | 756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.520248 |
https://cstheory.stackexchange.com/questions/25509/edit-distance-in-sublinear-space | 1,716,283,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00863.warc.gz | 153,764,422 | 40,773 | # Edit distance in sublinear space
What is the best known complexity for computing the exact edit distance between two strings of the same length using working space which is sublinear in the size of the input? I assume the input is stored in some read-only format. Is this a previously studied problem?
To make the question a little more specific, how about $\Theta(\sqrt{n})$ space where $n$ is the length of each input string.
Edit. Following the answer of David Eppstein, it seems a good question is simply if the edit distance can be found in polynomial time and $\Theta(\sqrt{n})$ space. Any lower bounds would also be interesting.
• Regarding the edit: I think you misunderstand something. David Eppstein’s answer shows the problem is solvable in NL, hence also in P. Aug 17, 2014 at 22:42
• ... Actually, the original Wagner–Fischer algorithm already does that. Aug 17, 2014 at 22:54
• I assume the edited version intended to ask for algorithms that were both sublinear space and polynomial time. Aug 17, 2014 at 23:03
• @DavidEppstein Yes, exactly. I have edited again for clarification.
– Simd
Aug 18, 2014 at 6:51
• BTW, assuming the standard pricing model of 1 per midmatch/delete/insert, then if the edit distance is l, then the path realizing the shortest path in the edit distance matric is going in distance at most l from the main diagonal, and then the edit distance be computed using O(l) space. Thus, with sqrt(n) space, you can compute the edit distance if it is small (i.e., smaller than sqrt(n)). It is only if it is large that this seems hard. Of course, in this case, arguably, you should care less. Aug 19, 2014 at 21:40
## 1 Answer
Just to get things going, rather than trying to close out this problem: there is an obvious nondeterministic algorithm using logarithmically many bits of space (search for a single path through the dynamic programming matrix) so by Savitch's theorem there is a deterministic algorithm with space $O(\log^2 n)$. Its time must be of the form $n^{O(\log n)}$, quasi-polynomial rather than exponential.
There are some space lower bounds for edit distance in http://arxiv.org/abs/1106.4412 but I don't think they match your version of the problem.
• How do you verify the path you have found is optimal?
– Simd
Aug 16, 2014 at 18:48
• Binary search or sequential search for the smallest distance for which a path can be found, i.e., nothing beyond the standard equivalence of decision and search problems. This doesn't affect the forms of either the space or time bound. Aug 16, 2014 at 20:00
• @David I think you are correct so I have deleted my answer. Aug 24, 2014 at 20:48
• Is it even computable in log space?
– Simd
Aug 26, 2014 at 8:52 | 684 | 2,704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-22 | latest | en | 0.935586 |
https://www.mrexcel.com/board/threads/list-without-blank-and-if-condition.1092262/ | 1,653,243,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00409.warc.gz | 1,069,735,842 | 16,399 | # List without blank and if condition
#### mba_110
##### New Member
Hi everyone,
I got some difficulty in sorting out the below formula, it is a complex one with array.
Code:
``=IFERROR(INDEX(Payroll!\$C\$9:\$C\$95,SMALL(IF(ISTEXT(Payroll!\$C\$9:\$C\$95),ROW(Payroll!\$C\$8:\$C\$94),""),ROW(Payroll!C9))),"")``
formula return with no value (The value i am calling is text)
Now what i am trying to do?
I am trying to make list without blanks its for my payroll sheet and i am trying to extract the full list without blank and with a condition that if salary amount cell which is Payroll!AB9 is not null then only that employees name should appear in list otherwise list serial should skip his name and go to next, same for all,
in above i manage to get only the code that make list without blank which is also not giving the value and in addition to that i want the add if(Payroll!AB9<>"",Payroll!C9,"") but first above code should work properly.
Any help here will be appreciated.
Thanks.
### Excel Facts
Author John Walkenbach was Mr Spreadsheet until his retirement in June 2019.
#### James006
##### Well-known Member
Hello,
To make your life easier, you should create a named range for your Payroll!C9:C95 area ... such as rng ... and have Array Formula :
Code:
``=IF(ROWS(\$2:2)<=COUNTA(rng),INDEX(rng,SMALL(IF(rng<>"",ROW(rng)-MIN(ROW(rng))+1),ROWS(\$2:2))),"")``
Hope this will help
#### Fluff
##### MrExcel MVP, Moderator
Another option, this is a normal formula, rather than an array formula
=IFERROR(INDEX(Payroll!\$C\$9:\$C\$95,AGGREGATE(15,6,ROW(Payroll!\$C\$9:\$C\$95)-ROW(Payroll!\$C\$9)+1/(Payroll!\$C\$9:\$C\$95<>""),ROWS(\$1:1))),"")
Last edited:
#### mba_110
##### New Member
Thanks fluff your formula is working perfect, however as i said i want to add the if(Payroll!AB9<>"",Payroll!C9,"") if employee's salary is zero on payroll!AB9 starting from first row then his name should not appear in list and go to next name i try to amend your formula like below but its not showing correct result, still showing the blank cells if payroll!AB9 is zero or null.
Code:
``=IFERROR(IF(Payroll!AB9<>"",INDEX(Payroll!\$C\$9:\$C\$95,AGGREGATE(15,6,ROW(Payroll!\$C\$9:\$C\$95)-ROW(Payroll!\$C\$9)+1/(Payroll!\$C\$9:\$C\$95<>""),ROWS(\$1:1))),""),"")``
Where i am making mistake ?
Last edited:
#### Fluff
##### MrExcel MVP, Moderator
=IFERROR(INDEX(Payroll!\$C\$9:\$C\$95,AGGREGATE(15,6,ROW(Payroll!\$C\$9:\$C\$95)-ROW(Payroll!\$C\$9)+1/((Payroll!\$C\$9:\$C\$95<>"")*(Payroll!\$AB\$9:\$AB\$95<>0)),ROWS(\$1:1))),"")
#### mba_110
##### New Member
That is great again....thumps up genius
Now i am taking it to final level, i got some employees to whom i am not paying directly and i have another list in summary showing there names as in payroll sheet and i want to exclude them from this list as well whether there is an amount on payroll!AB9 or not.
the rage for excluding employee is in sheet Summary!B24:B30
I hope this can be possible.
#### Fluff
##### MrExcel MVP, Moderator
Whilst it's almost certainly possible, that's beyond my knowledge.
Hopefully one of the Formula folk step in & help.
Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle.
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Go back | 1,218 | 4,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.904593 |
https://github.com/sympy/sympy/issues/10290 | 1,547,730,896,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658981.19/warc/CC-MAIN-20190117123059-20190117145059-00348.warc.gz | 540,371,477 | 19,388 | # Computing series where the free variable is not just a symbol is broken #10290
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### MooVI commented Dec 20, 2015
```In [1]: sin(x).series(x) Out[1]: x - x**3/6 + x**5/120 + O(x**6) In [2]: sin(sin(x)).series(sin(x)) Out[2]: sin(x) - sin(x)**3/6 + sin(x)**5/120 + O(x**6) #Correct, but odd notation In [3]: sin(exp(x)).series(exp(x)) Out[3]: exp(5*x)/120 - exp(3*x)/6 + exp(x) #Wrong, and also missing big O In [4]: sin(cos(x)).series(cos(x)) TypeError: zip argument #2 must support iteration``` Arguably series should just complain and quit if it's not given an symbol for the free variable. This odd behaviour is related to what `Order` does under `subs`: ```In [5]: Order(x).subs(x, exp(x)) Out[5]: 0 In [6]: Order(x).subs(x, cos(x)) TypeError: zip argument #2 must support iteration``` The error is occurring because `solveset` cannot solve an equation Order gives it in `.eval_subs`, and there is no code handling no solution.
### skirpichev added a commit to diofant/diofant that referenced this issue Dec 20, 2015
``` Add a quick exit for Expr.series if x is not a Symbol ```
`Fixes sympy/sympy#10290`
``` f53c9b3 ```
### skirpichev added a commit to diofant/diofant that referenced this issue Dec 20, 2015
``` Add a quick exit for Expr.series if x is not a Symbol ```
`Fixes sympy/sympy#10290`
``` 840f7f7 ```
### skirpichev added a commit to diofant/diofant that referenced this issue Dec 20, 2015
``` Add a quick exit for Expr.series if x is not a Symbol ```
`Fixes sympy/sympy#10290`
``` 9e835bd ``` | 500 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-04 | latest | en | 0.798563 |
http://www.brightstorm.com/tag/scalar-multiple-vector/ | 1,432,343,739,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207926964.7/warc/CC-MAIN-20150521113206-00071-ip-10-180-206-219.ec2.internal.warc.gz | 343,072,636 | 11,344 | • #### The Vector Equation of a Line - Concept
##### Math›Precalculus›Vectors and Parametric Equations
How to write the vector equation of a line.
• #### Vector Operations in 3D - Problem 2
##### Math›Precalculus›Vectors and Parametric Equations
How to find the angle between two vectors in three dimensions.
• #### Vector Operations in 3D - Problem 1
##### Math›Precalculus›Vectors and Parametric Equations
How to find the angle between two vectors in 3D, and how to tell whether two vectors are parallel or perpendicular.
• #### The Angle Between Vectors - Problem 2
##### Math›Precalculus›Vectors and Parametric Equations
How to determine whether two vectors are perpendicular or parallel.
• #### Perpendicular, Parallel and Skew Lines in Space - Problem 1
##### Math›Precalculus›Vectors and Parametric Equations
How to find the equation of a line parallel to and perpendicular to a given line in space.
• #### Perpendicular, Parallel and Skew Lines in Space - Concept
##### Math›Precalculus›Vectors and Parametric Equations
How to determine whether two lines in space are parallel or perpendicular.
• #### The Angle Between Planes - Problem 1
##### Math›Precalculus›Vectors and Parametric Equations
How to determine if two planes are parallel or perpendicular.
• #### The Angle Between Planes - Concept
##### Math›Precalculus›Vectors and Parametric Equations
How to find the angle between planes, and how to determine if two planes are parallel or perpendicular. | 337 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2015-22 | longest | en | 0.680882 |
https://netlib.org/lapack/explore-html/d1/d92/dorml2_8f_source.html | 1,696,423,006,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00370.warc.gz | 469,233,510 | 8,185 | LAPACK 3.11.0 LAPACK: Linear Algebra PACKage
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dorml2.f
Go to the documentation of this file.
1*> \brief \b DORML2 multiplies a general matrix by the orthogonal matrix from a LQ factorization determined by sgelqf (unblocked algorithm).
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8*> \htmlonly
10*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/dorml2.f">
11*> [TGZ]</a>
12*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/dorml2.f">
13*> [ZIP]</a>
14*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/dorml2.f">
15*> [TXT]</a>
16*> \endhtmlonly
17*
18* Definition:
19* ===========
20*
21* SUBROUTINE DORML2( SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC,
22* WORK, INFO )
23*
24* .. Scalar Arguments ..
25* CHARACTER SIDE, TRANS
26* INTEGER INFO, K, LDA, LDC, M, N
27* ..
28* .. Array Arguments ..
29* DOUBLE PRECISION A( LDA, * ), C( LDC, * ), TAU( * ), WORK( * )
30* ..
31*
32*
33*> \par Purpose:
34* =============
35*>
36*> \verbatim
37*>
38*> DORML2 overwrites the general real m by n matrix C with
39*>
40*> Q * C if SIDE = 'L' and TRANS = 'N', or
41*>
42*> Q**T* C if SIDE = 'L' and TRANS = 'T', or
43*>
44*> C * Q if SIDE = 'R' and TRANS = 'N', or
45*>
46*> C * Q**T if SIDE = 'R' and TRANS = 'T',
47*>
48*> where Q is a real orthogonal matrix defined as the product of k
49*> elementary reflectors
50*>
51*> Q = H(k) . . . H(2) H(1)
52*>
53*> as returned by DGELQF. Q is of order m if SIDE = 'L' and of order n
54*> if SIDE = 'R'.
55*> \endverbatim
56*
57* Arguments:
58* ==========
59*
60*> \param[in] SIDE
61*> \verbatim
62*> SIDE is CHARACTER*1
63*> = 'L': apply Q or Q**T from the Left
64*> = 'R': apply Q or Q**T from the Right
65*> \endverbatim
66*>
67*> \param[in] TRANS
68*> \verbatim
69*> TRANS is CHARACTER*1
70*> = 'N': apply Q (No transpose)
71*> = 'T': apply Q**T (Transpose)
72*> \endverbatim
73*>
74*> \param[in] M
75*> \verbatim
76*> M is INTEGER
77*> The number of rows of the matrix C. M >= 0.
78*> \endverbatim
79*>
80*> \param[in] N
81*> \verbatim
82*> N is INTEGER
83*> The number of columns of the matrix C. N >= 0.
84*> \endverbatim
85*>
86*> \param[in] K
87*> \verbatim
88*> K is INTEGER
89*> The number of elementary reflectors whose product defines
90*> the matrix Q.
91*> If SIDE = 'L', M >= K >= 0;
92*> if SIDE = 'R', N >= K >= 0.
93*> \endverbatim
94*>
95*> \param[in] A
96*> \verbatim
97*> A is DOUBLE PRECISION array, dimension
98*> (LDA,M) if SIDE = 'L',
99*> (LDA,N) if SIDE = 'R'
100*> The i-th row must contain the vector which defines the
101*> elementary reflector H(i), for i = 1,2,...,k, as returned by
102*> DGELQF in the first k rows of its array argument A.
103*> A is modified by the routine but restored on exit.
104*> \endverbatim
105*>
106*> \param[in] LDA
107*> \verbatim
108*> LDA is INTEGER
109*> The leading dimension of the array A. LDA >= max(1,K).
110*> \endverbatim
111*>
112*> \param[in] TAU
113*> \verbatim
114*> TAU is DOUBLE PRECISION array, dimension (K)
115*> TAU(i) must contain the scalar factor of the elementary
116*> reflector H(i), as returned by DGELQF.
117*> \endverbatim
118*>
119*> \param[in,out] C
120*> \verbatim
121*> C is DOUBLE PRECISION array, dimension (LDC,N)
122*> On entry, the m by n matrix C.
123*> On exit, C is overwritten by Q*C or Q**T*C or C*Q**T or C*Q.
124*> \endverbatim
125*>
126*> \param[in] LDC
127*> \verbatim
128*> LDC is INTEGER
129*> The leading dimension of the array C. LDC >= max(1,M).
130*> \endverbatim
131*>
132*> \param[out] WORK
133*> \verbatim
134*> WORK is DOUBLE PRECISION array, dimension
135*> (N) if SIDE = 'L',
136*> (M) if SIDE = 'R'
137*> \endverbatim
138*>
139*> \param[out] INFO
140*> \verbatim
141*> INFO is INTEGER
142*> = 0: successful exit
143*> < 0: if INFO = -i, the i-th argument had an illegal value
144*> \endverbatim
145*
146* Authors:
147* ========
148*
149*> \author Univ. of Tennessee
150*> \author Univ. of California Berkeley
151*> \author Univ. of Colorado Denver
152*> \author NAG Ltd.
153*
154*> \ingroup doubleOTHERcomputational
155*
156* =====================================================================
157 SUBROUTINE dorml2( SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC,
158 \$ WORK, INFO )
159*
160* -- LAPACK computational routine --
161* -- LAPACK is a software package provided by Univ. of Tennessee, --
162* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
163*
164* .. Scalar Arguments ..
165 CHARACTER SIDE, TRANS
166 INTEGER INFO, K, LDA, LDC, M, N
167* ..
168* .. Array Arguments ..
169 DOUBLE PRECISION A( LDA, * ), C( LDC, * ), TAU( * ), WORK( * )
170* ..
171*
172* =====================================================================
173*
174* .. Parameters ..
175 DOUBLE PRECISION ONE
176 parameter( one = 1.0d+0 )
177* ..
178* .. Local Scalars ..
179 LOGICAL LEFT, NOTRAN
180 INTEGER I, I1, I2, I3, IC, JC, MI, NI, NQ
181 DOUBLE PRECISION AII
182* ..
183* .. External Functions ..
184 LOGICAL LSAME
185 EXTERNAL lsame
186* ..
187* .. External Subroutines ..
188 EXTERNAL dlarf, xerbla
189* ..
190* .. Intrinsic Functions ..
191 INTRINSIC max
192* ..
193* .. Executable Statements ..
