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https://math.stackexchange.com/questions/2635356/solve-truth-table-task/2635368 | 1,560,683,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00174.warc.gz | 520,186,796 | 33,369 | I am solving next task: $$\left(x \vee y\right) \rightarrow \bar{x}$$
I assume that line above $\bar{x}$ makes it opposite of $x$ and my solving is this: $$\begin{array}{|cc|c|c|c|} \hline x &y & x\lor y &x & \left(x \vee y\right) \rightarrow \bar{x} \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \end{array}$$ Does my calculations correct? If not why?
EDITED
• Do you mean the claim $$(\bar{x} \vee y) \rightarrow x$$ ? Please note, that you can use \bar{x} in LATEX to generate \bar{x}. – 3nondatur Feb 4 '18 at 10:30 • @3nondatur I will EDIT – IntoTheDeep Feb 4 '18 at 10:31 ## 2 Answers \begin{align} &x &y &&x\lor y &&\bar x &&(x \vee y) \to\bar x \\ &T &T &&T &&\color{red}F &&F \\ &T &F &&T &&F &&F\\ &F &T &&T &&T &&T\\ &F &F &&F &&T &&T \end{align}(x \vee y)$is your hypothesis, the conditional statement is only false when the hypothesis$(x \vee y)$is true, and your conclusion ($\bar x\$) is false. When you hypothesis is false, the conditional statement is true by default.
• Can you please check edited answer – IntoTheDeep Feb 4 '18 at 10:33
• Noticed you have edited the question. Have edited my answer and provided an explanation for you as well. :) – Icycarus Feb 4 '18 at 10:37
\begin{align} &x &y &&x\lor y &&\bar{x} &&(x \vee y) \rightarrow \bar{x} \\ &T &T &&T &&F &&F \\ &T &F &&T &&F &&F\\ &F &T &&T &&T &&T\\ &F &F &&F &&T &&T \end{align}
Think of it this way: If the premise is true, then the conclusion must be true as well. The premise being true is necessary for the conclusion to be true. For your argument to be valid, if the premise is true then the conclusion cannot be false.
• Why this answer is different that @lcycarus's – IntoTheDeep Feb 4 '18 at 10:39
• I've noticed your edit of the the fourth column. I've edited and our answers match. – OGC Feb 4 '18 at 10:42 | 712 | 2,002 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-26 | latest | en | 0.66655 |
http://blog.fountainheadschools.org/index.php/author/fatema-topiwalafountainheadschools-org/ | 1,540,359,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583519859.77/warc/CC-MAIN-20181024042701-20181024064201-00085.warc.gz | 45,807,838 | 24,365 | Homework
## Homework
Homework
Solve the following:
a) There are 365 days in one year. How many days are there in 36 years?
b) 2560 students are going to the zoo. They have to be divided into groups so that each teacher has one group. There are 25 teachers. Each teacher will have to manage how many students?
c) John can run one block in 30 seconds. How far can he run in 5 minutes?
d) What is the area of a square-shaped room with length 25m?
Stars of the day: Ananya Agrawal and Nyassha Fudhnawala (For becoming a leader and helping her peers during Playtime)
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Highlights:
Math: Students understood the survey as a means of problem-solving or information gathering. Further, they understood the following terms in the context of survey- purpose, population, sample and sample size.
UOI: Students started working on their models to explain how different energies work.
Homework:-
UOI: Arrange for the materials you need to bring from home and prepare yourself to explain the model in front of your peers.
Math: Click on the link and solve the mixed bag: Math Mix bag- 19-10-2018
Stars of the day: Prisha Nangalia (for taking up the role of a teacher) and Bhomik Gotecha (for being a helpful peer).
Note: Dear parents, a gentle reminder to send the tuition policy undertaking form on Monday.
Quote of the day:
Regards,
Fatema Topiwala
Aarna, Pratham and Krishaa have encashed 30 points from their account for homework.
Homework
## Homework
Homework
Math
Solve the following:
1) Shalu shoots an arrow 5 meters far. Nitin shoots his arrow three times far as compared to Shalu. How far did Nitin’s arrow go?
2) A baseball team has 12 players. In a tournament, there are 7 such teams. How many players are there in total?
3) 500kg = _______ mg
4) 56l 35 ml=__________ml
5) Find the area and perimeter of a square shaped board game whose length is 35 cm.
[Unit Of Inquiry] Complete the preparations for your science fair project.
Here are a few sites you may browse to look for ideas if you are stuck unsure.
https://www.sciencebuddies.org/
https://www.instructables.com/https://www.sciencebuddies.org/
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Math
Click on the link and complete the worksheet:
Cycle 14 math mixed bag
Language
Click on the link and complete the worksheet
Lang mix bag 13-10-2018
[Unit Of Inquiry]
Students will brainstorm on the model/experiment they wish to create for the upcoming science fair.
Create 2 columns: Write down what do you wish to create and why in under the model column. Write down a list of material that you will require. You may bring all the materials that you have collected already on Monday to school.
Important: Monday is Day 5 which mean you must bring your Library books and ID cards.
Model Material
Stars of the day: Kunjjan Patel(showing efforts towards the feedback given), Jinay Shah (using metacognition to identify what helps him grow faster) and Aarna Choudhary(for becoming a confident speaker).
Food for thought:
We planted a few seeds this week in class. All our plants are growing at a different pace. Plants have their different need like water, sunlight, warmth(temperature) and fertilizers. We, humans, are similar in so many ways. Some students identified their fertilizer to as motivation by teachers and other said it’s daily activities or having fun. This all helps them grow.
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Homework
Language:
• Frame 5 sentences using the present continuous tense. (Example: I am going home after a long tiring and humid day a school.)
• Frame 5 sentences using the simple present tense. (Example: Pauline does not want to share the pie with her classmates as she loves it way too much)
Quote of the day: https://mindgridperspectives.com/2018/04/17/self-discipline/
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Homework
Math
Online practice:
1. Area of a square is 81 sq. cm. Find the perimeter of the square.
2. Create a square or a rectangle such that the perimeter is 48 cm and it is painted, the cost incurred shall be minimum. The cost of painting 1 sq. cm is Rs 10.
3. Rhea wants to fence her rectangular garden. The length of the garden is 25 meters and breadth is 20 meters. Find how much fencing material will she require?
4. The perimeter of a square is 36 cm. What will be the area of the square?
Fill in the blanks:
1) 8 kg and 250 g = ___________________________ g.
2) 51 l and 65 ml = ____________________________ml.
3) 85 m and 11 cm = ___________________________mm.
Important note: Students can dress up from home for their assembly tomorrow but they have to carry the uniform with them. Bring the required resources if any.
Stars of the day:
Priyanshi Bhavsar (for being a caring peer to sheer her resources)
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Homework
Language:
Open the link and complete the worksheet.
6-10-18 lang mix bag
Math:
Open the link and solve the worksheet.
Mixed bag – cycle 13
Stars of the day: Jinay Shah (showing good presentation skills) Ananya Agrawal and Prisha Nangalia( for taking up additional responsibilities).
Quote of the day: https://monstrousgame.com/self-reflection-quotes/self-reflection-quotes-stunning-reflection-quotes-unique-self-reflection-quotes-and-self-love-quotes
With Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Language
Students have been given an Alliteration worksheet and a helping verb.
They are expected to complete the worksheet and frame 2 sentences with the helping verb they had picked in class.
Link of the worksheet for those who are absent: http://englishlinx.com/alliteration/fun-with-alliteration.html
Stars of the day: Jinay Shah (for showing leadership skills), Tatsat Mashruwala and Ahaan Arora (for demonstrating cooperation).
Quote of the day:
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Homework
Math –
Complete the Math Mixbag Cycle 13 worksheet in your Math HW notebook.
Language –
• Complete the Language mixbag cycle 13 worksheet in your Language notebook.
• Create a storyboard for the book review on October 1.
• Research about experiments/models you want to create.
Important note: Parents are requested to be a part of the upcoming book review session as this helps build a child motivation to read and present.
Stars of the day: Tatsat Mashruwala (researched about energy forms independently) and Kunjjan Patel (Peer- checking was done with a lot of responsibility).
Quote of the day: https://inspirationformoms.porch.com/fridays-fantastic-finds-140/
Regards,
Fatema Topiwala
Homework
## Homework
Dear parents,
Homework
To be done in your E.E notebook
Stars of the day: Bhomik Gotecha (being a critical thinking during the scientific experiment presentations) | 1,668 | 6,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-43 | latest | en | 0.94459 |
https://electrical-riddle.com/en/forums/topic/60/ | 1,720,918,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00671.warc.gz | 205,805,642 | 25,169 | Home » Topics » English riddles » General Subjects » Electrical riddle no.7 – Over size diode problem
# Electrical riddle no.7 – Over size diode problem
• Creator
Topic
• #60
Hamid
There are two sensitive auxiliary relay in one portion of DC control circuit (relay no.1 & no.2) separated by one semiconductor diode. Two relays (R1, R2), their two run up contacts (C1, C2) and that mentioned diode bridge (D) is located in "H" form.
When relay no.1 picks up via related N.O. dry contact, the other relay will pick up via the same contact and diode that is directly biased too.
In opposite, the dry contact specified for relay no.2 only picks up one of the two relays because of diode inverse biased activity. In this case diode blocks applied voltage to relay no.2 .
After some time the diode fails due to overheating and it must be replaced by a similar diode. But maintenance man can't find the diode with the same characteristics in market, and he decides to use an over size option. The replaced diode had the same rating of voltage but the current-carrying capability of it was more than the last diode.
It seems all things are okay. But sometimes the circuit can not operate correctly. For example in some condition the auxiliary relay no.1 doesn't drop out after one temporary close-open command.
Indeed when two dry contacts (C1, C2) run up related relay and only the contact C1 return to first position, the relay no.1 doesn't drop out and remains in holding position.
How can you explain the reason of this problem?
Viewing 5 replies - 1 through 5 (of 5 total)
• Author
Replies
• #1180
Haajee
1. R1 picked-up & held by its self-holding contact. 2. R2 is picked-up via C1 & Fwd-biased D & held by its self-holding contact C2 3. If now C1 is opened only R1 will open but R2 will remain held by its self-holding contact C2. 4. Now D is reverse-biased so R1 cannot pick-up via C2 & D. 5. Now if D fails it is not the hi-current as no fwd-current is flowing thru it, but it will be the Reverse-Break-Down of D as it is not rated to block the suppliy voltage. 6. Instead of replacing it with a higher-current rated diode but with a higher-voltage PIV rated diode. 7. You should know that diodes for almost 50v to 1200v are made in the same package size, like 1N-4000 series comes 4001 <100v piv> to 4012 <1200v piv>
#1184
MDShunk
MDShunk Moderator Join Date: Jan 2007 Posts: 1,823 ——————————————————————————– I’d say you’ve welded the dry contacts in relay #1 by some means every once in a while. (possibly the same overcurrent situation that is taking out the diode). Have you checked the current draw on a recording basis, and compared it with the rating of the relay? Since relay #2 is unaffected, I think it’s safe to say that D1 isn’t the issue. Now, on the other hand, if relay #1 only failes to drop out when relay #2 is powered, then I’d guess that you have a “leaky” diode on your hands, passing enough holding current the wrong direction to keep relay #1 closed even after its contact is opened.
#1185
Hamid
——————————————————————————– Quote: ——————————————————————————————————- Originally Posted by MDShunk Since relay #2 is unaffected, I think it’s safe to say that D1 isn’t the issue. Now, on the other hand, if relay #1 only failes to drop out when relay #2 is powered, then I’d guess that you have a “leaky” diode on your hands, passing enough holding current the wrong direction to keep relay #1 closed even after its contact is opened. ———————————————————————————————————- Thank you very much. Your answer is correct completely. Indeed in semiconductor diode, the reverse leakage current is usually specified at a voltage and temperature. Also in general, the conclusion of higher the current carrying capability of the diode is the higher the leakage current. In this case the higher reverse leakage current can to hold auxiliary relay no.1 via contact C2.
#1186
Haajee
Quote: “I’d say you’ve welded the dry contacts in relay #1 by some means every once in a while. (possibly the same over-current situation that is taking out the diode). Have you checked the current draw on a recording basis, and compared it with the rating of the relay? ” You mean dry-contact by Dry-switch operated by its own coil & not the contact of R2? I took it as the NO contact of R@. Regards
#1187
Haajee
You call it “When relay no.1 picks up via related N.O. dry contact,” while it is called “Reed-Swith” which is operated by its own magnet/ Electro-magnet. by calling it “When relay no.1 picks up via related N.O. dry contact,” means it is aux-contact of relay #1, which means it a self-holding contact of relay #1, So once held it can only be de-energised by breaking the path between C1 & R1. Will you please use the standard Terms & Standard Drawing-formats
Viewing 5 replies - 1 through 5 (of 5 total)
• You must be logged in to reply to this topic. | 1,232 | 4,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.948748 |
https://docs.enthought.com/mayavi/mayavi/auto/example_spherical_harmonics.html | 1,719,352,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00063.warc.gz | 175,069,150 | 3,685 | # Spherical harmonics example¶
Plot spherical harmonics on the surface of the sphere, as well as a 3D polar plot.
This example requires scipy.
In this example we use the mlab’s mesh function: `mayavi.mlab.mesh()`. For plotting surfaces this is a very versatile function. The surfaces can be defined as functions of a 2D grid.
For each spherical harmonic, we plot its value on the surface of a sphere, and then in polar. The polar plot is simply obtained by varying the radius of the previous sphere.
Python source code: `spherical_harmonics.py`
```
# Author: Gael Varoquaux <gael.varoquaux@normalesup.org>
# Copyright (c) 2008, Enthought, Inc.
from mayavi import mlab
import numpy as np
from scipy.special import sph_harm
# Create a sphere
r = 0.3
pi = np.pi
cos = np.cos
sin = np.sin
phi, theta = np.mgrid[0:pi:101j, 0:2 * pi:101j]
x = r * sin(phi) * cos(theta)
y = r * sin(phi) * sin(theta)
z = r * cos(phi)
mlab.figure(1, bgcolor=(1, 1, 1), fgcolor=(0, 0, 0), size=(400, 300))
mlab.clf()
# Represent spherical harmonics on the surface of the sphere
for n in range(1, 6):
for m in range(n):
s = sph_harm(m, n, theta, phi).real
mlab.mesh(x - m, y - n, z, scalars=s, colormap='jet')
s[s < 0] *= 0.97
s /= s.max()
mlab.mesh(s * x - m, s * y - n, s * z + 1.3,
scalars=s, colormap='Spectral')
mlab.view(90, 70, 6.2, (-1.3, -2.9, 0.25))
mlab.show()
``` | 452 | 1,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.665738 |
https://metanumbers.com/12116 | 1,675,631,717,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00158.warc.gz | 419,123,641 | 7,551 | # 12116 (number)
12,116 (twelve thousand one hundred sixteen) is an even five-digits composite number following 12115 and preceding 12117. In scientific notation, it is written as 1.2116 × 104. The sum of its digits is 11. It has a total of 4 prime factors and 12 positive divisors. There are 5,568 positive integers (up to 12116) that are relatively prime to 12116.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 11
• Digital Root 2
## Name
Short name 12 thousand 116 twelve thousand one hundred sixteen
## Notation
Scientific notation 1.2116 × 104 12.116 × 103
## Prime Factorization of 12116
Prime Factorization 22 × 13 × 233
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 6058 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 12,116 is 22 × 13 × 233. Since it has a total of 4 prime factors, 12,116 is a composite number.
## Divisors of 12116
1, 2, 4, 13, 26, 52, 233, 466, 932, 3029, 6058, 12116
12 divisors
Even divisors 8 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 22932 Sum of all the positive divisors of n s(n) 10816 Sum of the proper positive divisors of n A(n) 1911 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 110.073 Returns the nth root of the product of n divisors H(n) 6.34014 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 12,116 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 12,116) is 22,932, the average is 1,911.
## Other Arithmetic Functions (n = 12116)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5568 Total number of positive integers not greater than n that are coprime to n λ(n) 696 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1456 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 5,568 positive integers (less than 12,116) that are coprime with 12,116. And there are approximately 1,456 prime numbers less than or equal to 12,116.
## Divisibility of 12116
m n mod m 2 3 4 5 6 7 8 9 0 2 0 1 2 6 4 2
The number 12,116 is divisible by 2 and 4.
• Arithmetic
• Deficient
• Polite
## Base conversion (12116)
Base System Value
2 Binary 10111101010100
3 Ternary 121121202
4 Quaternary 2331110
5 Quinary 341431
6 Senary 132032
8 Octal 27524
10 Decimal 12116
12 Duodecimal 7018
20 Vigesimal 1a5g
36 Base36 9ck
## Basic calculations (n = 12116)
### Multiplication
n×y
n×2 24232 36348 48464 60580
### Division
n÷y
n÷2 6058 4038.67 3029 2423.2
### Exponentiation
ny
n2 146797456 1778597976896 21549493088071936 261093658255079576576
### Nth Root
y√n
2√n 110.073 22.9678 10.4916 6.5565
## 12116 as geometric shapes
### Circle
Diameter 24232 76127.1 4.61178e+08
### Sphere
Volume 7.45017e+12 1.84471e+09 76127.1
### Square
Length = n
Perimeter 48464 1.46797e+08 17134.6
### Cube
Length = n
Surface area 8.80785e+08 1.7786e+12 20985.5
### Equilateral Triangle
Length = n
Perimeter 36348 6.35652e+07 10492.8
### Triangular Pyramid
Length = n
Surface area 2.54261e+08 2.0961e+11 9892.67
## Cryptographic Hash Functions
md5 9df2afabcdd08dbe0db7a49e509403c9 774b7ecdfa8ed5e9956629d700d812985483aa72 ae9166fe6d4f2f75ee8869b84caeab30cb559b9f858bfb371525a7d599b7043a 1581557c99a0d7edddc466d922fd41319b7fb827a96dfe6577e126da8936d2007920e71c690fb9bd07c6c3c0f5792bc4bd473ed9735602993f64834622c0689d a5b26c295a584138c152f5260bfc5266621ee2a1 | 1,446 | 4,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-06 | latest | en | 0.80434 |
http://theory.uwinnipeg.ca/physics/twodim/node6.html | 1,505,990,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687740.4/warc/CC-MAIN-20170921101029-20170921121029-00229.warc.gz | 339,008,772 | 1,967 | ## Instantaneous Velocity
Definition:
=
Interpretation: When t 0, the point Q in the figure gets closer and closer to the point P and the direction of approaches the direction of a tangent to the curve at point P. Thus the instantaneous velocity is parallel to the tangent and in the same direction as the motion. | 67 | 316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | latest | en | 0.902485 |
https://theportal.wiki/wiki/Decoding_the_Graph-Wall-Tome_Connection | 1,643,141,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00679.warc.gz | 603,050,650 | 11,455 | # Decoding the Graph-Wall-Tome Connection
Information Graph-Wall-Tome Connection Graph, Wall, Tome Aardvark 21 April 2020 Link Link Drive Doc Homepage Invite All Projects
An important aspect of the prompt is that neither the Graph, nor the Wall or the Tome are that important on their own. Only together does their explanatory power exceed the limits of any part alone. The common threads that run through all of them are what matter: what is common among them is what should be expanded, and what is particular to one should be promoted and developed if it is a part of fundamental physics toolkit, or removed if it isn't.
The goals of this project are to:
• Identify the common threads (the "unifying idea") in the Graph, Wall, and Tome. These will indicate how to Edit the Graph and Deface the Wall.
• Create and collect resources that make it easy to understand them, thus contributing to Rewriting the Tome.
## Guiding Questions and Comments by Eric Weinstein
What is $$F_A$$ geometrically?
$$F_A$$ is the curvature tensor associated with the connection or vector potential $$A$$.
What are $$R_{\mu v}$$ and $$R$$ geometrically?
$$R$$ is a scalar value, describing the "curvature of the spacetime manifold" at each point along the manifold. It's based on a concept of 'parallel transport', where you move a vector around some path on the manifold.
$$R$$ can be computed at each point on the manifold, and describes the difference in the vector's angle after following an infinitesimally small path around the neighborhood of that point, vs. what it was originally. The video does a great job of visualizing when and why that vector angle change would happen, with flat vs. curved manifolds.
In the video, they focus first on the curvature of space. Hopefully they incorporate back in curvature in time, because that's less obvious.
The same video then proceeds to explain $$R_{\mu v}$$. It progresses through some concepts.
Further thoughts on the meaning of R
Computing length in non-orthogonal bases
First, just describing the length of a vector on a curved space is hard. It is given by:
$$Length^{squared} = g_{11}dX^{1}dX^{1} + g_{12}dX^{1}dX^{2} + g_{21}dX^{2}dX^{1} + g_{22}dX^{2}dX^{2}$$
Some notes:
• This is not Pythagorean theorem, because $$dX^{1}$$ and $$dX^{2}$$ are not perpendicular.
• Instead, looks like a formula to get the diagonal from two opposite vertices in a parallelogram.
• If $$dX^{1}$$ and $$dX^{2}$$ are perpendicular, then $$g_{12}$$ and $$g_{21}$$ would be 0, and we would get $$Length^{squared} = g_{11}(dX^{1})^{2} + g_{22}(dX^{2})^{2}$$
• See: the video @ 14m27s
Computing vector rotation due to parallel transport
Then, they show parallel transport when following a parallelogram, but over a curved 3D manifold. To compute the vector rotation by components, they show:
$$dV^{1} = dX^{1}dX^{2} (V^{1}R^{1}_{112} + V^{2}R^{1}_{212} + V^{3}R^{1}_{312})$$
$$dV^{2} = dX^{1}dX^{2} (V^{1}R^{2}_{112} + V^{2}R^{2}_{212} + V^{3}R^{2}_{312})$$
$$dV^{3} = dX^{1}dX^{2} (V^{1}R^{3}_{112} + V^{2}R^{3}_{212} + V^{3}R^{3}_{312})$$
or, using $$i$$ to summarize across all 3 components (difference vectors):
$$dV^{i} = dX^{1}dX^{2} (V^{1}R^{i}_{112} + V^{2}R^{i}_{212} + V^{3}R^{i}_{312})$$
or , using $$j$$ to index over all 3 components (original vector):
$$dV^{i} = dX^{1}dX^{2} \Sigma_{j} [(V^{j}R^{i}_{j12}]$$
Open questions:
• Why a parallelogram?
• How to properly overlay the parallelogram onto the 3d manifold, in order to do the parallel transport?
• How does this relate to the length computation above?
Putting it all together
Now, moving to 4D, we can compute $$R_{\mu v}$$ as:
$$R_{00} = R^{0}_{000} + R^{1}_{010} + R^{2}_{020} + R^{3}_{030}$$
$$R_{10} = R^{0}_{100} + R^{1}_{110} + R^{2}_{120} + R^{3}_{130}$$
$$R_{01} = R^{0}_{001} + R^{1}_{011} + R^{2}_{021} + R^{3}_{030}$$
etc.
Indexing i over all 4 component vectors / dimensions, we get:
$$R_{00} = \Sigma_{i} R^{i}_{0i0}$$
$$R_{10} = \Sigma_{i} R^{i}_{1i0}$$
$$R_{01} = \Sigma_{i} R^{i}_{0i1}$$
etc.
Summarizing on $$\mu$$, we get:
$$R_{\mu 0} = \Sigma_{i} R^{i}_{\mu i0}$$
$$R_{\mu 1} = \Sigma_{i} R^{i}_{\mu i1}$$
etc
Summarizing on $$v$$, we get:
$$R_{\mu v} = \Sigma_{i} R^{i}_{\mu iv}$$
Open questions:
• If we hadn't moved from 3D to 4D, what would this all have looked like?
• What does this have to do with the parallelogram?
• Why are there two indices?
How do they relate?
What does this have to do with Penrose Stairs?
We’ve heard Eric talk about Penrose stairs and spinors - essentially phenomena where you cannot return to the original state through a 360 degree rotation, but require a 720 degree rotation.
What are “Horizontal Subspaces” and what do they have to do with Vector Potentials or Gauge fields?
From theplebistocrat:
Generally, we're wanting to understand how fermions arise from - or are embedded within / upon - topological "spaces" that have distinct rules which govern operations within those topological spaces, and then how those rules produce higher dimensional operations in corresponding spaces.
Just intuitively, and geometrically speaking, the image that I'm getting when describing all of this and trying to hold it in my head is the image of a sort of Penrose Tower of Babel, where the fundamental underlying structures reach upwards (but also downwards and inwards?) before reaching a critical rotation that corresponds to a collapse of structure into a higher dimensional fiber bundle.
But doesn't this require the symmetry break? How is left and right rotation in a subspace transformed into verticality? This is a crazy rabbit hole, friends. Keep your chins up. Let me know if this was helpful or leading astray.
"The source code of the universe is overwhelmingly likely to determine a purely geometric operating system written in a uniform programming language." - Eric Weinstein
## Direct Connections between the Graph, the Wall, and the Tome
### Connections between the Graph and the Wall
Graph->Wall connections suggested by Eric Weinstein
Alternative representation of the Graph->Wall connections suggested by Eric Weinstein
Alternative Graph->Wall Connections
The following figure connects the Graph to elements on the Wall. There are certain holes since there aren't visualizations or formulas for all concepts mentioned in the Graph.
Alternative representation of the connections between the Graph and the Wall
The following figure connects the Graph to elements on the Wall. There are certain holes since there aren't visualizations or formulas for all concepts mentioned in the Graph.
Alternative representation of the connections between the Graph and the Wall
On the left-hand side, we have the Graph. On the right-hand side, there is a list of all elements that appear on the Wall.
Interactive representation of the connections between the Graph and the Wall
An interactive version of the Wall that shows direct connections to the graph is available here.
Connections between an updated Graph and updated Wall
In the following figure, the holes in the wall have been filled with appropriate illustrations and equations.
### Connections between the Wall and the Tome
Connections between Wall and Tome
On the left-hand side, we have a list of all elements that appear on the Wall and on the right-hand side we have the table of contents of the Tome. Only connections between concepts that are also mentioned in the Graph are shown.
### Connections between the Graph and the Tome
Connections between Graph and Tome
On the left-hand side, we have the Graph and on the right-hand side we have the table of contents of the Tome. | 2,014 | 7,651 | {"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} | 3.640625 | 4 | CC-MAIN-2022-05 | latest | en | 0.944868 |
https://stats.stackexchange.com/questions/145323/why-are-second-order-derivatives-useful-in-convex-optimization | 1,723,341,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00890.warc.gz | 423,455,727 | 47,221 | # Why are second-order derivatives useful in convex optimization?
I guess this is a basic question and it has to do with the direction of the gradient itself, but I'm looking for examples where 2nd order methods (e.g. BFGS) are more effective than simple gradient descent.
• Is it too simplistic to just observe that "find the vertex of a paraboloid" is a much better approximation to the "find a minimum" problem than "find the minimum of this linear function" (which, of course, has no minimum because it's linear)?
– user41979
Commented Apr 8, 2015 at 21:31
Here's a common framework for interpreting both gradient descent and Newton's method, which is maybe a useful way to think of the difference as a supplement to @Sycorax's answer. (BFGS approximates Newton's method; I won't talk about it in particular here.)
We're minimizing the function $f$, but we don't know how to do that directly. So, instead, we take a local approximation at our current point $x$ and minimize that.
Newton's method approximates the function using a second-order Taylor expansion: $$f(y) \approx N_x(y) := f(x) + \nabla f(x)^T (y - x) + \tfrac12 (y - x)^T \, \nabla^2 f(x) \, (y - x) ,$$ where $\nabla f(x)$ denotes the gradient of $f$ at the point $x$ and $\nabla^2 f(x)$ the Hessian at $x$. It then steps to $\arg\min_y N_x(y)$ and repeats.
Gradient descent, only having the gradient and not the Hessian, can't just make a first-order approximation and minimize that, since as @Hurkyl noted it has no minimum. Instead, we define a step size $t$ and step to $x - t \nabla f(x)$. But note that \begin{align} x - t \,\nabla f(x) &= \arg\max_y \left[f(x) + \nabla f(x)^T (y - x) + \tfrac{1}{2 t} \lVert y - x \rVert^2\right] \\&= \arg\max_y \left[f(x) + \nabla f(x)^T (y - x) + \tfrac12 (y-x)^T \tfrac{1}{t} I (y - x)\right] .\end{align} Thus gradient descent minimizes a function $$G_x(y) := f(x) + \nabla f(x)^T (y - x) + \tfrac12 (y-x)^T \tfrac{1}{t} I (y - x).$$
Thus gradient descent is kind of like using Newton's method, but instead of taking the second-order Taylor expansion, we pretend that the Hessian is $\tfrac1t I$. This $G$ is often a substantially worse approximation to $f$ than $N$, and hence gradient descent often takes much worse steps than Newton's method. This is counterbalanced, of course, by each step of gradient descent being so much cheaper to compute than each step of Newton's method. Which is better depends entirely on the nature of the problem, your computational resources, and your accuracy requirements.
Looking at @Sycorax's example of minimizing a quadratic $$f(x) = \tfrac12 x^T A x + d^T x + c$$ for a moment, it's worth noting that this perspective helps with understanding both methods.
With Newton's method, we'll have $N = f$ so that it terminates with the exact answer (up to floating point accuracy issues) in a single step.
Gradient descent, on the other hand, uses $$G_x(y) = f(x) + (A x + d)^T y + \tfrac12 (x - y)^T \tfrac1t I (x-y)$$ whose tangent plane at $x$ is correct, but whose curvature is entirely wrong, and indeed throws away the important differences in different directions when the eigenvalues of $A$ vary substantially.
• This is similar to @Aksakal's answer, but in more depth. Commented Apr 9, 2015 at 3:19
• (+1) This is a great addition!
– Sycorax
Commented Apr 9, 2015 at 3:22
• I think what I don't understand is why it results in better optimization of $f$ using $N$ (quadratic approximation). Somehow saying $N$ approximates $f$ better feels to vague to really understand why exactly I prefer one vs the other. Commented Sep 8, 2020 at 18:02
• Isn't $||x-y||^2 = (x-y)^T (x-y)$ ? What is the function $I$? Commented Oct 18, 2022 at 15:10
• @newandlost $I$ here is the identity matrix (so it’s exactly what you wrote, just writing $I$ in to emphasize the difference versus the approximation $N_x$). Commented Oct 19, 2022 at 3:54
Essentially, the advantage of a second-derivative method like Newton's method is that it has the quality of quadratic termination. This means that it can minimize a quadratic function in a finite number of steps. A method like gradient descent depends heavily on the learning rate, which can cause optimization to either converge slowly because it's bouncing around the optimum, or to diverge entirely. Stable learning rates can be found... but involve computing the hessian. Even when using a stable learning rate, you can have problems like oscillation around the optimum, i.e. you won't always take a "direct" or "efficient" path towards the minimum. So it can take many iterations to terminate, even if you're relatively close to it. BFGS and Newton's method can converge more quickly even though the computational effort of each step is more expensive.
To your request for examples: Suppose you have the objective function $$F(x)=\frac{1}{2}x^TAx+d^Tx+c$$ The gradient is $$\nabla F(x)=Ax+d$$ and putting it into the steepest descent form with constant learning rate $$x_{k+1}= x_k-\alpha(Ax_k+d) = (I-\alpha A)x_k-\alpha d.$$
This will be stable if the magnitudes of the eigenvalues of $$I-\alpha A$$ are less than 1. We can use this property to show that a stable learning rate satisfies $$\alpha<\frac{2}{\lambda_{max}},$$ where $$\lambda_{max}$$ is the largest eigenvalue of $$A$$. Choosing a learning rate which is too large will overshoot the minimum, and the optimization will diverge.
The steepest descent algorithm's convergence rate is limited by the largest eigenvalue and the routine will converge most quickly in the direction of its corresponding eigenvector. Likewise, it will converge most slowly in directions of the eigenvector of the smallest eigenvalue. When there is a large disparity between large and small eigenvalues for $$A$$, gradient descent will be slow. Any $$A$$ with this property will converge slowly using gradient descent.
In the specific context of neural networks, the book Neural Network Design has quite a bit of information on numerical optimization methods. The above discussion is a condensation of section 9-7.
• Great answer! I'm accepting @Dougal 's answer as I think it provides a simpler explanation.
– Bar
Commented Apr 9, 2015 at 9:11
• why do we care about the eigenvectors of $I - \alpha A$? i.e. how does it affect the optimization or even better, each step for getting $x_{k+1}$? Commented Sep 8, 2020 at 17:59
• @CharlieParker If the learning rate is too large, the optimization can diverge. If the learning rate is too small, progress towards the minimum will be slow.
– Sycorax
Commented Sep 8, 2020 at 18:29
In convex optimization you are approximating the function as the second degree polynomial in one dimensional case: $$f(x)=c+\beta x + \alpha x^2$$
In this case the the second derivative $$\partial^2 f(x)/\partial x^2=2\alpha$$
If you know the derivatives, then it's easy to get the next guess for the optimum: $$\text{guess}=-\frac{\beta}{2\alpha}$$
The multivariate case is very similar, just use gradients for derivatives.
• is $f$ the approximation or the original function we want to minimize? sorry I find your prose hard to follow. Commented Sep 8, 2020 at 18:05 | 1,915 | 7,144 | {"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": 9, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.860441 |
https://lists.gnu.org/archive/html/igraph-help/2013-07/msg00056.html | 1,713,628,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00493.warc.gz | 328,151,656 | 3,697 | igraph-help
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## Re: [igraph] MCS and result of "igraph_get_subisomorphisms_vf2"
From: Wen Cheng Subject: Re: [igraph] MCS and result of "igraph_get_subisomorphisms_vf2" Date: Sun, 21 Jul 2013 19:01:49 -0500
Hello Tamas,
Regarding result of "igraph_get_subisomorphism_vf2()" I found mapped the second graph to the first one in an inverse way. For example, for mapping "4, 2, 3, 1, 0", I mapped vertex 4 in the second graph to vertex 0 in the first graph, mapped vertex 2 in the second graph to vertex 1 in the first graph, vertex 3 in the second graph to vertex 2 in the first graph and so on. Now I know how to map according to your answer.
With respect to MCS problem, I will study how to build the modular product manually. One more concern is that my dataset has 140 graphs and I need to find MCS for each pair of graphs. I wonder whether building modular product manually is practical or not. However, I will try to do it and see how long it may take.
Again, thank you very much for your time and effort on my questions!
Best regards,
Wen
On Fri, Jul 19, 2013 at 4:39 PM, Tamás Nepusz wrote:
Hello,
> Firstly, I wonder whether igraph can help find maximum common subgraph (or just common subgraphs) of two or more graphs. If so, could you please let me know which function can do that?
There is no implementation for the MCS problem in igraph yet. If your graphs are not too large, you can try building the modular product manually (see http://en.wikipedia.org/wiki/Modular_product_of_graphs) and then find the largest clique in the modular product graph -- this will correspond to a solution of the MCS problem. Finding largest cliques is implemented in igraph_largest_cliques.
> 0, 1, 2, 3, 4
> 0, 1, 3, 2, 4
> 4, 2, 1, 3, 0
> 4, 2, 3, 1, 0
>
> The first three mappings make sense but the last one seems weird. I wonder whether you have any comment on it.
Let's see. The edges in your first graph are:
0-1, 0-3, 1-2, 1-3, 2-3, 2-4, 3-4
The edges in the second graph are:
0-1, 1-2, 1-3, 2-3, 2-4, 3-4
The mapping vector is [4, 2, 3, 1, 0]. This describes the mapping _from_ the vertices of the _second_ graph _to_ the vertices of the _first_ graph. In other words, vertex 0 in the second graph is mapped to vertex 4 of the first graph, vertex 1 of the second graph is mapped to vertex 2 of the first graph and so on. So, let's take the edge list of the second graph and do all these replacements at once:
Replace 0 with 4; 1 with 2; 2 with 3; 3 with 1 and 4 with 0.
The remapped edge list is:
4-2, 2-3, 2-1, 3-1, 3-0, 1-0
Let's swap the order of vertices where the smaller vertex index comes second because I used the convention of writing the smaller index first when I wrote down the edge list of the first graph. This gives us:
2-4, 2-3, 1-2, 1-3, 0-3, 0-1
As you can see, this is a subset of the edge list of the first graph; only edge 3-4 is missing. Therefore, the mapping describes a proper sub-isomorphism.
Cheers,
Tamas
_______________________________________________
igraph-help mailing list | 906 | 3,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.898377 |
https://www.instructables.com/Magnet-Trampoline/ | 1,720,785,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00387.warc.gz | 512,498,614 | 33,451 | ## Introduction: Magnet Trampoline
While sometimes a bit polarizing, magnets enjoy a good party, too.
Whether you're feeling repulsive or like sticking together, the magnet trampoline is a very simple introductory project to teach some of the basics of forces and magnetism. Slap it together, take a bounce, and see what variations you can come up with!
• What: Magnet Trampoline
• Where: both the north and south poles, really
• Concepts: magnetism, forces, gravity, potential energy, springs
• Time: ~ 5 minutes to make
• Cost: ~\$0.75-\$1.25 (and re-usable)
• Materials:
• Ceramic Disc Magnets (donut-shaped, like these ones)
• Wood Dowel (match size of magnet donut hole)
• Wood block (piece of 2x4 works great)
• Tools:
• Drill
Let's bounce!
## Step 1: Choose and Cut a Dowel
Let's make a stick!
First up is to find a dowel that works with your magnets. Check the diameter for one where the magnets can slide easily but is otherwise fairly snug. Cut a length of about 8-12".
## Step 2: Drill and Place Your Dowel
You know the drill!
Find a drill bit that is the same size as your dowel and drill a hole halfway into a block of wood, keeping the drill positioned as vertically up and down as you can (a drill press works great if you have one). Twist to fit the dowel snugly in.
You are so so close.
## Step 3: Tower Time!
Place your magnets on your dowel reversing the orientation of the poles each time. This can be with anywhere between 3 and 12 magnets (or more, too). You want each of them to repel one another and to stay suspended in a springlike state. Check out the interesting variation in spacing between, and give push them down to have them spring back!
## Step 4: Noticing and Going Further
The joy in this project is its initial simplicity, but there is much more to do and notice. I am always amazed at what students come up with for exploration questions, but some prompts can include:
• What happens when you have some magnets that attract? What happens when you add magnets to the chain?
• What happens if you move that stack of magnets to the middle? Or the bottom?
• What is the distance between the lowest and the second lowest magnet? What about the second highest and the highest? Can you graph the distances between them all?
