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https://rehabilitationrobotic.com/how-many-protons-are-in-the-nucleus-of-an-atom/ | 1,653,169,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00455.warc.gz | 560,158,807 | 11,044 | Close
Table of Contents
## How many protons are in the nucleus of an atom?
Some nitrogen atoms have an atomic mass number of 15 (A=15). A is the number of neutrons plus protons in the nucleus. However, we already know that there are 7 protons….
Map of contents Close
Calling names The Big Bang
Counting in the nucleus Atoms at last
Neutrons and isotopes
Putting things in order
## What element has 4 protons 4 electrons and 6 neutrons?
Atomic Number
Name Protons Neutrons
Lithium 3 4
Beryllium 4 5
Boron 5 6
Carbon 6 6
## How many protons and electrons does a neutral atom of beryllium have?
(Round the atomic mass from the periodic table to the nearest whole number to get part of your answer.) From the periodic table, beryllium has an atomic number of 4 and a mass of almost 9 amu. Therefore, a neutral beryllium atom has 4 protons, 4 electrons, and 5 neutrons (# neutrons = mass # – # protons).
## How many protons does beryllium 11 contain?
Both atoms have the same mass number, namely 7. However, the nucleus of the lithium atom has three protons and four neutrons, while the nucleus of the beryllium atom has four protons and three neutrons.
## Why does Cu show positive electrode potential?
Copper has lower tendency than hydrogen to form ions, so if the standard hydrogen electrode is cconnected to the copper half-cell, the copper will be relatively less negative. The copper electrode has relatively lower number of electrons. so it has positive electrode potential.
## Why is E value of Cu positive?
Copper has a high energy of atomisation but a low hydration energy due to which the E0 value for Cu is positive.
## Why is e value for mn3 +/ mn2+?
Because Mn3+ has the outer electronic configuration of 3d4 and Mn2+ has the outer electronic configuration of 3d5. Thus, the conversion of Mn3+ to Mn2+ will be a favourable reaction since 3d5 is a very stable configuration as it is half filled configuration. Hence, Eo value for Mn3+ / Mn2+ couple is positive.
## What does negative electrode potential mean?
So, if an element or compound has a negative standard electrode reduction potential, it means it forms ions easily. The more negative the value, the easier it is for that element or compound to form ions (be oxidised, and be a reducing agent).
## Why iron has higher enthalpy of atomization than copper?
Copper has outer electronic configuration of 3d104s1 and 1 unpaired electron. Iron has more number of unpaired electrons than copper. Hence, Iron has higher enthalpy of atomisation than that of copper.
## Which element has highest enthalpy of atomisation in 3d series?
Sc & zn belongs to 3rd group pf periodic table. The extent ofmetallic bonding an elementundergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization.
## Why enthalpy of atomisation of zinc is lowest?
The enthalpy of atomization depends on the extensive metallic bonding which in turn depends on the number of unpaired electrons. Due to the absence of these unpaired electrons in Zn. The interatomic bonding is weak in Zn and hence the enthalpy of atomization of Zn is the lowest.
[Ar] 3d6 4s2
2021-06-17 | 787 | 3,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.888552 |
https://www.biostars.org/p/18238/ | 1,716,694,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00435.warc.gz | 581,800,369 | 7,949 | R: Manhattan Plot Of Fst Values Instead Of -Log(P)
3
3
Entering edit mode
12.2 years ago
Abdel ▴ 410
Hi all,
I have a tab delimited file with the following columns:
CHR SNP bp Fst
where Fst is the fixation index. Does anyone know how to make a Manhattan plot of these Fst values in R? Many thanks!
r fst visualization • 16k views
6
Entering edit mode
12.2 years ago
John ★ 1.5k
I found the following blog useful, you need tweak the codes to fit F-value not -log10(p) value - http://gettinggeneticsdone.blogspot.com/2011/04/annotated-manhattan-plots-and-qq-plots.html
Edits:
If you would like to use lattice or ggplot, we can create similar plot:
# just an example
Fst <- rnorm(10000, 10, 5)
chr <- c(rep(1, 2500), rep(2, 2500), rep(3, 2500), rep(4, 2500))
BP <- c(1:2500, 1:2500, 1:2500, 1:2500)
mydf <- data.frame (Fst, chr, BP)
require(ggplot2)
qplot(BP, Fst, facets = . ~ chr, col = factor(chr), data = mydf)
I like bw theme better
qplot(BP, Fst, facets = . ~ chr, col = factor(chr), data = mydf) + theme_bw()
You need to reshape plot area or change arrangement of graph to make all facets in single line as Manhattan plot. The graph is pretty close but not exactly as in above blog code.
0
Entering edit mode
You just have to replace bp by BP and Fst by P and it should work.
3
Entering edit mode
12.2 years ago
I have done the exact plot you want for FST.
here is where [R] becomes awesome:
dat<-read.csv("your.data.txt", header=FALSE, sep="\t")
plot(dat$FST, col=as.factor(dat$CHR))
If you need the true coordinates ie BP I will send you a script I wrote to take that into account. Otherwise the code above will suffice.
Pretty much R gives the X values a number 1:length(dat$FST) and then colors the points by the dat$CHR
0
Entering edit mode
Hi, I tried this code for data in the same format as above, am I missing something or should these two R commands be sufficient for plotting? Error in plot.window(...) : need finite 'xlim' values. Thanks
0
Entering edit mode
please post head(dat)
0
Entering edit mode
Hi all,
I need to do the same plot as described above but I am facing the same problem.
plot(dat$FST, col=as.factor(dat$CHR))
Error in plot.window(...) : need finite 'xlim' values
1: In min(x) : no non-missing arguments to min; returning Inf
2: In max(x) : no non-missing arguments to max; returning -Inf
3: In min(x) : no non-missing arguments to min; returning Inf
4: In max(x) : no non-missing arguments to max; returning -Inf
what I have to do? what do you mean by head(dat)?
0
Entering edit mode
Would you mind posting the script that you wrote to take BP into account? Cheers! :)
0
Entering edit mode
0
Entering edit mode
First of all thanks for your help. I just need to use this code only
plot(dat$FST, col=as.factor(dat$CHR))
but it give me the prevoius error. I dont need the true coordinates.
0
Entering edit mode
9.5 years ago
CB ▴ 10
I want to create a Manhattan plot to plot Fst values of each SNP. I used John's first code to plot Fst values for 3 different chromosomes and it worked well. But the size of chromosomes are fixed, and I want to set the maximum size for each chromosome. Is it possible? The code that I used was:
Fst <- rnorm(48167,0.13, 0.2)
chr <- c(rep(1, 14552), rep(2, 33582), rep(10, 33))
BP <- c(1:14552, 1:33582, 1:33)
data <- data.frame (Fst, chr, BP) qplot(BP, Fst, facets = . ~ chr, col = factor(chr), data = data)
I tried to use the package qqman to plot Fst values instead of P values and did not work.
Or I need to use the perl script that Zev.Kronenberg mentioned? Convert chrmosomes positions into genomic coordiantes?
0
Entering edit mode
With ggplot2 you can facet on the chromosome: something like +facet_grid(.~CHR). You will have to play with the margins on the facets, but it does work. If you use my script it make the process easier as you don't have to facet or fiddle with the coordinates. | 1,116 | 3,914 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.781779 |
https://www.paul-nguyen.com/teaching/calculus-ii/module-3-additional-integration-techniques-and-limits/3-4-improper-integrals/ | 1,722,985,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00457.warc.gz | 744,804,409 | 15,787 | # 3.4 Improper Integrals
In this lesson, you will learn about improper integrals.
### Lesson Objectives
• Recognize an improper integral: integration over an infinite interval or integration over an infinite discontinuity.
• Recognize that an improper integral is a limit of a definite (proper) integral.
• Understand the difference between convergence and divergence of improper integrals.
• Evaluate improper integrals by constructing and evaluating appropriate limits.
• Use the basic comparison test to establish the convergence or divergence of improper integrals.
### Lesson Content
View all of the following instructional videos. These will help you master the objectives for this module.
1. YouTube video: Improper Integral – Basic Idea and Example
2. YouTube video: Improper Integral – More Complicated Example
3. YouTube video:Improper Integral – Infinity in Upper and Lower Limits
4. YouTube video: Comparison Test for Improper Integrals
5. YouTube video: Improper Integrals
6. YouTube video: Improper Integral with Infinite Discontinuity at Endpoint
7. YouTube video: Improper Integral with Infinite Discontinuity at the Middle
The following required readings cover the content for this module. As you go through each reading, pay close attention to the content that will help you learn the objectives for this module.
1. Improper Integrals
2. Improper Integrals (type 1 improper integrals, type 2 improper integrals, direct comparison test, limit comparison test)
3. Improper Integrals
4. Improper Integrals (by Dr. Marcel B. Finan from Arkansas Tech University)
### Lesson Practice Exercises/Activities
Make your way through each of the practice exercises. This is where you will take what you have learned from the lesson content and lesson readings and apply it by solving practice problems.
1. Improper Integrals (27 problems with hint and answer)
2. Improper Integrals (answers and solutions provided) | 407 | 1,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.740103 |
https://discussions.unity.com/t/rotating-gameobject-around-player-with-mouse-y-position/102237 | 1,721,713,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00370.warc.gz | 188,965,113 | 6,345 | # Rotating GameObject Around Player With "Mouse Y" Position
Good Morning/Afternoon/Evening … Night
So I’ve currently in the process of scripting out a simplistic game mechanic, whereupon the main player can pick up/throw and rotate the desired gameobject at will. And that’s all going relatively well.
Only I also want the desired Gameobject to be able to rotate around the character depending on where the Mouse cursor is currently centered. So essentially, the Gameobject should seem as if it’s following the Y-Axis (Or camera angle) of the Mouse Cursor, by rotating in an circular arc around the main player.
And I do merely apologise if this is a relatively naive question to ask, I’m still in initial learning stages unfortunately. So do know any feedback/criticism or casual insults would be much appreciated
(Also, If you’re wanting the pickup/throw/rotate code for your own scripts, do merely ask, and I’ll be sure to send it across)
This worked for me for rotating objects on the y axis only. Not sure how to help you on the other parts. Sorry I am new too.
``````#pragma strict
var damping : int=200;
var target : Transform;
function LateUpdate()
{
var lookPos = target.position - transform.position;
lookPos.y = 0;
var rotation = Quaternion.LookRotation(lookPos);
transform.rotation = Quaternion.Slerp(transform.rotation, rotation, Time.deltaTime * damping);
}
``````
Something like this
`````` var multiple : float = 1.0;
var turnSpeedMultiplier : float = 2.0;
var turnSpeed : float = 0.0;
//// in update add following code
var targetTurnSpeed : float = 0.0;
targetTurnSpeed= Input.GetAxis("Mouse X");
targetTurnSpeed *= Mathf.Pow(turnSpeedMultiplier,3);
targetTurnSpeed *= multiple ;
turnSpeed = Mathf.Lerp(turnSpeed, targetTurnSpeed, Time.deltaTime * 25.0);
transform.rotation.eulerAngles.y += turnSpeed * Time.deltaTime;
`````` | 429 | 1,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.798489 |
https://www.jiskha.com/members/profile/posts.cgi?name=joker | 1,503,020,859,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00135.warc.gz | 921,010,552 | 2,857 | # Posts by joker
Total # Posts: 9
Maths
Evaluate: 5 + 8 + 11 + 14 + … … +395 + 398.
calculus
use a sign chart to solve inequality. Express answers in interval notation. x^2+6<2x
math
The top of a billboard is 30 feet above ground. The billboard casts a 235 feet long shadow. What is the angle of elevation from the tip of the shadow to the sun?
AP Statistics
12) Which of the following statements is true? a) The sampling distribution of the difference between two proportions will always be normal. b) When comparing two population proportions, either sample proportion can be used as the unbiased estimate of the true population ...
Same here | 158 | 650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.841497 |
http://www.jiskha.com/display.cgi?id=1305819379 | 1,493,508,985,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123632.58/warc/CC-MAIN-20170423031203-00349-ip-10-145-167-34.ec2.internal.warc.gz | 578,147,168 | 3,705 | # Algebra 1
posted by on .
If you rent a car for one day and drive it for 100 miles, the cost is \$40.00. If you drive it 200 miles, the cost is \$46.00. What are the data points given?
• Algebra 1 - ,
45 | 64 | 208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.838706 |
https://scholars.duke.edu/display/pub692006 | 1,618,370,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076454.41/warc/CC-MAIN-20210414004149-20210414034149-00268.warc.gz | 590,253,351 | 5,116 | # Sample size determination for comparing several survival curves with unequal allocations.
Journal Article (Journal Article)
Ahnn and Anderson derived sample size formulae for unstratified and stratified designs assuming equal allocation of subjects to three or more treatment groups. We generalize the sample size formulae to allow for unequal allocation. In addition, we define the overall probability of death to be equal to one minus the censored proportion for the stratified design. This definition also leads to a slightly different definition of the non-centrality parameter than that of Ahnn and Anderson for the stratified case. Assuming proportional hazards, sample sizes are determined for a prespecified power, significance level, hazard ratios, allocation of subjects to several treatment groups, and known censored proportion. In the proportional hazards setting, three cases are considered: (1) exponential failures--exponential censoring, (2) exponential failures--uniform censoring, and (3) Weibull failures (assuming same shape parameter for all groups)--uniform censoring. In all three cases of the unstratified case, it is assumed that the censoring distribution is the same for all of the treatment groups. For the stratified log-rank test, it is assumed the same censoring distribution across the treatment groups and the strata. Further, formulae have been developed to provide approximate powers for the test, based upon the first two or first four-moments of the asymptotic distribution. We observe the following two major findings based on the simulations. First, the simulated power of the log-rank test does not depend on the censoring mechanism. Second, for a significance level of 0.05 and power of 0.80, the required sample size n is independent of the censoring pattern. Moreover, there is very close agreement between the exact (asymptotic) and simulated powers when a sequence of alternatives is close to the null hypothesis. Two-moment and four-moment power series approximations also yield powers in close agreement with the exact (asymptotic) power. With unequal allocations, our simulations show that the empirical powers are consistently above the target value of prespecified power of 0.80 when 50 per cent of the patients are allocated to the treatment group with the smallest hazard.
### Cited Authors
• Halabi, S; Singh, B
### Published Date
• June 15, 2004
• 23 / 11
• 1793 - 1815
• 15160409
• 15160409
• 0277-6715
### Digital Object Identifier (DOI)
• 10.1002/sim.1771
• eng
• England | 549 | 2,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.902179 |
https://pt.scribd.com/doc/5375906/nr310206-optimization-techniques-set1 | 1,566,267,363,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00176.warc.gz | 608,367,715 | 73,398 | Você está na página 1de 2
# Code No: NR310206 NR
## III B.Tech I Semester Supplementary Examinations, November 2006
OPTIMIZATION TECHNIQUES
(Electrical & Electronic Engineering)
Time: 3 hours Max Marks: 80
All Questions carry equal marks
?????
1. (a) Determine the maximum and minimum values of the function: [8]
12x5 -45x4 +40x3 +5
(b) A d.c. generator has internal resistance of R ohms and develops an open
circuit voltage of ‘V’ volts. Find the value of load resistance ‘r’ for which the
power developed by the generator will be maximum. [8]
2. (a) State and explain the necessary and sufficient conditions for existence of rel-
ative optima in case of multivariable optimization with constraints. [8]
(b) Find the dimensions of a rectangular parallelepiped with largest volume whose
sides are parallel to the coordinate planes, to be inscribed in the ellipsoid. [8]
## 3. (a) State and explain the standard form of LPP. [8]
(b) Explain the significance of slack, surplus and artificial variables of LPP. [8]
## 4. Show that the following LPP has unbounded solution [16]
maximize Z = 3x1 + 2x2
subject to x1 − x2 ≤ 1
3x1 − 2x2 ≤ 6
x 1 , x2 ≥ 0
5. (a) If
Pall theP sources are emptied and all the destinations are filled, show that
ai = bj is a necessary and sufficient condition for the existence of a
feasible solution to a transportation problem [8]
(b) Prove that there are only m+n-1 independent equations in a transportation
problem, m and n being the no. of origins and destinations and that any one
equation can be dropped as the redundant equation. [8]
6. Draw the flowchart of Powell’s method. Explain about each block. [16]
## 7. Consider the problem:
Minimize f(x1 , x2 ) = (x1 − 1)2 + (x2 − 2)2
Subject to 2x1 − x2 = 0
and x1 ≤ 10
Construct φK function according to the interior penalty function approach and
complete the minimization of φK . [16]
1 of 2
Code No: NR310206 NR
8. Determine the value of u1 , u2 , u3 so as to maximize (u1 .u2 .u3 ) ,
Subject to, u1 + u2 + u3 = 10 and u1 , u2 , u3 ≥ 0
[16]
?????
2 of 2 | 605 | 2,048 | {"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} | 3.046875 | 3 | CC-MAIN-2019-35 | latest | en | 0.820446 |
lsbailey.tripod.com | 1,560,819,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998600.48/warc/CC-MAIN-20190618003227-20190618025227-00037.warc.gz | 102,016,106 | 14,363 | Soils
Here we go again.....
Practice Questions
1. Given that 1 kcal of heat is required to increase the temperature of 1 kg of water by 1°C:
a. How many kcals would be required to heat 200 kg of water by 20°C for a bath?
b. How many joules is this?
c. How many Btus?
d. If your water heater can supply 50 kBtu/h, how long will it take to heat this water?
2. a. Given that 1 kWh = 3.6 MJ and that 1 Btu = 1055 J, show that 1 kWh = 3412 Btu.
b. Why would it be incorrect to use this conversion factor directly to determine the amount of coal required to generate electricity in a power plant?
3. A typical home in the northern U.S. might require 120 MBtu of heat for the average winter.
a. If this heat were supplied by a natural gas furnace operating at 50 percent efficiency, how many cubic feet of gas would need to be purchased?
b. At a cost of \$0.90/ccf, what would it cost to heat this house for one season?
c. If a new 80 percent efficient furnace could be installed at a cost of \$4,000, how long would it take to pay back the cost of this furnace assuming gas prices remained the same?
4. Suppose the house in question 3 is located in Cleveland where the annual average solar flux is 160 W/m22. If 15 m2 of solar panels operating at 20 percent efficiency were installed on this house to collect and store solar energy in the form of hot water:
a. How much energy could be gained in one year in this manner?
b. What fraction of the annual heating requirement is this?
c. Using the hot-water heating requirements for a bath from question 1(c), how many hot baths would this energy supply in one year?
5. The annual average solar flux in Tucson is 250 W/m2. Suppose 10 m2 of solar electric panels operating at 15 percent efficiency were installed on a home there.
a. How many kWh of electricity could be collected by these panels in one year?
b. What fraction of the annual electrical requirement of 10,000 kWh for the average home does this represent?
c. How many square meters of solar panels would be required to supply 10,000 kWh per year?
6. Solar energy is converted naturally into wood biomass with an efficiency of about 0.1 percent. Suppose a wood lot of 100 hectares (106 m2) is located in Missouri, where the average annual solar flux is 200 watts/m2. Given that the heat value for wood is 12 MBtu/ton, how many tons of wood can be produced by this property each year?
7. With moderate winds, a modern large wind turbine can generate about 500 kW of electricity, whereas a large nuclear power plant can generate 1,000 MW.
a. How many wind turbines would be required to give the same output as one nuclear power plant?
b. Discuss some of the advantages and disadvantages to providing electrical power by each method.
8. Batteries are usually rated in terms of ampere-hours, indicating the current that the cell is capable of delivering for a specified time. A typical D-cell flashlight battery, for instance, might be rated at 3 ampere-hours. The total electrical energy available from such a battery is found by multiplying the ampere-hour rating by the battery voltage. Thus this same 1.5 volt D cell could deliver 4.5 watt-hours of electrical energy.
Convert this energy to kWh and compare the cost of electrical energy derived in this manner to that of standard "grid-based" electricity. Assume that the battery costs \$1.00 and that electricity from the power company is available at \$0.15/kWh.
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USA | 994 | 4,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.959945 |
https://programmer.group/1-introduction-to-logistic-regression.html | 1,721,700,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00594.warc.gz | 413,055,897 | 6,026 | # 1: Introduction to logistic regression
Logistic Regression is a classification model in machine learning. Logistic Regression is a classification algorithm. Although the name contains regression, it has a certain relationship with regression. Because the algorithm is simple and efficient, it is widely used in practice.
## 1 application scenario of logistic regression
• Is it spam
• Are you sick
• financial fraud
• False account number
Looking at the above example, we can find the characteristic, that is, the judgment between the two categories. Logistic regression is a sharp tool to solve the problem of binary classification
## 2 principle of logistic regression
To master logistic regression, you must master two points:
What is the input value in logistic regression
How to judge the output of logistic regression
### 2.1 input
The input of logistic regression is the result of linear regression.
### 2.2 activation function
• sigmoid function
• Judgment criteria
• The result of the regression is input into the sigmoid function
• Output result: a probability value in the [0, 1] interval. The default value is 0.5, which is the threshold value
• The final classification of logistic regression is to judge whether it belongs to a category through the probability value of belonging to a category, and this category is marked as 1 (positive example) by default, and the other category will be marked as 0 (negative example). (to facilitate loss calculation)
Interpretation of output results (important): it is assumed that there are two categories A and B, and our probability value is the probability value belonging to category A(1). Now there is a sample input to the logistic regression output result of 0.6, then the probability value exceeds 0.5, which means that the result of our training or prediction is category A(1). On the contrary, if the result is 0.3, the training or prediction result is category B(0).
So next, let's recall the previous linear regression prediction results. We use the mean square error to measure. If the logistic regression prediction results are wrong, how to measure the loss? Let's look at such a picture
So how to measure the difference between the predicted results of logistic regression and the real results?
## 3 loss and optimization
### 3.1 loss
The loss of logistic regression is called log likelihood loss, and the formula is as follows:
• Separate categories:
How to understand a single formula? This should be understood according to the function image of log
• Integrated complete loss function
Seeing this formula is actually similar to our information entropy.
Next, let's take the above example to calculate, and then we can understand the meaning.
We already know that the larger the value of log (P), the smaller the result, so we can analyze this loss formula
### 3.2 optimization
The gradient descent optimization algorithm is also used to reduce the value of the loss function. In this way, the weight parameters of the previous corresponding algorithm of logistic regression are updated to improve the probability of originally belonging to category 1 and reduce the probability of originally belonging to category 0.
# 4. Introduction to logistic regression api
• sklearn.linear_model.LogisticRegression(solver='liblinear', penalty='l2', C = 1.0)
• solver optional parameters: {liblinear ',' sag ',' saga ',' Newton CG ',' lbfgs'},
• Default: 'liblinear'; Algorithms for optimization problems.
• "liblinear" is a good choice for small data sets, while "sag" and "saga" are faster for large data sets.
• For multi class problems, only 'Newton CG', 'sag', 'saga' and 'lbfgs' can handle multiple losses; "liblinear" is limited to the "one verse rest" category.
• penalty: type of regularization
• C: Regularization strength
By default, those with a small number of categories are taken as positive examples
The LogisticRegression method is equivalent to sgdclassifier (loss = "log", penalty = ""), which implements a common random gradient descent learning. Instead, LogisticRegression is used (sag is implemented)
# 5: case: cancer classification prediction - benign / malignant breast cancer prediction
• Data introduction
data description
(1) 699 samples, a total of 11 columns of data, the first column is the search id, and the last 9 columns are related to tumor
The last column represents the value of tumor type.
(2) Contains 16 missing values, use "?" Mark.
## 1 Analysis
1.get data
2.Basic data processing
2.1 Missing value processing
2.2 Determine eigenvalue,target value
2.3 Split data
3.Characteristic Engineering(Standardization)
4.machine learning(logistic regression )
5.Model evaluation
## 2 code
import pandas as pd
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import LogisticRegression
#1: get data
names = ['Sample code number', 'Clump Thickness', 'Uniformity of Cell Size', 'Uniformity of Cell Shape',
'Marginal Adhesion', 'Single Epithelial Cell Size', 'Bare Nuclei', 'Bland Chromatin',
'Normal Nucleoli', 'Mitoses', 'Class']
names=names)
#2: Basic data processing
#2.1 Missing value processing
# print(pd.isnull(data)) # Determine whether there are missing values
# print(data.query(''))
data = data.replace(to_replace='?',value=np.NaN)
data = data.dropna() #Delete missing values
x = data.iloc[:,1:10] # characteristic value
y = data["Class"] # target value
# print(y)
#2.2 Split data
x_train,x_test,y_train,y_test = train_test_split(x,y,random_state=22)
#3: Feature Engineering
transfer = StandardScaler()
x_train = transfer.fit_transform(x_train)
x_test = transfer.fit_transform(x_test)
# 4: Machine learning logistic regression
estimator = LogisticRegression()
estimator.fit(x_train,y_train)
#5: Model evaluation
y_predict = estimator.predict(x_test)
print(y_predict)
print(estimator.score(x_test, y_test)) | 1,308 | 5,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.868879 |
https://www.enotes.com/homework-help/how-an-elements-atomic-mass-determined-472479 | 1,490,364,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188132.48/warc/CC-MAIN-20170322212948-00501-ip-10-233-31-227.ec2.internal.warc.gz | 901,959,391 | 19,215 | # How is an element's atomic mass determined?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Mass is a physical property of matter and is expressed in SI units as kilogram. The atomic mass of elements is not actually the mass of their atoms in kilograms, rather it is a value giving the ratio of the atom's mass to that of the mass of a carbon atom which has been fixed at 12 a.m.u (atomic mass unit).
There are three particles that make up an atom, proton, neutron and electron. The mass of protons and neutrons is considered to be equal though in reality neutrons are heavier than protons; the mass of electrons is very less and ignored in the calculation of atomic mass.
An atom of Carbon has 6 protons and 6 neutrons, it is given an atomic mass of 12 a.m.u which makes each proton and neutron in the nucleus of any element contribute 1 a.m.u to the atom's atomic mass.
The number of protons is unique for each element but they can have different numbers of neutrons which leads to elements having isotopes with different atomic mass. The atomic mass of an element is the weighted average of the atomic mass of its isotopes. This requires a knowledge of the relative abundance of each isotope of the element.
aishukul | Student, Grade 10 | (Level 3) Honors
Posted on
The atomic mass of an element is the element's number of protons and neutrons added together. You can add the protons and neutrons together to calculate the atomic mass. You can also look at the periodic table of elements where at the bottom of each element's symbol there is the atomic mass number of the specific element you are looking for. Hope this helps!
Sources:
atyourservice | Student, Grade 11 | (Level 3) Valedictorian
Posted on
The atomic mass is found by adding the amount of neutrons and protons in an element. The amount of protons can be told by looking at the atomic number.
malkaam | Student, Undergraduate | (Level 1) Valedictorian
Posted on
Atomic mass is an element's mass that is the sum of masses of protons and neutrons forming an element. The easiest way to determine the atomic mass is to look up in the periodic table. In a periodic table, the atomic masses increase as you go across the table. The atomic mass is the decimal figures given therein. All you need to know is the element symbol. Atomic mass is in terms of grams per mole or g/mol.
Another way to calculate the atomic mass of an element is to add up the masses of protons and neutrons.
Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons. The atomic mass of the atom is the mass of the protons plus the mass of the neutrons, 6 + 7, or 13.
Moreover, the atomic mass of an element is a weighted average of all the element's isotopes based on their natural abundance. It is simple to calculate the atomic mass of an element with these steps:
1. The first step is to multiply the mass of each isotope by its abundance. But of the abundance is a percent, divide your answer by 100.
2. After values have been derived add them up.
The answer is the atomic mass or the atomic weight of the element.
Example: You are given a sample containing 98% carbon-12 and 2% carbon-13. What is the relative atomic mass of the element?
First convert the percentages to decimal values by dividing each percentage by 100. The sample becomes 0.98 carbon-12 and 0.02 carbon-13. (Tip: You can check your math by making certain the decimals add up to 1. 0.98 + 0.02 = 1.00)
Next, multiply the atomic mass of each isotope by the proportion of the element in the sample:
0.98 x 12 = 11.76
0.02 x 13 = 0.26
11.76 + 0.26 = 12.02 g/mol
Sources:
chrisyhsun | Student, College Freshman | (Level 1) Salutatorian
Posted on
The mass of an atom naturally comes from the summation of the masses of the atom's individual parts. Every atom is composed of protons, electrons, and neutrons. However, the mass of each electron is so miniscule, even in comparison to the already tiny protons and electrons, that it is just disregarded as a negligible component. As such, the mass of an atom can be simplified to just being the mass of its component protons and neutrons. The mass of an individual proton is essentially equivalent to the mass of a single neutron. Based on this observation, each proton and each neutron is figured as one atomic mass unit. Therefore, any element's atomic mass will simply be the sum of its protons and neutrons. The number of protons in any given element can be found on the periodic table as its atomic number. In normal, regular cases, the number of neutrons in the atom will be the same as the number of protons, so the atomic mass is simply double the atomic number. However, isotopes of the same element contain different number of neutrons, in which case doubling the atomic number will not give the correct answer. For instance, there are several isotopes of carbon, including carbon-12 (6 protons and 6 neutrons) and carbon-14 (6 protons, 8 neutrons).
sid-sarfraz | Student, Graduate | (Level 2) Salutatorian
Posted on
Element
A genuine chemical matter consisting of a particular type of atom well-known by its atomic number is known as an element. Elements are divided into metals, metalloids, and nonmetals.
Mass
Mass is a fundamental, physical property of any substance.
Atom
The atom is the smallest unit that defines the chemical elements and their isotopes.
Atomic Mass
The Mass of an atom is referred to as atomic mass of any particular chemical element.
The average atomic mass can be determined by multiplying the sum of an elements isotope with their natural abundance on earth. It can also be stated as sum of the masses of protons, neutrons and electrons in an atom. Hence the equation is;
Atomic Mass = (mass of isotope x relative abundance) + (mass of isotope x relative abundance)
The atomic mass is used as a relative unit to find out the mass of molecules involved in chemical reactions. It can also show the number of atoms or molecules in a mass of substance.
For Example;
In a sample of 400 lithium atoms, it is found that 30 atoms are lithium-6 (6.015 g/mol) and 370 atoms are lithium-7 (7.016 g/mol). Calculate the average atomic mass of lithium.
Solution:
1) Calculate the percent abundance for each isotope:
Li-6: 30/400 = 0.075
Li-7: 370/400 = 0.925
2) Calculate the average atomic weight:
x = (6.015) (0.075) + (7.016) (0.925)
x = 6.94 g/mol
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rachellopez | Student, Grade 12 | (Level 1) Valedictorian
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Atomic mass is the mass of an atom, particle, or molecule. The atomic mass is determined by the number of protons and neutrons in the atom. For example, Oxygen has 8 protons (as seen by the atomic number) and 8 neutrons which gives oxygen an atomic mass of 16. However, different isotopes of oxygen (with a different amount of neutrons) can have different atomic masses.
Sources: | 1,724 | 7,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-13 | longest | en | 0.915994 |
http://en.m.wikibooks.org/wiki/Physics_Course/Relativity | 1,397,881,186,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00214-ip-10-147-4-33.ec2.internal.warc.gz | 79,950,353 | 6,163 | # Physics Course/Relativity
## Relativity
According to Newton Mass of any moving object remains the same regardless of speed travel . According to Einstein any moving object has Mass changes with speed travel
## Relativity Theory
### Change in Mass relate to speed travels
Speed of light is alsway constant , v = C . When matter moves at speed of light, matter no longer has mass . Observed fron Plank in experiment with Black Body Radiation
$m = 0$ when $v = C$
All matter moves at speed less than speed of light has a mass of mo
$m = m_o$ when $v < C$
Therefore for any matter moving at speed less than speed of light but not equal to the speed of light
$m = m_o (1 + \frac{v}{C})$
### Wave length of a Momentum
Any momentum travels at speed of light C carries quantum energy $E = h \frac{C}{\lambda}$ will have a wavelength
$\lambda = \frac{h}{p} = \frac{h}{C} \frac{1}{m}$
Which can be easily proved
$pC = h \frac{C}{\lambda}$
$\lambda = \frac{h}{p} = \frac{h}{C} \frac{1}{m}$ | 273 | 995 | {"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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | longest | en | 0.806323 |
https://glowing.com/community/topic/196719/7-months | 1,571,302,766,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673250.23/warc/CC-MAIN-20191017073050-20191017100550-00506.warc.gz | 522,735,063 | 8,991 | # 7 months...
Glow is saying in 7 months pregnant and I'm 27weeks 1day but I thought 7 months would be at 28weeks.
4w=1m
8w=2m
12=3m
16=4m
20w=5m
24w=6m
28w=7m
32w=8m
36w=9m
Or am I wrong on how I'm calculating this?
You | 101 | 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} | 3.125 | 3 | CC-MAIN-2019-43 | longest | en | 0.923368 |
http://slideplayer.com/slide/2501254/ | 1,571,524,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00283.warc.gz | 151,488,984 | 18,382 | # Lecture 10: RL & RC Circuits Nilsson & Riedel 7.1-7.6 ENG17 (Sec. 2): Circuits I Spring 2014 1 May 1, 2014.
## Presentation on theme: "Lecture 10: RL & RC Circuits Nilsson & Riedel 7.1-7.6 ENG17 (Sec. 2): Circuits I Spring 2014 1 May 1, 2014."— Presentation transcript:
Lecture 10: RL & RC Circuits Nilsson & Riedel 7.1-7.6 ENG17 (Sec. 2): Circuits I Spring 2014 1 May 1, 2014
Recap – Inductors / Capacitors 2
Overview Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 3
Natural Response 4 Currents and voltages that arise from an inductor or capacitor immediately after being disconnected from a DC (direct current) source.
Current in RL Circuit 5 After Open Switch
Time Constant 6
RL Circuit Example 7
Voltage in RC Circuit 8 After Switching
RC Circuit Example 9
Overview Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 10
Step Response 11 Currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application of a dc voltage or current source.
Step Response in RL Circuit 12
Step Response in RC Circuit 13
Overview Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 14
Generalized Solution 15
Overview Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 16
Sequential Switching 17 Defined: whenever switching occurs more than once in a circuit
Overview Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 18
Unbounded Response Defined: A circuit response that grows, rather than decays, exponentially with time. Possible with dependent sources 19
Recap Natural Response –RL Circuits –RC Circuits Step Response –RL Circuits –RC Circuits General Solution Sequential Switching Unbounded Response 20
Download ppt "Lecture 10: RL & RC Circuits Nilsson & Riedel 7.1-7.6 ENG17 (Sec. 2): Circuits I Spring 2014 1 May 1, 2014."
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EL630 HW4
# EL630 HW4 - EL630 Homework 4 1(Book 4-9 Find f x if F x...
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1 EL630: Homework 4 1. (Book: 4-9) Find ( ) if ( ) (1 ) ( ). x f x F x e u x c α = (Hint: For any function ( ), h x we have ( ) ( ) ( ) ( ), h x x a h a x a δ δ = which is called the sampling property of the impulse function.) 2. (Book: 4-10) If is (0,4) find (a) {1 2} and (b) {1 2| 1}. N P P X X X X 1 2 c x ( ) F x
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2 (Hint: (0,4) N means 2 2 ( ) / 2 2 1 ( ) 2 x f x e µ σ πσ = X with 2 0, 2 µ σ = . Then we have {1 2} (2) (1) P F F = X X X . 2 2 ( ) / 2 2 1 ( ) ( ) ( ) 1/ 2 2 x y x x F x e dy G erf µ σ µ µ σ σ πσ −∞ = = = X . Then use Table 4-1.) 3. (Book: 4-18) Show that ( ) ( | ) ( ) ( | )[1 ( )]. P A P A x F x P A x F x = + > X X 4. (Book: 4-19) Show that ( | ) ( ) ( | ) . ( ) x x P A x F x F x A P A = X 5. (Book: 4-27) A coin is tossed an infinite number of times. Show that the probability that k heads are observed at the n th toss but not earlier equals 1 . 1 n k n k p q k [See also book (4-63).] 6. (Book: 4-30) The probability that a driver will have an accident in 1 month equals 0.02. Find the probability that in 100 months he will have three accidents.
3 7. (Book: 4-31) A fair die is rolled five times. Find the probability that “one” shows twice,
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https://inforwriters.com/exercise-1-economic-order-quantity/ | 1,709,255,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00778.warc.gz | 299,349,715 | 34,792 | # Exercise 1 Economic Order Quantity
By Support
Exercise 1 Economic Order Quantity
BCO214-Production Management
Exercise 1: Economic Order Quantity. (35 points)
One of the Toyota plants must decide how many of a particular type to order for manufacturing automobiles.
The company plans to produce 300,000 cars in the next year. All cars planned for production use the same headlights (2 units per car); therefore, demand for the headlights for the next year is known to be 600,000 units. The purchasing agent wants to know how many headlights to buy at one time. Historically, headlights have been received half a day (lead time) after they were ordered. It costs € 50 to order lights, and the holding-cost fraction used by the auto company is 22 % per year. The headlights cost € 25 each.
• Compute the Economic Order Quantity.
• What is the number of orders per year?
• What is the frequency of orders? Assume the company works 365 days a year.
• What is the
• The company works with a Safety Stock of 0.75 days. What is the level of this stock?
• What is the Average Stock?
• What is the total Holding Cost?
• What is the total Ordering Cost?
• What is the total Cost of Stock Management?
• Draw the evolution of stock in time (the stock chart).
Material Requirement Planning (MRP). (35 points)
Exercise 3: Understand quality management and efficiency control and how to measure. (30 points)
3.1) “Quality begins with the design of a product in accordance with the customer specification further it involved the established” Discuss why and give examples.