194*
195* Test the input arguments
196*
197 info = 0
198 left = lsame( side, 'L' )
199 notran = lsame( trans, 'N' )
200*
201* NQ is the order of Q
202*
203 IF( left ) THEN
204 nq = m
205 ELSE
206 nq = n
207 END IF
208 IF( .NOT.left .AND. .NOT.lsame( side, 'R' ) ) THEN
209 info = -1
210 ELSE IF( .NOT.notran .AND. .NOT.lsame( trans, 'T' ) ) THEN
211 info = -2
212 ELSE IF( m.LT.0 ) THEN
213 info = -3
214 ELSE IF( n.LT.0 ) THEN
215 info = -4
216 ELSE IF( k.LT.0 .OR. k.GT.nq ) THEN
217 info = -5
218 ELSE IF( lda.LT.max( 1, k ) ) THEN
219 info = -7
220 ELSE IF( ldc.LT.max( 1, m ) ) THEN
221 info = -10
222 END IF
223 IF( info.NE.0 ) THEN
224 CALL xerbla( 'DORML2', -info )
225 RETURN
226 END IF
227*
228* Quick return if possible
229*
230 IF( m.EQ.0 .OR. n.EQ.0 .OR. k.EQ.0 )
231 \$ RETURN
232*
233 IF( ( left .AND. notran ) .OR. ( .NOT.left .AND. .NOT.notran ) )
234 \$ THEN
235 i1 = 1
236 i2 = k
237 i3 = 1
238 ELSE
239 i1 = k
240 i2 = 1
241 i3 = -1
242 END IF
243*
244 IF( left ) THEN
245 ni = n
246 jc = 1
247 ELSE
248 mi = m
249 ic = 1
250 END IF
251*
252 DO 10 i = i1, i2, i3
253 IF( left ) THEN
254*
255* H(i) is applied to C(i:m,1:n)
256*
257 mi = m - i + 1
258 ic = i
259 ELSE
260*
261* H(i) is applied to C(1:m,i:n)
262*
263 ni = n - i + 1
264 jc = i
265 END IF
266*
267* Apply H(i)
268*
269 aii = a( i, i )
270 a( i, i ) = one
271 CALL dlarf( side, mi, ni, a( i, i ), lda, tau( i ),
272 \$ c( ic, jc ), ldc, work )
273 a( i, i ) = aii
274 10 CONTINUE
275 RETURN
276*
277* End of DORML2
278*
279 END
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:60
subroutine dlarf(SIDE, M, N, V, INCV, TAU, C, LDC, WORK)
DLARF applies an elementary reflector to a general rectangular matrix.
Definition: dlarf.f:124
subroutine dorml2(SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC, WORK, INFO)
DORML2 multiplies a general matrix by the orthogonal matrix from a LQ factorization determined by sge...
Definition: dorml2.f:159 | 2,530 | 7,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-40 | longest | en | 0.436199 |
https://www.ginac.de/ginac.git/?p=ginac.git;a=blobdiff_plain;f=doc/tutorial/ginac.texi;h=f69d45107968ec0d2943d05179d43817aeb61efc;hp=f044d2c944ae00ceb6f5f1cdf1d146e3379a90d5;hb=a6bb52b00bf185271774e7d56215923700a3ec40;hpb=296ca653543d7b9ab1af2c4bbaa29f0783cee46e | 1,632,740,262,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00330.warc.gz | 797,068,906 | 2,281 | X-Git-Url: https://www.ginac.de/ginac.git//ginac.git?p=ginac.git;a=blobdiff_plain;f=doc%2Ftutorial%2Fginac.texi;h=f69d45107968ec0d2943d05179d43817aeb61efc;hp=f044d2c944ae00ceb6f5f1cdf1d146e3379a90d5;hb=a6bb52b00bf185271774e7d56215923700a3ec40;hpb=296ca653543d7b9ab1af2c4bbaa29f0783cee46e diff --git a/doc/tutorial/ginac.texi b/doc/tutorial/ginac.texi index f044d2c9..f69d4510 100644 --- a/doc/tutorial/ginac.texi +++ b/doc/tutorial/ginac.texi @@ -1105,8 +1105,16 @@ gamma(x+1/2) -> gamma(x+1/2) gamma(15/2) -> (135135/128)*Pi^(1/2) @end example -Most of these functions can be differentiated, series expanded and so -on. Read the next chapter in order to learn more about this. +@cindex branch cut +For functions that have a branch cut in the complex plane GiNaC follows +the conventions for C++ as defined in the ANSI standard. In particular: +the natural logarithm (@code{log}) and the square root (@code{sqrt}) +both have their branch cuts running along the negative real axis where +the points on the axis itself belong to the upper part. + +Besides evaluation most of these functions allow differentiation, series +expansion and so on. Read the next chapter in order to learn more about +this. @node Relations, Important Algorithms, Built-in functions, Basic Concepts @@ -1425,10 +1433,14 @@ When you run it, it produces the sequence @code{1}, @code{-1}, @code{5}, @cindex Laurent expansion Expressions know how to expand themselves as a Taylor series or (more -generally) a Laurent series. Similar to most conventional Computer -Algebra Systems, no distinction is made between those two. There is a -class of its own for storing such series as well as a class for storing -the order of the series. A sample program could read: +generally) a Laurent series. As in most conventional Computer Algebra +Systems, no distinction is made between those two. There is a class of +its own for storing such series as well as a class for storing the order +of the series. As a consequence, if you want to work with series, +i.e. multiply two series, you need to call the method @code{ex::series} +again to convert it to a series object with the usual structure +(expansion plus order term). A sample application from special +relativity could read: @example #include @@ -1436,25 +1448,28 @@ using namespace GiNaC; int main() @{ - symbol x("x"); - numeric point(0); - ex MyExpr1 = sin(x); - ex MyExpr2 = 1/(x - pow(x, 2) - pow(x, 3)); - ex MyTailor, MySeries; + symbol v("v"), c("c"); + + ex gamma = 1/sqrt(1 - pow(v/c,2)); + ex mass_nonrel = gamma.series(v, 0, 10); + + cout << "the relativistic mass increase with v is " << endl + << mass_nonrel << endl; + + cout << "the inverse square of this series is " << endl + << pow(mass_nonrel,-2).series(v, 0, 10) << endl; - MyTailor = MyExpr1.series(x, point, 5); - cout << MyExpr1 << " == " << MyTailor - << " for small " << x << endl; - MySeries = MyExpr2.series(x, point, 7); - cout << MyExpr2 << " == " << MySeries - << " for small " << x << endl; // ... @} @end example +Only calling the series method makes the last output simplify to +@math{1-v^2/c^2+O(v^10)}, without that call we would just have a long +series raised to the power @math{-2}. + @cindex M@'echain's formula -As an instructive application, let us calculate the numerical value of -Archimedes' constant +As another instructive application, let us calculate the numerical +value of Archimedes' constant @tex $\pi$ @end tex | 1,009 | 3,432 | {"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} | 3.234375 | 3 | CC-MAIN-2021-39 | latest | en | 0.833951 |
https://stage.geogebra.org/m/xeem2pfm | 1,712,921,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00283.warc.gz | 480,653,269 | 28,984 | GeoGebra Classroom
# Ulug Beg madrassa - spandrel - gp
Topic:
Geometry
Ulug Beg, the grandson of Timur Lenk, buit the Ulug Beg madrassa in 1417-1420. together with theTillya Kari madrassa and the Shir Dor madrassa it forms the impressive historical architectonic whole of the Registan square in Samarkand. The iwan of the main measures 35 m, and is decorated with a geometric pattern and an inscription in kufic schript: "Thise magnificent façade is of such a height it is twice the heavens and of such weight that the spine of the earth is about to crumble." The spandrel is decorated with a geometric pattern of remarkable 16-pointed stars. Mostly the greater stars go together with stars with half as much stars, often connected to the principle star by a wreath of polygons (often pentagons). This decoration shows 8-pointed stars, but the transistion is created in two steps.
• The 16-pointed stars are surrounded by a wreath of irregular hexagons.
• The 8-pointed stars are surrounded by a wreath of smaller irregular hexagons.
• Both wreaths define around the 16- and 8-pointed stars a wreath of 5-pointed stars.
In the tiling these transitions are realised by irregular pentagons of different shape, depending whather they link two hexadecagons or a hexadecagon and an octagon.
Zooming in unto the pattern you cannotice that the design doesn't stop with the creation of a line pattern. A wide range of decorating filling remains. Most remarkable is the colourful filling of the dark 16-pointed stars and the relief given to the white irregular hexagons. Expoiting relief in decoration is also used on the side walls of the entrance. White ties, realised in relief surround flat 10-pointed stars that are colourfully painted. | 416 | 1,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.933162 |
http://mathhelpforum.com/algebra/46864-algebra-2-please-help-fast.html | 1,511,114,668,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00287.warc.gz | 191,203,047 | 11,477 | Hey... Any one know what roster notation is? The notes i was given had a example of {-5,-4,-3,-2,-1,0...6} but... is that just listing all the numbers on a number line? or is it listing the bold dots on the number line?
also on set-builder notation how do i change a set builder notation into a interval notation? like how do i know witch side would have a ( or a [ for the equation? The problem for this is... {x|x > 10} The > has a line under it. my answer i put was (infinity (positive) , 10] ... Did i do this right? Thankz for the help XD ...
Order the given numbers from least to greatest. Then classify each number by the subsets of the real numbers to which it belongs. 2/3 , 6.17, sqrt28, -3 1/8, -4.9
i just don't really under stand the second part of the question "Then classify each number by the subsets of the real numbers to which it belongs."
2. Originally Posted by shawns15
Hey... Any one know what roster notation is?
Symbollically, we use two common methods to write sets. The roster notation is a complete or implied listing of all the elements of the set. So and are examples of roster notation defining sets with 4 and 20 elements respectively. The ellipsis, '', is used to mean you fill in the missing elements in the obvious manner or pattern, as there are too many to actually list out on paper.
The set-builder notation is used when the roster method is cumbersome or impossible. The set B above could be described by . The vertical bar, |'', is read as such that'' so this notation is read aloud as the set of x such that x is between 2 and 40 (inclusive) and x is even.'' (Sometimes a colon is used instead of |.) In set-builder notation, whatever comes after the bar describes the rule for determining whether or not an object is in the set. For the set the roster notation would be impossible since there are too many reals to actually list out, explicitly or implicitly.
The notes i was given had a example of {-5,-4,-3,-2,-1,0...6} but... is that just listing all the numbers on a number line? or is it listing the bold dots on the number line?
also on set-builder notation how do i change a set builder notation into a interval notation? like how do i know witch side would have a ( or a [ for the equation? The problem for this is... {x|x > 10} The > has a line under it. my answer i put was (infinity (positive) , 10] ... Did i do this right?
Not exactly. Interval notation for that would be $(10, +\infty)$
Thankz for the help XD ...
Order the given numbers from least to greatest. Then classify each number by the subsets of the real numbers to which it belongs. 2/3 , 6.17, sqrt28, -3 1/8, -4.9
Can you order these numbers from least to greatest. Not so hard!
i just don't really under stand the second part of the question "Then classify each number by the subsets of the real numbers to which it belongs."
Which of the following sets of numbers does each of those numbers belong? Here are your choices:
Real Numbers
Imaginary Numbers
Rational Numbers
Irrational Numbers
Integers (Pos/Neg Whole numbers and 0)
Natural Numbers
You may even have sets called Positive Whole Numbers and Negative Whole Numbers.
For example, 2/3 would belong to sets Real and Rational.
3. Originally Posted by masters | 786 | 3,250 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.958239 |
http://www.jiskha.com/display.cgi?id=1178113432 | 1,493,292,930,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122159.33/warc/CC-MAIN-20170423031202-00445-ip-10-145-167-34.ec2.internal.warc.gz | 568,367,550 | 3,438 | # Physics HELP!!!!!!!!
posted by on .
The fundamental frequency of a string fixed at both ends is 325 Hz. How long does it take for a wave to travel the length of this string?
The time for one cycle is 1/325 sec. The time to go to one end (1/2 wavelength away) is one half that time.
yes it is | 76 | 297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-17 | latest | en | 0.948748 |
https://www.excelhow.net/calculate-years-between-dates-in-ms-excel.html | 1,713,238,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00052.warc.gz | 712,249,950 | 23,180 | # Calculate Years Between Dates In Ms Excel
If you are an avid Ms Excel user, then you might have come across a task in which you needed to calculate the years between the dates; you might take it easy and do this task manually, which is also feasible but only for calculating the years between 3 to 4 dates. But when it comes to hundreds of dates, it becomes a cumbersome and time taking task to calculate the years between them manually.
But don’t worry since this post will teach you the simplest technique to calculate years between dates in Microsoft Excel. All you have to do is thoroughly read the article.
So without any further ado, let’s dive into it
## General Formula
`=YEARFRAC(start_date,end_date)`
### Explanations Of Syntax:
You must understand the syntax used in this formula in order to apply it to complete your task.
• Parenthesis (): The primary function of this symbol is to group the elements.
• The Comma Symbol (,): This acts as a separator.
• YEARFRAC: It computes the year’s percentage represented by the number of entire days between two dates (start date and end date).
## Summary
The YEARFRAC function, which returns a decimal value indicating the fraction of a year between two dates, may be used to compute the number of years between two dates. The formula in D6 in the example is:
`=YEARFRAC(A2, B2)`
### Explanation
YEARFRAC provides a decimal value that represents the fractional years between two dates. As an example:
`=YEARFRAC("January 1, 2020","January 1, 2022") / yields 2`
Here are a few instances of YEARFRAC’s calculated results:
Start Date End Date YEARFRAC result 2002/1/10 2022/6/1 20.39167 2003/3/20 2019/4/20 16.08333 1999/1/20 2019/7/10 20.47222
The formula in C2 in the example is: `=YEARFRAC(A2, B2)` / returns 20 Rounding outcome
You can round the number after you obtain the decimal value. For example, you might use the ROUND function to round to the nearest whole number:
`=ROUND(YEARFRAC(A2,B2),0)`
Only Whole Years
You may also wish to maintain only the integer component of the result, with no fractional values, to ensure that you only count entire years. In such instance, merely use the INT function to enclose YEARFRAC:
`=INT(YEARFRAC(A2,B2))`
Example #2
Assuming you already have the data established in the excel sheets, you do not need to repeat the first step. For those doing it for the very first time, open Excel on your laptop, create a new or empty excel sheet, and record your chosen date values as shown below.
We’ll use the YEARFRAC function to calculate the years between the dates on the excel mentioned above. This is the function that computes the number of years between two dates. The YEARFRAC function has the ability to compute in decimal; for example, it can display 1.5 years to imply one and a half years.
Years are where we anticipate recording our outcomes in the column with the title. To compute the years between the dates in cells A2 and B2,
Enter the formula =YEARFRAC (A2, B2) in the formula bar.
Apply the same procedure to the remaining cells, or hold and drag the cursor to compute the remaining years automatically.
Enter this formula in cell D2 and drag the fill handle to get the additional computations.
To round the decimal or fractional value to the closest whole number, use the ROUND function, such as =ROUND(YEARFRAC(A2, B2),0)
the above example shows how to round a fractional value to the nearest whole number.
If we want to return the integer component of the year’s value and not the fractional part, wrap the YEARFRAC function inside the INT function, as in =INT(YEARFRAC(B2, C2)). Figure 4: Obtaining the Year Value Absent Fractional Value
Note: There is an optional third input to the YEARFRAC function that determines how days are tallied when computing fractional years. The default approach is to calculate days between two dates using a 360-day year, with each month having 30 days.
### Related Functions
• Excel INT function
The Excel INT function returns the integer portion of a given number. And it will rounds a given number down to the nearest integer.The syntax of the INT function is as below:= INT (number)…
• Excel Round function
The Excel INT function rounds a number to a specified number of digits. You can use the ROUND function to round to the left or right of the decimal point in Excel.The syntax of the ROUND function is as below:=ROUND (number, num_digits)… | 1,036 | 4,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | longest | en | 0.875028 |
https://community.airtable.com/t5/formulas/viewing-financials-over-time/td-p/182854 | 1,716,370,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00087.warc.gz | 146,968,071 | 50,758 | # Viewing financials over time
Topic Labels: Data Extensions Formulas
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4 - Data Explorer
Hi,
I'm using AirTable to track potential sales opportunities and would like to visualise on a chart the revenue per month. This is similar to the Gantt chart view but instead would show the size of revenue rather than simply a project.
For example, if an opportunity lands on the 15th April that will last 2 months, completing on the 15th June and worth £100k, this would be spread over those months. I.e. £50k in April, £100k in May, and £50k in June. Then a bar chart would total all these by month.
In Excel this is relatively easy - you create a column for each month that calculates how much of the revenue in each month (can just about do that here but it's not easy). Then you create a bar chart for the table and it shows it easily. Can't find a way to do that in AirTable or with an app.
Any and all help is much appreciated!!
Dan
3 Replies 3
18 - Pluto
Yeah, you're going to need to create a new table just for that chart I believe
The variable amount based on the number of days makes this pretty tricky, and it seems like you'd need a table just for that kind of calculations I think. So for your example of 15 Apr - 15 June, you'd need a table that had 3 records, one for each month, and you'd probably need a script to help you figure out the allocations if you want this done automated
You'd then need to create a table just for this chart where each record in this table would be a single month-year, e.g. "Apr 2024", "May 2024", and you'd link these appropriately to the records in the previous table and have a rollup to sum up the values per month, and you'd create the chart for this table
4 - Data Explorer
Thank you for this. That makes sense.
I believe I can create columns that calculate the amount of revenue for each month. Does that need to be in a separate table?
For the separate table for the chart, each line needs to be a month year. Is there a way to full automate this (like in Excel)? As a nice-to-have, ideally I'd like a stacked chart which then shows 'in progress', 'secured', 'likely'; is that feasible?
Thank you again!
18 - Pluto
re: I believe I can create columns that calculate the amount of revenue for each month. Does that need to be in a separate table?
As long as you're able to populate the chart data table (i.e. the one where each record is a month-year) with the appropriate data you're good to go, really. I couldn't think of how to do that dynamically and so defaulted to having a separate table for it; I'd love to know how you do it!
---
re: For the separate table for the chart, each line needs to be a month year. Is there a way to full automate this (like in Excel)?
Hm, like pre-create all of the month year combos for you? Might need a script for that really. Depending on how you're populating the chart data table your automations might handle this for you
re: As a nice-to-have, ideally I'd like a stacked chart which then shows 'in progress', 'secured', 'likely'; is that feasible?