• Try drilling into the block of wood at different angles? What is the spacing like between those magnets? Are they farther apart or closer together than your vertical one? Why do you think that is?
• Why does a magnet spring up when you hold it down? Can you make it shoot higher?
Let me know your variations in the comments below. Enjoy, repel, stick together, and as always, keep exploring. | 619 | 2,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.925584 |
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Michael Strickland 844162 Physics 111 Lab September 14 2010 Kileigh Peturis Activity 5 Acceleration and Free Fall
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Introduction In everyday life we experience a variety of physical factors that affect everything and everyone. In the following lab experiment we will cover the properties referred to as acceleration and free fall. We will demonstrate this using three experiments that will have slightly different factors and trigonometrical properties that will affect each in similar and separate ways. The first experiment will use a fan cart set up on a measured, flat track and determine acceleration when it is constant using the kinematics equation: - = + x xo vot 12at2 . The second experiment consisted of a similar set up, but the fan was removed from the cart and the track set at an angle. This experiment was calculated trigonometrically by measuring the set up from the right angle it formed. Acceleration was determined in a similar fashion and it must be written as a = ms2 . (a =Acceleration)
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## This note was uploaded on 11/03/2010 for the course PHY 111 taught by Professor Sirola during the Spring '10 term at University Of Southern Mississippi .
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phy 3 - M ichael Strickland 844162 Physics 111 Lab...
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# Serway_PSE_quick_ch25 - Physics for Scientists and...
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Physics for Scientists and Engineers, 6e Chapter 25 – Electric Potential
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In the figure below, two points A and B are located within a region in which there is an electric field. The potential difference Δ V = V B V A is 1 2 3 33% 33% 33% 1 2 3 4 5 1. positive 2. negative 3. zero
When moving straight from A to B , E and d s in Equation 25.3 both point toward the right. Thus, the dot product E · d s is positive and Δ V is negative.
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In this figure, a negative charge is placed at A and then moved to B . The change in potential energy of the charge–field system for this process is 1 2 3 33% 33% 33% 1 2 3 4 5 1. positive 2. negative 3. zero
From Equation 25.3, Δ U = q 0 Δ V , so if a negative test charge is moved through a negative potential difference, the potential energy is positive. Work must be done to move the charge in the direction opposite to the electric force on it.
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The labeled points of the figure below are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves along the following transitions.
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Jill Tulane University ‘16, Course Hero Intern | 635 | 2,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-22 | latest | en | 0.909598 |
http://tiresizr.com/vehicle/1994/porsche/928 | 1,722,714,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00092.warc.gz | 30,677,720 | 44,986 | #### GTS
###### F: 225/45ZR17R: 255/40ZR17
(click for a tire code explanation)
### Front Tires
#### P ... Passenger Vehicle
When a tire size begins with a P, it signifies the tire is a p-metric size that was designed to be fitted on vehicles that are primarily used as passenger vehicles. This includes cars, minivans, sport utility vehicles and light duty pickup trucks (typically 1/4- and 1/2-ton load capacity). The use of p-metric sizes began in the late 1970s and they are the most frequently used type of tire size today.
#### 225 millimeters
Following the letter(s) that identify the type of vehicle and/or type of service for which the tire was designed, the three-digit numeric portion identifies the tire's section width/cross section in millimeters.
The 225 indicates that this tire is 225 millimeters across from the widest point when mounted and measured on a specified width wheel. Because many people think of measurements in inches, the 225mm can be converted to inches by dividing the width in millimeters by 25.4 (the number of millimeters per inch).
225mm / 25.4 = 8.86"
#### 45%
The ratio of the height of the tire's cross-section to it's width. 45 means that the height is equal to 45% of the tire's width.
Typically, following the three digits identifying the tire's width in millimeters is a two-digit number that identifies the tire's profile or aspect ratio.
The 45 indicates that this tire size's sidewall height (from rim to tread) is 50% of its width. The measurement is the tire's section height, and also referred to as the tire's series, profile or aspect ratio. The higher the number, the taller the sidewall. The lower the number, the lower the sidewall. We know that this tire size's section width is 225mm and that its section height is 50% of 225mm. By converting the 225mm to inches (225 / 25.4 = 8.86") and multiplying it by 50% (.50) we confirm that this tire size results in a tire section height of4.43".
###### Construction Method
The R in the 225/45ZR17 size identifies that the tire has a radial construction in which the tire's plies "radiate" out from the center of the wheel. Radial tires are by far the most popular type of tire today representing over 98% of all tires sold.
If the R in the size was replaced with a D (225/45ZR17), it would identify that the internal tire body plies crisscross on a diagonal and that the tire has a "bias ply" construction. Tires using this construction are for light truck and spare tire applications.
#### 17" wheel
The 17 indicates the tire and wheel diameter designed to be matched together.
Tires that have a rim diameter expressed in inches (225/45ZR17, as well as8, 10, 12, 13,14,15, 17, 18, 19,20, 22, 23, 24,26 and 28) are called "inch rim" sizes. These are the most common types of tire sizes and are used on most cars, minivans, vans, sport utility vehicles and light duty light trucks.
Tires and wheels with unique rim diameters should never be combined with traditional "inch rim" tires and wheels.
#### Z ... over 240mph
The maximum speed that the tire can sustain for 10 minutes.
Today, the only tires that continue to include the speed rating "in" the tire size (225/45ZR17) are Z-speed rated tires. In this case, following the two digits used to identify the aspect ratio are the letters ZR to identify the tire's speed rating (Z) and its internal construction (R). Since 1991, all other speed ratings are identified in the tire's Service Description.
### Rear Tires
#### P ... Passenger Vehicle
When a tire size begins with a P, it signifies the tire is a p-metric size that was designed to be fitted on vehicles that are primarily used as passenger vehicles. This includes cars, minivans, sport utility vehicles and light duty pickup trucks (typically 1/4- and 1/2-ton load capacity). The use of p-metric sizes began in the late 1970s and they are the most frequently used type of tire size today.
#### 255 millimeters
Following the letter(s) that identify the type of vehicle and/or type of service for which the tire was designed, the three-digit numeric portion identifies the tire's section width/cross section in millimeters.
The 255 indicates that this tire is 255 millimeters across from the widest point when mounted and measured on a specified width wheel. Because many people think of measurements in inches, the 255mm can be converted to inches by dividing the width in millimeters by 25.4 (the number of millimeters per inch).
255mm / 25.4 = 10.04"
#### 40%
The ratio of the height of the tire's cross-section to it's width. 40 means that the height is equal to 40% of the tire's width.
Typically, following the three digits identifying the tire's width in millimeters is a two-digit number that identifies the tire's profile or aspect ratio.
The 40 indicates that this tire size's sidewall height (from rim to tread) is 50% of its width. The measurement is the tire's section height, and also referred to as the tire's series, profile or aspect ratio. The higher the number, the taller the sidewall. The lower the number, the lower the sidewall. We know that this tire size's section width is 255mm and that its section height is 50% of 255mm. By converting the 255mm to inches (255 / 25.4 = 10.04") and multiplying it by 50% (.50) we confirm that this tire size results in a tire section height of5.02".
###### Construction Method
The R in the 255/40ZR17 size identifies that the tire has a radial construction in which the tire's plies "radiate" out from the center of the wheel. Radial tires are by far the most popular type of tire today representing over 98% of all tires sold.
If the R in the size was replaced with a D (255/40ZR17), it would identify that the internal tire body plies crisscross on a diagonal and that the tire has a "bias ply" construction. Tires using this construction are for light truck and spare tire applications.
#### 17" wheel
The 17 indicates the tire and wheel diameter designed to be matched together.
Tires that have a rim diameter expressed in inches (255/40ZR17, as well as8, 10, 12, 13,14,15, 17, 18, 19,20, 22, 23, 24,26 and 28) are called "inch rim" sizes. These are the most common types of tire sizes and are used on most cars, minivans, vans, sport utility vehicles and light duty light trucks.
Tires and wheels with unique rim diameters should never be combined with traditional "inch rim" tires and wheels.
#### Z ... over 240mph
The maximum speed that the tire can sustain for 10 minutes.
Today, the only tires that continue to include the speed rating "in" the tire size (255/40ZR17) are Z-speed rated tires. In this case, following the two digits used to identify the aspect ratio are the letters ZR to identify the tire's speed rating (Z) and its internal construction (R). Since 1991, all other speed ratings are identified in the tire's Service Description.
Dr. Ing. h.c. F. Porsche AG, usually shortened to Porsche, is a German automobile manufacturer specializing in high-performance sports cars, SUVs and sedans. Porsche AG is headquartered in Stuttgart, and is owned by Volkswagen AG, which is itself majority-owned by Porsche Automobil Holding SE. Porsche's current lineup includes the 718 Boxster/Cayman, 911, Panamera, Macan and Cayenne.
Wikipedia | 1,763 | 7,270 | {"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} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.937909 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3328/2/b/m/ | 1,656,573,075,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00480.warc.gz | 912,502,092 | 57,257 | # Properties
Label 3328.2.b.m Level $3328$ Weight $2$ Character orbit 3328.b Analytic conductor $26.574$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$3328 = 2^{8} \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3328.b (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$26.5742137927$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ x^2 + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 26) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + i q^{3} + 3 i q^{5} + q^{7} + 2 q^{9}+O(q^{10})$$ q + i * q^3 + 3*i * q^5 + q^7 + 2 * q^9 $$q + i q^{3} + 3 i q^{5} + q^{7} + 2 q^{9} - 6 i q^{11} + i q^{13} - 3 q^{15} - 3 q^{17} + 2 i q^{19} + i q^{21} - 4 q^{25} + 5 i q^{27} + 6 i q^{29} - 4 q^{31} + 6 q^{33} + 3 i q^{35} + 7 i q^{37} - q^{39} + i q^{43} + 6 i q^{45} + 3 q^{47} - 6 q^{49} - 3 i q^{51} + 18 q^{55} - 2 q^{57} + 6 i q^{59} + 8 i q^{61} + 2 q^{63} - 3 q^{65} + 14 i q^{67} + 3 q^{71} - 2 q^{73} - 4 i q^{75} - 6 i q^{77} + 8 q^{79} + q^{81} + 12 i q^{83} - 9 i q^{85} - 6 q^{87} + 6 q^{89} + i q^{91} - 4 i q^{93} - 6 q^{95} - 10 q^{97} - 12 i q^{99} +O(q^{100})$$ q + i * q^3 + 3*i * q^5 + q^7 + 2 * q^9 - 6*i * q^11 + i * q^13 - 3 * q^15 - 3 * q^17 + 2*i * q^19 + i * q^21 - 4 * q^25 + 5*i * q^27 + 6*i * q^29 - 4 * q^31 + 6 * q^33 + 3*i * q^35 + 7*i * q^37 - q^39 + i * q^43 + 6*i * q^45 + 3 * q^47 - 6 * q^49 - 3*i * q^51 + 18 * q^55 - 2 * q^57 + 6*i * q^59 + 8*i * q^61 + 2 * q^63 - 3 * q^65 + 14*i * q^67 + 3 * q^71 - 2 * q^73 - 4*i * q^75 - 6*i * q^77 + 8 * q^79 + q^81 + 12*i * q^83 - 9*i * q^85 - 6 * q^87 + 6 * q^89 + i * q^91 - 4*i * q^93 - 6 * q^95 - 10 * q^97 - 12*i * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + 2 q^{7} + 4 q^{9}+O(q^{10})$$ 2 * q + 2 * q^7 + 4 * q^9 $$2 q + 2 q^{7} + 4 q^{9} - 6 q^{15} - 6 q^{17} - 8 q^{25} - 8 q^{31} + 12 q^{33} - 2 q^{39} + 6 q^{47} - 12 q^{49} + 36 q^{55} - 4 q^{57} + 4 q^{63} - 6 q^{65} + 6 q^{71} - 4 q^{73} + 16 q^{79} + 2 q^{81} - 12 q^{87} + 12 q^{89} - 12 q^{95} - 20 q^{97}+O(q^{100})$$ 2 * q + 2 * q^7 + 4 * q^9 - 6 * q^15 - 6 * q^17 - 8 * q^25 - 8 * q^31 + 12 * q^33 - 2 * q^39 + 6 * q^47 - 12 * q^49 + 36 * q^55 - 4 * q^57 + 4 * q^63 - 6 * q^65 + 6 * q^71 - 4 * q^73 + 16 * q^79 + 2 * q^81 - 12 * q^87 + 12 * q^89 - 12 * q^95 - 20 * q^97
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/3328\mathbb{Z}\right)^\times$$.
$$n$$ $$261$$ $$769$$ $$1535$$ $$\chi(n)$$ $$-1$$ $$1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1665.1
− 1.00000i 1.00000i
0 1.00000i 0 3.00000i 0 1.00000 0 2.00000 0
1665.2 0 1.00000i 0 3.00000i 0 1.00000 0 2.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
8.b even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3328.2.b.m 2
4.b odd 2 1 3328.2.b.j 2
8.b even 2 1 inner 3328.2.b.m 2
8.d odd 2 1 3328.2.b.j 2
16.e even 4 1 26.2.a.a 1
16.e even 4 1 832.2.a.d 1
16.f odd 4 1 208.2.a.a 1
16.f odd 4 1 832.2.a.i 1
48.i odd 4 1 234.2.a.e 1
48.i odd 4 1 7488.2.a.g 1
48.k even 4 1 1872.2.a.q 1
48.k even 4 1 7488.2.a.h 1
80.i odd 4 1 650.2.b.d 2
80.k odd 4 1 5200.2.a.x 1
80.q even 4 1 650.2.a.j 1
80.t odd 4 1 650.2.b.d 2
112.l odd 4 1 1274.2.a.d 1
112.w even 12 2 1274.2.f.p 2
112.x odd 12 2 1274.2.f.r 2
144.w odd 12 2 2106.2.e.b 2
144.x even 12 2 2106.2.e.ba 2
176.l odd 4 1 3146.2.a.n 1
208.l even 4 1 2704.2.f.d 2
208.m odd 4 1 338.2.b.c 2
208.o odd 4 1 2704.2.a.f 1
208.p even 4 1 338.2.a.f 1
208.r odd 4 1 338.2.b.c 2
208.s even 4 1 2704.2.f.d 2
208.be odd 12 2 338.2.e.a 4
208.bh even 12 2 338.2.c.a 2
208.bj even 12 2 338.2.c.d 2
208.bl odd 12 2 338.2.e.a 4
240.bb even 4 1 5850.2.e.a 2
240.bf even 4 1 5850.2.e.a 2
240.bm odd 4 1 5850.2.a.p 1
272.r even 4 1 7514.2.a.c 1
304.j odd 4 1 9386.2.a.j 1
624.u even 4 1 3042.2.b.a 2
624.bi odd 4 1 3042.2.a.a 1
624.bm even 4 1 3042.2.b.a 2
1040.be even 4 1 8450.2.a.c 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
26.2.a.a 1 16.e even 4 1
208.2.a.a 1 16.f odd 4 1
234.2.a.e 1 48.i odd 4 1
338.2.a.f 1 208.p even 4 1
338.2.b.c 2 208.m odd 4 1
338.2.b.c 2 208.r odd 4 1
338.2.c.a 2 208.bh even 12 2
338.2.c.d 2 208.bj even 12 2
338.2.e.a 4 208.be odd 12 2
338.2.e.a 4 208.bl odd 12 2
650.2.a.j 1 80.q even 4 1
650.2.b.d 2 80.i odd 4 1
650.2.b.d 2 80.t odd 4 1
832.2.a.d 1 16.e even 4 1
832.2.a.i 1 16.f odd 4 1
1274.2.a.d 1 112.l odd 4 1
1274.2.f.p 2 112.w even 12 2
1274.2.f.r 2 112.x odd 12 2
1872.2.a.q 1 48.k even 4 1
2106.2.e.b 2 144.w odd 12 2
2106.2.e.ba 2 144.x even 12 2
2704.2.a.f 1 208.o odd 4 1
2704.2.f.d 2 208.l even 4 1
2704.2.f.d 2 208.s even 4 1
3042.2.a.a 1 624.bi odd 4 1
3042.2.b.a 2 624.u even 4 1
3042.2.b.a 2 624.bm even 4 1
3146.2.a.n 1 176.l odd 4 1
3328.2.b.j 2 4.b odd 2 1
3328.2.b.j 2 8.d odd 2 1
3328.2.b.m 2 1.a even 1 1 trivial
3328.2.b.m 2 8.b even 2 1 inner
5200.2.a.x 1 80.k odd 4 1
5850.2.a.p 1 240.bm odd 4 1
5850.2.e.a 2 240.bb even 4 1
5850.2.e.a 2 240.bf even 4 1
7488.2.a.g 1 48.i odd 4 1
7488.2.a.h 1 48.k even 4 1
7514.2.a.c 1 272.r even 4 1
8450.2.a.c 1 1040.be even 4 1
9386.2.a.j 1 304.j odd 4 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(3328, [\chi])$$:
$$T_{3}^{2} + 1$$ T3^2 + 1 $$T_{5}^{2} + 9$$ T5^2 + 9 $$T_{7} - 1$$ T7 - 1 $$T_{11}^{2} + 36$$ T11^2 + 36
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2} + 1$$
$5$ $$T^{2} + 9$$
$7$ $$(T - 1)^{2}$$
$11$ $$T^{2} + 36$$
$13$ $$T^{2} + 1$$
$17$ $$(T + 3)^{2}$$
$19$ $$T^{2} + 4$$
$23$ $$T^{2}$$
$29$ $$T^{2} + 36$$
$31$ $$(T + 4)^{2}$$
$37$ $$T^{2} + 49$$
$41$ $$T^{2}$$
$43$ $$T^{2} + 1$$
$47$ $$(T - 3)^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2} + 36$$
$61$ $$T^{2} + 64$$
$67$ $$T^{2} + 196$$
$71$ $$(T - 3)^{2}$$
$73$ $$(T + 2)^{2}$$
$79$ $$(T - 8)^{2}$$
$83$ $$T^{2} + 144$$
$89$ $$(T - 6)^{2}$$
$97$ $$(T + 10)^{2}$$ | 3,610 | 6,726 | {"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} | 2.796875 | 3 | CC-MAIN-2022-27 | latest | en | 0.222256 |
https://www.aqua-calc.com/one-to-one/area/square-yard/square-mile/1 | 1,582,480,735,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00439.warc.gz | 636,542,873 | 6,752 | # 1 square yard [yd²] in square miles
## yd² to mi² unit converter of area
1 square yard [yd²] = 3 × 10-7 square mile [mi²]
### square yards to square miles area conversion cards
• 1
through
25
square yards
• 1 yd² to mi² = 3 × 10-7 mi²
• 2 yd² to mi² = 6 × 10-7 mi²
• 3 yd² to mi² = 1 × 10-6 mi²
• 4 yd² to mi² = 1.3 × 10-6 mi²
• 5 yd² to mi² = 1.6 × 10-6 mi²
• 6 yd² to mi² = 1.9 × 10-6 mi²
• 7 yd² to mi² = 2.3 × 10-6 mi²
• 8 yd² to mi² = 2.6 × 10-6 mi²
• 9 yd² to mi² = 2.9 × 10-6 mi²
• 10 yd² to mi² = 3.2 × 10-6 mi²
• 11 yd² to mi² = 3.6 × 10-6 mi²
• 12 yd² to mi² = 3.9 × 10-6 mi²
• 13 yd² to mi² = 4.2 × 10-6 mi²
• 14 yd² to mi² = 4.5 × 10-6 mi²
• 15 yd² to mi² = 4.8 × 10-6 mi²
• 16 yd² to mi² = 5.2 × 10-6 mi²
• 17 yd² to mi² = 5.5 × 10-6 mi²
• 18 yd² to mi² = 5.8 × 10-6 mi²
• 19 yd² to mi² = 6.1 × 10-6 mi²
• 20 yd² to mi² = 6.5 × 10-6 mi²
• 21 yd² to mi² = 6.8 × 10-6 mi²
• 22 yd² to mi² = 7.1 × 10-6 mi²
• 23 yd² to mi² = 7.4 × 10-6 mi²
• 24 yd² to mi² = 7.7 × 10-6 mi²
• 25 yd² to mi² = 8.1 × 10-6 mi²
• 26
through
50
square yards
• 26 yd² to mi² = 8.4 × 10-6 mi²
• 27 yd² to mi² = 8.7 × 10-6 mi²
• 28 yd² to mi² = 9 × 10-6 mi²
• 29 yd² to mi² = 9.4 × 10-6 mi²
• 30 yd² to mi² = 9.7 × 10-6 mi²
• 31 yd² to mi² = 1 × 10-5 mi²
• 32 yd² to mi² = 1.03 × 10-5 mi²
• 33 yd² to mi² = 1.07 × 10-5 mi²
• 34 yd² to mi² = 1.1 × 10-5 mi²
• 35 yd² to mi² = 1.13 × 10-5 mi²
• 36 yd² to mi² = 1.16 × 10-5 mi²
• 37 yd² to mi² = 1.19 × 10-5 mi²
• 38 yd² to mi² = 1.23 × 10-5 mi²
• 39 yd² to mi² = 1.26 × 10-5 mi²
• 40 yd² to mi² = 1.29 × 10-5 mi²
• 41 yd² to mi² = 1.32 × 10-5 mi²
• 42 yd² to mi² = 1.36 × 10-5 mi²
• 43 yd² to mi² = 1.39 × 10-5 mi²
• 44 yd² to mi² = 1.42 × 10-5 mi²
• 45 yd² to mi² = 1.45 × 10-5 mi²
• 46 yd² to mi² = 1.49 × 10-5 mi²
• 47 yd² to mi² = 1.52 × 10-5 mi²
• 48 yd² to mi² = 1.55 × 10-5 mi²
• 49 yd² to mi² = 1.58 × 10-5 mi²
• 50 yd² to mi² = 1.61 × 10-5 mi²
• 51
through
75
square yards
• 51 yd² to mi² = 1.65 × 10-5 mi²
• 52 yd² to mi² = 1.68 × 10-5 mi²
• 53 yd² to mi² = 1.71 × 10-5 mi²
• 54 yd² to mi² = 1.74 × 10-5 mi²
• 55 yd² to mi² = 1.78 × 10-5 mi²
• 56 yd² to mi² = 1.81 × 10-5 mi²
• 57 yd² to mi² = 1.84 × 10-5 mi²
• 58 yd² to mi² = 1.87 × 10-5 mi²
• 59 yd² to mi² = 1.9 × 10-5 mi²
• 60 yd² to mi² = 1.94 × 10-5 mi²
• 61 yd² to mi² = 1.97 × 10-5 mi²
• 62 yd² to mi² = 2 × 10-5 mi²
• 63 yd² to mi² = 2.03 × 10-5 mi²
• 64 yd² to mi² = 2.07 × 10-5 mi²
• 65 yd² to mi² = 2.1 × 10-5 mi²
• 66 yd² to mi² = 2.13 × 10-5 mi²
• 67 yd² to mi² = 2.16 × 10-5 mi²
• 68 yd² to mi² = 2.2 × 10-5 mi²
• 69 yd² to mi² = 2.23 × 10-5 mi²
• 70 yd² to mi² = 2.26 × 10-5 mi²
• 71 yd² to mi² = 2.29 × 10-5 mi²
• 72 yd² to mi² = 2.32 × 10-5 mi²
• 73 yd² to mi² = 2.36 × 10-5 mi²
• 74 yd² to mi² = 2.39 × 10-5 mi²
• 75 yd² to mi² = 2.42 × 10-5 mi²
• 76
through
100
square yards
• 76 yd² to mi² = 2.45 × 10-5 mi²
• 77 yd² to mi² = 2.49 × 10-5 mi²
• 78 yd² to mi² = 2.52 × 10-5 mi²
• 79 yd² to mi² = 2.55 × 10-5 mi²
• 80 yd² to mi² = 2.58 × 10-5 mi²
• 81 yd² to mi² = 2.61 × 10-5 mi²
• 82 yd² to mi² = 2.65 × 10-5 mi²
• 83 yd² to mi² = 2.68 × 10-5 mi²
• 84 yd² to mi² = 2.71 × 10-5 mi²
• 85 yd² to mi² = 2.74 × 10-5 mi²
• 86 yd² to mi² = 2.78 × 10-5 mi²
• 87 yd² to mi² = 2.81 × 10-5 mi²
• 88 yd² to mi² = 2.84 × 10-5 mi²
• 89 yd² to mi² = 2.87 × 10-5 mi²
• 90 yd² to mi² = 2.91 × 10-5 mi²
• 91 yd² to mi² = 2.94 × 10-5 mi²
• 92 yd² to mi² = 2.97 × 10-5 mi²
• 93 yd² to mi² = 3 × 10-5 mi²
• 94 yd² to mi² = 3.03 × 10-5 mi²
• 95 yd² to mi² = 3.07 × 10-5 mi²
• 96 yd² to mi² = 3.1 × 10-5 mi²
• 97 yd² to mi² = 3.13 × 10-5 mi²
• 98 yd² to mi² = 3.16 × 10-5 mi²
• 99 yd² to mi² = 3.2 × 10-5 mi²
• 100 yd² to mi² = 3.23 × 10-5 mi²
#### Foods, Nutrients and Calories
PRIVATE SELECTION, TORTILLA CHIPS, BLUE CORN, UPC: 011110808561 contain(s) 536 calories per 100 grams or ≈3.527 ounces [ price ]
LECOUR'S, PECAN SHORTBREAD COOKIES, UPC: 051933053540 contain(s) 531 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Selenium, Se, foods low in Selenium, Se, and Recommended Dietary Allowances (RDAs) for Selenium
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Bimini Pink weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
seawater weighs 1 024 kg/m³ (63.92623 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-401B, liquid (R401B) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
A cubic picometer (pm³) is a derived metric SI (System International) measurement unit of volume with sides equal to one picometer (1pm)
The radiation absorbed dose is a measurement of radiation, in energy per unit of mass, absorbed by a specific object, such as human tissue.
ch³ to Ml conversion table, ch³ to Ml unit converter or convert between all units of volume measurement.
#### Calculators
Volume of a rectangular box calculator. Definition and formulas. | 2,796 | 5,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.146774 |
http://geekswithblogs.net/MarkPearl/archive/2010/08/10/a-solid-example-of-the-difference-between-a-functional-amp.aspx | 1,610,989,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00713.warc.gz | 41,798,077 | 11,437 | Geeks With Blogs
News
A few months ago I had a really frustrating debate with my younger brother. He had come up to JHB to come for a visit and we decided to talk about programming. Of course I thought I would put a good pitch in for F#, but just couldn’t seem to do it any justice.
Eventually his point was as follows - “What really is the difference between declaring a functional solution vs an iterative solution. Sure, in F# you have something like the Seq.map function, but isn’t it just a shorthand for a for loop or something like that – isnt it just syntactical sugar?”
Of course he was wrong, but at the time I just battled to give him a concrete and yet simple enough example of how a functional solution is at a higher level than an iterative solution.
Then a few weeks ago I found a classic example which I think might illustrate the point a bit better. It all happened while I was looking into tail-recursive functions through reflector.
Lets say you have a recursive solution in F# that looks like the following…
```let rec fooNonTail n =
match n with
| 0 -> 0
| _ -> 2 + fooNonTail (n-1) ```
If you were to go and look at the equivalent C# code generated via reflector (by examining the IL code) you would see something like the following…
```public static int fooNonTail(int n)
{
switch (n)
{
case 0:
return 0;
}
return (2 + fooNonTail(n - 1));
}```
That pretty much makes sense… we defined a recursive function in F#, and the IL came out as a recursive function in C#…
Now it gets interesting… let’s say instead of having the original F# fooNonTail function we replaced it with a tail recursive function that looked like the following in F#…
```let fooTailRec n =
let rec innerfooTailRec acc n =
match n with
| 0 when n <=0 -> acc
| _ -> innerfooTailRec (acc+2) (n-1)
innerfooTailRec 0 n ```
Not to much of a difference from the fooNonTail function. We have moved things around a bit, but in essence it is still a recursive function.
This is where the fun begins… suddenly it gets interesting when we look at what has happened behind the scenes in the IL code.
Instead of getting a similar method to the first IL code we have to look around a bit more for the actual implementation, and when you find it, it looks something like the following…
```public override int Invoke(int acc, int n)
{
while (true)
{
switch (n)
{
case 0:
{
int num = n;
int num2 = 0;
if (num > num2)
{
break;
}
return acc;
}
}
n--;
acc += 2;
}
}```
That’s not recursive at all??? What happened?
What has happened here is somewhere, something recognized that instead of expressing the function in IL code as recursive, it could instead convert it to a normal iterative while loop… which has several operational/performance benefits.
To me that’s amazing and is a classic example of part of the abstraction layer that a functional language has over an iterative language. It works at just a slightly higher level. In this situation, because I was “solving” the problem, instead of telling the compiler how to iterate each step to get to a solution, when the compiler came across a situation where it recognized it could reformat for optimizations, it did it behind the scenes and wala…
So while this is not a complete blog post on the power of functional languages and F# in particular, I believe it at least illustrates that F# works at a slightly higher level than C# (iteratively) and that it is well worth examining the IL code in reflector to see exactly how your F# applications are translated.
What does worry me a bit about such an example is that it is subtle… meaning the optimization could easily be missed unless you kept an eye out for it.
Well, I look forward to your comments / crits if you have anything to add… or examples to give - just keep it nice ;-)
Related Posts on Geeks With Blogs Matching Categories
Comments on this post: A solid example of the difference between a Functional & Iterative Language. F# style..
# re: A solid example of the difference between a Functional & Iterative Language. F# style..
And of course, the IL or even the byte code it gets turned into is just a higher level abstraction of what goes on in the registers. As you point out, the nice thing about F# is that -- having a wider range of abstraction metaphors -- it lets one craft code for which the solution resembles the problem. And this important because a problem and its solution are really just two sides of the same coin.
Left by TechNeilogy on Aug 10, 2010 9:35 PM
# re: A solid example of the difference between a Functional & Iterative Language. F# style..
Yes... I agree 100% - with me the challenge has been once you are used to a certain level of granualirity (i.e. iterative) - its hard to move to a higher level of abstraction... I guess it is kind of what a c++ programmer experienced when moving to a managed language like c#
Left by MarkPearl on Aug 11, 2010 7:39 AM | 1,138 | 4,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-04 | latest | en | 0.959574 |
https://support.stockcharts.com/doku.php?id=scans:troubleshooting:equality_scans | 1,685,609,541,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00023.warc.gz | 599,347,158 | 9,428 | # Troubleshooting Equality Scans
Equals signs (=) are often used in scan clauses when you are looking for a specific symbol, country, exchange, etc. The equals sign can also be used to look for a specific numeric value in a scan clause, but you may not always get the results you expect. Let's look at an example:
In the above chart, both the top MACD Line (12,26,9) and the bottom MACD Line (20,36,9) for HBI appear to have the same value: 0.800.
## Scanning for Equality
Now let's run a scan to look for symbols where those two MACD Line values are equal to each other:
```[type is stock] and [country is US] and [sma(20,volume) > 40000]
and [close > 21]
and [MACD Line(12,26,9) = MACD Line(20,36,9)]```
If we ran this scan on the same date as the chart, the scan would net us only one result:
Despite the fact that the HBI chart showed identical values for these two indicators, HBI does not appear in our scan results.
The reason for this discrepancy is that the values displayed on the chart are rounded to a certain number of decimal points (typically 2 or 3) for display purposes. The values used by the Scan Engine are not rounded to this degree - the more precise values used by the Scan Engine are often not exactly equal. If the longer-term MACD Line (20,36,9) is 0.8003 and the shorter-term MACD Line (12,26,9) is 0.8004, the Scan Engine does not consider those two values to be equal.
How can we fix our scan so we get the results we expect?
## Scanning for "Close Enough"
When comparing one calculated value to another, it is better to replace the one “equals” clause with two “close enough” clauses, like this:
```and [MACD Line(12,26,9) > [MACD Line(20,36,9) - 0.01]]
and [MACD Line(12,26,9) <= [MACD Line(20,36,9) + 0.01]]```
This tells the Scan Engine to return results where one MACD Line is within one cent above or below the other - not identical values, but extremely close to each other. When we run the updated scan, we see that HBI now appears in the scan results:
For even greater precision, you could scan for values within one tenth of a cent of each other:
```and [MACD Line(12,26,9) > [MACD Line(20,36,9) - 0.001]]
and [MACD Line(12,26,9) <= [MACD Line(20,36,9) + 0.001]]```
As long as the values are within that small range defined by the two clauses, they are “close enough” to be considered “equal” by our scan.
## Conclusion
The equals sign is great for non-numeric comparisons, but should be used with caution when comparing numeric values. When you do need to compare numbers, it is better to use “close enough” clauses, defining a small range of values so similar that they are essentially equal. | 710 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.878658 |
https://support.office.com/en-us/article/logest-function-f27462d8-3657-4030-866b-a272c1d18b4b?ui=en-US&rs=en-US&ad=US | 1,553,201,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.29/warc/CC-MAIN-20190321193403-20190321215403-00313.warc.gz | 637,908,050 | 22,888 | LOGEST function
This article describes the formula syntax and usage of the LOGEST function in Microsoft Excel.
Description
In regression analysis, calculates an exponential curve that fits your data and returns an array of values that describes the curve. Because this function returns an array of values, it must be entered as an array formula.
The equation for the curve is:
y = b*m^x
or
y = (b*(m1^x1)*(m2^x2)*_)
if there are multiple x-values, where the dependent y-value is a function of the independent x-values. The m-values are bases corresponding to each exponent x-value, and b is a constant value. Note that y, x, and m can be vectors. The array that LOGEST returns is {mn,mn-1,...,m1,b}.
Syntax
LOGEST(known_y's, [known_x's], [const], [stats])
The LOGEST function syntax has the following arguments:
• Known_y's Required. The set of y-values you already know in the relationship y = b*m^x.
• If the array known_y's is in a single column, then each column of known_x's is interpreted as a separate variable.
• If the array known_y's is in a single row, then each row of known_x's is interpreted as a separate variable.
• Known_x's Optional. An optional set of x-values that you may already know in the relationship y = b*m^x.
• The array known_x's can include one or more sets of variables. If only one variable is used, known_y's and known_x's can be ranges of any shape, as long as they have equal dimensions. If more than one variable is used, known_y's must be a range of cells with a height of one row or a width of one column (which is also known as a vector).
• If known_x's is omitted, it is assumed to be the array {1,2,3,...} that is the same size as known_y's.
• Const Optional. A logical value specifying whether to force the constant b to equal 1.
• If const is TRUE or omitted, b is calculated normally.
• If const is FALSE, b is set equal to 1, and the m-values are fitted to y = m^x.
• Stats Optional. A logical value specifying whether to return additional regression statistics.
• If stats is TRUE, LOGEST returns the additional regression statistics, so the returned array is {mn,mn-1,...,m1,b;sen,sen-1,...,se1,seb;r 2,sey; F,df;ssreg,ssresid}.
• If stats is FALSE or omitted, LOGEST returns only the m-coefficients and the constant b.
Remarks
• The more a plot of your data resembles an exponential curve, the better the calculated line will fit your data. Like LINEST, LOGEST returns an array of values that describes a relationship among the values, but LINEST fits a straight line to your data; LOGEST fits an exponential curve. For more information, see LINEST.
• When you have only one independent x-variable, you can obtain y-intercept (b) values directly by using the following formula:
Y-intercept (b):
INDEX(LOGEST(known_y's,known_x's),2)
You can use the y = b*m^x equation to predict future values of y, but Microsoft Excel provides the GROWTH function to do this for you. For more information, see GROWTH function.
• Formulas that return arrays must be entered as array formulas.
Note: In Excel Online you cannot create array formulas.
• When entering an array constant such as known_x's as an argument, use commas to separate values in the same row and semicolons to separate rows. Separator characters may be different depending on your regional settings.
• You should note that the y-values predicted by the regression equation may not be valid if they are outside the range of y-values you used to determine the equation.
Example 1
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Month Units 11 33100 12 47300 13 69000 14 102000 15 150000 16 220000 Formula Description Result =LOGEST(B2:B7,A2:A7, TRUE, FALSE) Note The formula in the example must be entered as an array formula in the Excel program. After copying the example to a blank worksheet, select the range C9:D9 starting with the formula cell. Press F2, and then press CTRL+SHIFT+ENTER. If the formula is not entered as an array formula, the single result is 1.4633. Note: If you have a current version of Office 365, then you can simply enter the formula in the top-left-cell of the output range, then press ENTER to confirm the formula as a dynamic array formula. Otherwise, the formula must be entered as a legacy array formula by first selecting the output range, entering the formula in the top-left-cell of the output range, and then pressing CTRL+SHIFT+ENTER to confirm it. Excel inserts curly brackets at the beginning and end of the formula for you. For more information on array formulas, see Guidelines and examples of array formulas. 1.4633 495.3048
Example 2
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Month Units 11 33,100 12 47,300 13 69,000 14 102,000 15 150,000 16 220,000 Formula Result =LOGEST(B2:B7,A2:A7,TRUE,TRUE) 1.4633 | 1,286 | 5,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.837189 |
http://mathhelpforum.com/algebra/62177-dividing-print.html | 1,529,586,094,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00143.warc.gz | 211,943,835 | 2,611 | dividing
• Nov 29th 2008, 10:34 AM
Mccoy31
dividing
(6y4 + 15y3 + 28y + 6) ÷ (y + 3)
• Nov 29th 2008, 10:54 AM
Mccoy31
• Nov 29th 2008, 12:29 PM
masters
Quote:
Originally Posted by Mccoy31
(6y4 + 15y3 + 28y + 6) ÷ (y + 3)
Use synthetic division
Code:
-3 ] 6 15 0 28 6 -18 9 -27 -3 ------------------------------- 6 -3 9 1 3
$\displaystyle 6y^3-3y^2+9y+1+\frac{3}{y+3}$ | 205 | 434 | {"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} | 3.09375 | 3 | CC-MAIN-2018-26 | latest | en | 0.180314 |
https://math.stackexchange.com/questions/1851140/binary-integer-variables-in-linear-programming | 1,652,790,031,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00630.warc.gz | 455,698,358 | 67,774 | # Binary integer variables in linear programming
Could someone please explain the concept of switch variables (binary integer decision variables) in linear programming?
This example has two alternative constraints
$$\begin{array}{ll} \text{maximize} & 1.5x_1 + 2x_2\\ \text{subject to} & x_1, x_2 \leq 300\\ & x_1 = 0 \quad \mbox{XOR} \quad x_1 \geq 10\end{array}$$
I have seen examples of solutions for such tasks by applying something like following:
$$x_1+My_1 = 0\\x_1 - My_1 \geq 10+M$$
Does someone know and understand this approach and can explain it to me?