3.2) Why is necessary Inspection in production?
100% inspection or sampling inspection. When do we use each and what are the reasons for it?
3.3) Why do we define tools for measuring and controlling quality. Give the list of basic tools in Quality Management, explain, and give examples.
3.4) What are the most usual “wastes” in a production system? How Lean Management can reduce or eliminate these wastes and improve the efficiency in a manufacturing process?
Students are required to use the Harvard Referencing System. The report needs to be uploaded onto Turnitin on Moodle. The assignment hand-in will be created by the course tutor.
Formalities:
• Word document.
• Wordcount: 2500 words.
• Font: Arial 12,5 pts.
• Text alignment: Justified.
• The in-text References and the Bibliography have to be in Harvard’s citation style.
Submission:
• Week 12: Via Moodle (Turnitin).
• Deadline: 8th January 2023 at 23:59 CEST. | 561 | 2,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.932504 |
https://nrich.maths.org/public/leg.php?code=-478&cl=3&cldcmpid=6155 | 1,477,530,142,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721027.15/warc/CC-MAIN-20161020183841-00455-ip-10-171-6-4.ec2.internal.warc.gz | 853,685,016 | 7,937 | # Search by Topic
#### Resources tagged with STEM - physical world similar to Natural Shapes:
Filter by: Content type:
Stage:
Challenge level:
### There are 35 results
Broad Topics > Applications > STEM - physical world
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Bigger or Smaller?
##### Stage: 4 Challenge Level:
When you change the units, do the numbers get bigger or smaller?
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
##### Stage: 4 Challenge Level:
Which units would you choose best to fit these situations?
### Does This Sound about Right?
##### Stage: 3 Challenge Level:
Examine these estimates. Do they sound about right?
### How Do You React?
##### Stage: 4 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Avalanche!
##### Stage: 2 and 3 Challenge Level:
Investigate how avalanches occur and how they can be controlled
### Reaction Timer
##### Stage: 3 Challenge Level:
This problem offers you two ways to test reactions - use them to investigate your ideas about speeds of reaction.
### Make Your Own Solar System
##### Stage: 2, 3 and 4 Challenge Level:
Making a scale model of the solar system
### Pinhole Camera
##### Stage: 3 Challenge Level:
Make your own pinhole camera for safe observation of the sun, and find out how it works.
### Stemnrich - the Physical World
##### Stage: 3 and 4 Challenge Level:
PhysNRICH is the area of the StemNRICH site devoted to the mathematics underlying the study of physics
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Global Warming
##### Stage: 4 Challenge Level:
How much energy has gone into warming the planet?
### Guessing the Graph
##### Stage: 4 Challenge Level:
Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from?
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Big and Small Numbers in the Physical World
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### Electric Kettle
##### Stage: 4 Challenge Level:
Explore the relationship between resistance and temperature
### Speed-time Problems at the Olympics
##### Stage: 4 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Heavy Hydrocarbons
##### Stage: 4 and 5 Challenge Level:
Explore the distribution of molecular masses for various hydrocarbons
### Chemnrich
##### Stage: 4 and 5 Challenge Level:
chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . .
### Perfect Eclipse
##### Stage: 4 Challenge Level:
Use trigonometry to determine whether solar eclipses on earth can be perfect.
### Far Horizon
##### Stage: 4 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Construct the Solar System
##### Stage: 4 and 5 Challenge Level:
Make an accurate diagram of the solar system and explore the concept of a grand conjunction.
### Playground Snapshot
##### Stage: 2 and 3 Challenge Level:
The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone?
### Physnrich
##### Stage: 4 and 5 Challenge Level:
PhysNRICH is the area of the StemNRICH site devoted to the mathematics underlying the study of physics
### Carbon Footprints
##### Stage: 4 Challenge Level:
Is it really greener to go on the bus, or to buy local?
### Temperature
##### Stage: 3 Challenge Level:
Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same?
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Observing the Sun and the Moon
##### Stage: 2 and 3 Challenge Level:
How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions.
### Helicopters
##### Stage: 2, 3 and 4 Challenge Level:
Design and test a paper helicopter. What is the best design? | 1,253 | 5,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.836028 |
https://brainmass.com/math/optimization/linear-optimization-model-toys-527109 | 1,656,938,699,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00511.warc.gz | 200,040,072 | 74,828 | Explore BrainMass
# Linear Optimization Model: Toys
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Super Toys Company makes to radio-controlled cars, Fast and SuperFast. They can sell all they make. Both models have the same components. Two of these can be obtained only from a single supplier. For next month, the supply of these is limited to 4000 of component X and 3500 of component Y. The following table provides details on the number of each component required for each product and the profit per unit.
....................................Components Required/Unit........................
.....................................A......................B........................Profit/Unit
Fast.............................18.....................6..............................\$24
SuperFast....................12.....................10............................\$40
1. What are the decision variable, objective function and constraints
2. Mathematically formulate a linear optimization model.
https://brainmass.com/math/optimization/linear-optimization-model-toys-527109
## SOLUTION This solution is FREE courtesy of BrainMass!
1. What are the decision variable, objective function, and constraints
The decision variables are the number of units that must be produced for Fast and SuperFast. Mathematically there are usually expressed as single digits. Let F and SF be the number of units, of Fast and SuperFast, produced respectively.
The objective function is the mathematical function which must be satisfied. The function could be either to maximize or minimize. From the question, the objective is to maximize the profit.
..................................................................
Please see attachments for complete solution.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! | 370 | 2,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-27 | longest | en | 0.829896 |
http://www.completepowerelectronics.com/circuit-analysis-pspice/ | 1,519,522,308,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00141.warc.gz | 410,339,143 | 10,759 | # Circuit Analysis Using Pspice - An Introduction
### Circuit Analysis by Pspice:
Using Pspice tool, we can do various types of analysis in an electrical circuit.
They are
• DC analysis,
• AC analysis also known as Sinusoidal steady state linear analysis,
• Transient analysis (also known as Time domain linear/non linear analysis).
In addition to this, Pspice will do the
• Sensitivity analysis/ Worst Case analysis,
• Optimizer ( component value optimization),
• Smoke (Part over stress analyzer) and
• Monte Carlo(statistical) analysis.
In this post we will see each analysis in brief:
DC Sweep Analysis:
The DC Sweep Analysis of electrical and electronic circuits can be employed to determine the dc operating point.
It is also possible to sweep a voltage or current source to obtain a dc(static) transfer function.
AC Sweep Analysis:
The AC Sweep Analysis can be employed to determine frequency response ( ex. Bode Plots) and to determine the impedances and admittances.
Pspice AC analysis can be used to perform a noise analysis.
Transient analysis:
The Transient analysis can be employed to generate the time domain response.
By inspecting the output waveforms the system distortion is obtained.
The Pspice postprocessor looks like an oscilloscope.
Smoke analysis:
The Smoke analysis can be employed to examine the maximum voltage, currents and power dissipation.
Based on these values we can find which components may be over stressed.
Monte Carlo analysis:
The Monte Carlo analysis can be employed to determine the typical, minimum and maximum (ie, worst case) values based on the device and lot tolerances.
It is a statistical analysis.
Sensitivity analysis/ Worst Case analysis:
The sensitivity and Worst Case analysis can be employed to determine the relative effect of each component on the measurement point on the circuit.
It is obtained from the bar graph generated by the analysis. | 392 | 1,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.830647 |
https://freezingblue.com/flashcards/172149/preview/ecology-chapter-8 | 1,722,702,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00507.warc.gz | 216,219,501 | 4,714 | # Ecology Chapter 8
Concept 8.1: Populations are dynamic entities that vary in size over time and space. Concept 8.2: The distributions and abundances of organisms are limited by habitat suitability, historical factors, and dispersal. Concept 8.3: Many species have a -------------of populations across their geographic range. patchy distribution Concept 8.4: The dispersion of individuals within a population depends on the (3 things) location of essential resources, dispersal, and behavioral interactions. Distribution: Geographic area over which individuals of a species occur. Abundance: The number of individuals in a specific area Ecologists often wish to understand what factors determine the -------------and ------------- of species. distribution and abundance Biomass The total weight of individuals in a specific area Density Number of individuals per unit area (land) or volume (density) Mark recapture experiments M/N=m/n M/N=m/n what do letters stand for? M = number of marked individuals releasedN = actual size of populationm = number of marked individuals captured the 2nd timen = the total number of individuals 2nd time Distribution – can have 2 definitions 1 1The geographic range of an organism(Determined on the MACRO scale usually by abiotic factors) Distribution – can have 2 definitions 2 2The spatial arrangement of individuals in a local population (Determined on the MICRO scale, usually by both abiotic and biotic factors) Abundance can be reported as -------- or ----------- population size (# individuals), or density (# individuals per unit area). Individuals can be defined as products of a single fertilization: The aspen grove would be one individual, a genet. Members of a genet may be independent physiologically, so members of a genet are called -------- ramets. Abiotic features of the environment include (5) moisture, temperature, pH, sunlight, nutrients, etc. disturbance- events that kill or damage some individuals, creating opportunities for other individuals to grow and reproduce. Individuals are uniformly spaced through the environment Cause? (2) 1Antagonistic interactions between individuals2Local depletion of resources An individual has an equal probability of occurring anywhere in an area Cause?(2) 1Neutral interactions between individuals and the environment2Neutral interactions between individuals Individuals live in areas of high local abundance separated by areas of low abundance Cause? (2) 1Attraction between individuals2Attraction of individuals to a common resource Dispersion is the spatial arrangement of individuals within a population: Dispersion types (3) 1Regular (uniform)—individuals are evenly spaced.2Random—individuals scattered randomly.3Clumped (aggregated)—the most common pattern The ecological niche: The physical and biological conditions that a species needs to grow, survive, and reproduce. A niche model is a predictive tool that models the environmental conditions occupied by a species based on the conditions at localities it is known to occupy. Authorzzto ID172149 Card SetEcology Chapter 8 DescriptionEcology Chapter 8 Updated2012-09-25T12:36:28Z Show Answers | 633 | 3,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.876997 |
https://www.physicsforums.com/threads/calculation-of-the-efficiency-of-an-engine.950768/ | 1,618,927,246,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00526.warc.gz | 1,067,051,002 | 18,301 | # Calculation of the efficiency of an engine
• Automotive
We know efficiency (η) = Woutput/Winput
Now in any cycle, for instance, the Otto cycle, let the total work output be WO, total heat added be WH & the total work done on the gas to compress it be WC.
Then, will efficiency be equal to (η) = (WO-WC)/WH ?
Or will it be equal to (η) = WO/(WH+WC) ?
anorlunda
Staff Emeritus
There are lots of internal things that happen in an engine. Compression is one. Friction is another.
The only useful definition of efficiency is useful work out divided by work in. Out and in refer to things external to the engine. Useful is also necessary, because for example heat leaves the engine and warms up the environment. That's energy out, but it is not useful.
In other words, efficiency is a human measure that you define to be useful for your purposes. In that respect, it is less about physics, and more about your opinions.
russ_watters
Mentor
We know efficiency (η) = Woutput/Winput
Now in any cycle, for instance, the Otto cycle, let the total work output be WO, total heat added be WH & the total work done on the gas to compress it be WC.
Then, will efficiency be equal to (η) = (WO-WC)/WH ?
Or will it be equal to (η) = WO/(WH+WC) ?
It's neither. Work output already does not include the compression energy. You are subtracting it twice.
CWatters
Homework Helper
Gold Member
Defining the input power in the case of a liquid or gas fuel is not always straight forward. Here in the UK we have gas home heating boilers that are more than 100% efficient because of the way the input power was defined historically.
Randy Beikmann
Gold Member
For an Otto cycle, there is no work input. There is heat input (Q_h) from the fuel. Work and heat are thermodynamically different forms of energy.
Taking the work of compression as W_c and the work of expansion as W_e the thermal efficiency is (W_e - W_c)/Q_h.
In words, it is the net output work divided by the heat input. | 488 | 1,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-17 | longest | en | 0.936869 |
https://socratic.org/questions/572ad8da11ef6b3b53cd5a5c | 1,638,848,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363332.1/warc/CC-MAIN-20211207014802-20211207044802-00202.warc.gz | 574,901,314 | 5,995 | # Question #d5a5c
May 5, 2016
$x = 3$
#### Explanation:
Note that as $4 \equiv - 1 \text{ (mod 5)}$ we have
${2}^{2010} \equiv {\left({2}^{2}\right)}^{1005} \equiv {4}^{1005} \equiv {\left(- 1\right)}^{1005} \equiv - 1 \text{ (mod 5)}$
Thus we are actually searching for the least $x \in {\mathbb{Z}}^{+}$ such that
$3 x \equiv - 1 \text{ (mod 5)}$
From here, we can simply iterate $x$ starting from $x = 1$.
$3 \left(1\right) \equiv 3 \cancel{\equiv} - 1 \text{ (mod 5)}$
$3 \left(2\right) \equiv 6 \equiv 1 \cancel{\equiv} - 1 \text{ (mod 5)}$
$3 \left(3\right) \equiv 9 \equiv - 1 \text{ (mod 5)}$
Therefore the least such positive integer $x$ is $x = 3$ | 275 | 669 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.625005 |
https://studen.com/mathematics/18010406 | 1,713,045,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00270.warc.gz | 513,304,019 | 45,487 | : asked on jevanoff
31.01.2023
# Ken is making trail mix for his next hiking club meeting. He mixes 2 pounds of peanuts, 2 pounds of raisins, and 1 pound of chocolate chips. His little sister begs him for a snack, so he gives her 5 ounces of the trail mix. Then, he portions out the rest for his meeting. If Ken splits the remaining trail mix evenly among 25 snack bags, how many ounces of trail mix will each bag have? ounces
0
09.07.2023, solved by verified expert
Unlock the full answer
3 ounces of trail mix per bag
Step-by-step explanation:
1 pound = 16 ounces
2 pounds of peanuts
2 pounds of raisins
+ 1 pound of chocolate chips
= 5 pounds of trail mix
5 pounds = 80 ounces of trail mix
80 - 5 (for Ken's sister) = 75 ounces
75 ÷ 25 = 3 ounces
Hope this helps~!
It is was helpful?
### Faq
Mathematics
P Answered by Master
3 ounces of trail mix per bag
Step-by-step explanation:
1 pound = 16 ounces
2 pounds of peanuts
2 pounds of raisins
+ 1 pound of chocolate chips
= 5 pounds of trail mix
5 pounds = 80 ounces of trail mix
80 - 5 (for Ken's sister) = 75 ounces
75 ÷ 25 = 3 ounces
Hope this helps~!
Mathematics
P Answered by PhD
The answer is in the image
Mathematics
P Answered by PhD
The answer is in the image
Mathematics
P Answered by PhD
F=ma
where F=force
m=mass
a=acceleration
Here,
F=4300
a=3.3m/s2
m=F/a
=4300/3.3
=1303.03kg
Mathematics
P Answered by PhD
The answer is in the image
Mathematics
P Answered by PhD
Salesperson will make 6% of 1800
=(6/100)*1800
=108
Salesperson will make \$108 in \$1800 sales
Mathematics
P Answered by PhD
Here,
tip=18%of \$32
tip=(18/100)*32
=0.18*32
=\$5.76
Total payment=32+5.76=\$37.76
Mathematics
P Answered by PhD
The wood before starting =12 feet
Left wood=6 feet
Wood used till now=12-6=6 feet
Picture frame built till now= 6/(3/4)
=8 pieces
Therefore, till now 8 pieces have been made.
Mathematics
P Answered by PhD
The wood before starting =12 feet
Left wood=6 feet
Wood used till now=12-6=6 feet
Picture frame built till now= 6/(3/4)
=8 pieces
Therefore, till now 8 pieces have been made.
Mathematics
P Answered by PhD
Salesperson will make 6% of 1800
=(6/100)*1800
=108
Salesperson will make \$108 in \$1800 sales
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FREE | 737 | 2,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.896512 |
https://origin.geeksforgeeks.org/queries-for-decimal-values-of-subarray-of-a-binary-array/?ref=lbp | 1,675,009,542,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00711.warc.gz | 453,766,889 | 33,866 | # Queries for decimal values of subarrays of a binary array
• Difficulty Level : Medium
• Last Updated : 13 Jul, 2022
Given a binary array arr[], we to find the number represented by the subarray a[l..r]. There are multiple such queries.
Examples:
```Input : arr[] = {1, 0, 1, 0, 1, 1};
l = 2, r = 4
l = 4, r = 5
Output : 5
3
Subarray 2 to 4 is 101 which is 5 in decimal.
Subarray 4 to 5 is 11 which is 3 in decimal.
Input : arr[] = {1, 1, 1}
l = 0, r = 2
l = 1, r = 2
Output : 7
3```
A Simple Solution is to compute decimal value for every given range using simple binary to decimal conversion. Here each query takes O(len) time where len is length of range.
An Efficient Solution is to do per-computations, so that queries can be answered in O(1) time.
The number represented by subarray arr[l..r] is arr[l]*+ arr[l+1]*….. + arr[r]*
1. Make an array pre[] of same size as of given array where pre[i] stores the sum of arr[j]*where j includes each value from i to n-1.
2. The number represented by subarray arr[l..r] will be equal to (pre[l] – pre[r+1])/.pre[l] – pre[r+1] is equal to arr[l]*+ arr[l+1]*+……arr[r]*. So if we divide it by , we get the required answer
Flowchart
Flowchart
Implementation:
## C++
`// C++ implementation of finding number` `// represented by binary subarray` `#include ` `using` `namespace` `std;` `// Fills pre[]` `void` `precompute(``int` `arr[], ``int` `n, ``int` `pre[])` `{` ` ``memset``(pre, 0, n * ``sizeof``(``int``));` ` ``pre[n - 1] = arr[n - 1] * ``pow``(2, 0);` ` ``for` `(``int` `i = n - 2; i >= 0; i--)` ` ``pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));` `}` `// returns the number represented by a binary` `// subarray l to r` `int` `decimalOfSubarr(``int` `arr[], ``int` `l, ``int` `r,` ` ``int` `n, ``int` `pre[])` `{` ` ``// if r is equal to n-1 r+1 does not exist` ` ``if` `(r != n - 1)` ` ``return` `(pre[l] - pre[r + 1]) / (1 << (n - 1 - r));` ` ``return` `pre[l] / (1 << (n - 1 - r));` `}` `// Driver Function` `int` `main()` `{` ` ``int` `arr[] = { 1, 0, 1, 0, 1, 1 };` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``int` `pre[n];` ` ``precompute(arr, n, pre);` ` ``cout << decimalOfSubarr(arr, 2, 4, n, pre) << endl;` ` ``cout << decimalOfSubarr(arr, 4, 5, n, pre) << endl;` ` ``return` `0;` `}`
## Java
`// Java implementation of finding number` `// represented by binary subarray` `import` `java.util.Arrays;` `class` `GFG {` ` ``// Fills pre[]` ` ``static` `void` `precompute(``int` `arr[], ``int` `n, ``int` `pre[])` ` ``{` ` ``Arrays.fill(pre, ``0``);` ` ``pre[n - ``1``] = arr[n - ``1``] * (``int``)(Math.pow(``2``, ``0``));` ` ``for` `(``int` `i = n - ``2``; i >= ``0``; i--)` ` ``pre[i] = pre[i + ``1``] + arr[i] * (``1` `<< (n - ``1` `- i));` ` ``}` ` ``// returns the number represented by a binary` ` ``// subarray l to r` ` ``static` `int` `decimalOfSubarr(``int` `arr[], ``int` `l, ``int` `r,` ` ``int` `n, ``int` `pre[])` ` ``{` ` ``// if r is equal to n-1 r+1 does not exist` ` ``if` `(r != n - ``1``)` ` ``return` `(pre[l] - pre[r + ``1``]) / (``1` `<< (n - ``1` `- r));` ` ``return` `pre[l] / (``1` `<< (n - ``1` `- r));` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `arr[] = { ``1``, ``0``, ``1``, ``0``, ``1``, ``1` `};` ` ``int` `n = arr.length;` ` ``int` `pre[] = ``new` `int``[n];` ` ``precompute(arr, n, pre);` ` ``System.out.println(decimalOfSubarr(arr,` ` ``2``, ``4``, n, pre));` ` ``System.out.println(decimalOfSubarr(arr,` ` ``4``, ``5``, n, pre));` ` ``}` `}` `// This code is contributed by Anant Agarwal.`
## Python3
`# implementation of finding number` `# represented by binary subarray` `from` `math ``import` `pow` `# Fills pre[]` `def` `precompute(arr, n, pre):` ` ` ` ``pre[n ``-` `1``] ``=` `arr[n ``-` `1``] ``*` `pow``(``2``, ``0``)` ` ``i ``=` `n ``-` `2` ` ``while``(i >``=` `0``):` ` ``pre[i] ``=` `(pre[i ``+` `1``] ``+` `arr[i] ``*` ` ``(``1` `<< (n ``-` `1` `-` `i)))` ` ``i ``-``=` `1` `# returns the number represented by ` `# a binary subarray l to r` `def` `decimalOfSubarr(arr, l, r, n, pre):` ` ` ` ``# if r is equal to n-1 r+1 does not exist` ` ``if` `(r !``=` `n ``-` `1``):` ` ``return` `((pre[l] ``-` `pre[r ``+` `1``]) ``/` ` ``(``1` `<< (n ``-` `1` `-` `r)))` ` ``return` `pre[l] ``/` `(``1` `<< (n ``-` `1` `-` `r))` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ``arr ``=` `[``1``, ``0``, ``1``, ``0``, ``1``, ``1``]` ` ``n ``=` `len``(arr)` ` ``pre ``=` `[``0` `for` `i ``in` `range``(n)]` ` ``precompute(arr, n, pre)` ` ``print``(``int``(decimalOfSubarr(arr, ``2``, ``4``, n, pre)))` ` ``print``(``int``(decimalOfSubarr(arr, ``4``, ``5``, n, pre)))` `# This code is contributed by` `# Surendra_Gangwar`
## C#
`// C# implementation of finding number` `// represented by binary subarray` `using` `System;` `class` `GFG {` ` ``// Fills pre[]` ` ``static` `void` `precompute(``int``[] arr, ``int` `n, ``int``[] pre)` ` ``{` ` ``for` `(``int` `i = 0; i < n; i++)` ` ``pre[i] = 0;` ` ``pre[n - 1] = arr[n - 1] * (``int``)(Math.Pow(2, 0));` ` ``for` `(``int` `i = n - 2; i >= 0; i--)` ` ``pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));` ` ``}` ` ``// returns the number represented by ` ` ``// a binary subarray l to r` ` ``static` `int` `decimalOfSubarr(``int``[] arr, ``int` `l, ``int` `r,` ` ``int` `n, ``int``[] pre)` ` ``{` ` ``// if r is equal to n-1 r+1 does not exist` ` ``if` `(r != n - 1)` ` ``return` `(pre[l] - pre[r + 1]) / (1 << (n - 1 - r));` ` ``return` `pre[l] / (1 << (n - 1 - r));` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main()` ` ``{` ` ``int``[] arr = { 1, 0, 1, 0, 1, 1 };` ` ``int` `n = arr.Length;` ` ``int``[] pre = ``new` `int``[n];` ` ``precompute(arr, n, pre);` ` ``Console.WriteLine(decimalOfSubarr(arr,` ` ``2, 4, n, pre));` ` ``Console.WriteLine(decimalOfSubarr(arr,` ` ``4, 5, n, pre));` ` ``}` `}` `// This code is contributed by vt_m.`
## PHP
`= 0; ``\$i``--)` ` ``\$pre``[``\$i``] = ``\$pre``[``\$i` `+ 1] + ``\$arr``[``\$i``] * ` ` ``(1 << (``\$n` `- 1 - ``\$i``));` `}` `// returns the number represented by ` `// a binary subarray l to r` `function` `decimalOfSubarr(&``\$arr``, ``\$l``, ``\$r``, ``\$n``, &``\$pre``)` `{` ` ``// if r is equal to n-1 r+1 does not exist` ` ``if` `(``\$r` `!= ``\$n` `- 1)` ` ``return` `(``\$pre``[``\$l``] - ``\$pre``[``\$r` `+ 1]) / ` ` ``(1 << (``\$n` `- 1 - ``\$r``));` ` ``return` `\$pre``[``\$l``] / (1 << (``\$n` `- 1 - ``\$r``));` `}` `// Driver Code` `\$arr` `= ``array``(1, 0, 1, 0, 1, 1 );` `\$n` `= sizeof(``\$arr``);` `\$pre` `= ``array_fill``(0, ``\$n``, NULL);` `precompute(``\$arr``, ``\$n``, ``\$pre``);` `echo` `decimalOfSubarr(``\$arr``, 2, 4, ``\$n``, ``\$pre``) . ``"\n"``;` `echo` `decimalOfSubarr(``\$arr``, 4, 5, ``\$n``, ``\$pre``) . ``"\n"``;` `// This code is contributed by ita_c` `?>`
## Javascript
``
Output
```5
3```
Time complexity: O(n)
Auxiliary Space: O(n)
This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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# How to make a Robot Walk
## This information has been adapted from an excellent site by A Miller.
Introduction
This section describes how to make a walking robot. The simplest robots are called BEAM ROBOTS. And the simplest walkers are called BEAM WALKERS. They consist of just enough circuitry and mechanics to carry out the intended function, and in this case it is the operation of walking. The simplest circuit to get a robot to carry out the walking function has been designed and patented by one of the earliest developers of Robots, Mark W. Tilden. He gave the name MicroCore to the type of circuit that drives these basic designs. The concept of the MicroCore is pretty clever. Using just a handful of components, you can build the control circuitry for a walking robot that senses its environment and adjusts its gait accordingly. The clever part is the circuit is so simple! A MicroCore consists of a capacitor, a gate, and a resistor. This is called a 'neuron.' Each neuron has an input and an output. The components are connected so that the capacitor and resistor form a delay circuit. If we make the input HIGH, the output take a short time to react. Mouse-over the animation below to see how the Neuron circuit works.
The gate is actually an inverter and it also has the ability to strengthen the pulse. If we connect the output of one neuron to the input of another and take the output of the second to the input of the first we have a complete loop and we have created a BiCore. A circuit can consist of as many gates (inverters) as you want, hooked-up nose to tail. Every 'neuron' of a MicroCore consists of a capacitor, a gate, and a resistor. A circuit containing a number of 'neurons' will produce amazing things. This is what this discussion is all about. If you have a BiCore controlling two wheels, and want the robot to turn right, you have to make the motor on the left run for longer - simply change the resistance for that neuron. If you incorporate 2 photodiodes into the MicroCore, you can make a robot always zigzag its way towards light! This is really impressive and demonstrates how simple "complex behaviour" really is. In this discussion we are going to explain the operation of the . . .
MicroCore
The MicroCore circuit can be likened to a nervous circuit in that it gives life and realism to a robotic shape. The MicroCore circuit is a basic circuit used in the majority of BEAM Biomorphic walkers. To get an idea of the basics of this topic, the following paper was prepared by Mark Tilden:
Biomorphic Robotics and Nervous Net Research: A New Machine Control Paradigm
Mark W. Tilden, Biophysics Division, Los Alamos National Labs Submitted for publication to the EANN '95 Conference Proceedings "Special Track on Robotics." Nervous Net (Nv) technology is a non-linear analog control system that solves real time control problems normally quite difficult to handle with digital methods. Nervous nets are to Neural nets the same way peripheral spinal systems are to the brain. This work has concentrated on the development of Nv based robot mechanisms with electronic approximations of biologic autonomic and somatic systems. It has been demonstrated that these systems, when fed back onto themselves rather than through a computer-based pattern generator, can successfully mimic many of the attributes normally attributed to lower biological organisms. Using Nv nets, highly successful legged robot mechanisms have been demonstrated which can negotiate terrains of inordinate difficulty for wheeled or tracked machines. That non-linear systems can provide this degree of control is not so surprising as the part counts for successful Nv designs. A fully adept insect-walker, for example, can be fully controlled and operated with as little as 12 standard transistor elements. Since the start of research in the winter of 1994, development of this technology has advanced to solving currently difficult sensory and cognitive problems. It is hoped in the coming years Nv systems may do for robot vision (amongst other disciplines) what has been done for autonomous robot vehicles, namely the reduction of currently complex systems down to an inexpensive but robust minimum. Further efforts are also being made to apply this control strategy to the expanding nano-technology field. At the nanometer scale Nv's may prove more feasible than nano-computers for control of self-assembling micro structures. For now, however, Nv research concentrates on problems of scale invariance, proving by example (or exhaustion) this control system can work at all scales, types, and styles of robotic application. The Nv control method could be adapted to most types of machine control, but it has been applied to autonomous robots because of the difficulty conventional control systems have solving the seemingly simple task of negotiating undefined complex environments. The 80 or so "biomorphic" robots (from the terms BIOlogy and MORPHology, and the Latin for "living" and "form") built so far are not "workers" in the traditional sense, but "survivors," in that they fight to solve the immediate problems of existence rather than procedural condition (i.e. they do not follow the rules of an internal program that mimics the external world, but the world itself). Nv control architectures focus on adaptive survival rather than the performance of specific tasks. Once survivability is under control, goals can be superimposed and the machine used as a platform to carry sensors and conventional electronic intelligence. It is believed that these machines, although now in an early stage of development, can within a few years be brought to the point that they can serve as inexpensive, robust, and versatile carriers for a variety of instruments. A vast number of applications would then be possible, including the location and possible clearing of land mines from civilian areas, security, maintenance, medical and prosthetic applications (a cost-effective "walking wheelchair" for example), and even cars with onboard "survival" instincts to save themselves and their passengers, from damaging accidents. Though the Nv based legged devices built so far cannot go everywhere, they can certainly go places not currently accessible to wheeled or tracked vehicles of similar scale. It seems that for handling undefined environments, biomorphic designs are a very
efficient and cost-effective approach. Initially it was thought these devices avoided the problems of an internal world representation by using a reactive or behaviour-based technique. Recent work has shown however that Nv biomorphs instead take a chaotic map of their surroundings onto their process control hierarchy (that is, they dynamically and efficiently adapt to the fractal complexity of their surroundings). This is due to the analog-electronic nature of the devices, the adaptive hardware of their structure, and the topological orientation of their interconnections. The defining characteristic of this adaptation is continuously updated by the immediate fractal complexity of the environment. These devices are "soft" designs, in that the environmental dimension must be absorbed, modified, and acted on for the devices to make successful headway through a complex world. These devices do not use "feedback" in the standard sense, but rather "implex", as the driving forces are augmented by perceived load rather than by a separate regulating path. The result is highly compliant, animal-like machine motions that "negotiate" rather than "bully" their way through environments, resulting in minimal damage to both world and robot. We talk about these devices in the general sense because the precepts of their existence and subsequent design are based upon environmental macros, such as fluidity, turgidity, gravity, scale, materials strength, and many other factors. The power of biomorphic designs is that this information is used as the defining principles to shape appropriate survivor(s) for a particular environment. The machines that emerge are vastly different from any conventional robotic forms. We suspect, at least from the experimental evidence, that this technique embodies a new type of non-linear control paradigm, and at least an entirely new engineering discipline for the matching of competent machines to complex environments. Here, once the problems of existence are ratified, the devices can do unsupervised, long term work without human intervention (some devices have been in continuous operation for over 5 years). The potential for this control paradigm is vast, but it is far from linear, and requires integrated design attempts to pull a competent ability from the Nv nets. To this end, the use of this technology to "evolve" machines from a lesser to a higher operational state has resulted in not only a wide spectrum of devices, but even completely different "species" of creatures, all evolved from a primal "genotype;" the single "cell" creature known as Turbot 1.0. A further advantage is the speed at which this evolution has occurred, indicating that realworld Lamarckian evolution may match the success of many computer models yet seen. The diversity of this technique offers potential solutions to two main research fields, macro and micro robotics, and experimental work has been done to produce adept prototypes for both. The conclusions are that there may be some universal chaos-bounded concepts that bind survivor oriented designs, allowing for the creation and optimization of devices that can do work in any environment, under many situations, using chemically inert, and thereby relatively safe, control techniques (the idea of seeding a wheat field with pest-fighting silicon biomorphs to produce high yield, insecticide-free foodstuffs is an attractive example). Considering that biomorphs may last long enough to replace most forms of long-term damaging chemicals (i.e. pesticides, bleaches, medicines) the potential for the field really opens up. Deployed artificial chemicals perform a task in their immediate area of concentration, and then disperse into the environment where, after a time, they cannot be absorbed adequately. Biomorph machines, made from biodegradable silicon and trace elements, can be made gregarious so they do not of their own volition disperse, and can be absolutely controlled by conventional methods. Whether at micro or macro scales, these designs are not just capable, but competent. Furthermore, as they are self "programming" and non-reproductive, their behaviour is both contained and predictable. Nv biomorphs are something new with a demonstrable potential. Future work will concentrate on how this technology could fill in the cracks between science
fiction and reality by finding out what is feasible now, and how to logically proceed to marketable, capable machines. In the coming years, it is hoped to be possible to demonstrate real machines to assess feasibility for macro and nano robotic applications. Expansions of the fields of robo-biology, robo-morphology, and artificial ecologies will be studied and published, along with extensions of this field from self-repairing processors, new computational paradigms, and even nano-robotic surgeons-in-a-capsule. Biomorphics is new, but it is slowly gaining the maturity and acceptance necessary to become a valid work tool. The basic circuit for a MicroCore consists of a capacitor, a gate, and a resistor. The arrangement of the components is shown in the circuit below. This is called a NEURON.
Mouseover the NEURON circuit below to see how it works. The secret to the operation is this: The capacitor charges and the circuit changes state a short time later. The capacitor and resistor form an arrangement called a DELAY CIRCUIT. The LED has been added to the output of the gate so we can see how and when the circuit operates.
We are interested in the operation of the output AFTER the input has changed. The circuit is just like a spinal cord without anything attached. It is called an Nv circuit. You can't drive motors yet but until you understand the basic states of the MicroCore circuit, the operation of motors will make even less sense than the LEDs. Two Neuron circuits will work when connected head-to-tail and with the output of the second connected to the input of the first:
## The operation of the circuit is shown in this animation:
You will notice the capacitor charges via the 1M resistor as the input of the gate is a very high impedance. When the output of a gate goes low, the capacitor is discharged via a diode on the input line of the gate. This diode is a Schottky diode and is designed to prevent the input of the gate going below 0v. The change from one state to the other is almost instantaneous and only one LED is illuminated at a time. We have slowed down the animation to show the gates in action and the two LEDs appear to be off for an instant. This does not actually occur. If we extend this circuit to the 6 inverters of a hex Schmitt Trigger IC, and take the output of the last Nv circuit to the input of the first, we get a circulating pattern. The circuit will always start-up and each alternate LED will be illuminated.
When the circuit is connected to a Robot, the following effect will be produced:
This is known as the Saturated State. This means all Nv's are firing at Max rate. There is a Maximum of 3 Processes. A process is defined as one LED/Nv ON at any one time. Note: there are never two illuminated LEDs side-by-side at the same time. This is the Fermei exclusion feature of the MicroCore and an attribute shared by biological Nv nets. Saturation is the natural Power-on state. It is the crazy-go-nuts state for a Nv Net - like a bug that has had too much coffee. Saturation will occur when the MicroCore encounters a disruption in main power or when too much data is received from sensors or Nu/Nv nets. Fortunately it's easy to get a more stable useful state. Adding a Process Neutralization Circuit (PNC). Wire up a switch to short out the input bias resistor like this.....
Closing the switch destroys any re-circulating pulses. Hold it long enough and all processes are destroyed. This is called the Null State, Off State. This is a Nv net at rest.
Two Process State. By holding the switch closed for approx 2 seconds you should be able to
achieve this....
If all values are even, and no glitches or noise is received from the motor, the processes should fall 180 degrees out of phase with each other. This is seen when they appear to be running side by side with each other. Two processes are like Parkinson's disease. The Nv's are trying to act against each other and if they fall into this mode, you have "lock up." This state can be useful in some cases but in the case of a simple walker it's not much use.
Single Process State. By Holding down the PNC switch for another 3 seconds or so you should get
this. This is the stable One Process State that is the basis of most of the Nv walkers.
Adding a Process Initialise Circuit (PIN). By wiring up another switch from an input to Vcc (+ve) you can introduce processes to the MicroCore. By using the PNC and PIN switches you should be able to cycle through all the MicroCore's usable states.
Now build the circuit and play with it. In this discussion we are going to concentrate on a two motor four leg walker which although it is not the most flexible design, it is the easiest to build and has proven its reliability and capability in 35 existing machines.
THE MOTORS
This is probably the biggest consideration in a MicroCore Walker. The level of success you have with your walker is directly related to the type of motor used. The MicroCore itself gets an implicit feedback from the motors, this is what gives it the adaptability. What to look for in a Motor.... EFFICIENCY: This is REALLY important both from a power consumption standpoint (more-efficient motors require a smaller battery or main-storage electrolytic). High efficiency motors give you a better chance of success. You should look for a motor with at least a 35% efficiency rating, good cassette motors and pager motors typically fall into this range, Mabuchi hobby motors are way off (typically10%). Much higher efficiencies are possible (up to 88%) but this is usually found in expensive medical grade motors like "Escap" and "MicroMo." Keep your eyes open when perusing the surplus catalogues, these sometimes go on sale for as little as \$5. SIZE: For the most part, smaller is better but it's not as important as efficiency. As well, you may want to consider your own skill level. In the beginning, don't try to work with things that are really small. Make sure you have extra motors. They come in handy.