Yeah, but you'd need to set up your data so that your chart data table was able to distinguish which amounts are what status, does that make sense? | 781 | 3,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.933724 |
https://www.studyadda.com/sample-papers/mathematics-sample-paper-3_q18/414/320085 | 1,597,487,842,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00229.warc.gz | 855,391,681 | 18,085 | • # question_answer Radius of a cylinder is r and the height is h. Find the change in the volume if the (a) height is doubled (b) height is doubled and the radius is halved (c) height remains same and the radius is halved.
Volume of cylinder $=\pi {{r}^{2}}h$ (a) Height is doubled i.e., h' =2h Volume of cylinder $=\pi {{r}^{2}}h'$ $=\pi {{r}^{2}}(2h)$ $=2\pi {{r}^{2}}h$ (Double of the original) (b) $h'=2h$and$r'=\frac{r}{2}$ Then volume of cylinder $=\pi {{r}^{2}}h$ $=\pi {{\left( \frac{r}{2} \right)}^{2}}\times 2h$ $=\pi \,\times \frac{{{r}^{2}}}{4}\times 2h$ $=\frac{1}{2}\pi {{r}^{2}}h$ (Half of the original) (c) $r'=\frac{r}{2}$ unit Volume of cylinder $=\pi r{{'}^{2}}h$ $=\pi \,{{\left( \frac{r}{2} \right)}^{2}}h$ $=\frac{1}{4}\pi {{r}^{2}}h$cubic unit (One fourth of the original) | 318 | 816 | {"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} | 4.1875 | 4 | CC-MAIN-2020-34 | latest | en | 0.641242 |
https://transum.org/Software/SW/Starter_of_the_day/starter_November26.ASP | 1,726,111,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00219.warc.gz | 541,343,401 | 8,766 | Without a calculator match a pie slice to a calculation to an answer. The three letters will make a word. Arrange your six three letter words to make a sentence.
C.
S.
A.
S.
Y.
T.
D. 50% of 740A. 20% of 800E. 25% of 360
I. 75% of 100O. 10% of 840A. 5% of 420
D. 370N. 90X. 75
Y. 160U. 84N. 21
The pie slices represent percentages. The matching percentages can be found in the calculations. The result of the calculations will match the given answers.
## A Mathematics Lesson Starter Of The Day
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## Curriculum Reference
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Practice 03 Solutions
# Practice 03 Solutions - practice 03 GADHIA TEJAS Due 10:00...
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practice 03 – GADHIA, TEJAS – Due: Jan 31 2007, 10:00 pm 1 Question 1 part 1 of 1 10 points Initially (at time t = 0) a particle is moving vertically at 7 . 9 m / s and and horizontally at 0 m / s. The particle accelerates horizontally at 2 . 1 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . At what time will the particle be traveling at 54 with respect to the horizontal? Correct answer: 0 . 622518 s (tolerance ± 1 %). Explanation: Basic Concept Kinematic Equation v = v o + a t . Solution The vertical velocity is v y t = v y 0 g t . The horizontal velocity is v x t = v y 0 + a t = a t . v x t v y t v t 54 The vertical component is the opposite side and the horizontal component is the adjacent side to the angle, so tan θ = v y t v x t = v y 0 g t a t a t tan θ = v y 0 g t a t tan θ + g t = v y 0 t = v y 0 a tan θ + g = (7 . 9 m / s) (2 . 1 m / s 2 ) tan(54 ) + (9 . 8 m / s 2 ) = 0 . 622518 s . Question 2 part 1 of 1 10 points A particle moving at a velocity of 8 . 1 m / s in the positive x direction is given an accel- eration of 7 . 8 m / s 2 in the positive y direction for 8 s. What is the final speed of the particle? Correct answer: 62 . 9235 m / s (tolerance ± 1 %). Explanation: Let : v xf = v xi = 8 . 1 m / s . a y = 7 . 8 m / s 2 , and t = 8 s . The vertical velocity undergoes constant ac- celeration: v yf = v yi + a t = 0 + (7 . 8 m / s 2 )(8 s) = 62 . 4 m / s . Thus v f = radicalBig v 2 xf + v 2 yf = radicalBig (8 . 1 m / s) 2 + (62 . 4 m / s) 2 = 62 . 9235 m / s . Question 3 part 1 of 3 10 points A particle moves in the xy plane with con- stant acceleration. At time zero, the particle is at x = 2 m, y = 5 m, and has velocity vectorv o = (7 m / s) ˆ ı + ( 4 . 5 m / s) ˆ . The accelera- tion is given by vectora = (3 m / s 2 ) ˆ ı + (1 m / s 2 ) ˆ . What is the x component of velocity after 1 . 5 s? Correct answer: 11 . 5 m / s (tolerance ± 1 %). Explanation:
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practice 03 – GADHIA, TEJAS – Due: Jan 31 2007, 10:00 pm 2 Let : a x = 3 m / s 2 , v xo = 7 m / s , and t = 1 . 5 s . After 1 . 5 s, vectorv x = vectorv xo + vectora x t = (7 m / s) ˆ ı + (3 m / s 2 ) ˆ ı (1 . 5 s) = (11 . 5 m / s) ˆ ı . Question 4 part 2 of 3 10 points What is the y component of velocity after 1 . 5 s? Correct answer: 3 m / s (tolerance ± 1 %). Explanation: Let : a y = 1 m / s 2 and v yo = 4 . 5 m / s . vectorv y = vectorv yo + vectora y t = ( 4 . 5 m / s) ˆ + (1 m / s 2 ) ˆ (1 . 5 s) = ( 3 m / s) ˆ . Question 5 part 3 of 3 10 points What is the magnitude of the displacement from the origin ( x = 0 m, y = 0 m) after 1 . 5 s? Correct answer: 15 . 8873 m (tolerance ± 1 %). Explanation: Let : d o = (2 m , 5 m) , v o = (7 m / s , 4 . 5 m / s) , and a = (3 m / s 2 , 1 m / s 2 ) . From the equation of motion, vector d = vector d o + vectorv o t + 1 2 a t 2 = bracketleftBig (2 m) ˆ ı + (5 m) ˆ bracketrightBig + [(7 m / s) ˆ ı + ( 4 . 5 m / s) ˆ ] (1 . 5 s) + 1 2 bracketleftBig (3 m / s 2 ) ˆ ı + (1 m / s 2 ) ˆ bracketrightBig (1 . 5 s) 2 = (15 . 875 m) ˆ ı + ( 0 . 625 m) ˆ , so | vector d | = radicalBig d 2 x + d 2 y = radicalBig (15 . 875 m) 2 + ( 0 . 625 m) 2 = 15 . 8873 m . Question 6 part 1 of 1 10 points A cannon fires a 0 . 278 kg shell with initial velocity v i = 11 m / s in the direction θ = 56 above the horizontal.
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http://www.valuewalk.com/2016/10/us-not-low-tax-jurisdiction/ | 1,511,373,755,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806620.63/warc/CC-MAIN-20171122175744-20171122195744-00408.warc.gz | 548,010,232 | 18,561 | Trump has been running his mouth about how the US is one of the highest-taxed countries in the world. This really makes the journalists upset. They say that it’s false. Every time he says this, they have a conniption.
How can it be true? How can we have higher taxes than Sweden—where taxes are so high that everything is free, there are ponies everywhere, and everyone is happy?
It depends on how you measure it.
The Wall Street Journal just posted a piece about this, offering the statistic (that most journalists use) that the US collects about 26% of GDP in taxes, compared to an average of 34.4% for other industrialized countries.
First, the fact that we collect 26% percent of GDP in taxes is crazy. For years, even decades, the total take we would end up with was around 20% of GDP—no matter how high or how low the tax rates went. Tax collection has become much more effective in the last 10 years, and a lot of what the IRS would call the “tax gap” has been closed.
But yes, the amount we pay in taxes collectively is lower than most industrialized countries.
But there’s another way to look at it.
### A math lesson on tax rates
What about tax rates? Who has the highest tax rates in the world?
The US is close to the top.
The top marginal income tax rate at the federal level is 39.6%.
Now, a lot of economists stop there. They say: US taxes are 39.6%, Sweden’s are 59.7%, so the US is a low-tax jurisdiction.
But you have to take into account state income taxes, too. California is the highest, at 13.3%. Some municipalities and counties also have income taxes. When you take into account New York’s state and city income taxes, it’s also about 13.3%.
So 39.6 + 13.3 = 52.9%. Catching up to Sweden!
We’re not done yet. We now pay a 3.8% surtax on investment income, which was intended to fund Obamacare. That makes it 56.7%.
Only four countries in the world are higher (not counting Aruba): Sweden, Finland, Canada, and Belgium.
We’re not done yet!
We also pay payroll taxes of 6.2% for the employee and 6.2% for the employer on the first \$118,500 of income, plus Medicare taxes of 2.9%, 1.45% for the employee and 1.45% for the employer. Economically speaking, the employee pays both. So add 2.9% to the total, and then you get to 59.6%.
We’re not done yet!
In some localities (like New York), there is something called an “unincorporated business tax,” or UBT, so if you have an LLC or sole proprietorship, you pay another 4% on your net business income.
63.6%.
Not done yet!
We haven’t yet discussed property taxes. In high-tax jurisdictions like New York, New Jersey, Connecticut, or Illinois, you can easily have a tax burden of \$20,000 annually on a middle-class home. I have heard that property tax bills of \$50,000 to \$70,000 are pretty common.
Nobody takes this into account in the global comparisons.
Even where I’m from, in the impoverished eastern part of Connecticut, it’s not uncommon to see \$4,000 property tax bills on houses that are worth about \$130,000.
Now we’re done.
So is the US a high-tax jurisdiction or a low-tax jurisdiction? Or a better question might be: How can we pay so little compared to other countries if our tax rates are so high?
### The most progressive tax code in the world
I think if I were an economics PhD student, the topic I would pick for my thesis would be to measure the progressivity of income taxes around the world. Like, how much people in low brackets pay compared to people in high brackets. This is an important question.
A lot of people spout off about taxes without really knowing what they are talking about. So let’s pull up the latest tax table from the IRS:
Source: Tax Foundation
##### Low-Tax Jurisdiction
So you might recall the Mitt Romney biff from the last election… when he said that half the country pays no income taxes. People thought this was a very tone-deaf thing to say. If you look at the table above, you can see that everyone pays at least some tax. So what gives?
Two points:
1. Generally, people in the lower brackets get a lot of deductions and credits (like the EITC) that completely eliminate their tax liability, or even create a negative one. That’s right—not only do a lot of these folks pay no income taxes, they actually receive money from the government.5
2. People still pay payroll tax to fund Social Security and Medicare.
Our tax code is very progressive—the effective tax rate for millions of people is zero or negative, while the effective tax rate for rich filers is in the thirties (or much higher when you add in state income taxes and other items that we discussed before).
Here are Sweden’s tax brackets, in USD (from Wikipedia):
0% \$0 to \$2,690
31% \$2,690 to \$62,140
51% \$62,140 to \$88,180
56% \$88,180+
So as you can see in Sweden, everyone pays a decent amount of tax. The rates are high, but the tax code is not all that progressive.
And this is how the OECD reports Sweden has a much higher tax burden than the US: the tax code is flatter, and everyone pays. Lower class, middle class, upper class—everyone.
Progressive tax codes are the worst things in the world.
### It all comes down to incentives
With a progressive tax code, you can divide people into groups and turn them against each other. The rich aren’t paying their fair share. The poor aren’t either. There is a lot of hate and discontent.
A flat tax (or nearly so) solves these problems: everyone has skin in the game, and also, a flat tax is progressive anyway. If there is a flat tax of 20%, and you earn \$100,000, you pay \$20,000. And if I earn \$1 million, I pay \$200,000—10 times as much.
But really, it all comes down to incentives. High marginal tax rates create a disincentive to working hard and producing cool stuff.
If Joe Shlabotnik makes \$25,000 a year and his marginal tax rate is 15%, he has no disincentive to go to work. He gets to keep 85 cents of every dollar.
But you don’t care about that guy. Actually, you do, but you don’t care about Joe Shlabotnik as much as Elon Musk.
Elon Musk lives in California, so he pays that 13.3% state income tax, and very likely he faces the 59.6% marginal rate that we discussed earlier. So is the 59.6% marginal rate keeping him from going to work? No.
What if it were 70%, like under Jimmy Carter?
What if it were over 90%, like Bernie Sanders wanted? Would he still go to work then? He might not.
You want guys like Elon Musk to keep going to work.
I haven’t said that these tax rates are unfair. They may very well be fair. I’m not getting into a political discussion here, nor am I complaining about my taxes.
I am saying two things:
1. The US is not a low-tax jurisdiction, at least in the way that it counts.
2. The top marginal rate is the one factor that is most (inversely) correlated to economic success.
I am no fan of Trump. And I’m not sure why he was complaining about high taxes with all those NOL carryforwards. But as you travel around the world, taxes here are not especially low. The US has this reputation as this Wild West capitalist fantasyland, but nothing could be further from the truth.
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Article by Jared Dillian | 1,809 | 7,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.969852 |
https://www.clutchprep.com/physics/practice-problems/94677/a-baseball-player-friend-of-yours-wants-to-determine-his-pitching-speed-you-have | 1,603,614,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888402.81/warc/CC-MAIN-20201025070924-20201025100924-00297.warc.gz | 671,491,399 | 35,932 | # Problem: A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.0 m above the ground. The ball lands 30 m away.What is his pitching speed?
###### Problem Details
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.0 m above the ground. The ball lands 30 m away.
What is his pitching speed? | 109 | 497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-45 | latest | en | 0.982341 |
https://goodboychan.github.io/python/datacamp/machine_learning/2020/08/06/02-Informed-Search.html | 1,656,475,178,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00399.warc.gz | 319,892,884 | 183,185 | import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
## Informed Search - Coarse to Fine
• Coarse to fine tuning
1. Random search
2. Find promising areas
3. Grid search in the smaller area
4. Continue until optimal score obtained
• Why Coarse to Fine?
• Utilizes the advantanges of grid and random search
• Wide search to begin with
• Deeper search once you know where a good spot is likely to be
• Better spending of time and computational efforts mean you can iterate quicker
### Visualizing Coarse to Fine
You're going to undertake the first part of a Coarse to Fine search. This involves analyzing the results of an initial random search that took place over a large search space, then deciding what would be the next logical step to make your hyperparameter search finer.
def visualize_hyperparameter(name):
plt.clf()
plt.scatter(results_df[name],results_df['accuracy'], c=['blue']*500)
plt.gca().set(xlabel='{}'.format(name), ylabel='accuracy', title='Accuracy for different {}s'.format(name))
plt.gca().set_ylim([0,100])
from itertools import product
max_depth_list = range(1, 6)
min_samples_leaf_list = range(3, 14)
learn_rate_list = np.linspace(0.01, 1.33, 200)
combinations_list = [list(x) for x in product(max_depth_list,
min_samples_leaf_list,
learn_rate_list)]
# Confirm the size fo the combinations_list
print(len(combinations_list))
# Sort the results_df by accuracy and print the top 10 rows
# Confirm which hyperparameters were used in this search
print(results_df.columns)
# Call visualize_hyperparameter() with each hyperparameter in turn
visualize_hyperparameter('max_depth')
11000
max_depth min_samples_leaf learn_rate accuracy
1 10 14 0.477450 97
4 6 12 0.771275 97
2 7 14 0.050067 96
3 5 12 0.023356 96
5 13 11 0.290470 96
6 6 10 0.317181 96
7 19 10 0.757919 96
8 2 16 0.931544 96
9 16 13 0.904832 96
10 12 13 0.891477 96
Index(['max_depth', 'min_samples_leaf', 'learn_rate', 'accuracy'], dtype='object')
visualize_hyperparameter('min_samples_leaf')
visualize_hyperparameter('learn_rate')
We have undertaken the first step of a Coarse to Fine search. Results clearly seem better when max_depth is below 20. learn_rates smaller than 1 seem to perform well too. There is not a strong trend for min_samples leaf though.
### Coarse to Fine Iterations
You will now visualize the first random search undertaken, construct a tighter grid and check the results.
def visualize_first():
for name in results_df.columns[0:2]:
plt.clf()
plt.scatter(results_df[name],results_df['accuracy'], c=['blue']*500)
plt.gca().set(xlabel='{}'.format(name), ylabel='accuracy', title='Accuracy for different {}s'.format(name))
plt.gca().set_ylim([0,100])
x_line = 20
if name == "learn_rate":
x_line = 1
plt.axvline(x=x_line, color="red", linewidth=4)
visualize_first()
def visualize_second():
for name in results_df2.columns[0:2]:
plt.clf()
plt.scatter(results_df[name],results_df['accuracy'], c=['blue']*500)
plt.gca().set(xlabel='{}'.format(name), ylabel='accuracy', title='Accuracy for different {}s'.format(name))
plt.gca().set_ylim([0,100])
x_line = 20
if name == "learn_rate":
x_line = 1
plt.axvline(x=x_line, color="red", linewidth=4)
max_depth_list = list(range(1, 21))
learn_rate_list = np.linspace(0.001, 1, 50)
visualize_second()
## Informed Search - Bayesian Statistics
• Bayes rule
• A statistical method of using new evidence to iteratively update our beliefs about some outcome $$P(A \vert B) = \frac{P(B \vert A) P(A)}{P(B)}$$
• LHS = the probability of A given B has occurred. B is some new evidence (Posterior)
• RHS = how to calculate LHS
• $P(A)$ is the 'prior'. The initial hypothesis about the event.