• Of course when $x_1=0,$ it cannot happen $x_1\ge10,$ so one can just use OR. Jul 6, 2016 at 16:31
Note that
$$\begin{array}{rl} x_1 = 0 \lor x_1 \geq 10 &\equiv (x_1 \geq 0 \land x_1 \leq 0) \lor x_1 \geq 10\\\\ &\equiv x_1 \geq 0 \land (x_1 \leq 0 \lor x_1 \geq 10)\end{array}$$
We can handle the disjunction $$x_1 \leq 0 \lor x_1 \geq 10$$ using the Big M method. We introduce binary variables $$z_1, z_2 \in \{0,1\}$$ such that $$z_1 + z_2 = 1$$, i.e., either $$(z_1,z_2) = (1,0)$$ or $$(z_1,z_2) = (0,1)$$. We introduce also a large constant $$M \gg 10$$ so that we can write the disjunction in the form
$$x_1 \leq M z_1 \land x_1 \geq 10 - M z_2$$
If $$(z_1,z_2) = (1,0)$$, we have $$x_1 \leq M$$ and $$x_1 \geq 10$$, which is roughly "equivalent" to $$x_1 \geq 10$$. If $$(z_1,z_2) = (0,1)$$, we have $$x_1 \leq 0$$ and $$x_1 \geq 10 - M$$, which is roughly "equivalent" to $$x_1 \leq 0$$.
Thus, we have a mixed-integer linear program (MILP)
$$\begin{array}{ll} \text{maximize} & 1.5x_1 + 2x_2\\ \text{subject to} & x_1, x_2 \leq 300\\ & x_1 \geq 0\\ & x_1 - M z_1\leq 0\\ & x_1 + M z_2 \geq 10\\ & z_1 + z_2 = 1\\ & z_1, z_2 \in \{0,1\}\end{array}$$
For a quick overview of MILP, read Mixed-Integer Programming for Control by Arthur Richards and Jonathan How.
• If you set $M$ to $300$ in the third constraint and $M$ to $10$ in the fourth constraint, and replace $z_2$ by $1-z_1$, you end up with exactly Erwin Kalvelagen's solution (his $\delta$ is your $z_1$). Jul 7, 2016 at 3:16
• Would it change anything if the first of the alternative constraints (x1 = 0) would have a value higher than zero at the right hand side? Jul 7, 2016 at 18:27
• @BastianSchoettle How much higher? Read my other answer to this question. If $x_1 = a$, where $a \in [10, 300]$, then the half-line is inside the polytope. Enlarging the feasible region cannot decrease the maximum. Jul 7, 2016 at 19:34
• @Kuifje The question alluded to the Big M method. Erwin's approach is much simpler, but it does not use any big M's. Jul 7, 2016 at 19:59
• @Rodrigo de Azevedo: Erwin's approach IS a big $M$ method, with specific values of $M$. As I mentioned above: if the first $M$ equals $300$, then you have $$x_1-300z_1\le 0$$ If the second equals $10$ then you have $$x_1+10z_2\ge 10$$ Now given that $z_1=1-z_1$, both equations are equivalent to $$10z_1\le x_1\le 300z_1$$ Jul 7, 2016 at 21:47
Using an extra binary variable $\delta$ we can write: \begin{align} & 10 \delta \le x_1 \le 300 \delta \\ &\delta \in \{0,1\} \end{align} $x_1$ is called a semi-continuous variable and some solvers support this directly without the need for extra binary variables.
Note that $x_1 = 0$ and $x_1 \geq 10$ are mutually exclusive.
Writing the inequality constraints in Disjunctive Normal Form (DNF), we obtain
$$\begin{array}{rl} & (x_1 \leq 300 \land x_2 \leq 300) \land (x_1 = 0 \lor x_1 \geq 10) \equiv\\\\ \equiv& (x_1 = 0 \land x_2 \leq 300) \lor (10 \leq x_1 \leq 300 \land x_2 \leq 300)\end{array}$$
Thus, the feasible region is the union of a half-line and a polytope. Hence, we solve two linear programs, namely,
$$\begin{array}{ll} \text{maximize} & 1.5x_1 + 2x_2\\ \text{subject to} & x_1 = 0\\ & x_2 \leq 300\end{array}$$
and
$$\begin{array}{ll} \text{maximize} & 1.5x_1 + 2x_2\\ \text{subject to} & x_1, x_2 \leq 300\\ & x_1 \geq 10\end{array}$$
and then take the maximum of the maxima of each linear program:
• over the half-line, the maximum is $600$, which is attained at $(0,300)$.
• over the polytope, the maximum is $1050$, which is attained at $(300,300)$.
• Thank you for the formatting and your explanation but I need to solve this task in a single model. I've updated my answer accordingly. Jul 6, 2016 at 19:10 | 1,621 | 4,301 | {"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": 17, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-21 | latest | en | 0.747524 |
https://oneclass.com/class-notes/ca/u-regina/biol/biol488/133709-lecture-17docx.en.html | 1,621,169,590,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00390.warc.gz | 456,032,647 | 145,107 | # BIOL488 Lecture Notes - Lecture 17: Nernst Equation, Resting Potential, Electrical Resistance And Conductance
19 views4 pages
7 Feb 2013
School
Department
Course
The Movement of Ions
Ionic movements through channels are influenced by two factors: diffusion and electricity
Diffusion
o Ions and molecules dissolved in water are in constant motion.
o This temperature-dependent, random movement will tend to distribute the ions
evenly throughout the solution
o There will be a net movement of ions from regions of high concentration to
regions of low concentration diffusion
o Diffusion will cause ions to be pushed through channels in the membrane
For example, NaCl is dissolved in fluid on one side of a permeable
membrane
The Na+ and Cl- ions will cross until they are evenly distributed in the
solutions on both sides
The difference in concentration on both sides is called a concentration
Thus, it is said that ions flow down a concentration gradient
o Driving ions across the membrane by diffusion happens when:
1) The membrane possess channels permeable to the ions, and
2) There is a concentration gradient across the membrane
Electricity
o Another way to induce a net movement of ions in a solution is to use an electrical
field
o Since opposite charges attract and like charges repel, there will be a net
movement of Na+ toward the negative terminal and of Cl- toward the positive
terminal.
o The movement of electrical charge is called electrical current (I) and is
measured in amperes
o Two important factors determine how much current will flow: electrical potential
and electrical conductance
o Electrical potential (voltage V) is the force exerted on a charged particle and it
reflects the difference in charge between the anode and the cathode
More current will flow as this difference is increased
o Electrical conductance (G) is the relative ability of an electrical charge to
migrate from one point to another
Depends on the number of particles available to carry electrical charge
and the ease with which these particles can travel through space
Electrical resistance (Ω) is simply the inverse of conductance and it is
the relative inability of an electrical charge to migrate
There is a simple relationship known as Ohm’s Law and it is I = gV.
If the conductance is zero, no current will flow even when the
potential difference is very large
The Ionic Basis of the Resting Membrane Potential
The membrane potential (Vm) is the voltage across the neuronal membrane at any
moment.
Vm can be measured by inserting a microelectrode into the cytosol
o A typical microelectrode is a thing glass tube with an extremely fine tip that will
penetrate the membrane of a neuron with minimal damage
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Class Notes
Textbook Notes | 634 | 2,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-21 | latest | en | 0.905726 |
https://kootenaychinwoo.com/debt-ratio-formula-examples-calculation/ | 1,586,195,805,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00416.warc.gz | 535,762,007 | 12,807 | hello everyone welcome to the channel of
Wallstreetmojo today we are going to discuss on the tutorial of debt ratio
the debt ratio as you can see in this yellow box debt ratio is one of the most
used solvency ratio by the investors so let’s directly jump on the debt ratio
formula it is one of the most used solvency ratio by the investor and it’s
pretty easy to calculate to let’s have a look at the formula of the debt ratio
and get into the nitty-gritty now the debt ratio formula is very simple
actually the debt ratio is equal to the total liabilities divided by the the
total asset it is as simple as that so basically we are comparing as you can
see the total liability in assets all you need to do is to look at the balance
sheet and find out where a firm has enough total assets to pay off its total
liability if you go in the explanation part of the debt ratio formula for
investors the financial statement are everything they look at all for statements and make their judgments one of the most important financial
statement is balance sheet by looking at the balance sheet the investors are able
to know what’s working for a come in what’s need to be improved so two of the
most important items on the balance sheets are assets and liabilities by
looking at the total assets and total liabilities investors are able to
understand with the form enough assets to pay off the liabilities and
that’s exactly what we call as debt ratio see by using this ratio we
calculate the proportionally total assets and total liabilities and by
looking at them we get to know the stands of the company at any stage so
what is the use of the debt ratio formula let’s look at that this ratio is
useful for two groups of people the first group is a top management of the
company who’s directly responsible for the
expansion or contraction of a company by using this ratio the top management sees
whether the company has enough resources to pay off its obligation the second
group is the investors who would like to see the position of the company before
they ever put in the money into the company and that’s why the investors
needs to know whether the form has enough assets to bear the expense of the
debt and other obligations so this ratio also measures the
financial leverage of the company and it also tells us the investors how leverage
the form is if the form has a higher level of liability is compared to the
assets then the form has more financial leverage and vice versa the case okay so
now let’s look at the example of the debt ratio formula let’s take a very
practical example to illustrate this ratio boom company has the following
details that we are going to discuss here the current asset non-current asset
details are I’m just trying to push the details over here the current
liabilities then we have the non current liabilities I’m just writing live over
here so this are all the details okay and let’s put the amount this is the
particulars and this are the amounts in dollars ok now the current assets is
close enough to 30 thousand then we have the non current assets as close enough
to 3 lakh dollars current liability let’s take as 40,000 and non current
liabilities as \$70,000 everything is in dollars now find out the debt ratio of
boom company this is we are discussing for boom company okay
in the above example we can see that we need to total the current and
non-current assets and also the current liabilities and non current liabilities
so the total assets are current assets + non current assets so the current
assets and the non current assets is 30000 + 3 lakh that is 3 lakh 30000 okay so I’m just quickly going to write over here as is equal to 30000 + 3 lakh so that’s the total of the total assets are current and
non-current now the total liabilities are current liabilities and non current
liabilities so let’s make the total 40,000 + 70,000 that’s 1 lakh 10000
okay so this are the two details okay now the debt ratio formula is the total
liability / the total asset so what is the total
liability over here as you can see one lakh ten okay so our final answer is
going to be is equal to the total liabilities divided by the the total
assets that is three lakh thirty so our debt ratio comes to 0.3333 okay the ratio of the boom company is close
enough to 0.33 so to know whether the disproportion between
the total liabilities and total asset is healthy or not we need to see the
similar companies under the same industry if the ratio of those companies
also in the similar range it means boom company is doing quite well in the
normal situation as lower as this ratio can be better as in terms of the
investment in solvency the debt ratio in terms of Excel with template you know we
can take a look at that and this is very simple I mean you need to provide the
two inputs of total liabilities in total asset and you can easily calculate the
ratio in the template as you can see in this particular template whatever we
have calculated in our excel sheet is over here you can see the current assets
non current the total assets three lakh thirty again the current liabilities
non-current liabilities forty plus seventy one hundred and one like ten
thousand so the total asset total liabilities and total asset which gives
us the debt ratio for the boom company now basically you can use this debt
ratio calculator to calculate your debt ratio okay so let’s put our total
liability as 2 lakh and a total asset as 1 lakh so that gives us our debt
ratio formula is two times so the debt is two times the total assets if we debt
our if we changed a debtor one over here and two over here we will get the debt
ratio formula is 0.5 X it is usually considered to as a greater it’s a good
number it’s a good way to calculate okay and if we put things over here as 5 and 2 or 5 and 1 over here you can see the debt ratio has increased to
5 see but 5 X is a very leveraged company like you know
companies like Jet Airways it has capital intensive company they have a debt close
enough to 12000 crores that is in terms of in INR the in the that’s the Indian currency INR will if you see that is a highly leveraged
company and that the debt is close enough to 12000 that means it’s that ratio is close enough to 4.5 X which is really high so such companies are really risking their cash flows their operations because
majority of their profits or funds are blocked in interest so the company
should not be highly leveraged or even if they are then they should set off
with some of the fixed deposits if they have any so that you know you can set
off the interest of the fixed deposit and the interest of the the interest of
the debt that you are going to pay so this is a debt calculator you can
calculate the debt ratio with the help of for this calculator I hope you have
got the concept regarding the debt ratio thank you | 1,502 | 6,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-16 | latest | en | 0.966444 |
http://www.ncatlab.org/nlab/show/torsion+of+a+metric+connection | 1,449,007,034,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2015-48/segments/1448398471441.74/warc/CC-MAIN-20151124205431-00242-ip-10-71-132-137.ec2.internal.warc.gz | 573,300,922 | 8,823 | # nLab torsion of a metric connection
### Context
#### Riemannian geometry
Riemannian geometry
## Applications
#### Differential geometry
differential geometry
synthetic differential geometry
## Applications
For other notions of torsion see there.
# Contents
## Definition
A (pseudo) Riemannian metric with metric-compatible Levi-Civita connection on a smooth manifold $X$ may be encoded by a connection with values in the Poincaré Lie algebra $\mathfrak{iso}(p,q)$.
This Lie algebra is the semidirect product
$\mathfrak{iso}(p,q) \simeq \mathfrak{so}(p,q) \ltimes \mathbb{R}^{p+q}$
of the special orthogonal Lie algebra and the abelian translation Lie algebra. Accordingly, a connection 1-form has two components
• $\Omega \in \Omega^1(U,\mathfrak{so}(p,q))$ (sometimes called the “spin connection”);
• $E \in \Omega^1(U,\mathbb{R}^{p+q})$ (sometimes called the “vielbein”).
The metric itself is
$g = \langle E \otimes E \rangle \,.$
Accordingly also the curvature 2-form has two components:
• $R = d \Omega + [\Omega \wedge \Omega] \in \Omega^2(U, \mathfrak{so}(p,q))$ – the Riemann curvature;
• $\tau = d E + [\Omega \wedge E]$ – the torsion.
This is the special case of the more general concept of torsion of a Cartan connection.
## Generalizations
In supergeometry a metric structure is given by a connection with values in the super Poincaré Lie algebra. The corresponding notion of torsion has an extra contribution from spinor fields: the super torsion?.
geometric contextgauge groupstabilizer subgrouplocal model spacelocal geometryglobal geometrydifferential cohomologyfirst order formulation of gravity
differential geometryLie group/algebraic group $G$subgroup (monomorphism) $H \hookrightarrow G$quotient (“coset space”) $G/H$Klein geometryCartan geometryCartan connection
examplesEuclidean group $Iso(d)$rotation group $O(d)$Cartesian space $\mathbb{R}^d$Euclidean geometryRiemannian geometryaffine connectionEuclidean gravity
Poincaré group $Iso(d-1,1)$Lorentz group $O(d-1,1)$Minkowski spacetime $\mathbb{R}^{d-1,1}$Lorentzian geometrypseudo-Riemannian geometryspin connectionEinstein gravity
anti de Sitter group $O(d-1,2)$$O(d-1,1)$anti de Sitter spacetime $AdS^d$AdS gravity
de Sitter group $O(d,1)$$O(d-1,1)$de Sitter spacetime $dS^d$deSitter gravity
linear algebraic groupparabolic subgroup/Borel subgroupflag varietyparabolic geometry
conformal group $O(d,t+1)$conformal parabolic subgroupMöbius space $S^{d,t}$conformal geometryconformal connectionconformal gravity
supergeometrysuper Lie group $G$subgroup (monomorphism) $H \hookrightarrow G$quotient (“coset space”) $G/H$super Klein geometrysuper Cartan geometryCartan superconnection
examplessuper Poincaré groupspin groupsuper Minkowski spacetime $\mathbb{R}^{d-1,1\vert N}$Lorentzian supergeometrysupergeometrysuperconnectionsupergravity
super anti de Sitter groupsuper anti de Sitter spacetime
higher differential geometrysmooth 2-group $G$2-monomorphism $H \to G$homotopy quotient $G//H$Klein 2-geometryCartan 2-geometry
cohesive ∞-group∞-monomorphism (i.e. any homomorphism) $H \to G$homotopy quotient $G//H$ of ∞-actionhigher Klein geometryhigher Cartan geometryhigher Cartan connection
examplesextended super Minkowski spacetimeextended supergeometryhigher supergravity: type II, heterotic, 11d
Revised on December 18, 2014 14:44:39 by Urs Schreiber (127.0.0.1) | 974 | 3,370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 35, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.695913 |
https://taccle2.eu/primary/by-subject/primary-maths/fraction-families-wales | 1,560,721,347,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998298.91/warc/CC-MAIN-20190616202813-20190616224813-00146.warc.gz | 620,542,931 | 18,066 | # Fraction Families
Age: 7+yrs
Ease ****
Overview
As you’ll see, this software is very adaptable. We strongly suggest you view the audio/visual demo outlining all the functions: http://www.visnos.com/demos/fraction-wall
This same activity can be used with different year groups so long as you choose the age-appropriate ‘family’ of fractions for your wall i.e. 2, 3 or 5. For older and/or more able learners, you can choose ‘large’ which gives you every fraction family or ‘mixed’ which gives you a random selection of fraction families. In this activity we refer to fractions as belonging ‘families’ i.e. The Twos would be halves, quarters, eighths and sixteenths and so on.
Description
Show the blank fraction wall to the class. Explain what it is and give a quick demo of how to use the mouse to highlight individual fractions. Explain to them that they are going to investigate fraction families, and how some fractions can actually have exactly the same value. Tell them that we call these equivalent fractions.
Ensure the ‘Eq’ is showing a green tick, not a red cross, and that you’ve selected the appropriate family, and then click on, say, ¼ on the fraction wall. Ask learners what they notice. Draw their attention to the ‘totals’ boxes running down the right hand side of the fraction wall. Ask them what these fractions tell us.
Ask learners to use the fraction wall to find as many equivalent fractions as possible. You may want them to write their answers like this ¼ = 2/8. Explain that they only have 5 minutes and when you call them back you’ll be discovering who found the most.
When you have done this (and checked the answers!), you can opt to move on to a more difficult wall, repeat the activity using the same wall but ask them to find equivalent decimals or percentages, or display a random wall and ask them to verbalise their observations as a means of consolidating what they’ve already learnt e.g. “One half is equal to two quarters”.
What do I need?
http://www.visnos.com/demos/fraction-wall work books or work sheets to record answers. Interactive whiteboard.
We used www.visnos.com because it is easy enough for less confident learners (and teachers!) to use.
The concept of ‘a fraction’ is difficult for many learners to understand – equivalent fractions even more so! This interactive fraction wall is very non-threatening yet demands that learners are completely engaged. Using it as we’ve done here, as an initial introduction, helps capture learners’ attention and is a great visual resource for learning a concept that can be abstract.
Hints and tips
You may decide (especially with older and/or more able learners) that learners would benefit from watching the online demo. The demo can be stopped or paused at any time so as not to overload them with information as to the all the functions.
We’ve referred to fractions as belonging to families, you must decide if this is appropriate for your learners. Either way, it doesn’t affect the learning outcomes of the lesson.
We’ve found that this activity works particularly well when learners work in pairs, thereby providing peer support and feedback throughout the task.
Safety
This software is 100% safe. There are no links to other sites and learners do not interact with other users. It was written by a teacher (Michael McDaid) for teachers!
Other opportunities to use the same software:
This post is also available in: Dutch, German, Italian, Spanish, Portuguese (Portugal), Romanian, Welsh
Tags: | 744 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-26 | latest | en | 0.924672 |
http://mathhelpforum.com/calculus/99124-solving-inequality-interval-notation.html | 1,526,860,892,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00226.warc.gz | 192,983,243 | 9,591 | # Thread: Solving Inequality in Interval Notation
1. ## Solving Inequality in Interval Notation
Solve the following inequality. Write the answer in interval notation. Note If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed, enter (infinity sign) as infinity and -(infinity sign) as -infinity.
x^3 - 16x </=0
I realize this might be easy for you guys but it isn't obvious to me and an explanation would be greatly appreciated!
2. It's $\displaystyle x(x+4)(x-4)\le 0.$
Study each factor for $\displaystyle (-\infty,-4),$ $\displaystyle (-4,0),$ $\displaystyle (0,4)$ and on $\displaystyle (4,\infty).$
For example $\displaystyle x$ is negative for the first interval, $\displaystyle x+4$ too and $\displaystyle x-4$ too, so we have three factors whose signs are negative, respectively, and hence its product is negative, so $\displaystyle x(x+4)(x-4)\le 0$ for the first interval. (Including the $\displaystyle -4,$ which is obviously a solution too.)
Do this procedure for the others factors and you'll get full solution set. | 281 | 1,094 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.87184 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.13/h/sean1.html | 1,675,044,494,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00705.warc.gz | 34,123,484 | 3,075 | SEARCH HOME
Math Central Quandaries & Queries
Question from Sean, a parent: Hi I was wondering how you would calculate how may 7 digit phone numbers there are with only odd digits?
Hi Sean,
There are 5 odd digits, 1, 3, 5, 7 and 9.
Suppose you are going to construct such a phone number starting from left to right. You have 5 choices for the first digit.
1
2
3
4
5
Regardless of which digit you chose for the first digit you have 5 choices for the second digit so the list of possible phone numbers is
11
12
13
14
15
21
22
.
.
.
55
So there are $5 \times 5$ possible 2 digit phone numbers.
Regardless of which 2 digit phone number you have you have 5 choices for the third number. How many possible 3 digit phone numbers are there?
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 219 | 861 | {"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} | 3.328125 | 3 | CC-MAIN-2023-06 | latest | en | 0.909982 |
http://us.metamath.org/mpeuni/xpsc0.html | 1,638,761,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00005.warc.gz | 82,494,287 | 7,740 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > xpsc0 Structured version Visualization version GIF version
Theorem xpsc0 16144
Description: The pair function maps 0 to 𝐴. (Contributed by Mario Carneiro, 14-Aug-2015.)
Assertion
Ref Expression
xpsc0 (𝐴𝑉 → (({𝐴} +𝑐 {𝐵})‘∅) = 𝐴)
Proof of Theorem xpsc0
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 xpsc 16141 . . . 4 ({𝐴} +𝑐 {𝐵}) = (({∅} × {𝐴}) ∪ ({1𝑜} × {𝐵}))
21fveq1i 6151 . . 3 (({𝐴} +𝑐 {𝐵})‘∅) = ((({∅} × {𝐴}) ∪ ({1𝑜} × {𝐵}))‘∅)
3 fnconstg 6052 . . . 4 (𝐴𝑉 → ({∅} × {𝐴}) Fn {∅})
4 vex 3189 . . . . . . . . . . . . 13 𝑥 ∈ V
5 fvi 6214 . . . . . . . . . . . . 13 (𝑥 ∈ V → ( I ‘𝑥) = 𝑥)
64, 5ax-mp 5 . . . . . . . . . . . 12 ( I ‘𝑥) = 𝑥
7 elsni 4167 . . . . . . . . . . . . 13 (𝑥 ∈ {𝐵} → 𝑥 = 𝐵)
87fveq2d 6154 . . . . . . . . . . . 12 (𝑥 ∈ {𝐵} → ( I ‘𝑥) = ( I ‘𝐵))
96, 8syl5eqr 2669 . . . . . . . . . . 11 (𝑥 ∈ {𝐵} → 𝑥 = ( I ‘𝐵))
10 velsn 4166 . . . . . . . . . . 11 (𝑥 ∈ {( I ‘𝐵)} ↔ 𝑥 = ( I ‘𝐵))
119, 10sylibr 224 . . . . . . . . . 10 (𝑥 ∈ {𝐵} → 𝑥 ∈ {( I ‘𝐵)})
1211ssriv 3588 . . . . . . . . 9 {𝐵} ⊆ {( I ‘𝐵)}
13 xpss2 5192 . . . . . . . . 9 ({𝐵} ⊆ {( I ‘𝐵)} → ({1𝑜} × {𝐵}) ⊆ ({1𝑜} × {( I ‘𝐵)}))
1412, 13ax-mp 5 . . . . . . . 8 ({1𝑜} × {𝐵}) ⊆ ({1𝑜} × {( I ‘𝐵)})
15 1on 7515 . . . . . . . . . 10 1𝑜 ∈ On
1615elexi 3199 . . . . . . . . 9 1𝑜 ∈ V
17 fvex 6160 . . . . . . . . 9 ( I ‘𝐵) ∈ V
1816, 17xpsn 6364 . . . . . . . 8 ({1𝑜} × {( I ‘𝐵)}) = {⟨1𝑜, ( I ‘𝐵)⟩}
1914, 18sseqtri 3618 . . . . . . 7 ({1𝑜} × {𝐵}) ⊆ {⟨1𝑜, ( I ‘𝐵)⟩}
2016, 17funsn 5899 . . . . . . 7 Fun {⟨1𝑜, ( I ‘𝐵)⟩}
21 funss 5868 . . . . . . 7 (({1𝑜} × {𝐵}) ⊆ {⟨1𝑜, ( I ‘𝐵)⟩} → (Fun {⟨1𝑜, ( I ‘𝐵)⟩} → Fun ({1𝑜} × {𝐵})))
2219, 20, 21mp2 9 . . . . . 6 Fun ({1𝑜} × {𝐵})
23 funfn 5879 . . . . . 6 (Fun ({1𝑜} × {𝐵}) ↔ ({1𝑜} × {𝐵}) Fn dom ({1𝑜} × {𝐵}))
2422, 23mpbi 220 . . . . 5 ({1𝑜} × {𝐵}) Fn dom ({1𝑜} × {𝐵})
2524a1i 11 . . . 4 (𝐴𝑉 → ({1𝑜} × {𝐵}) Fn dom ({1𝑜} × {𝐵}))
26 dmxpss 5526 . . . . . . 7 dom ({1𝑜} × {𝐵}) ⊆ {1𝑜}
27 sslin 3819 . . . . . . 7 (dom ({1𝑜} × {𝐵}) ⊆ {1𝑜} → ({∅} ∩ dom ({1𝑜} × {𝐵})) ⊆ ({∅} ∩ {1𝑜}))
2826, 27ax-mp 5 . . . . . 6 ({∅} ∩ dom ({1𝑜} × {𝐵})) ⊆ ({∅} ∩ {1𝑜})
29 1n0 7523 . . . . . . . 8 1𝑜 ≠ ∅
3029necomi 2844 . . . . . . 7 ∅ ≠ 1𝑜
31 disjsn2 4219 . . . . . . 7 (∅ ≠ 1𝑜 → ({∅} ∩ {1𝑜}) = ∅)
3230, 31ax-mp 5 . . . . . 6 ({∅} ∩ {1𝑜}) = ∅
33 sseq0 3949 . . . . . 6 ((({∅} ∩ dom ({1𝑜} × {𝐵})) ⊆ ({∅} ∩ {1𝑜}) ∧ ({∅} ∩ {1𝑜}) = ∅) → ({∅} ∩ dom ({1𝑜} × {𝐵})) = ∅)
3428, 32, 33mp2an 707 . . . . 5 ({∅} ∩ dom ({1𝑜} × {𝐵})) = ∅
3534a1i 11 . . . 4 (𝐴𝑉 → ({∅} ∩ dom ({1𝑜} × {𝐵})) = ∅)
36 0ex 4752 . . . . . 6 ∅ ∈ V
3736snid 4181 . . . . 5 ∅ ∈ {∅}
3837a1i 11 . . . 4 (𝐴𝑉 → ∅ ∈ {∅})
39 fvun1 6228 . . . 4 ((({∅} × {𝐴}) Fn {∅} ∧ ({1𝑜} × {𝐵}) Fn dom ({1𝑜} × {𝐵}) ∧ (({∅} ∩ dom ({1𝑜} × {𝐵})) = ∅ ∧ ∅ ∈ {∅})) → ((({∅} × {𝐴}) ∪ ({1𝑜} × {𝐵}))‘∅) = (({∅} × {𝐴})‘∅))
403, 25, 35, 38, 39syl112anc 1327 . . 3 (𝐴𝑉 → ((({∅} × {𝐴}) ∪ ({1𝑜} × {𝐵}))‘∅) = (({∅} × {𝐴})‘∅))
412, 40syl5eq 2667 . 2 (𝐴𝑉 → (({𝐴} +𝑐 {𝐵})‘∅) = (({∅} × {𝐴})‘∅))
42 xpsng 6363 . . . . 5 ((∅ ∈ V ∧ 𝐴𝑉) → ({∅} × {𝐴}) = {⟨∅, 𝐴⟩})
4342fveq1d 6152 . . . 4 ((∅ ∈ V ∧ 𝐴𝑉) → (({∅} × {𝐴})‘∅) = ({⟨∅, 𝐴⟩}‘∅))
44 fvsng 6404 . . . 4 ((∅ ∈ V ∧ 𝐴𝑉) → ({⟨∅, 𝐴⟩}‘∅) = 𝐴)
4543, 44eqtrd 2655 . . 3 ((∅ ∈ V ∧ 𝐴𝑉) → (({∅} × {𝐴})‘∅) = 𝐴)
4636, 45mpan 705 . 2 (𝐴𝑉 → (({∅} × {𝐴})‘∅) = 𝐴)
4741, 46eqtrd 2655 1 (𝐴𝑉 → (({𝐴} +𝑐 {𝐵})‘∅) = 𝐴)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 384 = wceq 1480 ∈ wcel 1987 ≠ wne 2790 Vcvv 3186 ∪ cun 3554 ∩ cin 3555 ⊆ wss 3556 ∅c0 3893 {csn 4150 ⟨cop 4156 I cid 4986 × cxp 5074 ◡ccnv 5075 dom cdm 5076 Oncon0 5684 Fun wfun 5843 Fn wfn 5844 ‘cfv 5849 (class class class)co 6607 1𝑜c1o 7501 +𝑐 ccda 8936 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-sep 4743 ax-nul 4751 ax-pow 4805 ax-pr 4869 ax-un 6905 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-ral 2912 df-rex 2913 df-reu 2914 df-rab 2916 df-v 3188 df-sbc 3419 df-dif 3559 df-un 3561 df-in 3563 df-ss 3570 df-pss 3572 df-nul 3894 df-if 4061 df-pw 4134 df-sn 4151 df-pr 4153 df-tp 4155 df-op 4157 df-uni 4405 df-br 4616 df-opab 4676 df-mpt 4677 df-tr 4715 df-eprel 4987 df-id 4991 df-po 4997 df-so 4998 df-fr 5035 df-we 5037 df-xp 5082 df-rel 5083 df-cnv 5084 df-co 5085 df-dm 5086 df-rn 5087 df-res 5088 df-ima 5089 df-ord 5687 df-on 5688 df-suc 5690 df-iota 5812 df-fun 5851 df-fn 5852 df-f 5853 df-f1 5854 df-fo 5855 df-f1o 5856 df-fv 5857 df-ov 6610 df-oprab 6611 df-mpt2 6612 df-1o 7508 df-cda 8937 This theorem is referenced by: xpscfv 16146 xpsfeq 16148 xpsfrnel2 16149 xpsff1o 16152 xpsle 16165 dmdprdpr 18372 dprdpr 18373 xpstopnlem1 21525 xpstopnlem2 21527 xpsxmetlem 22097 xpsdsval 22099 xpsmet 22100
Copyright terms: Public domain W3C validator | 3,301 | 5,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-49 | latest | en | 0.135769 |
http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Variational_Methods.htm | 1,709,490,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00001.warc.gz | 13,183,515 | 7,167 | # Variational Methods
Michael Fowler, UVa
### Introduction
So far, we have concentrated on problems that were analytically solvable, such as the simple harmonic oscillator, the hydrogen atom, and square well type potentials. In fact, we shall soon be confronted with situations where an exact analytic solution is unknown: more general potentials, or atoms with more than one electron. To make progress in these cases, we need approximation methods. The best known method is perturbation theory, which has proved highly successful over a wide range of problems (but by no means all). We shall soon be discussing perturbation methods at length. First, though, we shall review two other approximation methods: in this lecture, the variational method, then in the next lecture the semiclassical WKB method. The variational method works best for the ground state, and in some circumstances (to be described below) for some other low lying states; the WKB method is good for higher states.
### Variational Method for Finding the Ground State Energy
The idea is to guess the ground state wave function, but the guess must have an adjustable parameter, which can then be varied (hence the name) to minimize the expectation value of the energy, and thereby find the best approximation to the true ground state wave function. This crude sounding approach can in fact give a surprisingly good approximation to the ground state energy (but usually not so good for the wave function, as will become clear).
We’ll begin with a single particle in a potential, $H= p 2 /2m+V( r → )$. If the particle is restricted to one dimension, and we’re looking for the ground state in any fairly localized potential well, we can start with the family of normalized Gaussians, $| ψ,α〉= ( α π ) 1/4 e −α x 2 /2$: just find $〈 ψ,α|H| ψ,α〉$, differentiate the result with respect to $α$, setting this to zero (and checking that you have in fact found a minimum.) Not surprisingly, this gives the exact ground state for the simple harmonic oscillator potential, and for nothing else. What is perhaps surprising is that the result is only off by 30% or so for the attractive delta-function potential, even though the wave function looks a lot different (solved in detail in Griffiths, page 258). Obviously, the Gaussian family cannot be used if there is an infinite wall anywhere: one must find a family of wave functions vanishing at the wall.
To gain some insight into what we’re doing, suppose the Hamiltonian $H= p 2 /2m+V( r → )$ has the set of (unknown to us) eigenstates
$H|n〉= E n |n〉.$
Since the Hamiltonian is Hermitian, these states span the space of possible wave functions, including our variational family, so:
$| ψ,α〉= ∑ a n ( α )|n〉 .$
From this,
$〈 ψ,α|H| ψ,α〉 〈 ψ,α| ψ,α〉 = ∑ | a n | 2 E n ≥ E 0$
for any $| ψ,α〉$. (We don’t need the denominator if we’ve chosen a family of normalized wave functions, as we did with the Gaussians above.) Evidently, minimizing $〈 ψ,α|H| ψ,α〉 〈 ψ,α| ψ,α〉$ as a function of $α$ gives us an upper bound on the ground state energy, hopefully not too far from the true value.
We can see immediately that this will probably be better for finding for the ground state energy than for mapping the ground state wave function: suppose the optimum state in our family is actually $| α min 〉=N( |0〉+0.2|1〉 )$, with the normalization constant $N≅0.98$, a 20% admixture of the first excited state. Then the wave function is off by of order 20%, but the energy estimate will be too high by $0.04( E 1 − E 0 )$ usually a much smaller error.
To get some idea of how well this works, Messiah applies the method to the ground state of the hydrogen atom. We know it’s going to be spherically symmetric, so it amounts to a one-dimensional problem: just the radial wave function. Using standard notation,
$a 0 = ℏ 2 /m e 2 , E 0 =m e 4 /2 ℏ 2 , ρ=r/ a 0$
and for a trial wave function u
$E( u )=− E 0 ∫ u( d 2 d ρ 2 + 2 ρ )udρ ∫ u 2 dρ$
(we’re going to take u real).
Messiah tries three families:
$u 1 =ρ e −αρ u 2 = ρ α 2 + ρ 2 u 3 = ρ 2 e −αρ$
and finds $α min =1, π/4, 3/2$ respectively. The first family$, u 1 ,$ includes the exact result, and the minimization procedure finds it.
For the three families, then the energy of the best state is off by 0, 25%, 21% respectively.
The wave function error is defined as how far the square of the overlap with the true ground state wave function falls short of unity. For the three families, $ε=1− | 〈 ψ 0 | ψ var 〉 | 2$ = 0, 0.21, 0.05. Notice here that our hand waving argument that the energies would be found much more accurately than the wave functions comes unstuck. The third family has far better wave function overlap than the second, but only a slightly better energy estimate. Why?
A key point is that the potential is singular at the origin, there is a big contribution to potential energy from a rather small region, and the third family wave function is the least accurate of the three there. The second family functions are very inaccurate at large distances: the expectation value $〈 r 〉=1.5 a 0 ,∞,1.66 a 0$ for the three families. But at large distances, both kinetic and potential energies are small, so the result can still look reasonable. These examples reinforce the point that the variational method should be used cautiously.
### Variational Method for Higher States
In some cases, the approach can be used easily for higher states: specifically, in problems having some symmetry. For example, if the one dimensional attractive potential is symmetric about the origin, and has more than one bound state, the ground state will be even, the first excited state odd. Therefore, we can estimate the energy of the first excited state by minimizing a family of odd functions, such as $ψ( x,α )=( π /2 α 3/2 )x e −α x 2 /2$.
### Ground State Energy of the Helium Atom by the Variational Method
We know the ground state energy of the hydrogen atom is -1 Ryd, or -13.6 ev. The He+ ion has $Z=2,$ so will have ground state energy, proportional to $Z 2 ,$ equal to -4 Ryd. Therefore for the He atom, if we neglect the electron-electron interaction, the ground state energy will be -8 Ryd, -109 ev., the two electrons having opposite spins will both be in the lowest spatial state. Actually, experimentally, the He atom ground state energy is only -79 ev, because the repulsion between the electrons loosens things.
To get a better value for the ground state energy still using tractable wave functions, we change the wave functions from the ionic wave function $( Z 3 /π a 0 3 ) 1/2 e −Zr/ a 0$ with $Z=2$ to $( Z ′ 3 /π a 0 3 ) 1/2 e − Z ′ r/ a 0$ with $Z ′$ now a variable parameter. In other words, we are trying to allow for the electron-electron repulsion, which must push the wave functions out a bit, by keeping exactly the same shaped wave function but lessening the effective nuclear charge as reflected in the spread of the wave function from $Z$ to $Z ′ ,$ and we’ll determine $Z ′$ by varying it to find the minimum total energy, including the term from electron-electron repulsion.
To find the potential energy from the nuclear-electron interactions, we of course use the actual nuclear charge $Z=2,$ but the $Z ′$ wave function, so the nuclear P.E. for the two electrons is:
This could have been figured out from the formula for the one-electron ion, where the potential energy for the one electron is $−2 Z 2$ Ryd, one factor of $Z$ being from the nuclear charge, the other from the consequent shrinking of the orbit.
The kinetic energy is even easier: it depends entirely on the shape of the wave function, not on the actual nuclear charge, so for our trial wave function it has to be $Z ′ 2$ Ryds per electron.
The tricky part is the P.E. for the electron-electron interaction. This is positive.