THE GEARS
You cant build a walker without them. A motor alone doesn't put out enough torque and usually runs too fast. What to look for in a gearbox... Efficiency/Size/Numbers: For all the same reasons as above. Compliance: This is really critical. You should be able to grasp the output shaft with a pair of pliers and turn it to have the gears spin the motor. If you can't make the motor spin, then you have a gear train that is too inefficient (most likely) or too high a ratio. Worm drives are also OUT, they only go one way (motor to gear and not gear to motor) and they tend to choke under high loads. Output RPM: The ideal is about 30 RPM @ 5V. Higher RPM means you probably won't have enough
torque (and if you do, the robot will jump around so fast it's hard to figure out what it's doing). A lower RPM means the machine may be moving too slow to be of use. A high ratio may mean the legs will bend under the torque.
## INTERFACING MOTOR AND GEARS
I strongly suggest you find a factory motor/gearbox combination. If you have to build your own then bear a couple of things in mind...... Direct gear contact: belt drives, friction drives, flexible shafts etc all have big problems as far as efficiency and compliance are concerned. Keep everything clean: Glue, solder flux and metal fillings are deadly enemies to gearboxes.
MATERIALS
Solder is our friend, only use materials that can be easily soldered. This will make it easy to build a frame. Welding wire or filler rod is the best. Copper clad carbon steel rod 1/6" to 3/32"diameter is cheap and available at any welding supply place. Any nice shinny option is High Nickel rod used for TIG welding cast iron but it's MUCH more expensive. Brass tube and wire found at most hobby shops is good too. I suggest a solder with an organic/water soluble flux, "Hydro X" by Multicore is my favourite.
## BASIC FRAME LAYOUT
The diagram below shows the basic layout. You need to keep the motors and output shafts lined up front to back and the front motor should be tilted at 30 degrees. This means the front motor will supply lift and push but we'll discuss that more later. You should mount the motors far enough apart to fit all your electronics including batteries (usually about 4").
## More on . . How to make a Robot Walk
This information has been adapted from an excellent site by A Miller.
Leg shape and configuration will vary greatly between machines. A few things to bear in mind are: Contact point: This is the most important aspect of leg design. The shape of the leg is less important than where it touches the ground. By placing your robot on a sheet of graph paper as shown above you can get symmetrical contact points. Width: Try to make the legs at least 2/3 the length of your robot, this of course will depend on the available torque. It has also been shown that making the back legs slightly wider than the front helps stability. Connection: Make sure your legs are connected with something structurally sound, krazy glue doesnt cut it. If you can't solder the legs directly to the output shaft then try and find some sort of locking ring or set- screw that will fit. Look for brass gears or pulleys that have their own set-screw and then you can solder the legs directly to the brass. Angle: By angling the legs slightly forward they will have the ability to "ratchet" over obstacles. Your legs will change shape several times before you are finished so it's best to make a set of "test" legs that are easily recoupable before you use the good materials. 12 or 14 gauge household copper wire makes effective re-configurable "Gumby" legs.
MAKING IT WALK:
We are going to make a minor detour here (you may have noticed we don't have the MicroCore connected to anything yet).
MicroCore/Leg Interfacing
This section covers connecting the Nv net to the Muscles - the motors. The following is a Low Current solution. High current motors and drives present a whole new set of problems that, trust me, you don't want to deal with right now. Besides, the MicroCore does its best when driving through a low gain system.
## Rethinking the Nv net:
By now you should have familiarized yourself with the basic 6 Nv MicroCore circuit. But a few calculations will show that with two motors and two directions, 4 Nv's would seem appropriate. For basic walking functionally 4 of the 6 available Nv's in a 74C14 chip (40106) are needed in a two motor walker (although 6 makes for a whole new set of behaviours), so rebuild the basic circuit to get this....
## The ALS245 Driver:
The 74ALS245 is an Octal bus transceiver designed for data transmission. We are going to use it to drive Motors. If you've chosen your Motor/Gear combination properly, this won't be a problem. If you look up a TTL data book you'll see the '245 has 8 bi-directional non-inverting amplifiers each capable of driving 50mA, a direction pin (Dir) for selecting the direction of the signal through the chip and an enable pin (E). If you tie the Dir pin to +ve and the E pin to -ve, you can essentially think of the chip as 8 active drivers going from left to right like this graphic.... A data book will show you more but for the purposes of driving your legs this'll do......
The ALS version has been found to be the best for current, feedback etc... CMOS such as HCT, HC, C will work but they loose a little in the feedback. They do consume less power so for solar applications they are the better choice..... If you are building a battery walker use ALS where-ever you can and contrary to what some people will tell you, they are still available.
## Ganging up for Current
If you've tested your motors you will probably find they need a little more than 50mA to do anything useful. And with a few calculations, you'll see there are 2 drivers available for each Nv, so gang them together like this....
The result is four drives capable of 100mA each. Good, but not great.... So if you need more, double your output by stacking a few chips like this...
The connection
So now you have a 4 Nv loop set up and a 4 channel driver set up. The next step is to glue them together and add the motors. Like this . . .
Power Up
Power on the whole thing, stabilize the loop (use the PIN and PNC buttons to get a "One Process"), and watch this animation:
If things aren't right, change the polarity of one of the motors and it should work.
## Tuning for a walking Gate...
Convergence, or the subtle art of falling over.....
This is where we try and make your pile of wire and batteries walk. This is kind of the counter-intuitive part of the whole thing. Most people are of the misunderstanding that in order for a Robot to remain on its feet it has to be balanced at all times. This is the train of thought that leads to so many 6 leg walkers; three legs is a stable platform from which to move your other three. Although this is in some cases successful, it makes for a robot that doesn't adapt well. Walking should be thought of as controlled falling. Static Balance is not the key, rather it's Dynamic Imbalance.....
There are several ways to make your walker stumble around. All of which are in some way related to the center of gravity of your bug. None is more important than the others nor is it possible to make it walk without adjusting all of them.
Nv Time value: You'll note that in its raw state the Nv is just an RC time value (but remember it is not
a constant, it adjusts itself according to load). So at a base level you can change the Duration of Rotation of each of the motors. One Meg Ohm is the default value, it is just a good starting point. You can adjust the values as high as 20 Meg Ohm or as low as a few k Ohms. By changing the duration of the leg's movement you change where it stops. If you think of the Walker as a first class lever (there are three classes of levels and the first class is the see-saw class) and the feet as fulcrums, you'll see that the position of the leg when it stops is crucial to which direction it tips.
Weight Redistribution: This one is obvious. By moving the components from front to back you
change the Center of gravity. The batteries are a good candidate for this since they are invariably the heaviest thing on the bug.
Leg Shape: This is the thing most likely to change dramatically. By bending the legs back and forth
you change the fulcrum point and thus the balance. Remember that contact point is more critical than leg shape.
What it should be achieving: The easiest way to get an idea about what the legs should be doing
is to step through the motions manually .. This is where the compliance thing comes in. If you cant
move the legs manually then the motors are not going to provide an appropriate feedback to the MicroCore By twisting the front leg CW approx 45 off center and the back leg approx 30 CW you have what well call start position. The Walker should now be balanced with its front left foot in the air and be just on the verge of tipping forward onto it. This is where the dynamic imbalance thing comes in, the bot Literally falls over onto its front foot. You should be able to tell if it is at the tip point by giving its butt a little tap, it should tip to the front foot and stay tipped. If it doesnt, try moving the battery forward or back in order to find the balance point (leg configuration will come later). The next step (literally) may or may not be obvious. By moving the back legs CCW you will move the fulcrum back, thus making the front tip down and the rear right foot raise off the ground. Keep rotating it until you have moved it 60 or so. The front two feet and the back left one should be flat on the ground and the rear right will be just off the ground towards the front of the walker. The walker has just completed a half a cycle (two Nv processes). Now, by moving the front leg CCW 90 you will provide lift and drive with the front left foot and raise the right side of the front into the air just to the tipping point. Now it's time for the rear to produce the drive forward that tips the walker forward and raises the rear left foot while stepping. This is done by rotating the rear CW 60.
Finale! One walking cycle and a full loop around the MicroCore.
Now go through the process a few more times manually with the power off, and familiarize yourself with what it should be doing. You may have to move the battery around and change the legs a little. But remember that Symmetry is VERY important. In order for both sides to be doing the same thing you need to have the feet contacting the same place with respect to the body (how the leg gets there isnt as important). The body should also sit flat when all feet are down and the legs are straight out. If the body leans, then one leg is shorter than the others and your bug will limp. On the next page we will show how to connect a 4 Nv MicroCore circuit to two motors via a 74HC245 octal buffer chip.
## To search more than 40 BEAM sites, click BEAM ONLINE.
In this section we show how to connect a 4 Nv MicroCore circuit to two motors via a 74HC245 octal buffer chip. This is not a project with a kit. It is a Feature Article with references to sites on the web. You will need to go to hobby supply shops for the components - especially the hardware items and motors. Try the LINKS page of BEAM ONLINE for suppliers. The circuit has been taken from Chiu-Yuan Fang's excellent site. On it he has produced a BEAM Robot called "Walker" and has a number of photographs to show how it has been put together. The following are thumbnails of some of these shots:
## The electronics, batteries and motors
The underside
View 2 more pages of the excellent pictures of Chiu-Yuan Fang's Walker Version 2: Page-1 photos and Page-2 photos The circuit diagram for the Walker is shown below. The diagram has been laid out to show how the signal progresses though the circuit. It circulates around the four Schmitt trigger gates, while at the same time driving two motors, in either forward or reverse direction. The circuit turns on with a long delay via the gate between pins 1&2. Two 1M trim-pots provide "straight-line" motion.
The 74HC245 octal buffer (driver) chip has a 50mA capability per output and two outputs are joined in parallel in the diagram above to get 100mA per line. If you require more current, the following transistor H-Bridge can be used:
The circuit above will provide up to about 500mA drive-current for each motor and this is needed when a motor has to be started under load. As soon as the motor "starts", the current will drop, but it's the ability of the circuit to provide a high starting-current that prevents a "stalled condition." The LEDs provide indicators to show the operation of the circuit. | 6,281 | 29,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-34 | latest | en | 0.922377 |
https://socratic.org/questions/what-volume-of-1-0-m-cuso-4-would-you-need-to-make-750-l-of-a-20-m-solution | 1,576,071,591,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540531917.10/warc/CC-MAIN-20191211131640-20191211155640-00010.warc.gz | 539,804,798 | 6,138 | # What volume of 1.0 M CuSO_4 would you need to make .750 L of a .20 M solution?
May 13, 2016
$0.150 \cdot L$ are required.
#### Explanation:
$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$.
And thus $\text{Moles of solute } \left(n\right)$ $=$ $\text{Concentration "(mol*L^-1)xx"volume } \left(L\right)$
In the required volume there are $0.750 \cdot L \times 0.20 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.150 \cdot m o l$.
Thus, in the starting volume, we need $0.150 \cdot L$, which contains $0.150 \cdot m o l$. This volume is then diluted to $0.750 \cdot L$. | 205 | 589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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} | 3.890625 | 4 | CC-MAIN-2019-51 | longest | en | 0.724329 |
cn.antosh.in | 1,590,814,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00556.warc.gz | 309,239,509 | 36,325 | Sunday, January 14, 2018
IP addresses, when started a few decades ago, used the concept of classes. This architecture is called classful addressing.
In classful addressing, the IP address space is divided into five classes: A, B, C, D, and E. Each class occupies some part of the whole address space.
Recognizing Classes
We can find the class of an address when the address is given either in binary or dotted-decimal notation. In the binary notation, the first few bits can immediately tell us the class of the address; in the dotted-decimal notation, the value of the first byte can give the class of an address
Netid and Hostid
In classful addressing, an IP address in classes A, B, and C is divided into netid and hostid. These parts are of varying lengths, depending on the class of the address. The classes D and E are not divided into netid and hostid.
In class A, 1 byte defines the netid and 3 bytes define the hostid. In class B, 2 bytes define the netid and 2 bytes define the hostid. In class C, 3 bytes define the netid and 1 byte defines the hostid.
Class A
Since only 1 byte in class A defines the netid and the leftmost bit should be 0, the next 7 bits can be changed to find the number of blocks in this class. Therefore, class A is divided into 27 = 128 blocks that can be assigned to 128 organizations. Each organisation can have 224= 16,777,216 addresses, which means the organization should be a really large one to use all these addresses.
Class B
Since 2 bytes in class B define the class and the two leftmost bit should be 10 (fixed), the next 14 bits can be changed to find the number of blocks in this class. Therefore, class B is divided into 214= 16,384 blocks that can be assigned to 16,384 organizations, each block in this class contains 216= 65,536 addresses.
Class C
Since 3 bytes in class C define the class and the three leftmost bits should be 110 (fixed), the next 21 bits can be changed to find the number of blocks in this class. Therefore, class C is divided into 221= 2,097,152 blocks, in which each block contains 256 addresses, that can be assigned to 2,097,152 organizations. Each block contains 28=256 addresses.
Class D
There is just one block of class D addresses. It is designed for multicasting.
Each address in this class is used to define one group of hosts on the Internet. When a group is assigned an address in this class, every host that is a member | 587 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-24 | longest | en | 0.921462 |
http://www.ask.com/web?qsrc=6&o=102140&oo=102140&l=dir&gc=1&q=Calculating+Cubic+Feet | 1,464,511,082,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278417.79/warc/CC-MAIN-20160524002118-00022-ip-10-185-217-139.ec2.internal.warc.gz | 360,404,600 | 17,166 | Web Results
## Volume Calculator for Cubic Feet | Marjam
www.marjam.com/calculators/volume-calculator-cubic-feet
Volume Calculator for Cubic Feet. Simply enter the Length, Width and Height values and view the calculated total in Cubic feet. Calculator. Enter Length (Ft)
## 4 Ways to Find Cubic Feet - wikiHow
www.wikihow.com/Find-Cubic-Feet
Doing this requires "cubic measurement," which is just another way of ... If it doesn't measure well into whole feet, feel free to measure it in inches for now.
## Cubic Feet Calculator, Inches to Cubic Feet(CFT), volume and ...
www.ginifab.com/feeds/cbm/cbm_caluclator_inch_lb.html
Cargo is commonly calculated using a combination of volume (total space in cubic feet or meters) and payload (total weight). A cubic foot is the space occupied ...
Apr 23, 2012 ... This guide shows you How To Calculate Cubic Feet Watch This and Other Related films here: ...
## How to Calculate Cubic Yards Calculator | Today's Homeowner
www.todayshomeowner.com/cubic-yard-calculator/
27 cubic feet in one cubic yard (3' x 3' x 3'); 46,656 cubic inches in one cubic yard ... calculator below, then click “calculate” to find the number of cubic yards.
## Volume calculator - Math Central
mathcentral.uregina.ca/QQ/database/QQ.09.06/debbie1.html
How many cubic yards of cement do we need? Thanks ... 24 ft is 24/3 = 8 yards 30 ft is ... You can calculate this and similar volumes using our volume calculator.
## Cubic Feet conversion calculators, tables and forumas
www.metric-conversions.org/volume/cubic-feet-conversion.htm
There is no universally agreed symbol for the cubic foot/feet. ... To calculate the volume of a given item or space in cubic feet, measure the length, width and ...
## Cubic Inches to Cubic Feet conversion
www.metric-conversions.org/volume/cubic-inches-to-cubic-feet.htm
Cubic Inches to Cubic Feet (in³ to ft³) conversion calculator for Volume conversions with additional tables and formulas.
## How It Works - Stone and Mulch Calculators - Landscape Calculator
www.landscapecalculator.com/how-it-works/stone-mulch.aspx
So for our example: 0.25 foot depth times 500 square feet equals 125 cubic feet. If you are calculating 0.5, 2, or 3, cubic foot bags, simply divide 125 cubic feet by ...
## Mulch Calculator | How much mulch do I need?
www.landscapecalculator.com/calculators/mulch
Area in Square Feet: Depth in Inches. One Inch, Two Inches, Three Inches, Four Inches. Outputs. Cubic Yards of Mulch: Number of 2 Cubic Foot Bags: Number ...
A cubic foot is a unit of volume defined as the volume of a cube having sides one foot in lenght. It is equal to a little over 28 liters.
### Cubic Ft Calculator by Inches - CASCOM
www.cascom.army.mil
Volume Calculator. for. Cubic Feet in inches. Enter Length (In) Enter Width (In) Enter Height (In). Total Cubic Feet.
### Cubic Yards Calculator - Calculator Soup
www.calculatorsoup.com
Calculate cubic yards, cubic feet or cubic meters for landscape material, mulch, land fill, gravel, cement, sand, containers, etc. Enter measurements in US or ...
### FAQ: How do I convert Square Feet to Cubic Feet - Online Conversion
www.onlineconversion.com
Since cubic feet has an extra dimension, you could fit an unlimited number of square feet ... Notice that I convert 3 inches to 0.25 feet before using the formula. | 836 | 3,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | longest | en | 0.771807 |
https://research-portal.uea.ac.uk/en/publications/diffraction-of-flexural-gravity-waves-by-a-vertical-cylinder-of-n | 1,722,705,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00368.warc.gz | 388,692,731 | 11,457 | # Diffraction of flexural-gravity waves by a vertical cylinder of non-circular cross section
N. B. Dişibüyük, A. A. Korobkin, O. Yılmaz
Research output: Contribution to journalArticlepeer-review
7 Citations (Scopus)
## Abstract
The linear three-dimensional problem of flexural-gravity wave (hydro-elastic wave) diffraction by a vertical cylinder of an arbitrary smooth cross section is studied using an asymptotic approach combined with the vertical mode method for water of finite depth. The surface of the water is covered by an infinite, continuous elastic ice plate. The rigid cylinder extends from the sea bottom to the ice surface. The ice plate is frozen to the cylinder. The ice deflection is described by the equation of a thin elastic plate of constant thickness with clamped edge conditions at the cylinder. The flow under the ice is described by the linear theory of potential flows. The coupled problem of wave diffraction is solved in two steps. First, the problem is solved without evanescent waves similar to the problem of water waves diffracted by a vertical cylinder. This solution does not satisfy the edge conditions. Second, a radiation problem with a prescribed motion of the ice plate edge is solved by the vertical mode method. The sum of these two solutions solve the original problem. Both solutions are obtained by an asymptotic method with a small parameter quantifying a small deviation of the cylinder cross section from a circular one. Third-order asymptotic solutions are obtained by solving a set of two-dimensional boundary problems for Helmholtz equations in the exterior of a circle. Strains along the edge, where the ice plate is frozen to the cylinder, are investigated for nearly square and elliptic cross sections of the vertical cylinders depending on the characteristics of ice and incident wave. The strains are shown to be highest in the places of high curvatures of the cross sections. The derived asymptotic formulae can be used in design of vertical columns in ice. They directly relate the strains in ice plate to the shape of the column.
Original language English 102234 Applied Ocean Research 101 12 Jun 2020 https://doi.org/10.1016/j.apor.2020.102234 Published - Aug 2020
## Keywords
• Asymptotic approach
• Clamped edge conditions
• Hydro-elastic waves
• Ice cover
• Non-circular vertical cylinder
• Vertical mode method | 501 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.927281 |
http://onbiostatistics.blogspot.com/2009/04/least-squares-means-marginal-means-vs.html | 1,548,119,507,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00299.warc.gz | 169,171,480 | 18,823 | ## Sunday, April 05, 2009
### Least squares means (marginal means) vs. means
If you work with SAS, you probably heard and used the term 'least squares means' very often. Least squares means (LS Means) are actually a sort of SAS jargon. Least square means is actually referred to as marginal means (or sometimes EMM - estimated marginal means). In an analysis of covariance model, they are the group means after having controlled for a covariate (i.e. holding it constant at some typical value of the
covariate, such as its mean value).
I often find that it is neccessary to use a very simple example to illulatrate the difference between LS Means and Means to my non-statistician colleagues. I made up the data in Table 1 above. There are two treatment groups (treatment A and treatment B) that are measured at two centers (Center 1 and Center 2).
The mean value for Treatment A is simply the summation of all measures divided by the total number of observations (Mean for treatment A = 24/5 = 4.8); similarly the Mean for treatment B = 26/5 = 5.2. Mean for treatmeng A > Mean for treatment B.
Table 2 shows the calculation of least squares means. First step is to calculate the means for each cell of treatment and center combination. The mean 9/3=3 for treatment A and center 1 combination; 7.5 for treatment A and center 2 combination; 5.5 for treatment B and center 1 combination; and 5 for treatment B and center 2 combination.
After the mean for each cell is calculated, the least squares means are simply the average of these means. For treatment A, the LS mean is (3+7.5)/2 = 5.25; for treatment B, it is (5.5+5)/2=5.25. The LS Mean for both treatment groups are identical.
It is easy to show the simple calculation of means and LS means in the above table with two factors. In clinical trials, the statistical model often needs to be adjusted for multiple factors including both categorical (treatment, center, gender) and continuous covariates (baseline measures). The calculation of LS mean is not easy to demonstrate. However, the LS mean should be used when the inferential comparison needs to be made. Typically, the means and LS means should point to the same direction (while with different values) for treatment comparison. Occasionally, they could point to the different directions (treatment A better than treatment B according to mean values; treatment B better than treatment A according to LS Mean).
SAS procedure GLM has a nice discussion about the comparison of Least Square Means vs. Means. A small article "Means vs LS Means and Type I vs Type III Sum of Squares"by Dan may also help.
yun said...
Yes, you are right on lsmeans and means. Actually for balanced design, it the final data strueture is balanced, then mean=lsmean. But generally they differ.
Anonymous said...
Thank you very much for posting this blog. That was exactly the explanation I needed.
Anonymous said...
SAS folk have never understood experimental design. They've always had a bias of coming from the regression side of the coin.
These terms are unnecessary, and as you state, exist only in the minds of SAS.
What you describe is the addition of a second "blocking variable" in a design. You could describe it as a factor in a 2-way ANOVA, or control it out with ANCOVA.
But to make two different terms for something that has already existed for a hundred years or so, is SAS being SAS.
Furthermore, when I run a posthoc in JMP for a one-way ANOVA with more than 2 levels, "SAS" gives me LS Means as the group means, just because there's unequal 'n'. This is incorrect. I have to go through and generate descriptives to get the actual group means.
Anonymous said...
Brilliant explanation.
BAten said...
Thanks for this example. Do you have any showing when one is able to calculate a mean, but not a LSM?
Thx so much! This is exactly what I need!
Anonymous said...
Look like simple. How about for regression model? It seems lsmeans is defined only for effects not for covariates? It is right?
Thanks.
ss said...
got it but u mean to say that , while calculating lsmeans we r considering center effect ...
Anonymous said...
Thank you. Good explanation!
Joe Locascio said...
Yes, SAS's "LSMeans" are means adjusted for the covariate(s). In an imbalanced factorial anova design, the factors are essentially confounded "covariates" and the LSmeans are adjusting for that, giving you an average of cell averages, rather than just the marginal means blind to (and confounded with the other factor(s)). (This can be viewed from a regression/general linear model perspective, with categorical factors being dummy coded). Neither kind of means are right or wrong - they answer different questions. I typically request both in SAS. You can come up with all kinds of combinations of means, covariate means, and correlations of covariates with the dependent variable, resulting in covariate adjusted means being in the same or opposite ordinal relation as the raw descriptive means, or where the covariate adjusted means don't change the descriptive means at all. You can map these things graphically with little group ellipses representing scatterplots and their respective regression lines.
Anonymous said...
Great explanation. Clear and incorporates the use of a familiar concept, that most folks understand - the calculation of a mean score. Linking a new concept to an familiar concept is a great way to teach. Thanks!
Anonymous said...
thanks so much, made it so easy to understand!
Anonymous said...
Thank you for this explanation. Simple and easy to understand!
Anonymous said...
Thanks. It helped me a lot.
Anonymous said...
Thank you so much! It really helped me.
Angus McLean said...
Least square means is used in SAS for bioequivalence parameters such as peak drug concentrations (Cmax).
Can you outline in simple terms how it is calculated? Can I do the calculation in Excel? I know that for a balanced study with all subjects completing it is the geometric mean, but suppose one subject drops out.
Angus McLean said...
Can you outline for me in the most simple terms how the calculation for LS means is done in SAS as applies to bioequivalence parameters such as Cmax (peak drug concentration in plasma). The design to consider is the usual cross over design.
Anonymous said...
To Angus's question:
Please see a separate article "Cookbook SAS Codes for Bioequivalence Test in 2x2x2 Crossover Design"
http://onbiostatistics.blogspot.com/2012/04/cookbook-sas-codes-for-bioequivalence.html
Anonymous said...
great explanation - thanks
Anonymous said...
Can anyone explain what's the difference between fixed effects estimates and lsmeans in SAS output? In SAS, the highest level is the reference level for fixed effects estimates. It seems that the difference of the lsmeans estimates with the highest level(the same as the fixed effects) lsmeans is the fixed effects estimates. Is this right and why?
Anonymous said...
Thank you for your explanation! However, I still have a question.
If I want to compare the efficacy of treatment A and treatment B, which statistic I should choose: the mean or the LS-mean?
Web blog from Dr. Deng said...
depending on which statistical method you are using to do the comparison. For t-test, you will simply compare the means. for analysis of variance or analysis of covariance, you will likely compare the LS Mean.
Anonymous said...
Your explanation about the LS-means was incorrect as it does not account for the sample size (n) in each cell when you took the simple average of the two centers in Step 2 (Table 2). For example, if n=10000 in the cell Center_1/Treatment_A with each response=3, then the LS-mean for treatment A will be close to 3 as the data in the cell Center_2/Treatment_A are almost negligible. But it would still be 5.5 based on your method.
Anonymous said...
Take your example. Let the variables be TREATMNT, CENTER and VAL. In SAS, if the statements are "MODEL VAL=TREATMNT CENTER TREATMNT*CENTER; LSMEANS TREATMNT;", then the LSMEANs are 5.25, 5.25.
But if the model statement is "MODEL VAL=TREATMNT CENTER;", then the LSMEANs for the variable TREATMNT are 5 and 5. | 1,832 | 8,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-04 | latest | en | 0.952677 |
https://math.answers.com/Q/How_much_is_3_plus_4_minus_one | 1,701,859,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00560.warc.gz | 437,506,440 | 44,756 | 0
# How much is 3 plus 4 minus one?
Updated: 8/20/2019
Wiki User
11y ago
3+4-1=6
that is 3+4=7
7-1=6 | 54 | 107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-50 | latest | en | 0.91823 |
http://www.solutioninn.com/as-mentioned-in-exercise-733-among-college-students-who-hold | 1,498,659,264,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323682.21/warc/CC-MAIN-20170628134734-20170628154734-00654.warc.gz | 656,319,130 | 7,395 | # Question: As mentioned in Exercise 7 33 among college students who hold
As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a mean of 20.20 hours and a standard deviation of 2.60 hours. Find the probability that the average time spent working per week for 18 randomly selected college students who hold part-time jobs during the school year is
a. Not within 1 hour of the population mean
b. 20 to 20.50 hours
c. At least 22 hours
d. No more than 21 hours
View Solution:
Sales0
Views264 | 150 | 635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-26 | longest | en | 0.95618 |
https://www.shaalaa.com/question-bank-solutions/all-kings-queens-aces-are-removed-pack-52-cards-remaining-cards-are-well-shuffled-then-card-drawn-it-find-probability-that-drawn-card-red-card-basic-ideas-of-probability_74765 | 1,618,893,633,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039375537.73/warc/CC-MAIN-20210420025739-20210420055739-00018.warc.gz | 1,104,777,946 | 9,112 | # All Kings, Queens and Aces Are Removed from a Pack of 52 Cards. the Remaining Cards Are Well-shuffled and Then a Card is Drawn from It. Find the Probability that the Drawn Card is a Red Card. - Mathematics
Short Note
All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is a red card.
#### Solution
There are 4 kings, 4 queens and 4 aces. These are removed.
Thus, remaining number of cards = 52 − 4 − 4 − 4 = 40.
Number of red cards now = 26 − 6 = 20.
∴ P(getting a red card) =("Number of favourable outcomes")/"Number of all possible outcomes"
= 20/40 = 1/2
Thus, the probability that the drawn card is a red card is 1/2.
Is there an error in this question or solution?
#### APPEARS IN
RS Aggarwal Secondary School Class 10 Maths
Chapter 15 Probability
Exercise 15B | Q 23.2 | Page 698 | 260 | 924 | {"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} | 4.40625 | 4 | CC-MAIN-2021-17 | latest | en | 0.925375 |
https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/18270?l=1 | 1,506,149,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00511.warc.gz | 670,059,815 | 23,069 | ## Re: [PrimeNumbers] Re: Prime definitions that exclude 2
Expand Messages
• I can also give you a definition that will include 2 as a prime: A prime is a positive integer that has only two divisors, regardless whether it has smaller
Message 1 of 6 , Aug 7, 2006
I can also give you a definition that will include 2
as a prime:
A prime is a positive integer that has only two
divisors, regardless whether it has smaller number
that can be tested for divisibility. In this case,
even though 2 is much different from other odd primes,
we artificially treated it as the same as other
primes.
I am not a math specialist and so I dont quite follow
your definition. But to have a way of difining 2 to
be a prime cannot refute the fact that 2 is very
different from all other primes. We can easily
exclude 2 as a prime but we cannot do the same with
317. So there is no objective standard on 2 as there
is on 317. Thus ultimately, it is subjective human
conventions that decides wether 2 is a prime. Humans
should just be honest and say flatly that 2 is a prime
not because it is like 317 but because we want it to
be. We should just be as honest as we treated 1. We
say 1 is not a prime not because it is not but because
we dont want it to be.
Below is a honest quot from math world on prime
numbers.
As more simply noted by Derbyshire (2004, p. 33), "2
pays its way [as a prime] on balance; 1 doesn't."
Derbyshire, J. Prime Obsession: Bernhard Riemann and
the Greatest Unsolved Problem in Mathematics. New
York: Penguin, 2004.
So, for anyone to insist that 2 is a prime based on
objective truth, the same as 317, is not really being
honest. To attemp to justify our artificical
conventions by cleverly formulating a seemingly
objective definition that does include 2 is not being
totally honest. The honest thing to do is to say that
2 could easily be a prime or a non-prime, but it suits
our purpose better if it is treated as a prime. Just
like 1 could easily be a prime or non-prime, but it
suits our purpose better if it is treated as a
non-prime. But the purpose of today may not be
relevant to the objective truth.
--- jbrennen <jb@...> wrote:
> --- shuangtheman wrote:
> >
> > Can any one offer a list of definitions that would
> include
> > 2 as a prime?
>
> I already did, but let me rephrase it in a more
> concise way...
>
>
***************************************************************
> A natural number X is prime if the set of natural
> numbers not
> divisible by X is non-empty and is closed under
> multiplication.
>
***************************************************************
>
> So it says two things: divisibility by X is not an
> inherent
> property (so 1 is excluded), and divisibility by X
> cannot be
> "created out of non-divisibility" -- the only way to
> have a
> product divisible by X is to have one of the
> multiplicands
> divisible by X.
>
> This precisely describes the prime numbers. Explain
> to me how
> this definition is faulty for the number 2, but not
> for any
> odd primes.
>
>
>
>
>
>
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• Following a list of primes that starts with 2, the author Derbyshire wrote: “At this point, someone usually objects that 1 is not included in this or any
Message 2 of 6 , Aug 7, 2006
Following a list of primes that starts with 2, the
author Derbyshire wrote: At this point, someone
usually objects that 1 is not included in this or any
other list of primes. It fits the definition, doesnt
it? Well, yes, strictly speaking, it does, and if you
want to be a barrack-rood lawyer about it, you can
write in a 1 at the start of the list for your own
satisfacion. Including 1 in the primes, however, is a
major nuisance, and modern mathematicians just dont,
by common agreement. (The last mathematician of any
importance who did seems to have been Henri Lebesgue,
in 1899.) Even including 2 is a nuisance, actually.
Countless theorems begin with, Let p be any odd
prime
. However, 2 pays its way on balance; 1
doesnt, so we just leave it out.
from Derbyshire, J. Prime Obsession: Bernhard Riemann
and the Greatest Unsolved Problem in Mathematics. New
York: Penguin, 2004. Page 33.
So the above quote proves my point that 2 is treated
as a prime today the same way as 1 is not, by human
agreement rather than objective truth. 2 as prime
serves us better and so lets call it a prime. 1 as a
prime does serve us as well so lets ignore it.
Clearly serving us is not the same as serving God. If
God is uniqueness there is no way He would consider
uniqueness and in turn the related concept of oneness
to be anything other than a prime. There is no way He
would treat a follower of uniqueness/oneness, such as
2 following 1 or even following odd, to be a prime.
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• ... It does not. You re right that it s human agreement, but any selection of axioms (and postulates) and inference rules, for example is a human agreement. To
Message 3 of 6 , Aug 7, 2006
--- Shi Huang <shuangtheman@...> wrote:
> Following a list of primes that starts with 2, the
> author Derbyshire wrote: At this point, someone
> usually objects that 1 is not included in this or any
> other list of primes. It fits the definition, doesnt
> it? Well, yes, strictly speaking, it does, and if you
> want to be a barrack-rood lawyer about it, you can
> write in a 1 at the start of the list for your own
> satisfacion. Including 1 in the primes, however, is a
> major nuisance, and modern mathematicians just dont,
> by common agreement. (The last mathematician of any
> importance who did seems to have been Henri Lebesgue,
> in 1899.) Even including 2 is a nuisance, actually.
> Countless theorems begin with, Let p be any odd
> prime
. However, 2 pays its way on balance; 1
> doesnt, so we just leave it out.
>
> from Derbyshire, J. Prime Obsession: Bernhard Riemann
> and the Greatest Unsolved Problem in Mathematics. New
> York: Penguin, 2004. Page 33.
>
> So the above quote proves my point that 2 is treated
> as a prime today the same way as 1 is not, by human
> agreement rather than objective truth.
It does not.
You're right that it's human agreement, but any selection
of axioms (and postulates) and inference rules, for example
is a human agreement. To think otherwise shows a complete
disregard for how the foundations of modern mathematics are
defined.
However, there are fundamental differences between the issues
that '1' causes, and the issues that '2' causes which make them
not comparable.
> 2 as prime
> serves us better and so lets call it a prime. 1 as a
> prime does serve us as well so lets ignore it.
I assume that should read "doesn't". But it's still a gross
misrepresentation of the truth. Having a unit as a prime messes
up /almost everything/.
> Clearly serving us is not the same as serving God.
Obviously. Ockham's razor indicates that there's no need to have
brought up the latter at all, and persuades us that the simplest
set of rules is usually the better one. Having a unit as a prime
complicates almost every otherwise simple rule we have, and
therefore is unwarranted, and unwanted.
It appears that you don't understand _why_ 2 causes the problems
that it does in the situations where one needs to say "an odd prime".
It's usually not its primeness that causes the problem, but it's
_size_. It is the only prime for which x == -x (mod p) for all x,
which messes up assumptions about orders (see carmichael's lambda,
for example). Lack of divisibility by 2 messes things up too,
but lack of divisibility by 3 messes things up in elliptic curves
over GF(3^n), and there's no temptation to not call 3 a prime -
it wasn't the _primeness_ of 2 that was the problem, merely the fact
that 2 occured as a multiplier that one wanted to invert.
Coupled to this, it appears that you don't understand why having
1 causes the problems that it does too. The fact that it is a unit
messes up practically everything it touches.
The reason why units have been isolated in their own special
category is for a very simple reason - they behave fudamentally
differently from the other members of the multiplicative group.
To group them all together and then to have to separate them again
almost everywhere provides the mathematician with no perceptable
gain, and plenty of pain.
I notice that precisely _no_ sources for primes to concretely be
defined by a mathematician to exclude 2 have been cited yet, which
reinforces my previous post on the subject.
Phil
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• ... How is 2 expressible as an even number of sums of a single number that isn t 1 or 2? It can t. So your definition of prime that excludes 2 includes 2. Does
Message 4 of 6 , Aug 7, 2006
--- shuangtheman <shuangtheman@...> wrote:
> 1. A prime is a positive integer that cannot be expressed by the even number
> of sums of any single number except 1 and itself.
How is 2 expressible as an even number of sums of a single number that isn't 1
or 2? It can't. So your definition of prime that excludes 2 includes 2.
Does this depend on what the meaning on of 'is' is, or something?
I think the latter, and I recommend retreating.
Phil
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Your message has been successfully submitted and would be delivered to recipients shortly. | 2,567 | 10,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-39 | latest | en | 0.972629 |
http://www.romannumerals.co/date/april-3-2002-in-roman-numerals/ | 1,585,460,920,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00060.warc.gz | 303,600,180 | 14,036 | ## What is April 3, 2002 in Roman Numerals?
### A: IV・III・MMII
Your question is, "What is April 3, 2002 in Roman Numerals?", the answer is 'IV・III・MMII'. Here we will explain how to convert and write the date 4/3/2002 with the correct Roman numeral figures.
April 3, 2002 =
IV・III・MMII
## How is April 3, 2002 converted to Roman numerals?
To convert April 3, 2002 to Roman Numerals the conversion involves you to split the date into place values (ones, tens, hundreds, thousands), like this:
MonthDayYear
DateApril32002
Number Place Values432000 + 2
Numeral Place ValuesIVIIIMM + II
=IVIIIMMII
## How do you write April 3, 2002 in Roman numerals?
To write April 3, 2002 in Roman numerals correctly, combine the converted values together. The highest numerals must always precede the lowest numerals for each date element individually, and in order of precedence to give you the correct written date combination of Month, Day and Year, like this:
IV・III・MMII
## More from Roman Numerals.co
April 4, 2002
Learn how April 4, 2002 is translated to Roman numerals.