• $P(A\vert B)$ is the probability given new evidence
• $P(B)$ is the 'marginal likelihood'. It is the probability of observing this new evidence
• $P(B \vert A)$ is the likelihood which is the probability of observing the evidence, given the event we care about
• Bayes in Hyperparameter Tuning
• Pick a hyperparameter combination
• Build a model
• Get new evidence (the score of the model)
• Update our belief and chose better hyperparamters next round
### Bayes Rule in Python
In this exercise you will undertake a practical example of setting up Bayes formula, obtaining new evidence and updating your 'beliefs' in order to get a more accurate result. The example will relate to the likelihood that someone will close their account for your online software product.
These are the probabilities we know:
• 7% (0.07) of people are likely to close their account next month
• 15% (0.15) of people with accounts are unhappy with your product (you don't know who though!)
• 35% (0.35) of people who are likely to close their account are unhappy with your product
p_unhappy = 0.15
p_unhappy_close = 0.35
# Probability someone will close
p_close = 0.07
# Probability unhappy person will close
p_close_unhappy = (p_unhappy_close * p_close) / p_unhappy
print(p_close_unhappy)
0.16333333333333336
### Bayesian Hyperparameter tuning with Hyperopt
In this example you will set up and run a bayesian hyperparameter optimization process using the package Hyperopt. You will set up the domain (which is similar to setting up the grid for a grid search), then set up the objective function. Finally, you will run the optimizer over 20 iterations.
You will need to set up the domain using values:
• max_depth using quniform distribution (between 2 and 10, increasing by 2)
• learning_rate using uniform distribution (0.001 to 0.9)
Note that for the purpose of this exercise, this process was reduced in data sample size and hyperopt & GBM iterations. If you are trying out this method by yourself on your own machine, try a larger search space, more trials, more cvs and a larger dataset size to really see this in action!
Note: This session requires Hyperopt packages
from sklearn.model_selection import train_test_split
# To change categorical variable with dummy variables
credit_card = pd.get_dummies(credit_card, columns=['SEX', 'EDUCATION', 'MARRIAGE'], drop_first=True)
X = credit_card.drop(['ID', 'default payment next month'], axis=1)
y = credit_card['default payment next month']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, shuffle=True)
import hyperopt as hp
from sklearn.model_selection import cross_val_score
# Set up space dictionary with specified hyperparamters
space = {'max_depth': hp.hp.quniform('max_depth', 2, 10, 2),
'learning_rate': hp.hp.uniform('learning_rate', 0.001, 0.9)}
# Set up objective function
def objective(params):
params = {'max_depth': int(params['max_depth']),
'learning_rate': params['learning_rate']}
best_score = cross_val_score(gbm_clf, X_train, y_train,
scoring='accuracy', cv=2, n_jobs=4).mean()
loss = 1 - best_score
return loss
# Run the algorithm
best = hp.fmin(fn=objective, space=space, max_evals=20,
rstate=np.random.RandomState(42), algo=hp.tpe.suggest)
print(best)
100%|██████████| 20/20 [02:16<00:00, 6.83s/trial, best loss: 0.18080952380952375]
{'learning_rate': 0.0128515490384306, 'max_depth': 6.0}
## Informed Search - Genetic Algorithms
• Genetics in Machine Learning
1. Create some models (that have hyperparameter settings)
2. Pick the best (by our scoring function) : these are the ones that "survive"
3. Create new models that are similar to the best ones
4. Add in some randomness so we don't reach a local optimum
5. Repeat until we are happy!
• It allows us to learn from previous iterations, just like bayesian hyperparameter tuning
• Takes care of many tedious aspects of machine learning
### Genetic Hyperparameter Tuning with TPOT
You're going to undertake a simple example of genetic hyperparameter tuning. TPOT is a very powerful library that has a lot of features. You're just scratching the surface in this lesson, but you are highly encouraged to explore in your own time.
This is a very small example. In real life, TPOT is designed to be run for many hours to find the best model. You would have a much larger population and offspring size as well as hundreds more generations to find a good model.
You will create the estimator, fit the estimator to the training data and then score this on the test data.
For this example we wish to use:
• 3 generations
• 4 in the population size
• 3 offspring in each generation
• accuracy for scoring
Note: This session requires tpot packages
from tpot import TPOTClassifier
# Assign the values outlined to the inputs
number_generations = 3
population_size = 4
offspring_size = 3
scoring_function = 'accuracy'
# Create the tpot classifier
tpot_clf = TPOTClassifier(generations=number_generations, population_size=population_size,
offspring_size=offspring_size, scoring=scoring_function,
verbosity=2, random_state=2, cv=2)
# Fit the classifier to the training data
tpot_clf.fit(X_train, y_train)
# Score on the test set
print(tpot_clf.score(X_test, y_test))
Generation 1 - Current best internal CV score: 0.8204285714285714
Generation 2 - Current best internal CV score: 0.8204285714285714
Generation 3 - Current best internal CV score: 0.8204285714285714
Best pipeline: DecisionTreeClassifier(input_matrix, criterion=gini, max_depth=1, min_samples_leaf=10, min_samples_split=9)
0.8176666666666667
You can see in the output the score produced by the chosen model (in this case a version of Naive Bayes) over each generation, and then the final accuracy score with the hyperparameters chosen for the final model. This is a great first example of using TPOT for automated hyperparameter tuning.
### Analysing TPOT's stability
You will now see the random nature of TPOT by constructing the classifier with different random states and seeing what model is found to be best by the algorithm. This assists to see that TPOT is quite unstable when not run for a reasonable amount of time.
tpot_clf = TPOTClassifier(generations=2, population_size=4, offspring_size=3,
scoring='accuracy', cv=2, verbosity=2, random_state=42)
# Fit the classifier to the training data
tpot_clf.fit(X_train, y_train)
# Score on the test set
print(tpot_clf.score(X_test, y_test))
Generation 1 - Current best internal CV score: 0.8213809523809524
Generation 2 - Current best internal CV score: 0.8213809523809524
Best pipeline: XGBClassifier(input_matrix, learning_rate=0.001, max_depth=9, min_child_weight=7, n_estimators=100, nthread=1, subsample=0.45)
0.8195555555555556
tpot_clf = TPOTClassifier(generations=2, population_size=4, offspring_size=3,
scoring='accuracy', cv=2, verbosity=2, random_state=122)
# Fit the classifier to the training data
tpot_clf.fit(X_train, y_train)
# Score on the test set
print(tpot_clf.score(X_test, y_test))
Generation 1 - Current best internal CV score: 0.7811904761904762
Generation 2 - Current best internal CV score: 0.7811904761904762
Best pipeline: LogisticRegression(input_matrix, C=0.5, dual=False, penalty=l2)
0.7724444444444445
tpot_clf = TPOTClassifier(generations=2, population_size=4, offspring_size=3,
scoring='accuracy', cv=2, verbosity=2, random_state=99)
# Fit the classifier to the training data
tpot_clf.fit(X_train, y_train)
# Score on the test set
print(tpot_clf.score(X_test, y_test))
Generation 1 - Current best internal CV score: 0.7960476190476191
Generation 2 - Current best internal CV score: 0.8049523809523809
Best pipeline: GradientBoostingClassifier(input_matrix, learning_rate=0.5, max_depth=3, max_features=0.6000000000000001, min_samples_leaf=15, min_samples_split=17, n_estimators=100, subsample=0.6000000000000001)
0.812
You can see that TPOT is quite unstable when only running with low generations, population size and offspring. The first model chosen was a Decision Tree, then a K-nearest Neighbor model and finally a Random Forest. Increasing the generations, population size and offspring and running this for a long time will assist to produce better models and more stable results. | 3,129 | 12,227 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-27 | longest | en | 0.658205 |
https://www.astronomyclub.xyz/star-formation-2/info-kau.html | 1,568,911,362,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00046.warc.gz | 758,106,705 | 4,833 | ## Info
Time (days)
Radial velocity and light curves for 8 Cep, the prototypical Cepheid. (a) Apparent magnitude as a function of time within the period. (b) Temperature as a function of time. (c) Radius, relative to the minimum radius, as a function of time. (d) Radial velocity of the surface as a function of time. Note that the radial velocity is one quarter cycle (90°) out of phase with the radius.
ceding section. To see how a Cepheid oscillates, lets consider the oscillations of a normal star. These oscillations are radial. They involve inward and outward motions of the outer layers of the star. Suppose we are able to perturb a star by decreasing its radius R. The density then increases, and the pressure increases. The excess pressure will make the outer layers expand back. However, just as a swing overshoots its lowest point as it returns from its maximum height, the star can overshoot its equilibrium radius R0. Now the star is larger than its equilibrium radius, and the pressure decreases, allowing material to fall back. This process then repeats itself.
In the above analysis, we have ignored the effects of opacity. In a normal star, the opacity decreases as the temperature increases. Now, we again start the perturbation by reducing R below R0. This causes P and T to increase. The increase in T decreases the opacity. The reduction of the opacity allows some of the excess pressure to be relieved by allowing heat to flow out of the denser regions as radiation. This reduces the tendency of the star to overshoot. If we had started with a perturbation in which R > R0, P and T would have decreased. The opacity would have increased, and the tendency to fall back too fast would be reduced. The result of the opacity is to quench the oscillation.
For a narrow range of conditions, the opacity increases as the temperature increases. The source of the opacity is the ionization of He+ to form He+ +. If we now start with a perturbation in which R < R0, the pressure and temperature increase. Now the opacity also increases, so the excess pressure is not relieved, except by driving the star back. The tendency to overshoot is enhanced. Similarly, with R > R0, the pressure and temperature decrease. The opacity also decreases, reducing the pressure even further. The material falls back quickly and overshoots. This oscillation can continue indefinitely, rather than being quenched. These are the conditions that produce a Cepheid variable. | 534 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.942312 |
https://www.fmaths.com/tips/faq-what-is-mathematics-for-me.html | 1,632,770,404,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00686.warc.gz | 779,313,227 | 11,777 | # FAQ: What Is Mathematics For Me?
## What is Mathematics according to you?
Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. It is the building block for everything in our daily lives, including mobile devices, architecture (ancient and modern), art, money, engineering, and even sports.
## What is math in your own words?
The Merriam-Webster dictionary defines mathematics as “the science of numbers and their operations, interrelations, combinations, generalizations, and abstractions and of space configurations and their structure, measurement, transformations, and generalizations.”
## What is the full meaning of mathematics?
Mathematics (from Greek: μάθημα, máthēma, ‘knowledge, study, learning’) includes the study of such topics as quantity (number theory), structure (algebra), space (geometry), and change (analysis). It has no generally accepted definition.
## What is a math person?
Essentially, the aim is to provide students a meaningful experience of mathematics that solidifies their appreciation of the discipline regardless of their future studies or career trajectories.
You might be interested: What Is Bar Graph In Mathematics?
## Do we need mathematics everyday?
Math is vital in our world today. Everyone uses mathematics in our day to day lives, and most of the time, we do not even realize it. Without math, our world would be missing a key component in its makeup. “ Math is so important because it is such a huge part of our daily lives.
## Why is math so hard?
Math is a very abstract subject. For students, learning usually happens best when they can relate it to real life. As math becomes more advanced and challenging, that can be difficult to do. As a result, many students find themselves needing to work harder and practice longer to understand more abstract math concepts.
## What is math in simple words?
Mathematics is the study of numbers, shapes and patterns. The word comes from the Greek word “μάθημα” (máthema), meaning “science, knowledge, or learning”, and is sometimes shortened to maths (in England, Australia, Ireland, and New Zealand) or math (in the United States and Canada). Numbers: how things can be counted.
## How do we use math in everyday life?
10 Ways We Use Math Everyday
• Chatting on the cell phone. Chatting on the cell phone is the way of communicating for most people nowadays.
• In the kitchen. Baking and cooking requires some mathematical skill as well.
• Gardening.
• Arts.
• Keeping a diary.
• Planning an outing.
• Banking.
• Planning dinner parties.
## How is mathematics used in today’s world?
It gives us a way to understand patterns, to quantify relationships, and to predict the future. Math is a powerful tool for global understanding and communication. Using it, students can make sense of the world and solve complex and real problems.
## What is full form of Kiss?
Keep It Simple, Stupid. Internet » Chat. Rate it: KISS. Keep It Short And Simple.
## What is a good definition in math?
What about a good definition? A good definition for a parallelogram is a. quadrilateral with two pairs of parallel. congruent sides. Notice that we’re using words that.
## What is another name for mathematics?
What is another word for mathematics?
arithmetic calculation
calculus division
figures geometry
math multiplication
## How do you become a maths person?
Heller outlines some of those mindset-changing steps here:
1. Create opportunities for cooperative learning.
2. Give students the chance to productively struggle.
3. Encourage participation, even if the student doesn’t have the right answer yet.
4. Re-envision math as a language.
## Is there such a thing as a math person?
But experts suggest that there’s actually no such thing as a “ math person.” In fact, they argue that the myth of “ math people” makes students more anxious about math.
## What does 3 mean in math?
For example, consider the following expression: 3! In mathematics, the expression 3! is read as “three factorial” and is really a shorthand way to denote the multiplication of several consecutive whole numbers. | 879 | 4,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.947653 |
http://oeis.org/A275532 | 1,611,018,935,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00669.warc.gz | 71,940,801 | 3,992 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A275532 Possible values for sum of numbers in Collatz trajectory (A033493), sorted into ascending order. 1
1, 3, 7, 15, 31, 36, 46, 49, 55, 63, 66, 67, 91, 106, 119, 127, 139, 145, 148, 186, 190, 197, 214, 235, 248, 255, 259, 274, 281, 288, 301, 302, 316, 325, 330, 339, 346, 357, 386, 393, 399, 413, 427, 442, 452, 465, 497, 498, 500, 505, 509, 511, 519, 535, 540, 557 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Values a(n) such that a(n) = A033493(x) = A033493(y) for distinct numbers x and y: 442, 609, 633, 724, 904, 925, ... LINKS MATHEMATICA Take[#, 56] &@ Union@ Table[Total@ FixedPointList[Which[# == 1, 1, OddQ@ #, 3 # + 1, True, #/2] &, n] - 1, {n, 10^3}] (* Michael De Vlieger, Aug 02 2016 *) PROG (MAGMA) Set(Sort([&+[k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]: n in [1..2^10] | &+[k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]] le 2^10])) CROSSREFS Cf. A033493, A275531. Sequence in context: A320024 A098583 A275531 * A212315 A043729 A331503 Adjacent sequences: A275529 A275530 A275531 * A275533 A275534 A275535 KEYWORD nonn AUTHOR Jaroslav Krizek, Jul 31 2016 STATUS approved
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http://gmatclub.com/forum/i-feel-like-taking-the-test-38985-40.html?sort_by_oldest=true | 1,485,019,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00266-ip-10-171-10-70.ec2.internal.warc.gz | 112,292,971 | 51,375 | I feel like taking the test : General GMAT Questions and Strategies - Page 3
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# I feel like taking the test
Author Message
VP
Joined: 25 Jun 2006
Posts: 1172
Followers: 3
Kudos [?]: 148 [0], given: 0
### Show Tags
02 Dec 2006, 22:50
I tried GMATPrep 1 again after deleting the previous tests this morning. got 780( 51, 45).
Math got 1 question wrong, but I really saw some tough questions at the last 15 questions.
Verbal has a lot of repeats, but I uncovered some new questions as well.
Director
Joined: 05 Feb 2006
Posts: 898
Followers: 3
Kudos [?]: 107 [0], given: 0
### Show Tags
03 Dec 2006, 01:16
tennis_ball wrote:
I tried GMATPrep 1 again after deleting the previous tests this morning. got 780( 51, 45).
Math got 1 question wrong, but I really saw some tough questions at the last 15 questions.
Verbal has a lot of repeats, but I uncovered some new questions as well.
I think you should relax now
Really that is my advice, not to put any more pressure on your brain... Accumulate stamina for tuesday... Good luck!!!
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9
Kudos [?]: 149 [0], given: 0
### Show Tags
03 Dec 2006, 04:32
J-2.... Good luck for all you, guys
Cross fingers and touching wood for u .... Each of u has to come back with the good news :D
VP
Joined: 25 Jun 2006
Posts: 1172
Followers: 3
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### Show Tags
03 Dec 2006, 07:29
Yes. you are right. Relax.
but I still need to do some questions every day just to keep warm and be in the best mind state. but won't do too many.
Director
Joined: 05 Feb 2006
Posts: 898
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Kudos [?]: 107 [0], given: 0
### Show Tags
03 Dec 2006, 23:17
Fig wrote:
J-2.... Good luck for all you, guys
Cross fingers and touching wood for u .... Each of u has to come back with the good news :D
Thanks Fig! We will try not to dissapoint the club
VP
Joined: 25 Jun 2006
Posts: 1172
Followers: 3
Kudos [?]: 148 [0], given: 0
### Show Tags
03 Dec 2006, 23:57
all are done. and ready to go.
Manager
Joined: 20 Nov 2006
Posts: 213
Followers: 1
Kudos [?]: 10 [0], given: 0
### Show Tags
04 Dec 2006, 01:44
Good luck for you guys for your GMAT!!!
Simaq, Tennisball and karlfurt... Keep up the practice score in your Test also.
Manager
Joined: 25 Sep 2006
Posts: 152
Followers: 1
Kudos [?]: 5 [0], given: 0
### Show Tags
04 Dec 2006, 01:48
I have also planned to take IT in 2 weeks time.
Though, my Verbal is not as prepared as you guys. Nevertheless, I will try my best.