Each electron has a wave function $( Z ′ 3 /π a 0 3 ) 1/2 e − Z ′ r/ a 0$, a spherical charge probability distribution.
Denoting charge probability density by $ρ( r )$, we need
$I= ∫ ∫ d r → 1 d r → 2 ρ( r → 1 )ρ( r → 2 ) | r → 1 − r → 2 | =16 π 2 ∫ 0 ∞ r 1 2 d r 1 ∫ 0 ∞ r 2 2 d r 2 ρ( r 1 )ρ( r 2 ) r > , r > =max( r 1 , r 2 ).$ (see Messiah page 691).
Collecting terms, the total energy (for $Z=2$ ) is:
and this is minimized by taking $Z ′ =2− 5 16$, giving an energy of -77.5 ev, off the true value by about 1 ev, so indeed the presence of the other electron is taken care of as far as total energy is concerned by shielding the nuclear charge by an amount $( 5/16 )e.$ | 2,321 | 8,511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 42, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.901261 |
https://www.svtuition.org/2015/06/balance-sheet-vs-cash-flow-statement.html | 1,568,737,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573098.0/warc/CC-MAIN-20190917161045-20190917183045-00269.warc.gz | 950,771,869 | 94,432 | Balance sheet and cash flow statement are both important financial statements which are very necessary for any corporate to make. At the end of the year, both statements are prepared to show correct financial health.
There are many differences between balance sheet and cash flow statements which we can explain on the basis of following points.
1. Meaning
Balance sheet - Balance sheet is the statement which shows the assets and liabilities of any organisation. One side, we show all the assets and other side, we show all the liabilities, difference between this will be shareholders' equity capital.
Cash flow statement - Cash flow statement is the statement which shows the inflow and out flow of cash. In different activities, we either receive or pay the activities. Difference between inflow and outflow cash will show net balance of cash in hand.
2. Benefit
Balance sheet is made to find the real financial position of company. For example, A company has lots of long term debt but there is low amount of fixed asset and debt equity ratio is so high. From this point, we can estimate, there is not good financial position of company.
Cash flow statement is made to find the real capacity to pay bill, debt and expenses of company. If company's cash inflow is so high, company can do more business. Because with this, it can easily pay its bill. Even if company has big net profit but outflow of cash is high than inflow of cash, cash flow statement can tell us lots of net profit which is not in cash, are not getting fast. So, due to the shortage of cash, we can capture in big problem.
3. Method of Making
We can make balance sheet with vertical or horizontal method.
We can make cash flow statement with direct or indirect method.
4. Parts
Balance sheet has three major parts. One is assets, second is liabilities and third is equity capital.
Cash flow has also three major parts. Operating activities, financial activity and investing activities.
5. Formula
Balance sheet » Assets = Outside liabilities + Equity capital
Cash flow statement » Net cash flow from operating activities + Net cash flow from financial activities + Net cash inflow from investing activities.
Resource : Difference Between Similar Terms
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Accounting Education: Balance Sheet vs Cash Flow Statement
Balance Sheet vs Cash Flow Statement
http://1.bp.blogspot.com/-APXS41R3FpY/VXCK5R45eeI/AAAAAAAAC6A/PjSmcQaz5u4/s640/How-they-fit.png
http://1.bp.blogspot.com/-APXS41R3FpY/VXCK5R45eeI/AAAAAAAAC6A/PjSmcQaz5u4/s72-c/How-they-fit.png
Accounting Education
https://www.svtuition.org/2015/06/balance-sheet-vs-cash-flow-statement.html
https://www.svtuition.org/
http://www.svtuition.org/
http://www.svtuition.org/2015/06/balance-sheet-vs-cash-flow-statement.html
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UTF-8 | 691 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.953473 |
https://javarevisited.blogspot.com/2012/03/how-to-format-decimal-number-in-java.html | 1,531,943,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00331.warc.gz | 701,770,090 | 28,277 | How to format Decimal Number in Java - DecimalFormat Example
We often need to format decimal numbers in Java like formatting numbers upto 2 decimal places or 3 decimal places or we want to introduce leading zeros in front of numbers. Thankfully Java programming language provides many different ways to format numbers in Java like either using Math.round() or setScale() from BigDecimal but caveat is that they also do rounding of numbers i.e. 1.6 will be rounded on 2.0 if we use Match.round(). If we are just interested in formatting decimal numbers upto n decimal digits than DecimalFormat is way to go. java.text.DecimalFormat descends from NumberFormat and provides dynamic way of formatting numbers in Java. While creating instance of DecimalFormat you can pass an String pattern which describes on which format decimal number should be formatted and than DecimalFormat.format() method will do the rest for you. In this Java tutorial we will see how to format a decimal number in 2 decimal place, format upto 3 decimal place, using comma to separated 3 digits etc.
Java DecimalFormat Example
DecimalFormat in Java is defined in java.text package and its a subclass of NumberFormat. In order to format a decimal number in Java we need an instance of DecimalFormat with predefined pattern. Once you have instance of DecimalFormat, you can call DecimalFormat.format() method for converting any double or long number into needed format. here is code example of creating DecimalFormat and formatting numbers in Java:
import java.text.DecimalFormat;
public class DecimalFormatExample {
public static void main(String args[]) {
//formatting numbers upto 2 decimal places in Java
DecimalFormat df = new DecimalFormat("#,###,##0.00");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
//formatting numbers upto 3 decimal places in Java
df = new DecimalFormat("#,###,##0.000");
System.out.println(df.format(364565.14));
System.out.println(df.format(364565.1454));
}
}
Output:
364,565.14
364,565.15
364,565.140
364,565.145
If you look at the output of decimal format example you will see that when we used format upto 2 decimal number any floating point number which has more than 2 digits after decimal point numbers will be printed only up to two digits. same is true in case of formatting numbers upto 3 decimal digit. if decimal number doesn't contain 3 digit than zero will be per pended.
Careful while Formatting Decimal number using DecimalFormat
Though java.text.DecimalFormat is a nice utility class and allow you to dynamically format numbers in Java it has one problem that its not thread-safe or synchronized properly. So never share decimal format between multiple threads. Its also not advisable to cache DecimalFormat as static resource without proper synchronization.
That's all on how to format a decimal number in Java using DecimalFormat. There are other way also like I said but DecimalFormat is neat, clean and simple way of formatting numbers upto n number of decimal places.
Further Learning
Complete Java Masterclass
Java Fundamentals: The Java Language
Java In-Depth: Become a Complete Java Engineer!
Other Java Tutorials you may like
Omi Tewary said...
import javax.swing.JOptionPane;
import java.text.DecimalFormat;
public class Sample3 {
public static void main(String args[]){
double amount,iRate,monPay,totalPay;
int years;
String amountStr;
String irateStr;
//takes input for loan amount,rate and loan period
amountStr = JOptionPane.showInputDialog(null,"Enter loan amount \$ : ");
irateStr = JOptionPane.showInputDialog(null,"Enter % Annual Interest: ");
yearsStr = JOptionPane.showInputDialog(null,"Enter the loan Period; ");
amount = Double.parseDouble(amountStr); //convert input String into double
iRate = Double.parseDouble(irateStr); //convert input string into double
years = Integer.parseInt(yearsStr); //convert input string into integer
monPay = (amount * iRate / 1200) /
( 1 - Math.pow( 1 / ( 1 + iRate / 1200), 12 * years)) ;//calculate monthly payment
totalPay = monPay * 12 * years;
DecimalFormat df = new DecimalFormat("0.00");
System.out.println(df.format(monPay));
JOptionPane.showMessageDialog(null,"your monthly payment is: " + monPay + "\n" +
"your total payment is: " + totalPay );
}
}
Anonymous said...
@omi tewary .In which format you want your ouput. | 932 | 4,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-30 | longest | en | 0.598502 |
https://www.convertopedia.com/unit-converters/area-converter/hectare-to-acre-converter/ | 1,716,069,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00424.warc.gz | 649,055,353 | 24,057 | Use the below online Hectare to Acre converter tool to convert the land area measure of hectare to acre:
Hectare
Hectare is the unit of area used in the measurement of land. It is abbreviated as ha. One hectare is equal to a square with 100 meter sides or 10,000 m2. One hectare contains about 2.47 acres. There are 100 hectares in one square kilometer.
Acre
Acre is a unit of land area, which is approximately equal to 4047 square metres or 4840 square yards or 1/640 of a square mile. As a unit of measure, an acre has no prescribed shape. Any area of 43,560 square feet is an acre. In India, agricultural land is measured in acres.
How to convert Hectare to Acre
1 hectare = 107639.104 acre
Example 1:
Convert 15 Hectare to Acre
1 ha = 2.471053 acre
Therefore, 15 ha = 15 * 2.471053 = 37.0658 acre
Example 2:
Convert 50 Hectare to Acre
1 ha = 2.471053 acre
Therefore, 50 ha = 50 * 2.471053 = 123.552 acre | 278 | 917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.838408 |
http://mathhelpforum.com/algebra/76705-cube-root.html | 1,481,042,833,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541910.39/warc/CC-MAIN-20161202170901-00107-ip-10-31-129-80.ec2.internal.warc.gz | 177,334,898 | 9,685 | 1. ## Cube root
f (x) = (x − a)3 + (x − b)3 + (x − c)3 , a < b < c.
The number of real roots of f (x) = 0 is
(A) 3; (B) 2; (C) 1; (D) 0.
Please guide me with this question..
2. $\lim_{x\to -\infty}=-\infty, \ \lim_{x\to\infty}=\infty\Rightarrow$ f has at least one real root.
$f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2>0, \forall x\in\mathbf{R}\Rightarrow$ f is strictly increasing, so f is injective. Therefore, f has only one real root. | 184 | 433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2016-50 | longest | en | 0.674723 |
https://www.daniweb.com/programming/software-development/threads/494321/can-someone-help-me-with-my-logic-for-a-bill-program | 1,534,798,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217006.78/warc/CC-MAIN-20180820195652-20180820215652-00017.warc.gz | 882,688,164 | 12,697 | I am trying to calculate a cable bill for a residential and business customer can someone point me in the right direction for my logic?For residential customers it is 4.50 processing fee,30 basic service fee and 5 per channel. For business customers processing fee = 16.50 basicservice fee is 80 for the first 10 connections if connections greater than 10 then its 4 for each additional and 50 per channel when I hit calculate no matter how many connections or channels I choose for each category it gives me the same output which means my logic must be wrong can anyone make a suggestion?
``````Option Strict On
Public Class Main
Private Sub btnCalculate_Click(sender As Object, e As EventArgs) Handles btnCalculate.Click
If Data_Validated_ok() = False Then Exit Sub
Dim Connections As Integer = lstConnections.SelectedIndex
Dim totalDue As Decimal
lblTotal.Text = totalDue.ToString("C2")
End If
lblTotal.Text = totalDue.ToString("C2")
End If
End Sub
' make a function to calculate residential customers
Private Function CalcResidentialTotalDue(ByVal premiumChannels As Decimal) As Decimal
Const ResidentialProcessing As Decimal = 4.5D
Const ResidentialBasic As Integer = 30
Const ResidentialPremium As Integer = 5
End Function
' make a function to calculate business customers
) As Decimal
Const BusinessProcessing As Decimal = CDec(16.5)
Const BusinessServiceFee As Integer = 80
End Function
``````
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Last Post by Reverend Jim
I dont see an If statement that will determine if there more that 10 connections.
Try
``````If radResidential.Checked Then
totalDue = 4.5 + 30 + 5 * premiumChannels
Else
totalDue = 16.50 + 80 + 4 * Math.Max(premiumChannels - 10, 0)
End If
``````
Math.Max returns the maximum of the two values. If the number of channels is 10 or fewer it return 0, otherwise it returns the number of channels over 10. | 455 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-34 | longest | en | 0.729041 |
https://www.f1technical.net/forum/viewtopic.php?t=29489 | 1,675,782,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00647.warc.gz | 760,923,419 | 7,790 | ## Basic aero questions
All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
MarkBolton
MarkBolton
0
Joined: Mon Nov 16, 2020 2:32 pm
### Basic aero questions
Hi,
Regarding induced drag, I understand where it comes from-the horizontal element of the resultant lifting force due to the rear facing lift vector.
But what is this 'horizontal' element in relation to? The ground, mean chord line or the direction of travel?
What is the correct term for this plane?
Is it possible to have a 3D lifting aero shape (not something symmetrical) in which all the lift is perpendicular to the 'horizontal' plane, so that there is no induced drag?
Thank you,
gruntguru
gruntguru
548
Joined: Sat Feb 21, 2009 6:43 am
### Re: Basic aero questions
MarkBolton wrote:
Mon Nov 16, 2020 2:46 pm
Hi,
Regarding induced drag, I understand where it comes from-the horizontal element of the resultant lifting force due to the rear facing lift vector.
But what is this 'horizontal' element in relation to? The ground, mean chord line or the direction of travel?
What is the correct term for this plane?
Is it possible to have a 3D lifting aero shape (not something symmetrical) in which all the lift is perpendicular to the 'horizontal' plane, so that there is no induced drag?
Thank you,
You know there is an aero forum here? https://www.f1technical.net/forum/viewforum.php?f=6
je suis charlie
godlameroso
305
Joined: Sat Jan 16, 2010 8:27 pm
Location: Miami FL
### Re: Basic aero questions
MarkBolton wrote:
Mon Nov 16, 2020 2:46 pm
Hi,
Regarding induced drag, I understand where it comes from-the horizontal element of the resultant lifting force due to the rear facing lift vector.
But what is this 'horizontal' element in relation to? The ground, mean chord line or the direction of travel?
What is the correct term for this plane?
Is it possible to have a 3D lifting aero shape (not something symmetrical) in which all the lift is perpendicular to the 'horizontal' plane, so that there is no induced drag?
Thank you,
Turning the air requires energy, the manifestation of that energy conversion is drag. You cannot eliminate drag if the shape creates a pressure differential as it cuts across the air, you can minimize extra drag only, never zero drag. Only way to have no induced drag is to not have bodywork in the path of the air.
"Is it possible to have a 3D lifting aero shape (not something symmetrical) in which all the lift is perpendicular to the 'horizontal' plane, so that there is no induced drag?"
Yes to the former, no to the latter. It only matters which way the air is flowing on the trailing edge that determines the force vector on the surface.
Look at the bargeboards, they're a perfect example of this. The airflow hits it dead on, but is pushed out and up, so the force vector is still towards the ground. You can see the way the floor flicks up aft of the side pod undercut, to understand where the force vector of the bargeboards is.
The air flows freely under the bargeboards, while over them, the air turns out and up, which slows it down increasing its pressure which comes to a head when that air interacts with the sidepod undercut.
Saishū kōnā
Just_a_fan
Just_a_fan
561
Joined: Sun Jan 31, 2010 7:37 pm
### Re: Basic aero questions
If you generate "lift" then you generate induced drag.
Turbo says "Dumpster sounds so much more classy. It's the diamond of the cesspools." oh, and "The Dutch fans are drunk. Maybe" | 865 | 3,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.865629 |
https://questioncove.com/updates/4d54a6bace8eb764b25d841a | 1,685,546,050,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00608.warc.gz | 543,135,712 | 5,304 | Mathematics 56 Online
OpenStudy (anonymous):
write an equation in slope intercept form for the line that passes through the given point and is perpendicular to (4,-2); y=3
OpenStudy (anonymous):
What kind of line is y=3? You need to understand this first....?
OpenStudy (anonymous):
This is a horizontal line! (sketch it if not sure) A line perpendicular to this would be a vertical line...and this vertical line must go through the point (4-2). The equation for a vertical line is always x= "something" and so in this case x=4! | 129 | 533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-23 | longest | en | 0.921012 |
https://superbgrade.com/essays/the-combined-gas-law | 1,717,095,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00427.warc.gz | 456,864,483 | 63,219 | # The Combined Cas Law
2021-05-12
2Â pages
531 words
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Essay
This essay has been submitted by a student.
This is not an example of the work written by our professional essay writers.
The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law.
If this sample essay on"The Combined Cas Law " doesn’t help,
our writers will!
This law states: The ratio between the pressure-volume product divided by the temperature of a system remains constant.
This can be stated mathematically as:
Where: p is the pressure, V is the volume, T is the temperature measured in kelvins, and k is a constant (with units of energy divided by temperature).
Reminder: 1atm= 760 torr = 101.3 kPa & Celsius to Kelvin= add 273 and Kelvin to Celsius= subtract 273
If the problem does not state which unit to give the result in, then make sure that temperature is converted into Kelvin and for the Pressure and Volume just make sure you stay constant and use the same unit on both sides of the equation.
Combination of 3 Laws:
Boyle's Law states that the pressure-volume product is constant:
In other words as external pressure on a gas increases the volume decreases, and vice versa.
Charles's Law shows that the volume is proportional to absolute temperature:
In other words as temperature increases the volume increases, and vice versa.
Gay-Lussac's Law says that the pressure is proportional to the absolute temperature:
In other words as temperature increases the pressure increases, and vice versa.
Where P is the pressure, V the volume and T the absolute temperature and of an ideal gas.
By combining (1) and either of (2) or (3) we can gain a new equation with P, V and T. Equation (2) is used in this example, and the subscript on the constant is dropped so that k = k2. We get the Combined Gas Law! (No specific scientist invented this law; rather three laws were just combined mathematically to come up with this law)
The combined gas law can be used to explain the mechanics where pressure, temperature, and volume are affected. For example: air conditioners, refrigerators and the formation of clouds.
Calculations with the Combined Gas Law:
For example, the pressure and temperature of a gas are changed to STP (101.3 kPa/00C) from 22.00C and 30.8 kPa. What will be the new volume if the original volume was 205 mL?
Solution:
First, we must convert degrees Celsius to Kelvins. This means that our original temperature was 295 K and our target temperature is 273 K.
Plug values into the formula
Given:
Volume (V) = 205 mL, Temperature (T) = 295 K, Pressure (P) = 30.8 kPa
Temperature1 (T1) = 273 K, Pressure1 (P1) = 101.3 kPa, the unknown is V1. Our equation is:
30.8 kPa x 205 mL / 295 K = 101.3 kPa x V1 / 273 K
Manipulating the equation, we get:
V1 = 30.8 kPa x 205 mL / 295 K x 273 K / 101.3 kPa
V1 = 57.68 mL, However, because our problem only was to three digits of precision, the answer is:
V1 = 57.7 mL: Therefore the new volume of the gas is 57.7 mL.
A couple of examples are the internal combustion engine. For example, a four stroke engine like your car operates on the principle of taking a volume of gas/air mixture, compressing it, igniting it, and pushing the exhaust out. The movement of the pistons moves the drive shaft..... Also, weather balloons are launched daily from weather stations across the country. The balloon begins at the earth at a certain P, T, and V and upon its accent all three of these variables change in response to the surroundings.
If you want discreet, top-grade help, order a custom paper from our experts.
If you are the original author of this essay and no longer wish to have it published on the SuperbGrade website, please click below to request its removal: | 929 | 3,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.934234 |
https://www.topperlearning.com/answer/q1-a-plank-is-moving-on-a-ground-with-a-velocity-v-and-a-block-is-moving-on-the-plank-with-a-velocity-u-what-is-the-velocity-of-block-wrt-ground-v-u-t/wibhz522 | 1,620,912,524,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989916.34/warc/CC-MAIN-20210513111525-20210513141525-00197.warc.gz | 959,175,398 | 38,932 | # Q1)A plank is moving on a ground with a velocity v and a block is moving on the plank with a velocity u. What is the velocity of block wrt ground- v-u towards right, v-u towards left, u towards right Q2) A man can swim in still water with a speed of 2m/s. If he wants to cross a river flowing with a speed root3 m/s along the shortest possible path, then in which direction should he swim-120 deg to the water current, 150 deg to the water current, 90 deg to the water current Q3)A train 150 m length is going towards north direction at a speed of 10m/s.a birdflies at a speed of 5m/s towards south direction parallel to the railway track. The time taken by the bird to cross the train is equal to -12s, 8 s, 15s,10s
### Answered by | 15th Jun, 2013, 05:21: PM
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REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India. | 278 | 1,020 | {"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} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.927063 |
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# composite trapezoidal rule error term Castle Point, New York
The composite trapezoidal rule is given by $$T_n[f]:=h\left(\frac{f(a)+f(b)}{2}+\sum_{k=1}^{n-1}f(x_k)\right)\;\;\;\;\;\left(h:=\frac{b-a}{n},\;x_k:=a+kh\right)$$ First, I've shown, that it holds: $$\int_{[a,b]}f(x)\text{ dx}-T_1[f]=-\frac{(b-a)^3}{12}f''(\xi)$$ After that, It's easy to see, that it holds: \begin{split} \int_{[a,b]}f(x)\text{ dx}-T_n[f] &= \left|\sum_{k=1}^n\left(\int_{x_{k-1}}^{x_k}f(x)\text{ Please try the request again. Your cache administrator is webmaster. I'm asked to show, that if $f''\in L^2(a,b)$, then it holds: $$\left|\int_{[a,b]}f(x)\text{ dx}-T_n[f]\right|\le\frac{\sqrt{b-a}}{\sqrt{120}}\left\|f''\right\|_{L^2(a,b)}h^2$$ How can I conclude this error bound from my previous result?
Please try the request again. How would I pass the output of one command to multiple commands? Dungeons in a 3d space game Is "The empty set is a subset of any set" a convention? Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.6/ Connection
Why Rosetta probe has been programmed to "auto shutoff" at the moment of hitting the surface? more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science The system returned: (22) Invalid argument The remote host or network may be down. The system returned: (22) Invalid argument The remote host or network may be down.
Applying Cauchy's sum inequality in the third inequality, one has $$\left|\int_a^bf(x)\,dx-T_n(f) \right|$$$$\le\sum_{k=1}^n\,\left|\int_{x_{k-1}}^{x_k}f(x)\,dx-\frac{(x_k-x_{k-1})}2[f(x_{k-1})+f(x_k)]\right|$$$$\le \frac 1{\sqrt {120}}\sum_{k=1}^n \left[\sqrt{\left(\frac {b-a}n\right)^5}\cdot \sqrt {\int_{x_{k-1}}^{x_k}[f''(x)]^2\,dx} \right]$$$$\le \frac 1{\sqrt {120}}\sqrt {\sum_{k=1}^n\left(\frac {b-a}n\right)^5}\cdot\sqrt {\sum_{k=1}^n\int_{x_{k-1}}^{x_k}[f''(x)]^2\,dx}$$$$\le \frac 1{\sqrt {120}}\sqrt {n\left(\frac {b-a}n\right)^5}\sqrt What is missing from a non-afterburning engine to prohibit the use of afterburning? more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) share|cite|improve this answer edited Jan 24 '14 at 8:27 answered Jan 24 '14 at 6:56 Tony Piccolo 2,5202515 add a comment| Your Answer draft saved draft discarded Sign up or Your cache administrator is webmaster. Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection Are there any saltwater rivers on Earth? Please try the request again. Please try the request again. The system returned: (22) Invalid argument The remote host or network may be down. Suggestions for HDMI/aerial/audio socket Text editor for printing C++ code Should wires be tinned to under the insulation? One starts from the formula$$\int_a^b f(x)\,dx=\frac {(b-a)}2[f(a)+f(b)]-\frac 12\int_a^b(x-a)(b-x)f''(x)\,dx$$that will be proved later. Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection Guess the word Even sharper upper bound for prime product? Not the answer you're looking for? Red Herring Bonkers In The Red Herring Bunkers Can I reduce "couldn't find anything" to "nothing" in this sentence? Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.5/ Connection The system returned: (22) Invalid argument The remote host or network may be down. Why was Spanish Fascist dictatorship left in power after World War II? Is there a term referring to the transgression that often begins a horror film? Your cache administrator is webmaster. Is it ok to use a function to output the text domain name in a wordpress theme Natural Pi #0 - Rock Using the MOD expression on an SQL Feature Class Shortcode in shortcode: How to append variable? integration numerical-methods definite-integrals estimation share|cite|improve this question edited Jan 24 '14 at 9:07 Christian Blatter 116k372203 asked Jan 19 '14 at 15:10 0xbadf00d 1,1352718 Can you give your source Your cache administrator is webmaster. Your cache administrator is webmaster. Applying the inequality of the modulus of an integral and the Cauchy's integral inequality, one has$$\left|\int_a^b f(x)\,dx-\frac {(b-a)}2[f(a)+f(b)]\right|\le\frac 12\int_a^b(x-a)(b-x)|f''(x)|\,dx\le \frac 12 \sqrt {\int_a^b (x-a)^2(b-x)^2\,dx} \cdot\sqrt {\int_a^b[f''(x)]^2\,dx}=\frac 1{\sqrt {120}}\sqrt{(b-a)^5}\sqrt {\int_a^b[f''(x)]^2\,dx}Now the
current community blog chat Mathematics Mathematics Meta your communities Sign up or log in to customize your list. Generated Wed, 05 Oct 2016 03:10:42 GMT by s_hv972 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.7/ Connection Word play. The system returned: (22) Invalid argument The remote host or network may be down.
The system returned: (22) Invalid argument The remote host or network may be down. Please try the request again. Please try the request again. asked 2 years ago viewed 1350 times active 1 year ago Blog Stack Overflow Podcast #89 - The Decline of Stack Overflow Has Been Greatly… Linked 3 Correct or not? $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2+\ln^2(2\sin(x))}dx\stackrel?=\frac{\pi}{8}\left[\frac{\zeta(2)}{2}+\ln(2\pi)\right]$ | 1,868 | 6,061 | {"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} | 3.15625 | 3 | CC-MAIN-2018-51 | latest | en | 0.774109 |
http://irodovsolutionsmechanics.blogspot.com/2009/09/irodov-problem-1341.html | 1,638,880,664,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00087.warc.gz | 34,036,049 | 11,400 | ## Sunday, September 13, 2009
### Irodov Problem 1.341
a) Let ABC be the original triangle. Since all its sides are equal to a it is an equilateral triangle. Let CD be the perpendicular bisector along which the triangle moves with a velocity v w.r.t to frame K. From elementary trigonometry.
Let A'B'C' be the triangle as seen from a frame K. w.r.t to K, the triangle will shrink by a factor along the direction of motion (along CD). However, along the direction perpendicular to the motion of the triangle (along AB) all lengths will remain intact.
In other words,
AB = A'B' = a, AD = A'D' = a/2, C'D' would Thus, w.r.t to frame K, the triangle will no longer be equilateral since the sides A'C and C'B' would have shrunk. Hence, we have,
b) In this part the triangle is moving in the direction of one of the sides (along AB) and not the bisector. Thus, all measurements of the triangle in this direction shrink by a factor relative to frame K. The measurements in the perpendicular direction (along CD) however remain intact.
Thus, we have, | 261 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-49 | latest | en | 0.924377 |
https://www.physicsforums.com/threads/two-electrons-and-distance-equation.568798/ | 1,542,642,006,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745800.94/warc/CC-MAIN-20181119150816-20181119172816-00403.warc.gz | 983,997,125 | 12,140 | # Homework Help: Two Electrons and Distance Equation
1. Jan 18, 2012
### jfkline21
An electron (mass of 9.110×10-31kg and charge of -1.602×10-19C) is released at rest above the surface of the earth (where g = 9.81 m/s2). A second electron directly below it exerts an electrostatic force on the first electron that exactly cancels the gravitational force. What is the distance between the two?
2. Relevant equations
F=Eq
F=ma
F=kq1q2/r^2
3. The attempt at a solution
I honestly don't know at all how to solve this. If someone could explain to me how and maybe show me the initial steps, that would be great!!
2. Jan 18, 2012
### Delphi51
We aren't supposed to help before you make an attempt.
But I can give you a little hint. Just write the obvious:
force of gravity = electric force
Then look for the formula for each side of the equation and write the next step in formulas. Solve for the quantity you want to find. Put in the numbers and run it through your calculator. | 259 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-47 | latest | en | 0.903323 |
https://www.rwmansiononpeachtree.com/what-does-the-ljung-box-q-test-test-for/ | 1,723,721,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00275.warc.gz | 749,771,109 | 14,535 | # What does the Ljung-Box Q test test for?
What does the Ljung-Box Q test test for?
## What does the Ljung-Box Q test test for?
The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) are useful qualitative tools to assess the presence of autocorrelation at individual lags. The Ljung-Box Q-test is a more quantitative way to test for autocorrelation at multiple lags jointly [1].
### How do you interpret the p value in the Ljung-Box test?
You’ve interpreted the test wrong. If the p value is greater than 0.05 then the residuals are independent which we want for the model to be correct. If you simulate a white noise time series using the code below and use the same test for it then the p value will be greater than 0.05.
#### How do you do a Ljung-Box test in R?
To conduct a Ljung-Box test, we can use the Box-test function from the built in stats package. We pass our time series, a lag, and the type which will be Ljung . We choose a lag of 1, because we want to see if there is autocorrelation with each lag.
What is Jarque Bera test used for?
In statistics, the Jarque–Bera test is a goodness-of-fit test of whether sample data have the skewness and kurtosis matching a normal distribution. The test is named after Carlos Jarque and Anil K. Bera. The test statistic is always nonnegative.
What is McLeod Li test?
McLeod-Li test is a test for autoregressive conditional heteroskedasticity in either raw data or residuals from a conditional mean model (but not for residuals from a GARCH model; there Li-Mak test should be used instead).
## How do you interpret Arima results?
Interpret the key results for ARIMA
1. Step 1: Determine whether each term in the model is significant.
2. Step 2: Determine how well the model fits the data.
3. Step 3: Determine whether your model meets the assumption of the analysis.
### What is Jarque-Bera null hypothesis?
The null hypothesis of the Jarque-Bera test is a joint hypothesis of the skewness being zero and the excess kurtosis being zero. With a p-value >0.05, one would usually say that the data are consistent with having skewness and excess kurtosis zero.
#### How do you read Jarque-Bera value?
What the Results Mean. In general, a large J-B value indicates that errors are not normally distributed. For example, in MATLAB, a result of 1 means that the null hypothesis has been rejected at the 5% significance level. In other words, the data does not come from a normal distribution. | 602 | 2,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.814954 |
https://www.coursehero.com/file/6254379/ch04-p125/ | 1,487,542,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170253.67/warc/CC-MAIN-20170219104610-00043-ip-10-171-10-108.ec2.internal.warc.gz | 800,347,412 | 111,248 | ch04-p125
# ch04-p125 - 125. At maximum height, the y-component of a...
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v 0 y gt again but now “starting the clock” at the highest point so that v 0 y = 0 (and 1.0 s t = ). This leads to v y = –9.8 m/s and () 2 2 (10 m/s) 9.8 m/s 14 m/s +− = . (c) The x 0 value may be obtained from x = 0 = x 0 + (10 m/s)(1.0s), which yields 0 10m. x =− (d) With v 0 y = 9.8 m/s denoting the y -component of velocity one second before the top of the trajectory, then we have yy v t g t y == + 0 00 1 2 2 where t = 1.0 s. This yields 0 4.9 m. y (e) By using x x 0 = (10 m/s)(1.0 s) where x 0 = 0, we obtain x = 10 m. (f) Let t = 0 at the top with 0 y yv == . From 2 1 2 y y t g t −= , we have, for t = 1.0 s, 22 (9.8 m/s )(1.0 s) / 2 4.9 m. y 125. At maximum height, the y -component of a projectile’s velocity vanishes, so the
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
Ask a homework question - tutors are online | 417 | 1,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-09 | longest | en | 0.81179 |
https://www.mql5.com/en/forum/108382 | 1,540,088,048,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513548.72/warc/CC-MAIN-20181021010654-20181021032154-00544.warc.gz | 1,001,419,719 | 11,966 | # How to calculate profit from volume and pips
203
Hello
My question seems very simple, but I'm in doubt about the answer. If we have number of pips of profit or loss at the end of trade and also we have open and closing prices and also we have volume of trade in base currency using
OrderLots()*MarketInfo(OrderSymbol(),MODE_LOTSIZE)/AccountBalance()
How can we calculate how much is the percentage of profit or loss?
473
Hi AR, I know not maths - is blind spot for me...
BUT - since I have not practicable alternative approach to the unending mathmatical type issues in financial trading:
have you used Print() function to 'test out' your various theories and calculation variants in noddy script testbed??
A hammer and chisel approach using iterative slog with some visual o/p via Print() is a stepwise refinement sneak up on the answer idea I personally use daily - LOL
Anyway, just a thought...
'appy calculating AR
203
Well, in this special case, I prefer analytics. I think this may be more complicated than we reach the answer with try and error. I hope to use the same formula used by strategy tester or something like that.
Anyhow thanks
203
I wish I could get an answer and work on it during weekend. I'm in doubt we can not use simply number of pips multiplied by number of lots for answer because base currency and trade currency are different and pip values are also different. any Idea?
thanks
2653
How to calculate profit from volume and pips
profit = MarketInfo(Symbol(), MODE_TICKVALUE) * OrderLots() * pips;
How can we calculate how much is the percentage of profit or loss?
%profit = 100*(1-( AccountValue / (AccountValue - OrderProfit)));
203
phy:
How to calculate profit from volume and pips
profit = MarketInfo(Symbol(), MODE_TICKVALUE) * OrderLots() * pips;
How can we calculate how much is the percentage of profit or loss?
%profit = 100*(1-( AccountValue / (AccountValue - OrderProfit)));
Thanks
How about revers pairs(in cases like we are using dollar as base currency).
Or generally speaking, how about situations that TICKVALUE is different at the beginning and the end of trade?
and how to distinguish these situations when our base currency is not USD? and when we have symbols other than currency pairs; something like metals, stock market quta, and ...
and
sorry, I'm confused. Isn't MarketInfo(OrderSymbol(),MODE_LOTSIZE) used?
2653
How about revers pairs(in cases like we are using dollar as base currency).
TICKVALUE provides the conversion to your deposit currency.
Or generally speaking, how about situations that TICKVALUE is different at the beginning and the end of trade?
Your profit is calculated using the tick value at close of the trade. Dealer will tell you what your profit is, or you can estimate what he will tell you.
and how to distinguish these situations when our base currency is not USD?
TICKVALUE is in the deposit currency.
and when we have symbols other than currency pairs; something like metals, stock market quta, and ...
and sorry, I'm confused. Isn't MarketInfo(OrderSymbol(),MODE_LOTSIZE) used?
That tells you the "size" of a lot of a specified security on your Dealer's platform.
See examples below:
---------
Sample Market Info, Deposit Currency is USD
ODL Securities
ODL-MT4 Demo
Report for EURUSD
1.5428 Low day price.
1.5643 High day price.
2008.05.01 18:33:19 The last incoming tick time (last known server time).
1.5455 Last incoming bid price. For the current symbol, it is stored in the predefined variable Bid
1.5457 Last incoming ask price. For the current symbol, it is stored in the predefined variable Ask
0.0001 Point size in the quote currency. For the current symbol, it is stored in the predefined variable Point
4 Count of digits after decimal point in the symbol prices. For the current symbol, it is stored in the predefined variable Digits
10 Stop level in points.
100000 Lot size in the base currency.
10 Tick value in the deposit currency.
0.0001 Tick size in points.
0.3 Swap of the long position.
-0.4 Swap of the short position.
0 Market starting date (usually used for futures).
0 Market expiration date (usually used for futures).
1 Trade is allowed for the symbol.
0.1 Minimum permitted amount of a lot.
0.1 Step for changing lots.
1000 Maximum permitted amount of a lot.
0 Swap calculation method. 0 - in points; 1 - in the symbol base currency; 2 - by interest; 3 - in the margin currency.
0 Profit calculation mode. 0 - Forex; 1 - CFD; 2 - Futures.
0 Margin calculation mode. 0 - Forex; 1 - CFD; 2 - Futures; 3 - CFD for indices.
0 Initial margin requirements for 1 lot.
0 Margin to maintain open positions calculated for 1 lot.
0 Hedged margin calculated for 1 lot.
1545.7 Free margin required to open 1 lot for buying.
0 Order freeze level in points. If the execution price lies within the range defined by the freeze level, the order cannot be modified, cancelled or closed.
0.9448 Low day price.
0.9534 High day price.
2008.05.01 18:32:58 The last incoming tick time (last known server time).
0.949 Last incoming bid price. For the current symbol, it is stored in the predefined variable Bid
0.9497 Last incoming ask price. For the current symbol, it is stored in the predefined variable Ask
0.0001 Point size in the quote currency. For the current symbol, it is stored in the predefined variable Point
4 Count of digits after decimal point in the symbol prices. For the current symbol, it is stored in the predefined variable Digits
10 Stop level in points.
100000 Lot size in the base currency.
9.81643271 Tick value in the deposit currency.
0.0001 Tick size in points.
0 Swap of the long position.
0 Swap of the short position.
0 Market starting date (usually used for futures).
0 Market expiration date (usually used for futures).
1 Trade is allowed for the symbol.
0.1 Minimum permitted amount of a lot.
0.1 Step for changing lots.
1000 Maximum permitted amount of a lot.
0 Swap calculation method. 0 - in points; 1 - in the symbol base currency; 2 - by interest; 3 - in the margin currency.
0 Profit calculation mode. 0 - Forex; 1 - CFD; 2 - Futures.
0 Margin calculation mode. 0 - Forex; 1 - CFD; 2 - Futures; 3 - CFD for indices.
0 Initial margin requirements for 1 lot.
0 Margin to maintain open positions calculated for 1 lot.
0 Hedged margin calculated for 1 lot.
932.15 Free margin required to open 1 lot for buying.
0 Order freeze level in points. If the execution price lies within the range defined by the freeze level, the order cannot be modified, cancelled or closed.
Report for SLV ( Silver)
15.992 Low day price.
17.032 High day price.
2008.05.01 18:33:43 The last incoming tick time (last known server time).
16.182 Last incoming bid price. For the current symbol, it is stored in the predefined variable Bid
16.222 Last incoming ask price. For the current symbol, it is stored in the predefined variable Ask
0.001 Point size in the quote currency. For the current symbol, it is stored in the predefined variable Point
3 Count of digits after decimal point in the symbol prices. For the current symbol, it is stored in the predefined variable Digits
10 Stop level in points.
5000 Lot size in the base currency.
5 Tick value in the deposit currency.
0.001 Tick size in points.
-1.65 Swap of the long position.