Dates in Roman Numbers
Select another date to convert in to Roman Numbers | 316 | 1,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.81243 |
https://math.answers.com/movies-and-television/Does_two_one_thirds_equal_one_half | 1,713,731,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00738.warc.gz | 331,122,093 | 48,108 | 0
# Does two one thirds equal one half?
Updated: 10/26/2022
Wiki User
13y ago
No it doesn't it equals two thirds which is greater than one half.
Wiki User
13y ago
Wiki User
12y ago
yes
Earn +20 pts
Q: Does two one thirds equal one half?
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Related questions
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11/6
### What is one-half of two-thirds?
One-half of two-thirds is one-third.
### Does one third equal one half?
No, one third is in fact less than half. For example, if you had two pizzas and cut one pizza into two equal peices, that would have halves. Then, if you cut the other into three equal peices, you would have thirds. The thirds would be smaller than the halves.
### What does one half plus two thirds equal?
seven sixths so 1whole and a sixth
no two thirds is
### What is four sixths rounded to the nearest half?
Four sixths, which is equal to two thirds, if rounded to the nearest half is approximately one half. In terms of a decimal analysis, two thirds is very close to .67, and .67 - .5 = .17, so two thirds is just .17 more than a half, but if we compare it to the number one, two thirds is fully a third (or .333...) less than one. And that is more than .17 so two thirds is closer to a half than it is to one.
### What plus what equals two-thirds?
1 third plus one third =============== There are many other fraction combinations which equal two-thirds. One, for example, is one-half plus one-sixth.
### What is two thirds multiplied by one half?
One half multiplied by two-thirds is two-sixths and if you reduce it would be one-third.
### What is six thirds minus one half?
Six thirds is two. Two minus one half is one and a half. 2 - 0.5 - 1.5
### What is one and one half divided by two thirds?
One and one half divided by two thirds = 9/4, or 2 and 1/4
### Two and two thirds plus two and two thirds equal?
five and one third | 478 | 1,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.980899 |
https://www.wazeesupperclub.com/how-do-you-write-an-equation-in-general-form/ | 1,653,791,170,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00434.warc.gz | 1,234,176,841 | 11,503 | # How do you write an equation in general form?
## How do you write an equation in general form?
If we wanted to write an equation with a positive value first, we could write 4x – 3y = – 24. To graph an equation in general linear form, compute the x-intercept (a, 0) and the y-intercept (0, b): a = and b = . Then connect the intercepts with a straight line and extend the line on both sides.
## How do you write 0.0021 in scientific notation?
1 Answer. Kalyanam S. ⇒2.1⋅10−3 is the scientific notation.
## How do I write an equation in slope intercept form?
Step 1: Locate the y-intercept. Step 2: Locate another point that lies on the line. Step 3: Calculate the slope from the y-intercept to the second point. Step 4: Write an equation in slope intercept form given the slope and y-intercept.
## How do you write an equation in slope intercept form given two points?
Once you know m and b, you can write the equation of the line.
1. Step 1: Find the slope (m) The slope of the line through two points (x1,y1) and (x2,y2) can be found by using the formula below.
2. Step 2: Find the y-intercept (b)
3. Step 3: Write the equation in slope-intercept form (y = mx + b)
4. Step 4: Check Your Equation.
## How do you write a number in scientific notation on a calculator?
How to Convert a Number to Scientific Notation
1. Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point.
2. Count how many places you moved the decimal point.
3. If you moved the decimal to the left b is positive.
## How do you know if a function is in standard form?
The quadratic function f(x) = a(x – h)2 + k, a not equal to zero, is said to be in standard form. If a is positive, the graph opens upward, and if a is negative, then it opens downward. The line of symmetry is the vertical line x = h, and the vertex is the point (h,k).
## Is general and standard form the same?
General form will typically be in the form “y=mx+b”. M is the slope of the graph, x is the unknown, and b is the y-Intercept. this means that the product of a number and x added to b will equal y. Standard form will always be “x+y= a number value.” so, let’s get some practice.
## How do you solve system of equations?
Here’s how it goes:
1. Step 1: Solve one of the equations for one of the variables. Let’s solve the first equation for y:
2. Step 2: Substitute that equation into the other equation, and solve for x.
3. Step 3: Substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.
## How do you write in scientific notation?
A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, can be written in scientific notation as 6.5 ✕ 10^8.
## How do you write an equation in point slope form on a calculator?
The point slope form equation is: y – y1 = m * (x – x1) , where: x1 , y1 are the coordinates of a point, and. m is the slope.
## How do u write 0.0003 in scientific notation?
To write the number 0.0003 in scientific notation, 0.0003 is repeatedly multiplied by 10 until the first non-zero number (in this case a 3) is just to the left of the decimal place: Step 1 – multiply by 10: 0.0003 x 10 = 0.003.
## How do you convert an equation to a line?
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## How do you solve standard form?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3.
## How do you write 8 in scientific notation?
Why is 8 written as 8 x 100 in scientific notation?
## How do you graph a standard form equation?
First, find the intercepts by setting y and then x equal to zero. This is pretty straightforward since the line is already in standard form. Plot the x and y-intercepts, which in this case is (9, 0) and (0, 6) and draw the line on the graph paper!
## How do you graph a slope equation?
Graphing a Linear Equation Using Slope Intercept Form
1. y = mx + b.
2. Graph the equation: y = 2x + 4.
3. Slope = 2 or 2/1.
4. Y-intercept = 4 or (0,4)
5. Step 1: Plot the y-intercept on your graph.
6. Step 2: From the y-intercept (0,4) use the slope to plot your next point.
7. The slope is 2, so you will rise 2 (up) and run 1 (to the right). | 1,195 | 4,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2022-21 | latest | en | 0.897518 |
http://forums.wolfram.com/mathgroup/archive/2009/Feb/msg00405.html | 1,590,537,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00405.warc.gz | 50,366,423 | 7,965 | Re: Shortest Path Problem
• To: mathgroup at smc.vnet.net
• Subject: [mg96395] Re: Shortest Path Problem
• From: dh <dh at metrohm.com>
• Date: Thu, 12 Feb 2009 06:43:29 -0500 (EST)
• References: <gmu8kl\$ggb\$1@smc.vnet.net>
```
Hi Antonio,
assume we have a weight matrix w. From w we create a graph: g. We then
serach the shortest path from vertex 1 to 4. Finally we display the
costs of all shortest paths:
w = {{3, 7, 2, 6}, {7, 9, 3, 1}, {3, 2, 7, 6}, {9, 3, 6, 10}};
g = FromAdjacencyMatrix[w, EdgeWeight, Type -> Directed];
ShortestPath[g, 1, 4 ]
hope this helps, Daniel
ntonio wrote:
> Dear Mathematica Users,
>
> I am not familiar with Graph theory and hope that some Mathematica
> users might help me. I am having a Shortest path problem that I hope
> to solve using Mathematica.
>
> My input is a Grid defind as,
>
> MyGrid = Table[RandomInteger[{1, 5}], {i, 1, 10}, {j, 1, 10}]
>
> in this 10x10 grid i'd like the shortest path from point A, let's say
> MyGrid[[10, 10]] to point B MyGrid[[1, 1]]. For every point inside
> this square grid I have 8 possible directions or lines
> (up,down,left,right and the 4 diagonals). The weight of each step is
> given inside the MyGrid cell, i.e. let MyGrid[[2, 3]]=1 and MyGrid[[2,
> 4]]=3
> So in going from coordinate (2,3) to (2,4) it takes 3 times as long as
> if going from (2,4) to (2,3). So all directions are possible but they
> are asymetrical in the sense that they have diferent weights if going
> foward or backward.
>
> I tried reading Mathematica help but it is very poor with no examples.
> All I know is that I have to use the Combinatorica package and the
> ShortestPath[] command, but as a start I have no idea in how to create
> a Graph needed as input to this command from MyGrid.
> | 575 | 1,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.852538 |
https://www.briangwilliams.us/health-risks/valuesolid-for-sheffer-stroke-1.html | 1,582,342,280,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00119.warc.gz | 685,737,111 | 8,535 | ## Valuesolid For Sheffer Stroke
SIERPINSKI'S GASKET (above left) can be approximated using mathematical logic. Plugging four-digit binary numbers into the truth table for the Sheffer stroke yields a color diagram in which the green squares form a gasketlike shape (left). The figure can also be seen in the value solid for the Sheffer stroke (above).
4 1 1 ,„y.ii„i. XII! * ° »"IX WEX EjVLE ___
w
X 35SO orb X X o ^xc o liixx 3XC o Oxk okÈ X X o X o
X jÉO mAmWS, ix X !IXX "™o oHo ilXX "OO X O X xlxp 3XX axe O o X X o X o
It is also possible to change 0 to 1 and 1 to 0 by applying the operator NOT: that is, NOT P is true if P is false, and vice versa. There are 16 possible truth tables for
ALL POSSIBLE GAMES of tic-tac-toe can be depicted in a quasi-fractal diagram. The squares in the 3-by-3 grid are divided into smaller grids that show all the opening moves (right). Subsequent moves are illustrated in still smaller grids created by subdividing the unoccupied squares. A sample game can be viewed by repeatedly magnifying sections of the diagram ( far right).
possible moves for O. If we just put O's in each of those subsquares, though, we would have nowhere to put X's second move. Instead we repeat the trick already used for the opening move: We divide each of the eight unmarked subsquares into a 3-by-3 grid of sub-subsquares, getting eight small tic-tac-toe boards. We put an X in the top left corner of each, to represent X's opening move. Then we put one of O's eight possible moves into each of the small tic-tac-toe boards.
We can continue in this fashion, recording all the possible moves in subgrids of ever smaller size. At every stage, all the unoccupied squares are subdivided into 3-by-3 grids, and all moves previous to that stage are copied into the cells of those grids. The final figure has a quasi-fractal structure because the rules of the game are recursive: the possible moves at each stage are determined by the moves made before. The geometry of fractals is also recursive: similar shapes repeat on ever smaller scales. The tic-tac-toe figure is a quasi-frac-tal rather than a true fractal because the game ends after a finite number of moves.
Now we turn to logic. The simplest area of conventional mathematical logic, prop-ositional calculus, is concerned with statements whose "truth-value" is either 1, representing true, or 0, representing false. For example, the statement P = "pigs can fly" has a truth-value of 0, whereas Q = "Africa is a continent" has a truth-value of 1. Statements can be combined using various logical operators, such as AND and OR. If P and Q are as above, the statement P AND Q is "pigs can fly, and Africa is a continent." This statement is false, so the truth-value of P AND Q is 0. The results of applying AND to statements can be summed up in a truth table:
It is also possible to change 0 to 1 and 1 to 0 by applying the operator NOT: that is, NOT P is true if P is false, and vice versa. There are 16 possible truth tables for c o
two statements, representing all the possible ways to put 0's and 1's in the table's final column. We can denote them with successive four-digit binary numbers: 0000, 0001, 0010, 0011 and so on, up to 1111. (In decimal notation, these numbers are 0, 1, 2, 3, ... , 15.) This list leads to another quasi-fractal. To draw it, sketch a 16-by-16 array of squares and add a border above the top row that identifies each column with one of the binary numbers [see illustration on page 86]. Then add a similar border down the left side of the array to enumerate the rows. Choose 16 different colors to correspond to the 16 binary numbers and color the border squares accordingly. Next, choose a logical operator: for example, the Sheffer stroke, which is represented by the symbol I. In computer engineering, the Sheffer stroke is known as NAND, because P I Q = NOT (P AND Q). Its truth table is:
P Q P I Q 0 0 1 (0 1 1 1 0 1 1 1 0
P I Q truth table and put the square's column number in the table's second column. Then perform the NAND operations and put the resulting truth-values in the table's final column. This yields another four-digit binary number. Find the color that corresponds to this number and use it to mark the square in the 16-by-16 array. For instance, consider the square in row 5, column 11. In binary notation, these numbers are 0101 and 1011. Plugging them into the truth table for P I Q yields:
P Q P I Q 0 1 1 1 0 1 0 1 1 1 1 0
The number in the final column is 1110, or 14 in decimal notation. So the square in row 5, column 11 is given the color corresponding to 14.
The final product of this laborious process is shown in the illustration on page 86. Notice that the green squares, corresponding to the binary number 1111, form a shape very similar to Sierpinski's gasket! Instead of color-coding the picture, one can also graph the value of each square in a third dimension, as a height given by its decimal number divided by 16. For example, the height of the square in row 5, column 11 would be 14/16 = 0.875. These graphs are called value solids. In the value solid for the Sheffer stroke, a gasketlike shape can clearly be seen. The explanation is simple: any formal logical system that involves recursion—whether a game or a truth table—can provide a recipe for drawing quasi-fractals. E3
Now, for each of the squares in the 16-by-16 array, put the square's four-digit row number in the first column of the | 1,403 | 5,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-10 | longest | en | 0.923209 |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-3-logic-3-1-statements-negations-and-quantified-statements-exercise-set-3-1-page-120/31 | 1,590,421,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00534.warc.gz | 697,030,004 | 13,000 | ## Thinking Mathematically (6th Edition)
Published by Pearson
# Chapter 3 - Logic - 3.1 Statements, Negations, and Quantified Statements - Exercise Set 3.1 - Page 120: 31
#### Answer
a) An equivalent way can be: "At least one student is a business major." b) A negation can be: "No student is a business major.".
#### Work Step by Step
a) An equivalent way can be: "At least one student is a business major." b) A negation can be: "No student is a business major.".
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 162 | 632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-24 | latest | en | 0.847454 |
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# Funny Short Math Jokes and Puns, Math is Fun!
A mathematical joke is a form of humor which relies on aspects of mathematics or a stereotype of mathematicians to derive humor. The humor may come from a pun, or from a double meaning of a mathematical term, or from a lay person's misunderstanding of a mathematical concept.
Instead of good-bye we say Calc-U-later
• Why should you not mix alcohol and calculus? Because you should never drink and derive.
• Write the expression for the volume of a thick crust pizza with height "a" and radius "z". The formula for volume is π·(radius)**2·(height). In this case, pi·z·z·a.
• How do you make seven even? Just remove the “s.
Q: What is a proof?
A: One-half percent of alcohol.
Q: What is gray and huge and has integer coefficients?
A: An elephantine equation.
Q: Why do truncated Maclaurin series fit the original function so well?
A: Because they are “Taylor” made.
Q: What is gray and huge and has integer coefficients?
A: An elephantine equation.
Q: What’s a polar bear?
A: A rectangular bear after a coordinate transform.
Q: What do you get if you cross a mosquito with a mountain climber?
A: You can’t cross a vector with a scalar.
Theorem. 3=4.
Proof. Suppose a + b = c
This can also be written as: 4a − 3a + 4b − 3b = 4c − 3c
After reorganizing: 4a + 4b − 4c = 3a + 3b − 3c
Take the constants out of the brackets: 4(a + b − c) = 3(a + b − c)
Remove the same term left and right:
4=3
A mathematician and an engineer are on a desert island. They find two palm trees with one coconut each. The engineer shinnies up one tree, gets the coconut, and eats it. The mathematician shinnies up the other tree, gets the coconut, climbs the other tree and puts it there. “Now we’ve reduced it to a problem we know how to solve.
There are a mathematician and a physicist and a burning building with people inside. There are a fire hydrant and a hose on the sidewalk. The physicist has to put the fire out…so, he attaches the hose to the hydrant, puts the fire out, and saves the house and the family. Then they put the people back in the house, set it on fire, and ask the mathematician to solve the problem. So, he takes the hose off the hydrant and lays it on the sidewalk. “Now I’ve reduced it to a previously solved problem” and walks away.
Three men are in a hot-air balloon. Soon, they find themselves lost in a canyon somewhere. One of the three men says, “I’ve got an idea. We can call for help in this canyon and the echo will carry our voices far.” So he leans over the basket and yells out, “Helloooooo! Where are we?” (They hear the echo several times.) Fifteen minutes later, they hear this echoing voice: “Hellooooo! You’re lost!!” One of the men says, “That must have been a mathematician.” Puzzled, one of the other men asks, “Why do you say that?” The reply: “For three reasons: (1) He took a long time to answer, (2) he was absolutely correct, and (3) his answer was absolutely useless.
Infinitely many mathematicians walk into a bar. The first says, "I'll have a beer." The second says, "I'll have half a beer." The third says, "I'll have a quarter of a beer." Before anyone else can speak, the barman fills up exactly two glasses of beer and serves them. "Come on, now,” he says to the group, “You guys have got to learn your limits.”
Scientists caught a physicist and a mathematician and locked them in separate rooms so both could not interact with each other. They started studying their behavior. The two were assigned a task to remove a hammered nail from inside the wall. The only tools they had were a hammer and a nail-drawer. After some muscular effort, both solved the tasks similarly by using the nail-drawer. Then there was a second task, to remove the nail that was barely touching the wall with its sharp end. The physicist simply took the nail with his hand. The mathematician hammered the nail inside the wall with full force and proudly announced: the problem has been reduced to the previous one!
A mathematician organizes a raffle in which the prize is an infinite amount of money paid over an infinite amount of time. Of course, with the promise of such a prize, his tickets sell like hot cake.
When the winning ticket is drawn, and the jubilant winner comes to claim his prize, the mathematician explains the mode of payment:
"1 dollar now, 1/2 dollar next week, 1/3 dollar the week after that..."
Sherlock Holmes and Watson travel on a balloon. They were hidden in clouds, so they didn’t know which country they flew above. Finally they saw a guy below between clouds, so they asked.
“Hey, you know where we are?” “Yes” “Where?” “In a balloon”.
And the guy was hidden by clouds again.
Watson:”Goddamn, what a stupid idiot!”
Holmes:”No my friend, he’s a mathematician”.
Watson:”How can you know that, Holmes?”
Holmes:”Elementary, my dear Watson. He responded with an absolutely correct and absolutely useless answer”.
• My girlfriend is the square root of -100. She’s a perfect 10, but purely imaginary.
• How do mathematicians scold their children? "If I've told you n times, I've told you n+1 times..."
• What’s the best way to woo a math teacher? Use acute angle.
• What do you call a number that can't keep still? A roamin' numeral.
• Take a positive integer N. No wait, N is too big; take a positive integer k.
• A farmer counted 196 cows in the field. But when he rounded them up, he had 200.
• Why should you never argue with decimals? Because decimals always have a point.
When someone once asked Professor Eilenberg if he could eat Chinese food with three chopsticks, he answered, "Of course," according to Professor Morgan.
How are you going to do it?
I'll take the three chopsticks, I'll put one of them aside on the table, and I'll use the other two.
A statistics professor is going through security at the airport when they discover a bomb in his carry-on. The TSA officer is livid. "I don't understand why you'd want to kill so many innocent people!" The professor laughs and explains that he never wanted to blow up the plane; in fact, he was trying to save them all. "So then why did you bring a bomb?!" The professor explains that the probability of a bomb being on an airplane is 1/1000, which is quite high if you think about it, and statistically relevant enough to prevent him from being able to fly stress-free. "So what does that have to do with you packing a bomb?" the TSA officer wants to know, so the professor explains. "You see, if there's 1/1000 probability of a bomb being on my plane, the chance that there are two bombs is 1/1000000. So if I bring a bomb, the chance there is another bomb is only 1/1000000, so we are all much safer."
The great probabilist Mark Kac (1914-1984) once gave a lecture at Caltech, with Feynman in the audience.
When Kac finished, Feynman stood up and loudly proclaimed, "If all mathematics disappeared, it would set physics back precisely one week."
To that outrageous comment, Kac shot back with that yes, he knew of that week; it was "Precisely the week in which God created the world."
An experimental physicist meets a mathematician in a bar and they start talking. The physicict asks, "What kind of math do you do?" to which the mathematician replies, "Knot theory."
The physicist says, "Me neither!"
A poet, a priest, and a mathematician are discussing whether it's better to have a wife or a mistress.
The poet argues that it's better to have a mistress because love should be free and spontaneous.
The priest argues that it's better to have a wife because love should be sanctified by God.
The mathematician says, "I think it's better to have both. That way, when each of them thinks you're with the other, you can do some mathematics."
Three mathematicians walk into a bar.
Bartender asks:”Will all of you guys have beer?”
The first mathematician: “I don’t know”.
The second mathematician: “I don’t know”.
The third one: ”Yes”.
A mathematician is attending a conference in another country and is sleeping at a hotel. Suddenly, there is a fire alarm and he rushes out in panic. He also notices some smoke coming from one end of the corridor. As he is running, he spots a fire extinguisher. “Ah!”, he exclaims, “A solution exists!” and comes back to his room and sleeps peacefully.
Two statisticians go to hunt a bear. After roaming the woods for a while, they spot a lone grizzly. The first statistician takes aim and shoots, but it hits three feet in front of the bear. The second one shoots next, and it hits three feet behind the bear. They both agree that they have shot the bear and go to retrieve it..
• Parallel lines have so much in common. It’s a shame they’ll never meet.
• I just saw my math teacher with a piece of graph paper. I think he must be plotting something.
• Are monsters good at math? No, unless you Count Dracula.
• My girlfriend is the square root of -100. She's a perfect 10, but purely imaginary.
• Q: Why is a math book depressed? A: Because it has so many problems.
• How do you stay warm in an empty room? Go into the corner where it is always 90 degrees.
• There are three kinds of people in the world: those who can count and those who can't.
• Q: Why did I divide sin by tan? A: Just cos.
• Q: Where's the only place you can buy 64 watermelons and nobody wonders why? A: In an elementary school math class.
• 60 out of 50 people have trouble with fractions.
• But why did 7 eat 9? Because you’re supposed to eat 3 squared meals a day.
• Q: Why is the obtuse triangle depressed? A: Because it is never right.
• Q: Why did the 30-60-90 degree triangle marry the 45-45-90 degree triangle? A: Because they were right for each other.
• Q: Why didn't the Romans find algebra very challenging? A: Because they always knew X was 10.
• Two statisticians went out hunting and they found a deer. The first one overshoots by 5 meters. The second one undershoots by 5 meters. They both hug each other and shout out “We Got It!”
An astronomer, a physicist and a mathematician are on a train traveling from England to Scotland. It is the first time for each of them.
Some time after the train crosses the border, the three of them notice a sheep in a field.
“Amazing!” says the astronomer. “All the sheep in Scotland are black!”.
“No, no” responds the physicist. “Some sheep in Scotland are black!”
The mathematician closes his eyes pityingly, and intones: “In Scotland, there is at least one field, containing at least one sheep, at least one side of which is black.”
An engineer, a physicist and a mathematician go to a hotel.
The boiler malfunctions in the middle of the night and the radiators in each room set the curtains on fire.
The engineer sees the fire, sees there is a bucket in the bathroom, fills the bucket with water and throws it over the fire.
The physicist sees the fire, sees the bucket, fills the bucket to the top of his mentally calculated error margin and throws it over the fire.
The mathematician sees the fire, sees the bucket, see the solution and goes back to sleep. | 2,682 | 11,078 | {"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} | 3.84375 | 4 | CC-MAIN-2024-30 | latest | en | 0.911251 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Book%3A_A_First_Course_in_Electrical_and_Computer_Engineering_(Scharf)/6%3A_Linear_Algebra/6.02%3A_Vectors | 1,596,676,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735990.92/warc/CC-MAIN-20200806001745-20200806031745-00421.warc.gz | 284,048,913 | 23,762 | 6.02: Vectors
For our purposes, a vector is a collection of real numbers in a one- dimensional array.1 We usually think of the array as being arranged in a column and write
$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ | \\ x_n \end{bmatrix}$
Notice that we indicate a vector with boldface and the constituent elements with subscripts. A real number by itself is called a scalar, in distinction from a vector or a matrix. We say that $$x$$ is an n-vector, meaning that $$x$$ has $$n$$ elements. To indicate that $$x_1$$ is a real number, we write
$x_1 \;∈ \;\mathbb{R}$
meaning that $$x_1$$ is contained in $$\mathbb{R}$$, the set of real numbers. To indicate that $$x$$ is a vector of $$n$$ real numbers, we write
$x\;∈\;\mathbb{R}^n$
meaning that $$x$$ is contained in $$R^n$$, the set of real n-tuples. Geometrically, $$R^n$$ is n-dimensional space, and the notation $$x\;∈\;\mathbb{R}^n$$ means that $$x$$ is a point in that space, specified by the $$n$$ coordinates $$x_1,x_2,...,x_n$$. The figure shows a vector in $$R^3$$, drawn as an arrow from the origin to the point $$x$$. Our geometric intuition begins to fail above three dimensions, but the linear algebra is completely general.
We sometimes find it useful to sketch vectors with more than three dimensions in the same way as the three-dimensional vector of the figure. We then consider each axis to represent more than one dimension, a hyperplane, in our n-dimensional space. We cannot show all the details of what is happening in n-space on a three-dimensional figure, but we can often show important features and gain geometrical insight.
Vectors with the same number of elements can be added and subtracted in a very natural way:
$x+y = \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \\ x_3+y_3 \\ | \\ x_n+y_n \end{bmatrix}; x−y = /begin{bmatrix} x_1−y_1 \\ x_2−y_2 \\ x_3−y_3 \\ | \\ x_n−y_n \end{bmatrix}$
Exercise $$\PageIndex{1}$$
The difference between the vector $$x=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ and the vector $$y=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ is the vector $$z=x−y=\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$$. These vectors are illustrated in the Figure. You can see that this result is consistent with the definition of vector subtraction in the Equation. You can also picture the subtraction in the Figure by mentally reversing the direction of vector $$y$$ to get $$−y$$ and then adding it to $$x$$ by sliding it to the position where its tail coincides with the head of vector $$x$$. (The head is the end with the arrow.) When you slide a vector to a new position for adding to another vector, you must not change its length or direction.
Exercise $$\PageIndex{2}$$
Compute and plot x+y and x−y for each of the following cases:
1. $$x=\begin{bmatrix}1\\3\\2\end{bmatrix},y=\begin{bmatrix}1\\2\\3\end{bmatrix}$$
2. $$x=\begin{bmatrix}−1\\3\\−2\end{bmatrix},y=\begin{bmatrix}1\\2\\3\end{bmatrix}$$
3. $$x=\begin{bmatrix}1\\−3\\2\end{bmatrix},y=\begin{bmatrix}1\\3\\2\end{bmatrix}$$
Scalar Product
Several different kinds of vector multiplication are defined.2 We begin with the scalar product. Scalar multiplication is defined for scalar $$a$$ and vector $$x$$ as
$ax=\begin{bmatrix}ax_1 \\ ax_2 \\ ax_3 \\ | \\ ax_n\end{bmatrix}$
If $$|a|<1$$, then the vector $$ax$$ is "shorter" than the vector $$x$$; if $$|a|>1$$, then the vector $$ax$$ is ‚"longer" than $$x$$. This is illustrated for a 2-vector in the Figure.
Exercise $$\PageIndex{3}$$
Compute and plot the scalar product ax when $$x=\begin{bmatrix}1 \\ 1/2 \\ l/4 \end{bmatrix}$$ for each of the following scalars:
1. $$a=1$$
2. $$a=−1$$
3. $$a=−1/4$$
4. $$a=2$$
Exercise $$\PageIndex{4}$$
Given vectors $$x,y,z∈\mathbb{R}^n$$ and the scalar $$a∈\mathbb{R}$$, prove the following identities:
1. $$x+y=y+x$$. Is vector addition commutative?
2. $$(x+y)+z=x+(y+z)$$. Is vector addition associative?
3. $$a(x+y)=ax+ay$$. Is scalar multiplication distributive over vector addition?
Footnotes
1. In a formal development of linear algebra, the abstract concept of a vector space plays a fundamental role. We will leave such concepts to a complete course in linear algebra and introduce only the functional techniques necessary to solve the problems at hand.
2. The division of two vectors is undefined, although three different “divisions” are defined in MATLAB. | 1,295 | 4,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-34 | latest | en | 0.836593 |
http://energyadvocate.com/fw49.htm | 1,725,944,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00529.warc.gz | 10,659,154 | 2,516 | # Energy in Capacitors and in Batteries
Energy Storage Batteries, Energy in Capacitors, (condensors) Energy in What limits capacitor's ability to store energy Winner: battery! Batteries store energy and so do capacitors. However, the principles are much different, and batteries can store much more energy. Batteries actually store chemical energy. One can make a battery, for example, by placing a piece of zinc and a piece of carbon (such as the graphite in a lead pencil) into a lemon (which serves as an electrolyte). The battery will work until the zinc has been used up (providing the electrolyte has not dried out). However the battery is made, the total energy E that it can produce is the voltage V of the battery multiplied by the current I it produces multiplied by the time t. (E = VIt) The product It is the total amount of electric charge Q that moves through the wires when electric current flows, to the energy is E = VQ. In a capacitor, the energy comes from the fact that the stored charge on one plate repels like charge on the same plate and attracts unlike charge on the other. When an external circuit is provided, the charge flows from one plate to the other, each tending to neutralize the other. If half the charge becomes neutralized, the voltage is half of its starting value. (In a battery, the voltage remains constant instead.) In terms of the charge Q stored on the capacitor and the starting voltage V, the energy E = (1/2)QV. There are many ways to make capacitors, but the important considerations are the surface area of the plates and the distance between them. If you want a capacitor to be able to hold a large voltage V, the distance between the plates must be large; however, the amount of charge that can be stored at a given voltage is diminished. Overall, the energy that can be stored in a capacitor depends mainly upon the overall volume of the capacitor. For a given volume, a fairly mundane battery can typically store several hundred times as much energy as a capacitor. Sorry, it's not likely you'll be able to replace your car battery with a capacitor! | 471 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.924189 |
https://math.stackexchange.com/questions/532426/a-question-of-h-g-wells-mathematics/532746 | 1,709,048,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00778.warc.gz | 391,913,623 | 41,920 | # A question of H.G. Wells' mathematics
H.G Wells' short story The Plattner Story is about a man who somehow ends up being "inverted" from left to right. So his heart has moved from left to right, his brain, and any other asymmetries belonging to him. Then H.G Wells' goes on a slight metaphysical exposition:
There is no way of taking a man and moving him about in space, as ordinary people understand space, that will result in our changing his sides. Whatever you do, his right is still his right, his left his left. You can do that with a perfectly thin and flat thing, of course. If you were to cut a figure out of paper, any figure with a right and left side, you could change its sides by simply lifting and turning it over. But with a solid it is different. Mathematical theorists tell us that the only way in which the right and left sides of a solid body can be changed is by taking that body clean out of space as we know it, - taking it out of ordinary existence, that is, and turning it somewhere outside space. This is a little abstruse, no doubt, but anyone with a slight knowledge of mathematical theory will assure the reader of its truth. To put the thing in technical language, the curious inversion of Plattner's right and left sides is proof that he has moved out of our space into what is called the Fourth Dimension, and that he has returned again to our world.
This is leaving me a bit perplexed. A reflection is an odd isometry after all, and rotations/translations are direct. So it shouldn't be possible for the man to be reflected using continuous,direct transformations, regardless of how many dimensions he is using. But then again, his argument with the 2/3D analogue is convincing... Thoughts and explanations?
• If the person lived in a non-orientable "space," you could get a reflection. For example, if the person lived in $K\times \mathbb R$, then it might be possible to invert the person left-to-right. Oct 19, 2013 at 19:13
• Your typo "There is now way" is unfortunate! Should it be "There is no way" or "There is now a way"? Oct 19, 2013 at 20:37
• @TonyK Highly unfortunate! Typo edited. Oct 19, 2013 at 23:52
Wells’s explanation seems perfectly correct to me. Think of $\mathbb R^3$ as embedded into $\mathbb R^4$ by $(x,y,z)\mapsto(x,y,z,0)$. Then apply, in four-space, the rigid rotation $$R_\theta\colon\quad \pmatrix{1&0&0&0\\0&\cos\theta&0&-\sin\theta\\0&0&1&0\\0&\sin\theta&0&\cos\theta}\,.$$ Here $\theta=0$ gives the identity, and $\theta=\pi$ gives a $180^\circ$-rotation in four-space, which sends $(x,y,z,0)$ to $(x,-y,z,0)$, thus exchanging left and right.
• Yes, this is akin to flipping a asymmetrical polygon through the third dimension. Oct 19, 2013 at 21:21
• @ThomasAndrews, just so. Oct 19, 2013 at 21:23
• This is simply rotating the $y{-}w$ plane $180^\circ$. It reverses $w$ as well as $y$, but we limited roundlanders do not see that. (+1)
– robjohn
Oct 20, 2013 at 9:37
• @robjohn, quite right, and it seems to me that this is precisely what Wells was trying to say. The important thing is that in four-space, the motion is orientation-preserving. Oct 20, 2013 at 17:03
• Or as I mention in this answer, where I cite your answer here, it can be orientation preserving, as there are some liftings that do and some that do not preserve orientations.
– robjohn
Oct 20, 2013 at 17:12
I think the key here is that a flat object living purely in the $xy$-plane is not affected by reflection in $z$. Thus, an apparent reflection across the $x$-axis can be achieved by reflecting across both the $x$ and $z$ axes — this is a net rotation and hence can be achieved by continuous transformation without breaking orientability.
The same principle would apply to a 3D object embedded in $\mathbb R^4$. Having no width along the fourth dimension makes one invariant under reflections across that dimension.
• Having no width along the fourth dimension […] — What an enlightening choice of words!
– Wolf
Nov 7, 2022 at 9:58
This reminds me of this answer of mine, which explores chirality in higher (and lower) dimensions.
Firstly, let's use the definition of dimension1 of an object as the last value of $n$ such that the object or a rotated copy of it can be said to cover an open subset of $\mathbb R^n$.
Now, the concept of "chirality" is where a reflection cannot be composed of rotations.
One can show that if an $n$ dimensional object is taken and place/projected in $m$ dimensional space for $n\neq m$ and viewed from the larger of the two, the object is achiral.
Yes, there are chiral objects in 2D and 1D. Take the following sequence of dots:
... .. .
If it is living in a 1D space, it cannot be "turned around". This means that these two mirror images:
... .. .|. .. ...
are not the same object.
In two dimensions, take a look at these two mirror images:
X | X
H-+ | +-H
T | T
Note that one cannot rotate one into the other if one is confined to two dimensions (i.e., we cannot make the object "jump" out of the page).
Now, to get a bit more mathematical. This is not going to be rigorous, but I'll try to be convincing.
I'm only going to consider $180^\mathrm{o}$ rotations along a coordinate axes ($R_{x_i}$) and reflections along coordinate planes ($F_{x_i}$), but this can be generalized. Let us note these things of the two operations:
• A rotation $R_{x_i,x_j}$ or $R_{ij}$ has the following properties (check these out for yourself):
• It takes the object and flips the signs of $x_i$ and $x_j$ of all of the points on the object.
• It is commutative (note: I'm only talking about 180 degree rotations here)
• Two rotations of the same kind lead to an identity operation $I$
• $R_{ij}R_{jk} = R_{jk}$
• A flip $F_{x_i}$ or $F_i has the following properties: - It flips the sign of$x_i$only - It is commutative - Two flips of the same kind lead to an identity operation -$F_iF_j = R_{ij}$Now, for an object to be chiral, we should not be able to chalk up any flip as a composition of rotations. Let us take an$n$dimensional object in$m$dimensional space. If$n=m$, for an asymmetric object, the above is not necessary. Note that any composition of rotations can be reduced down to something where there is at most one of each kind. In three dimensions, this gives us the options$I, R_{xy},R_{yz},R_{zx}, R_{xy}R_{yz},R_{yz},R_{zx}R_{yz},R_{xy}R_{zx}, R_{xy}R_{yz}R_{zx}$, or$8$combinations. However, using$R_{ij}R_{jk} = R_{jk}$we can reduce the rotations to the$4$compositions$I, R_{xy},R_{yz},R_{zx}$However, the total number of sign flip combinations (i.e. all the ordered tuples obtained by selectively flipping the signs of coordinates) is$2^3=8$. But we can obtain all$8$by composing flips and rotations. Given a sign flip combination: If there are an even number of coordinates with flipped signs, we just compose the$R_{ij}$s corresponding to pairs of coordinates which have their sign flipped. If there are an odd number, we ignore one of the sign flips, do the same thing as in the even sign flip case, and append a$F_{j}$around the originally ignored coordinates. This means that there are objects belonging to the set formed by composing flips and rotations that do not belong to the set formed by just composing rotations. Chirality exists. This can be easily generalized, we can show that in any number of dimensions, we can form all sign flip combinations ($2^n$) by composing flips and rotations, but only half by composing only rotations as, for any composition of$\left\lfloor\frac{n}{2}\right\rfloor +1$or more rotation objects, there will be at least a pair of objects sharing an index 2, so we can always use$R_{ij}R_{jk} = R_{jk}$to reduce the composition to one of less than or equal to$\frac{n}{2}objects. However, if the object is lesser-dimensional, this no longer works. In this case, we can always align one of the "extra" axes of the object withx_1$. Now,$x_1=0$for all points on the object. This makes the operation$F_{1}$impotent (and equivalent to the identity operation$I$). Now, we had shown that we can make the$2^N$unique sign flip combinations by composing a number of rotations and possibly a single flip. Now, we can show that the composition of a number of rotations and a flip is simply a composition of rotations, as we can multiply the former by$F_{1}=I$and then reduce the composition of two flips to the composition of multiple rotations. Thus, a lower dimensional object in a higher dimensional space cannot be chiral. 1. There are many ways to define dimension which don't always give the same result. My favorite is the Lebesgue dimension which is a pretty beautiful concept, but it's not what I need here. 2. By pigeon hole principle — there are$n$possible indices and$n+1$(if$n$is odd) or$n+2$total indices from the composition of$\left\lfloor\frac{n}{2}\right\rfloor +1$rotation objects each with a nonequal pair of indices • Nice, generalised explanation, thanks! Just being pedantic but do you mean open subset in your second paragraph? Oct 20, 2013 at 6:25 • @KieranCooney Yeah, I did, thanks :) The real generalized explanation comes from group theory, but I'm not fluent enough with it to present it. But it involves playing with the same types of group elements -- rotations and flips. Oct 20, 2013 at 6:27 • As it happens I have just started reading Stillwell's Naive Lie Theory, so I can see how these things would tie in. It's so nice to see all these connections pop up. Oct 20, 2013 at 6:31 • @KieranCooney Heads up, I asked for a more complete explanation of essentially the same problem from a group theoretical POV here. Oct 20, 2013 at 10:47 • I have a correction. The fourth rotation bullet point should be something like$\prod_{i=1}^n R_i=(-1)^{n-1}I$. Oct 20, 2013 at 11:01 Martin Gardner's The Ambidextrous Universe discusses this and related issues (mathematical, physical, chemical, anatomical...) in detail, with Gardner's inimitably readable style. Jeffrey Weeks's The Shape of Space investigates the geometry and topology of$3$-manifolds in more mathematical detail, yet in a friendly, engaging way. As Thomas Andrews notes, it's possible for a traveller to get "reflected" by traversing a closed loop in a non-orientable universe. The impossibility of doing so in$\mathbf{R}^3$(or in any orientable$3$-manifold) amounts to disconnectedness of the set of linear frames and the definition of orientability. I am not too sure about the analogy in 2/3D. If I consider the example given all that he is doing is changing the normal direction of the 2D object embedded in 3D space, i.e. 'up' and 'down' (in 3D) are transposed, and so 'left' and 'right' are transposed in 2D. This means that we are still in 2D, but have changed the 'direction' of the space in which this 2D space is embedded. For an object in the 2D space to 'observe' an object as having left and right transposed both the 2D space with normal vector in$z$and the 2D space with normal vector in$-z\$ need to co-exist, which is not physical. The way I would see this, then, is that it is possible to transpose left and right, but that it is not possible from within a given space to observe that this is the case
Does this seem sensible?