Director
Joined: 05 Feb 2006
Posts: 898
Followers: 3
Kudos [?]: 107 [0], given: 0
### Show Tags
04 Dec 2006, 02:31
tennis_ball wrote:
all are done. and ready to go.
Good luck to you Tennisball! You are starting 1st... hope all is well!
Director
Joined: 02 Mar 2006
Posts: 580
Location: France
Followers: 1
Kudos [?]: 99 [0], given: 0
### Show Tags
04 Dec 2006, 03:01
Good luck to you guys, and I am eager to know your results. It will be your first or 2nd time? In my case, it will be the 2nd and I think the pressure will be higher than it was 5 months ago.
VP
Joined: 25 Jun 2006
Posts: 1172
Followers: 3
Kudos [?]: 148 [0], given: 0
### Show Tags
04 Dec 2006, 23:47
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1
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05 Dec 2006, 07:50
Good luck everyone - Tennisball, Simaq, Karlfurt, MBAlad... Did i miss anyone? Who else is taking this montH? I guess alot of us have tests scheduled this week or next to make it in time for the 2nd round deadline or just to get it over before before year-end.
I hope we all come out happy and contented with our scores!
Best of luck!!!
_________________
Impossible is nothing
VP
Joined: 28 Mar 2006
Posts: 1381
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Kudos [?]: 31 [0], given: 0
### Show Tags
05 Dec 2006, 17:49
Guys how many test do you take per week??
VP
Joined: 25 Jun 2006
Posts: 1172
Followers: 3
Kudos [?]: 148 [0], given: 0
### Show Tags
05 Dec 2006, 18:55
trivikram wrote:
Guys how many test do you take per week??
I take 2, Sat and Sun. but i agree that too many is not necessary.
VP
Joined: 28 Mar 2006
Posts: 1381
Followers: 2
Kudos [?]: 31 [0], given: 0
### Show Tags
05 Dec 2006, 19:31
tennis_ball wrote:
trivikram wrote:
Guys how many test do you take per week??
I take 2, Sat and Sun. but i agree that too many is not necessary.
Thanks ! What are the colleges you wre refrring to some N... Are they in Singapore?
VP
Joined: 25 Jun 2006
Posts: 1172
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Kudos [?]: 148 [0], given: 0
### Show Tags
05 Dec 2006, 21:18
trivikram wrote:
tennis_ball wrote:
trivikram wrote:
Guys how many test do you take per week??
I take 2, Sat and Sun. but i agree that too many is not necessary.
Thanks ! What are the colleges you wre refrring to some N... Are they in Singapore?
ya. i cannot get into a good MBA, but I also don't wanna get into a mediocre MBA, so I am taking a finance masters to change my career path first.
Manager
Joined: 04 Feb 2004
Posts: 71
Location: India
Followers: 1
Kudos [?]: 25 [0], given: 0
### Show Tags
06 Dec 2006, 01:38
Below are the results for PR online
Online Test GMAT 8 11/20/2006 V 35 Q48 670
Online Test GMAT 9 11/21/2006 V31 Q51 660
Online Test GMAT 10 11/30/2006 V30 Q48 640
Online Test GMAT 11 12/6/2006 V33 Q47 650
not able to cross the 700 deadline (target 700+)- not sure if I would schedule the test this Saturday
Manager
Joined: 25 Sep 2006
Posts: 152
Followers: 1
Kudos [?]: 5 [0], given: 0
### Show Tags
06 Dec 2006, 01:57
deep wrote:
Below are the results for PR online
Online Test GMAT 8 11/20/2006 V 35 Q48 670
Online Test GMAT 9 11/21/2006 V31 Q51 660
Online Test GMAT 10 11/30/2006 V30 Q48 640
Online Test GMAT 11 12/6/2006 V33 Q47 650
not able to cross the 700 deadline (target 700+)- not sure if I would schedule the test this Saturday
Have u tried Gmatprep ? It is an official test and the questions feel exactly the same as the real test. Use that to better gauge estimate score.
_________________
livin in a prison island...
06 Dec 2006, 01:57
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# I feel like taking the test
Moderators: WaterFlowsUp, HiLine
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,407 | 7,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.920558 |
https://www.nagwa.com/en/videos/487180372649/ | 1,680,268,559,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00150.warc.gz | 1,001,072,314 | 12,681 | # Question Video: Adding Two Fractions with Different Denominators Mathematics • 5th Grade
Calculate (1/4) + (1/3).
02:23
### Video Transcript
Calculate one-fourth plus one-third.
We need to think about the steps that we need to follow when we add and subtract fractions. First, we check to see if the fractions have common denominators, and if they don’t, we need to rename our fractions with a common denominator. One-fourth and one-third do not share common denominator, which means we need to rename them so that they do have a common denominator. 12 is a multiple of four and three: three times four equals 12 and four times three equals 12.
But remember if we multiply the denominator by something, we also need to multiply the numerator by the same amount. In this case, we multiply one times three, and one-fourth becomes three-twelfths. And then we multiply one by four; one-third becomes four-twelfths. Everything we did in this step was part of finding the common denominator.
It’s time for step two: to add or subtract. In our case, we’re adding, and now we’re adding three-twelfths plus four-twelfths. Three plus four equals seven. And when we’re adding and subtracting fractions, the denominator does not change. So we bring 12 across. This was our adding step.
Our third and final step is to simplify this number or reduce it if that’s possible. We asked the question, can this value, seven-twelfths, be reduced? The answer here is no. Seven-twelfths is already in its simplest form. This means that one-fourth plus one-third equals seven-twelfths. Seven-twelfths is your final answer. | 369 | 1,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.940308 |
https://mathexamination.com/lab/pinch-point.php | 1,611,802,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00482.warc.gz | 436,320,271 | 8,628 | ## Do My Pinch Point Lab
As stated above, I made use of to compose a simple and also straightforward mathematics lab with only Pinch Point Nonetheless, the easier you make your lab, the simpler it ends up being to obtain stuck at completion of it, after that at the start. This can be very aggravating, and all this can happen to you because you are utilizing Pinch Point and/or Modular Equations inaccurately.
With Modular Formulas, you are currently utilizing the incorrect equation when you obtain stuck at the start, otherwise, then you are most likely in a stumbling block, as well as there is no feasible way out. This will only become worse as the issue ends up being extra complex, but after that there is the question of how to proceed with the issue. There is no chance to effectively deal with addressing this type of mathematics issue without being able to immediately see what is going on.
It is clear that Pinch Point and Modular Equations are challenging to discover, and it does take method to develop your very own sense of intuition. Yet when you want to address a math issue, you need to use a device, and the tools for finding out are used when you are stuck, and also they are not made use of when you make the wrong move. This is where lab Assist Service is available in.
As an example, what is wrong with the inquiry is incorrect suggestions, such as obtaining a partial worth when you do not have enough working parts to finish the whole work. There is a great factor that this was wrong, as well as it refers reasoning, not intuition. Reasoning permits you to comply with a step by step treatment that makes sense, and also when you make a wrong move, you are generally forced to either try to move forward and also correct the blunder, or try to go backward and do a backwards action.
Another instance is when the student does not comprehend an action of a procedure. These are both rational failings, and there is no other way around them. Also when you are stuck in a location that does not permit you to make any kind of step, such as a triangle, it is still crucial to comprehend why you are stuck, so that you can make a better relocation and go from the action you are stuck at to the next location.
With this in mind, the very best way to solve a stuck circumstance is to just take the progression, rather than trying to step. Both procedures are various in their strategy, yet they have some standard similarities. Nevertheless, when they are tried with each other, you can promptly tell which one is better at solving the issue, and also you can additionally tell which one is much more effective.
Let's discuss the very first instance, which connects to the Pinch Point math lab. This is not also complicated, so allow's first go over exactly how to start. Take the complying with procedure of affixing a part to a panel to be made use of as a body. This would certainly call for three measurements, as well as would certainly be something you would certainly need to attach as part of the panel.
Currently, you would have an extra measurement, however that does not mean that you can simply maintain that dimension as well as go from there. When you made your very first step, you can conveniently forget about the dimension, and then you would certainly need to go back and retrace your steps.
Nevertheless, rather than keeping in mind the additional measurement, you can use what is called a "mental faster way" to assist you keep in mind that extra dimension. As you make your first step, visualize on your own taking the dimension and attaching it to the component you wish to connect to, and after that see just how that makes you really feel when you repeat the procedure.
Visualisation is a very powerful technique, and is something that you ought to not skip over. Visualize what it would certainly feel like to in fact connect the part and be able to go from there, without the dimension.
Now, let's consider the second instance. Let's take the exact same process as previously, but now the student has to bear in mind that they are going to move back one action. If you tell them that they need to return one step, but then you remove the idea of having to move back one step, after that they will not know how to wage the issue, they won't recognize where to look for that action, as well as the procedure will be a mess.
Instead, make use of a mental shortcut like the mental representation to mentally reveal them that they are mosting likely to return one action. and also put them in a placement where they can move on from there. without having to think about the missing an action.
## Pay Me To Do Your Pinch Point Lab
" Pinch Point - Need Aid With a Math lab?" Unfortunately, several trainees have actually had a trouble understanding the ideas of linear Pinch Point. The good news is, there is a brand-new layout for straight Pinch Point that can be used to educate linear Pinch Point to pupils who struggle with this idea. Trainees can use the lab Aid Solution to help them discover new methods in straight Pinch Point without facing a hill of troubles as well as without needing to take an examination on their concepts.
The lab Aid Service was created in order to aid battling students as they relocate from college and also secondary school to the college and task market. Many trainees are unable to manage the stress of the discovering procedure as well as can have really little success in grasping the concepts of straight Pinch Point.
The lab Aid Service was established by the Educational Screening Service, that supplies a range of different online tests that students can take and also practice. The Test Help Solution has actually assisted many students boost their ratings as well as can help you improve your ratings as well. As students relocate from college as well as senior high school to the university as well as task market, the TTS will certainly assist make your students' shift easier.
There are a couple of various ways that you can make the most of the lab Assist Service. The main manner in which students utilize the lab Aid Solution is through the Answer Managers, which can aid trainees learn techniques in straight Pinch Point, which they can use to help them succeed in their programs.
There are a number of issues that trainees experience when they first utilize the lab Assist Solution. Students are commonly overwhelmed as well as do not understand how much time they will need to commit to the Solution. The Response Supervisors can assist the students review their idea discovering as well as help them to evaluate all of the material that they have actually currently learned in order to be gotten ready for their following program job.
The lab Assist Service functions similarly that a teacher carries out in terms of aiding students realize the concepts of direct Pinch Point. By providing your trainees with the tools that they need to discover the crucial principles of direct Pinch Point, you can make your trainees much more effective throughout their researches. As a matter of fact, the lab Assist Service is so efficient that several pupils have switched over from conventional mathematics class to the lab Aid Solution.
The Job Manager is developed to aid pupils manage their homework. The Job Manager can be set up to schedule how much time the pupil has offered to complete their assigned research. You can additionally establish a customized period, which is a great feature for pupils that have an active timetable or a very hectic secondary school. This feature can help trainees prevent feeling bewildered with mathematics tasks.
An additional useful function of the lab Help Service is the Student Aide. The Trainee Aide helps trainees handle their work and provides a location to post their homework. The Student Assistant is helpful for students that do not want to get bewildered with answering several questions.
As trainees get even more comfy with their tasks, they are encouraged to connect with the Task Supervisor as well as the Pupil Aide to get an on the internet support group. The online support group can aid pupils keep their focus as they address their projects.
Every one of the tasks for the lab Aid Solution are included in the bundle. Trainees can login as well as finish their designated work while having the student aid available behind-the-scenes to help them. The lab Aid Solution can be a wonderful help for your trainees as they start to browse the challenging university admissions and also work searching waters.
Trainees must be prepared to get used to their assignments as quickly as feasible in order to reach their primary objective of entering into the college. They need to strive sufficient to see results that will enable them to stroll on at the following degree of their studies. Obtaining made use of to the procedure of finishing their tasks is really important.
Students have the ability to find different ways to help them find out how to make use of the lab Aid Service. Learning exactly how to use the lab Aid Solution is vital to trainees' success in university and job application.
## Pay Someone To Take My Pinch Point Lab
Pinch Point is made use of in a great deal of colleges. Some instructors, however, do not utilize it very effectively or use it incorrectly. This can have an adverse effect on the trainee's understanding.
So, when assigning projects, use an excellent Pinch Point aid service to help you with each lab. These solutions give a range of practical services, including:
Projects may need a great deal of reviewing and looking on the computer. This is when using a help solution can be a great benefit. It enables you to get even more work done, enhance your understanding, and also stay clear of a great deal of anxiety.
These kinds of homework services are a fantastic way to begin collaborating with the most effective sort of help for your requirements. Pinch Point is among one of the most tough subjects to master for trainees. Working with a service, you can ensure that your needs are satisfied, you are shown appropriately, and you understand the product appropriately.
There are many ways that you can show yourself to function well with the class and achieve success. Utilize an appropriate Pinch Point help service to lead you and obtain the job done. Pinch Point is just one of the hardest classes to learn yet it can be quickly understood with the right aid.
Having a homework service also assists to boost the trainee's qualities. It allows you to include extra credit report along with increase your GPA. Getting extra credit is usually a huge benefit in several colleges.
Pupils that don't make the most of their Pinch Point class will end up continuing of the remainder of the class. The good news is that you can do it with a quick as well as easy service. So, if you wish to move ahead in your course, make use of an excellent assistance service. One thing to remember is that if you actually want to increase your grade degree, your program job requires to obtain done. As high as feasible, you need to comprehend as well as work with all your troubles. You can do this with a great aid service.
One benefit of having a homework solution is that you can assist yourself. If you do not feel great in your capability to do so, then a great tutor will be able to aid you. They will certainly have the ability to resolve the troubles you face and assist you recognize them in order to get a far better grade.
When you graduate from secondary school and also go into university, you will need to work hard in order to stay ahead of the other pupils. That implies that you will certainly require to work hard on your homework. Using an Pinch Point solution can aid you get it done.
Maintaining your grades up can be hard due to the fact that you usually require to examine a whole lot and also take a lot of examinations. You don't have time to work on your qualities alone. Having a good tutor can be a fantastic aid since they can aid you as well as your homework out.
An aid service can make it simpler for you to handle your Pinch Point course. In addition, you can find out more about on your own and also help you be successful. Find the most effective tutoring solution and also you will certainly be able to take your research study abilities to the next degree. | 2,474 | 12,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.975992 |
https://math.answers.com/basic-math/What_is_10_percent_off_7000_dollars | 1,721,831,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00778.warc.gz | 328,215,922 | 47,555 | 0
# What is 10 percent off 7000 dollars?
Updated: 4/28/2022
Wiki User
12y ago
10% of 7000 dollars is \$700 so 7000-700= \$6300
Wiki User
12y ago | 59 | 151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.774866 |
https://math.icalculator.info/arithmetic/decimal-number-system/video-tutorial.html | 1,675,351,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00644.warc.gz | 401,967,641 | 5,810 | # Decimal Number System and Other Numbering Systems
In addition to the video tutorial for Decimal Number System and Other Numbering Systems on this page, you can also access the following Arithmetic learning resources for Decimal Number System and Other Numbering Systems
Arithmetic Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
1.7Decimal Number System and Other Numbering Systems
In this Math video tutorial on Arithmetic we look into the field of Decimal Number System and Other Numbering Systems and cover the following points:
• What is the decimal number system? Why it is called "decimal"?
• How are the numbers written in the decimal system?
• What is the decomposed form of a number in the decimal system?
• What are the place values and classes of a number?
• What is the binary system of numbers? What are the commonalities/differences with the decimal system?
• What is the hexadecimal system of numbers? What are the commonalities/differences with the hexadecimal system of numbers.
• What other symbols we use in the hexadecimal system to represent digits?
• How do we use the basic operations in various number systems?
## Whats next?
Enjoy the "Decimal Number System and Other Numbering Systems" video tutorial? People who liked the "Decimal Number System and Other Numbering Systems" video tutorial found the following resources useful:
1. Video Tutorial Feedback. Helps other - Leave a rating for this video tutorial (see below)
2. Arithmetic Math tutorial: Decimal Number System and Other Numbering Systems. Read the Decimal Number System and Other Numbering Systems math tutorial and build your math knowledge of Arithmetic
3. Arithmetic Revision Notes: Decimal Number System and Other Numbering Systems. Print the notes so you can revise the key points covered in the math tutorial for Decimal Number System and Other Numbering Systems
4. Arithmetic Practice Questions: Decimal Number System and Other Numbering Systems. Test and improve your knowledge of Decimal Number System and Other Numbering Systems with example questins and answers
5. Check your calculations for Arithmetic questions with our excellent Arithmetic calculators which contain full equations and calculations clearly displayed line by line. See the Arithmetic Calculators by iCalculator™ below.
6. Continuing learning arithmetic - read our next math tutorial: Rounding and Significant Figures | 462 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-06 | latest | en | 0.811217 |
https://discourse.threejs.org/t/are-spherical-coordinates-flipped/58467 | 1,709,537,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00428.warc.gz | 210,663,610 | 9,897 | # Are Spherical coordinates flipped?
Q: Are Spherical coordinates theta and phi flipped in Vector3 source code?
Im walking through class Spherical
https://threejs.org/docs/?q=spher#api/en/math/Spherical
And getting the results back to a position vector from method setFromSpherical of which is setFromSphericalCoords
In my mini test in testing to get back Cartesian coords
And according to the wikipedia article and threejs’s notes
Theta Should be Up to Down
And Phi should be around
Now I know its in Z up axis. But I would assume from reading the source it’s converting it properly.