1.45 Swap of the short position.
0 Market starting date (usually used for futures).
0 Market expiration date (usually used for futures).
1 Trade is allowed for the symbol.
0.1 Minimum permitted amount of a lot.
0.1 Step for changing lots.
1000 Maximum permitted amount of a lot.
0 Swap calculation method. 0 - in points; 1 - in the symbol base currency; 2 - by interest; 3 - in the margin currency.
1 Profit calculation mode. 0 - Forex; 1 - CFD; 2 - Futures.
4 Margin calculation mode. 0 - Forex; 1 - CFD; 2 - Futures; 3 - CFD for indices.
0 Initial margin requirements for 1 lot.
0 Margin to maintain open positions calculated for 1 lot.
0 Hedged margin calculated for 1 lot.
811.1 Free margin required to open 1 lot for buying.
0 Order freeze level in points. If the execution price lies within the range defined by the freeze level, the order cannot be modified, cancelled or closed.
Report for GLD
847.1 Low day price.
881.65 High day price.
2008.05.01 18:33:43 The last incoming tick time (last known server time).
851.2 Last incoming bid price. For the current symbol, it is stored in the predefined variable Bid
851.7 Last incoming ask price. For the current symbol, it is stored in the predefined variable Ask
0.01 Point size in the quote currency. For the current symbol, it is stored in the predefined variable Point
2 Count of digits after decimal point in the symbol prices. For the current symbol, it is stored in the predefined variable Digits
10 Stop level in points.
100 Lot size in the base currency.
1 Tick value in the deposit currency.
0.01 Tick size in points.
-8.25 Swap of the long position.
7.25 Swap of the short position.
0 Market starting date (usually used for futures).
0 Market expiration date (usually used for futures).
1 Trade is allowed for the symbol.
0.1 Minimum permitted amount of a lot.
0.1 Step for changing lots.
1000 Maximum permitted amount of a lot.
0 Swap calculation method. 0 - in points; 1 - in the symbol base currency; 2 - by interest; 3 - in the margin currency.
1 Profit calculation mode. 0 - Forex; 1 - CFD; 2 - Futures.
4 Margin calculation mode. 0 - Forex; 1 - CFD; 2 - Futures; 3 - CFD for indices.
0 Initial margin requirements for 1 lot.
0 Margin to maintain open positions calculated for 1 lot.
0 Hedged margin calculated for 1 lot.
851.7 Free margin required to open 1 lot for buying.
0 Order freeze level in points. If the execution price lies within the range defined by the freeze level, the order cannot be modified, cancelled or closed.
203
Thanks Phy
my last question:
shouldn't I use avrage of tick value at beginning and end?
2653
Why?
203
phy:
Why?
Once you told me that tick value at the end is important, but once more I wanted to confirm, regarding reverse pairs.
263
Phy
Once again your concise and accurate answer helped me. I was struggling with converting a risk amount in pips to the account currency. What I did not realise from the documentation is that the TICKVALUE is for one lot, but your answer to AR78's question provided the spark of realisation ;-)
Thanks
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# x>1, y>1, is x>y?
Author Message
TAGS:
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12740
Location: United States (WA)
GMAT 1: 750 Q49 V42
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WE: Information Technology (Hospitality and Tourism)
Followers: 2612
Kudos [?]: 12389 [0], given: 3831
x>1, y>1, is x>y? [#permalink] 19 May 2003, 15:41
Expert's post
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Can anybody solve this one?
x>1, y>1, is x>y?
i) squ x > y
ii) squ y < x
bb
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12740
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2612
Kudos [?]: 12389 [0], given: 3831
Expert's post
brstorewala wrote:
I would say A
i am using this problem for the Question bank and it is an old one. I can't understand commander's last comment here
http://www.gmatclub.com/phpbb/viewtopic.php?t=368&highlight=another+hard
Any thoughts?
Manager
Joined: 28 Feb 2003
Posts: 100
Followers: 1
Kudos [?]: 3 [0], given: 0
from statement 1....
square root of any number x (where x>1), is less than the number itself.......
so if the square root of x is greater than y, then obviously x >y always.....so statement 1 is sufficient....
from statement 2.......
Assume y = 9 and x = 4........in this case
sqrt 9 < 4 ( i.e sqrt y < x) and y > x
Assume y = 9 and x = 12.....in this case
sqrt 9 < 12 ( i.e sqrt y < x) and x > y
so statement 2 is insufficient
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12740
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2612
Kudos [?]: 12389 [0], given: 3831
Expert's post
brstorewala wrote:
from statement 1....
square root of any number x (where x>1), is less than the number itself.......
so if the square root of x is greater than y, then obviously x >y always.....so statement 1 is sufficient....
from statement 2.......
Assume y = 9 and x = 4........in this case
sqrt 9 < 4 ( i.e sqrt y < x) and y > x
Assume y = 9 and x = 12.....in this case
sqrt 9 < 12 ( i.e sqrt y < x) and x > y
so statement 2 is insufficient
Sorry for taking up your time. I was doing this one late at night and just did not understand the question. It is a one line explanation; sorry about that
bb
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statement -2 is confusiing
surareroot y<x
please expl. why this is not sufficient
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"A" it is . From statment 1 we have x> y^2.....and for all x and y > 1 , x has to be > y...Suff
From statement 2 we have y < x^2...for x = 2 and y = 3, it is true, so x<y in this case , but for x = 4 and y = 3, above exp is still true but x > y...thus insuff.
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Press of Hudson-Kimberly Publishing Company, 1898 - Geometry - 406 pages
### Contents
Problems 110 Measurement Ratio 113 Limits 129 Proportional Lines 135 Similar Figures Problems 141 Exercises Problems 161 AREAS 167 Relations of Homologous Parts of Similar Figures 179 Triangles Areas 185 Moulding of Polygons Exercises 199 Problems 209
Pyramids 289 Regular Polyedrons Problems Eulers Theorem 308 The Cylinder 316 The Cone 328 The Sphere 340 Spherical Angles and Polygons 353 Spherical Measurements 368 Exercises 381 Tables of Denominate Numbers 388 Index 399 Copyright
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Page 126 - Proposition 7. 289. ff four quantities are in proportion, they are in proportion by composition and division; that is, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference.
Page 233 - Prove that the area of a triangle is equal to half the product of its perimeter by the radius of the inscribed circle.
Page 390 - Weight is used in weighing gold, silver, jewels, liquors, &c., and is generally adopted in philosophical experiments. 24 grains (gr.) make 1 pennyweight, marked pwt. 20 pennyweights " 1 ounce " oz. 12 ounces " 1 pound,
Page 234 - When will they be equal, and when supplementary ? 2. An angle formed by two chords intersecting within the circumference of a circle is measured by one-half the sum of the intercepted arcs.
Page 33 - A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center. The curve which bounds the circle is called the circumference Any portion of the circumference is called an arc.
Page 20 - If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.
Page 391 - CUBIC MEASURE 1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 27 cubic feet = 1 cubic yard (cu. yd.) 128 cubic feet = 1 cord (cd... | 728 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.815359 |
https://www.teacherspayteachers.com/Product/5th-Grade-Math-Jeopardy-Game-Divide-Unit-Fractions-by-Whole-Numbers-5NF7-1979581 | 1,526,939,521,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00447.warc.gz | 832,987,783 | 18,281 | # 5th Grade Math Jeopardy Game - Divide Unit Fractions by Whole Numbers 5.NF.7
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Fifth Grade Common Core Math Jeopardy Game - 5 NF.7 Divide Unit Fractions by Whole Numbers 5.NF.7 Practice provides two ways for students to practice and show mastery of their ability to divide unit fractions by whole numbers and whole numbers by unit fractions. 5.NF.B.7 5.NF.B.7a 5.NF.B.7b 5.NF.B.7c
Jeopardy board includes 25 distinct problems and utilizes hyperlinks to award/deduct money from your student teams! (79 slides in all!) Board is automatically updated to show which questions have already been utilized. Also includes an Answer Key if you wish to use questions as Quiz/Test. The PowerPoint file can be used on computers, or Promethean and Smart boards.
Take a look at the preview file and buy today for your students benefit!
Standard 5.NF.7 Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
Apply and extend previous understandings of division to divide unit
fractions by whole numbers and whole numbers by unit fractions.
a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context
for (1/3) ÷ 4, and use a visual fraction model to show the quotient.
Use the relationship between multiplication and division to explain
that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3.
b. Interpret division of a whole number by a unit fraction, and
compute such quotients. For example, create a story context for
4 ÷ (1/5), and use a visual fraction model to show the quotient. Use
the relationship between multiplication and division to explain that
4 ÷ (1/5) = 20 because 20 × (1/5) = 4.
c. Solve real world problems involving division of unit fractions by
non-zero whole numbers and division of whole numbers by unit
fractions, e.g., by using visual fraction models and equations to
represent the problem. For example, how much chocolate will each
person get if 3 people share 1/2 lb of chocolate equally? How many
1/3-cup servings are in 2 cups of raisins?
Total Pages
79 pages
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Teaching Duration
1 hour
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\$3.50 | 569 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-22 | latest | en | 0.878135 |
https://studysoup.com/tsg/1042286/single-variable-calculus-early-transcendentals-8-edition-chapter-2-2-problem-3 | 1,632,671,248,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00505.warc.gz | 588,115,118 | 12,546 | ×
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# Explain the meaning of each of the following. (a) lim xl23 fsxd (b) lim xl41 fsxd 2
ISBN: 9781305270336 484
## Solution for problem 3 Chapter 2.2
Single Variable Calculus: Early Transcendentals | 8th Edition
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Problem 3
Explain the meaning of each of the following. (a) lim xl23 fsxd (b) lim xl41 fsxd 2
Step-by-Step Solution:
Step 1 of 3
1) a. Match the following groups with its description: Metazoa mouth develops first Deuterostomes true tissues Bilateria multicellularity/tissues Protostomes bilateral symmetry, triploblastic Eumetazoa anus develops first, then mouth b. Organize the above developments chronologically through evolutionary time by placing them on the proper number on the tree: Choanoflagella tes 1 (1)
Step 2 of 3
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##### ISBN: 9781305270336
Single Variable Calculus: Early Transcendentals was written by and is associated to the ISBN: 9781305270336. This full solution covers the following key subjects: . This expansive textbook survival guide covers 95 chapters, and 5427 solutions. This textbook survival guide was created for the textbook: Single Variable Calculus: Early Transcendentals, edition: 8. The full step-by-step solution to problem: 3 from chapter: 2.2 was answered by , our top Calculus solution expert on 03/19/18, 03:29PM. The answer to “Explain the meaning of each of the following. (a) lim xl23 fsxd (b) lim xl41 fsxd 2” is broken down into a number of easy to follow steps, and 18 words. Since the solution to 3 from 2.2 chapter was answered, more than 232 students have viewed the full step-by-step answer.
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Explain the meaning of each of the following. (a) lim xl23 fsxd (b) lim xl41 fsxd 2 | 530 | 1,955 | {"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} | 2.78125 | 3 | CC-MAIN-2021-39 | latest | en | 0.860472 |
https://arduino.stackexchange.com/questions/61748/arduino-max-power-supply-limit | 1,715,992,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00419.warc.gz | 85,171,635 | 42,483 | # Arduino max power supply limit
What is the maximum current that can be supplied to an Arduino Mega board before frying/breaking?
Basically i have several components connected to my board which draw a total current of 1.6A (which i calculated through a table), and thinking of using a 9V 2A power supply. I know that the arduino voltage limit is 7-12V but just wondering about the current as I can't find it on the datasheets?
• I think you mean "supplied from"; you can supply infinite amps to it. If you components can use vin, then you're not really limited, other than to the pins/traces. Feb 18, 2019 at 20:17
• yes sorry i meant i will have an external 9V 2A power supply to power the arduino and my components. as i will be connecting this through the japan socket, the max current input limit would be 1A as suggested by @CrossRoads? Or is this also infinite? Thanks! Feb 18, 2019 at 21:39
1A coming in thru the barrel jack connecor, then you risk blowing the 1A rated reverse polarity protection diode. The 5V regulator will overheat at high currents above 7.5V. The chip itself can have 800mA put thru if if properly cooled and the IO limits of current per port are respected.
So 1.6A, no way.
• thanks for the info! so im assuming the max current that can be provided through the japan jack is 1A? Feb 18, 2019 at 21:09
• Thru the barrel jack, yes ,1A. Thru the USB jack, 0.5A. Feb 18, 2019 at 21:10
From the ATmega328P datasheet.
• One should specify that these are the limits of the MCU unit. But the current does not have to go through the MCU. For a 5V pin the limit comes from the power supply, 500 mA for USB and 1000 mA for the linear voltage regulator (barrel jack). In both cases one must subtract the power consumption of the board. Also note that USB power is disconnected as soon as you put voltage through the regulator: they have no chance to work concurrently. Dec 3, 2021 at 8:15
Your question isn't clear. Are you talking about drawing regulated 5V from the Arduino power supply? I believe the built-in regulator has a hard limit if 1A total output, but how much you can draw from it will depend on various factors (mostly how much waste heat it generates. If you drive it from 9V it will generate quite a bit of heat, since a linear regulator simply dissipates the excess voltage as heat. Better to use a 7V supply)
You won't be able to draw 1.6A of 5V from the Arduino. You should probably get a separate 2A 5V regulated power supply for your external components. Again though, your question isn't very clear so I'm not sure I'm answering what you're asking.
• Thanks for your reply. So the idea is to have an external power supply to power the arduino board along with its components. I re-did my calculations, and the total current and power dissipated by the components and arduino board would be 3.8W and 0.6A. Do you reckon this power supply would be enough amazon.co.uk/gp/product/B00HG9PT4K/… ? It's rated at 9V/650mA and says its for the arduino. The alternative solution would be to have a 9V/2A if this is pushing on the limit Feb 18, 2019 at 21:10
• Is that what your external components need? 9V? Most solid state components need regulated 5V. If your external components need 9V then you might get away with driving them from the 9V supply and also powering the Arduino into it's barrel connector using the same 9V supply. The Arduino regulator should handle fluctuations in input voltage as long as you don't draw so much current from the 9V supply that its voltage drops below 7V or so. Feb 18, 2019 at 21:15
• I have 2 8V stepper motors, a 3V LED light and HC05 bluetooth module Feb 18, 2019 at 21:25
• You should not cut it that close. A 2A supply would be a much better choice. You'd be better off using a separate low power 5V regulated supply for the Arduino. Keeping the power supplies separate will avoid big swings in voltage as your power components (motor and bluetooth module) turn on and off. Feb 19, 2019 at 1:55 | 1,032 | 3,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.935961 |
https://www.fact-archive.com/encyclopedia/Straightedge | 1,623,763,910,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00594.warc.gz | 696,846,937 | 3,424 | ## Online Encylopedia and Dictionary Research Site
Online Encyclopedia Search Online Encyclopedia Browse
# Straightedge
For the subculture, see Straight edge.
A straightedge is a tool similar to a ruler, but without markings.
An straightedge is used in ruler-and-compass constructions. It may be used to:
• Given two points, draw the line connecting them.
• Given a point and a circle, draw either tangent.
• Given two circles, draw any of their common tangents.
It may not be marked or used together with the compass so as to transfer the length of one segment to another.
It is possible to do all straightedge-and-compass constructions without the straightedge. That is, it is possible, using only a compass, to find the intersection of two lines given two points on each, and to find the tangent points to circles. It is not, however, possible to do all constructions using only a straightedge. It is possible to do them by drawing one circle and its center, then using only the straightedge.
## Making a straightedge
A real-world straightedge is a practical tool that approximates the idealized mathematical object described above.
Making a practical straightedge without access to a pre-existing straightedge or modern machine tools is an interesting problem in mechanical engineering. One way of doing this is to make a high-quality flat surface first, using engineer's blue and polishing techniques, and then shape the straightedge so as to lie flat along the flat surface.
The method of producing accurate flat surfaces using three trial surfaces was invented by Sir Joseph Whitworth in 1830.
Another way of doing this is described in the link at the end of the article.
## References
• Wayne R. Moore, Foundations of Mechanical Accuracy, Moore Special Tool Company, Bridgeport, CT (1970) | 370 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-25 | latest | en | 0.878482 |
http://www.cfd-online.com/Forums/openfoam-programming-development/115932-zero-equation-turbulence-models.html | 1,480,728,466,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00205-ip-10-31-129-80.ec2.internal.warc.gz | 382,360,092 | 17,451 | # Zero Equation Turbulence models
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April 9, 2013, 11:56 Zero Equation Turbulence models #1 New Member Stefan Gracik Join Date: Oct 2011 Location: Penn State Posts: 8 Rep Power: 7 Has anyone implemented any zero equation turbulence models to OpenFOAM? That is, a turbulence model which uses algebraic equations to calculate turbulent viscosity rather than PDE's (k, ε etc.). The classic example of this is the Prandtl mixing length model. I'm trying to implement a specialized model and am having trouble modifying the 2 and 1 eq models to remove the PDE's, so I figured having a zero equation model already implemented would help me understand what I need to do. Thanks
April 16, 2013, 11:25 #2 New Member Stefan Gracik Join Date: Oct 2011 Location: Penn State Posts: 8 Rep Power: 7 I've made some progress on this, but am stuck again. It is an equation designed for modeling external airflow over buildings. The turbulence model I'm trying to implement is depended on this distance to wall, where H is the average height of the building (implemented as a dimensionedScalar) for walldist <= 1.3*H, nut = const*exp(walldist/H)*U and for walldist >1.3*H, nut = const*walldist ... etc. I thought the best way to write this would be with an if statement. Currently I have written it like this (where d_ is the walldist) Code: ``` if (d_ <= (1.3*H_)) { nut_ = a_*zh_*exp(-b_*d_/H_)*U_*pow((d_/H_),2); nut_.correctBoundaryConditions(); } else { nut_ = 0.16*Uh_*(d_ + z0_)/Foam::log((d_ + z0_)/z0_); nut_.correctBoundaryConditions(); }``` but receive a pretty long error Code: ```wmakeLnInclude: linking include files to ./lnInclude Making dependency list for source file mykEpsilon.C SOURCE=mykEpsilon.C ; g++ -m64 -Dlinux64 -DWM_DP -Wall -Wextra -Wno-unused-parameter -Wold-style-cast -Wnon-virtual-dtor -O3 -DNoRepository -ftemplate-depth-100 -I/opt/openfoam211/src/turbulenceModels -I/opt/openfoam211/src/transportModels -I/opt/openfoam211/src/finiteVolume/lnInclude -I/opt/openfoam211/src/meshTools/lnInclude -I/opt/openfoam211/src/turbulenceModels/incompressible/RAS/lnInclude -IlnInclude -I. -I/opt/openfoam211/src/OpenFOAM/lnInclude -I/opt/openfoam211/src/OSspecific/POSIX/lnInclude -fPIC -c \$SOURCE -o Make/linux64GccDPOpt/mykEpsilon.o mykEpsilon.H: In constructor ‘Foam::incompressible::RASModels::mykEpsilon::mykEpsilon(const volVectorField&, const surfaceScalarField&, Foam::transportModel&, const Foam::word&, const Foam::word&)’: mykEpsilon.H:81:31: warning: ‘Foam::incompressible::RASModels::mykEpsilon::H_’ will be initialized after [-Wreorder] mykEpsilon.H:77:31: warning: ‘Foam::dimensionedScalar Foam::incompressible::RASModels::mykEpsilon::Uh_’ [-Wreorder] mykEpsilon.C:47:1: warning: when initialized here [-Wreorder] mykEpsilon.C:160:19: error: no match for ‘operator<=’ in ‘((Foam::incompressible::RASModels::mykEpsilon*)this)->Foam::incompressible::RASModels::mykEpsilon::d_ <= Foam::operator*(const Foam::dimensioned&, const Foam::dimensioned&) [with Type = double]((*(const Foam::dimensioned*)(&((Foam::incompressible::RASModels::mykEpsilon*)this)->Foam::incompressible::RASModels::mykEpsilon::H_)))’ mykEpsilon.C:160:19: note: candidates are: /opt/openfoam211/src/OpenFOAM/lnInclude/UList.C:224:6: note: bool Foam::UList::operator<=(const Foam::UList&) const [with T = double] /opt/openfoam211/src/OpenFOAM/lnInclude/UList.C:224:6: note: no known conversion for argument 1 from ‘Foam::dimensioned’ to ‘const Foam::UList&’ /opt/openfoam211/src/OpenFOAM/lnInclude/VectorSpaceI.H:693:13: note: template bool Foam::operator<=(const Foam::VectorSpace&, const Foam::VectorSpace&) mykEpsilon.C:162:50: error: no match for ‘operator=’ in ‘((Foam::incompressible::RASModels::mykEpsilon*)this)->Foam::incompressible::RASModels::mykEpsilon::nut_ = Foam::operator*(const Foam::tmp >&, const Foam::tmp >&) [with Type = Foam::Vector, PatchField = Foam::fvPatchField, GeoMesh = Foam::volMesh]((*(const Foam::tmp >*)(& Foam::pow(const Foam::tmp >&, const scalar&) [with PatchField = Foam::fvPatchField, GeoMesh = Foam::volMesh, Foam::scalar = double]((* &2.0e+0)))))’ mykEpsilon.C:162:50: note: candidates are: /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1083:6: note: void Foam::GeometricField::operator=(const Foam::GeometricField&) [with Type = double, PatchField = Foam::fvPatchField, GeoMesh = Foam::volMesh] /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1083:6: note: no known conversion for argument 1 from ‘Foam::tmp, Foam::fvPatchField, Foam::volMesh> >’ to ‘const Foam::GeometricField&’ /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1108:6: note: void Foam::GeometricField::operator=(const Foam::tmp >&) [with Type = double, PatchField = Foam::fvPatchField, GeoMesh = Foam::volMesh] /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1108:6: note: no known conversion for argument 1 from ‘Foam::tmp, Foam::fvPatchField, Foam::volMesh> >’ to ‘const Foam::tmp >&’ /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1144:6: note: void Foam::GeometricField::operator=(const Foam::dimensioned
&) [with Type = double, PatchField = Foam::fvPatchField, GeoMesh = Foam::volMesh] /opt/openfoam211/src/OpenFOAM/lnInclude/GeometricField.C:1144:6: note: no known conversion for argument 1 from ‘Foam::tmp, Foam::fvPatchField, Foam::volMesh> >’ to ‘const Foam::dimensioned&’ make: *** [Make/linux64GccDPOpt/mykEpsilon.o] Error 1 stefan@stefan-OpenFOAM:~/OpenFOAM/stefan-2.1.1/src/turbulenceModels/incompressible/RAS/mykEpsilon\$``` Any idea what the issue is? I have a feeling it has something to do with comparing walldist to a dimensionedScalar, but I'm really not sure.
April 17, 2013, 03:33 #3 Member cosimo bianchini Join Date: Mar 2009 Location: Florence, Tuscany, Italy Posts: 88 Rep Power: 9 If I did not misunderstand you should use your if statement within a loop on the internal cells of your volScalarField nut_. What the error is saying is that it makes no sense a statement like <= for a volScalarField in fact for some cells it will be > for others <. In order to make it work write something like forAll(nut_,cellI) { if(d_[cellI] <= (1.3*H_)) { nut_[cellI] = a_*zh_*exp(-b_*d_[cellI]/H_)*mag(U_)*pow((d_[cellI]/H_),2); } else { nut_[cellI] = 0.16*Uh_*(d_[cellI] + z0_)/Foam::log((d_[cellI] + z0_)/z0_); } } assuming that all other variables except for d_ and U_ are of type scalar this should work. Another thing you should pay attention to is in the use of vector variables (I guessed U_ it is velocity vector) in a statement with scalars on the LHS, I added the operator mag(U_) to evaluate the magnitude of velocity but I'm not sure this is what you want. Change the operator accordingly to your needs but check that it returns a scalar. __________________ Cosimo Bianchini Ergon Research s.r.l. Via Panciatichi, 92 50127 Florence - ITALY Tel: +39 055 0763716 Mob: +39 320 9460153 e-mail: cosimo.bianchini@ergonresearch.it URL: www.ergonresearch.it
April 17, 2013, 14:12 Thanks #4 New Member Stefan Gracik Join Date: Oct 2011 Location: Penn State Posts: 8 Rep Power: 7 Thanks for the help!! but now I have a new error. d_ is generated from walldist.H, so it's the distance to the nearest wall. All of the other constants I've defined as dimensionedScalar Code: ```a_ ( dimensioned::lookupOrAddToDict ( "a", coeffDict_, 0.09 ) ), b_ ( dimensioned::lookupOrAddToDict ( "b", coeffDict_, 0.09 ) ), H_ ( dimensioned//::lookupOrAddToDict ( "H", dimensionSet(0, 1, 0, 0, 0, 0, 0), 1.92 ) ), Uh_ ( dimensioned//::lookupOrAddToDict ( "Uh", dimensionSet(0, 1, -1, 0, 0, 0, 0), 1.44 ) ), zh_ ( dimensioned//::lookupOrAddToDict ( "zh", dimensionSet(0, 1, 0, 0, 0, 0, 0), 1.92 ) ), z0_ ( dimensioned//::lookupOrAddToDict ( "z0", dimensionSet(0,1,0,0,0,0,0), 1.3 ) ),``` but I still receive the error "mykEpsilon.C:180:83: error: cannot convert ‘Foam::dimensioned’ to ‘double’ in assignment" Anyone have any idea why this could be?
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A246584 Number of overcubic partitions of n. 7
1, 2, 6, 12, 26, 48, 92, 160, 282, 470, 784, 1260, 2020, 3152, 4896, 7456, 11290, 16836, 24962, 36556, 53232, 76736, 110012, 156384, 221156, 310482, 433776, 602200, 832224, 1143696, 1565088, 2131072, 2890266, 3902344, 5249356, 7032576 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Convolution of A001935 and A002513. - Vaclav Kotesovec, Aug 16 2019 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..10000 Michael D. Hirschhorn, A note on overcubic partitions, New Zealand J. Math., 42:229-234, 2012. Bernard L. S. Lin, Arithmetic properties of overcubic partition pairs, Electronic Journal of Combinatorics 21(3) (2014), #P3.35. James A. Sellers, Elementary proofs of congruences for the cubic and overcubic partition functions, Australasian Journal of Combinatorics, 60(2) (2014), 191-197. FORMULA G.f.: Product_{k>=1} (1+x^k) * (1+x^(2*k)) / ((1-x^k) * (1-x^(2*k))). - Vaclav Kotesovec, Aug 16 2019 a(n) ~ 3^(3/4) * exp(sqrt(3*n/2)*Pi) / (2^(19/4)*n^(5/4)). - Vaclav Kotesovec, Aug 16 2019 MAPLE # to get 140 terms: ph:=add(q^(n^2), n=-12..12); ph:=series(ph, q, 140); g1:=1/(subs(q=-q, ph)*subs(q=-q^2, ph)); g1:=series(g1, q, 140); seriestolist(%); # second Maple program: with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)*add(d* `if`(irem(d, 4)=2, 3, 2), d=divisors(j)), j=1..n)/n) end: seq(a(n), n=0..40); # Alois P. Heinz, Aug 17 2019 MATHEMATICA nmax = 50; CoefficientList[Series[Product[(1+x^k) * (1+x^(2*k)) / ((1-x^k) * (1-x^(2*k))), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 16 2019 *) nmax = 50; CoefficientList[Series[Product[(1+x^(2*k)) / (1-x^k)^2, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 16 2019 *) CROSSREFS Trisections: A246585, A246586, A246587. Sequence in context: A136515 A141347 A300120 * A054454 A084170 A245264 Adjacent sequences: A246581 A246582 A246583 * A246585 A246586 A246587 KEYWORD nonn AUTHOR N. J. A. Sloane, Sep 03 2014 STATUS approved
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Last modified January 23 22:36 EST 2020. Contains 331177 sequences. (Running on oeis4.) | 1,015 | 2,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-05 | latest | en | 0.516156 |
http://forums.wolfram.com/mathgroup/archive/2006/Apr/msg00597.html | 1,723,149,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00368.warc.gz | 11,235,434 | 7,786 | Re: Setting up equations (Revision)
• To: mathgroup at smc.vnet.net
• Subject: [mg66019] Re: [mg65990] Setting up equations (Revision)
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 27 Apr 2006 02:26:08 -0400 (EDT)
• References: <200604260837.EAA02689@smc.vnet.net> <E1D0F0B5-4CE6-4740-9A3D-C17A255794F8@cox.net>
• Sender: owner-wri-mathgroup at wolfram.com
```My first response had some unnecessary steps.
eqn=5 x+6 y+7 z==a x+b y+c z;
Solve[(Flatten[CoefficientList[#,{x,y,z}]]&/@
eqn),{a,b,c}]
{{a -> 5, b -> 6, c -> 7}}
Bob Hanlon
hanlonr at cox.net
On Apr 26, 2006, at 7:35 AM, Bob Hanlon wrote:
> eqn=5 x+6 y+7 z==a x+b y+c z;
>
> Solve[Equal@@
> (Flatten[CoefficientList[#,{x,y,z}]]&/@
> List@@eqn),{a,b,c}]
>
> {{a -> 5, b -> 6, c -> 7}}
>
>
> Bob Hanlon
> hanlonr at cox.net
>
>
>
> On Apr 26, 2006, at 4:37 AM, Yaroslav Bulatov wrote:
>
>> I'm trying to do things of the form
>> Solve[5 x + 6 y + 7 z == a x + b y + c z, {a, b, c}]
>>
>> But since x,y,z are variables, what I really mean is
>> Solve[5==a && 6==b && 7==c], so I need to convert to this form
>>
>> If I only have one variable, the following does what I need
>>
>> LogicalExpand[a*x + b*x^2 + O[x]^3 == 2*x + 3*x^2 + O[x]^3]
>>
>> But what to do if I have several variables?
>>
>
```
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• Next by thread: Re: Setting up equations | 571 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-33 | latest | en | 0.770419 |
https://www.hackmath.net/en/word-math-problems/cube-root?tag_id=86 | 1,611,288,382,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00339.warc.gz | 786,529,031 | 9,398 | # Diagonal + cube root - math problems
#### Number of problems found: 7
• Cube diagonal
Determine the length of the cube diagonal with edge 37 mm.
• Body diagonal
Calculate the cube volume, whose body diagonal size is 75 dm. Draw a picture and highlight the body diagonal.
• Cube diagonals
Calculate the length of the side and the diagonals of the cube with a volume of 27 cm3.
• Cube - wall
V kocke ABCDEFGH je ?. Aký je povrch kocky?
• Cube and sphere
Cube with the surface area 150 cm2 is described sphere. What is sphere surface?
• Cube in a sphere
The cube is inscribed in a sphere with a volume 7253 cm3. Determine the length of the edges of a cube.
• Body diagonal
The cuboid has a volume of 32 cm3. Its side surface area is double as one of the square bases. What is the length of the body diagonal?
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Diagonal - math word problems. Cube root - math word problems. | 321 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.892875 |
https://www.abcadda.com/21c-to-f/ | 1,725,731,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00637.warc.gz | 607,666,139 | 20,285 | 0 Celsius = 32 Fahrenheit
celsius
fahrenheit
Answer:
0 Celsius = 32 Fahrenheit
(0 c = 32 f)
21c to f
Table of Contents
21 degrees celsius converted to 69.8 fahrenheit
21c to f °C = 69.8 °F
°F = ( 21 x 1.8 ) + 32 = 69.8 °F
21c to f in Canada, the UK, and several other European countries, temperatures are measured in degrees Celsius (° C).
Luckily, it’s easy to convert between the 21c to f celsius to fahrenheit when needed. Just plug the known temperature into the appropriate equation to convert it like 21 c to f.
21Celsius to Fahrenheit is the conversion of temperature from one unit Celsius to one Fahrenheit unit.
Temperature is measured with a thermometer. While Kelvin is the SI unit of temperature, we regularly use 21 Celsius or Celsius (° C) and Fahrenheit (° F) to measure temperature.
Let us see this article to calculate 21degrees celsius to fahrenheit
Definition of 21c to f / Celsius To Fahrenheit
The temperature in 21 Celsius is given in (° C). For example, the normal body temperature is 21 ° C.
The temperature in Fahrenheit is given in ° F. For example, a normal body temperature is 98.6 ° F.
21 Celsius vs. 21 Fahrenheit: Key Difference
Before we explain how 21 Celsius is converted to Fahrenheit (and Fahrenheit to Celsius), let’s look at the main differences between the 21 celsius in Fahrenheit temperature scales.
• What is 21 degrees celsius converted to fahrenheit
• 21 degrees celsius is equal to what fahrenheit
• What is 21 degrees in fahrenheit converted to celsius
• What is 21 degrees celsius to fahrenheit
• Minus 21 degrees celsius converted to what temperature fahrenheit
• What is -21c equal to in f N/A
• -21c equal to what degree in f?
• How to turn 21 f to 21c
• What is f equal to -21c
• 21c to what f
• What is 21c equal to in f
How do I convert 21 degrees celsius to fahrenheit?
The formula ° F = ° C × (9/5) + 32 is used to convert Celsius to Fahrenheit. For example,
You want to convert 21c in Fahrenheit. When giving your answer in ° C to ° F, you should know that the temperature scale is Celsius and Fahrenheit.
It doesn’t matter what your final answer is, but if you’ve ever been expected to spell a name, it’s good to know. Conversion is very easy:
Formula for converting 21 celsius to fahrenheit
Multiply the temperature by ° C by 1.8. Add 32 to this number. Here is the answer in ° F.
° F = (° C × 9/5) + 32
Converting Fahrenheit to Celsius is that easy.
° C = (° F – 32) × 5/9
Example of converting ° C to ° F.
For example, to convert 26 ° C to ° F (warm daytime temperature):
° F = (° C × 9/5) + 32
° F = (26 × 9/5) + 32
° F = (46.8) + 32
The two different types of units or scales used to measure temperature are Celsius and Fahrenheit. In other words, they are two different scales for measuring temperature.
21 degrees c to f
Answer: 21c Fahrenheit is the same as 105 ° Fahrenheit.
Let’s take a closer look at the transformations between the 21° celsius to Fahrenheit
Explanation:
The formula for converting 21°c to f / celsius to fahrenheit is 21c to Fahrenheit = ° C × (9/5) + 32
21c in f = [C × (9/5) + 32]
Against this background, C = 21 c to f
21 c to f = 21 × (9/5) + 32
21 degrees C to F = 72 + 32
21 c in F = 69.8
So 21c to Fahrenheit is equal to 105 ° F.
Conclusion
The temperature change is performed utilizing an alternate equation relying upon the two temperature scales you are changing over.
For example, to convert 21 degrees Celsius to Fahrenheit, add our numbers to a formula like this: F = C * 9/5 + 32
As with math calculations and transformations, it’s a good idea to double-check your results.
If you use this formula for your own conversion, you can check your result with a calculator from 21 degrees C in F. | 1,012 | 3,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.677961 |
http://mathhelpforum.com/pre-calculus/81523-please-check-my-work.html | 1,503,041,883,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00009.warc.gz | 264,329,685 | 11,108 | Can you please check my work
Can you please tell me where I went wrong becasue my teacher said that answer is 13720
2. Originally Posted by supersaiyan
Can you please check my work
Can you please tell me where I went wrong becasue my teacher said that answer is 13720
Let me get this straight.
You have a function $f(x) = 400e^{\frac{\ln(2.375)}{2}x}$...
And you have to find out what $f'(10)$ is?
3. Originally Posted by supersaiyan
Can you please check my work
Can you please tell me where I went wrong becasue my teacher said that answer is 13720
4. Yes, i have, he said check your work again
5. Originally Posted by Mush
Let me get this straight.
You have a function $f(x) = 400e^{\frac{\ln(2.375)}{2}x}$...
And you have to find out what $f'(10)$ is?
Yes
6. Originally Posted by supersaiyan
Can you please check my work
Can you please tell me where I went wrong becasue my teacher said that answer is 13720
$f(x) = 400 e^{kx}$ where $k = \frac{1}{2} \ln (2.375)$.
$f'(x) = 400 k e^{kx}$.
$f'(10) = 400 k e^{10k}$.
$10 k = 5 \ln (2.375) = \ln (2.375)^5 \Rightarrow e^{10k} = 2.375^5$.
So $f'(10) = 200 [\ln (2.375)] \, 2.375^5$.
Now use a calculator if a decimal approximation is required. | 396 | 1,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.936666 |
http://www.reference.com/browse/+quality | 1,435,702,028,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00000-ip-10-179-60-89.ec2.internal.warc.gz | 685,686,908 | 18,937 | Definitions
Related Questions
# quality
Quality-adjusted life years, or QALYs, is a way of measuring disease burden, including both the quality and the quantity of life lived, as a means of quantifying in benefit of a medical intervention. The QALY model requires utility independent, risk neutral, and constant proportional tradeoff behaviour.
They are based on the number of years of life that would be added by the intervention. Each year in perfect health is assigned the value of 1.0 down to a value of 0 for death. If the extra years would not be lived in full health, for example if the patient would lose a limb, or be blind or be confined to a wheelchair, then the extra life-years are given a value between 0 and 1 to account for this.
The meaning and usefulness of QALY is debated. Perfect health is hard, if not impossible, to define. Some argue that there are health states worse than death, and that therefore there should be negative values possible on the health spectrum (indeed, some health economists have incorporated negative values into calculations). Determining the level of health depends on measures that some argue place disproportionate importance on physical pain or disability over mental health. The effects of a patient's health on the quality of life of others - caregivers, family etc. also does not figure into these calculations.
The "weight" values between 0 and 1 are usually determined by methods such as:
• Time-trade-off (TTO) - In this method, respondents are asked to choose between remaining in a state of ill health for a period of time, or being restored to perfect health but having a shorter life expectancy.
• Standard gamble (SG) - In this method, respondents are asked to choose between remaining in a state of ill health for a period of time, or choosing a medical intervention which has a chance of either restoring them to perfect health, or killing them.
• Visual analogue scale (VAS) - In this method, respondents are asked to rate a state of ill health on a scale from 0 to 100, with 0 representing death and 100 representing perfect health. This method has the advantage of being the easiest to ask, but is the most subjective.
Another way of determining the weight associated with a particular health state is to use standard descriptive systems such as the EuroQol EQ-5D questionnaire, which categorise health states according to the following dimensions: mobility, self-care, usual activities (e.g. work, study, homework or leisure activities), pain/discomfort and anxiety/depression.
However, the weight assigned to a particular condition can vary greatly, depending on the population being surveyed. Those who do not suffer from the affliction in question will, on average, overestimate the detrimental effect on quality of life, compared to those who are afflicted.