• Yes, but I'll admit, only partially. Your concerns seem to me to be exactly the same concerns I was having about flipping the man in 3d using 4D. But I find the example in 2/3D to be relatively intuitive, unless I'm missing something. Oct 20, 2013 at 0:01 | 2,981 | 11,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.973102 |
https://blocking.net/3518/getting-started-with-blockchain-what-is-a-51-power-attack/ | 1,696,304,454,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00366.warc.gz | 156,864,163 | 9,787 | # Getting started with blockchain | What is a 51% power attack?
Concerned about the bitcoin's small partners, I must have heard 51% of the power to attack the word, what does it mean, what kind of "bad things" can be done after launching this kind of attack? Today, I will briefly introduce the 51% of the power attack related matters.
In the Bitcoin white paper, there has been such a statement that the sum of the honest node control power is greater than the sum of the attacker's computing power of the cooperative relationship, and the system is safe.
In other words, when the computing power controlled by the malicious node in the system exceeds the computing power controlled by the honest node, the system is at risk of being attacked. This attack, initiated by a malicious node that controls more than 50% of the power, is called a 51% attack (51% Attack).
Is it possible that all cryptocurrency systems are at risk of a 51% power attack? In fact, it is not. Only the cryptocurrency based on the PoW (workload proof) consensus mechanism has 51% computational attacks, such as bitcoin, bit cash, and current Ethereum; instead of the cryptocurrency of the PoW consensus algorithm. There is no 51% power attack, such as EOS, TRON, etc. based on the DPoS (trusted equity) consensus mechanism.
After learning about the 51% power attack, you must be curious about what bad things can be done with this attack.
1. Double Spending . Double flower means that a "money" has been spent twice or even many times.
How does the 51% power attack strike a double flower? Suppose Xiaohe has 666BTC. He also pays the coins and sends them to his other wallet address. In other words, a small black money is transferred to two people at the same time. In the end, the deal sent to Dabai was first confirmed and packaged in a block with a block height of N.
At this time, it controlled the black of more than 50% of the power, and launched a 51% power attack. By reassembling the Nth block, he packaged the deal into his block and continued to extend the block on the chain. Due to the power of computing power, this quantity will become the longest legal chain . This small black 666BTC double flower success, the 666BTC in the big white wallet "missed".
In addition, what can 51% of the power attack do? It can also suppress the transmission/reception of bitcoin at an address.
Xiao Hei and Da Bai quarreled, and Xiao Hei relied on 51% of his computing power. He knew that the big white bitcoin address could make the transaction related to Dabai unrecognizable. For example, in order to express his respect to Nakamoto, Dabai wants to send a bitcoin to the "Creation Address". The black that controls more than half of the computing power will not pack the deal, not only that, but also the other miners will not pack the deal. How did Xiaohe do it?
If the new block dug by other miners packs the deal, Xiaohe will choose not to continue mining after this block. He will choose to rebuild the new block after the last block and reject the deal. Relying on your own computing power, this chain of small blacks will become the longest legal chain.
Under this circumstance, other miners will not have to package the transactions related to Dabai, otherwise the excavated blocks will be isolated by Xiaohei, and the rewards will also be invalidated.
If you have more than 50% of your computing power, you can do whatever you want. What can you do with bad things? Actually not. Even if you control more than 50% of the calculation power, you can't transfer other people's coins (stolen coins), because this operation requires a private key to sign. If you want to forge a signature to "steal the currency," this behavior is an honest miner can't. Tolerance, this will subvert the system consensus. In this case, other miners will not continue to expand the blockchain after the blocks he has dug, and will actively fork out the legal blockchain, and the blocks excavated by the “stolen coins” miners will be isolated.
In addition, the system's block rewards are modified. For example, the block reward is changed from the current 12.5BTC to 50BTC. This behavior is also a subversion of the system consensus. Honest miners will refuse to branch out a new chain in such illegal blocks. .
——End——
Author | Yan Wenchun
Produced|Baihua blockchain (ID: hellobtc)
『Declaration : This series of content is only for the introduction of blockchain science, and does not constitute any investment advice or advice. If there are any errors or omissions, please leave a message.
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EigenDA is a secure, high-throughput, decentralized data availability (DA) service on Ethereum. | 1,299 | 5,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.975079 |
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2. has left the building
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I think 41...*prepares to be totally wrong*
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I think its 41. But I could be wrong.....Something rings a bell from back at school about doing the multiplication bit first??
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It's 1. 0x anything is 0 and 0+1 is 1
5. crap i didnt see the +1, I would have thought it was 1
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Originally Posted by Bonkers
It's 1. 0x anything is 0 and 0+1 is 1
but what about the 40+ part at the start? Please explain, i am hopeless at maths. I thought it would be 41 using bodmas so it would be 40+0+1..
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I'm on my phone and can't see the poll!
Originally Posted by FindMyLunch
I'm on my phone and can't see the poll!
Me too
9. 40 + 40 x 0 + 1 = ?
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81
41
1
10. wishes she was a glow worm. A glow worm's never glum, 'cos how can you be grumpy when the sun shines out of your bum?
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Two hundred twenty-one pounds is equal to fifteen stones and eleven pounds (221lbs = 15st 11lb). | 57 | 235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-50 | latest | en | 0.790938 |
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All Right Reserved. Copyright 2005-2018 | 1,718 | 7,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-09 | latest | en | 0.840611 |
https://www.tutorialguruji.com/python/how-to-find-common-elements-from-a-list-of-lists-such-that-order-of-occurrence-is-maintained/ | 1,632,386,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00118.warc.gz | 1,028,591,110 | 22,015 | # How to find common elements from a list of lists such that order of occurrence is maintained?
I have a list of lists where the length of the lists are same. I need to find the common elements from them with the order of occurrence maintained. For example:
Suppose the list of lists is `[['a','e','d','c','f']['e','g','a','d','c']['c','a','h','e','j']]` The output list should contain `['a','e','c']` Priority should be given to elements which occur earlier in most of the lists. In this example ‘a’ occurs earlier, then ‘e’ and so on. How to proceed with this?
you could find common items first then sorted it
```from collections import defaultdict
data = [['a','e','d','c','f'],['e','g','a','d','c'],['c','a','h','e','j']]
common = set(data[0])
for line in data:
common = common.intersection(set(line))
res = defaultdict(int)
for line in data:
for idx, item in enumerate(line):
if item in common:
res[item] += idx
[item[0] for item in sorted(res.items(), key=lambda x: x[1])]
```
output:
```['a', 'e', 'c']
``` | 278 | 1,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.799629 |
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It is currently 23 May 2013, 23:45
# If the two-digit integers M and N are positive and have the
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If the two-digit integers M and N are positive and have the [#permalink] 20 May 2007, 04:20
If the two-digit integers M and N are positive and have the same digits, but in the reverse order, which CANNOT be the sum of M and N?
181
165
121
99
44
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the sum should be a multiple of 11.
Intern
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You are right:)
Can you detail some of your thinking?
Wendy
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I agree with Grad's explanation. Edited my post.
Last edited by vshaunak@gmail.com on 21 May 2007, 03:27, edited 1 time in total.
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Let digits be a and b
Let M = 10a + b, hence N = 10b + a (reverse order)
M + N = 10a + b + 10b + a => 11(a+b)
Hence choice A
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If the two-digit integers M and N are positive and have the 3 31 Aug 2008, 10:53
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https://gomathanswerkey.com/go-math-grade-3-answer-key-chapter-6-understand-division-extra-practice/ | 1,718,581,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00273.warc.gz | 246,500,091 | 54,344 | # Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice
Students who are willing to practice more questions can Download Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice. It is very important to learn the basics or fundamentals to become master in maths. Students of Grade 3 can easily understand the basics of division provided in the Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice.
## Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice
The topics covered in this chapter are Repeated subtraction, Equal groups, Number line, related multiplication, and division facts. Check the topics and start your preparation for the exam. Get the step by step procedure to understand Division.
### Common Core – Page No. 123000
Lessons 6.1–6.3 Make equal groups.
Complete the table.
Counters Number of Equal Groups Number in Each Group 1. 18 9 ________ 2. 24 ________ 8 3. 12 6 ________ 4. 35 7 ________ 5. 32 ________ 4 6. 25 ________ 5
Counters Number of Equal Groups Number in Each Group 1. 18 9 2 2. 24 3 8 3. 12 6 2 4. 35 7 5 5. 32 8 4 6. 25 5 5
Explanation:
1. Number of counters = 18
Number of equal groups = 9
Number in each group = x
x × 9 = 18
x= 18/9 = 2
Therefore number in each group = 2
2. Number of counters = 24
Number in each group = 8
Number of equal groups = x
x × 8 = 24
x = 24/8 = 3
Thus the number of equal groups = 3
3. Number of counters = 12
Number of equal groups = 6
Number in each group = x
x × 6 = 12
x = 12/6 = 2
So, the number in each group = 2
4. Number of counters = 35
Number of equal groups = 7
Number in each group = x
x × 7 = 35
x = 35/7 = 5
x = 5
Therefore number in each group = 5
5. Number of counters = 32
Number of equal groups = x
Number in each group = 8
x × 8 = 32
x = 32/8 = 4
Thus the number of equal groups = 4
6. Number of counters = 25
Number of equal groups = x
Number in each group = 5
x × 5 = 25
x = 25/5 = 5
So, the number of equal groups = 5
Lesson 6.4
Write a division equation for the picture.
Question 1.
Type below:
__________
Answer: 27 ÷ 3 = 9 or 27 ÷ 9 = 3
Explanation:
Total number of counters = 27
Number of equal groups = 3
Number in each group = 9
The division equation is
Number of counters by number of groups = 27 ÷ 3 = 9
or
Number of counters by number in each group = 27 ÷ 9 = 3
Question 2.
Type below:
__________
Answer: 24 ÷ 4 = 6 or 24 ÷ 6 = 4
Explanation:
Total number of counters = 24
Number of equal groups = 4
Number in each group = 6
The division equation is
Number of counters by number of groups = 24 ÷ 4 = 6
or
Number of counters by number in each group = 24 ÷ 6 = 4
Lesson 6.5
Write a division equation.
Question 3.
Type below:
__________
Answer: 3 groups, 15 ÷ 5 = 3
Explanation:
Step 1:
Starts at 15
Step 2:
Count back by 5s as many times as you can.
Step 3:
Count the number of times you jumped back 5.
You jumped back by 15 three times
There are 3 jumps of 5 in 15.
Question 4.
Type below:
__________
Answer: 24 ÷ 6 = 4
Explanation:
Step 1:
Begins at 24
Step 2:
Subtract with 6 until you get 0.
Step 3:
Count the number of times you subtract with 6.
You subtract 4 times
There are 4 groups of 6 with 24
So, 24 ÷ 6 = 4
### Common Core – Page No. 124000
Lesson 6.6
Make an array. Then write a division equation.
Question 1.
12 tiles in 4 rows
______ ÷ ______ = ______
Answer: 12 ÷ 4 = 3
Explanation:
â– â– â–
â– â– â–
â– â– â–
â– â– â–
Total number of tiles = 12
Number of rows = 4
Number of tiles in each row = x
Divide the number of tiles by number of rows = 12 ÷ 4 = 3
Question 2.
18 tiles in 3 rows
______ ÷ ______ = ______
Answer: 18 ÷ 3 = 6
Explanation:
â– â– â– â– â– â–
â– â– â– â– â– â–
â– â– â– â– â– â–
Total number of tiles = 18
Number of rows = 3
Number of tiles in each row = y
Divide the number of tiles by no. of rows = 18 ÷ 3 = 6
Question 3.
35 tiles in 5 rows
______ ÷ ______ = ______
Answer: 35 ÷ 5 = 7
Explanation:
â– â– â– â– â– â– â–
â– â– â– â– â– â– â–
â– â– â– â– â– â– â–
â– â– â– â– â– â– â–
â– â– â– â– â– â– â–
Total number of tiles = 35
Number of rows = 5
Number of tiles in each row = p
Divide the number of tiles by number of rows = 35 ÷ 5 = 7
Question 4.
28 tiles in 7 rows
______ ÷ ______ = ______
Answer: 28 ÷ 7 = 4
Explanation:
â– â– â– â–
â– â– â– â–
â– â– â– â–
â– â– â– â–
â– â– â– â–
â– â– â– â–
â– â– â– â–
Total number of tiles = 28
Number of rows = 7
Number of tiles in each row = x
Divide the number of tiles by number of rows = 28 ÷ 7 = 4
Lesson 6.7
Complete the equations.
Question 5.
8 × ______ = 40 40 ÷ 8 = ______
Explanation:
Let x be the unknown factor
8 × x = 40
x = 40/8
x = 5
Check whether the related multiplication and division facts are the same or not.
40 ÷ 8 = 5
The related facts of 40 and 8 are 5.
Question 6.
6 × ______ = 36 36 ÷ 6 = ______
Explanation:
Let y be the unknown factor
6 × y = 36
y = 36/6 = 6
Check if the related multiplication and division facts are the same or not.
36 ÷ 6 = 6
The related facts of 36 and 6 are 6.
Question 7.
3 × ______ = 21 21 ÷ 3 = ______
Explanation:
Let x be the unknown factor
3 × x = 21
x = 21
Check whether the related facts are the same or not.
21 ÷ 3 = 7
The quotient is 7.
Question 8.
2 × ______ = 18 18 ÷ 2 = ______
Explanation:
Let b be the unknown factor
2 × b = 18
b = 18/2 = 9
Check the related multiplication and division facts
18 ÷ 2 = 9
The related facts of 18 and 2 are 9.
Lesson 6.8 (pp. 239–243)
Write the related facts for the array.
Question 9.
â– â– â– â– â–
â– â– â– â– â–
â– â– â– â– â–
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______
3 × 5 = 15
5 × 3 = 15
15 ÷ 3 = 5
15 ÷ 5 = 3
Explanation:
Total number of tiles = 15
Number of equal rows = 3
Number of rows in each group = 5
So, the related 5, 3 and 15 is 5× 3 = 15, 3×5 = 15, 15 ÷ 3 = 5 and 15÷ 5 = 3
Question 10.
â– â– â– â– â– â–
â– â– â– â– â– â–
â– â– â– â– â– â–
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______
3 × 6 = 18
6 × 3 = 18
18 ÷ 3 = 6
18 ÷ 6 = 3
Explanation:
Total number of tiles = 18
Number of equal rows = 3
Number of rows in each group = 6
So, the related 18, 3 and 6 is 3 × 6 = 18, 6 × 3 = 18, 18 ÷ 3 = 6 and 18 ÷ 6 = 3
Question 11.
â– â– â– â– â–
â– â– â– â– â–
______ × ______ = ______
______ × ______ = ______
______ ÷ ______ = ______
______ ÷ ______ = ______
2 × 5 = 10
5 × 2 = 10
10 ÷ 2 = 5
10 ÷ 5 = 2
Explanation:
Total number of tiles = 10
Number of equal rows = 2
Number of rows in each group = 5
So, the related 2, 5 and 10 is 2 × 5 = 10, 5 × 2 = 10, 10 ÷ 2 = 5 and 10 ÷ 5 = 2
Lesson 6.9
Find the quotient.
Question 12.
7 ÷ 1 = ______
Explanation:
Any number divided by 1 will be the same number. Thus the quotient is 7.
Question 13.
4 ÷ 4 = ______
Explanation:
The number divided by the same number will be always 1. Thus the quotient is 1.
Question 14.
9 ÷ 1 = ______
Explanation:
Any number divided by 1 will be always the same number. So, the quotient is 9.
Question 15.
0 ÷ 1 = ______
Explanation:
0 divided by any number is always 0. So, the quotient is 0.
Question 16.
Anton has 8 flower pots. He plants 1 seed in each pot. How many seeds does Anton use?
______ seeds
Explanation:
Anton has 8 flower pots.
He plants 1 seed in each pot.
Number of seeds Anton used = x
x × 1 = 8
x = 8/1
x = 8
Therefore there are 8 seeds in 8 flower pots.
The problems in the Go Math Grade 3 Answer Key Chapter 6 Understand Division Extra Practice helps the students to learn the concepts and for exam preparation. If you have any doubts in this chapter you can refer Go Math Grade 3 Answer Key Chapter 6 Understand Division. The students of Grade 3 can get all chapters solution in our Go Math Answer Key.
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Scroll to Top | 2,909 | 8,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-26 | latest | en | 0.844273 |
http://research.stlouisfed.org/fred2/series/PLOGINVNA623NUPN?rid=258 | 1,432,933,508,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930423.94/warc/CC-MAIN-20150521113210-00044-ip-10-180-206-219.ec2.internal.warc.gz | 209,776,969 | 19,996 | # Price Level of Government Consumption for Vietnam
2010: 13.83747 PPP of Government Consumption over Exchange Rate (+ see more)
Annual, Not Seasonally Adjusted, PLOGINVNA623NUPN, Updated: 2012-09-17 10:33 AM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: pg
Source: University of Pennsylvania
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(a) Price Level of Government Consumption for Vietnam, PPP of Government Consumption over Exchange Rate, Not Seasonally Adjusted (PLOGINVNA623NUPN)
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: pg
Price Level of Government Consumption for Vietnam
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Name: Email: | 880 | 3,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-22 | latest | en | 0.866416 |
https://www.codingdrills.com/tutorial/introduction-to-dynamic-algorithms/emerging-trends-in-dp | 1,716,231,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00246.warc.gz | 623,183,224 | 76,728 | Emerging Trends and Innovations in Dynamic Programming
Dynamic Algorithms: Future Trends
Dynamic programming is a powerful technique used to solve complex problems by breaking them down into simpler overlapping subproblems. Over the years, dynamic programming has evolved and given rise to several future trends, which we will explore in this post. We will delve into the emerging innovations and advancements that have the potential to further enhance the effectiveness of dynamic algorithms.
Trend 1: Parallel Processing
As the demand for faster computations increases, parallel processing has emerged as a significant trend in dynamic programming. With the advent of multi-core processors, leveraging parallel computing techniques can significantly speed up the execution of dynamic algorithms. By distributing the workload across multiple cores, programmers can harness the power of parallel processing to solve computationally intensive problems in real-time.
Let's consider an example to understand how parallel processing can be applied in dynamic programming.
``````import multiprocessing as mp
def fibonacci(n):
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci(n-2)
def parallel_fibonacci(n):
if n <= 1:
return n
pool = mp.Pool(mp.cpu_count())
result = pool.map_async(fibonacci, [n-1, n-2])
return result.get()[0] + result.get()[1]
print(parallel_fibonacci(10))
``````
In this example, we use the `multiprocessing` module to parallelize the computation of the Fibonacci sequence. By utilizing multiple processes, we can concurrently calculate the results for `fibonacci(n-1)` and `fibonacci(n-2)`, significantly reducing the overall execution time.
Trend 2: Approximation Algorithms
Another emerging trend in dynamic programming is the use of approximation algorithms. These algorithms provide near-optimal solutions to complex problems when an exact solution is computationally infeasible or too time-consuming. Approximation algorithms sacrifice optimality in favor of efficiency, allowing programmers to tackle larger problem instances.
Let's consider the Traveling Salesman Problem (TSP) as an example. The TSP aims to find the shortest possible route for a salesman to visit a set of cities and return to the starting city. Solving TSP optimally becomes exponentially difficult as the number of cities increases. However, approximation algorithms can provide solutions that are within a reasonable range of optimality.
``````def tsp_approximation(cities):
# Approximation algorithm to solve the Traveling Salesman Problem
# Implementation details omitted for brevity
return best_route
cities = ["A", "B", "C", "D", "E"]
best_route = tsp_approximation(cities)
print(best_route)
``````
The `tsp_approximation` function implements an approximation algorithm to solve the TSP. Although the returned route may not be optimal, it provides a good approximation that can be used in practice.
Trend 3: Reinforcement Learning
Reinforcement learning has gained momentum as an innovative approach to dynamic programming. Combining concepts from artificial intelligence and dynamic programming, reinforcement learning allows algorithms to learn through trial and error while interacting with an environment. This approach has shown promising results in solving complex problems that were previously challenging for traditional dynamic programming techniques.
One popular reinforcement learning algorithm is Q-learning, which uses a Q-value table to determine the optimal action to take in a given state. By continuously updating and refining the Q-values based on rewards and penalties, the algorithm learns to make decisions that maximize the expected return.
``````# Q-learning example
# Implementation details omitted for brevity
def q_learning():
# Q-table initialization
q_table = np.zeros((num_states, num_actions))
while not converged:
current_state = get_current_state()
action = choose_action(current_state)
new_state, reward = take_action(current_state, action)
q_value = (1 - learning_rate) * q_table[current_state][action] + learning_rate * (reward + discount_factor * np.max(q_table[new_state]))
q_table[current_state][action] = q_value
return q_table
optimal_q_table = q_learning()
print(optimal_q_table)
``````
In this example, we demonstrate the Q-learning algorithm, an essential component of reinforcement learning. The algorithm iteratively updates the Q-values based on the rewards obtained for different state-action pairs, ultimately converging to an optimal Q-table.
In conclusion, dynamic programming continues to evolve with new trends and innovations. Parallel processing, approximation algorithms, and reinforcement learning are some of the emerging trends that hold the potential to make dynamic algorithms more efficient and powerful. By staying informed and adopting these future trends, programmers can enhance their problem-solving capabilities and tackle complex computational challenges effectively. | 956 | 4,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.85609 |
https://gamesmissing.com/what-does-tick-speed-does-minecraft | 1,653,041,722,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00125.warc.gz | 344,337,750 | 13,478 | What does tick speed does minecraft?
1
Date created: Tue, Sep 7, 2021 6:14 PM
Date updated: Thu, May 19, 2022 4:07 PM
Content
Top best answers to the question «What does tick speed does minecraft»
• Simply put, a game tick is the unit of time in Minecraft. The default tick speed in Minecraft is 20 ticks per second. It means that one tick occurs every 0.05 seconds. The advancements or updates in Minecraft happen based on ticks. The day in the game lasts for 24000 ticks. In real life, that would be 20 minutes.
FAQ
Those who are looking for an answer to the question «What does tick speed does minecraft?» often ask the following questions:
🎮 What is minecraft random tick speed?
• The default random tick speed in Minecraft is 3. To speed things up, let's change the random tick speed to 50000 with the following /gamerule command: 3. Turtle Eggs will crack for the first time. Now, you need to patiently wait for the turtle eggs to hatch three times.
🎮 What is the default tick speed minecraft?
• The default tick speed in Minecraft is 20 ticks per second. It means that one tick occurs every 0.05 seconds. The advancements or updates in Minecraft happen based on ticks. The day in the game lasts for 24000 ticks.
🎮 What is the highest tick speed in minecraft?
In Java Edition, the maximum number of scheduled ticks per game tick is 65,536. In Bedrock Edition, the maximum number of scheduled ticks in a chunk per game tick is 100.
We've handpicked 24 related questions for you, similar to «What does tick speed does minecraft?» so you can surely find the answer!
How long does one tick last in minecraft?
• This implies that one Minecraft tick will last for 0.5 seconds. The length of a day in the Minecraft world is not equivalent to a real one- spanning 24 hours. Each Minecraft in-game day lasts for a total duration of 20 minutes. During the course of the single gaming day, a player traverses through 24,000 gaming ticks.
How long is a minecraft tick?
Game tick. A gametick is where Minecraft's game loop runs once. The game normally runs at a fixed rate of 20 ticks per second, so one tick happens every 0.05 seconds (50 milliseconds, or five hundredths of a second), making an in-game day last exactly 24000 ticks, or 20 minutes.
How long does minecraft take to tick in real time?
• Minecraft time to real time Minecraft time Minecraft ticks Real time 1 second 0.2 7 0.013 8 seconds 1 minute 16. 6 0.8 3 seconds 1 hour 1,000 50 seconds 1 day 24,000 20 minutes 2 more rows ...
How does jumping in minecraft increase movement speed?
• Jumping can be combined with sprinting to increase the player's movement speed. A single jump while sprinting costs four times as much hunger as a normal jump. Most land-based mobs are able to jump when they are walking up blocks.
How does the attack speed decrease in minecraft?
• For the in-game default level III, attack speed decreases by 30% (attack speed cooldown attribute decreases by 0.6), and it takes ~370 times longer than usual to break a block with the proper tool for that block. Negative levels increase attack speed, but still act as level 4 with respect to mining.
How much time is one tick in minecraft?
• A Minecraft game tick is 1/20s: Minecraft’s game loop normally runs at a fixed rate of 20 ticks per second, so one tick happens every 0.05 seconds.
Why do i get tick lag in minecraft?
• If your suffering from “tick lag”, you’ll notice blocks come back after breaking, your movement and the movement of mobs will be jerky. This occurs in singleplayer because, when in single player, Minecraft still runs a server, along with the client. And that server is run on your machine.
How long does it take for a minecraft block to get a tick?
• A block of copper (or any of its non-oxidized variants) may advance one stage in oxidation. [upcoming: JE 1.17] In Java Edition, because random block ticks are granted randomly, there is no way to predict when a block can receive its next tick. The median time between ticks is 47.30 seconds (946.03 game ticks).
Which is an example of a tick in minecraft?
• One good example of ticks is the daylight cycle in Minecraft. Each day is 24,000 ticks. This means 1 day in Minecraft equals 20 minutes in real life. Anything you are doing in game will be put in the same loop to be synchronized with the rest of the world.
What is the fastest horse speed in minecraft?
Movement speed
A horse's maximum speed is 14.23 blocks/second, and the average horse speed is about 9 blocks/sec.
Does chess improve processing speed?
A number of studies have established that chess learning clearly improves cognitive functioning and academic performance. It is likely that an increase in processing speed is one of the basic factors that supports these gains.
What are the different levels of speed in minecraft?
• There are levels of Speed such as Speed II, Speed III, Speed IV and so on. The higher the level of Speed, the faster your movement.
What does game speed actually do in titan quest?
• Chr/enemies/everyone moves faster, attacks faster. It's the equivalent of enabling speed hack on Cheat Engine. All it does is make the gameplay twice as fast on very fast. Some people can find the old style RPGs to be a bit too slow so this is probably for them. It's the equivalent of enabling speed hack on Cheat Engine.
Does graphic card affect game speed?
Graphics card do not usually affect computer performance but it really boils down to what you aim to do with… Having a higher-end video card can increase your FPS or Frames Per Second resulting in faster performance. It also allows you to bump up your graphical settings to make your game visually prettier.
How long does speed chess last?
Tournament Time Standards
Many tournaments will have faster games at 1 hour per player, 30 minutes (Action Chess), 15 minutes (Quick Chess), or 5 minutes (Speed or Blitz Chess). With a friend, a fast game is 5 minutes per player. A slower game is 15 to 30 minutes, and a long game is an hour or more.
How to set repeaters to full 4 tick delay in minecraft?
• Place a note block behind the last repeater, and right-click it 13 times. To conserve more space, the next 3 repeaters actually go to the right of the note block. Place a repeater to the right of the note block, and then 2 more to the right of that repeater. All of the repeaters should be set to full 4-tick delay.
What is the attack speed of an axe in minecraft?
• Stone and diamond axes now both do 9 damage, instead of the previous 8 and 10 respectively. Axes now have attack speed based on the tier, with wooden and stone having a speed of 0.8, iron having a speed of 0.9, and diamond and gold having a speed of 1.
What is the average speed of a horse in minecraft?
• -Speed for naturally spawning horse is 0.1125 to 0.3375. Note: The player has a walking speed of .1. Note: You can spawn a horse with any speed you want but I would not suggest going over 2.5 for the horse is to fast to handle. Note: Donkeys normally have a movement speed of 0.175.
What is the command to get 1000 speed in minecraft?
8 Answers. You need the command: Speed boost: /effect @p 1 100 10. Jump boost: /effect @p 8 100 5.
What is the command to speed up time in minecraft?
Putting /time add 1 in a repeating command block will make the day go by twice as fast as usual, and /time add 2 will be three times as fast.
What is poker speed cloth?
Description. Enhance the look, feel, and performance of your poker table with water-resistant, suited speed cloth. Preferred by professionals, made from high-quality, 100% polyester, this speed cloth offers the best card slide available.
What is pokerstars regular speed?
Hyper (3 min levels) Turbo (5 min) Regular (10 min) Slow (15 min)
Is it better to have high or low tick rates in minecraft?
• If you are a hardcore gamer who likes to play with many mods and texture packs, you would want lower tick rates so as not to overload your computer. On the other hand, if you like playing on survival mode from time to time without any mods or resource packs, higher tick speeds will save wear and tear on your CPU.
How long does speed chess game last?
Blitz chess, also known as Speed Chess, is a single, time-controlled game of 1-10 minutes. Playing Blitz chess creates a flurry of moves, as both players hustle to complete their respective halves of a chess game, depending on which rule set they're playing under. | 1,978 | 8,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.918385 |
https://prezi.com/_ogpl5uvf5ff/my-favorite-number-project/ | 1,540,089,700,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513548.72/warc/CC-MAIN-20181021010654-20181021032154-00327.warc.gz | 783,432,180 | 21,210 | ### Present Remotely
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# My favorite number project!!
No description
by
## Meredith Randall
on 19 April 2017
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#### Transcript of My favorite number project!!
The number I chose to do for the project is 41.
Why I chose to do 41...
I chose this number because it's my jersey number in basketball!!
Comparing!!
Is it a square number?
No. Its not a square number.
First 10 multiples of 41!
41, 82, 123, 164, 205, 246, 287, 328, 369, 410
Interesting facts
Year 41 B.C. was the year that the Julian calender is believed to have been created (the calender that we use today!)
My favorite number project!!
By: Meredith Randall
Homeroom: Mrs Barnard
Color group:
Orange
The roman numeral for 41 is XLI.
The Binary number for 41 is 101001
Would it be a good first move in
the Factor Game?
I think it would be a good first move in the Factor Game
but only if you go first. Its a prime number so the amount of points the other player gets is 1 point to start off with.(prime numbers are numbers that the only factors are 1 and itself. A composite number is the opposite. The number of factors in a composite number are always above 2 factors.)
My number, 41, isn't
divisible
by anything other that itself and 1. Thus forth, no number when multiplied can get a
product
of 41 (Except when you multiply 1 and 41 together).
Factors of my number.
The factors of my number are only 1 and 41.
41 is odd. I know its odd because 1 is odd and since 1 is in 41, its odd.
The number I chose to compare 41 with is 38. The
GCF
of 41 and 38 is 1
Prime factorization
The prime factorization of 41 is 1 times 41= 41
What is a square number?
A square number is a number that has a set of factors that is the same exact number. Example: 9. 3x3=9. 9 is a square number!
Is it odd or even??
Full transcript | 571 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-43 | latest | en | 0.930566 |
http://phoebe-project.org/docs/2.0/tutorials/ecc/ | 1,521,580,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00744.warc.gz | 236,369,411 | 11,002 | # PHOEBE 2.0 Documentation
## 2.0 Docs
Prev: Potentials Next: Apsidal Motion
.
# Eccentricity (Volume Conservation)¶
## Setup¶
As always, let’s do imports and initialize a logger and a new Bundle. See Building a System for more details.
%matplotlib inline
import phoebe
from phoebe import u # units
import numpy as np
import matplotlib.pyplot as plt
logger = phoebe.logger()
b = phoebe.default_binary()
WARNING: Constant u'Gravitational constant' is already has a definition in the u'si' system [astropy.constants.constant]
WARNING: Constant u'Solar mass' is already has a definition in the u'si' system [astropy.constants.constant]
WARNING: Constant u'Solar radius' is already has a definition in the u'si' system [astropy.constants.constant]
WARNING: Constant u'Solar luminosity' is already has a definition in the u'si' system [astropy.constants.constant]
/usr/local/lib/python2.7/dist-packages/astropy/units/quantity.py:782: FutureWarning: comparison to None will result in an elementwise object comparison in the future.
return super(Quantity, self).__eq__(other)
## Relevant Parameters¶
print b.get(qualifier='ecc')
Parameter: ecc@binary@component
Qualifier: ecc
Description: Eccentricity
Value: 0.0
Constrained by:
Constrains: ecosw@binary@component, esinw@binary@component, pot@primary@component, pot@secondary@component
Related to: per0@binary@component, ecosw@binary@component, esinw@binary@component, rpole@primary@component, q@binary@component, syncpar@primary@component, sma@binary@component, pot@primary@component, rpole@secondary@component, syncpar@secondary@component, pot@secondary@component
print b.get(qualifier='ecosw', context='component')
Parameter: ecosw@binary@component
Qualifier: ecosw
Description: Eccentricity times cos of argument of periastron
Value: 0.0
Constrained by: ecc@binary@component, per0@binary@component
Constrains: None
Related to: ecc@binary@component, per0@binary@component
print b.get(qualifier='esinw', context='component')
Parameter: esinw@binary@component
Qualifier: esinw
Description: Eccentricity times sin of argument of periastron
Value: 0.0
Constrained by: ecc@binary@component, per0@binary@component
Constrains: None
Related to: ecc@binary@component, per0@binary@component
## Relevant Constraints¶
b.filter(qualifier='pot', context='constraint')
<ParameterSet: 2 parameters | components: primary, secondary>
print b.get(qualifier='pot', component='primary', context='constraint')
Constrains (qualifier): pot
Expression in SI (value): rocherpole2potential({rpole@primary@component}, {q@binary@component}, {ecc@binary@component}, {syncpar@primary@component}, {sma@binary@component}, 1)
Current Result (result): 6.28266165375
print b.get(qualifier='ecosw', context='constraint')
Constrains (qualifier): ecosw
Expression in SI (value): {ecc@binary@component} * (cos({per0@binary@component}))
Current Result (result): 0.0
print b.get(qualifier='esinw', context='constraint')
Constrains (qualifier): esinw
Expression in SI (value): {ecc@binary@component} * (sin({per0@binary@component}))
Current Result (result): 0.0
## Influence on Meshes (potentials, volumes)¶
b.add_dataset('mesh', times=np.linspace(0,1,11))
<ParameterSet: 2 parameters | contexts: compute, dataset>
b.set_value('ecc', 0.2)
b.run_compute()
<ParameterSet: 662 parameters | components: primary, secondary>
print b['pot@primary@model']
ParameterSet: 11 parameters
0.0@pot@primary@latest@model: 6.22827408689
0.1@pot@primary@latest@model: 6.1432940846
0.2@pot@primary@latest@model: 6.10376571797
0.3@pot@primary@latest@model: 6.10376571797
0.4@pot@primary@latest@model: 6.1432940846
0.5@pot@primary@latest@model: 6.22827408689
0.6@pot@primary@latest@model: 6.36091908169
0.7@pot@primary@latest@model: 6.49345186737
0.8@pot@primary@latest@model: 6.49345186737
0.9@pot@primary@latest@model: 6.36091908169
1.0@pot@primary@latest@model: 6.22827408689
ax, artists = b['mesh01'].plot(x='times', y='pot')
print b['rpole@primary@model']
ParameterSet: 11 parameters
axs, artists = b['mesh01'].plot(x='times', y='rpole')
print b['volume@primary@model']
ParameterSet: 11 parameters
ax, artists = b['mesh01'].plot(x='times', y='volume')
b.remove_dataset('mesh01')
b.add_dataset('rv', times=np.linspace(0,1,51))
<ParameterSet: 15 parameters | contexts: compute, dataset>
b.run_compute()
<ParameterSet: 4 parameters | components: primary, secondary>
axs, artists = b.plot()
b.remove_dataset('rv01')
## Influence on Light Curves (fluxes)¶
b.add_dataset('lc', times=np.linspace(0,1,51))
<ParameterSet: 15 parameters | contexts: compute, dataset>
b.run_compute()
<ParameterSet: 2 parameters | qualifiers: fluxes, times>
axs, artists = b.plot()
Prev: Potentials Next: Apsidal Motion
.