But when just sliding Theta I would expect up down, instead its reverse Phi is up down, Theta is around.
And when I dig through other papers is found this
Convert from spherical coordinates to rectangular coordinates
These equations are used to convert from spherical coordinates to rectangular coordinates.
• 𝑥 = ρ sin φ cosθ
• 𝑦 = ρ sin φ sinθ
• 𝑧 = ρ cos φ
compared to THREEs
``````const sinPhiRadius = Math.sin( phi ) * radius;
this.x = sinPhiRadius * Math.sin( theta );
this.y = Math.cos( phi ) * radius;
this.z = sinPhiRadius * Math.cos( theta );
``````
Maybe im just ignorant in this space still
It does not help that all discussions of this coordinate system talk of different orientations in each persons example instead of a unified mode. right hand left hand kdjfni3roiwjeff
2 Likes
You may be right? But also fixing it could break the entire thrinternet. Might be worth noting in the docs or something tho…
also like… greek letters for roll/pitch/yaw has always bugged me.
whoops oh!! hmmm not sure if i should try a bug post
In your link, z is the vertical axis. In 3js, y is the vertical axis. In my opinion this is the natural convention since every xy diagram uses y as the vertical axis. If z is added there is only room for this in the plain. Unless you exchange it for y. z is the depth value. In addition I prefer the spherical coordinates with the equator as reference and theta with -90 to +90 degrees instead of from the north pole 0 to south pole 180 degrees. But that is purely subjective.
On wiki the representation like in your link is described as inconsistent with the polar coordinates.
It depends heavily on the intended use. I also like to swap x and z because the camera is oriented in minus z by default. That looks like this, for example:
``````function SphericalToCartesian(r, theta, phi){
const x = r * Math.cos(theta) * Math.sin(phi);
const y = r * Math.sin(theta);
const z = r * Math.cos(theta) * Math.cos(phi);
return new THREE.Vector3(x, y, z);
}
// theta = 90
// +y -z phi = +/- 180
// | /
// | /
// phi = -90 -x ________|/________ +x phi = +90
// /|
// / |
// / |
// phi = 0 +z -y
// theta = -90
``````
As a physicist, I am very mathematically inclined and I wonder why people like to swap y with z as soon as it becomes three-dimensional. I personally find this confusing. I think it’s logical that the z axis is in the plane.
If you want theta and phi to be swapped, then suggestions would be the right topic.
But z and y will certainly not be swapped and I think that’s a good thing because in my opinion it’s good and natural as it is.
1 Like
Here’s a reason why I flipped the Y and Z axis: my game is a top-down game where the world is rendered in front of the camera rather than underneath it. This allows the world to just use X and Y coordinates (which I find more intuitive) and eliminates gimbal lock.
1 Like
Then all I would suggest is to reword the line
The poles (phi) are at the positive and negative y axis. The equator (theta) starts at positive z.
on the docs page three.js docs
So it’s more clear of the developer decision to flip.
Cause to me it feels like its being redundant for a good reason when done looking at the article.
That said, THREE is no stranger to breaking things to fix things in a next version and just giving a slight warning in the changelog
Yup… it’s a debate with 3 choices, but we all agree that X up is an abomination.
1 Like
I do not agree
X-up is perfectly normal to me. At least it is as normal as Y-up and Z-up. That’s why Suica can be set into any orientation or handedness:
https://boytchev.github.io/suica/examples/suica-orientation.html
As for spherical coordinates, their conversion into Cartesian can be done with different sets of equations – it depends on the direction of axis and on angles (where is their origin and direction). So, for me the coordinates are not flipped – it is just that there are several possible sets.
1 Like
Ok I’m switching to X up.
You are cruel. If you start using X-up, I will lose my uniqueness…
4 Likes
Hmm. ok… I’ll try W up.
5 Likes
Thats funny
2 Likes
3 Likes
You can tell it’s Friday
3 Likes
This is hurting my brain.
1 Like
It has 4 dimensions with the quaternions
Quaternion - Axis, Angle Visualization
Incidentally, the 3-dimensional space is only a special case of an n-dimensional vector space.
And then there is the curvature of space in the universe and things like that …
1 Like | 1,256 | 5,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.862327 |
https://math.answers.com/other-math/What_will_be_the_answer_if_you_have_1_square_and_in_it_36_squares_what_is_the_number_of_squares | 1,686,263,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00578.warc.gz | 446,240,172 | 51,106 | 0
# What will be the answer if you have 1 square and in it 36 squares what is the number of squares?
Wiki User
2010-12-13 15:02:44
You really should do your own homework - this is a question designed to make you analyse number patterns and devise a method to predict the answer that can be applied to grids of differing size.
If we start with a square cut into a 3x3 grid, we can count the nine single (1x1) squares in the grid, the one 3x3 square, and then four 2x2* squares, making a total of 14.
Try it out, then work your way up to 6x6 (a 36 square grid) by way of 4x4 and 5x5, looking to see how the grid's dimensions correlate to the number of varying-sized squares that can be counted.
As a tip- in a 6x6 grid, you will have one 6x6 square, thirty-six 1x1 squares, and how many 2x2, 3x3, 4x4, and 5x5 squares?
*The squares can overlap, obviously.
Wiki User
2010-12-13 15:02:44
Study guides
20 cards
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See all cards
3.8
2571 Reviews | 292 | 951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | longest | en | 0.933142 |
https://brainly.com/question/364871 | 1,485,067,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281353.56/warc/CC-MAIN-20170116095121-00529-ip-10-171-10-70.ec2.internal.warc.gz | 803,366,082 | 9,468 | 2015-03-21T15:25:56-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
A perfect square is any integer that is the square of another integer.
For instance 16=4*4 hence 16 is a perfect square, however 8 isn't one.
hi hippa iiiii'mmm baaackk!
i'm the same guy as CyberCerebrum! some person deleted my other account so i made a new one! i'm over the thing where i was doing caps on people. :)
thanks!
2015-03-21T15:26:38-04:00
This is a number you can square root and come out with a whole number. 4 is a perfect square because it can be square rooted to 2.
Another way to think about it is if you are trying to make squares composed of other little squares. If I want my big square to have four little squares on each side, I will have 16 little squares to make the square.
We can do the reverse by taking the big whole square and dividing it by the number of little square on one side. When we do so, we will come out with an answer that matches the number you were dividing with. :) I hope this didn't confuse you too much.
An AP Math student
soo hard to choose the best... | 332 | 1,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.953813 |
https://file.scirp.org/Html/17-7400823_19884.htm | 1,580,043,217,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251688806.91/warc/CC-MAIN-20200126104828-20200126134828-00405.warc.gz | 438,490,818 | 19,469 | Monty Hall Problem and the Principle of Equal Probability in Measurement Theory
Applied Mathematics
Vol. 3 No. 7 (2012) , Article ID: 19884 , 7 pages DOI:10.4236/am.2012.37117
Monty Hall Problem and the Principle of Equal Probability in Measurement Theory
Shiro Ishikawa
Department of Mathematics, Faculty of Science and Technology, Keio University, Yokohama, Japan
Email: ishikawa@math.keio.ac.jp
Received April 28, 2012; revised May 28, 2012; accepted June 5, 2012
Keywords: Linguistic Interpretation; Quantum and Classical Measurement Theory; Philosophy of Statistics; Fisher Maximum Likelihood Method; Bayes’ Theorem
ABSTRACT
In this paper, we study the principle of equal probability (i.e., unless we have sufficient reason to regard one possible case as more probable than another, we treat them as equally probable) in measurement theory (i.e., the theory of quantum mechanical world view), which is characterized as the linguistic turn of quantum mechanics with the Copenhagen interpretation. This turn from physics to language does not only realize the remarkable extension of quantum mechanics but also establish the method of science. Our study will be executed in the easy example of the Monty Hall problem. Although our argument is simple, we believe that it is worth pointing out the fact that the principle of equal probability can be, for the first time, clarified in measurement theory (based on the dualism) and not the conventional statistics (based on Kolmogorov’s probability theory).
1. Introduction
1.1. Monty Hall Problem
The Monty Hall problem is well-known and elementary. Also it is famous as the problem in which even great mathematician P. Erdös made a mistake (cf. [1]). The Monty Hall problem is as follows:
Problem 1 [Monty Hall problem 1]. You are on a game show and you are given the choice of three doors. Behind one door is a car, and behind the other two are goats. You choose, say, door 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that () the door 3 has a goat.
And further, He now gives you the choice of sticking with door 1 or switching to door 2? What should you do?
In the framework of measurement theory [2-12], we shall present two answers of this problem in Sections 3.1 and 4.2. Although this problem seems elementary, we assert that the complete understanding of the Monty Hall problem can not be acquired within Kolmogorov’s probability theory [13] but measurement theory (based on the dualism).
1.2. Overview: Measurement Theory
As emphasized in refs. [7,8], measurement theory (or in short, MT) is, by a linguistic turn of quantum mechanics (cf. Figure 1: ③ later), constructed as the scientific theory formulated in a certain C*-algebra A (i.e., a norm closed subalgebra in the operator algebra composed of all bounded operators on a Hilbert space H, cf. [14,15]). MT is composed of two theories (i.e., pure measurement theory (or, in short, PMT] and statistical measurement theory (or, in short, SMT). That is, it has the following structure:
(A) MT (measurement theory)
where Axiom 2 is common in PMT and SMT. For completeness, note that measurement theory (A) (i.e., (A1) and (A2)) is not physics but a kind of language based on “the (quantum) mechanical world view”. As seen in [9], note that MT gives a foundation to statistics. That is, roughly speaking(B) it may be understandable to consider that PMT and
Figure 1. The development of the world views from our standing point. For the explanation of (①-⑧), see [8,10].
SMT is related to Fisher’s statistics and Bayesian statistics respectively.
Also, for the position of MT in science, see Figure 1, which was precisely explained in [8,10].
When, the C*-algebra composed of all compact operators on a Hilbert space H, the (A) is called quantum measurement theory (or, quantum system theory), which can be regarded as the linguistic aspect of quantum mechanics. Also, when A is commutative (that is, when A is characterized by, the C*-algebra composed of all continuous complex-valued functions vanishing at infinity on a locally compact Hausdorff space (cf. [16])), the (A) is called classical measurement theory. Thus, we have the following classification:
(C)
The purpose of this paper is to clarify the Monty Hall problem in the classical PMT and classical SMT.
2. Classical Measurement Theory (Axioms and Interpretation)
2.1. Mathematical Preparations
Since our concern is the Monty Hall problem, we devote ourselves to classical MT in (C). Throughout this paper, we assume that is a compact Hausdorff space. Thus, we can put, which is defined by a Banach space (or precisely, a commutative C*-algebra) composed of all continuous complex-valued functions on a compact Hausdorff space, where its norm is defined by. Let be the dual Banach space of. That is, is a continuous linear functional on, and the norm is defined by such that. The bi-linear functional is also denoted by, or in short.
Define the mixed state such that and for all such that. And put
Also, for each, define the pure state
such that
. And put
which is called a state space. Note, by the Riesz theorem (cf. [16]), that is a signed measure on and is a measure on such that. Also, it is clear that is a point measure at, where. This implies that the state space can be also identified with (called a spectrum space or simply, spectrum) such as
(1)
Also, note that is unital, i.e., it has the identity I (or precisely,), since we assume that is compact.
According to the noted idea (cf. [17]) in quantum mechanics, an observable in is defined as follows:
(D1) [Field] X is a set, , the power set of X) is a field of X, that is, “”, “”.
(D2) [Additivity] F is a mapping from to satisfying: 1): for every, is a non-negative element in such that, 2): and, where 0 and I is the 0-element and the identity in respectively. 3): for any, such that, it holds that.
For the more precise argument (such as countably additivity, etc.), see [7,9].
2.2. Classical PMT in (A1)
In this section we shall explain classical PMT in (A1).
With any system S, a commutative C*-algebra can be associated in which the measurement theory (A) of that system can be formulated. A state of the system S is represented by an element and an observable is represented by an observable in. Also, the measurement of the observable O for the system S with the state is denoted by or more precisely, . An observer can obtain a measured value by the measurement .
The AxiomP 1 presented below is a kind of mathematical generalization of Born’s probabilistic interpretation of quantum mechanics. And thus, it is a statement without reality.
AxiomP 1 [Measurement]. The probability that a measured value obtained by the measurement belongs to a set is given by.
Next, we explain Axiom 2 in (A). Let be a tree, i.e., a partial ordered set such that “and” implies “or” In this paper, we assume that T is finite. Also, assume that there exists an element, called the root of T, such that holds. Put. The family is called a causal relation (due to the Heisenberg picture), if it satisfies the following conditions (E1) and (E2).
(E1) With each, a C*-algebra is associated.
(E2) For every, a Markov operator is defined (i.e., , ). And it satisfies that holds for any,.
The family of dual operators
is called a dual causal relation (due to the Schrödinger picture). When
holds for any, the causal relation is said to be deterministic.
Here, Axiom 2 in the measurement theory (A) is presented as follows:
Axiom 2 [Causality]. The causality is represented by a causal relation
.
For the further argument (i.e., the W*-algebraic formulation) of measurement theory, see Appendix in [7].
2.3. Classical SMT in (A2)
It is usual to consider that we do not know the state when we take a measurement. That is because we usually take a measurement in order to know the state. Thus, when we want to emphasize that we do not know the the state, is denoted by. Also, when we know the distribution of the unknown state, the is denoted by.
The AxiomS 1 presented below is a kind of mathematical generalization of AxiomP 1.
AxiomS 1 [Statistical measurement] The probability that a measured value obtained by the measurement belongs to a set is given by
.
Remark 1. Note that two statistical measurements and can not be distinguished before measurements. In this sense, we consider that, even if, we can assume that
(2)
2.4. Linguistic Interpretation
Next, we have to answer how to use the above axioms as follows. That is, we present the following linguistic interpretation (F) [= (F1) – (F3)], which is characterized as a kind of linguistic turn of so-called Copenhagen interpretation (cf. [7,8]). That is, we propose:
(F1) Consider the dualism composed of “observer” and “system (= measuring object)”. And therefore, “observer” and “system” must be absolutely separated.
(F2) Only one measurement is permitted. And thus, the state after a measurement is meaningless since it can not be measured any longer. Also, the causality should be assumed only in the side of system, however, a state never moves. Thus, the Heisenberg picture should be adopted.
(F3) Also, the observer does not have the space-time. Thus, the question: “When and where is a measured value obtained?” is out of measurement theory, and so on. This interpretation is, of course, common to both PMT and SMT.
Remark 2. Note that quantum mechanics has many interpretations (i.e., several Copenhagen interpretation, many worlds interpretation, statistical interpretation, etc.). On the other hand, we believe that the interpretation of measurement theory (A) is uniquely determined as in the above. This is our main reason to propose the linguistic interpretation of quantum mechanics. We believe that this uniqueness is essential to the justification of Heisenberg’s uncertainty principle (cf. [10,18]).
2.5. Preliminary Fundamental Theorems
We have the following two fundamental theorems in measurement theory:
Theorem 1 [Fisher’s maximum likelihood method (cf. [9])]. Assume that a measured value obtained by a measurement belongs to . Then, there is a reason to infer that the unknown state is equal to, where is defined by
Theorem 2 [Bayes’ method (cf. [9])]. Assume that a measured value obtained by a statistical measurement
belongs to.
Then, there is a reason to infer that the posterior state (i.e., the mixed state after the measurement) is equal to vpost, which is defined by
The above two theorems are, of course, the most fundamental in statistics. Thus, if we believe in Figure 1, we can answer to the following problem (cf. [4,9]):
(G) What is statistics? Or, where is statistics in science? which is certainly the most essential problem in the philosophy of statistics.
3. The First Answer to Monty Hall Problem
3.1. Fisher’s Method (The First Answer)
In this section, we present the first answer to Problem 1 (Monty-Hall problem) in classical PMT. Put
with the discrete topology. Assume that each state means
(3)
Define the observable in such that
(4)
where it is also possible to assume that,. Thus we have a measurement, which should be regarded as the measurement theoretical representation of the measurement that you say “door 1”. Here, we assume that
1) “measured value is obtained by the measurementThe host says “Door 1 has a goat”;
2) “measured value is obtained by the measurementThe host says “Door 1 has a goat”;
3) “measured value is obtained by the measurementThe host says “Door 1 has a goat”.
Recall that, in Problem 1, the host said “Door 3 has a goat”. This implies that you get the measured value “3”
by the measurement. Therefore, Theorem 1 (Fisher’s maximum likelihood method) says that you should pick door number 2. That is because we see that
(5)
and thus, there is a reason to infer that. Thus, you should switch to door 2. This is the first answer to Problem 1 (the Monty-Hall problem 1).
3.2. Bayes’ Method (Answer to Modified Monty Hall Problem 2)
In the sense mentioned in Remark 3 later, the following modified Monty Hall problem (Problem 2) is completely different from Problem 1 (the Monty Hall problem 1). However, it is worth examining Problem 2 for the better understanding of Problem 3 later.
Problem 2 [Modified Monty Hall problem 2]. Suppose you are on a game show, and you are given the choice of three doors (i.e., “number 1”, “number 2”, “number 3”). Behind one door is a car, behind the others, goats. You pick a door, say number 1. Then, the host, who set a car behind a certain door, says
(#1) the car was set behind the door decided by the cast of the distorted dice. That is, the host set the car behind the k-th door (i.e., “number k”) with probability pk (or, weight such that,).
And further, the host says, for example() the door 3 has a goat.