QALYs are used in cost-utility analyses to calculate the ratio of cost to QALYs saved for a particular health care intervention. This is then used to allocate healthcare resources, with an intervention with a lower cost to QALY saved ratio being preferred over an intervention with a higher ratio. This method is controversial because it means that some people will not receive treatment as it is calculated that cost of the intervention is not warranted by the benefit to their quality of life. However, its supporters argue that since health care resources are inevitably limited, this method enables them to be allocated in the way that is most beneficial to society. | 706 | 3,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.951045 |
https://www.urionlinejudge.com.br/judge/en/profile/165995 | 1,603,821,703,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00550.warc.gz | 960,036,442 | 6,472 | # PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1066 Even, Odd, Positive and... 02464º 9109522 Python 3 0.016 1/24/18, 9:38:28 AM
1060 Positive Numbers 02837º 9105375 Python 3 0.016 1/23/18, 1:52:37 PM
1059 Even Numbers 01949º 9105325 Python 3 0.012 1/23/18, 1:44:18 PM
1048 Salary Increase 01462º 9102089 Python 3 0.016 1/22/18, 6:50:54 PM
1006 Average 2 04594º 9100723 Python 3 0.016 1/22/18, 2:14:26 PM
1005 Average 1 04892º 9100703 Python 3 0.016 1/22/18, 2:06:44 PM
1004 Simple Product 05697º 9100585 Python 3 0.016 1/22/18, 1:40:47 PM
1003 Simple Sum 05920º 9100572 Python 3 0.016 1/22/18, 1:37:59 PM
1002 Area of a Circle 06336º 9100472 Python 3 0.016 1/22/18, 1:11:42 PM
1001 Extremely Basic 06683º 9018880 Python 0.012 1/5/18, 6:59:13 PM
1175 Array change I 07278º 8880956 C 0.000 12/6/17, 4:33:21 PM
1174 Array Selection I 07830º 8879461 C 0.000 12/6/17, 1:02:28 PM
1173 Array fill I 08609º 8872808 C 0.000 12/5/17, 1:24:28 PM
1172 Array Replacement I 09218º 8872392 C 0.000 12/5/17, 12:08:25 PM
1026 To Carry or not to Carry 02645º 8865937 C 0.092 12/4/17, 4:54:36 PM
1165 Prime Number 06014º 8830122 C 0.004 11/29/17, 3:48:28 PM
1164 Perfect Number 04612º 8829860 C 0.000 11/29/17, 3:40:26 PM
1160 Population Increase 07184º 8829450 C 0.008 11/29/17, 1:54:00 PM
1159 Sum of Consecutive Even... 04209º 8820193 C 0.000 11/28/17, 1:57:09 PM
1158 Sum of Consecutive Odd... 03764º 8812335 C 0.000 11/27/17, 6:24:21 PM
1157 Divisors I 05036º 8811409 C 0.000 11/27/17, 5:40:46 PM
1156 S Sequence II 03603º 8811356 C 0.000 11/27/17, 5:35:47 PM
1155 S Sequence 04131º 8810378 C 0.000 11/27/17, 3:50:25 PM
1154 Ages 04979º 8810082 C 0.000 11/27/17, 3:35:17 PM
1153 Simple Factorial 06543º 8809660 C 0.000 11/27/17, 2:46:44 PM
1151 Easy Fibonacci 06158º 8809568 C 0.000 11/27/17, 2:36:12 PM
1150 Exceeding Z 03480º 8803488 C 0.000 11/26/17, 7:04:50 PM
1149 Summing Consecutive Integers 03711º 8643004 C 0.000 10/18/17, 3:56:05 PM
1146 Growing Sequences 02691º 8439228 C 0.128 10/15/17, 4:32:00 PM
1145 Logical Sequence 2 03776º 8438909 C 0.012 10/14/17, 6:06:04 PM
1 of 4 | 1,043 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.316365 |
http://www.aptech.com/resources/tutorials/introduction-to-gauss-graphing-data/ | 1,561,418,225,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999779.19/warc/CC-MAIN-20190624231501-20190625013501-00304.warc.gz | 206,719,403 | 19,316 | # Introduction to GAUSS: Graphing Data
### Introduction
This exercise uses the data generated in Exercise One and Exercise Two. If you have not completed these exercises, STOP and complete the exercises. The model used in this exercise follows the data generating process below:
$$y_i = 2 + 3.5x_i + \epsilon_i$$
## Step One: Load the saved x and y data
In the last part of this tutorial series, we saved two matrices x and y. You can load them by entering:
load x, y;
This command will look for the files x.fmt and y.fmt in your current working directory. It will then create matrices x and y in your GAUSS workspace and fill them with the data from the corresponding file.
If keep your data in a separate directory that you do not want to be your working directory, you can specify a separate 'load' path. When you specify a 'load' path, GAUSS will look in that directory when you load data with the load command.
load path = C:\gauss19\data;
load x, y;
This code will attempt to load x and y from the directory C:\gauss19\data.
## Step Two: Graph the data
### Create a new file
Open a new program file by clicking the New File toolbar button, or using the hotkey CTRL+N.
### Save the file
Save the file by either clicking the Save File toolbar button or by using the hotkey CTRL+S. Give the file the name plotexample.gss.
### Add commands to the program file
Add the load command from above. Next, add the command:
plotScatter(x, y);
As you start typing plot… GAUSS will provide you with an autocomplete menu listing all functions that begin with plot. You may use the arrow buttons to scroll to one of the functions in this dropdown list and hit enter to have GAUSS enter this command into your file.
You may also notice that when you type the opening parenthesis, GAUSS will give you a tooltip listing the inputs for the function.
### Draw the graph
To run the file, either use the hotkey CTRL+R or select Current File from the drop-down menu to the right of the Run button.
After your file is run, you should see a graph that is similar to the image below.
While your graph will look quite similar to the graph above, it may show a different color for the points or have the grid turned on for example. All of these settings come from your graphics preference settings.
### Change graphics preferences
To open your graphics preferences, select Tools > Preferences from the main menu. This will open the preferences dialog menu. Select Graphics from the left side of the preferences menu.
Towards the top of the Graphics preferences window, under Profiles is a list of graph types. In the above image, the current profile is Scatter (Default).
The Graph Settings are below. This section contains the settings for the currently selected graph profile. The Series one through five, control the settings for each column of data that you draw.
To change the color, or symbol for the one column of data that you are graphing in this example, click Series 1 as shown in the above image and customize its settings.
After you have made some changes to the graph, save your changes by clicking Apply button and then OK. Now run your file again and see your changes reflected in the new graph.
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Get help from our friendly experts. | 730 | 3,343 | {"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": 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} | 2.765625 | 3 | CC-MAIN-2019-26 | latest | en | 0.800583 |
armantheparman.medium.com | 1,685,269,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00085.warc.gz | 132,389,883 | 37,683 | # The Maths of Ethereum Price
--
What I’m about to spell out will change what you think about the future of Ethereum. The conclusion of this article is counterintuitive, but irrefutable. I have never heard anyone else state these facts, and can not draw from other articles.
Warning: I use the terms cost and price frequently, and it’s easy to confuse the two and get lost in the language; they are not interchangeable. Cost is an amount to be paid in total. Price is an amount per unit.
If demand for ETH goes up, then price will go up as well, all else being equal.
The following will show that even a massive increase in transaction volume will not increase demand for ETH, and therefore will not increase price. This sounds crazy — at first.
We need to recognise that there is a range of market price that the public will tolerate for a particular smart contract execution, measured in USD.
For example, the public is prepared to pay around \$50 USD to put their house title on the Ethereum blockchain, or something like that, it might be too cheap, but it doesn't really matter. The important part is that there is going to be a number give or take a few per cent that will be generally acceptable. Much more than that, and people will shy away. Much less than that, and it will be more attractive.
Next, look at these equations to see how everything is connected:
The first equation in diagram 1 shows that the overall amount of ETH the public will need to buy (and consume) to execute smart contracts is proportional to how much ETH is consumed per average transaction (red brace in diagram 1), and proportional to the overall number of smart contract transactions (Tx volume).
The amount of ETH consumed per transaction (red brace), is equal to the cost of gas per transaction (eg \$50 USD needed, blue arrow). The average cost of gas (in USD) is proportional to the average amount of gas consumed in an average transaction. The cost of gas is also proportional to the price of gas. Price of gas is the amount of ETH consumed per gas consumed (This price of gas can be converted to USD consumed per gas consumed using ETH price)
We can therefore rewrite the first equation like this:
The component within the blue brace in diagram 2 comes from the blue arrow in diagram 1, and I said earlier that it is relatively fixed.
So if transaction volume goes up (Tx volume), then Volume ETH demanded goes up, so ETH price goes up. Hooray!
Unfortunately, everyone stops here.
I kept digging because something intuitively doesn’t feel right.
What happens to the transaction costs (blue brace) when ETH price goes up?
Looking at diagram 2, Tx volume can be increased, increasing demand for ETH (left hand side of equation), which increases ETH price, which increases gas price. But gas price is within the blue brace! This is not allowed. Blue brace total must stay the same, because the public will not tolerate too much of an increase in cost for an average transaction.
To compensate, the gas price falls, and less ETH is consumed per gas consumed. That’s right. When more ETH is demanded because of more transactions, less ETH is demanded per transaction, because gas price falls. This is necessary, otherwise transaction costs will escalate an Ethereum will not be used.
This proves that the expected massive increase in ETH transaction volume does not increase the demand of ETH, or price of ETH.
What matters to ETH price is the average cost per transaction (in USD, but paid in ETH) that an average user is willing to pay. Trust me, it won’t be an astronomical amount.
So what makes ETH price go up? Speculative demand. (Or reduced supply)
If speculative demand is high, price goes up, gas price goes down, \$50 transaction costs (as per the original example) remains unchanged.
Speculation drives ETH price. Not utility. Similarly, the industrial properties of gold have not driven it to its current price of \$1580 per oz. Speculation of its monetary properties and SOV properties has done this.
For ETH to moon, it needs to compete against Bitcoin as money, and it can not. It has poor monetary properties compared to Bitcoin. Utility in the form of smart contracts is not a monetary property. Demand for utility, as I have shown, does not increase ETH price.
ETH is staying on Earth. Bitcoin is going to moon. | 904 | 4,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | longest | en | 0.951648 |
https://www.brightstorm.com/math/precalculus/linear-equations-and-inequalities/applied-linear-equations-investment-problems/ | 1,675,490,974,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00444.warc.gz | 696,229,774 | 28,536 | ###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
##### Thank you for watching the video.
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# Applied Linear Equations: Investment Problem - Concept
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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In Algebra II and in the real world, sometimes we need to solve investment math problems by using linear equations. When solving word problems using linear equations, we first need to pull out the relevant information and put it into equation form. When working with investment math problems, we are usually asked to calculate amounts earned by interest on original principal amounts.
This example we're going to be dealing with linear equations in a Investment problem. So we have here is I have \$3,000 to invest I put some of it to an account that pays 6 percent and the rest into an account that pays 8 percent. These are both simple interest which basically means that you take interest once a year. So later in your units and your years you're going to talk about sort of more complicated interest. This is just we take it once you're done with it. So invest some at 6 some at 8 and after a year I hope to make \$216. How much should I invest in each account?
Okay so let's take a look at this. As I do with a lot of my problems is I actually take the word problem and then turn it to some sort of diagram that I can sort all the information out. So we have one account, that's the first one we'll call this one being the 6 percent. I'm investing a certain amount in this account, we don't know what it is so let's call it x. Okay we then have another account, this one is at 8 percent and we don't know how much we actually have to invest in this one. But we don know that together they are \$3,000 so if I invested 1,000 here I'll know that I would have to invest 2,000 here. If I invested 500 I'll have to invest 2,500 put together they have to add up to \$3,000.
So this one is just going to be 3,000 minus whatever I invested over here, and together we end up with \$216 so the interest from one account plus the interest from the other account is equal to 216. Okay so once we have our diagram we can go ahead and make our equation, so we get 6 percent from this amount we invested. We invested \$100 we get 6 percent as interest, we'll get 6 percent of a 100, \$6. So basically we just multiply 6 percent as the decimal times the amount we invested. So this turn to 0.06x, 6 percent remember to change your decimal over there, okay this one is just 8 percent times the amount we invested, same idea we 0.08 times 3000 minus x and that's going to equal the total amount of interest we make \$216.
Okay we have a linear equation so solve it as we would any other equation, distribute this through and again if you want you can multiply by 100 to clear our decimals, for this example I'm going to leave it in there either way it's just fine. So 06x stays the same 0.8 times 3000 it's a little bit big so we can use our calculator. 0.08 times 3000 is 240 0.08 times x and then that's 216 is still there. So in your equation combine like terms find 0.06 minus 0.08 will end up with -0.02x is 216. Can bring our 240 around -0.02x equals 216 minus 240, 216 minus 240 I should be able to do that in my head but just to make sure you end up with -24 divide by -0.02 to isolate that variable and -24 divided by 0.02 is \$1200 okay make sure we always answer our questions correctly.
So the questions is asking for how much that I invested in each account so x is here was the amount we invested at 6 percent, so we know that we need to get 1200 at 6 percent out total amount of money was \$3000 that 3000 minus this 12000 we invested will be 1800, 1800 at 8 percent. So in order to get \$216 interest you'll have to invest 1200 at 6 percent and 1800 at 8 percent. | 1,016 | 4,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.968457 |
https://discuss.leetcode.com/topic/109965/simple-solution-after-finding-the-rule-in-zigzag-conversion-problem | 1,521,671,794,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647706.69/warc/CC-MAIN-20180321215410-20180321235410-00356.warc.gz | 569,837,054 | 8,698 | # Simple solution after finding the rule in ZigZag Conversion problem
• This is very simple as long as we find the rule.
when numRows=3, the index for each position in the string s would be:
1 5 9
2 4 6 8 10
3 7 11
So the output index from string s for each row would be:
row0= [0:4:end]
row1= [1:2:end]
row2= [2:4:end]
when numRows=4, the index for each position in the string s would be:
1 7 13 19
2 5 8 11 14 17 20
3 6 9 12 15 18 21
4 10 16 22
So the output index from string s for each row would be:
row0=[0:6:end]
row1=[1:3:end]
row2=[2:3:end]
row3=[3:6:end]
So we can easily find the rule, which would be for numRows, we will have N_loop items for each loop (a vertical line and a cross line), where N_loop = numRows + (numRows-2). Meanwhile, for the rows on top or bottom, the repeated index would be N_loop because there is no repetition between them, while for rows in middle, the repeated index would be (numRows-1).
So the output index from string s for each row would be:
row0=[0:N_loop:end]
row1=[1:(numRows-1):end]
row2=[2:(numRows-1):end]
...
row numRows-1=[numRows-1:N_loop:end]
Python implementation will be:
``````class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
new_s = [];
Nloop = numRows + (numRows-2);
#row 1:
idx = 0;
while (idx<len(s)-1):
new_s.append(s[idx]);
idx +=Nloop;
# middle rows
for r in range(1,numRows-1):
idx = r;
while (idx<len(s)-1):
new_s.append(s[idx]);
idx +=numRows-1;
#last rows
idx = numRows-1;
while (idx<len(s)-1):
new_s.append(s[idx]);
idx +=Nloop;
return new_s;``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 535 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.738855 |
https://lessonplanet.com/lesson-plans/applying-mathematical-reasoning/42 | 1,701,910,232,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00597.warc.gz | 415,155,296 | 47,273 | # Applying Mathematical Reasoning Teacher Resources
Make sense of the world around you by using mathematical reasoning to analyze situations, understand relationships, and draw conclusions. These are fundamental skills that should be in every student mathematician's toolkit.
Showing 3,899 resources
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#### Math Problems, Calculations and Mirror Images
For Students 4th - 5th
In this online math exercises worksheet, students complete a variety of assignments including practicing eight math calculations, two word problems and drawing three mirror images. Students check their work online as they go.
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#### Yertle the Turtle
For Students 2nd - 5th
In this Dr. Seuss story problems learning exercise, students read through the Dr. Seuss story problems and solve them. There are 10 story problems on the page and they are all very basic math.
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#### Number Sense and Operations
For Students 4th - 6th
In this math story problem worksheet, learners read the 9 story problems and select the correct answer from the 4 choices available. The top of this worksheet indicates that it is appropriate for Grade 5.
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#### Real Numbers
For Students 6th - 8th
In this math worksheet, students evaluate the expressions that use integers. They also differentiate between the values of two integers as greater or less than.
Lesson Plan
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#### Addition Of Real Numbers
For Teachers 6th - 8th
In this math worksheet, middle schoolers practice combining the terms using the operation of addition. They pay special attention to the signs.
Lesson Plan
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#### Multiplication and Division of Real Numbers
For Teachers 6th - 8th
In this math worksheet, middle schoolers use the operations of division and multiplication to simplify the expressions. They must be sure to apply the order of operations correctly.
Worksheet
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#### Multi Step Operations
For Students 4th - 6th
In this math operations worksheet, students read the story problem and solve the problem using multiple step operations. There are 8 story problems on this worksheet and the answers are on the second page.
Worksheet
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#### Single Step Operations
For Students 4th - 6th
In this math operations worksheet, students solve the math story problems by solving single step operations. There are 10 problems on this worksheet and the answers are on the second page.
Worksheet
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#### Choose the Operation and Method
For Students 3rd - 6th
In this math operations and methods learning exercise, students read the story problems and then fill in the table with the appropriate information. The table asks for operation, method, how and answer. There are 3 problems on this...
Worksheet
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#### Number Stories: Ice Cream
For Students K - 2nd
For this number stories worksheet, students solve 3 problems that require them to subtract the number 4 using math counter pictures. A number line is provided
Worksheet
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#### Zoo groups
For Students 5th - 8th
In this zoo animals learning exercise, students read about a particular zoo animal and answer the math word problems about the amount of meals, pounds, herds, and more. Students complete 6 problems.
Lesson Plan
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#### Money Matters
For Teachers 2nd - 5th
Students practice spending money by completing math word sentences. In this economics lesson, students utilize problem solving strategies to complete word problems written on the board involving making purchases with money. Students...
Worksheet
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#### Factoring Application
For Students 9th
In this factoring application instructional activity, 9th graders solve 10 different problem related to various factoring applications. First, they determine the length of a room given the area of a rectangle and square footage. Then,...
Worksheet
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#### Valentine Application of the Quadratic Formula
For Students 8th - 10th
For this quadratic formula application, students write a quadratic equation that models the situation and then solves the equation to determine the number of students, women, and men that are in a class. The solution is provided.
Lesson Plan
Curated OER
#### Using Negative Numbers
For Teachers 6th
Sixth graders practice using negative numbers by completing number equations. In this math problem solving lesson, 6th graders discuss the different ways negative numbers can be used in a math problem and practice completing equations...
Worksheet
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#### Draw a Picture
For Students 4th - 5th
For this problem solving strategies worksheet, students read and analyze the instructions and examples for drawing a picture to solve a math problem. Students solve 1 problem pertaining to students' positions in a line.
Worksheet
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#### Later And Earlier 2
For Students 3rd - 4th
In this math activity, students examine clocks and figure elapsed time . For each of 20 clocks, students write: the time the clock shows now, 15 minutes later, 45 minutes later and 90 minutes later. There is very little room for students...
Worksheet
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#### It's Time for Time
For Students 4th - 5th
In this math worksheet, students solve 4 problems pertaining to telling time and elapsed time. All are word or story problems. Students show their work on the page.
Lesson Plan
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#### Applications and Problem Solving Using Quadratic Equations
For Teachers 6th - 8th
In this math worksheet, middle schoolers solve problems in the form of quadratic equations after translating into expressions from words.
Lesson Plan
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#### Application Problems with Fractions
For Teachers 7th - 8th
In this math worksheet, students compare fractions from a chart. They simplify six fractions. Students tell the fraction of a day they spend doing various activities. Each fraction is simplified. Three problems are multiple choice problems.
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#### Geometric Sequences - Bacterial Growth
For Students 8th - 10th Standards
Bring algebra to life with scientific applications. Math minded individuals calculate and graph the time it takes a bacterium to double. They discuss geometric sequences and use a chart to graph their findings. There are 38 questions all...
Assessment
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#### Graphs (2007)
For Teachers 9th - 12th Standards
Challenge the class to utilize their knowledge of linear and quadratic functions to determine the intersection of the parent quadratic graph and linear proportional graphs. Using the pattern for the solutions, individuals develop a...
Interactive3:28
1
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#### Study Jams! Creating Equations from Word Problems
For Students 4th - 7th Standards
Word problems are full of information waiting to be deciphered. This animated video explains how to look for the important parts and how to translate those word cues into the correct signs. Go through a problem from start to finish and...
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#### A Boxer named Bobo
For Teachers 8th - 11th
Bobo the Boxer needs a pen and algebra learners need to find the maximum area that can be created from a fixed amount of fence. A graphing calculator is used to assist in their exploration. | 1,565 | 7,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.921651 |
http://math.stackexchange.com/questions/215350/properties-of-borel-sets-and-lebesgue-measure | 1,469,752,501,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00049-ip-10-185-27-174.ec2.internal.warc.gz | 157,631,874 | 18,537 | # Properties of Borel sets and Lebesgue measure
Let $A$ be a set, let $B$ be a Borel set such that $B \subseteq A$.
Because $B$ is a Borel set, can I automatically say that I can represent it as a countable union of closed sets. Thus I can demonstrate that $m(A\setminus B)=0$.
Alternatively, let $C$ is a set and $D$ is a Borel set such that $C \subseteq D$. Then, since $D$ is Borel we can write it as a countable intersection of open sets. So $m(D \setminus C)=0$.
I'm trying to find a relationship between Borel sets and the Inner/Outer approximation theorems.
-
Sadly, we can do nothing so simple. Countable unions (intersections) of closed (open) sets are called $F_\sigma$ $(G_\delta)$ sets. Countable unions (intersections) of $G_\delta$ $(F_\sigma)$ sets are called $G_{\delta\sigma}$ $(F_{\sigma\delta})$ sets. Similarly, we have $G_{\delta\sigma\delta}$ and $F_{\sigma\delta\sigma}$ sets, $G_{\delta\sigma\delta\sigma}$ and $F_{\sigma\delta\sigma\delta}$ sets, and so on. All of these are Borel sets (along with basic open and closed sets)--and in fact comprise the entirety of the collection of Borel sets. For a more explicit description of this transfinitely recursive construction of the Borel heirarchy--in $\omega_1$ steps, not countably many (Thanks, Trevor, for pointing that out!)--see here.
Taking $A$ to be the overlying set, $B'$ to be any Borel subset of $A$ of positive measure, and $B=A\smallsetminus B'$, we have that $B$ is a Borel set and $A\smallsetminus B=B'$, and furthermore $m(A\smallsetminus B)=m(B')>0$. On the other hand, take $D$ to be any Borel set of positive measure, and let $C=\emptyset$, so that $D\smallsetminus C=D$, and so $m(D\smallsetminus C)=m(D)>0$.
It is worth noting that we can approximate any measurable set from without (within) by some $G_\delta$ $(F_\sigma)$ set, in exactly the way you described. We just can't do it with every $G_\delta$ $(F_\sigma)$ superset (subset). Also, we can't necessarily do this for arbitrary sets, which may not be measurable.
-
It is perhaps worth mentioning that the "and so on" in the first paragraph means "for $\omega_1$ many steps" and not "for countably many steps". – Trevor Wilson Oct 17 '12 at 3:52
Interesting, Trevor! I'd never heard or suspected that. Can you link me to a source? – Cameron Buie Oct 17 '12 at 4:05
Wait...I fumbled my positioning of $B$ and $A$ in the set differences. – emka Oct 17 '12 at 4:39
Ah! In that case, you run into a similar problem, taking $A$ to be the overlying set, and $B$ to be a set not of full measure. I'll correct my answer to reflect that. – Cameron Buie Oct 17 '12 at 5:06
@CameronBuie See en.wikipedia.org/wiki/Borel_hierarchy. The reason is that each of the finite stages you mentioned properly contains the previous ones, so we can pick for each $n$ a set that doesn't appear until stage $n$. Then we can take a single set coding these countably many sets, and this set will not appear at any finite stage. – Trevor Wilson Oct 17 '12 at 15:55 | 874 | 2,994 | {"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} | 3.390625 | 3 | CC-MAIN-2016-30 | latest | en | 0.895865 |
https://paperzz.com/doc/9127578/lesson-9-%E2%80%93-surface-area-of-cones-example-exercises | 1,638,166,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00624.warc.gz | 488,141,702 | 5,131 | ### Lesson 9 – Surface Area of Cones Example Exercises
```Homework
Geometry & Probability 8
Unit 1 – Volume & Surface Area
Name
Date
Period
Lesson 9 – Surface Area of Cones
The lateral area L.A. of a cone is π times the radius times the slant height, or L.A. = πrℓ.
The total surface area of a cone with slant height ℓ and radius r is the lateral area plus the area of the base, or
2
2
S.A. = L.A. + πr or S.A. = πrℓ + πr .
Example
Find the lateral and total surface areas of the cone. Round to the nearest tenth.
Lateral Surface Area
L.A. = πrℓ
L.A. = π • 3 • 5 r = 3, ℓ = 5
L.A. ≈ 47.1
Total Surface Area
The lateral and total surface areas of the cone are about 47.1 and 75.4 square centimeters.
Exercises
Find the lateral and total surface area of each cone. Use the pi key on your calculator. Show
your work. Round to the nearest hundredth.
1. LA = _________________
3. LA = _________________
2. SA = _________________
4. SA =_________________
5. _________________
Audrey uses a metal scoop to measure the correct amount of food to give to her horse. The scoop is
shaped like a cone with a diameter of 6 inches and a slant height of 8.5 inches. What is the lateral area
of the cone?
Formula:
``` | 502 | 1,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.833181 |
https://forum.azimuthproject.org/plugin/viewcomment/22684 | 1,632,541,305,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00416.warc.gz | 317,537,765 | 2,830 | Coriolis is complex rather than truly constant because variable geophysical flows interact with constant rotation to comprise the Effect (see below). These flows vary in velocity and direction, and the latent Force itself varies greatly with latitude and altitude (Vertical Coriolis). Only in unphysical approximation is it idealized as a constant waiting to happen. Its also not helpful to cite that its relatively constant overall, on-average, when dealing with subscale cases like ENSO or QBO, one by one. Coriolis is extra variable at the planetary surface, where cloud-mediated solar-driven convection causes increased inertial interaction of flow with rotation.
Gross energy misestimation controversies are still quite common in modern geophysics, like in Wind Energy, where recent total estimates of wildly varying magnitude cannot be all correct ("Coastline of England" fractal-scales uncertainty is a factor). I think Laplace greatly underestimated Coriolis in his Tidal Equations. [Keeling & Whorf 2000] seem to mostly neglect vertical Coriolis Effect energy input as such, in badly overestimating the role of Tidal Energy forcing. They also seem too much to discover what they wish in noisy data.
There may be a Three Jesus Fallacy at work with some of the controversies. Psychologists [Rokeach 64] put together three patients who each claimed to be Jesus, expecting conflict. Instead they got along rather well as their outlier Confirmation Bias was supported; that it was reasonable to claim to be Jesus, even as each demurred as to the other two being mistaken, while seeing themself as the one true Jesus. Beware such a dynamic in the extreme Lunisolar Forcing holdout faction. Let's stop short of asserting "lunacy", as we are all by definition Jesuses to some degree O:-)
Here is an approximate simplified Horizontal Coriolis Force formula for a moving mass. As is evident, there are multiple variables (a, m, v) and one constant (w).
F = 2 * m * v * w * sin(a)
F is force
m is object mass
v is object velocity
w is Earth angular velocity
a is latitude | 447 | 2,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.945 |
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# Table 1.40 shows mens and womens world records for swimming distances from 50 meters to
ISBN: 9780470484753 315
## Solution for problem 1.6.12 Chapter 1-6
Functions Modeling Change: A Preparation for Calculus | 4th Edition
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Problem 1.6.12
Table 1.40 shows mens and womens world records for swimming distances from 50 meters to 1500 meters.22 (a) What values would you add to Table 1.40 to represent the time taken by both men and women to swim 0 meters? (b) Plot mens time against distance, with time t in seconds on the vertical axis and distance d in meters on the horizontal axis. It is claimed that a straight line models this behavior well. What is the equation for that line? What does its slope represent? On the same graph, plot womens time against distance and find the equation of the straight line that models this behavior well. Is this line steeper or flatter than the mens line? What does that mean in terms of swimming? What are the values of the vertical intercepts? Do these values have a practical interpretation? (c) On another graph plot the womens times against the mens times, with womens times, w, on the vertical axis and mens times, m, on the horizontal axis. It should look linear. How could you have predicted this linearity from the equations you found in part (b)? What is the slope of this line and how can it be interpreted? A newspaper reporter claims that the womens records are about 8% slower than the mens. Do the facts support this statement? What is the value of the vertical intercept? Does this value have a practical interpretation?
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Early Development 6/8/17 11:25 AM -Developmental Psych’s Major Issue examine physical, cognitive, and social development across lifespan 1. physical o how you grow 2. cognitive o how your thinking changes o how your attention...
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Reset your password | 630 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-50 | latest | en | 0.91404 |
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Chapter: Problem:
Write a spreadsheet program that computes the trajectory of a rubber-bandlaunched projectile. Your spreadsheet calculations should have cells in which you can enter key parameters of the problem such as dimensions, launch angle, and amount of rubber-band stretch. Use an array of cells to indicate the position of the projectile at various points along its trajectory. Choose any appropriate values for the projectile mass m and spring constant k.
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Velocity of the projectile will be found by equating potential energy due to rubber stretching and the kinetic energy developed due to motion of the projectile. The equation is given by,
Here m is the mass of the projectile, v is the velocity, K is the stiffness of the rubber band and l is the stretching of rubber band
Parameter Value Notation Stiffness of rubber (k) 500N/m \$C\$3 Projectile mass (m) 2 kg \$C\$4 Rubber stretch (l) 0.02m \$C\$5 Acceleration due to gravity (g) 9.8 m/s2 \$C\$7 Launch Angle ? Degree 30 \$C\$8 Range ( R) 0.008828678 none
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Corresponding Textbook
Design Concepts for Engineers | 4th Edition
9780136069553ISBN-13: 013606955XISBN: Mark N HorensteinAuthors:
Alternate ISBN: 9780133002485 | 372 | 1,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.806547 |
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Anonymous asked in Food & DrinkCooking & Recipes · 12 months ago
# If replacing 1 cup = 1.5 cups, does that mean replacing 1tsp = 1.5tsp?
Is it a 1:1.5 ratio?
I am replacing flour with oat flour, and according to a chart, 1 cup of flour is to be replaced by 1.5 cups of oat flour
But the recipe only calls for a tsp, not a whole cup
Relevance
• Alex
Lv 7
12 months ago
Correct. The ratio is 1 to 1.5.
So 1 tsp would be 1.5 tsp, 2 cups would be replaced by 3 cups, etc.
• 12 months ago
Your math is sound, it should be fine to substitute 1 tsp with 1.5 tsp
• CB
Lv 7
12 months ago
NO! You are substituting ingredients (Flour for Oat flour) you are not replacing 1 cup of flour with 1.5 cups flour to increase the amount of the recipe.
• 12 months ago
That little bit will not matter much but you are correct 1 and 1/2 teaspoons would be the replacement. | 265 | 875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-34 | latest | en | 0.911599 |
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10 months ago | 1,476 | 5,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-05 | latest | en | 0.75265 |
https://backgroundimagepro.com/qa/why-is-it-called-horsepower.html | 1,627,640,193,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00584.warc.gz | 129,553,441 | 8,075 | # Why Is It Called Horsepower?
## Is 1 HP equal to a horse?
Does one horsepower equal one horse.
Not quite.
It’s a common misconception that one horsepower is equal to the peak power production of a horse, which is capable of a maximum of around 14.9 horsepower.
By comparison, a human being is capable of approximately five horsepower at peak power production..
## How many horses does it take to make 1 horsepower?
See, a common misconception is that one horsepower is the same as the peak power output of an actual horse, the truth is that it’s closer to 14.9 horses per horsepower.
## Is it better to have more torque or horsepower?
Horsepower is how rapidly the vehicle can perform that work. … Of course, horsepower is tied to torque mathematically. Horsepower equals torque multiplied by rpm, divided by a constant. Because there is generally a limit on how fast you can spin an engine, having higher torque allows for greater horsepower at lower rpms.
## Does higher horsepower mean faster?
The power produced by an engine is called its horsepower. For cars, horsepower translates into speed. So if you want to go faster, and get up to speed quicker, you need more horsepower.
## Is 300 a lot of horsepower?
Yes. 300 horsepower is indeed a lot. Of course it all depends on how heavy of a vehicle you are driving. But for an average car, 300 is quite a good amount of power.
## How much horsepower is enough?
Usually somewhere around 250 is more than enough for normal people or beginners so they don’t kill themselves. For seasoned enthusiasts 400 or 500 may be fine. If you have no sense of self preservation and just want to almost die a whole lotta times. 1000 HP is for you.
## What does horsepower mean?
Horsepower refers to the power an engine produces. It’s calculated through the power needed to move 550 pounds one foot in one second or by the power needs to move 33,000 pounds one foot in one minute. The power is gauged by the rate it takes to do the work.
## How many cc’s are in 1 horsepower?
14Many people have asked for a relationship between horsepower and cc or how many cc in a hp. The short answer is about 14 to 17cc = 1 hp or about 1 cu.in. = 1 bhp for an ordinary, basic car.
## How many horsepower is a human?
When considering human-powered equipment, a healthy human can produce about 1.2 hp (0.89 kW) briefly (see orders of magnitude) and sustain about 0.1 hp (0.075 kW) indefinitely; trained athletes can manage up to about 2.5 hp (1.9 kW) briefly and 0.35 hp (0.26 kW) for a period of several hours.
## How many watts does it take to make 1 horsepower?
746 wattsOne electric horsepower is equal to exactly 746 watts.
## How was horsepower determined?
Each horse pushed with a force that Watt estimated at 180 pounds. From this, Watt calculated that one horsepower was equivalent to one horse doing 33,000 foot-pounds of work in one minute.
## Why is 1hp equal to 746 watts?
1Hp = 745.69 Newton meter per second. A 1 hp motor cannot use less than 745.7 Watts so if it uses 1 kW it is about 75% efficient, which sounds about right for a motor of that size. One mechanical horsepower of 550 foot-pounds per second is equivalent to 745.7 watts.
## How much electricity does a 1hp motor use?
1 unit = 1 kilowat-hour, i.e. when an electrical appliance consumes 1 killowat of Power for one hour . Here, we have a 1 HP motor. So, 1 HP for 24 hour = 746×24= 17904 Watt-hour which is equal to 17904/1000= 17.904 units.
## What is the most horsepower in a car?
Nine of the most powerful production cars on saleWelcome to what is, quite simply, a list of very powerful cars you can actually buy. … Koenigsegg Regera – 1,479bhp. … Bugatti Chiron – 1,479bhp. … Rimac Concept S – 1,384bhp. … Nio EP9 – 1,341bhp. … Dodge Demon – 840bhp. … Ferrari 812 Superfast – 789bhp. … Lamborghini Aventador S – 740bhp.More items…
## Which is equal to 1 HP in metric?
The electrical equivalent of one horsepower is 746 watts in the International System of Units (SI), and the heat equivalent is 2,545 BTU (British Thermal Units) per hour. Another unit of power is the metric horsepower, which equals 4,500 kilogram-metres per minute (32,549 foot-pounds per minute), or 0.9863 horsepower. | 1,060 | 4,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.942202 |
http://stacks.math.columbia.edu/tag/02JS | 1,498,164,123,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319902.52/warc/CC-MAIN-20170622201826-20170622221826-00033.warc.gz | 376,172,583 | 3,944 | # The Stacks Project
## Tag 02JS
Lemma 28.27.2. Let $f : X \to Y$ and $g : Y \to S$ be morphisms of schemes. Let $x \in X$ and set $y = f(x)$, $s = g(y)$. Assume $f$ and $g$ locally of finite type. Then $$\dim_x(X_s) \leq \dim_x(X_y) + \dim_y(Y_s).$$ Moreover, equality holds if $\mathcal{O}_{X_s, x}$ is flat over $\mathcal{O}_{Y_s, y}$, which holds for example if $\mathcal{O}_{X, x}$ is flat over $\mathcal{O}_{Y, y}$.
Proof. Note that $\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(y)}(\kappa(x)) + \text{trdeg}_{\kappa(s)}(\kappa(y))$. Thus by Lemma 28.27.1 the statement is equivalent to $$\dim(\mathcal{O}_{X_s, x}) \leq \dim(\mathcal{O}_{X_y, x}) + \dim(\mathcal{O}_{Y_s, y}).$$ For this see Algebra, Lemma 10.111.6. For the flat case see Algebra, Lemma 10.111.7. $\square$
The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4916–4928 (see updates for more information).
\begin{lemma}
Let $f : X \to Y$ and $g : Y \to S$ be morphisms of schemes.
Let $x \in X$ and set $y = f(x)$, $s = g(y)$.
Assume $f$ and $g$ locally of finite type.
Then
$$\dim_x(X_s) \leq \dim_x(X_y) + \dim_y(Y_s).$$
Moreover, equality holds if $\mathcal{O}_{X_s, x}$ is flat
over $\mathcal{O}_{Y_s, y}$, which holds for example if $\mathcal{O}_{X, x}$
is flat over $\mathcal{O}_{Y, y}$.
\end{lemma}
\begin{proof}
Note that $\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(y)}(\kappa(x)) + \text{trdeg}_{\kappa(s)}(\kappa(y))$.
Thus by Lemma \ref{lemma-dimension-fibre-at-a-point} the statement
is equivalent to
$$\dim(\mathcal{O}_{X_s, x}) \leq \dim(\mathcal{O}_{X_y, x}) + \dim(\mathcal{O}_{Y_s, y}).$$
For this see Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-total}.
For the flat case see
Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}.
\end{proof}
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## Add a comment on tag 02JS
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 748 | 2,124 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-26 | longest | en | 0.632847 |
http://betcasino.top/page/8/ | 1,568,595,076,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572439.21/warc/CC-MAIN-20190915235555-20190916021555-00532.warc.gz | 28,864,096 | 7,673 | ## Gambling Strategy & System: What’s The Difference?
When it comes to gambling, the terms strategy and system are used interchangeably. There aren’t really any accepted definitions of these terms in a gambling sense, as people interpret them in their own way. Some people view systems and strategies as pretty much the same thing, while others view them as completely separate. Continue reading “Gambling Strategy & System: What’s The Difference?”
## Probability to win at roulette
On the surface, the best probability for the roulette player to be ahead is in one trial (spin): 48.6% to win (versus 51.4% to lose), as far as even-money betting is concerned. I don’t agree that it is the best strategy (betting all your bankroll on one spin).