Last update: 06/07/2017 11:30 a.m. (CET) | 1,416 | 4,781 | {"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} | 2.59375 | 3 | CC-MAIN-2018-13 | longest | en | 0.595903 |
https://number.academy/198 | 1,679,859,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00440.warc.gz | 488,929,567 | 13,487 | # Number 198
Number 198 spell 🔊, write in words: one hundred and ninety-eight . Ordinal number 198th is said 🔊 and write: one hundred and ninety-eighth. The meaning of the number 198 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 198. What is 198 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 198.
## What is 198 in other units
The decimal (Arabic) number 198 converted to a Roman number is CXCVIII. Roman and decimal number conversions.
The number 198 converted to a Mayan number is Decimal and Mayan number conversions.
#### Weight conversion
198 kilograms (kg) = 436.5 pounds (lbs)
198 pounds (lbs) = 89.8 kilograms (kg)
#### Length conversion
198 kilometers (km) equals to 123.031 miles (mi).
198 miles (mi) equals to 318.650 kilometers (km).
198 meters (m) equals to 318.650 feet (ft).
198 feet (ft) equals 60.351 meters (m).
198 centimeters (cm) equals to 78.0 inches (in).
198 inches (in) equals to 502.9 centimeters (cm).
#### Temperature conversion
198° Fahrenheit (°F) equals to 92.2° Celsius (°C)
198° Celsius (°C) equals to 388.4° Fahrenheit (°F)
#### Power conversion
198 Horsepower (hp) equals to 145.61 kilowatts (kW)
198 kilowatts (kW) equals to 269.24 horsepower (hp)
#### Time conversion
(hours, minutes, seconds, days, weeks)
198 seconds equals to 3 minutes, 18 seconds
198 minutes equals to 3 hours, 18 minutes
### Codes and images of the number 198
Number 198 morse code: .---- ----. ---..
Sign language for number 198:
Number 198 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
#### Number 198 infographic
King Mithridates V of Parthia died in battle against the Kushans. He was succeeded by Orodes II.
### Gregorian, Hebrew, Islamic, Persian and Buddhist Year (Calendar)
Gregorian year 198 is Buddhist year 741.
Buddhist year 198 is Gregorian year 345 a. C.
Gregorian year 198 is Islamic year -437 or -436.
Islamic year 198 is Gregorian year 813 or 814.
Gregorian year 198 is Persian year -425 or -424.
Persian year 198 is Gregorian 819 or 820.
Gregorian year 198 is Hebrew year 3958 or 3959.
Hebrew year 198 is Gregorian year 3562 a. C.
The Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is the official calendar in Iran and Afghanistan.
## Mathematics of no. 198
### Multiplications
#### Multiplication table of 198
198 multiplied by two equals 396 (198 x 2 = 396).
198 multiplied by three equals 594 (198 x 3 = 594).
198 multiplied by four equals 792 (198 x 4 = 792).
198 multiplied by five equals 990 (198 x 5 = 990).
198 multiplied by six equals 1188 (198 x 6 = 1188).
198 multiplied by seven equals 1386 (198 x 7 = 1386).
198 multiplied by eight equals 1584 (198 x 8 = 1584).
198 multiplied by nine equals 1782 (198 x 9 = 1782).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 198
Half of 198 is 99 (198 / 2 = 99).
One third of 198 is 66 (198 / 3 = 66).
One quarter of 198 is 49,5 (198 / 4 = 49,5 = 49 1/2).
One fifth of 198 is 39,6 (198 / 5 = 39,6 = 39 3/5).
One sixth of 198 is 33 (198 / 6 = 33).
One seventh of 198 is 28,2857 (198 / 7 = 28,2857 = 28 2/7).
One eighth of 198 is 24,75 (198 / 8 = 24,75 = 24 3/4).
One ninth of 198 is 22 (198 / 9 = 22).
show fractions by 6, 7, 8, 9 ...
### Calculator
198
#### Is Prime?
The number 198 is not a prime number. The closest prime numbers are 197, 199.
The 198th prime number in order is 1213.
#### Factorization and factors (dividers)
The prime factors of 198 are 2 * 3 * 3 * 11
The factors of 198 are 1 , 2 , 3 , 6 , 9 , 11 , 18 , 22 , 33 , 66 , 99 , 198
Total factors 12.
Sum of factors 468 (270).
#### Powers
The second power of 1982 is 39.204.
The third power of 1983 is 7.762.392.
#### Roots
The square root √198 is 14,071247.
The cube root of 3198 is 5,828477.
#### Logarithms
The natural logarithm of No. ln 198 = loge 198 = 5,288267.
The logarithm to base 10 of No. log10 198 = 2,296665.
The Napierian logarithm of No. log1/e 198 = -5,288267.
### Trigonometric functions
The cosine of 198 is -0,996829.
The sine of 198 is -0,079579.
The tangent of 198 is 0,079832.
### Properties of the number 198
More math properties ...
## Number 198 in Computer Science
Code typeCode value
Unix timeUnix time 198 is equal to Thursday Jan. 1, 1970, 12:03:18 a.m. GMT
IPv4, IPv6Number 198 internet address in dotted format v4 0.0.0.198, v6 ::c6
198 Decimal = 11000110 Binary
198 Decimal = 21100 Ternary
198 Decimal = 306 Octal
198 Decimal = C6 Hexadecimal (0xc6 hex)
198 BASE64MTk4
198 SHA2241c536d78d9de04e5446c1b226e7e30b8a25d3baf81eba66e7185d6e0
198 SHA256a4e00d7e6aa82111575438c5e5d3e63269d4c475c718b2389f6d02932c47f8a6
198 SHA38470fb78c99d0db0c0ba382a31f254e4ec58320b5ed3ca68723bc5578bd8cf3f6f88599da0fecfcef240ec09589b048b3c
More SHA codes related to the number 198 ...
If you know something interesting about the 198 number that you did not find on this page, do not hesitate to write us here.
## Numerology 198
### The meaning of the number 9 (nine), numerology 9
Character frequency 9: 1
The number 9 (nine) is the sign of ideals, Universal interest and the spirit of combat for humanitarian purposes. It symbolizes the inner Light, prioritizing ideals and dreams, experienced through emotions and intuition. It represents the ascension to a higher degree of consciousness and the ability to display love for others. He/she is creative, idealistic, original and caring.
More about the the number 9 (nine), numerology 9 ...
### The meaning of the number 8 (eight), numerology 8
Character frequency 8: 1
The number eight (8) is the sign of organization, perseverance and control of energy to produce material and spiritual achievements. It represents the power of realization, abundance in the spiritual and material world. Sometimes it denotes a tendency to sacrifice but also to be unscrupulous.
More about the the number 8 (eight), numerology 8 ...
### The meaning of the number 1 (one), numerology 1
Character frequency 1: 1
Number one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.
More about the the number 1 (one), numerology 1 ...
## Interesting facts about the number 198
### Asteroids
• (198) Ampella is asteroid number 198. It was discovered by A. L. N. Borrelly from Marseille Observatory on 6/13/1879.
### Areas, mountains and surfaces
• The total area of Trangan is 830 square miles (2,149 square km). Country Indonesia (Maluku). 198th largest island in the world.
### Distances between cities
• There is a 198 miles (318 km) direct distance between Ahvaz (Iran) and Isfahan (Iran).
• There is a 198 miles (318 km) direct distance between Baghdad (Iraq) and Erbil (Iraq).
• There is a 198 miles (318 km) direct distance between Belgrade (Serbia) and Budapest (Hungary).
• There is a 198 miles (318 km) direct distance between Cochabamba (Bolivia) and Santa Cruz de la Sierra (Bolivia).
• More distances between cities ...
• There is a 198 miles (318 km) direct distance between Ecatepec (Mexico) and León (Mexico).
### Games
• Pokémon Murkrow (Yamikarasu) is a Pokémon number # 198 of the National Pokédex. Murkrow is sinister and flying type Pokémon in the second generation. It is a Flying Egg group Pokémon. Murkrow's other indexes are Kalos Centro index 051 - Highland Pokédex, Sinnoh index 074 , Johto index 208 .
### History and politics
• United Nations Security Council Resolution number 198, adopted 18 December 1964. Extending peacekeeping force in Cyprus for 3 months. Resolution text.
### Mathematics
• 198 is the smallest number n such that the sum of its proper divisors minus the product of its digits equals n
• 198 = 11 + 99 + 88.
## № 198 in other languages
How to say or write the number one hundred and ninety-eight in Spanish, German, French and other languages.
Spanish: 🔊 (número 198) ciento noventa y ocho German: 🔊 (Nummer 198) einhundertachtundneunzig French: 🔊 (nombre 198) cent quatre-vingt-dix-huit Portuguese: 🔊 (número 198) cento e noventa e oito Hindi: 🔊 (संख्या 198) एक सौ, अट्ठानवे Chinese: 🔊 (数 198) 一百九十八 Arabian: 🔊 (عدد 198) مائةثمانية و تسعون Czech: 🔊 (číslo 198) sto devadesát osm Korean: 🔊 (번호 198) 백구십팔 Danish: 🔊 (nummer 198) ethundrede og otteoghalvfems Hebrew: (מספר 198) מאה תשעים ושמנה Dutch: 🔊 (nummer 198) honderdachtennegentig Japanese: 🔊 (数 198) 百九十八 Indonesian: 🔊 (jumlah 198) seratus sembilan puluh delapan Italian: 🔊 (numero 198) centonovantotto Norwegian: 🔊 (nummer 198) en hundre og nitti-åtte Polish: 🔊 (liczba 198) sto dziewięćdzisiąt osiem Russian: 🔊 (номер 198) сто девяносто восемь Turkish: 🔊 (numara 198) yüzdoksansekiz Thai: 🔊 (จำนวน 198) หนึ่งร้อยเก้าสิบแปด Ukrainian: 🔊 (номер 198) сто дев'яносто вiсiм Vietnamese: 🔊 (con số 198) một trăm chín mươi tám Other languages ...
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If you know something interesting about the number 198 or any other natural number (positive integer), please write to us here or on Facebook. | 2,865 | 9,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.729669 |
http://math.stackexchange.com/questions/160272/subset-of-mathbbi-cap-0-1-irrationals-in-0-1-that-is-closed-in-math | 1,469,442,093,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.33/warc/CC-MAIN-20160723071024-00188-ip-10-185-27-174.ec2.internal.warc.gz | 152,598,791 | 19,164 | # Subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has measure $\epsilon \in (0,1)$
Measure theory guarantees that every Lebesgue finite measurable set $E$ has a closed subset $F$ such that $m(E \backslash F)<\epsilon$ for small $\epsilon$.
But today I saw in some text that for all small $\epsilon >0$ there exist a subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has Lebesgue measure less than $\epsilon$? Note the set is required to be closed in $\mathbb{R}$, not in $\mathbb{I}\cap [0,1]$. Can someone give me an example / a proof that such thing does not exist?
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What is $\mathbb{I}$? – Chris Eagle Jun 19 '12 at 10:28
@ChrisEagle Irrationals. I have fixed it. – Polymorpher Jun 19 '12 at 10:29
Such a thing does indeed exist. See this answer, for example. – Dejan Govc Jun 19 '12 at 10:44
@DejanGovc Thanks! Can you make it an answer so that people can see this question is answered? – Polymorpher Jun 19 '12 at 11:09
In what text did you see this and what page? – T. Eskin Jun 19 '12 at 11:55
In fact, we can choose a set which have exactly measure $\varepsilon$.
For a fixed $\delta>0$, consider the set $S_{\delta}:=\bigcup_{n\in\mathbb N}(q_n-2^{-n}\delta,q_n+\delta 2^{-n})$, where $\{q_n,n\in\Bbb N\}$ is an enumeration of the rationals of $[0,1]$. Then $S_{\delta}$ is open and dense in $[0,1]$, since it contains all the rationals of this interval. The maps $f\colon\delta\mapsto \lambda(S_{\delta}\cap (0,1))$ is Lipschitz-continuous. Indeed, if $\delta_1\leq\delta_2$, we have \begin{align*}f(\delta_2)-f(\delta_1)&=\lambda(S_{\delta_2}\setminus S_{\delta_1})\\\ &\leq \lambda\left((0,1)\cap \bigcup_{n=0}^{+\infty}(q_n-2^{-n}\delta_2,q_n+\delta_2 2^{-n})\setminus (q_n-2^{-n}\delta_1,q_n+\delta_1 2^{-n})\right)\\\ &\leq \sum_{n=0}^{+\infty}\lambda((q_n-2^{-n}\delta_2,q_n+\delta_2 2^{-n})\setminus (q_n-2^{-n}\delta_1,q_n+\delta_1 2^{-n}))\\\ &=2(\delta_2-\delta_1)\sum_{n=0}^{+\infty}2^{-n}. \end{align*} Now we use the intermediate value theorem to pick $\delta$ such that $\lambda(S_{\delta}\cap (0,1))=1-\varepsilon$, and we consider the complement in $[0,1]$ of $S_{\delta}$.
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And with a bit more work we can find a more "constructive" construction of such a set (at least, one that doesn't rely on the intermediate value theorem). Let $U_0 = \emptyset$, and $U_n = U_{n-1} \cup (q_n - r, q_n + r)$ where $r$ is chosen so that $\lambda(U_n) = (1-2^{-n})(1-\epsilon)$. Since $U_{n-1}$ is a finite union of intervals, $\lambda(U_n)$ is a piecewise-linear function, so this can be done explicitly. – Robert Israel Jun 24 '12 at 6:58
Let $\epsilon > 0$. Enumerate the rationals in $[0,1]$ with the sequence $\{q_n\}_{n=1}^\infty$. Now for each $n$, choose an open interval $I_n\subseteq [0,1]$ containing $q_n$ with length less than $\epsilon/2^n$ If $q_n = 0$ or $q_n = 1$, choose a half-open interval. Said half-open intervals are still relatively open in $[0,1]$.
Put $G = \bigcup_n I_n$. The Lebesgue measure of $G$ is at most $\epsilon$. Now put $F = [0,1] - G.$
The set $G$ is open. The set $F$ is closed. And its measure is at least $1 - \epsilon$. Since $G$ contains all of the rationals in $[0,1]$, $F$ contains only irrationals.
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Sorry I messed up a little bit in my question. What I really meant was for all small $\epsilon$, the subset can be made to have measure less than $\epsilon$. The measure of F being constructed is bounded below by $\epsilon$, not above. – Polymorpher Jun 19 '12 at 11:45 | 1,210 | 3,526 | {"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} | 3.125 | 3 | CC-MAIN-2016-30 | latest | en | 0.76027 |
http://math.stackexchange.com/questions/113580/example-of-a-subalgebra-of-infinite-generated-algebra-which-cannot-be-extended-t/113608 | 1,469,378,419,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00242-ip-10-185-27-174.ec2.internal.warc.gz | 155,378,114 | 19,019 | # Example of a subalgebra of infinite-generated algebra which cannot be extended to a maximal subalgebra
First of all, sorry for my English, especially mathematical one. The problem is that I know how to call such things in Ukrainian but unfortunately did not manage to find proper translations to English. So, I will give some brief definitions of terms in order to not confuse you with the names.
So, an algebra (maybe, partial) is a set of elements together with a set of operations defined on these elements $(A, \Omega)$.
Let $B \subset A$. Then a closure $[B]_f$ of $A$ by an $n$-ary operation $f \in \Omega$ is a set defined by two rules:
1. $B \subseteq [B]_f$
2. $\forall (a_1, a_2, \ldots, a_n) \subset [B]_f$ if $f(a_1, a_2, \ldots, a_n)$ is defined then $f(a_1, a_2, \ldots, a_n) \in [B]_f$
A closure $[B]$ of $A$ is $B^0 \cup B^1 \cup B^2 \ldots$ where $B^0 = B$, $B^{i+1} = \bigcup_{f \in \Omega}[B^i]_f$.
The definition of subalgebra and therefore algebra extension is obvious, I think.
$B$ is a system of generators (I believe, it is close to the basis) for algebra $(A,\Omega)$ if $[B] = A$.
And, finally, subalgebra $(B,\Omega)$ is called maximal subalgebra of $(A,\Omega)$ if there is no subalgebra $(C,\Omega)$ such that $B \subset C \subset A, B \neq A, C \neq A, C \neq B$.
It can be proved that for a algebra with existing finite system of generators each its subalgebra can be extended to some maximal subalgebra. But for algebras with infinite systems of generators, this statement is not always true. And I need a counter-example. Therefore, I need an example of infinite-based algebra and a subalgebra which is impossible to extend to some maximum subalgebra.
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I thought about infinite-based vector spaces and replacements of all vectors with i-th coordinate equal to some number (maybe, to all numbers <=i) for each i, but I didn't manage to come up with something. – ikostia Feb 26 '12 at 13:07
Unfortunately, every proper subspace $W$ of a vector space $V$ is contained in a maximal (I assume by this you also mean it doesn't equal the whole thing) subspace; just take a vector $v \in V$ not in $W$, let $B$ be a basis of $W$, extend $B \cup \{ v \}$ to a basis of $V$, and consider the span of this basis minus $v$. – Qiaochu Yuan Feb 26 '12 at 13:10
Your "creators" are usually called "generators". – g.castro Feb 26 '12 at 13:25
Anyway, I would recommend you try thinking about groups instead of vector spaces. In fact there is a relatively straightforward example of an abelian group which has no maximal proper subgroups whatsoever. – Qiaochu Yuan Feb 26 '12 at 13:26
@g.castro: thanks – ikostia Feb 26 '12 at 13:31
The answer is an abelian group of rational numbers $Q$. It has no maximal subgroups at all. Proof can be found here. Thanks, Qiaochu Yuan, for the direction.
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EDITED: Hint: Consider the operation $f(n) = n-1$ on the natural numbers (with $f(0)$ arbitrary).
The original hint (Consider an algebra without any operations, or with only a trivial operation) was completely wrong.
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Hm, identity operation cannot give you any new elements. This means, that if you have a system B, then the closure of this system by identity operation would give you B again. This in turn means that adding new elements to B means nothing but adding new elements, so if you just take any subset B of algebra base A, you can add all but one of the A\B elements to B and you will get a maximal subalgebra. I believe, that very alike thought have place with constant operation. Where am I wrong? – ikostia Feb 26 '12 at 13:51
Thank you. I sketched a new construction. – g.castro Feb 26 '12 at 20:25
This one seems to be really cool and easy to understand! Thanks! – ikostia Feb 27 '12 at 9:33
I'm willing to edit your answer in purpose to add a full proof. Please, respond if it is okay. Or, if you would like, you can do it by yourself. – ikostia Feb 27 '12 at 9:40
Sure, go ahead. – g.castro Feb 27 '12 at 10:07 | 1,129 | 3,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-30 | latest | en | 0.877613 |
https://www.siemensstemday.com/educators/activities?c=4 | 1,660,632,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572221.38/warc/CC-MAIN-20220816060335-20220816090335-00373.warc.gz | 846,584,403 | 10,171 | # Activities
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## A Ball In Motion
• Energy
• 9–12
Level: 4
Students will investigate the relationship between quadratic functions and the parabolic path traveled by a ball in motion. Students will analyze data to understand the mathematical relationships that exist along the path of a ball in flight.
• Math
• Science
.pdf
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## All Geared Up!
• Manufacturing
• 9–12
Level: 3
Learn about gears, how they work, and differences in gear size as well as develop an understanding of angular speed. Students will analyze a variety of situations by applying arc length and other trigonometric functions to determine degrees of rotation.
• Math
• Technology
.pdf
## 3D Printing Robots
• IT
• 9–12
Level: 3
Students will investigate the application of 3D printing to space technology.
• Math
• Science
• Technology
.pdf
## Credit Card Debt
• IT
• 9–12
Level: 4
Description: Students consider the benefits and tradeoffs of using credit and learn about the role interest plays in using credit cards. Then, they create an equation that describes the length of time it takes to pay off a debt.
• Math
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## Hole-in-One
• Manufacturing
• K–5
Level: 3
In this activity, students will design their own miniature golf hole and then determine the path along which a ball would travel to score a "hole-in-one."
• Math
.pdf
## How Does Your City Grow
• IT
• 9–12
Level: 4
Working in groups, students will select a city and then use U.S. government census data to develop an algebraic relationship between time and population size.
• Math
.pdf
## Laser Beaming
• Manufacturing
• K–5
Level: 3
In this activity, students will use a laser pointer and flat mirrors to explore the law of reflection.
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## Modeling Constraints
• IT
• 9–12
Level: 2
Students will work in small groups to investigate constraints of starting a business. Each group will be assigned a specific constraint. Small groups will then write a linear inequality and graph the inequality on a group capture sheet.
• Math
.pdf
## Pixel Art
• IT
• 9–12
Level: 3
Students work together to build an algorithm for drawing a pixel picture using coordinate directions and color assignments on graph paper. Then, students follow the algorithms of other groups to create the images on a larger scale with post-it notes.
• Math
.pdf
• IT | 550 | 2,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-33 | latest | en | 0.871425 |
https://app.leg.wa.gov/wac/default.aspx?cite=458-57-125 | 1,601,415,436,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00107.warc.gz | 259,714,337 | 8,214 | HTML has links - PDF has Authentication
458-57-115 << 458-57-125 >> 458-57-135
Apportionment of tax when out-of-state property is included in the gross estate of a decedent.
(1) Introduction. This rule applies to deaths occurring on or after May 17, 2005, and discusses how to apportion the estate tax when there is out-of-state property included in the gross estate. The estate tax rule on apportionment of estate tax for deaths occurring on or before May 16, 2005, can be found in WAC 458-57-025.
(2) Calculation of apportioned tax. Apportionment of the tax is allowed for estate property located outside of Washington, even if the other state where the out-of-state property is located does not impose an estate tax. The amount of tax is determined by multiplying the preapportioned tax using Table W by a fraction. The numerator of the fraction is the value of the property included in the decedent's gross estate that is located in Washington. The denominator of the fraction is the value of the decedent's gross estate. Intangible property is located in Washington if the decedent was a resident of this state at death. Property qualifying for the farm deduction is excluded from the numerator and denominator of the fraction. See WAC 458-57-155, Farm deduction, for additional information.
(a) Example - Washington resident decedent. A widow dies during 2014 leaving a gross estate of \$4.1 million. The decedent was a Washington resident at death. Decedent's primary residence is located in Seattle, Washington. The decedent also owned a second home in Arizona valued at \$300,000 and unimproved real property in South Dakota valued at \$750,000. The estate had \$100,000 in expenses deductible for federal estate tax purposes. The applicable exclusion amount for 2014 after adjustment for inflation is \$2,012,000.
Under the facts of this example the estate owes Washington estate tax on a Washington taxable estate of \$1,988,000, computed as shown below:
Gross estate: \$4,100,000 Less allowable deductions: (\$100,000) Less applicable exclusion amount: (\$2,012,000) Washington taxable estate: \$1,988,000
The preapportionment Washington estate tax for this estate, using the table provided in WAC 458-57-115 (3)(a), equals \$238,320, computed as follows: \$100,000 + (\$988,000 x 14%) = \$238,320.
Because the decedent owned out-of-state property, a house in Arizona and unimproved real property in South Dakota that are not subject to Washington estate tax, the tax due to Washington is calculated by multiplying the amount of preapportionment tax computed above by the fraction described in this subsection (2). Also, because the decedent was a Washington resident at death, the numerator of the fraction is the value of all property included in the decedent's gross estate that is located in this state, including the decedent's intangible personal property. The denominator of the fraction is the value of the decedent's gross estate. Using the facts in our example, the tax owed to Washington equals \$177,287, computed as follows: ((\$4,100,000 - \$1,050,000)/\$4,100,000) x \$238,320 = \$177,287.
(b) Example – Nonresident decedent. A widow dies during 2013 leaving a gross estate of \$6 million. The decedent was a Colorado resident at death and all of the decedent's property is located in that state, except for a vacation home located in Washington valued at \$650,000. The estate had \$100,000 in expenses deductible for federal estate tax purposes. The applicable exclusion amount for 2013 is \$2,000,000.
Under the facts of this example, the estate owes Washington estate tax on a Washington taxable estate of \$3,900,000, computed as shown below:
Gross estate: \$6,000,000 Less allowable deductions: (\$100,000) Less applicable exclusion amount: (\$2,000,000) Washington taxable estate: \$3,900,000
The preapportionment Washington estate tax for this estate, using the table provided in WAC 458-57-115 (3)(a), equals \$534,000, computed as follows: \$390,000 + (\$900,000 x 16%) = \$534,000.
Because the decedent owned property located outside Washington, the tax due to Washington is calculated by multiplying the amount of preapportionment tax computed above by the fraction described in this subsection (2). Also, because the decedent was not a Washington resident at death, the numerator of the fraction does not include the value of decedent's intangible personal property. The denominator of the fraction is the value of the decedent's gross estate. Using the facts in this example, the tax owed to Washington equals \$57,850, computed as follows: (\$650,000/\$6,000,000) x \$534,000 = \$57,850.
(3) When is property located in Washington? The location of property owned by the decedent is determined at the time of death.
(a) All real property physically situated in this state, with the exception of federal trust lands, and all interests in such property, is located in Washington. Interests in real property include, but are not limited to:
(i) Mineral interests;
(ii) Decedent's beneficial interest in real property held in trust; and
(iii) Decedent's interest in jointly owned property (e.g., tenants in common, joint with right of survivorship).
(b) Tangible personal property of a decedent is located in Washington if:
(i) At the time of death the property is situated in Washington; and
(ii) It is present for a purpose other than transiting the state.
(c) Intangible personal property of a decedent is located in Washington if the decedent was a resident of this state at death.
(d) Example. A nonresident decedent was a construction contractor doing business as a sole proprietor. The decedent was constructing a large building in Washington. At the time of death, any of the decedent's equipment that was located at the job site, such as tools, earthmovers, bulldozers, trucks, etc., is located in Washington for estate tax purposes because that property was present in the state for a purpose other than transiting the state.
[Statutory Authority: RCW 83.100.200, 82.32.300, 82.01.060(2), 83.100.020, 83.100.040, 83.100.047, 83.100.048, 83.100.120, and 83.100.210. WSR 14-14-075, § 458-57-125, filed 6/27/14, effective 7/28/14. Statutory Authority: RCW 83.100.047 and 83.100.200. WSR 06-07-051, § 458-57-125, filed 3/9/06, effective 4/9/06.]
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Translate | 1,584 | 6,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.95863 |
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# Analysis and Decision Tree
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Techware Incorporated is considering the introduction of two new software products to the market. In particular, the company has four options regarding these two proposed products: introduce neither product, introduce product 1 only, introduce product 2 only, or introduce both products. Research and development costs for products 1 and 2 are \$180,000 and \$150,000, respectively. Note that the first option entails no costs because research and developments efforts have not yet begun.
The success of these software products depends on the trend of the national economy in the coming year and on the consumers' reaction to these products. The company's revenues earned by introducing product 1 only, product 2 only, or both products in various states of the national economy are given below. The probabilities of observing a strong, fair, and weak trend in the national economy in the coming year are 0.30, 0.50, and 0.20, respectively.
Revenue table for Techware's decision problem
Trend in national economy
Strong Fair Weak EMV
Decision
Introduce neither product \$0 \$0 \$0 \$0
Introduce product 1 only \$500,000 \$260,000 \$120,000 \$304,000
Introduce product 2 only \$420,000 \$230,000 \$110,000
Introduce both products \$820,000 \$390,000 \$200,000
Probability 0.3 0.5 0.2
a) Calculate the EMV of the two latter alternatives and write results in cells above.
b) Construct a decision tree of the product manager's decision and identify the course of action that maximizes EMV. | 394 | 1,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | longest | en | 0.905455 |
http://mathhelpforum.com/discrete-math/36807-proof-countable-print.html | 1,527,481,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870771.86/warc/CC-MAIN-20180528024807-20180528044807-00338.warc.gz | 193,439,109 | 2,602 | # proof of countable
• May 1st 2008, 11:23 AM
natewalker205
proof of countable
Prove that the set {1,2,3}xnatural numbers is countable.
• May 1st 2008, 11:30 AM
Isomorphism
Quote:
Originally Posted by natewalker205
Prove that the set {1,2,3}xnatural numbers is countable.
Hint:
if $\displaystyle X = \{1,2,3\} \times \mathbb{N}, \text{consider } f:X \to \mathbb{N}$ defined as $\displaystyle f((i,n)) = 3(n-1)+i$
P.S: Though I have tailored this definition for this problem, any cartesian product of countable sets is countable | 181 | 531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | latest | en | 0.869966 |
http://www.avrocks.com/excel-match-data-multiple-columns.html | 1,568,681,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572980.56/warc/CC-MAIN-20190917000820-20190917022820-00367.warc.gz | 243,366,479 | 7,110 | # Excel Match Data Multiple Columns
So I have two sheets with same type of data but for different months from different systems. I want to see if both systems have same data for each ID and if for each SK ID the org ID and Entity ID matches.
First sheet has 50,000 columns and second one is 150,000.
Columns :- A :- SK ID B :-Org ID C :- Entity ID
So SK ID in Sheet 1 should match SK ID in Sheet 2
AND For each SK ID that matches, does the Org ID and Entity ID associated with it also matches for both sheets? If not, then what doesn't match? Do all three differ or just Org ID is different rest matches, etc. There are various duplicates for IDs so that I'll take care of next.
One sheet has less rows than other one, so I'll be using the sheet with less rows to do the matching.
I tried using index, match, lookups, if statements, nothing seems to be working for some reason.
If it's possible for the output to be "Match" or "No Match" or maybe something where I can have output from the other table and then I can put a if statement to see if A1 matches B1 to C1, etc.
Again, I want to see if A1,B1,and C1 would match with ANY cell from A1-A150,000, B1-B150000, and C1-C150000, in other sheet. And if all match then maybe say "Match" and if one or other doesn't match then list or tell what doesn't?
Replay
You should be able to do this with a formula like this:
``````=IF(ISNA(MATCH(A1,Sheet2!\$A\$1:\$A\$150000,0)+MATCH(B1,Sheet2!\$B\$1:\$B\$150000,0)+MATCH(C1,Sheet2!\$C\$1:\$C\$150000,0)),"No Match","Match")
```
```
If it can't find one of them you get "No Match". If it finds all three you get "Match".
Edit: If you're checking that the are all on the same row, then you would use something like this:
``````=IF(ISNA(MATCH(A1&B1&C1,Sheet2!\$A\$1:\$A\$150000&Sheet2!\$B\$1:\$B\$150000&Sheet2!\$C\$1:\$C\$150000,0)),"No Match","Match")
```
```
This is an array formula so it needs to be entered using Ctrl+Shift+Enter.
Category: microsoft excel Time: 2016-07-29 Views: 0 | 577 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-39 | latest | en | 0.888408 |
https://www.cocoandlowe.com/what-is-the-average-pension-of-a-federal-employee-7b188d08/ | 1,709,267,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00760.warc.gz | 703,722,981 | 18,450 | # What is the average pension of a federal employee?
The average civilian federal employee who retired in FY 2016 was 61.5 years old and had completed 26.8 years of federal service. he average monthly annuity payment to workers who retired under CSRS in FY 2018 was \$4,973. Workers who retired under FERS received an average monthly annuity of \$1,834.
>> Click to read more <<
## Correspondingly, how many years do you have to work for the federal government to get a pension?
5 years
Also to know is, can I retire after 20 years of federal service? Immediate Retirement
If you retire at the MRA with at least 10, but less than 30 years of service, your benefit will be reduced by 5 percent a year for each year you are under 62, unless you have 20 years of service and your benefit starts when you reach age 60 or later.
## Similarly, how much does a GS 13 make in retirement?
How much does a GS 13 make in retirement? Payment for a GS-12, Step 10, Rest of the US, is \$ 95,388 in 2018. Using that as a maximum of 3, and with 30 years and under 62, that equates to an income of 28,616 \$ (\$ 25,754 with survivor benefit). At age 62 or older, it would be \$ 31,478 (\$ 28,330).
## How do I calculate my federal retirement income?
FERS (Immediate or Early)
FERS annuities are based on high-3 average pay. Generally, the benefit is calculated as 1 percent of high-3 average pay multiplied by years of creditable service. For those retiring at age 62 or later with at least 20 years of service, a factor of 1.1 percent is used rather than 1 percent.
## How much will my FERS pension be?
How much does this equal in guaranteed pension income? FERS Pension = 1% x high-3 salary x years worked. FERS Pension = 1.1% x high-3 salary x years worked. This equals 1% – 1.1% of your highest annual salary for every year of federal service.
## Can I retire after 30 years of federal service?
Normally, an employee is eligible to retire from federal service when the employee has at least 30 years of service and is at least age 55 under the Civil Service Retirement System or 56 and two months under the Federal Employees Retirement System; has at least 20 years of service and is at least age 60; or has at …
## Can I retire after 10 years of federal service?
An employee who has reached a minimum retirement age is entitled to immediate benefits after 10 to 30 years of service. Again, if they have less than 30 years in service, benefits are reduced by 5% for each year they are under age 62 unless they’ve reached 20 years of service and retire at age 60 or older.
## Can you retire after 10 years of work?
Since you can earn 4 credits per year, you need at least 10 years of work that subject to Social Security to become eligible for Social Security retirement benefits. | 686 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.96982 |
http://www.chegg.com/homework-help/introduction-to-mathematical-statistics-6th-edition-chapter-3.2-solutions-9780130085078 | 1,474,938,216,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660916.37/warc/CC-MAIN-20160924173740-00253-ip-10-143-35-109.ec2.internal.warc.gz | 380,294,325 | 16,459 | Introduction to Mathematical Statistics (6th Edition) View more editions Solutions for Chapter 3.2
• 1025 step-by-step solutions
• Solved by publishers, professors & experts
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Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
100% (4 ratings)
If the random variable X has a Poisson distribution such that P(X = 1) =P(X = 2), find P(X = 4).
SAMPLE SOLUTION
Chapter: Problem:
100% (4 ratings)
• Step 1 of 2
If, we have
For the Poisson distribution the density function is given by,
and
The two probabilities are equal, then.
• Step 2 of 2
Now, we have to find
Hence,
Therefore, the required probability is 0.090.
Corresponding Textbook
Introduction to Mathematical Statistics | 6th Edition
9780130085078ISBN-13: 0130085073ISBN: Authors: | 227 | 834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.770755 |
https://jsxgraph.uni-bayreuth.de/wiki/index.php?title=Epidemiology:_The_SEIR_model&diff=prev&oldid=2337 | 1,643,186,569,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00449.warc.gz | 395,298,469 | 7,912 | # Difference between revisions of "Epidemiology: The SEIR model"
Jump to navigationJump to search
For many important infections there is a significant period of time during which the individual has been infected but is not yet infectious himself. During this latent period the individual is in compartment E (for exposed).
Assuming that the period of staying in the latent state is a random variable with exponential distribution with parameter a (i.e. the average latent period is $a^{-1}$), and also assuming the presence of vital dynamics with birth rate equal to death rate, we have the model:
$\frac{dS}{dt} = \mu N - \mu S - \beta \frac{I}{N} S$
$\frac{dE}{dt} = \beta \frac{I}{N} S - (\mu +a ) E$
$\frac{dI}{dt} = a E - (\gamma +\mu ) I$
$\frac{dR}{dt} = \gamma I - \mu R.$
Of course, we have that $S+E+I+R=N$.
The lines in the JSXGraph-simulation below have the following meaning:
* Blue: Rate of susceptible population
* Black: Rate of exposed population
* Red: Rate of infectious population
* Green: Rate of recovered population (which means: immune, isolated or dead) | 284 | 1,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-05 | latest | en | 0.887295 |
http://www.ques10.com/p/7833/engineering-graphics-question-paper-jun-2015-fir-1/ | 1,560,670,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997801.20/warc/CC-MAIN-20190616062650-20190616084650-00241.warc.gz | 297,458,198 | 9,036 | Question Paper: Engineering Graphics : Question Paper Jun 2015 - First Year Engineering (Semester 1) | Gujarat Technological University (GTU)
0
Engineering Graphics - Jun 2015
First Year Engineering (Semester 1)
TOTAL MARKS: 100
TOTAL TIME: 3 HOURS
(1) Question 1 is compulsory.
(2) Attempt any four from the remaining questions.
(3) Assume data wherever required.
(4) Figures to the right indicate full marks.
Objective Question
Attempt all questions 1 (a) 1 When the plane cuts the cone parallel to one of the generator, the curve trace out by section is
(a) Ellipse (b) Parabola (c) Hyperbola (d) Involute
(1 marks)
1 (a) 2 In first angle projection method The Left hand side view is placed on
(a) Above elevation (b) Right side of elevation (
C) below elevation (d) Left side of elevation
(1 marks)
1 (a) 3 A French curve is used to draw
(a) Circles
(b)Smooth freeform curve (c) Right circular cone (d) orthographic projection
(1 marks)
1 (a) 4 If line is inclined to vertical plane and parallel to Horizontal plane, then which one of the following is always correct?