He says to you, “Do you want to pick door number 2?” Is it to your advantage to switch your choice of doors?
In what follows we study this problem. Let and be as in Section 3.1. Under the hypothesis (#1), define the mixed state such that:
(6)
Thus we have a statistical measurement
. Note that
1) “measured value is obtained by the statistical measurementThe host says “Door 1 has a goat”;
2) “measured value is obtained by the statistical measurementThe host says “Door 2 has a goat”;
3) “measured value is obtained by the statistical measurementThe host says “Door 1 has a goat”.
Here, assume that, by the statistical measurement , you obtain a measured value 3which corresponds to the fact that the host said “Door 3 has a goat”. Then, Theorem 2 (Bayes’ theorem) says that the posterior state is given by
(7)
That is,
(8)
Particularly, we see that (H) if, then it holds that , , , and thus, you should pick Door 2.
Remark 3. The difference between Problem 1 and Problem 2 should be remarked. Since the (#1) in Problem 2 is the information from the host to you, Problem 1 and Problem 2 are completely different. Although the above (H) may be generally regarded as the proper answer of the Monty Hall problem, we do not admit that the (H) is proper. That is, we consider that the (H) is not the second answer to the Monty Hall problem.
4. The Second Answer to Monty Hall Problem
In this section, we shall present the second answer. However, before it, we have to prepare the principle of equal probability (i.e., unless we have sufficient reason to regard one possible case as more probable than another, we treat them as equally probable). For completeness, note that measurement theory urges us to use only Axioms 1 and 2.
4.1. The Principle of Equal Probability
Put with the discrete topology. And consider any observable in.
Define the bijection such that
and define the observable in such that
where and
.
Let be a non-negative real number such that.
(I) For example, fix a state. And, by the cast of the distorted dice, you choose an observable with probability pk. And further, you take a measurement
.
Here, we can easily see that the probability that a measured value obtained by the measurement (I) belongs to is given by
(9)
which is equal to. This implies that the measurement (I) is equivalent to a statistical measurement:
.
Note that the (9) depends on the state. Thus, we can not calculate the (9) such as the (8).
However, if it holds that, we see that is independent of the choice of the state. Thus, putting, we see that the measurement (I) is equivalent to the statistical measurement, which is also equivalent to (from the formula (2) in Remark 1).
Thus, under the above notation, we have the following theorem.
Theorem 3 [The principle of equal probability (i.e., the equal probability of selection)]. If , the measurement (I) is independent of the choice of the state. Hence, the (I) is equivalent to a statistical measurement
.
It should be noted that the principle of equal probability is not “principle” but “theorem” in measurement theory.
Remark 4. This theorem was also discussed in [5,6], where we missed the formula (2) in Remark 1. Thus, the argument in [5,6] was too abstract. And thus, it might be regarded as ambiguous and vague. In fact, we must admit that the explanation in [5,6] is not yet accepted generally. Therefore, we recommend readers to read [5,6] after the understanding of the concrete explanation (I) in the linguistic interpretation (F). Also, note that Theorem 3 is independent of Axiom 2. And further, for the principle of equal (a priori) probabilities in equilibrium statistical mechanics, see [11], in which how to use measurement theory (and thus statistics) in statistical mechanics is explained.
4.2. The Second Answer to Monty Hall Problem (i.e., Modified Monty Hall Problem 3)
As an application of Theorem 3, we consider the following modified Monty-Hall problem:
Problem 3 [Modified Monty Hall problem 3]. Suppose you are on a game show, and you are given the choice of three doors (i.e., “number 1”, “number 2”, “number 3”). Behind one door is a car, behind the others, goats.
(#2) You choose a door by the cast of the fair dice, i.e., with probability 1/3.
According to the rule (#2), you pick a door, say number 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that
() the door 3 has a goat.
He says to you, “Do you want to pick door number 2?” Is it to your advantage to switch your choice of doors?
[Answer]. Consider and O1 as in Section 3.1. Then, Theorem 3 says that the answer of Problem 3 is the same as the (H). Thus, you should pick the door 2.
Remark 5. The difference between the (#1) in Problem 2 and the (#2) in Problem 3 is clear in the dualism (F). The former is host’s selection, but the latter is your selection (i.e., observer’s selection). That is, in Problem 3, the information from host to you is only the (). This situation is the same as that of Problem 1. In this sense, we think that Problems 1 and 3 are similar. That is, we can conclude that Problem 1 [resp. Problem 3] is the Monty Hall problem in PMT [resp. SMT]. Also, our recent report [19] will promote a better understanding of measurement theory.
5. Conclusions
In the conventional statistics based on Kolmogorov’s probability theory, Problem 3 may be unconsciously confused with Problem 2. On the other hand, as mentioned in Remark 5, the difference between Problems 2 and 3 can be clearly described in measurement theory (based on the dualism (F)). This is the merit of measurement theory.
What we executed in this paper may be merely the translation from “ordinary language” to “scientific language”, that is,
We believe that this translation is just “the mechanical world view” or “the method of science” (at least, science in the series L of Figure 1). That is, ordinary science (at least, its basic statements) should be described in terms of measurement theory. For example, for the translation of equilibrium statistical mechanics and the Zeno’s paradoxes, see [11] and [12] respectively. Probably, we refrained from the publication of [12], if we were not sure of “MT = the method of science (or the form of scientific thinking)”.
In this paper (as well as [9]), we showed one of advantages of the measurement theoretical foundation of statistics through the examination of the Monty Hall problem. Also, recall that measurement theory possesses a great power to answer to the problem (G). However, our methodology should be tested from various points of view, because the classic statistics methodology (based on Kolmogorov’s probability theory) can be good applied in many fields. We hope that our approach will be examined from various view points.
REFERENCES
1. P. Hoffman, “The Man Who Loved Only Numbers, the Story of Paul Erdös and the Search for Mathematical Truth,” Hyperion, New York, 1998,
2. S. Ishikawa, “Fuzzy Inferences by Algebraic Method,” Fuzzy Sets and Systems, Vol. 87, No. 2, 1997, pp. 181-200. doi:10.1016/S0165-0114(96)00035-8
3. S. Ishikawa, “A Quantum Mechanical Approach to Fuzzy Theory,” Fuzzy Sets and Systems, Vol. 90, No. 3, 1997, pp. 277-306. doi:10.1016/S0165-0114(96)00114-5
4. S. Ishikawa, “Statistics in Measurements,” Fuzzy Sets and Systems, Vol. 116, No. 2, 2000, pp. 141-154. doi:10.1016/S0165-0114(98)00280-2
5. S. Ishikawa, “Mathematical Foundations of Measurement Theory,” Keio University Press Inc., 2006. http://www.keio-up.co.jp/kup/mfomt/
6. S. Ishikawa, “Monty Hall Problem in Unintentional Random Measurements,” Far East Journal of Dynamical Systems, Vol. 3, No. 2, 2009, pp. 165-181.
7. S. Ishikawa, “A New Interpretation of Quantum Mechanics,” Journal of Quantum Information Science, Vol. 1, No. 2, 2011, pp. 35-42. doi:10.4236/jqis.2011.12005
8. S. Ishikawa, “Quantum Mechanics and the Philosophy of Language: Reconsideration of Traditional Philosophies,” Journal of Quantum Information Science, Vol. 2, No. 1, 2012, pp. 2-9. doi:10.4236/jqis.2012.21002
9. S. Ishikawa, “A Measurement Theoretical Foundation of Statistics,” Applied Mathematics, Vol. 3, No. 3, 2012, pp. 183-192.
10. S. Ishikawa, “The Linguistic Interpretation of Quantum Mechanics,” 2012. http://arxiv.org/pdf/1204.3892.pdf
11. S. Ishikawa, “Ergodic Hypothesis and Equilibrium Statistical Mechanics in the Quantum Mechanical World View,” World Journal of Mechanics, Vol. 2, No. 2, 2012, pp. 125-130. doi:10.4236/wjm.2012.22014
12. S. Ishikawa, “Zeno’s Paradoxes in the Mechanical World View,” 2012. http://arxiv.org/pdf/1205.1290.pdf
13. A. Kolmogorov, “Foundations of the Theory of Probability (Translation),” Chelsea Pub Co. Second Edition, New York, 1960.
14. G. J. Murphy, “C*-Algebras and Operator Theory,” Academic Press, Boston, 1990.
15. J. von Neumann, “Mathematical Foundations of Quantum Mechanics,” Springer Verlag, Berlin, 1932.
16. K. Yosida, “Functional Analysis,” 6th Edition, SpringerVerlag, Berlin, 1980.
17. E. B. Davies, “Quantum Theory of Open Systems,” Academic Press, London, 1976.
18. S. Ishikawa, “Uncertainty Relation in Simultaneous Measurements for Arbitrary Observables,” Reports on Mathematical Physics, Vol. 29, No. 3, 1991, pp. 257-273. doi:10.1016/0034-4877(91)90046-P
19. S. Ishikawa, “What Is Statistics? The Answer by Quantum Language,” arXiv:1207.0407v1 [physics.data-an], 2012. http://arxiv.org/abs/1207.0407v1 | 5,615 | 22,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-05 | longest | en | 0.919268 |
https://genstat.kb.vsni.co.uk/knowledge-base/chipermt/ | 1,718,642,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00272.warc.gz | 235,609,326 | 11,975 | 1. Home
2. CHIPERMTEST procedure
CHIPERMTEST procedure
Performs a random permutation test for a two-dimensional contingency table (L.H. Schmitt, M.C. Hannah & S.J. Welham).
Options
`PRINT` = string tokens Output required (`summary`, `observed`, `expected`); default `summ` What to plot (`histogram`); default `hist` Method for calculating chi-square (`pearson`, `maximumlikelihood`); default `pear` Number of permutations to make; default 999 Seed for the random number generator used to make the permutations; default 0 continues from the previous generation or (if none) initializes the seed automatically
Parameters
`DATA` = tables Table containing observed data Saves the observed chi-square value Saves the chi-square values from the permuted data sets Saves the probability value from the test
Description
The `CHIPERMTEST` procedure uses a permutation test to calculate the significance probability for a chi-square test of the independence of rows and columns in a two-dimensional contingency table. This provides a nonparametric alternative to the more usual chi-square test of independence (see the `CHISQUARE` procedure). The usual test depends upon the fact that the distribution of its so-called “chi-square” test statistic becomes a chi-square distribution as the numbers of observations become infinite. (Technically, we would say that the distribution is asymptotically chi-square.) However, the test is unreliable with smaller numbers, especially when the expected number in any cell of the table is less than five.
The permutation test simulates the random distribution of table values that may occur in tables that have the same overall distribution of numbers over the columns, and over the rows, as in the original table. We can assess the significance of the chi-square statistic that we can calculate from the observed table, by seeing where it lies in the distribution of statistics that we obtain from the permuted data.
The `NTIMES` option specifies how many permutations are done (default 999). The `SEED` option supplies the seed that is used in the `RANDOMIZE` directive to generate the permutations. The default of zero continues the existing sequence of random numbers if `RANDOMIZE` has already been used in the current Genstat job. If `RANDOMIZE` has not yet been used, Genstat picks a seed at random.
The `DATA` parameter supplies the observed data values, in a table with two classifying factors. The `CHISQUARE` can save the chi-square statistic calculated from the `DATA` table (in a scalar). The `CHIPERMUTED` parameter can save the chi-square statistics calculated from the permuted data sets (in a variate), and the `PROBABILITY `parameter can save the significance probability from the permutation test (in a scalar).
The `PRINT` option controls the output, with the following settings:
`summary` prints a summary, containing the chi-square statistic, the minimum and maximum statistics calculated from the permuted data sets, and the probability (default); prints the `DATA` table; and prints the expected values for tables with the same overall distribution of numbers over rows and over columns, but no interaction between the row and column factors (i.e. in a table where the rows and columns are independent).
By default, `CHIPERMTEST` plots a histogram showing the distribution of statistics obtained from the permuted data sets, with the chi-square statistic from the observed data superimposed as a vertical line. You can suppress this by setting option `PLOT=*`.
The `METHOD` option controls how the chi-square statistic is calculated. The default is to use the usual Pearson approximation (see the Method section), but you can set `METHOD=likelihood` to calculate it by maximum likelihood instead (using the Genstat facilities for generalized linear models).
Options: `PRINT`, `PLOT`, `METHOD`, `NTIMES`, `SEED`.
Parameters: `DATA`, `CHISQUARE`, `CHIPERMUTED`, `PROBABILITY`.
Method
The Pearson statistic is calculated as
chi-square = sum( (oe) × (oe) / e ),
where o = observed, and e = expected. The alternative, maximum-likelihood method takes the deviance from fitting a generalized linear model with a log link and a Poisson distribution.
The permutations are constructed using the method Roff & Bentzen (1989).
Reference
Roff, D.A. & Bentzen, P. (1989). The statistical analysis of mitochondrial DNA polymorphisms: χ2 and the problem of small samples. Mol. Biol. Evol., 6, 539-545.
Procedure: `CHISQUARE`.
Commands for: Basic and nonparametric statistics, Regression analysis.
Example
```CAPTION 'CHIPERMTEST example','Data from Roff & Bentzen (1988)';\
STYLE=meta,plain
FACTOR [LEVELS=14; LABELS=!t(A,B,C,D,E,F,G,H,I,J,K,L,M,N)] River
FACTOR [LEVELS=2] Gene2
TABLE [CLASSIFICATION=River,Gene2; VALUES=\
13,16,8,10,8,5,11,6,9,11,12,10,11,8,\
17,4,10,1,12,7,6,4,12,5,16,5,7,0] B2
CHISQUARE B2
CHIPERMTEST [PRINT=summary,observed,expected; SEED=301453] B2
```
Updated on March 8, 2019 | 1,159 | 4,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.791814 |
http://www.jiskha.com/display.cgi?id=1324172125 | 1,462,384,428,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860123840.94/warc/CC-MAIN-20160428161523-00220-ip-10-239-7-51.ec2.internal.warc.gz | 587,834,509 | 3,448 | Wednesday
May 4, 2016
Homework Help: math
Posted by latesha on Saturday, December 17, 2011 at 8:35pm.
Solve the following equation.
4m^2 = 2 − 7m
m = 1 (smaller value)
m = 2 (larger value) | 75 | 191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-18 | longest | en | 0.941552 |
https://scifi.stackexchange.com/questions/227448/short-story-about-friends-talking-about-parallel-universes-and-probability | 1,718,890,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00079.warc.gz | 435,181,318 | 39,193 | # Short story about friends talking about parallel universes and probability
Trying to find a short story where two people talk about how the "many worlds" theories (or possibly the idea that if the universe is infinite everything will eventually happen) with the example of a monster suddenly appearing. They go over a hill and are eaten by a monster that suddenly appeared.
Trying to find a short story
"The Other Tiger", a short story by Arthur C. Clarke, first published in Fantastic Universe, June-July 1953, available at the Internet Archive. You may have read it in one of these compilations.
where two people talk about how the "many worlds" theories (or possibly the idea that if the universe is infinite everything will eventually happen)
It's the "infinite universe", not the "many worlds" theory:
"Well, let's be perfectly logical and see where it gets us. Our only assumption, remember, is that the universe is infinite."
"Right. Personally I don't see what else it can be."
"Very well. That means there must be an infinite number of stars and planets. Therefore, by the laws of chance, every possible event must occur not merely once but an infinite number of times. Correct?"
"I suppose so."
"Then there must be an infinite number of worlds exactly like Earth, each with an Arnold and Webb on it, walking up this hill just as we are doing now, saying these same words."
with the example of a monster suddenly appearing.
"Yet in some universe those — what shall I call them ? — twins of ours will walk around that corner and meet anything, absolutely anything that imagination can conceive. For as I said at the beginning, if the cosmos is infinite, then all possibilities must arise."
"So it’s possible," said Arnold, with a laugh that was not quite as light as he had intended, "that we may walk into a tiger or something equally unpleasant."
They go over a hill and are eaten by a monster that suddenly appeared.
Yet of course it was not totally inconceivable that during the night the rain-sodden hillside had caved inward to reveal an ominous cleft leading down into the subterranean world. As for what had laboriously climbed up that cleft, drawn towards the unknown light of day — well, it was really no more unlikely than the giant squid, the boa-constrictor or the feral lizards of the Jurassic jungle. It had strained the laws of zoological probability but not to the breaking-point.
Webb had spoken the truth. In an infinite cosmos everything must happen somewhere — including their singularly bad luck. For it was hungry — very hungry — and a tiger or a man would have been a small yet acceptable morsel to any one of its half dozen gaping mouths.
• That's it, THANK YOU!!! Commented Feb 16, 2020 at 4:31
• You're welcome! Commented Feb 16, 2020 at 4:45 | 609 | 2,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.974501 |
http://www.answers.com/Q/What_is_the_algorithm_of_generating_unique_random_numbers_like_in_mobile_recharge_coupons | 1,508,284,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00815.warc.gz | 404,156,173 | 47,985 | What would you like to do?
# What is the algorithm of generating unique random numbers like in mobile recharge coupons?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
luhn algo
1 person found this useful
Thanks for the feedback!