Theoretically, no bankroll will put a player ahead guaranteed, IF flat-betting and playing very long consecutive sessions. There are moments, however, when the roulette player can be ahead by at least one betting unit. Even in even-money bets, the player has a good chance to be ahead by at least one unit after 5, or 10, or even 100 spins. But more than 20 spins are NOT recommended; the probability (odds) to lose go(es) above 50%! Think about it! Continue reading “Probability to win at roulette”
## Roulette Systems to Avoid
SA1: Straight-Up Bet on One Favorite Roulette Number
Only the payoff is enticing in this case (35-to-1), but you are assured of losing at a rate of 5.26% of your total bets. The roulette software you get with this system (SuperRoulette) shows lengthy reports on each roulette number. The roulette wheel does not favor any number whatsoever. There is a discrepancy among numbers, but there is no strong bias towards a particular one — unless the wheel is worn out. Continue reading “Roulette Systems to Avoid” | 399 | 1,788 | {"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} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.939019 |
https://stats.stackexchange.com/questions/593782/neither-logistic-nor-exponential-function-fit-this-data-which-function-would | 1,708,521,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00393.warc.gz | 587,808,256 | 41,187 | # Neither logistic nor exponential function fit this data - which function would?
I have data that follows a very obvious trend but I am failing to find the right function to fit the data. I've tried a logistic as well as an exponential function. However, both functions do not seem to fit the data very well given how clean they look.
Below is the Python code to reproduce the plot - what kind of function would provide a better fit?
import numpy as np
from matplotlib import pyplot as plt
from scipy.optimize import curve_fit
y = np.array([0.00814809, 0.01093976, 0.01294903, 0.01403736,
0.01579931, 0.0171861 , 0.02268682, 0.02785176,
0.03149701, 0.03315134, 0.03389025, 0.03420652,
0.03465968, 0.03477675, 0.03535143])
x = np.array([10, 20, 30, 40, 50, 60, 150, 300, 600,
900, 1200, 1500, 1800, 2100,3000])
def logistic(x, L, x0, k, b):
y = L / (1 + np.exp(-k * (x - x0))) + b
return y
def exponential(x, a, b, c):
return a * x**b + c
plt.scatter(x, y, color="black")
xrange = np.linspace(x.min(), x.max(), 1000)
# fit a logistic function
popt, pcov = curve_fit(logistic, x, y, method="dogbox", maxfev=20000)
plt.plot(xrange, logistic(xrange, popt[0], popt[1], popt[2], popt[3]), label="logistic")
# fit an exponential function
popt, pcov = curve_fit(exponential, x, y, maxfev=10000)
plt.plot(xrange, exponential(xrange, popt[0], popt[1], popt[2]), label="exponential")
plt.legend()
plt.show()
• "Predictive accuracy" is usually measured in terms of an error and grows smaller as the amount of training data increases. Your numbers don't do that, so please tell us how they are computed.
– whuber
Oct 27, 2022 at 21:36
• It's certainly quite possible to choose simple-ish functions that get close to those points, but beware function-hunting; it's better to use understanding of the variables and the way they should relate in this situation to inform the model than just find something that 'fits'. (That's a by-eye fit, not any attempt to optimize coefficients or anything, since it's largely just there to illustrate that it's doable, but not necessarily useful.) Oct 27, 2022 at 23:17
• It's very important to think about the purpose for which you need this function. e.g. if you're mostly trying to interpolate other values within the range of the data, identifying an explicit parametric function is not necessary - a suitable spline fit should work just fine for that (though a simple transformation of the variables might make it a little easier). If the aim is to extrapolate beyond x=3000, function-hunting is particularly dangerous. ... Oct 27, 2022 at 23:38
• That was a fair while back now. To my recollection I did a by-eye identification of an approximate upper bound on y (erring a tad on the high side), did a logit type transformation of y and a log transformation on x to get very rough linearity and then backed out an approximate y=f(x) curve from that. It would not be responsible to use that function to try to identify/compare the size and location of major features, since it's an inadequate description of the curve. I see no reason why things like inflexion points couldn't be taken from spline fits. Oct 29, 2022 at 0:22
A preliminary inspection consists in ploting the data with various scales. For example :
We observe that the curve with ln(x) on the horizontal axis looks like a logistic curve. So, we try a logistic regression but with ln(x) instead of x.
The result is :
I agree with most of the comments to obi's question and I upvoted several.
• The OP seems to have tacitly assumed linearity in the predictor (linearity inside the logistic function). A natural approach it seems to me would have been to use a regression spline in the predictor up front. Oct 30, 2022 at 12:33
• Thank you, I also ended up doing the log-tranform you suggested. @Frank, could you elaborate what you mean by using a regression spline in the predictor?
– obi
Oct 31, 2022 at 18:32
• Details are in my RMS course notes. Oct 31, 2022 at 19:35 | 1,082 | 3,978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.1875 | 3 | CC-MAIN-2024-10 | latest | en | 0.813769 |
https://physics.stackexchange.com/questions/474183/how-much-in-the-past-are-we-viewing-the-accretion-disk-of-a-black-hole | 1,719,238,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00327.warc.gz | 408,801,236 | 43,921 | # How much in the past are we viewing the accretion disk of a black hole?
Sine the gravity near a black hole is so high, the time moves slower than the observer(in our case we from the earth).In that case, imagine if are viewing a black hole's accretion disk(the innermost rim) a million light years away and if we are observing it now then since it took the light a million years to reach us, we are looking at the past state of the black hole(how it was a million years before). But light is traveling in time that is faster than it used to the moment light escapes the event horizon.
So if we imagine one photon at a time, the first photon reaches us and then as we observe the second photon, we have moved so much into the future as our time travels faster compared to the time the photon originated.
Are millions of years old black holes actually even more older than they appear to be?
• "Sine the gravity near a black hole is so high, the time moves slower" - You have the cause and result reversed. Gravity near a black hole is so high, because time there moves slower. Not the other way around. Gravity is a consequence of a spacetime curvature, more specifically, of time curvature, which is known as time dilation (time moving slower). Commented Apr 23, 2019 at 3:50
• @safesphere Not quite; it's a case where more than one approach is mathematically coherent and complete. You can start from spacetime and geometry, or you can start from the concept of a field with a Lagrangian. You end up with the same predictions for observable quantities. Commented Apr 23, 2019 at 13:03
• @AndrewSteane In General Relativity you first find the spacetime metric from the gravitational equations. Then the metric defines the Lagrangian. So the concept of a field you are referring to is a direct consequence of the spacetime geometry. The only case where your comment would make sense, professor, is the Newtonian gravity, but (1) it does not explain the nature of gravity and (2) doesn't correctly define black holes, which are a subject here. Commented Apr 23, 2019 at 14:51
• @safesphere If you want to find out more about this, look up "Einstein Hilbert action" in the context of Lagrangian mechanics. General relativity can itself be derived from a suitable Lagrangian. Commented Apr 27, 2019 at 22:22
• @ Andrew Steane Shouldn't we just say that gravitation and time dilation go together, rather than saying that one causes the other? (I think I can infer that from your 1st comment above.) Commented Jun 24, 2019 at 19:00
Consider the following. You ask a young volunteer to come to live near a black hole, where time passes much slower. You observe them to age much less than you. Are they "stuck in the past"? It might seem so, since you were their biological age years ago. But this is not the correct interpretation, because at one point they can get on a rocket and come greet you. Surely they have not "arrived from the past", since they could have come to you at any point, and you could have come to them as well. So it really makes sense to talk about things happening synchronously, at the same time at your location as near the black hole. So really, the most accurate thing to say is that time passes slower near a black hole, but still we can synchronize!
Another, somewhat more everyday example. The Earth has formed some 4 billion years ago. The difference in the gravitational field at one position on the surface of the Earth and another one about one meter higher causes time to flow very very slightly differently. To be precise, the difference is about one second less flows one meter lower in $$10^{16}$$ seconds. But guess what, 4 billion years is $$\sim 10^{17}$$ seconds, so some $$\sim 10$$ seconds less have elapsed per every meter lower you go since the birth of Earth.
But now take a look at your feet. Are you looking ten seconds in the past? Are your feet traveling in and out of the past as you swing on a swing? Of course not. You are looking at things that happen (almost) synchronously in every physically meaningful way. So once again, slower flow of time means exactly that, clocks tick slower, but the notion of synchronicity can stay!
The additional question you are asking can be formulated in the following way: How longer will it take for a photon fired off from near the black hole to arrive then when a photon would be fired off from the "same position" without the black hole nearby? Notice that I put the "same position" in quotation marks, because the translation dictionary between a position near a black hole and a position in the space-time without the black hole will not be unique. We are after all in curved space-time, and the geometry is simply different from Euclidean space!
For instance, you can define the position near the black hole by infinity-signaling distance - the distance that is defined by how much time it takes for a light signal to reach infinity. Naturally, points at the same infinity-signaling distance in the black hole space-time as in flat Minkowski will take the same time to reach infinity. So in this operational definition of position the presence of the black hole does not push the signal "further into the past".
There are other definitions that can be used. For instance, you define distances by how much string of vanishing mass you have to use to reach a certain point. Then you would get that signals from the "same points" in the black hole geometry do arrive later than in a Minkowski one. The precise numbers will depend on your translation dictionary, but for reasonable choices and for light sources orbiting at astrophysically realistic distances from the black hole, the time delays will be of the order of $$10 \mu {\rm s} (M/M_\odot)$$, where $$M$$ is the mass of the black hole, and $$M_\odot$$ is the solar mass. If you plug in the mass of M87 observed by the Event horizon telescope, you get a delay of the order of days.
• Thanks for the response, it does help but I have a double, but what if I use a very powerful telescope to view that person (it's hypothetical as the telescope size should be so massive but if I somehow manage to do so), I look at the person's past. I want to know the difference between that person placed near a black hole and not near a black hole, somewhere else where gravity is not as strong. How does the black hole's massive gravity affect what we see? Commented Apr 21, 2019 at 19:29
• @dhayanandbaskar You don't see their past. You just see them moving slow. Good answer +1. Commented Apr 22, 2019 at 8:11
• @safesphere. By past I meant - when the light reaches us after millions of years, we are looking at millions of year old state, not what is currently happening. Can you please elaborate on how we see them move slow? Commented Apr 22, 2019 at 10:52
• @dhayanandbaskar I was responding to your comment above, not answering your question. In the comment you have two observers, one near and one away from a black hole. No "millions of years" were mentioned. As you look at these observers, you'd see the one closer to the black hole moving slower, like in a slow motion film. Commented Apr 23, 2019 at 3:45
If the true shape and texture of the accretion disk (viewed from 45°) around a rotating black hole at the time when the photon whose origin is closest to the observer is emitted is this:
and the rings of which the disk is composed of are rotating with the prograde orbital velocity, taking the light travel time and the counterclockwise rotation of the disk components into account the observed distortion would be
As you can see the inner parts of the disk are observed farther in the past than the outer parts, therefore the clockwise distortion. If the texture of the disk was static in a way that the radial white stripes would represent constant φ, the observed distortion would look like
That's also how it would look if all the photons would need the same time to reach the observer, but still took the curved paths. The images were generated with the relativistic raytracer from here. The redshift is not shown above, since some colors would shift outside the visible spectrum:
The white line in the green area marks the parts which are neither redshifted nor blueshifted. If there is enough space between the atoms that light can shine through, you would see some even older images of the disk getting bent around the photon sphere:
For the correspondig texture renderings with a transparent disk see here.
• Thanks a lot for taking time to answer. I really appreciate the amount of effort you have spent on this. This exactly answers the question. Commented Jun 26, 2019 at 11:36
• What is the scale on these pictures? An accretion disc does not extend below about $3 r_s$. Commented Jul 1, 2019 at 6:58
• @Rob Jeffries - Who told you that, the ISCO for a maximally rotating Kerr black hole (a=1) goes all the way down to r=1; set r(6r-r²-9Q²+3a²)+4Q²(Q²-a²)-8a(r-Q²)^(3/2)=0 and solve for r. In this scenario we have a Kerr Newman black hole with a=0.95, Q=0.3 where the prograde ISCO is at r=1.313818. If the scale you are asking for is the field of view, that is 30°×24° with the observer at a distance of r=50 and an outer disk radius of r=10, in units of G=M=c=kB=1. For other parameters and angles see here: notizblock.yukterez.net/viewtopic.php?p=904#p904 Commented Jul 1, 2019 at 8:02 | 2,201 | 9,372 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.943569 |
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## Algebra Calculator - MathPapa
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Algebra Calculator shows you the step-by-step solutions! ... If you would like to create your own math expressions, here are some symbols that the calculator ...
## Simplify any Algebraic Expression - WebMath
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## How to Solve an Algebraic Expression: 10 Steps (with Pictures)
www.wikihow.com/Solve-an-Algebraic-Expression
How to Solve an Algebraic Expression. An algebraic expression is a mathematical phrase that contains numbers and/or variables. Though it cannot be solved ...
Sep 20, 2007 ... Mr. Karaba, a Math and Science teacher, strives to help the student and parent population learn how to solve simple algebraic equations.
## Algebraic expressions | Algebra basics | Khan Academy
The core idea in algebra is using letters to represent relationships between numbers without specifying what those numbers are!
Apr 18, 2011 ... Beginning Algebra & Solving Algebraic Equations. Bill Witte ... Basic Algebra Lessons for Beginners -- Get the FREE course Today ...
## SOLVING EQUATIONS - SOS Math
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This sections illustrates the process of solving equations of various forms. It also shows you how to check your answer three different ways: algebraically, ...
## Solve equations and simplify expressions (Algebra 2, Equations and ...
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In algebra 1 we are taught that the two rules for solving equations are the addition rule and the multiplication/division rule. The addition rule for equations tells us ...
## Model Algebra Equations | MathPlayground.com
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Write, read, and evaluate expressions in which letters stand for numbers. Understand that a variable can represent an unknown number. Understand solving an ...
## Writing Algebraic Equations - Math Goodies
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Writing Algebraic Equations is presented by Math Goodies. Learn to translate open ... Solution: Let x represent the amount of money Jeanne needs. Then the ... | 483 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-04 | latest | en | 0.830106 |
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JMP Sandwich Estimator
To generate a sandwich estimate of a SE in JMP do the following:
1. Run the multiple regression like usual
2. Save the leverage plots
• this generates 2 columns for each of your leverage plots
• One column is the Y of the plot the other is the X of the plot
• Check this by doing a simple regression.
3. Center the X and Y for the variable you want to estimate
• You will have to create new columns
• Just subtract off the mean
• Now you can compute the beta-hat by averaging the product of these two columns divided by the average of the square of the X column. Check this by creating a product column and the square column averaging and dividing.
4. Compute the variance by sum(LX2 LY2)/sum(LX2)
• Use the product column you create above, call it P.
• sum(p2) is simple the mean squared plus the variance of P.
• A simillar argument works for LX.
• Alternatevly you can use a formula: sum(p2)sum(LX2) | 234 | 950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.796293 |
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9 years ago | 458 | 1,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-25 | latest | en | 0.804859 |
https://www.physicsforums.com/threads/help-me-with-the-set-up-the-distance-speed-formulation.398795/ | 1,519,455,306,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815435.68/warc/CC-MAIN-20180224053236-20180224073236-00373.warc.gz | 892,043,307 | 15,351 | # Help me with the set up! (The distance/speed formulation)
1. Apr 26, 2010
### koosy
1. The problem statement, all variables and given/known data
crossing a river from point A to a point B directly across point A. The river R is 1 mile wide, the swimmer swims at s=5mph and the river is flowing westbound at f=1mph. Find the optimal direction theta to swim in order to reach point B the fastest. This is an optimization problem, but I just need some pointers, as I think m basic formulation is all wrong. Thanks
2. Relevant equations
3. The attempt at a solution
minimize T(theta)=R/(sin(theta)*5)*f-cos(theta)*s*(R/(sin(theta)*s))
2. Apr 26, 2010
### pat666
this is just relative velocity, convert all your info to si units(unnecessary but a good habit) the shortest distance is obviously straight across. Draw your vector diagrams and find the resultant!
3. Apr 26, 2010
### pat666
I get 4.9mp/h at 78.5degrees.
4. Apr 26, 2010
### pat666
just curios what are s and f
5. Apr 27, 2010
### koosy
s is the swimmer's speed and f is the drift of the river, flowing at a westward direction.
6. Apr 27, 2010
### koosy
Actually' I've just realized it really isn't relative velocity. This link gave me good pointers. Thanks for your help, though!
http://www.janrik.net/mathexpl/swimwalk.html
7. Apr 27, 2010
### pat666
the how to and answers i gave you should be correct it may not technically fit under the relative velocity heading but the same techniques are used.
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The ultraspherical polynomials are solutions to the Ultraspherical Differential Equation for Integer and . They are generalizations of Legendre Polynomials to -D space and are proportional to (or, depending on the normalization, equal to) the Gegenbauer Polynomials , denoted in Mathematica (Wolfram Research, Champaign, IL) GegenbauerC[n,lambda,x]. The ultraspherical polynomials are also Jacobi Polynomials with . They are given by the Generating Function
(1)
and can be given explicitly by
(2)
where is a Jacobi Polynomial (Szegö 1975, p. 80). The first few ultraspherical polynomials are
(3) (4) (5) (6)
In terms of the Hypergeometric Functions,
(7) (8) (9)
They are normalized by
(10)
Derivative identities include
(11) (12) (13) (14) (15) (16) (17) (18)
(Szegö 1975, pp. 80-83).
(19)
for , 3, ....
Special double- Formulas also exist
(20) (21) (22) (23)
Special values are given in the following table.
Special Polynomial Legendre Polynomial 1 Chebyshev Polynomial of the Second Kind
Koschmieder (1920) gives representations in terms of Elliptic Functions for and .
See also Birthday Problem, Chebyshev Polynomial of the Second Kind, Elliptic Function, Hypergeometric Function, Jacobi Polynomial
References
Abramowitz, M. and Stegun, C. A. (Eds.). Orthogonal Polynomials.'' Ch. 22 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 771-802, 1972.
Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, p. 643, 1985.
Iyanaga, S. and Kawada, Y. (Eds.). Gegenbauer Polynomials (Gegenbauer Functions).'' Appendix A, Table 20.I in Encyclopedic Dictionary of Mathematics. Cambridge, MA: MIT Press, pp. 1477-1478, 1980.
Koschmieder, L. Über besondere Jacobische Polynome.'' Math. Zeitschrift 8, 123-137, 1920.
Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 547-549 and 600-604, 1953.
Szegö, G. Orthogonal Polynomials, 4th ed. Providence, RI: Amer. Math. Soc., 1975. | 616 | 2,072 | {"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} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.721253 |
https://www.coursehero.com/file/5994139/PS12/ | 1,493,195,052,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00146-ip-10-145-167-34.ec2.internal.warc.gz | 885,029,140 | 102,772 | PS12 - Problem Set # 12 Solutions Chapter 12 #4 a) A...
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Problem Set # 12 Solutions Chapter 12 #4 a) A depreciation of the currency makes American goods more competitive. This is because a depreciation means that the same price in dollars translates into fewer units of foreign currency. That is, in terms of the foreign currency, American goods become cheaper so that foreigners buy more of them. For example, suppose that the exchange rate between yen and dollars falls from 200 yen/dollar to 100 yen/dollar. If an American can of tennis balls costs \$2.50, its price in yen falls from 500 yen to 250 yen. This fall in price increases the quantity of American-made tennis balls demanded in Japan. That is, American tennis balls are more competitive. b) Consider first the case of floating exchange rates. We know that the position of the LM curve determines output (is ‘superpotent’ for floaters with capital mobility). Hence, we know that we want to keep the money supply fixed. As shown in the figure below, we want to use fiscal policy to shift the IS curve to the left to cause the exchange rate to fall (depreciate). We can do this by reducing government spending or increasing taxes. Using the model from class: Fiscal contraction shifts the IS curve to the left. At β , i’<i* inducing capital outflows. The decreased demand for the domestic currency depreciates the real exchange rate ( ε falls). The real depreciation makes domestic goods more competitive, increasing exports and increasing net exports. This causes the IS curve to shift back until the current account and capital account are back in balance, at α . Here, we are back to are original level of Y e , but the exchange rate has been depreciated as desired. Now suppose that the exchange rate is fixed. If we want to increase competitiveness, we need to
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Ask a homework question - tutors are online | 534 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-17 | longest | en | 0.9238 |
https://www.wolframalpha.com/examples/mathematics/calculus-and-analysis/sequences/ | 1,524,448,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00179.warc.gz | 937,415,069 | 7,256 | Browse examples
## Examples for
Sequences
Sequences are lists of numbers, oftentimes adhering to a pattern or rule. Wolfram|Alpha has faculties for working with and learning about commonly occurring sequences like the Fibonacci sequence, the Lucas sequence, arithmetic sequences and geometric sequences, in addition to others.
Sequences
Investigate the properties of sequences, perform convergence tests or evaluate their limits.
Analyze a sequence:
Compute the limit of a sequence:
More examples
Sequence Recognition
Find formulas for incompletely specified sequences. Sum or multiply incompletely specified infinite sequences or series.
Compute a possible formula and continuation for a sequence:
Sum an incompletely specified sequence of terms:
Sum an incompletely specified infinite series:
Multiply an incompletely specified sequence of terms:
More examples | 157 | 876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.863544 |
xuzesalytu.dellrichards.com | 1,601,154,397,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00060.warc.gz | 1,108,379,250 | 4,136 | # How to write a conclusion for a maths project
Mental arithmetic also simplifies other computations and estimations. A number of states are moving toward state-funded preschool education to provide early education and care for these children.
## How to write a conclusion for a maths project
Developing Proportional Reasoning The concept of ratio is much more difficult than many people realize. Substantial time should be devoted to mathematics instruction each school day, with enough time devoted to each unit and topic to enable stu dents to develop understanding of the concepts and procedures involved. We make the following recommendation concerning the rational numbers: The curriculum should provide opportunities for students to develop a thorough understanding of rational numbers, their various representa tions including common fractions, decimal fractions, and percents, and operations on rational numbers. The relations core consists of such skills as constructing the relations more than, less than, and equal to. This support requires the provision of time and resources. Conceptual supports objects or diagrams that show the magnitude of the quantities and connect them to the number names and written numerals have been found to help children acquire insight into the base number system. Step 4 Cite all references used and include additional information, charts, graphs and data in appendices. By emphasizing both the relationships among quantities and ways of representing these relationships, instruction can introduce students to the basic ideas of algebra as a generalization of arithmetic. Solving Problems as a Context for Learning An important part of our conception of mathematical proficiency involves the ability to formulate and solve problems coming from daily life or other domains, including mathematics itself. Studies in almost every domain of mathematics have demonstrated that problem solving provides an important context in which students can learn about number and other mathematical topics. Step 3 Write a conclusion. Recommendation 2: Mathematics experiences in early childhood set- tings should concentrate on 1 number which includes whole num- ber, operations, and relations and 2 geometry, spatial relations, and measurement, with more mathematics learning time devoted to num- ber than to the other topics. In addition, the general and specific math- ematical process goals see Chapter 2 must be integrated with the content in order to allow children to make connections between mathematical ideas and deepen their mathematical reasoning abilities. Educational television programming and software, for example, can teach children about mathematics.
Although it is true that young children are more competent in math- ematics than many early childhood teachers, parents, and the general public believe, there are limits to what they can do in mathematics. Mastery of that system does not come easily, however. Further, the development of proportional reasoning can be supported by having students explore proportional situations in a variety of problem contexts using concrete materials or through data collection activities.
## Conclusion of maths in daily life
If children are not encouraged to use the mental computational procedures they have when entering school, those procedures will erode. About 24 percent of early childhood workers are in center-based settings, 28 percent are in regulated home-based settings, and about 48 percent work in informal care arrangements outside both of these systems. Whether or not students are performing a written algorithm, they can use mental arithmetic to simplify certain operations with numbers. The research-based principles and mathematics teaching-learning paths de- scribed in this report can also reduce the disparity in educational outcomes between children from low-SES backgrounds and their higher SES peers. Mastery of that system does not come easily, however. Conclusion 8: In the context of each of these content areas, young chil- dren should engage in both general and specific mathematical thinking processes as described above and in Chapter 2. Ways to engage younger children in meaningful uses of negative integers should be devel oped and tested. For adults the simplicity of calculating with single-digit numbers often masks the complexity of learning those combinations and the many different methods children can use in carrying out such calculations. Remember, in a math report, your findings could be that the initial thesis was wrong. Mathematical proficiency as we have defined it cannot be developed unless regular time say, one hour each school day is allocated to and used for mathematics instruction in every grade of elementary and middle school. Each stage of the project should support or disprove your initial thesis statement. The curriculum has to be organized within and across grades so that time for learning is used effectively.
In this chapter, we present conclusions and recommendations to help move the nation toward the change needed in school mathematics. Preverbal number knowledge is shared by humans from diverse cul- tural backgrounds as well as by other species. The community of people concerned with mathematics education will need to pay continued attention to studies of the effectiveness of new programs and will need to examine the available data carefully.
Ways to engage younger children in meaningful uses of negative integers should be devel oped and tested.
Rated 9/10 based on 2 review | 977 | 5,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-40 | latest | en | 0.94002 |
http://www.perlmonks.org/index.pl?node_id=458882 | 1,516,350,072,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00523.warc.gz | 536,421,333 | 9,010 | Syntactic Confectionery Delight PerlMonks
### Typoglycemia Fun
by reasonablekeith (Deacon)
on May 20, 2005 at 09:09 UTC Need Help??
Description: There is an eamil which fatlos anorud from time to tmie saying that, beacuse of the way the haumn brian redas wrdos as a wohle, taht the oderr of inividudal letetrs isn't taht imatrpont. The only caveat benig taht the the frist and lsat leettrs must be in the correct potision. Awyany, I thugoht I'd konck up a perl sprcit to test tihs, and what do you know, it works. But only to a pinot. The exmpale eaiml is clarely tuend, and any hrad to raed wrods seem to have been pleyad arnuod wtih. Tihs is more noibctale for the lngoer wdors, whree a two chaatrcer laed-in and lead-out smees to hlep maesivsly. The oehtr thnig I noctied was that if you ceatre a new wrod afetr the shfulifng, then you only see that new word, wihch makes tngihs a bit cofnisung. Ejony. PS. I wdolun't try and run peldroc thruogh it, that gets _very_ coufinsng.
```#!/perl -w
use strict;
# shuffle function from List::Util
sub shuffle (@) {
my @a=\(@_);
my \$n;
my \$i=@_;
map {
\$n = rand(\$i--);
(\${\$a[\$n]}, \$a[\$n] = \$a[\$i])[0];
} @_;
}
while (<DATA>) {
my \$line = \$_;
my \$line_copy = \$line;
while (\$line =~ m/(\w+)/g) {
next if length(\$1) < 4;
my \$start_offset = pos(\$line) - length(\$1) +1;
my \$word_middle_length = length(\$1) - 2;
if (\$word_middle_length > 4) {
\$word_middle_length -= 2;
\$start_offset++;
}
my \$word_middle = substr(\$line, \$start_offset, \$word_middle_le
+ngth);
my \$shuffled_word_middle = join('', shuffle( split(//, \$word_m
+iddle) ) );
substr(\$line_copy, \$start_offset, \$word_middle_length) = \$shuf
+fled_word_middle;
}
print \$line_copy;
}
__DATA__
There is an email which floats around from time to time saying that, b
+ecause of the way
the human brain reads words as a whole, that the order of individual l
+etters isn't that
important. The only caveat being that the the first and last letters m
+ust be in the
correct position. Anyway, I thought I'd knock up a perl script to test
+ this, and what do
you know, it works. But only to a point. The example email is clearly
+tuned, and any
hard to read words seem to have been played around with. This is more
+noticable for the
longer words, where a two character lead-in and lead-out seems to help
+ massively.
The other thing I noticed was that if you create a new word after the
+shuffling, then
you only see that new word, which makes things a bit confusing. Enjoy.
PS. I wouldn't try and run perldoc through it, that gets _very_ confus
+ing.
```
Replies are listed 'Best First'.
Re: Typoglycemia Fun
by pelagic (Priest) on May 20, 2005 at 09:14 UTC
vrey fnnuy idened!
pelagic
Re: Typoglycemia Fun
by tcf03 (Deacon) on May 21, 2005 at 02:25 UTC
or for the web-venturous
```#!/usr/bin/perl -T
use strict;
use warnings;
#use diagnostics;
use CGI qw/:standard/;
use CGI::Carp qw(fatalsToBrowser);
# Clean up our UNIX environment
delete @ENV{qw(IFS CDPATH ENV BASH_ENV)};
# Flush the output buffer
\$|++;
my @text;
# shuffle function from List::Util
sub shuffle (@) {
my @a=\(@_);
my \$n;
my \$i=@_;
map {
\$n = rand(\$i--);
(\${\$a[\$n]}, \$a[\$n] = \$a[\$i])[0];
} @_;
}
print start_html(-title=>"Fun with Typoglycemia",
-BGCOLOR=>"#cccc99" ), "\n";
print "<tt><center>\n";
print h3("Fun", "With", "Typoglycemia"),"\n";
print "</center>\n";
print hr,"\n";
print "\n\n<table border=\"0\" cellspacing=\"0\" cellpadding=\"1\" wid
+th=\"70%\" align=\"center\">\n\n";
print "<tr align=\"center\" bgcolor=\"#999966\">\n
<th>Typoglycemic Text</th>\n
</tr>\n";
print "<td>\n";
if ( defined (param('TEXT')) )
{
# \$/=undef;
@text=param('TEXT');
for (@text) {
my \$line = \$_;
my \$line_copy = \$line;
while (\$line =~ m/(\w+)/g) {
next if length(\$1) < 4;
my \$start_offset = pos(\$line) - length(\$1) +1;
my \$word_middle_length = length(\$1) - 2;
if (\$word_middle_length > 4) {
\$word_middle_length -= 2;
\$start_offset++;
}
my \$word_middle = substr(\$line, \$start_offset, \$word_middle_le
+ngth);
my \$shuffled_word_middle = join('', shuffle( split(//, \$word_m
+iddle) ) );
substr(\$line_copy, \$start_offset, \$word_middle_length) = \$shuf
+fled_word_middle;
}
print "\$line_copy\n";
}
}
print "</td>\n";
print "</table>\n";
print br,br;
print "\n\n<table border=\"0\" cellspacing=\"0\" cellpadding=\"1\" wid
+th=\"70%\" align=\"center\">\n\n";
print "<tr align=\"center\" bgcolor=\"#999966\">\n
</tr>\n";
print "</table>\n";
print br,br;
print start_form;
print "<center>\n";
print textarea('TEXT',"", 15, 80), "\n",br;
print submit;
print end_form;
print "</center>\n",br;
print "\n\n<table border=\"0\" cellspacing=\"0\" cellpadding=\"1\" hei
+ght=\"20\" width=\"70%\" align=\"center\">\n\n";
print "<tr align=\"center\" bgcolor=\"#999966\">\n
<td> </td>\n
</tr>\n";
print "</table>\n",br;
print end_html;
I couldn't resist...
Ted
--
"That which we persist in doing becomes easier, not that the task itself has become easier, but that our ability to perform it has improved."
--Ralph Waldo Emerson
Re: Typoglycemia Fun
by jZed (Prior) on May 21, 2005 at 03:48 UTC
For a scholarly analysis of the "jumbled spelling" urban legend which has been floating around the net for a few years, see this article on jumbled spelling from a Cambridge linguist which also includes a few other scripts which do similar things to yours. The issue has come up here at PerlMonks before and there were a few other scripts posted. I tried some searches but wasn't able to find it.
update Ah, found it. (The previous PM posting on a jumbled spelling generator).
Re: Typoglycemia Fun
by shenme (Priest) on May 20, 2005 at 20:07 UTC
I'd seen references to the idea, and had had no problems reading the shuffles, until 'fatlos', which totally threw me. I'm thinking that words that have two internal vowels together suffer much more than other words. Even 'smees' is kinda hard. Who knows what _could_ have happened to 'caveat' if it had been shuffled - 'cvaaet'?
Yeah, I had trouble reading "fatlos" too. I'm thinking it's probably because the word is "float", so it should probably be shuffled as such, so it should start with an 'f' and end with a 'ts', not just an 's'. (No, I'm not saying the code should be changed or anything, that's just why I think 'fatlos' was so hard for me to read (actually, I just looked at the cleartext, 'cause I didn't understand it right away.))
-Bryan
I wonder which would be better? Keeping closing consanant groups together (e.g. ts, ly, ck, etc.) or keeping the syllable count the same? Or both?
Hmmm. Something to play with this weekend.
Jack
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http://gpuzzles.com/mind-teasers/22-24-usa-president/ | 1,492,941,550,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118519.29/warc/CC-MAIN-20170423031158-00307-ip-10-145-167-34.ec2.internal.warc.gz | 163,302,421 | 12,246 | • Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : 22nd and 24th USA President Riddle
Difficulty Popularity
The Parents of 22nd and 24th president of USA is same.
They are born on the same day but are not twins.
They are not brothers.
None of them are adopted.
Both of them are male.
Explain?
Discussion
Suggestions
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Logical Marble Riddle
Difficulty Popularity
There is a box which have 33 yellow marbles and 35 green marbles. You also have 34 green marbles outside the box.
Randomly remove two marbles from the box.
* If they are of different colors, put the yellow one back in the box.
* If they are the same color, take them out and put a green marble back in the box.
Repeat this until only one marble remains in the box.
What is the color of the sole marble left in the box ?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Trick Statement Puzzle
Difficulty Popularity
Can you find out if the following statement is true?
It does not matter how much older the sibling is, their younger one will eventually be half as old as the elder one.
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Water Measurement Riddle
Difficulty Popularity
A barman is having a 12 liters jug full of beer. He needs to divide or split that beer into two equal parts. All he has is two empty jugs of capacity 8 liters and 5 liters.
How will he do it using them?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Awesome Logic Problem
Difficulty Popularity
I went for an official tour in a big city where there are over 100 buildings.
Building-1 is named first building.
Building-2 is named second building.
Building-3 is named third building.
A visitors decides to walk through all the buildings, he finds all the buildings except building-62.
Visitor later founds that the local of the city have given it another name.
What is the name of the Building ?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Truth Or Lie Brain Teaser
Difficulty Popularity
Zoe and Joe are twin brothers. One of them is a falsifier however the other one always speaks truth. I asked one of the, “Does Zoe lie?” The person replied with a yes.
Whom did I talk with? Zoe or Joe?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Best Maths Brain Teaser
Difficulty Popularity
A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below.
Name 1 2 3 4 5
Gerry c b True True False
Billy c c True True True
Clark a c False True True
Peeta b a True True False
Jonathan a b True False True
None of the two students gave the same number of correct answers. Can you find out the correct answers to the question? Also, calculate the individual score of all the five guys.
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Matchsticks Relationship Puzzle - Know Your Alphabets
Difficulty Popularity
What is the relationship between these matchsticks figures below?
Hint: You must know your alphabets.
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Tricky Picture Death Riddle
Difficulty Popularity
AS you can see the picture, all you have to do is analyze it and tell who all from the pictured people will die if the person at E pushes the round object to the slide on the slope. Keep in mind all the physics and the terrain while you analyze the things.
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Famous Send More Money Equation Puzzle
Difficulty Popularity
Below equation is one the most popular equation.Can you solve the below equation by replacing alphabet with a digit such that the equation holds as well?
SEND
+ MORE
======
MONEY
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Murder Mystery Brain Teaser
Difficulty Popularity
A homicide detective is called upon a crime scene. He finds that a body is lying on the ground in front of a multistory building. By all the means, it looks like a simple suicide case. But there are doubts in his mind.
He goes to the first floor and moves into the room facing the direction of the body. He opens the window in that direction and looks down towards his team. The he goes to the second floor and again moves into the room facing that direction, opens the window and looks down at his team.
He continues with the same process till the top floor. After that, he returns back to where his team is standing. He tells them that it is a suicide.
How did he come to such a conclusion?
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In a football game , Chelsea defeated Ar... | 1,304 | 5,387 | {"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} | 3.40625 | 3 | CC-MAIN-2017-17 | latest | en | 0.943716 |
https://playtaptales.com/numbers/centillioctingentillion/ | 1,627,219,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00322.warc.gz | 475,926,051 | 5,402 | ## Centillioctingentillion
A Centillioctingentillion (1 Centillioctingentillion) is 10 to the power of 302403 (10^302403). This is an immensely astronomical number!
## How many zeros in a Centillioctingentillion?
There are 302,403 zeros in a Centillioctingentillion.
## What's before Centillioctingentillion?
A Centillinovenonagintaseptingentillion is smaller than a Centillioctingentillion.
## What's after Centillioctingentillion?
A Centilliunoctingentillion is larger than a Centillioctingentillion.
## Centillioctingentillionaire
A Centillioctingentillionaire is someone whos assets, net worth or wealth is 1 or more Centillioctingentillion. It is unlikely anyone will ever be a true Centillioctingentillionaire. If you want to be a Centillioctingentillionaire, play Tap Tales!
## Is Centillioctingentillion the largest number?
Centillioctingentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
## Centillioctingentillion written out
Centillioctingentillion is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00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## Big Numbers
This is just one of many really big numbers! | 201,920 | 404,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.682058 |
https://www.experts-exchange.com/questions/26846015/Adding-conditions-to-a-formula-in-excel.html | 1,495,760,097,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.80/warc/CC-MAIN-20170525233846-20170526013846-00222.warc.gz | 870,804,806 | 31,162 | Solved
# Adding conditions to a formula in excel
Posted on 2011-02-24
280 Views
I have a formula in a cell on a worksheet that is =SUM(D18:D46)/COUNT(D18:D46) and it give me a pecentage. Basically I am using this as a quality rating. What I would like to know how to do it make it to where cell reads incomplete until there is data in all the fields it looks at. I would like to have it, if the certain cell is not relevent for a particular area to be able to put N/A in the cell and have the formula ignor it but if the field is empty to treat it like a 0 so it will have an impact on the final percentage. Quality-RCI-Assessment-Tool-ES.xls
0
Question by:jlcannon
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LVL 81
Expert Comment
ID: 34973894
=IF(COUNT(D18:D46)=0,"Incomplete",SUM(D18:D46)/COUNT(D18:D46))
Kevin
0
LVL 81
Expert Comment
ID: 34973908
0
LVL 50
Expert Comment
ID: 34973938
I think that if you want to count blanks as zero then you can divide by the count of non N/A cells, try this version in D14 copied across
=IF(COUNTA(D18:D46),SUM(D18:D46)/COUNTIF(D18:D46,"<>N/A"),"")
regards, barry
0
LVL 50
Expert Comment
ID: 34973983
...my suggestion will give you 85% for column D because it's counting blanks as zeroes.....if you want "incomplete" rather than a blank then put that in place of "" in my suggestion, i.e.