(a) True length = Plan length
(b)True length = Elevation length
(c)True length ≠ Plan length
(d)None of the above
(1 marks)
1 (a) 5 To obtain the true shape of the section of solid, an auxiliary plane is set at
(a) Parallel to cutting plane
(b)Perpendicular to cutting plane (c ) Parallel to Ground plane
(d)Perpendicular to XY Plane
(1 marks)
1 (a) 6 A square plane is inclined to HP & perpendicular to VP its elevation appears as
(a) Rhombus
(b)Square (c) Straight line
(d)Rectangle
(1 marks)
1 (a) 7 Length (L) of line in Isometric view will be equal to
(a) 0.707 L
(b)0.815 L (c) True length L (d) 0.866 L
(1 marks)
Attempt all questions
1 (b) 1 For scale, which one is not correct
(a) 1:2
(b)1:20
(c)1:1/2
(d)1/2
(1 marks)
1 (b) 2 In Isometric view length, Width, and Height are inclined at
(a) 30°
(b)60°
(c)90°
(d)120°
(1 marks)
1 (b) 3 When a point is below HP & in Front of VP it is in
(1 marks)
1 (b) 4 In orthographic view the lines Perpendicular to arrow X are drawn as (1) (1) Parallel to XY in Plan (2) Parallel to XY in elevation (3) Perpendicular to XY in Elevation
(a) 1
(b)2
(c)3
(d)1&2
(1 marks)
1 (b) 5 When the cone, resting on base on V.P., is cut by section plane parallel to V.P. then the true shape is __________ and can be seen in __________ view.
(a) Circle, Front
(b)Ellipse, Front
(c)Ellipse, Top
(d)Circle, Top.
(1 marks)
1 (b) 6 The isometric view of a vertical line is represented at an angle of ____ in front view and having a length _________ the original length of line.
(a) 30° , Same as
(b)30° , Less than
(c)90°, Same as
(d)90° , Less than
(1 marks)
1 (b) 7 While drawing the isometric view of the sphere, its diameter is taken as
(a) Equal to actual diameter
(b)11/9 times of the actual diameter
(c)21/9 times of the actual diameter
(d)none of the above
(1 marks)
2 (a) Construct the ellipse if the distance between the focus and the directrix is 50 mm. & The eccentricity is 2/3. Draw the tangent and the normal to the ellipse at given point.(7 marks) 2 (b) The projectors of the ends of a line AB are 50mm apart. The end A is 20mm above the H.P. and 30mm in front of the V.P. The end B is 10mm below the H.P. and 40mm behind the V.P. Determine the true length of line AB, its inclinations with H.P. and V.P. and apparent angles also(7 marks) 3 (a) Construct the Involute of circle of 30 mm diameter for one turn. Draw tangent and normal to the Involute at any point on it..(7 marks) 3 (b) A line AB, 100 mm long, is inclined at 45° to HP. The end A is 10 mm above the HP and is 65 mm in front of the VP Draw projections of the line if its Front View measures 90 mm and find the inclination of the line with the VP.(7 marks) 4 (a) A hexagonal pyramid, side of the base 25 mm long and height 70 mm resting on HP on its side, has one of its triangular faces perpendicular to the HP and inclined at 60° to VP. Draw its projections.(7 marks) 4 (b) Draw an Archimedean spiral of 1.5 convolutions, the greatest and least radii being 115 mm and 25 mm respectively. Draw tangent and normal to the spiral at any point on the curve.(7 marks) 5 (a) A hexagonal Prism, side of base 30 mm and height 60 mm, is standing upright with base on H.P. one side of the base and axis are parallel to V.P. It is cut by section plane making an angle of 60° to H.P. and crossing the axis 10 mm from the top. Draw top view, sectional front view, sectional left hand side view and true shape of section.(7 marks) 5 (b) A rhombus of negligible thickness is having its diagonals 100 mm and 50 mm long. Draw the projections of the rhombus when the longer diagonal is inclined at 30° to the Horizontal Plane and 30° to Vertical Plane.(7 marks) 6 (a) 1, In orthographic projection why second and fourth angle projection method are not used?
2. Draw the symbol of First angle projection method.
3. Give the dimension of Title block and list the information given in it.
(7 marks)
6 (b) Using the Third angle projection method, draw the following view for the FIGURE-1. Give the dimensions using the Aligned dimensioning method. (i) front view (ii) Top view (iii) Full Sectional Right Hand Side End View. (7 marks) 7 (a) Construct a diagonal scale of representative fraction = (1/36) showing yard, foot and inch. Scale should be long enough to measure 5 yard.(7 marks) 7 (b) The orthographic views of an object using the first angle projection method are shown in the FIGURE-2 . Draw the isometric projection (7 marks) | 1,573 | 5,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-26 | latest | en | 0.80726 |
https://www.education.com/worksheets/divisibility-rules/ | 1,670,319,523,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711077.50/warc/CC-MAIN-20221206092907-20221206122907-00063.warc.gz | 792,815,015 | 43,258 | # Search Printable Divisibility Rule Worksheets
18 filtered results
18 filtered results
Divisibility Rules
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Worksheet
Introduce your students to divisibility rules with this handy worksheet that goes over the rules and asks students to fill in examples.
Math
Worksheet
Math Review Part 3: Let's Soar in Grade 4
Worksheet
Math Review Part 3: Let's Soar in Grade 4
In this review assessment, your students will flex their math muscles as they tackle division word problems.
Math
Worksheet
Division: Equal Groups (Part One)
Worksheet
Division: Equal Groups (Part One)
After looking at a sample model, write a division sentence to represent each drawing.
Math
Worksheet
Divisibility Rules: Dividing by 4
Worksheet
Divisibility Rules: Dividing by 4
Learners review the divisibility rule for 4Â with this fun and helpful practice worksheet!
Math
Worksheet
Divisibility Puzzle
Worksheet
Divisibility Puzzle
Challenge your students to think deeply and apply what they have learned about divisibility rules.
Math
Worksheet
Divisibility Rules: Dividing by 3
Worksheet
Divisibility Rules: Dividing by 3
Help math learners review the divisibility rule for 3Â with this fun practice worksheet!
Math
Worksheet
Division: Find the Total Number of Groups
Worksheet
Division: Find the Total Number of Groups
Use division to figure out the total number of groups! Colorful illustrations make this exercise fun and engaging.
Math
Worksheet
Divisibility Rules: Dividing by 2
Worksheet
Divisibility Rules: Dividing by 2
Help math learners review the divisibility rule for 2 with this fun practice worksheet!
Math
Worksheet
Matching Factor Tree Cards
Worksheet
Matching Factor Tree Cards
Students will match a product to its factor tree and list of prime factors in this interactive activity.
Math
Worksheet
Division: Find the Total Number of Objects
Worksheet
Division: Find the Total Number of Objects
You can use division to find the total number of objects in each group.
Math
Worksheet
Is it Divisible by 10?
Worksheet
Is it Divisible by 10?
Ease some of the difficulty of division and improve your fourth grader's number knowledge with this worksheet that reviews divisibility by 10.
Math
Worksheet
Is it Divisible by... 2?
Worksheet
Is it Divisible by... 2?
Show your fourth grader how to tell if a number is divisible by two, with this helpful worksheet.
Math
Worksheet
Glossary: The Factor Tree Strategy
Worksheet
Glossary: The Factor Tree Strategy
Use this glossary with the EL Support Lesson: The Factor Tree Strategy.
Math
Worksheet
Number Categorization
Worksheet
Number Categorization
Does your fourth grader need help with times tables? See if she can pick out which number does not belong in each group by finding a common multiple.
Math
Worksheet
Vocabulary Cards: The Factor Tree Strategy
Worksheet
Vocabulary Cards: The Factor Tree Strategy
Use these vocabulary cards with the EL Support Lesson: The Factor Tree Strategy.
Math
Worksheet
Divisibility Rules: Dividing by 5
Worksheet
Divisibility Rules: Dividing by 5
Practice applying the divisibility rule for 5 with this fun and helpful division worksheet!
Math
Worksheet
Divisibility Rules: Dividing by 6
Worksheet
Divisibility Rules: Dividing by 6
Introduce learners to the divisibility rule for 6 with this helpful practice worksheet!
Math
Worksheet
Divisibility Rules: Dividing by 8
Worksheet
Divisibility Rules: Dividing by 8
Learners explore the divisibility rule for 8 in this friendly practice worksheet!
Math
Worksheet
### Search Printable Divisibility Rule Worksheets
It’s time to talk about divisibility rules! Dive into the nuts and bolts of division with these educator-created worksheets that explore some key tips and tricks to elementary division. This curated collection covers the basics of divisibility rules, number categorization, factor trees, and division word problems for your child to practice. | 870 | 3,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.868395 |
https://www.dreamencyclopedia.net/number | 1,675,304,844,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00173.warc.gz | 761,857,878 | 11,136 | What does it mean to see an number in a dream?
## Number Dream Meaning: From 4 Different Sources
### Number
Symbol: Ever since the times of Pythagoras and Plato, numbers have played a vital role in explaining the order of the Universe.
Depth Psychology: Dreaming about a specific number: the dream number might prove to be the lucky number you need to win the lottery. See the chapter “Numbers in Dreams.” Number One See the chapter “Numbers in Dreams.”
### Number
1. Personal, individual power (note which numbers). Possible difficulties regarding financial or business affairs.
A measurement of time (as in “days are numbered”).
### Number
Dreams of numbers represent order, logic, and a desire to figure things out, solve a problem, assign a value, classify, and/or organize. Each number has its own significance; look up the number in this dictionary.
If the number in your dream is more than a single digit, then add them together to get a single digit.
For example 196 would be 7 (1+9+6=16, 1+6=7). See One-Ten, Twenty-One, Eleven, 24/7, 365 and Math.
### Number
A specific number can be significant itself (such as “1” representing being alone, or “4” representing stability), or it might represent something specific (such as an age, a year, or a number of people or objects).
Dreaming about numbers or math can also mean that your logical mind was working hard, perhaps trying to understand or solve a challenge in your life.
For more clues, consider the number’s context, your feelings about it during the dream, and any personal meaning it has for you.
Sometimes numbers show up when you’re dropping off to sleep or waking up that can seem important at the time, although they often have no significant meaning.
## 15 dream interpretations related to the symbols you see in your dreams.
### Numbers
To dream of numbers, denotes that unsettled conditions in business will cause you uneasiness and dissatisfaction. See Figures. ... numbers dream meaning
### Telephone Number
Dreams of a telephone number symbolize the importance of making contact or getting your message through to a particular person.
If you are unable to dial the numbers or unable to establish a connection, this symbolizes a challenge/breakdown in communication with this person. Consider the numbers you dial and whether or not you are able to get through.
If you do make contact, then this can signify your ability to telepathically communicate with this person. See Number and Phone.... telephone number dream meaning
### One (number)
1. Most important.
2. The problem can be reduced to one thing. ... one (number) dream meaning
### Seven, Number
Luck is on the way (lucky 7). ... seven, number dream meaning
### Six, Number
1. Disguised reference to “sex.” 2. Be reasonable. ... six, number dream meaning
### Eating A Specific Number Of Grapes
If a person sees himself eating specific number of grapes (ie he counts them while eating them) it suggest that he will either be given the same number of lashes or the same number of pimples will appear on his body.... eating a specific number of grapes dream meaning
### Other Numbers
Ten: a new beginning; the male and female together—ten to one. Eleven: eleventh hour. Twelve: a year, time; a full cycle or wholeness, as in the zodiac or 12 disciples. Thirteen: bad luck. Big numbers: a lot; impressive; much of oneself involved. ... other numbers dream meaning
### What About The Unlucky Number 13?
Many people have a deeply rooted aversion against—even a fear of—the number Thirteen. Ancient cultures assigned great importance to it, considering it more positive than negative.
It is stated in old writings that “he who knows the meaning of the number 13 has the key to power and control”!
Christianity, however, was opposed to any kind of occultism and had a lot to do with giving this number such a bad reputation. They insisted that 13 was an unlucky number because there were 13 people sitting at the table of the Last Supper. This gave rise, for instance, to the belief that when 13 people sat at a table, one of them would die in the same year. And to this day, there are hotels where no room is numbered 13. Theaters in Italy don’t have a seat numbered 13. But this suspicion is rare in the rest of the world.
It is only prevalent in places where the Christian Church is very influential.
Cheiro, in The Book of Numbers, wrote:
In the Indian pantheon there are 13 Buddhas.
The mystical discs which surmount Indian and Chinese pagodas are 13 in number. Enshrined in the Temple of Atsusa, in Japan, is a sacred sword with 13 objects of mystery forming its hilt. Turning westward, 13 was the sacred number of the Mexicans. They had 13 snake gods.
The original states that formed the American Union were 13; its motto, E Pluribus Unum, has 13 letters, the American eagle has 13 feathers in each wing, and when George Washington raised the Republican standard he was saluted with 13 guns.
The sum of the number 13 is 4 (1+3=4), the number of “radicals,” because Four-people often feel misunderstood and unconsciously invite secret envy and enemies. They are not inclined to recognize authorities who act as if the power is theirs alone and often misuse it. Challenging conventional standards, laws, and the powerful—and speaking out—has never been popular with the general public, least of all with the ruling authorities.
The number 13 is Four on a higher level and has thereby more gravity, increasing the intensity of any revolutionary conviction even more—including the struggle to bring about social reform and justice.
13 is a symbol of your whole person and your entire life. Don’t let others drive you crazy—13 is not an unlucky number! On the contrary, it seeks to “revolutionize” in the sense of reforming a world that is in dire need of it.... what about the unlucky number 13? dream meaning
### Figures Or Numbers
This omen is a difficult one to judge, as the importance of the dream depends upon the Figures involved, and these again depend upon the circumstances of the dream. As a rough guide, low Figures are fortunate, high Figures are bad, while medium, in-between Figures show difficulties that can be surmounted if you exert yourself. Obviously, high Figures for a girl or for a working-man would only be low Figures for the wealthy individual, or prosperous business man, to whom a few thousands would mean little or nothing.... figures or numbers dream meaning
### Phone Number
Dreams that feature your phone number are about a desire for intimacy and connection.... phone number dream meaning
Frequency or rate of vibration. ... badge number dream meaning
### Numbers 1-12
Fragment personalities with construct.... numbers 1-12 dream meaning
### Numbers, Symbols, Miscellaneous
\$10(in pocket): God-Mind always available • 20: Anger • 60: End of series of cycles • 50’s: Centering yourself.
• 8: Angelic frequency.
• 911: Emergency situation; call for help.
• 220: Changing something in the environment • 3 fetuses: Ideas that were never given a chance for completion • 3 shoe boxes: Feeling all ideas about future support are closed off • 300 pounds: Creating twinning program... numbers, symbols, miscellaneous dream meaning
### Master Number
There are three double digit numbers that require special emphasis. These are the numbers eleven, twenty-two and thirty-three. They are called master numbers because they are thought to possess huge potential. Eleven is the most intuitive of all the numbers and it suggests powerful illumination. Twenty-two suggests the realization of your goals. Thirty-three is a powerful symbol of personal wisdom and understanding.
MINUS NUMBER
If negative or minus numbers appear in your dream, they are not necessarily negative symbols. In some instances they may contain a message to look back to a past time or situation and evaluate it more objectively. Perhaps you need to take a few steps back from a situation or to revisit things, as glitches may still need to be sorted out. On the other hand, negative numbers may reveal a sense of negativity and could be a wake-up call for you to start thinking more positively.... master number dream meaning
### Lucky Numbers
What did you dream of last night?
What did you see and whathappened in those dreams?
Convert your dreams into Lucky Numbers to take on this weeks draws, they could help you get lucky!
1. King, Human blood, White man, Left eye
2. Monkey, Native, A Spirit Chief, Copper, Money, Jockey
3. Sea Water, Accident Frog, Sailor, Sex
4. Dead man, Turkey, Small fortune, Bed
5. Tiger, Fight, Strong man
6. Ox Blood, Gentleman, Milk
7. Lion, ‘Thief, Big stick, Chickens
8. Pig, Drunken man, Loafer Fat man, Chinese king
9. Moon, Baby, Hole, Owl, Devil, Pumpkin, Anything round
10. Eggs, Train, Boat Grave, Anything oval
11. Carriage, Wood, Tree, Furniture, Bicycle, Flowers
12. Dead woman, Ducks, Small fire, Chinese Queen
13. Big fish, Ghosts, Spirits
14. Old woman, Fox, Detective, Nurse, Native woman
15. Bad woman, Prostitute, Canary, White horse, Small knife
16. Small house, Coffin, Pigeon, Young woman, Paper money, Letter
17. Diamond woman, Queen. Pearls, Diamond, Stars, White woman;
18. Silver money, Servant girl, Right eye, Butterfly, Hook, Rain
19. Little girl, Smoke, Bread, Big bird, Left hand
20. Cat Sky, Handkerchief, Body, Music, Minister, Naked woman
21. Old man, Stranger, Fisherman, Elephant, Knife, Nose, Teeth
22. Rats, Motor car Big ship, Left foot, Shoes
23. Horse, Doctor, Head, Hair, Crown
24. Mouth, Wild cat, Vixen, Lioness, Hole, Purse
25. Big house, Church, Boxer, Hospital
27. Dog, Policeman, Newborn baby, Medicine, Sad news 28 Sardines, Small fish, Thief, Right foot Surprise, Small child
29. Small water, Coffin, Rain, Tears, Big knife, Right hand
30. Fowl, Graveyard, Sun, Throat, Indian, Forest
31. Big fire, Bishop, Big spirit Feathers, Fight Woman
32. Gold money, Dirty woman, Snake
33. Lithe boy, Spider
34. Meat, Human dung, Anything dirty, Cripple, Tramp
35. Clothes, Sheep, Big hole, Big grave
36. Shrimp, Stick, Admiral, Cigars, Gum
37. Arrow, Lawyer, Treasure, Cooking, Stream
38. Crocodile, Balloons, Sjhambok, Fireworks, Stadium
39. Sangoma, Soccer team, Tattoos, Bloodshed, Teacher
40. Birth, Clock, Snail, Dwarf, River Traditional healer
41. Cattle, Planets, Cave, Desert, Monster.
42. Tornado, Spear Umbrella, Camel, Door
43. Army, Thunder Astronaut Rabbit Turtle
44. Shark, Stud farm, Body builder, Injury, Mud
45. Football, Computers, Jewellery, Wrestler, Storm
46. Ambulance, Beard, Sea sand, Scissors, Key
47. Stallion, Kite, TV, Lightening, Carnival, Hut
48. Clown, Rainbow, Nightmare, Whale, Wealth
49. Shebeen. Circus, Chocolate, Space ship
... lucky numbers dream meaning | 2,527 | 10,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.90536 |
https://community.quickbase.com/communities/community-home/digestviewer/viewthread?GroupId=103&MessageKey=1b640823-ae54-4217-8454-b77da42f4e82&CommunityKey=d860b0f8-6a48-487b-b346-44c47a19a804&tab=digestviewer&ReturnUrl=%2Fcommunities%2Fcommunity-home%2Fdigestviewer%3Ftab%3Ddigestviewer%26CommunityKey%3Dd860b0f8-6a48-487b-b346-44c47a19a804 | 1,576,270,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569146.17/warc/CC-MAIN-20191213202639-20191213230639-00122.warc.gz | 319,341,635 | 31,274 | # Discussions
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## Dynamic form rules stopped at first condition
• #### 1. Dynamic form rules stopped at first condition
Posted 24 days ago
Hi all,
I have a dynamic form rule as below
When Requester is equal to Finance
and all of the following condition are true
Estimated price <= 2000
Action display a message You're over budget estimated
The problem is the rule is kind of skipping the second condition "and all of the following condition are true
Estimated price <= 2000"
The display message just appears after I choose the requester field
What I've missed here?. Can you please show me some direction?
The idea is to show the message if the requester is finance and estimated price is <=2000
Many thanks
------------------------------
Syaeful Bahri
------------------------------
• #### 2. RE: Dynamic form rules stopped at first condition Best Answer
Posted 24 days ago
maybe try adding a second condition that
Estimated price is not equal to (blank).
Maybe that will prevent the message from firing prematurely, when the user has not yet entered the estimated price and the field is still blank.
------------------------------
Mark Shnier (YQC)
Quick Base Solution Provider
http://QuickBaseCoach.com
markshnier2@gmail.com
------------------------------
• #### 3. RE: Dynamic form rules stopped at first condition
Posted 23 days ago
Hi Mark,
Problem solved, thank you
------------------------------
Syaeful Bahri
------------------------------
• #### 4. RE: Dynamic form rules stopped at first condition
Posted 23 days ago
:) thx for letting us know!
------------------------------
Mark Shnier (YQC)
Quick Base Solution Provider
http://QuickBaseCoach.com
markshnier2@gmail.com
------------------------------
• #### 5. RE: Dynamic form rules stopped at first condition
Posted 24 days ago
Edited by Adam Keever 24 days ago
Mark is correct.
Here another way to achieve your goal. Create a formula checkbox field and use this formula:
`If([Requester]="Finance" and [Estimated price]<=2000,true,false)`
Then use this dynamic form rule where [message] is the name of the formula check box field:
Here is the result:
------------------------------
------------------------------
• #### 6. RE: Dynamic form rules stopped at first condition
Posted 23 days ago | 496 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-51 | longest | en | 0.795238 |
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Fall10Pre1
Fall10Pre1 - Math 1920 Prelim I 7:30 PM to 9:00 PM You are...
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Unformatted text preview: Math 1920, Prelim I September 30, 2010. 7:30 PM to 9:00 PM You are NOT allowed calculators, the text or any other boolc or'notes. SHOW ALL WORK! Writing clearly and legibly will improve your chances of receiving the maximum credit that your sblution deserves. Please label the questions in you answer booklet clearly. Write your name and section number on each booklet you use. You may leave when you have finished, but if you have not handed in your exam booklet and~left the room by 8:45pm, please remain in your seat so as not to disturb others who are still working. ' 1. Consider the vectors V = i+ Zj + ak and W = i+j + k. (a) (10 pts) Find all values of the number a (if any) such that V is perpendicular to w. (b) (10 pts) Find all the values of the number a (if any) such that the area of the parallelogram determined by v and W is equal to V0. 2. (10 pts) Find the plane through the origin perpendicular to the plane 22: + 2y + z = 1 and perpendicular to the vector V = (1,1, —4). 3; Consider the function 97(55, y) = My? - 932. (a) (9 pts) Sketch the level curves g(ac,y) = c, for c =071, 2. (Ware-a rs itsdomam D of g? (c) (4 pts) What is the bonndary of D? (d) (4 pts) ls D a closed set, an open set, both or neither? '1} 4. (9 pts) The wave equation, where a2 is constant, is given by It describes the motion of a waveform; examples of such waves could include fluid, sound, «light. Suppose u(a:, t) represents the displacement of a vibrating guitar string at time t at a distance a: from one end of the string. If u(:z:, t) = sin(a° — at), show that it satisfies the wave equation. in: r 5. Calculate each of the following limits or show it doesth exist. , — l (a) (10 pts) lim —‘/————— “4+. (am—44.3) a: — y — 1 x7+y+1 \,. y ‘ b\ 10 pts hm ‘ ——-————. < ’ < ) (mm—40.0) x2 + y2 6. Consider the force vector field given by F = sci +ryj + zk. (a) (10 pts) Calculate the work done on a particle by the force F when the particle . moves along the conical helix r(9) =, (6.00s o9)i + {6’ sin 0)j + 6k from 9 =7 0 to ’ 0 = 27r. ,, , (b) (10 pts) Calculate the work done on a particle by F when the particle moves along / " “‘ "my” 'a‘str’aight‘iinerfrom the*'origin"tamer”point“with‘co-‘orriinatesf‘Zir:Oflfl.’13088133?” A‘M’H" work done along these two paths depend on the path taken? ...
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https://justaaa.com/physics/1210021-a-20-m-long-200-n-uniform-ladder-rests-with-one | 1,726,255,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00306.warc.gz | 296,100,153 | 9,641 | Question
# A 2.0 m long, 200 N uniform ladder, rests with one end against a smooth wall...
A 2.0 m long, 200 N uniform ladder, rests with one end against a smooth wall making an angle of 30 degrees with the wall. The coefficient of friction between the ground and the lower end of the ladder is .5. With how large a force can you push the upper end of the ladder parallel to the wall before it starts to slip down?
Please rate it up thanks :)
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Coins can be redeemed for fabulous gifts. | 121 | 505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.909552 |
https://homeworkmules.com/how-do-you-find-the-values-of-sin-2theta-and-cos-2theta-when-cos-theta-12-13/ | 1,718,494,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00478.warc.gz | 274,895,092 | 17,233 | # How do you find the values of ##sin 2theta## and ##cos 2theta## when ##cos theta = 12/13##?
Get first the value of the of the opposite side fo get ##sin## and use the Double Angle Formula.
##costheta = 12/13##
##c^2=a^2+b^2##
##13^2=a^2 + 12^2##
##a^2=13^2-12^2= 169-144= 25##
##c= sqrt25 =5##
##sintheta = 5/13##
##sin2theta = 2sinu cosu =2(5/13)(12/13)= 120/169##
##cos2theta=cos^2u – sin^2u = (12/13)^2 – (5/13)^2= (144/169) – (5/169)= (144 – 25)/169 = 119/169##
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Customers referred by a friend | 882 | 3,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.86085 |
http://mathhelpforum.com/trigonometry/78004-i-need-help-trig.html | 1,524,319,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00447.warc.gz | 203,069,063 | 9,931 | # Thread: I need help with trig
1. ## I need help with trig
IV BEEN WORKING ON THESE QUESTIONS FOR A VERY LONG TIME AND I CANT GET THE ANSWERS.
A) CALCULATE SIN225+COS120-TAN(-330)
B) (COS30)(COS30)=(SIN30)(SIN30)
I DONT KNOW HOW BUT LEFT HAS TO BE EQUAL TO RIGHT FOR PART B
2. Hello, Viviana!
A) Calculate: .$\displaystyle \sin225^o+\cos120^o-\tan(\text{-}330^o)$
You're expected to know the trig values of "standard angles."
. . $\displaystyle \begin{array}{ccccc}\sin225^o &=&\text{-}\sin45^o &=& \text{-}\frac{\sqrt{2}}{2}\\ \\[-4mm] \cos 120^o &=&\text{-}\cos60^o &=&\text{-}\frac{1}{2} \\ \\[-4mm] \tan (\text{-}330^o) &=&\tan30^o &=&\frac{\sqrt{3}}{3} \end{array} \quad\hdots\;\text{ etc.}$
$\displaystyle B)\;\;(\cos30^o)(\cos30^o) \:=\:(\sin30^o)(\sin30^o)$ . ??
I DONT KNOW HOW BUT LEFT HAS TO BE EQUAL TO RIGHT FOR PART B
That is quite impossible . . . equivalent to: $\displaystyle 3 = 1$ | 342 | 910 | {"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} | 3.640625 | 4 | CC-MAIN-2018-17 | latest | en | 0.403789 |
https://byjus.com/question-answer/what-is-the-final-velocity-of-an-electron-accelerating-through-a-potential-of-1600-v-1/ | 1,679,612,265,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00314.warc.gz | 188,414,437 | 29,343 | Question
# What is the final velocity of an electron accelerating through a potential of 1600 V, if its initial velocity is zero.
A
2.37 × 107 m s1
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2.37 × 108 m s1
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2.37 × 106 m s1
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2.37 × 105 m s1
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Solution
## The correct option is C 2.37 × 107 m s−1As we know,12mv2=eVv=√2eVm =√[2×1.6×10−19×16009.1×10−31]v=2.37×107 m/s.
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Explore more | 256 | 703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.775325 |
http://stackoverflow.com/questions/2838727/how-do-i-get-the-sums-of-the-digits-of-a-large-number-in-haskell | 1,429,454,096,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246639121.73/warc/CC-MAIN-20150417045719-00305-ip-10-235-10-82.ec2.internal.warc.gz | 257,054,628 | 20,742 | # How do I get the sums of the digits of a large number in Haskell?
I'm a C++ Programmer trying to teach myself Haskell and it's proving to be challenging grasping the basics of using functions as a type of loop. I have a large number, 50!, and I need to add the sum of its digits. It's a relatively easy loop in C++ but I want to learn how to do it in Haskell.
I've read some introductory guides and am able to get 50! with
sum50fac.hs::
fac 0 = 1
fac n = n * fac (n-1)
x = fac 50
main = print x
Unfortunately at this point I'm not entirely sure how to approach the problem. Is it possible to write a function that adds (mod) x 10 to a value and then calls the same function again on x / 10 until x / 10 is less than 10? If that's not possible how should I approach this problem?
Thanks!
-
By the way, a slightly nicer version would be fac n = product [1..n]. Explicit recursion is unfashionable in Haskell circles. – C. A. McCann May 15 '10 at 4:15
While that is true, you need to learn it. – Paul Johnson May 15 '10 at 6:28
Why not just
sumd = sum . map Char.digitToInt . show
-
I find going via strings and characters is really unnatural since there is nothing in the problem that is concerned with strings. It's just numbers and the solution should reflect that. – svenningsson May 15 '10 at 10:42
sumd 0 = 0
sumd x = (x `mod` 10) + sumd (x `div` 10)
Then run it:
ghci> sumd 2345
14
UPDATE 1:
This one doesn't generate thunks and uses accumulator:
sumd2 0 acc = acc
sumd2 x acc = sumd2 (x `div` 10) (acc + (x `mod` 10))
Test:
ghci> sumd2 2345 0
14
UPDATE 2:
Partially applied version in pointfree style:
sumd2w = (flip sumd2) 0
Test:
ghci> sumd2w 2345
14
I used flip here because function for some reason (probably due to GHC design) didn't work with accumulator as a first parameter.
-
Writing your recursive functions in this manner is going to use unbounded stack. Much better to recurse in terms with an accumulator, so no thunks are generated. – jrockway May 15 '10 at 3:11
Best to wrap sumd2 so the user doesn't have to put in the initial accumulation value. – Thomas Eding May 15 '10 at 22:12
This is just a variant of @ony's, but how I'd write it:
import Data.List (unfoldr)
digits :: (Integral a) => a -> [a]
digits = unfoldr step . abs
where step n = if n==0 then Nothing else let (q,r)=n`divMod`10 in Just (r,q)
This will product the digits from low to high, which while unnatural for reading, is generally what you want for mathematical problems involving the digits of a number. (Project Euler anyone?) Also note that 0 produces [], and negative numbers are accepted, but produce the digits of the absolute value. (I don't want partial functions!)
If, on the other hand, I need the digits of a number as they are commonly written, then I would use @newacct's method, since the problem is one of essentially orthography, not math:
import Data.Char (digitToInt)
writtenDigits :: (Integral a) => a -> [a]
writtenDigits = map (fromIntegral.digitToInt) . show . abs
Compare output:
> digits 123
[3,2,1]
> writtenDigits 123
[1,2,3]
> digits 12300
[0,0,3,2,1]
> writtenDigits 12300
[1,2,3,0,0]
> digits 0
[]
> writtenDigits 0
[0]
In doing Project Euler, I've actually found that some problems call for one, and some call for the other.
## About . and "point-free" style
To make this clear for those not familiar with Haskell's . operator, and "point-free" style, these could be rewritten as:
import Data.Char (digitToInt)
import Data.List (unfoldr)
digits :: (Integral a) => a -> [a]
digits i = unfoldr step (abs i)
where step n = if n==0 then Nothing else let (q,r)=n`divMod`10 in Just (r,q)
writtenDigits :: (Integral a) => a -> [a]
writtenDigits i = map (fromIntegral.digitToInt) (show (abs i))
These are exactly the same as the above. You should learn that these are the same:
f . g
(\a -> f (g a))
And "point-free" means that these are the same:
foo a = bar a
foo = bar
Combining these ideas, these are the same:
foo a = bar (baz a)
foo a = (bar . baz) a
foo = bar . baz
The laster is idiomatic Haskell, since once you get used to reading it, you can see that it is very concise.
-
I am unable to find what what the .show and .abs are doing in the writtenDigits function. Could you explain what are they doing? thanks – Tim May 15 '10 at 15:39
".show"? . doesn't work like it does in C/C++/Java/Python/PHP, see my expanded answer for details. The reason for abs is that I want the functions to be defined for negative numbers, rather than crash or loop (which they will without the abs). That makes them total functions, not partial functions, which in Haskell (or any language really) good. – MtnViewMark May 15 '10 at 17:42
To sum up all digits of a number:
digitSum = sum . map (read . return) . show
show transforms a number to a string. map iterates over the single elements of the string (i.e. the digits), turns them into a string (e.g. character '1' becomes the string "1") and read turns them back to an integer. sum finally calculates the sum.
-
Just to make pool of solutions greater:
miterate :: (a -> Maybe (a, b)) -> a -> [b]
miterate f = go . f where
go Nothing = []
go (Just (x, y)) = y : (go (f x))
sumd = sum . miterate f where
f 0 = Nothing
f x = Just (x `divMod` 10)
-
The 'miterate' function is really just 'unfoldr'. – svenningsson May 15 '10 at 10:44
Yep. I haven't looked good enough (i.e. in Data.List). Why it wasn't imported in Prelude?... – ony May 15 '10 at 13:41
Well, one, your Haskell function misses brackets, you need fac (n - 1). (oh, I see you fixed that now)
Two, the real answer, what you want is first make a list:
listdigits n = if n < 10 then [n] else (listdigits (n `div` 10)) ++ (listdigits (n `mod` 10))
This should just compose a list of all the digits (type: Int -> [Int]).
Then we just make a sum as in sum (listdigits n). And we should be done.
Naturally, you can generalize the example above for the list for many different radices, also, you can easily translate this to products too.
-
(++) is slow however, I wouldn't recommend to use it in other algorithms. – Yasir Arsanukaev May 15 '10 at 3:38
Although maybe not as efficient as the other examples, here is a different way of approaching it:
import Data.Char
sumDigits :: Integer -> Int
sumDigits = foldr ((+) . digitToInt) 0 . show
Edit: newacct's method is very similar, and I like it a bit better :-)
- | 1,817 | 6,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | latest | en | 0.906474 |
https://www.physicsforums.com/threads/resistor-is-a-current-limiting-element.628168/ | 1,669,790,708,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00163.warc.gz | 1,001,356,497 | 16,946 | # Resistor is a current limiting element.?
harsha2591
In a simple circuit having battery (6v) connected to two resistors (1K each) in series, 3mA would flow through the entire circuit.
1) If resistor is a current limiting element, current following the first resistor should reduce its value (from 3mA to some other) before entering the another resistor right??
2) If current reduces, then conservation of charges is violated right?
3) As per ohm's law, V=IR. In the above circuit, voltage gets dropped (3V)across the first resistor. Ohm's law implies voltage varies linearly with current if resistance is constant. So this would mean drop in voltage across resistor should limit currewnt from 3mA to some other value right?
if resistor is not a current limiter, then how does a resistor in screwdriver tester limits huge current to flow through our body?
Staff Emeritus
Hi harsha2591! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
Current does not diminish as it progresses through a circuit, it is the same at every point. Charge is conserved! 3mA through each resistor, and 3v across each.
Last edited by a moderator:
Gold Member
2021 Award
In a simple circuit having battery (6v) connected to two resistors (1K each) in series, 3mA would flow through the entire circuit.
Yes, correct
1) If resistor is a current limiting element, current following the first resistor should reduce its value (from 3mA to some other) before entering the another resistor right??
No, the overall resistance of the circuit ( that is the load across the battery/other PSU) sets the amount of current flowing in the circuit
in your above circuit say there was just one 1k Ohm resistor then the current flowing in the circuit would be 6mA. You double the resistance and you will halve the current flowing. It doesn't matter if those two 1k resistors are separate ones or just a single 2 k resistor the current flowing in the circuit will be 3mA
You mite have 10 resistors of different values in a circuit some in series some in parallel, you need to work out or measure the total value of resistance to find out what the current flowing will be
Dave
Gold Member
if resistor is not a current limiter, then how does a resistor in screwdriver tester limits huge current to flow through our body?
What you need to remember is that the situation as described in the above (and other circuit problems) is what happens after the system has settled down. On a long wire, some charge may, indeed, flow through the nearer components before the voltage pulse (at switch on) arrives at a further component and before it 'can know about' that distant resistor. But all this is not relevant to basic circuit theory, which starts to kick in within a few nanoseconds of switch on, by which time, all the components are interacting with each other and all the 'rules' start to apply, such as sharing of the Volts along a series chain.
It's not like cars on a motorway, where a hold up, 20 miles away is reducing the flow of traffic there but not having any effect on the ones which are just joining it. Although, even in this case, the motorway would eventually jam up totally, given enough time, and the total number that could get onto the start section would be limited by the rate that they could exit at the other end.
I_am_learning
Resisters are current limiting things, not current dropping things.
You connect 2K to 6V. The current is limited at 3mA. If you now hook 6K to 6V, current is further limited at 1mA. Thats it.
If there were no resistance, <a direct short between battery terminals> then huge currents will flow (theoritically infinite), i.e. the current has no limit. But in practice, even if you short out the terminals, the resistance in the battery (internal resistance) and the shorting wire will limit the current within few Amperes.
Gold Member
Kirchoffs current law.
Current entering a node equals current leaving a node.
Current leaving the source EQUALS current going back to the source.
Gold Member
But that doesn't tell you what currents may be flowing in parts of the circuit in between. It's a simplification which works. Think of the current in a parallel resonant circuit, compared with the current into and out of the generator that's connected.
Gold Member
But that doesn't tell you what currents may be flowing in parts of the circuit in between. It's a simplification which works. Think of the current in a parallel resonant circuit, compared with the current into and out of the generator that's connected.
Ok...well, once the circuit reaches steady state, the capacitor and inductor oscillate between each other (the source sees and open circuit at the L and C) and all the current from the source goes thru the resistor and all the current returns to the source. Simplified...indeed.
Kholdstare
It depends on how you energize your system. If you use a current source, resistance might be called "voltage drop incrementing" element.
Gold Member
Kirchoffs current law.
Current entering a node equals current leaving a node.
Current leaving the source EQUALS current going back to the source.
All that K1 really says is that the sum of the current at a node is Zero. It doesn't, except by implication, say what you are saying - only if you can identify just two nodes, one each end of the power source.