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You need to store the last number used somewhere - for example, in a variable that doesn't disappear after closing the function. Then, every time the function "Serial Number" | 805 | 3,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.830567 |
https://discourse.julialang.org/t/memory-allocation-usage-high-with-power-compared-with-x-x/1239 | 1,550,309,989,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480240.25/warc/CC-MAIN-20190216085312-20190216111312-00095.warc.gz | 558,749,864 | 7,614 | # Memory allocation usage high with power (^) compared with x*x
#1
I’m running v0.5 memory allocation analysis of the following code and cannot understand the allocations occurring on the line with ^ compared with using *
``````function myTest(n)
a=zeros(Float64,100)
for i=1:100
a[i]=2.5^2.
end
end
function myTest1(n)
a=zeros(Float64,100)
for i=1:100
a[i]=2.5*2.5
end
end
myTest(1)
myTest1(1)
Profile.clear_malloc_data()
myTest(1)
myTest1(1)
``````
Memory allocation output as follows:
`````` - function myTest(n)
- a=zeros(Float64,100)
1440 for i=1:100
6400 a[i]=2.5^2.
- end
- end
-
- function myTest1(n)
- a=zeros(Float64,100)
1440 for i=1:100
0 a[i]=2.5*2.5
- end
- end
-
- myTest(1)
- myTest1(1)
- Profile.clear_malloc_data()
-
- myTest(1)
- myTest1(1)
-
-``````
#2
Further information. Output given from memory allocation was with --inline=no. When not including the inline flag there is no allocation on the ^ line. So interested to know what is the correct answer?
Using v0.5.
#3
I looked at the @code_llvm, and in the first function I see
`%17 = call double @"julia_^_70720"(double 2.500000e+00, double 2.000000e+00)`
and in the second
`store double 6.250000e+00, double* %23, align 8`
For some reason the product operation gets optimized into a constant number, but the power doesn’t here.
I would note if you left out the dot on the 2., so it was just 2.5^2, then they become the same. The optimization seems to work for integer powers but not float powers.
#4
This is due to the fact that constants inline, and function calls whose values can be determined by type information will inline the values as well (since then the values are determined not at runtime, but at compile time), but function calls whose output depends on the input values cannot inline. So Julia actually will compile away:
``````a=zeros(Float64,100)
for i=1:100
a[i]=2.5*2.5
end
``````
to be the value since that value can be computed during compilation, whereas
``````for i=1:100
a[i]=2.5^2.
end
``````
is not computable at compilation time.
So yes, what you’re actually measuring here isn’t `*` vs `^`, rather it’s compiler optimizations on constant values. This is why you should always double check microbenchmarks to make sure you don’t have everything constant in a way the compiler can just “solve” it. For example, square using a random number instead of `2.5`.
As an extra little note, `^` actually has a branch for low powers to do iteration. Here’s the source for it: https://github.com/JuliaLang/julia/blob/master/base/intfuncs.jl#L159
So other than the branch (the conditional: in very very performant code you can sometimes measure a difference due to branch predictions), the code exactly just multiplying when the power is 2.
#5
Thanks. My main concern is the memory allocation. I’m running more complex calculation with powers and the memory allocation is strangely large on the line - when I expected no memory allocation.
#6
That sounds like a type-instability.
#7
Heres the equation
`````` s.Sf[i]= g^2*(2.*pi)^-4* s.f[i]^-5* exp(-5/4.*(s.f[i]/fp).^-4)*
Gamma^(exp(-1 * (((s.f[i]-fp)^2) /
(2*(SigmaA*(s.f[i]<fp) + SigmaB*(s.f[i]>=fp))^2*fp^2)) ))
``````
What do you think?
#9
``````s.Sf[i]= g^2*(2.*pi)^-4* s.f[i]^-5* exp(-5/4.*(s.f[i]/fp).^-4)*
Gamma^(exp(-1 * (((s.f[i]-fp)^2) /
(2*(SigmaA*(s.f[i]<fp) + SigmaB*(s.f[i]>=fp))^2*fp^2)) ))
``````
Are any of these arrays? e.g. why are you using `.^` rather than `^`? “Vectorized” operations on arrays generally allocate new arrays for their results.
It could also be a type instability, as someone mentioned, e.g. maybe you’ve changed the type of `fp` or `Gamma` somewhere (or they are non-`const` globals, etc.). `@code_warntype` is useful here to detect whether type inference has succeeded.
#10
Only s.f is any array which I’m looping over. I’ve run code_warntype - see below on the full function. It seems to be ok. I’ve also pulled out the boolean selectors and used an if statement on the SigmaA and SigmaB, but that didn’t change anything.
Should a line like this have any memory allocations?
Full loop
`````` for i=1:nf
if s.f[i]<fp
sigma=SigmaA
else
sigma=SigmaB
end
s.Sf[i]= g^2*(2.*pi)^-4* s.f[i]^-5* exp(-5/4.*(s.f[i]/fp).^-4)*
Gamma^(exp(-1 * (((s.f[i]-fp)^2) /
(2*sigma^2*fp^2)) ))
end
``````
``````Variables:
#self#::MetoceanTools.#makeSpecJONSWAP
inParm::Array{Float64,1}
Hs::Float64
Tp::Float64
Gamma::Float64
SigmaA::Float64
SigmaB::Float64
#temp#@_8::Int64
s::MetoceanTools.waveSpec
g::Float64
fp::Float64
nf::Int64
sumS::Float64
alpha::Float64
#temp#@_15::Int64
i@_16::Int64
#temp#@_17::Int64
i@_18::Int64
#temp#@_19::Int64
sigma::Float64
i@_21::Int64
``````
.mem output of loop
`````` 0 nf=length(s.f)
266831936 s.Sf=zeros(nf)
0 for i=1:nf
0 if s.f[i]<fp
0 sigma=SigmaA
- else
0 sigma=SigmaB
- end
2050585152 s.Sf[i]= g^2*(2.*pi)^-4* s.f[i]^-5* exp(-5/4.*(s.f[i]/fp).^-4)*
- Gamma^(exp(-1 * (((s.f[i]-fp)^2) /
- (2*sigma^2*fp^2)) ))
- end
``````
NB. All happening in functions - not in global scope
#11
What’s this type? Could you please share its definition? Is there a reason why it’s not parameteric? I would assume it would be parameterized in order to be strictly typed, and the typing of its fields might be the issue.
#12
You should `.` broadcast this as much as possible, like `exp.()` instead of `exp` so that way it will fuse (and it will work on v0.6). On v0.5, not all of these will fuse though, so this is a case where moving to v0.6 might give a performance boost (unless you use things like (^).(x,y) a bunch of places, which just looks nasty)
#13
Type defintion
``````type waveSpec
f::Array{Float64,1}
df::Array{Float64,1}
th::Array{Float64,1}
dth::Array{Float64,1}
Sf::Array{Float64,1}
Sth::Array{Float64,1}
S::Array{Float64,2}
function waveSpec()
t=new()
t.f=makeWaveSpecFreq()
t.df=makeWaveSpecDelFreq(t.f)
return t
end
end
``````
Constructors
``````function makeWaveSpecFreq()
f=[1/x for x=30:-0.05:2]
return f
end
function makeWaveSpecDelFreq(f::Array{Float64,1})
df=copy(f)
nf=length(f)
df[1]=f[2]-f[1]
df[2:(nf-1)]=( f[3:nf]-f[2:(nf-1)] )/2 +
( f[2:(nf-1)]-f[1:(nf-2)] )/2
df[nf]=f[nf]-f[nf-1]
return df
end
``````
#14
I see. For future reference I would make this a little less strict:
``````type waveSpec{T<:Number}
f::Vector{T}
df::Vector{T}
th::Vector{T}
dth::Vector{T}
Sf::Vector{T}
Sth::Vector{T}
S::Matrix{T}
function waveSpec()
t=new()
t.f=makeWaveSpecFreq()
t.df=makeWaveSpecDelFreq(t.f)
return t
end
end
``````
since if you don’t restrict to `Float64` everywhere, then your code will naturally work with other numbers (arbitrary precision, etc). That’s why I found it odd there was no type parameter on it when it was holding numbers.
But okay, this is strictly typed so there is no instability here. It must be the temporary allocations due to broadcasting then.
#15
It sounds like this is not an array operation (the computations of `s.Sf[i]` are purely scalar), in which case admonitions about `broadcast` are inapplicable.
#17
`^` between scalars does not cause any allocations in normal use. It happens to allocate when you turn off inlining but why do you worry about that?
#18
Oh, didn’t even realize this was part of it.
Compiler inlining is an optimization which will pretty much always be done to stop this exact problem. So you should profile with inlining on because that’s what users will see.
#19
Ok. That is clear. I’m running with version v0.5 and with inlining turned on its causes memory allocations to pop up on lines that are not correct (this came up in one of my previous posts). So I guess I have to wait until I can get a version 0.6 working for which I understand this issue has been corrected. | 2,429 | 7,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | latest | en | 0.796477 |
http://www.solutioninn.com/acorner-construction-lot-has-the-shape-of-a-right-triangle | 1,506,049,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688158.27/warc/CC-MAIN-20170922022225-20170922042225-00099.warc.gz | 579,654,558 | 7,232 | # Question: Acorner construction lot has the shape of a right triangle
Acorner construction lot has the shape of a right triangle. If the two sides perpendicular to each other are 37 m long and 42.3 m long, what is the length of the hypotenuse?
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https://www.teacherspayteachers.com/Product/Comparing-Fractions-33H-4NF1-2495091 | 1,508,545,563,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824471.6/warc/CC-MAIN-20171020230225-20171021010225-00653.warc.gz | 982,794,595 | 22,930 | Total:
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# Comparing Fractions (3.3H, 4.NF.1)
Subject
Common Core Standards
Product Rating
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PDF (Acrobat) Document File
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Product Description
Goal:
This was created to help teachers and student compare fractions using area linear models and justify comparisons.
This activity aligns with the Texas Education Agency
STAAR Released Items for 3.3H.
Teacher Preparation:
Student Worksheets – Copy and staple on regular white paper
Teacher Key – Copy in color and staple on regular white paper
Directions
Use the various fraction models to compare fractions and justify your comparisons.
TEKS
3.3H compare two fractions having the same numerator or denominator in problems by reasoning about their sizes and justifying the conclusion using symbols, words, objects, and pictorial models.
CCSS
3.NF.3d Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols
>, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Notice:
This activity contains 4 non-student pages (cover sheet, instructions, TpT info and credits).
Total Pages
10 pages
Included
Teaching Duration
45 minutes
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 362 | 1,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.873383 |
https://www.convert-measurement-units.com/convert+Meters+per+minute+to+Feet+per+hour.php | 1,638,550,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00264.warc.gz | 748,400,999 | 12,745 | Convert m/min to fph (Meters per minute to Feet per hour)
## Meters per minute into Feet per hour
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Meters+per+minute+to+Feet+per+hour.php
## How many Feet per hour make 1 Meters per minute?
1 Meters per minute [m/min] = 196.850 393 700 79 Feet per hour [fph] - Measurement calculator that can be used to convert Meters per minute to Feet per hour, among others.
# Convert Meters per minute to Feet per hour (m/min to fph):
1. Choose the right category from the selection list, in this case 'Velocity'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Meters per minute [m/min]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Feet per hour [fph]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '855 Meters per minute'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Meters per minute' or 'm/min'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Velocity'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '47 m/min to fph' or '22 m/min into fph' or '77 Meters per minute -> Feet per hour' or '63 m/min = fph' or '8 Meters per minute to fph' or '91 m/min to Feet per hour' or '40 Meters per minute into Feet per hour'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(2 * 49) m/min'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '855 Meters per minute + 2565 Feet per hour' or '35mm x 38cm x 78dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.295 999 988 206 4×1024. For this form of presentation, the number will be segmented into an exponent, here 24, and the actual number, here 1.295 999 988 206 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.295 999 988 206 4E+24. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 1 295 999 988 206 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 936 | 3,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.777234 |
https://www.dataunitconverter.com/megabit-per-hour-to-pebibyte-per-minute | 1,708,468,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00133.warc.gz | 753,961,154 | 17,338 | # Mbit/Hr to PiB/Min → CONVERT Megabits per Hour to Pebibytes per Minute
expand_more
info 1 Mbit/Hr is equal to 0.0000000000018503717077085942340393861134 PiB/Min
Input Megabits per Hour (Mbit/Hr) - and press Enter.
Mbit/Hr
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
## Megabits per Hour (Mbit/Hr) Versus Pebibytes per Minute (PiB/Min) - Comparison
Megabits per Hour and Pebibytes per Minute are units of digital information used to measure storage capacity and data transfer rate.
Megabits per Hour is a "decimal" unit where as Pebibytes per Minute is a "binary" unit. One Megabit is equal to 1000^2 bits. One Pebibyte is equal to 1024^5 bytes. There are 9,007,199,254.740992 Megabit in one Pebibyte. Find more details on below table.
Megabits per Hour (Mbit/Hr) Pebibytes per Minute (PiB/Min)
Megabits per Hour (Mbit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Megabits that can be transferred in one Hour. Pebibytes per Minute (PiB/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Pebibytes that can be transferred in one Minute.
## Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min) Conversion - Formula & Steps
The Mbit/Hr to PiB/Min Calculator Tool provides a convenient solution for effortlessly converting data rates from Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Megabit) and target (Pebibyte) data units.
Source Data Unit Target Data Unit
Equal to 1000^2 bits
(Decimal Unit)
Equal to 1024^5 bytes
(Binary Unit)
The conversion from Data per Hour to Minute can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min) can be expressed as follows:
diamond CONVERSION FORMULA PiB/Min = Mbit/Hr x 10002 ÷ (8x10245) / 60
Now, let's apply the aforementioned formula and explore the manual conversion process from Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Pebibytes per Minute = Megabits per Hour x 10002 ÷ (8x10245) / 60
STEP 1
Pebibytes per Minute = Megabits per Hour x (1000x1000) ÷ (8x1024x1024x1024x1024x1024) / 60
STEP 2
Pebibytes per Minute = Megabits per Hour x 1000000 ÷ 9007199254740992 / 60
STEP 3
Pebibytes per Minute = Megabits per Hour x 0.000000000111022302462515654042363166809 / 60
STEP 4
Pebibytes per Minute = Megabits per Hour x 0.0000000000018503717077085942340393861134
Example : By applying the previously mentioned formula and steps, the conversion from 1 Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min) can be processed as outlined below.
1. = 1 x 10002 ÷ (8x10245) / 60
2. = 1 x (1000x1000) ÷ (8x1024x1024x1024x1024x1024) / 60
3. = 1 x 1000000 ÷ 9007199254740992 / 60
4. = 1 x 0.000000000111022302462515654042363166809 / 60
5. = 1 x 0.0000000000018503717077085942340393861134
6. = 0.0000000000018503717077085942340393861134
7. i.e. 1 Mbit/Hr is equal to 0.0000000000018503717077085942340393861134 PiB/Min.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Megabits per Hour to Pebibytes per Minute using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Megabit ?
A Megabit (Mb or Mbit) is a decimal unit of digital information that is equal to 1,000,000 bits and it is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of mebibit (Mibit) is used instead.
arrow_downward
#### What is Pebibyte ?
A Pebibyte (PiB) is a binary unit of digital information that is equal to 1,125,899,906,842,624 bytes (or 9,007,199,254,740,992 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'pebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'petabyte' (PB). It is widely used in the field of computing as it more accurately represents the storage size of high end servers and data storage arrays.
## Excel Formula to convert from Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min)
Apply the formula as shown below to convert from 1 Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min).
A B C
1 Megabits per Hour (Mbit/Hr) Pebibytes per Minute (PiB/Min)
2 1 =A2 * 0.000000000111022302462515654042363166809 / 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Megabits per Hour (Mbit/Hr) to Pebibytes per Minute (PiB/Min) Conversion
You can use below code to convert any value in Megabits per Hour (Mbit/Hr) to Megabits per Hour (Mbit/Hr) in Python.
megabitsperHour = int(input("Enter Megabits per Hour: "))
pebibytesperMinute = megabitsperHour * (1000*1000) / (8*1024*1024*1024*1024*1024) / 60
print("{} Megabits per Hour = {} Pebibytes per Minute".format(megabitsperHour,pebibytesperMinute))
The first line of code will prompt the user to enter the Megabits per Hour (Mbit/Hr) as an input. The value of Pebibytes per Minute (PiB/Min) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Pebibytes(PiB) are there in a Megabit(Mbit)?expand_more
There are 0.000000000111022302462515654042363166809 Pebibytes in a Megabit.
#### What is the formula to convert Megabit(Mbit) to Pebibyte(PiB)?expand_more
Use the formula PiB = Mbit x 10002 / (8x10245) to convert Megabit to Pebibyte.
#### How many Megabits(Mbit) are there in a Pebibyte(PiB)?expand_more
There are 9007199254.740992 Megabits in a Pebibyte.
#### What is the formula to convert Pebibyte(PiB) to Megabit(Mbit)?expand_more
Use the formula Mbit = PiB x (8x10245) / 10002 to convert Pebibyte to Megabit.
#### Which is bigger, Pebibyte(PiB) or Megabit(Mbit)?expand_more
Pebibyte is bigger than Megabit. One Pebibyte contains 9007199254.740992 Megabits.
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http://www.answers.com/topic/peter-lorre | 1,555,680,108,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527720.37/warc/CC-MAIN-20190419121234-20190419142029-00018.warc.gz | 207,260,869 | 18,645 | ## Results for: Peter-lorre
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# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow ( Full Answer )
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# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo ( Full Answer )
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# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
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# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
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# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
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# What is the answer for 5c equals -75?
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# How many pixels does the iPhone 5c have?
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# What is minus 5c in Fahrenheit?
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# How many inches is a iPhone 5c?
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In Uncategorized
# What was Peter Lorre?
An Austro-Hungarian-American actor (1904-1964) Famous for acting in The Maltese Falcon and Casablanca amongst others | 556 | 1,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.901626 |
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