=IF(COUNTA(D18:D46),SUM(D18:D46)/COUNTIF(D18:D46,"<>N/A"),"incomplete")
barry
0
Author Comment
ID: 34974174
@ barryhoudini when I use =IF(COUNTA(D18:D46),SUM(D18:D46)/COUNTIF(D18:D46,"<>N/A"),"incomplete") it returns "TRUE" in the box
@zorvek this still returns 100% for me. I am looking to count a blank cell as a 0 and not count an n/a.
0
Author Comment
ID: 34974203
sorry guys my last post is inaccurate. If there is any blank cells in the column I want it to retunr an incomplete so it forces then to either choose 1 or 0 or n/a and if its n/a i want it to ignore the cell and not have it factor into the %
0
LVL 81
Expert Comment
ID: 34974224
Try this:
=IF(COUNT(D18:D44)=0,"Incomplete",SUM(D18:D44)/(ROWS(D18:D44)-COUNTIF(D18:D44,"N/A")))
See attached.
Kevin
Quality-RCI-Assessment-Tool-ES.xls
0
LVL 81
Accepted Solution
zorvek (Kevin Jones) earned 500 total points
ID: 34974281
Then use this:
=IF(COUNTBLANK(D18:D47)>0,"Incomplete",SUM(D18:D47)/(ROWS(D18:D47)-COUNTIF(D18:D47,"N/A")))
See attached.
Kevin
Quality-RCI-Assessment-Tool-ES.xls
0
Author Comment
ID: 34974289
@zorvek, this is comming close but it is showing me 85% but since there are blanks it should say incomplete.
0
Author Comment
ID: 34974446
@zorvek the post directly after the one giving me 85% worked perfect. thank you.
0
Author Closing Comment
ID: 34974453
Thank you. this was the exact solution I hoped to find.
0
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Graphs within dashboards are meant to be dynamic, representing data from a period of time that will change each time the dashboard is updated with new data. Rather than update each graph to point to a different set within a static set of data, t… | 1,076 | 3,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-22 | longest | en | 0.865474 |
https://www.oreilly.com/library/view/beginning-pre-calculus-for/9781598632910/9781598632910_ch12lev1sec9.html | 1,576,148,144,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00255.warc.gz | 805,914,714 | 10,334 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
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### Plotting Cosine Values
As Figure 12.17 illustrates, when you plot cosine values using a Cartesian plane, plot angle values on the x axis, and move to the right to show increasing values of angles, the resulting y coordinate value is 1 when the angle of your circle is 0. When you reach (90°), the value of the y coordinate value drops to 0. From there, as you move past , the value of the cosine decreases, reaching −1 when you are at π. As you move past π, the value of the cosine begins to increase, reaching 0 again at . After that, the value begins ...
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No credit card required | 209 | 921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.808242 |
https://socratic.org/questions/an-object-previously-at-rest-slides-3-m-down-a-ramp-with-an-incline-of-pi-3-and- | 1,571,881,103,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00548.warc.gz | 698,634,355 | 6,109 | # An object, previously at rest, slides 3 m down a ramp, with an incline of (pi)/3 , and then slides horizontally on the floor for another 25 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?
Feb 23, 2016
0.098
#### Explanation:
By conservation of energy, all the gravitational potential energy converted by friction to heat.
Delta "GPE" = mg (sin(pi/3)*3"m")
"Work done by friction" = Sigma ("Friction" * "distance")
$= m g \cos \left(\frac{\pi}{3}\right) \mu \left(3 \text{m") + mg mu(25"m}\right)$
So, equating them gives
$m g \left(\sin \left(\frac{\pi}{3}\right) \cdot 3 \text{m") = mgcos(pi/3) mu(3"m") + mg mu(25"m}\right)$
sin(pi/3) * 3"m" = mu(cos(pi/3)*(3"m") + (25"m"))
mu = frac{sin(pi/3) * 3"m"}{cos(pi/3)*(3"m") + (25"m")}
= frac{sqrt3/2*3"m"}{1/2*(3"m") + (25"m")}
$\approx 0.098$ | 316 | 868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.704113 |
https://www.teacherspayteachers.com/Product/NEW-Fractions-on-a-Number-Line-HANDS-ON-ACTIVITY-Great-for-grades-4th-6th-3032793 | 1,490,767,793,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190183.80/warc/CC-MAIN-20170322212950-00580-ip-10-233-31-227.ec2.internal.warc.gz | 951,683,717 | 25,978 | Total:
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# NEW Fractions on a Number Line HANDS-ON ACTIVITY Great for grades 4th-6th!
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This hands-on activity reviews proper fractions, improper fractions, mixed numbers, comparing fractions, and equivalent fractions. Students will use their knowledge to determine where each fraction goes on the number line.
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\$3.00 | 568 | 2,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-13 | longest | en | 0.73544 |
https://www.hpacademy.com/forum/how-to-degree-a-cam-7/show/true-tdc | 1,563,556,414,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00155.warc.gz | 715,791,877 | 32,546 | Forum » How to Degree a Cam » True Tdc
# True Tdc
### How to Degree a Cam
Discussion and questions related to the course How to Degree a Cam
Page 1
I followed the process as displayed but I'm still curious as to after I find these values. Add them together then divide by 2. Once the pointer is moved to face the sum value of those 2 numbers..Is true tdc that value in the operating direction of the crank? Or do you move it back to the markings on the sprocket to begin?
I'm not sure I follow your question? The key point to understand is that you'll end up with two marks on your degree wheel from the two places that the piston contacts the stop. True TDC is exactly in between these two marks. By adding the two degree values together and dividing by two, you'll find this value. At this point you can move your pointer or you can loosen off the degree wheel and rotate it to correct any error. If I haven't explained this clearly enough, please let me know how else I can help.
This also confused me at first .....
i think what kl1ne308 means is that after you have divided your two numbers and you Bend your wire to your required point on the degree wheel, it comes across in the course as if that point is TDC when in actual fact TDC is at 0 on the degree wheel. It’s confusing because you don’t actually move the wheel to zero you leave it at the point where you bend the wire.
took me a minute or two to work it out
Yup, move the wheel and/or pointer until the stop is contacted an even number of degrees with the crankshaft BTDC and ATDC.
Same principle if using a DTI - you want the same amount of travel of the guage at the same amount of degrees of rotation. | 387 | 1,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.943553 |
http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2010/10/04/Investigation_for_elliptical_beam_(crystalaser)&oldid=459974 | 1,462,351,432,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122533.7/warc/CC-MAIN-20160428161522-00151-ip-10-239-7-51.ec2.internal.warc.gz | 214,653,944 | 6,220 | # User:Pranav Rathi/Notebook/OT/2010/10/04/Investigation for elliptical beam (crystalaser)
Today I investigated the possibility of beam (crystalaser) to be elliptical, and I did it because the outer periphery of the trap in the video looks horizontally elliptical (major axis to be horizontal in CCD frame).
• In the process first I explored; how the beam orientation changes from laser to CCD? At the laser output beam front/ phase front have some spatial profile in transverse plan (circular or elliptical). If the profile is circular than the spot size (beam diameter) is same horizontally and vertically. If it’s elliptical (major axis might have any orientation, for simplification I consider it to be horizontal or vertical) than the spot size along the major axis should be greater or in other words the measurement of the spot size should be different horizontally and vertically. This is what I have investigated; if they are significantly different at two different locations, than it will be proved that our beam is elliptical.
• The orientation of the beam front interchanges as it’s reflected by the mirrors. The result of the interchange can be seen in the picture. Horizontal axis at the laser becomes vertical at the camera and vice versa. This means if the beam looks vertically elliptical at the CCD than its horizontally elliptical at the laser. The reason we still see the beam horizontally elliptical in the video is the orientation of the camera. The camera is installed (on its side) such that the horizontal axis of CCD (corresponding horizontal axis in frame) is vertical and vice versa. So the beam is actually vertically elliptic at the CCD and horizontally elliptic at the laser.
## Result
• I measured the beam diameter vertically and horizontally at two different locations. First 64 inches away from the laser: Horizontal diameter= 21.39-20.35=1.04*2=2.08mm
• Vertical diameter=14.15-13.15=1*2=2mm; difference=2.08-2.00=.08mm (3.84% of 2.08)
At 128 inches : Horizontal ( after periscope mirrors horizontal becomes vertical) diameter=12.15-9.45=2.7*2=5.4mm
• Vertical (after periscope mirrors vertical becomes horizontal) diameter=10.16-7.62=2.54*2=5.08mm; difference=5.4-5.08=.32mm (5.9% of 5.4)
• So at different locations the beam diameter is average 5% more horizontally. At the objective the beam is 5.9% more elongated of perfectly circular. So do we see an elliptical trap due to this elongation??????
• Steve Koch 22:17, 4 October 2010 (EDT): This seems like it could be the case. I don't think the beam was promised to have the exact same waist x & y anyway, right? Do you think it's good to go?
• pranav rathi Yes that’s true, even it is seen that in solid state lasers beam is mostly elliptical. In our case it is not that bad, so I think we are good to go.
• Steve Koch 05:56, 5 October 2010 (EDT):Good! | 713 | 2,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | latest | en | 0.898025 |
https://www.britannica.com/video/185522/interpretation-thought-experiment-cat-Schrodinger | 1,708,792,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00302.warc.gz | 675,060,737 | 16,238 | Learn about the quantum mechanical interpretation of the Schrödinger's cat thought experiment
Learn about the quantum mechanical interpretation of the Schrödinger's cat thought experiment
© MinutePhysics (A Britannica Publishing Partner)
Transcript
This wouldn't be a YouTube channel without a cat video. So without further ado, we present Schrodinger's cat. I'm sure you've heard some version of this famous thought experiment.
You put a cat in a bunker with some unstable gunpowder that has a 50% chance of blowing up in the next minute and a 50% chance of doing nothing. The gunpowder is Einstein's version. Schrodinger preferred poisonous gas. But whatever.
So until we look in the bunker, we don't know whether the cat is dead or alive. And when we do look, it is either dead or alive. So if we repeat the experiment enough times with enough cats and bunkers and gunpowder, we'll see that half the time kitty survives and half the time kitty goes bye bye.
The quantum mechanical interpretation is that before we look, the cat is in a superposition. It's both dead and alive, and our act of looking forces nature's decision. So our curiosity kills the cat. But what about the cat's perspective?
Well, the cat either sees the gunpowder explode or not. So inside the bunker, we actually have these two possibilities. The powder explodes and the cat sees it explode, or the powder doesn't explode and the cat doesn't see it explode. There's no option the powder explodes and the cat doesn't see it explode. So the cat's reality becomes entangled with the outcome of the experiment, and it's our observation of the experiment that forces nature to collapse to one option or the other.
But we're like the cat, too. Either the cat dies and we see it dead, or the cat lives and we see it alive. So who's observing us to force nature to collapse to one reality? Or do both possibilities happen in parallel within a larger multiverse? This collapsing to one reality problem is one of the biggest unanswered questions in quantum physics. So for kitty's sake, can I haz answer, pleez? | 454 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.920889 |
https://fr.scribd.com/document/24255172/Applied-Physics-Question-Paper-51 | 1,571,324,711,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00110.warc.gz | 496,434,214 | 76,150 | Vous êtes sur la page 1sur 1
# 1.
## The kind of force between atoms in a solid
(a) neutral forces
(b) repulsive forces only
(c) attractive forces only
(d) both attractive and repulsive forces
(a) ductile
(b) weak
(c) soft
(d) strong
## 3. The relation for the potential energy U(r) of a diatom is
(where ‘r’ is the inter nuclear separation, n, m are positive numbers, and a, b are positive constants.)
## (a) U(r) = arm - brn
(b) U(r) = -arm + brn
(c) U(r) = (-a/rm ) + (b/rn )
(d) U(r) = (-a/rm ) - (b/rn )
## 4. The structure of unit cell for most of the crystals is
(a) Parallelopiped
(b) Elliptical
(c) Spherical
(d) Cylindrical
## 5. The lattice parameters of Rhombohedral system are
(a) a = b = c, α = β = γ 6= 900
(b) a = b = c, α = β 6= γ = 900
(c) a 6= b = c, α = β = γ 6= 900
(d) a = b 6= c, α = β = γ = 900
6. The relation between density ρ, lattice parameter a, molecular weight M, number of atoms per unit cell n and
(a) M/N = ρa3
(b) ρa3 = nM/N
(c) MN = na
(d) ρa3 = nN/M
(a) 0.68
(b) 0.52
(c) 0.74
(d) 0.34
## 8. The coordination number of polonium is
(a) 12
(b) 6
(c) 8
(d) 4
9. If ‘a’ is the side of a BCC cell, then the nearest neighbour distance is
(a) a
(b) a 2
√
(c) a 2
√
(d) a 3 2
10. Highly close packed structure is
(a) Polinium
(b) SC
(c) BCC
(d) FCC
## 11. The coordination number of diamond is
(a) five
(b) four
(c) six
(d) seven
12. The number of zinc atoms in the unit cell of zinc sulphide crystal is
(a) 2
(b) 4
(c) 8
(d) 1
(a) 1.75
(b) 1.58
(c) 1.85
(d) 5.17
## (a) Base centered
(b) face centred
(c) simple cubic
(d) body centered
15. The plane in a crystalline solid intercepts the crystal axes at (2a, b, c) then what are it’s Miller indices?
(a) (2 1 2)
(b) (1 2 2)
(c) (1 2 2)
(d) (2 1 1)
(a) [1 1 3]
(b) [1 1 1]
(c) [1 1 2]
(d) [1 1 0]
## 17. In a simple cubic lattice d100 : d110 : d111 is
√ √
(a) 1 : 1 2 : 1 3
(b) 3 : 2 : 1
(c) 6 : 3 : 2
(d) 1 : √12 : √2
3
## (a) The velocity of sound at 0o C in air
(b) The velocity of sound at 100o C in air
(c) The velocity of light
(d) The velocity of sound at 30o C in air
## (a) number of lines on grating
(b) Order of the diffraction pattern
(c) Grating element
(d) Line of the spectrum
## (a) in Laue method monochromatic X-ray beam is used
(b) in rotating crystal method monochromatic X-ray beam is used
(c) in powder method monochromatic X-ray beam is used
(d) in Laue method white X-radiation is used | 947 | 2,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.597199 |
https://realanswers-ph.com/math/the-total-of-two-numbers-is-2720-th-2382945 | 1,680,439,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00569.warc.gz | 531,068,044 | 33,585 | , 28.10.2019 15:29 aimeedelacruz24
The total of two numbers is 2720. the bigger number is 4 times as much as the smaller number. what is twice of the smaller number?
Another question on Math
Math, 28.10.2019 18:29
Which of the following is equivalent to the complex number i^51? choose 1 (choice a) 1 (choice b) i (choice c) -1 (choice d) -i
Math, 28.10.2019 19:29
If a+b /b = 11/8 what is the value of a/b show your solution | 148 | 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} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.877147 |
https://socratic.org/questions/how-do-you-evaluate-1-t-1-9-t-9-as-t-approaches-9#311110 | 1,725,750,792,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00604.warc.gz | 515,916,068 | 5,739 | # How do you evaluate (1/t-1/9)/(t-9) as t approaches 9?
$- \frac{1}{81}$
$\frac{\frac{1}{t} - \frac{1}{9}}{t - 9} = \frac{\frac{9 - t}{9 t}}{t - 9} = - \frac{1}{9 t}$ then
${\lim}_{t \to 9} \frac{\frac{1}{t} - \frac{1}{9}}{t - 9} = - \frac{1}{81}$ | 130 | 249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.46888 |
http://123doc.org/doc_search_title/3811025-data-interpretation-workbook-version-2.htm | 1,505,840,870,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685912.14/warc/CC-MAIN-20170919164651-20170919184651-00691.warc.gz | 1,914,735 | 13,330 | # Data interpretation workbook version 2
## Data interpretation workbook version 2
... Chennai 20 01 35145 65139 451 92 51 124 37346 20 02 1 726 4 5 824 8 523 14 5 024 8 489 32 2003 24 800 63309 56469 523 68 51406 20 04 28 316 70316 7 125 3 54169 523 15 20 05 36503 6 929 4 696 32 58360 554 92 2006 29 129 5 921 6 ... centres over the years Mumbai 12 10 15 11 13 14 16 Delhi 24 28 21 27 23 20 19 Kolkata 18 12 23 19 16 21 24 Hyderabad 17 21 25 24 23 19 20 Chennai 12 10 13 11 14 Ques Approximately, what was the difference ... www.BankExamsToday.com Data Interpretation Workbook – v2 Perimeter of E = 2 r = × 22 / × 10 = 440 / m Cost of fencing of E = 440 / × 22 = Rs.13 82. 85 Area of C = (15 )2 = 22 5m2 So, cost of flooring of C = 22 5 ×...
• 68
• 17
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## the SketchUp Version 5Student Workbook phần 2 pot
... 23 and draw a line to connect the ends of the remaining segments 27 Push/Pull the first arc down to the bottom of the box Then right-click the second arc and select Convert to Polygon 24 The ... Move first, then move the point.) 39 the SketchUp Workbook Version 10 Then hover over the lower edge (do not click) 13 The midpoint of the top edge can now be moved, but (again) the move direction ... selected first Therefore, select the wing first, and then activate Rotate Place the center of the protractor on the top face, along the intersection with the other wing The next click sets the rotation...
• 30
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## Ebook Hacking Credit Card Version 2
... *************************************** %20 or %20 1=(select %20 fieldname %20 from %20 configuration %20 where %20 left(fieldna me,10)='xadminpage' %20 and %20 left(fieldvalue,1)='a') ...
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## Sách Pre Intensive Writing Task 2-Version 2
... DS1 .2: SS2: DS2.1: DS2 .2: Paragraph (Body paragraph 2) TS: SS1: DS1.1: SS2: DS2.1: DS2 .2: ... DS1 .2: SS2: DS2.1: DS2 .2: Paragraph Advantages TS: disadvantages/ reasons against (Body paragraph 2) Agree SS1: Argument for DS1.1: SS2: DS2.1: ... Supporting 2: Supporting 2: Developing 2. 1: ……………………… Developing 2. 1: ……………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… Developing 2. 2: ……………………… Developing 2. 2:...
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## Dragon In Flight (Version 2.5)
... layers in front, fold the sides to the center line Repeat behind 21.) Fold into a preliminary base Watch the X 23.) Fold the point down to meet the edges formed in the last step 24.) Sink the ... Inside reverse fold the wing outward Opening the wingtips as you The leading edges will line up with line AB 42.) Fold the inner layer up on the front and back of each wing A B 43.) Fold down the ... reflect different wing positions Then curve the wingtip along the existing lines Repeat on the other wing Shaping them to taste 66b.) The wings are done 67.) Wrap the sides of the tail fin over the...
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## Dragonfly, version 2
... downwards Don’t crease 20 Reverse-fold, flipping over the rectangular single-ply Press flat and sharpen all creases not to corner 21 Repeat 16 -20 on left Turn over 22 -23 22 Flip flap over, and ... won’t lie flat 23 Bisect lower angle, swiveling at the upper edge Release the stretched layer, and flip the long flap back to the right 24 Repeat 22 -23 on left Turn over not to tip 25 Fold top point ... Fold top point to center Turn back over 26 Valley fold, creasing sharply 27 Bring one layer to front (closed sink) \$ Dragonfly, v2 (continued) Copyright 1998 -20 00 Stephen Hecht All Rights Reserved...
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## Nâng cao hiệu quả công tác thống kê, kiểm kê đất đai bằng việc xây dựng giao diện hỗ trợ cho phần mềm TK05 version 2 1
... ng kê, ki m ñ t ñai chưa cao ð nâng cao hi u q a công tác, th c hi n ñ tài: "Nâng cao hi u qu công tác th ng kê, ki m ñ t ñai b ng vi c xây d ng giao di n h tr cho ph n m m TK05 Version 2. 1 ... ng kê, ki m ñ t ñai 2. 1. 4 K t qu th ng kê, ki m ñ t ñai 2. 1. 5 Lưu tr , qu n lý cung c p d li u th ng kê, ki m ñ t ñai 2. 1. 6 Báo cáo k t qu th ng kê, ki m ñ t ñai 2. 1. 7 ... ng kê, ki m phiên b n Version 2. 0 TK05 V2 .1 Ph n m m qu n lý s d li u th ng kê, ki m ñ t ñai phiên b n Version 2. 1 01- TKðð Bi u ki m di n tích ñ t nông nghi p 02- TKðð Bi u th ng kê, ...
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## Giáo trình Internet - Tin học cơ sở Version 2
... giảng Tin học sở - Khoa Công nghệ thông tin - Trường Đại học Bách Khoa Hà Nội 22 Tìm kiếm thông tin với Google (tiếp) Trang hiển thị kết tìm kiếm Bài giảng Tin học sở - Khoa Công nghệ thông tin - ... virus Bài giảng Tin học sở - Khoa Công nghệ thông tin - Trường Đại học Bách Khoa Hà Nội 40 Phần mềm diệt virus BKAV (tiếp) Bài giảng Tin học sở - Khoa Công nghệ thông tin - Trường Đại học Bách Khoa ... giảng Tin học sở - Khoa Công nghệ thông tin - Trường Đại học Bách Khoa Hà Nội 31 Giao diện Yahoo! Mail (tiếp) Thư có tệp tin đính kèm Thư đọc Thư đọc Bài giảng Tin học sở - Khoa Công nghệ thông tin...
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## Giáo trình Flash video - Tin học cơ sở Version 2
... 11/1996: Macromedia Flash 6/1997: M Flash 5/1998: M Flash 6/1999: M Flash 8 /20 00: M Flash 3 /20 02: M Flash MX (6) 9 /20 03: M Flash MX 20 04 (7) 9 /20 05: M Flash (Basic & Pro) 4 /20 07: Adobe Flash CS3 Professional ... giảng Tin học sở - Khoa Công nghệ thông tin - Trường 12 học Bách Đại Không gian làm việc Flash (4) Vùng thiết kế Nơi đặt xếp đối tượng cần thiết kế cho chương trình Bài giảng Tin học sở - Khoa ... Projector) Bài giảng Tin học sở - Khoa Công nghệ thông tin - Trường 30 học Bách Đại Cấu hình cho loại định dạng Bài giảng Tin học sở - Khoa Công nghệ thông tin - Trường 31 học Bách Đại Các định...
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## Ebook Hacking Credit Card Version 2 - Hàm
... Hàm MAX(column) Hàm MAX trả giá trị lớn cột Các giá trị NULL không xét đến Ví dụ: SELECT MAX(Age) FROM Persons kết trả về: 45 Hàm MIN(column) Hàm MAX trả giá trị nhỏ cột ... MIN(Age) FROM Persons kết trả về: 19 Lưu ý: Hàm MIN MAX áp dụng cho cột có liệu chuỗi văn Dữ liệu cột so sánh theo thứ tự tăng dần từ điển Hàm SUM(column) Hàm SUM trả tổng giá trị cột Các giá trị ... trả về: 98 Ví dụ: Tìm tổng số tuổi tất người có tuổi lớn 20 : SELECT SUM(Age) FROM Persons WHERE Age > 20 kết trả về: 79 GROUP BY HAVING Các hàm tập hợp (ví dụ SUM) thông thường cần thêm chức mệnh...
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## Generic Nissan Ecu Sensor Or Data Register Table Ver 2
... TEMP/S 0x27 °CELCIUS WORD VAL/16 MODE DOOR PBR 0x28 VOLT WORD OBJ TEMP HEAD 0x29 °CELCIUS WORD VAL/16 OBJ TEMP FOOT 0x2A °CELCIUS WORD VAL/16 AIRMIX DOOR 0x2B °CELCIUS WORD VAL/16 AIRMIX DOOR 0x2C ... IN CAR SEN FT 0x 22 °CELCIUS WORD VAL/16 DEF DUCT SEN 0x23 °CELCIUS WORD VAL/16 VENT DUCT SEN 0x24 °CELCIUS WORD VAL/16 FLOOR DUCT SEN 0x25 °CELCIUS WORD VAL/16 SUNLOAD 0x26 KCAL WORD VAL/160 COOLANT ... 0x2C CELCIUS WORD VAL/16 MODE DOOR ANGLE 0x2D DEGREES WORD INTAKE DOOR ANGLE 0x2E DEGREES WORD BLOWER MOTOR 0x2F VOLT WORD VAL/16 SET TEMP ADJ 0x40 °CELCIUS WORD VAL/16 COMPRESSOR 0x41 ON/OFF...
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## Layer 3 MPLS VPN Enterprise Consumer Guide Version 2
... 65001 PE1 CE3 PE4 CE2 ASN: 100 eBGP4 update: N3 AS_PATH: 65001 Site -3 N3 CE4 141441 Site -2 eBGP4 update: N3 AS_PATH: 100 100 PE2 ASN: 65001 PE3 Layer MPLS VPN Enterprise Consumer Guide Version OL-8851-01 ... across the MPLS VPN backbone Layer MPLS VPN Enterprise Consumer Guide Version OL-8851-01 37 Connecting to an MPLS/ VPN Service Provider Figure 24 MPLS VPN Backbone Considered Area VPN- IPv4 Update ... Network X Layer MPLS VPN Enterprise Consumer Guide Version OL-8851-01 43 Connecting to an MPLS/ VPN Service Provider In Figure 32 , the enterprise customer subscribed to the L3 MPLS VPN service...
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## Welcome to Version 2.0 of theForger''''s Win32 API Tutorial
... file:///C|/dona/forgers -win32- tutorial/ tutorial/index.html (2 of 3) [7/8/2003 4:34:43 PM] theForger's Win32 API Tutorial I would like to thank the following for the contributions they've made: Yih Horng, Todd Troxell, ... message they mean to add it into the WndProc() of your window class as follows: file:///C|/dona/forgers -win32- tutorial/ tutorial/window_click.html (1 of 5) [7/8/2003 4:34:45 PM] Tutorial: Handling ... file:///C|/dona/forgers -win32- tutorial/ tutorial/menus.html (1 of 6) [7/8/2003 4:34:46 PM] Tutorial: Menus and Icons You will want to add the rc file to your project or makefile depending on what tools you...
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## utf-8''''''''PP DLNN Version 2
... Khoang ming ND2.I .2 NG M HC B mỏy cu õm ND2.I .2 NG M HC ND2.I .2 NG M HC Cỏch to õm S phỏt õm cỏch phỏt õm bng hi phi cỏch phỏt õm bng hi hng cỏch phỏt õm bng hi mc ND2.I.3 ND2.I NG M HC Thc ... ND1.VI .2 TNG QUAN V NGễN NG HC Quan h ca cỏc n v ngụn ng 2. 1 Quan h tuyn tớnh (quan h ng on) (syntagmatical relation) A+i + +i + ++ng + +y + x+a + x+a ND1.VI .2 TNG QUAN V NGễN NG HC 2. 2 Quan ... tng lai ND1.IV .2 TNG QUAN V NGễN NG HC Cỏch thc phỏt trin ca ngụn ng: Ngụn ng phỏt trin t t khụng t bin nhy vt Ngụn ng phỏt trin khụng ng u gia cỏc mt ng õm, t vng,ng phỏp ND1.IV .2 TNG QUAN V...
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Xem thêm | 3,074 | 8,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-39 | latest | en | 0.507788 |
https://sociology_en.en-academic.com/1511/multi-dimensional_scaling | 1,719,302,941,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00231.warc.gz | 458,439,252 | 10,887 | multi-dimensional scaling
multi-dimensional scaling
Often, a set of data (for example attitude items) cannot be represented in one dimension, such as in a unidimensional scale or factor analysis. The items may then be modified or selected, so that they can be so represented (as in item analysis and scale construction); or, alternatively, a representation can be sought in a space of two or more dimensions. The purpose of multi-dimensional scaling (MDS) is to seek as good a representation of the data as possible in as few dimensions as possible.
In the simplest case (non-metric Euclidean distance MDS), data are thought of as giving information on the similarity or dissimilarity between pairs of objects: for example, positive correlations can be interpreted as similarities, such that the higher the value the more similar are the variables. The purpose of MDS is to represent each variable as a point in a low-dimensional space so that the distance between the points accurately reflects the relative size of the data similarities and dissimilarities. Non-metric (or ordinal) MDS preserves only the rank order of the data; metric (linear or power) MDS preserves the quantitative information. Computer programs implementing MDS usually seek such a solution iteratively; that is, from a preliminary guess successively brought into closer conformity to the data, in a cycle of improvements.
MDS is a very useful and general family of procedures (a General Distance Model) appropriate to a wide variety of data, including (for example) two-way correlation matrices, rectangular individual by variable matrices, and three-way stacks of data; to different models (distance, scalar-products or factor, weighted distance); and different levels of measurement (such as monotonic/non-metric, linear/metric, and power transformations). It has been used successfully on a wide range of sociological and psychological data (see, Multidimensional Scaling, 1978).
Dictionary of sociology. 2013.
Look at other dictionaries:
• multi-dimensional scaling — MDS A technique used to assist understanding of a consumer s perception of competing brands or products. A perceptual map is often drawn showing how consumers perceive competitive products along certain dimensions or attributes. For example,… … Big dictionary of business and management
• multi-dimensional scaling — /ˌmʌltɪ daɪˌmenʃ(ə)nəl skeɪlɪŋ/ noun a method of carrying out market research, in which the respondents are given a scale (usually 1 to 5) on which they base their replies … Marketing dictionary in english
• scaling, multi-dimensional — See multi dimensional scaling … Dictionary of sociology
• scaling — scales, scaling A form of measurement technique based on the observation of supposed common cultural meanings or shared social interpretations. One common sociological use of this technique is the attempt to devise measures of social prestige or… … Dictionary of sociology
• multidimensional scaling — multi dimensional scaling … Dictionary of sociology
• Dimensional analysis — In physics and all science, dimensional analysis is a tool to find or check relations among physical quantities by using their dimensions. The dimension of a physical quantity is the combination of the basic physical dimensions (usually mass,… … Wikipedia
• Multi-core processor — Diagram of a generic dual core processor, with CPU local level 1 caches, and a shared, on die level 2 cache … Wikipedia
• Three-dimensional integrated circuit — In electronics, a three dimensional integrated circuit (3D IC, 3D IC, or 3 D IC) is a chip in which two or more layers of active electronic components are integrated both vertically and horizontally into a single circuit. The semiconductor… … Wikipedia
• Cluster analysis (in marketing) — Cluster analysis is a class of statistical techniques that can be applied to data that exhibit “natural” groupings. Cluster analysis sorts through the raw data and groups them into clusters. A cluster is a group of relatively homogeneous cases or … Wikipedia
• Locating engine — Articleissues refimprove=May 2008 essay=May 2008A locating engine performs the computing of coordinates of locations of objects and of persons based on methods of multilateration or triangulation. This applies mainly with real time locating… … Wikipedia | 870 | 4,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.912788 |
https://thatsmaths.com/tag/euler/ | 1,500,605,051,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423681.33/warc/CC-MAIN-20170721022216-20170721042216-00120.warc.gz | 686,939,752 | 34,482 | ## Posts Tagged 'Euler'
### Brun’s Constant and the Pentium Bug
Euclid showed by a deliciously simple argument that the number of primes is infinite. In a completely different manner, Euler confirmed the same result. Euler’s conclusion followed from his demonstration that the sum of the reciprocals of the primes diverges:
$\displaystyle \sum_{p\in\mathbb{P}} \frac{1}{p} = \infty$
Obviously, this could not happen if there were only finitely many primes.
### The Bridges of Paris
Leonhard Euler considered a problem known as The Seven Bridges of Königsberg. It involves a walk around the city now known as Kaliningrad, in the Russian exclave between Poland and Lithuania. Since Kaliningrad is out of the way for most of us, let’s have a look closer to home, at the bridges of Paris. [TM073: search for “thatsmaths” at irishtimes.com ]
### Clothoids Drive Us Round the Bend
The article in this week’s That’s Maths column in the Irish Times ( TM043 ) is about the mathematical curves called clothoids, used in the design of motorways.
* * *
### Experiment and Proof
Many mathematicians spend their time proving results. The (very old) joke is that they are machines for turning coffee into theorems. A theorem is a statement that has been shown, by a sequence of irrefutable steps, to follow logically from a set of fundamental assumptions known as axioms.
These axioms themselves may be self-evident, or may simply be assumed to be true. Given this, the statement contained in a theorem is known with certainty to be true.
### Euler’s Gem
This week, That’s Maths in The Irish Times ( TM032 ) is about Euler’s Polyhedron Formula and its consequences.
Euler’s Polyhedron Formula
The highlight of the thirteenth and final book of Euclid’s Elements was the proof that there are just five “Platonic solids”. Recall that a regular polygon is a plane figure with all sides and angles equal, for example a square. By joining identical polygons together, we can form solid bodies called regular polyhedra. Continue reading ‘Euler’s Gem’ | 477 | 2,058 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-30 | longest | en | 0.9376 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/How-to-do-instrumental-variables-analysis-in-SAS/td-p/177838?nobounce | 1,516,601,833,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00624.warc.gz | 669,099,526 | 27,391 | ## How to do instrumental variables analysis in SAS?
Occasional Contributor
Posts: 16
# How to do instrumental variables analysis in SAS?
Hi guys,
How to do instrumental variables analysis in SAS? I Just find STATA software can do it with IVREGRESS procedure. Is there any specified STAT procedure can do that in SAS?
If there is no such a procedure, how to do it with several data steps or procedures? Any examples?
Thanks a lot!
Hongqiu
Posts: 2,655
## Re: How to do instrumental variables analysis in SAS?
Look at PROC MODEL and PROC SYSLIN in the SAS/ETS documentation, and PROC CALIS in the SAS/STAT documentation. There should be examples for each of these (maybe more than you want, as PROC MODEL and PROC CALIS are very versatile in what they can do).
Steve Denham
Occasional Contributor
Posts: 16
## Re: How to do instrumental variables analysis in SAS?
Thanks for your suggestion. I'll find more details from documents.
Hongqiu
SAS Employee
Posts: 89
## Re: How to do instrumental variables analysis in SAS?
As Steve mentioned, PROC SYSLIN or PROC MODEL are the easiest ways to estimate a regression using two-stage least squares. Here is an example:
There are also more examples at SAS/ETS 13.1 User's Guide Example Programs
data in;
label q = "Quantity"
p = "Price"
s = "Price of Substitutes"
y = "Income"
u = "Unit Cost";
drop i e1 e2;
p = 0; q = 0;
do i = 1 to 60;
y = 1 + .05*i + .15*rannor(123);
u = 2 + .05*rannor(123) + .05*rannor(123);
s = 4 - .001*(i-10)*(i-110) + .5*rannor(123);
e1 = .15 * rannor(123);
e2 = .15 * rannor(123);
demandx = 1 + .3 * y + .35 * s + e1;
supplyx = -1 - 1 * u + e2 - .4*e1;
q = 1.4/2.15 * demandx + .75/2.15 * supplyx;
p = ( - q + supplyx ) / -1.4;
output;
end;
run;
/*-- OLS Estimation --*/
proc syslin data=in;
demand: model q = p y s;
supply: model q = p u;
run;
/*-- Two-Stage Least Squares Estimation --*/
proc syslin data=in 2sls;
endogenous p;
instruments y u s;
demand: model q = p y s;
supply: model q = p u;
run;
Occasional Contributor
Posts: 16 | 621 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.720093 |
https://cstheory.stackexchange.com/questions/46432/is-there-a-notion-of-indistinguishability-obfuscation-for-almost-equivalent-circ | 1,656,406,953,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00176.warc.gz | 232,145,086 | 65,949 | # Is there a notion of indistinguishability obfuscation for almost equivalent circuits?
In the definition of indistinguishability obfuscation (iO), we have a probabilistic algorithm Obs that receives as input a circuit C, such that $$i)$$ the output Obs(C) is a circuit with the same functionality as C and $$ii)$$ for any $$C_1$$ and $$C_2$$ with the same functionality, we have that Obs($$C_1$$) is indistinguishable from Obs($$C_2$$).
Has a similar notion (and candidate construction) for a similar concept where the security holds for $$C_1$$ and $$C_2$$ are almost the same (i.e., they agree on all except a negligible fraction of the inputs, and it is computationally hard to find the points where they differ).
• Wouldn't "regular" iO imply this? Since $\text{Obs}$ needs to be efficient, in particular we can't use its output to distinguish between "almost-the-same" $C_1,C_2$, and so the outputs $\text{Obs}(C_1)$ and $\text{Obs}(C_2)$ must still be (computationally) indistinguishable as well? Mar 5, 2020 at 6:20
• Not by the (text-book) definition of iO. The security definition only asks that of C_1 and C_2 have the same functionality, their obfuscation are indistinguishable and if C_1 is not implementing the same function as C_2, anything could happen. Answering your comment, I just realized that maybe what I am asking is equivalent for Virtual Black Box obfuscation for evasive functions (what is known to be impossible in the most general case) Mar 6, 2020 at 17:25
• I understand this is not guaranteed by the definition itself, but my point is that it could be implied by it — since $\textrm{Obs}$ is assumed efficient, it could otherwise be used to "break" the computational hardness of distinguishing $C_1$ and $C_2$. Otherwise: take $C_1\neq C_2$ (different functionality) computationally hard to distinguish. Pass them through $\textrm{Obs}$. If you can distinguish $\textrm{Obs}(C_1)$ from $\textrm{Obs}(C_2)$, then you have distinguished $C_1$ from $C_2$ (which was assumed hard). So $\textrm{Obs}(C_1)$, $\textrm{Obs}(C_2)$ must also be hard to distinguish. Mar 6, 2020 at 18:16
• (Unless I am botching something here, possibly two different notions of distinguishability?) Mar 6, 2020 at 18:19
• I can post a longer answer later, but a keyword you should look up for is differing input obfuscation. Mar 7, 2020 at 1:07
The definition of diO tells us: if there's a non-uniform PPT adversary that can distinguish the obfuscations of two circuits with "almost same" functionalities Obf($$C_1$$) and Obf($$C_2$$), then there exists a non-uniform PPT extractor that can extract an input(witness) $$x$$ where $$C_1(x) \neq C_2(x)$$. (the formal definition is on page 11)
If there are only polynomially many inputs where their outputs are different evaluated on $$C_1, C_2$$, then the standard iO definition also implies diO, using a binary search argument on the input space. | 790 | 2,904 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.907903 |
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