Gold Member
All that K1 really says is that the sum of the current at a node is Zero. It doesn't, except by implication, say what you are saying - only if you can identify just two nodes, one each end of the power source.
Great. We are in total agreement! | 1,271 | 5,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.929843 |
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# Scientific Method
TermDefinition
Dependent Variable The variable in an experiment that is measured.
Mass A balance measures this quantitative observation.
Line Graph A graph that is used when data has bee taken over time.
Hypothesis An educated guess about how one variable will affect another.
Purpose or Question The first step in the scientific method.
Control Variable The variable in an experiment that stays the same or is not changed.
Length A meter stick measures this quantitative observation.
Bar Graph A graph that is used to display descriptive data.
Independent Variable The variable in an experiment that is changed.
The 5 senses Qualitative observations are based on these.
Quantitative Observation An observation that is based on a measurement and contains numbers.
Temperature A thermometer measures this qualitative measurement.
Liters Metric unit that measures liquid.
Meters Metric unit that measures length.
Grams Metric unit that measure mass.
graduated cylinder and beaker Instruments that measure liquids.
Data Table An easy way to organize data.
Conclusion A factual summary of data.
Qualitative Observation Observations based on our five senses.
Research The step in the scientific method after asking a question.
Hypothesis The step in the scientific method after doing research.
Perform the Experiment The step in the scientific method after stating a hypothesis.
Collect Data What is done while performing an experiment.
Graph the Data What is done with data after completing an experiment.
Write a Conclusion The last step in the scientific method that gives you a chance to tell what you learned from doing the experiment.
Procedures Step by step instructions on how to complete an experiment.
Created by: charmon | 444 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-44 | latest | en | 0.886507 |
http://www.adras.com/sumif.t118258-5.html | 1,638,634,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362999.66/warc/CC-MAIN-20211204154554-20211204184554-00201.warc.gz | 93,512,984 | 2,971 | From: Seth.Schwarzkopf on 22 Mar 2010 16:37 i am using a sum if in a formula and i want to copy and paste it down my spread sheet. when i do excell changes all the cell references down 1 number. the only number in the sum if formula i want to have change is the critria, i want the range and sum range to be the same for all cells involved. i would just type it in each cell but i am looking at 300 pluss cells that i would have to change. is there a way to keep excell from changing all the cell references in a formulas From: Gary Brown on 22 Mar 2010 16:46 The '\$' makes formulas 'absolute. Example: =SUMIF(\$A\$1:\$A\$5,E1,\$B\$1:\$B\$5) only E1 (criteria) will change. The rest are LOCKED. \$A\$1 \$A means column A will stay Column A no matter what \$1 means row 1 will stay Row 1 no matter what You can combine these such as... A\$1 A means column A will change as you copy across columns \$1 means row 1 will stay Row 1 no matter what -- Hope this helps. If it does, please click the Yes button. Thanks in advance for your feedback. Gary Brown "Seth.Schwarzkopf" wrote: > i am using a sum if in a formula and i want to copy and paste it down my > spread sheet. when i do excell changes all the cell references down 1 number. > the only number in the sum if formula i want to have change is the critria, i > want the range and sum range to be the same for all cells involved. i would > just type it in each cell but i am looking at 300 pluss cells that i would > have to change. is there a way to keep excell from changing all the cell > references in a formulas From: Per Jessen on 22 Mar 2010 16:51 You are obviously using relative references, where you have to use absolute references. =A1 is a relative refence =\$A\$1 is a absolute reference Your formula could look like this: =SUMIF(\$A\$1:\$A\$500,"=" & A1) Hopes this helps. .... Per On 22 Mar., 21:37, Seth.Schwarzkopf wrote:> i am using a sum if in a formula and i want to copy and paste it down my > spread sheet. when i do excell changes all the cell references down 1 number. > the only number in the sum if formula i want to have change is the critria, i > want the range and sum range to be the same for all cells involved. i would > just type it in each cell but i am looking at 300 pluss cells that i would > have to change. is there a way to keep excell from changing all the cell > references in a formulas | Pages: 1 Prev: HELP: how to locate objects on userform from VBA codeNext: Local cube files | 661 | 2,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.876981 |
http://nrich.maths.org/public/leg.php?code=213&cl=3&cldcmpid=739 | 1,506,209,910,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689806.55/warc/CC-MAIN-20170923231842-20170924011842-00597.warc.gz | 256,515,961 | 6,146 | # Search by Topic
#### Resources tagged with Mean similar to Pairs:
Filter by: Content type:
Stage:
Challenge level:
### There are 15 results
Broad Topics > Handling, Processing and Representing Data > Mean
### Pairs
##### Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
### A Mean Tetrahedron
##### Stage: 3 Challenge Level:
Can you number the vertices, edges and faces of a tetrahedron so that the number on each edge is the mean of the numbers on the adjacent vertices and the mean of the numbers on the adjacent faces?
### The Mean Problem
##### Stage: 4 Challenge Level:
There are four unknown numbers. The mean of the first two numbers is 4, and the mean of the first three numbers is 9. The mean of all four numbers is 15. If one of the four numbers was 2, what were. . . .
### M, M and M
##### Stage: 3 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Wipeout
##### Stage: 3 Challenge Level:
Can you do a little mathematical detective work to figure out which number has been wiped out?
### Bat Wings
##### Stage: 3 Challenge Level:
Three students had collected some data on the wingspan of some bats. Unfortunately, each student had lost one measurement. Can you find the missing information?
### Archery
##### Stage: 3 Challenge Level:
Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?
### Converging Means
##### Stage: 3 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Unequal Averages
##### Stage: 3 Challenge Level:
Play around with sets of five numbers and see what you can discover about different types of average...
### Litov's Mean Value Theorem
##### Stage: 3 Challenge Level:
Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
### Searching for Mean(ing)
##### Stage: 3 Challenge Level:
Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
### For Richer for Poorer
##### Stage: 3 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality? | 746 | 3,190 | {"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} | 4.0625 | 4 | CC-MAIN-2017-39 | latest | en | 0.896311 |
http://www.naic.edu/~phil/optics/hornpos/hornpositions.html | 1,532,000,120,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00177.warc.gz | 507,772,867 | 10,213 | # Measured horn positions
#### 08may04
The horns are surveyed into position using a theodolite. A cap with targets is placed over the mouth of the horn. This cap has 5 targets mounted at known positions. The theodolite shoots the secondary to define the reference dome centerline coordinate system. The turret is moved to the correct position for this feed and then the targets on the horn cap are shot. This gives the cap location in dome centerline coordinates. The horn position is then computed using the distance from the horn cap to the horn phase center. (The positions are then converted to focus coordinate system where z is along the paraxial ray, y is toward the stairwell, and x is up the elevation rails.. 17apr18.. i think this sentence is incorrect.. i think things remain in dome centerline coordinates.). The table below shows the corrections that must be made to the horns to put them at the nominal positions (you move in the direction of the correction).
NOTES:
• When moving the turret, increasing the turret encoder angle moves a horn at focus towards the stairwell (+y).
• The Alfa receiver wants to be located -1.063 inches in x from the nominal position to balance the uphill and downhill horns. The values reported are relative to this position.
• KEY:
• Note: Lynn's memos show the error. The xxC are the corrections (or minus the errors).
• xxC Translation correction. Distance to move in domeCenterline coordinates to correct error.
• +X uphill,+Y toward stairwell, +Z up vertical (this used to say focus direction 19apr18).
• RxC Rotation correction(right handed). RYC + is clockwise around Y axis. RXC + is clockwise about +X axis.
• ==> lynns' memos define Xtilt,Ytilt error to be:
• ==> xtilt err + when moving from the +x axis in +z direction. This is a +err rotation about yaxis
• ==> ytilt err + when moving from the +y axis in +z direction . This is a -err rotation about the x axis.
• ==> so the mapping from lynn's error Tilt definitions to my correction rotation definitions:
• RYC = -xtiltErr
• RXC = +yiltErr
• 23apr18 correction:
• i had been mapping +ytiltErr->- RxC This was wrong. It should have been +RXC
• on 23apr18 i changed the sign of all of RXC entries so that values are consistent with the rotation definitions.
• <-- This marks the most recent survey for this receiver
• X + along azimuth arm. Y perpendicular to az arm + is toward the stairwell. Z is along the focus direction + makes the focal length longer.
• nominal: this is the nominal position for the measurement in the focus (ray trace coordinates). If you subtract the correction from the nominal, you will get the current reading.
•
Feed turDeg XC Inches YC Inches ZC Inches RYC Deg RXC Deg notes nominal raytrace -354.00 0. -234.0 1.74 0. Focus in ray trace coordinates nominal Dome Centerline -248.145 0. -389.822 .63 0 focus in dome center line coordinates. Note tht TYC of .63 is a positive (CW) rotation about the Y axis. See note about Lynn's "xtilt,ytilt definitions" LBN 28jan00 309.12 -.193 .104 .296 .099 -.7779 <--see #4. 21aug02 see #9 LBW 28jan00 285.5 -.124 -.922 -.048 -.225 -.9985 see #5. 11feb04 286.52 -.46 -1.36 1.00 .22 -.97 14feb04 285.42 -.47 -.03 1.00 .23 +.76 see #12 28apr04 285.42 -.93 -.07 .94 02nov12 285.477 -.794 .126 .896 .390 -.738 <-- see survey notes ALFA 28apr04 26.27 -.42* 0 -.51 see #15 02may02 26.64 see survey notes. 02nov12 26.403 .741 .07 -.517 .353 .129 <-- see survey notes SBN 28jan00 76.6 -.003 .292 .112 -.069 .1328 28jan00 76.6 -.038 .253 .079 -.057 .1620 see #1 04feb02 76.83 -.021 -.257 .158 -.028 .0743 29mar02 76.88 -.013 -.132 .123 -.032 .1558 #8 11feb04 76.83 -.13 -.35 .02 -.07 .05 14feb04 76.56 -.07 -.05 -.11 .04 -.20 see # 12 28apr04 76.555 -.59 -.02 -.26 02may04 76.90 -0.59 -0.01 -0.20 08may04 76.94 .01 see survey notes 31oct12 79.6143 -.236 +.01 +1.173 -.45 .51 <-- see survey notes turret later moved to 79.967 SBTX 28jan00 62.25 -.143 .093 -.067 -.086 -.3677 see #6 11feb04 62.16 -.25 -.24 -.19 .00 -.31 14feb04 61.97 -.20 -.06 -.27 .01 .32 see # 12 28apr04 61.96 -.62 -.01 -.43 02may04 62.29 -0.66 -0.00 -0.38 <-- 08may04 62.29 -.02 see survey notes 31oct12 61.9441 -.518 +.181 +.122 1.2 -.2 <-- see survey notes SBW 28jan00 103.8 .447 .141 -.379 -.821 - .3547 see #7 11feb04 103.88 -.14 -.28 -2.54 .25 -.5 14feb04 103.66 -.09 -.07 -2.64 .17 .55 see # 12,#14 28apr04 103.66 -.58 .01 -.269 02may04 104.03 -0.24 0.01 -2.80 02nov12 103.673 -.152 .085 -.021 .536 -.479 <-- see survey notes SBH 29mar02 124.99 -0.027 1.068 .703 .284 .6941 see #8 11feb04 125.83 -.12 -.36 -.91 .23 .56 14feb04 125.55 -.08 -.05 -.12 .31 -.59 see #12 ,#13 28apr04 125.55 -.53 0 -.23 02may04 125.94 -0.15 0.02 -0.08 02nov12 125.717 -.132 .076 .084 .279 .571 <-- see survey notes CB 28jan00 206.65 -.124 -.001 .010 .410 .1998 28jan00 206.65 -.165 .023 0.020 .396 .2013 repeated 28aug02 see #10 11feb04 206.65 -.17 -.34 -.11 .17 .42 14feb04 206.38 -.15 -.05 -.17 .32 -.43 see # 12 28apr04 206.38 -.66 .05 -.21 02may04 206.77 0.01 0.01 -0.29 02nov12 206.556 -.008 .039 -.214 .524 .489 <-- see survey notes CBH_1 11feb04 164.57 -.13 -.14 2.33 .27 -.96 14feb04 164.67 .23 0.00 .36 -.74 .34 28apr04 164.67 -.28 .01 .23 02may04 165.01 -0.22 0.00 0.05 <-- 08may04 165.03 .02 see survey notes 02nov12 164.711 -.428 .073 .159 .435 -.438 <-- see survey notes CBH_2 11feb04 158.45 .05 -.71 2.28 .27 .42 14feb04 158.01 .51 -.03 .34 -.69 -.86 02may04 158.38 <--see survey notes XB 04feb02 161.99 -.085 -.449 .441 .259 -.0412 see #2. 29mar02 162.39 -.066 -1.131 -.011 .097 .1208 see #3. 08jan04 309.94 see #11 11feb04 309.94 -.17 -.28 -.11 .1 .82 14feb04 309.72 -.15 .02 -.19 .22 -.90 see #12 28apr04 309.72 -.43 .07 -.15 02may04 310.06 -0.22 -0.00 -0.14 02nov12 309.800 -.185 .045 -.121 .385 .811 <-- see survey notes
1. The beam under the floor by the sbn receiver was removed to work on the seti IF box. This caused the floor to shift. The beam was replaced and then this receiver was remeasured. After the measurement, the turret position was moved to 76.83 deg.
2. After the measurement the horn was moved up .44 inches. The turret was moved 161.99 -> 162.34. This turned out to be the wrong turret direction to move. The next survey measured twice the error.
3. The large y error was because the turret was moved the wrong direction on 04feb02. After the measurement the turret was moved to 161.50.
4. After the measurement, the lbn turret position was moved to 310.14.
5. After the measurement the lbw turret position was changed to 286.52. This looks like it was the WRONG direction.
6. After the measurement, the sbtx turret position was moved to 62.16 deg. This was the WRONG direction
7. After the measurement the sbw turret position was moved to 103.8->103.88 degrees.
8. After the measurement (and getting the sign right!), the turret was moved, 124.99->125.827 degrees.
9. 21aug02 lbn horn moved down 3.84 inches to correct for phase center error.
10. 28aug02 cband horn moved down 1.08 inches to put phase center at focus.
11. jan04 receivers moved on floor to get ready for alfa installation.
12. 14feb04. The 11feb04 survey yCorrection was set to zero by moving the turret positions and then the 14feb04 survey was done.
13. 14feb04 sbhigh was moved down .91 inches before the survey to put the phase center at focus.
14. 14feb04 sbw was out of focus in z. It was not moved down because there was no more adjustment room on the mount. We need to make a new bracket.
15. Alfa x listed is measuredCorrection - 1.063 . Alfa wants to be 1.063 downhill from the nominal focus to balance the outer feeds.
#### Survey notes:
• 28apr04: after kevlar cables and alfa installed. The tilt values were small and were not reported.
• 02may04: Horn survey after horns moved (sbtx,sbrx,lbw,sbw(focus) were not moved). The corrections quoted are after the turret postions were updated. Previous surveys (11feb04 till 01may04) had been using an "old" secondary reference set for the secondary. On 01may04 we switched back to the "new" set. This caused a motion of about .45 inches y (.37 turret degrees). The turret position for cbh_2, lbw,alfa were also updated by .37 turret degrees because of the reference system change (even though they were not measured).
• 08may04:The turret encoder lost its position index on 06may04 (alfa cover broke encoder). I aligned the encoder tracking some lbw sources. Mike and Ellen then surveyed sbtx,sbn, and cbh_1. Using all of the measurements there was a -.027Deg (before-after) difference in the turret encoder. This is .034 inches or -1.199 arcseconds on the sky. Since this is small we'll continue to use the turret positions computed by lynn. If you wanted to correct for this, you would subtract .027 turret degrees from the offset i've put in the encoder offset of the little stars. The current offset in the little stars is 40843 enc counts. The correct value should be: (40843 - (.027TurDeg*477.867TurDegToEncCnts))=40830 encoder counts.
• 31oct12,02nov12 (see below)
## 31oct,02nov12 horn surveys (top)
A new sband narrow horn was installed on 28oct12. Lynn surveyed the sbn and sbtx horns on 31oct12. He surveyed the rest of the horns on 02nov12. An "old" (pre may04) secondary reference file was used for the theodolite shooting. This data set was used for the sbn and sbtx analysis. Lynn did a comparison with the old and "newer" secondary references and saw little difference. All of the other receivers were analyzed using the newer secondary reference derived from the 28jun04 videogrammetry.
On 02may04 this "old" secondary reference was also used. At that time it was thought that there was a .45" y error between the two system (see 02may04 above).
• sbn
• New receiver. Had to move the floor location since the new receiver was bumping up against the transmitter.
• turret position used for last measurement was79.6143.
• The errrors,corrections reported here are used the old secondary reference.
• in 02may04 this reference set was also used. later analysis showed a .45 inch y offset between old and new ref sets.
• after the survey was done, i moved the final sbn turret position to 79.967 deg (to correct for this .45 inch error).
• After taking into account the tilt errors, this may have not been the optimum solution.
• The sbn receiver was lifted in Z 1.2 inches on 26nov12 to correct for Z the measured z error. It was not resurveyed.
• sbtx:
• shot 3 times.
• final turret was 61.9441
• correction was .181 inches or .181*.7973(turdeg/inch) = +.1418 deg.
• this would give 62.086 deg for turret
• there would also be a possible .45" correction for the old secondary ref set
• giving 62.44 tur deg pos
• I'm leaving the tx turret position unchanged: 62.29 turret deg.
• sbw,lbw,cb,xb,cbh,alfa,sbh
• All analysis done with latest 28jun04 videogrametry survey data.
• The turret positions remained as they were prior to the 02nov12 survey.
• I did not correct the turret position to minimize the y error..
Table showing the corrections needed from the nov12 surveys.
• These are taken from the large table above.
• The numbers are corrections. You need to move the feed by this amount. This is the negative of what lynn reports.
rcvr turPos for measurement deg Xcor inches Ycor inches Zcor inches TYcor deg TZcor deg lbw 285.477 -.794 .126 .896 .390 .738 alfa 26.403 .741 .07 -.517 .353 -.129 sbn 79.6143 -.236 +.01 +1.173 -.45 -.51 sbtx 61.9441 -.518 +.181 +.122 1.2 .2 sbw 103.673 -.152 .085 -.021 .536 .479 sbh 125.717 -.132 .076 .084 .279 -.571 cb 206.556 -.008 .039 -.214 .524 -.489 cbh 164.711 -.428 .073 .159 .435 .438 xb 309.800 -.185 .045 -.121 .385 -.811
### Plotting the corrections for each receiver:
The plot shows the x,y,z errors (in dome centerline coordinates) from the survey (.ps) (.pdf):
• The data is in dome centerline coordinates.
• X positive up the azimuth arm
• Y perpendicular to azimuth.+ is towards the stairwell. This is the turret direction for small turret moves.
• Z focus. + is up.
• colors show the different errors:
• Black:X
• Red : Y
• solid line uses turret position of measurement
• dashed line uses current turret position (11jan13).
• Green: Z or focus.
• The data shows the amount to move the feed to correct for the error. (this is the negative of lynn's reported errors).
• Top: The Correction need in inches
• Bottom: The correction needed in lambdas (using the lambda for the center of the feed).
### Lynn's emails and reports:
processing: /share/megs/survey/121102/hornsurvey.pro, /share/megs/survey/121031/xxx
# Horn phase centers.
Many of our horns are kildall horns. A paper written by per simon kildall was used to calculate the phase center of the horns and position them. They computed the phase center for a horn that covered 9 to 17 Ghz. For this horn the phase center was z= 7 +/- 1.7 mm (see paper for definition of z=0 .. end of waveguide in horn). The phase center should scale with with the throatdiameter of the horn.
For 13 Ghz (assuming diameter = lambda) the focus is z= .31*throatDiameter. This was then used to position our kildall horns in the focus direction.
Focus curves were done (by moving the platform up and down) to try and find the optimum platform height. After doing this, all of the kildall horns needed a platform height that was lambda/2 higher than the none kildall horns (more info)
German cortes modeled the lbw kildall horn, and found that the phase center was about z=-3.05 cm inside the waveguide. After more research, i think it turned out that there was a sign error in the original kildall paper. The kildall horns were repositioned (scaling the lbw position by the throat diameter).
horn throatDiameter z focus from horn throat lbw 20.254 cm (7.975") -3.0401 cm sbw 12.253 cm (4.824") -1.845 cm sbh 7.945 cm (3.128") -1.197 cm cband 5.969 cm (2.350") -0.899 cm
note: lbw was switched from a kildall horn to a cortes horn design on feb2003
When surveying, we measure the locations of targets sitting on a plate connected to the mouth of the horn. You then need to translate these numbers up the z axis to the focus position. To do this we need to know :
• the target height above the plate (1")
• the plate thickness
• the distance from top of horn (where plate connects to the throat of the horn (z=0)
• the distance from throat of horn to horn focus.
The table below tries to summarize this info
horn type throat Diameter zFocus from horn throat hornThroat to horn end survey plate thickness targetTo focus lbw 1.115-1.760Ghz kildall cortes after feb03 20.254 cm (7.975") -3.0401 cm 2.813 cm sbn lynn? sbtx lynn? sbw 1.7-3 GHz kildall 12.253 cm (4.824") -1.845 cm 1.702 cm .665 cm (.262") sbh 3-4GHz kildall 7.945 cm (3.128") -1.197 cm 1.103 cm .665 cm (.262") cband 4-6GHz kildall 5.969 cm (2.350") -0.899 cm 0.829 cm .665 cm (.262") cbandWide 4-8GHz cortes xband 8-10GHz lynn
Notes:
• throatDiameter
• kildall horns.. may have scaled it from lbw
• zfocus
• kildall horns:
• used germans hfss modeling of lbw and then scaled by throat diameter for other kildall horns
• survey plate thickness
• measured.
` home_~phil` | 4,535 | 15,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-30 | latest | en | 0.838535 |
http://www.braingle.com/brainteasers/teaser.php?op=2&id=44413&comm=0 | 1,498,541,755,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320995.93/warc/CC-MAIN-20170627050500-20170627070500-00334.warc.gz | 480,252,873 | 6,240 | Browse Teasers
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## Two Ones
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #44413 Fun: (2.05) Difficulty: (2.11) Category: Probability Submitted By: shenqiang
Which is more likely to happen?
1) Roll a fair die 4 times and get at least one 1;
2) Roll 2 fair dice 24 times and get at least one "double 1".
Rolling a fair die once and failing to get a 1 has probability 5/6, and rolling 2 fair dice 6 times and failing to get a double 1 has probability (35^36)^6, which is slightly larger than 5/6.
Therefore, if you do each of the above 4 times and still fail to get a 1, the latter has a larger probability, therefore, the latter has a smaller succeeding probability than the former.
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## What Next?
See another brain teaser just like this one...
Or, just get a random brain teaser
If you become a registered user you can vote on this brain teaser, keep track of
which ones you have seen, and even make your own. | 255 | 1,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.923896 |
https://core.tcl-lang.org/tips/info/84e714bd8c6cb303 | 1,596,646,760,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735963.64/warc/CC-MAIN-20200805153603-20200805183603-00405.warc.gz | 267,107,643 | 9,775 | # Check-in [84e714bd8c]
Bounty program for improvements to Tcl and certain Tcl packages.
```42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 .. 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 ``` ```1 % string is wide 18446744073709551615 1 % string is wide 18446744073709551616 0 So, valid wide integers appear to range from -2^64+1 to +2^64-1. Most people learn in school that 64-bit integers range from -2^63 to 2^63-1. Are Tcl's wide integers 65-bit, but then excluding -18446744073709551616?
% expr int(2147483648) ; #on LP64/ILP64 platforms 2147483648 % expr int(2147483648) ; #on other platforms -2147483648 ................................................................................ * On all platforms, the int() math function is modified to do 64-bit truncation, as it already does on [LP64/ILP64](https://en.wikipedia.org/wiki/64-bit_computing#64-bit_data_models) systems (e.g. 64-bit Linux). int() will thus become synonym for wide(). The wide() function will become deprecated in Tcl 9.0, but it will not be removed yet. * The ranges of "string is integer" and "string is wideinteger" are changed to match the range of the int()/wide() math function. So these functions will report true (1) if the number is in the range -9223372036854775808..9223372036854775807. The "string is wideinteger" command will be deprecated in Tcl 9.0, but it will not be removed yet. * The C function Tcl\_GetIntFromObj() is changed to return TCL\_OK if the Tcl_Obj contains values in the range of -2147483648..4294967295. So, it succeeds if the number fits in either a platform "int", either a platform "unsigned int" type. * The C function Tcl\_GetLongFromObj() is changed to behave like Tcl\_GetIntFromObj() if sizeof(long) == sizeof(ing), and to behave like Tcl\_GetWideIntFromObj() if sizeof(long) == sizeof(Tcl_WideInt) # Implementation Currently, the proposed implementation is available in the [all-wideint branch] (https://core.tcl.tk/tcl/timeline?r=all-wideint). (WIP) # Copyright This document has been placed in the public domain. ``` ``` | | > > > | | ``` ```42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 .. 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 ``` ```1 % string is wide 18446744073709551615 1 % string is wide 18446744073709551616 0 | 728 | 2,362 | {"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} | 2.921875 | 3 | CC-MAIN-2020-34 | latest | en | 0.693873 |
https://www.civilengineeringx.com/structural-analysis/space-trusses/ | 1,606,602,886,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00623.warc.gz | 612,421,213 | 18,910 | ## Space Trusses
Space trusses, because of their shape, arrangement of members, or applied loading, cannot be subdivided into plane trusses for the purposes of analysis and must, therefore, be analyzed as three-dimensional structures subjected to three-dimensional force systems. As stated in Section 4.1, to simplify the analysis of space trusses, it is assumed that the truss members are connected at their ends by frictionless ball-and-socket joints, all external loads and reactions are applied only at the joints, and the centroidal axis of each member coincides with the line connecting the centers of the adjacent joints. Because of these simplifying assumptions, the members of space trusses can be treated as axial force members.
The simplest internally stable (or rigid) space truss can be formed by connecting six members at their ends by four ball-and-socket joints to form a tetrahedron, as shown in Fig. 4.28(a). This tetrahedron truss may be considered as the basic space truss element. It should be realized that this basic space truss is internally stable in the sense that it is a threedimensional rigid body that will not change its shape under a general three-dimensional loading applied at its joints. The basic truss ABCD of Fig. 4.28(a) can be enlarged by attaching three new members, BE;CE, and DE, to three of the existing joints B;C, and D, and by connecting them to form a new joint E, as depicted in Fig. 4.28(b). As long as the new joint E does not lie in the plane containing the existing joints B;C,
and D, the new enlarged truss will be internally stable. The truss can be further enlarged by repeating the same procedure (as shown in Fig. 4.28(c)) as many times as desired. Trusses constructed by this procedure are termed simple space trusses.
A simple space truss is formed by enlarging the basic tetrahedron element containing six members and four joints by adding three additional members for each additional joint, so the total number of members m in a simple space truss is given by
m = 6 + 3 (j-4) = 3j – 6 [eq 4.5]
in which j =total number of joints (including those attached to the supports).
# Reactions
The types of supports commonly used for space trusses are depicted in Fig. 4.29. The number and directions of the reaction forces that a support may exert on the truss depend on the number and directions of the translations it prevents.
As suggested in Section 3.1, in order for an internally stable space structure to be in equilibrium under a general system of threedimensional forces, it must be supported by at least six reactions that satisfy the six equations of equilibrium (Eq. (3.1)):
Because there are only six equilibrium equations, they cannot be used to determine more than six reactions. Thus, an internally stable space structure that is statically determinate externally must be supported by exactly six reactions. If a space structure is supported by more than six reactions, then all the reactions cannot be determined from the six equilibrium equations, and such a structure is termed statically indeterminate externally. Conversely, if a space structure is supported by fewer than six reactions, the reactions are not su‰cient to prevent all possible movements of the structure in three-dimensional space, and such a structure is referred to as statically unstable externally. Thus, if
As in the case of plane structures discussed in the previous chapter, the conditions for static determinacy and indeterminacy, as given in Eq. (4.6), are necessary but not sufficient. In order for a space structure to be geometrically stable externally, the reactions must be properly arranged so that they can prevent translations in the directions of, as well as rotations about, each of the three coordinate axes. For example, if the lines of action of all the reactions of a space structure are either parallel or intersect a common axis, the structure would be geometrically unstable.
# Static Determinacy, Indeterminacy, and Instability
If a space truss contains m members and is supported by r external reactions, then for its analysis we need to determine a total of m þ r unknown forces. Since the truss is in equilibrium, each of its joints must also be in equilibrium. At each joint, the internal and external forces form a three-dimensional concurrent force system that must satisfy the three equations of equilibrium, ∑Fx = 0, ∑Fy= 0, and ∑Fz = 0. Therefore, if the truss contains j joints, the total number of equilibrium equations available is 3j. If m + r = 3j, all the unknowns can be determined by solving the 3j equations of equilibrium, and the truss is statically determinate.
Space trusses containing more unknowns than the available equilibrium equations (m+r>3j) are statically indeterminate, and those with fewer unknowns than the equilibrium equations (m+r<3j) are statically unstable. Thus, the conditions of static instability, determinacy, and indeterminacy of space trusses can be summarized as follows:
In order for the criteria for static determinacy and indeterminacy, as given by Eq. (4.7), to be valid, the truss must be stable and act as a single rigid body, under a general three-dimensional system of loads, when attached to the supports.
# Analysis of Member Forces
The two methods for analysis of plane trusses discussed in Sections 4.5 and 4.6 can be extended to the analysis of space trusses. The method of joints essentially remains the same, except that three equilibrium equations (∑Fx = 0, ∑Fy = 0, and ∑Fz = 0) must now be satisfied at each joint of the space truss. Since the three equilibrium equations cannot be used to determine more than three unknown forces, the analysis is started at a joint that has a maximum of three unknown forces (which must not be coplanar) acting on it. The three unknowns are determined by applying the three equations of equilibrium. We then proceed from joint to joint, computing three or fewer unknown forces at each subsequent joint, until all the desired forces have been determined.
Since it is di‰cult to visualize the orientations of inclined members in three-dimensional space, it is usually convenient to express the rectangular components of forces in such members in terms of the projections of member lengths in the x; y, and z directions. Consider a member AB of a space truss, as shown in Fig. 4.30. The projections of its length LAB in the x; y, and z directions are xAB; yAB, and zAB, respectively, as shown, with
1. If all but one of the members connected to a joint lie in a single plane and no external loads or reactions are applied to the joint, then the force in the member that is not coplanar is zero.
2. If all but two of the members connected to a joint have zero force and no external loads or reactions are applied to the joint, then unless the two remaining members are collinear, the force in each of them is also zero.
The first type of arrangement is shown in Fig. 4.31(a). It consists of four members AB;AC;AD, and AE connected to a joint A. Of these, AB;AC, and AD lie in the xz plane, whereas member AE does not. Note that no external loads or reactions are applied to joint A. It should be obvious that in order to satisfy the equilibrium equation ∑Fy = 0, the y component of FAE must be zero, and therefore ∑AE = 0.
The second type of arrangement is shown in Fig. 4.31(b). It consists of four members AB;AC;AD, and AE connected to a joint A, of which AD and AE are zero-force members, as shown. Note that no external loads or reactions are applied to the joint. By choosing the orientation of the x axis in the direction of member AB, we can see that the equilibrium equations∑Fy = 0 and ∑Fz = 0 can be satisfied only if FAC = 0. Because the x component of FAC is zero, the equation ∑Fx = 0 is satisfied only if FAB is also zero.
As in the case of plane trusses, the method of sections can be employed for determining forces in specific members of space trusses. An imaginary section is passed through the truss, cutting the members whose forces are desired. The desired member forces are then calculated by applying the six equations of equilibrium (Eq. (3.1)) to one of the two portions of the truss. No more than six unknown forces can be determined from the six equilibrium equations, so a section is generally chosen that does not pass through more than six members with unknown forces.
Because of the considerable amount of computational efort involved, the analysis of space trusses is performed today on computers. However, it is important to analyze at least a few relatively small space trusses manually to gain an understanding of the basic concepts involved in the analysis of such structures.
# Example 4.12
Determine the reactions at the supports and the force in each member of the space truss shown in Fig. 4.32(a).
Solution
Static Determinacy The truss contains 9 members and 5 joints and is supported by 6 reactions. Because m+r=3j and the reactions and the members of the truss are properly arranged, it is statically determinate.
Member Projections The projections of the truss members in the x; y, and z directions, as obtained from Fig. 4.32(a), as well as their lengths computed from these projections, are tabulated in Table 4.1.
Zero-Force Members It can be seen from Fig. 4.32(a) that at joint D, three members, AD;CD, and DE, are connected. Of these members, AD and CD lie in the same ðxzÞ plane, whereas DE does not. Since no external loads or reactions are applied at the joint, member DE is a zero-force member. | 2,134 | 9,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | longest | en | 0.968435 |
https://ch.mathworks.com/matlabcentral/cody/problems/12-fibonacci-sequence/solutions/1061598 | 1,603,124,345,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00539.warc.gz | 271,751,990 | 16,872 | Cody
# Problem 12. Fibonacci sequence
Solution 1061598
Submitted on 26 Nov 2016 by Hiroshi Miyagawa
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; f = 1; assert(isequal(fib(n),f))
f = 1
2 Pass
n = 6; f = 8; assert(isequal(fib(n),f))
f = 8
3 Pass
n = 10; f = 55; assert(isequal(fib(n),f))
f = 55
4 Pass
n = 20; f = 6765; assert(isequal(fib(n),f))
f = 6765
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 208 | 626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.754935 |
http://mathandreadinghelp.org/math_homework_help_evaluating_radicals.html | 1,563,398,813,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00387.warc.gz | 98,783,698 | 7,844 | # Math Homework Help: Evaluating Radicals
In middle school, you'll evaluate radicals that have perfect squares or perfect cubes. Then, in high school, you'll simplify more complex radicals by factoring them. Read on to find out how!
The radical sign, which looks like a check mark with a horizontal line extending from the top, has a number on the inside called the radicand. When the radical sign is used alone, you'll need to find the square root of the radicand. If a small 'three' is written next to the symbol, you'll find the radicand's third root (also known as the cube root). If a four, five, six or any other number appears beside the radical sign, you'll find the radicand's corresponding root.
### Square Roots
A number's root can be multiplied by itself a certain number of times to produce the original number. The square root of a number equals that number when it's multiplied by itself once. For instance, (square root of 9) = 3 because 3 x 3 = 9. Likewise, (square root of 16) = 4 because 4 x 4 = 16. Here are a few more examples of square roots:
(square root of 49) = 7
(square root of 81) = 9
(square root of 144) = 12
### Cube Roots
A number's cube root has to be multiplied by itself twice to produce the original number. For example, since 2 x 2 x 2 = 8, the (cube root of 8) = 2. Three is the cube root of 27 because 3 x 3 x 3 = 27. These are some other cube root examples:
(cube root of 64) = 4
(cube root of 125) = 5
(cube root of 1,000) = 10
You'll also encounter radicals that aren't perfect squares or cubes, like (square root of 28). To simplify a radical like this, you'll need to apply the following property:
(square root of ab) = (square root of a) x (square root of b)
This means you can break a radical into the product of two different radicals by factoring the radicand, like this:
(square root of 28) = (square root of 4) x (square root of 7)
Hopefully, you'll be able to find at least one factor of the radicand that's a perfect square, as we did in this problem. Since (square root of 4) = 2, we can simplify the expression to 2(square root of 7). The original radical is now in its simplest form because (square root of 7) can't be simplified any further. Here's an example with cube roots:
(cube root of 40) = (cube root of 8) x (cube root of 5) = 2(cube root of 5)
Did you find this useful? If so, please let others know!
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Online Only | 1,073 | 4,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2019-30 | longest | en | 0.944313 |
https://www.diyaudio.com/community/threads/non-polarised-electrolytics.173357/ | 1,726,644,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00574.warc.gz | 678,812,931 | 26,024 | # non-polarised electrolytics
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
#### hitsware
When putting 2 electrolytics in series to
make a non-polarised cap .......
Is it preferable to put + to + or - to - ?
#### jam
I usually put (+) to (+) but in reality it should not make any difference......but.........
For small capacitors - say, 1 uF or less - a non-electrolytic type will very
likely be satisfactory if its size - these are usually much larger - is not a
problem.
There are several approaches to using normal polarized electrolytic capacitors
to construct a non-polarized type.
None of these is really great and obtaining a proper replacement would
be best. In the discussion below, it is assumed that a 1000 uF, 25 V
non-polarized capacitor is needed.
Here are three simple approaches:
* Connect two electrolytic capacitors of twice the uF rating and at least
equal voltage rating back-back in series:
- + + -
o----------)|-----------|(-----------o
2,000 uF 2,000 uF
25 V 25 V
It doesn't matter which sign (+ or -) is together as long as they match.
The increased leakage in the reverse direction will tend to charge up the
center node so that the caps will be biased with the proper polarity.
However, some reverse voltage will still be unavoidable at times. For
signal circuits, this is probably acceptable but use with caution in
power supply and high power applications.
* Connect two electrolytic capacitors of twice the uF rating and at least
equal voltage rating back-back in series. To minimize any significant
reverse voltage on the capacitors, add a pair of diodes:
+---|>|----+----|<|----+
| - + | + - |
o-----+----)|----+-----|(----+------o
2,000 uF 2,000 uF
25 V 25 V
Note that initially, the source will see a capacitance equal to the full
capacitance (not half). However, the diodes will cause the center node
to charge to a positive voltage (in this example) at which point the diodes
will not conduct in the steady state.
However, there will be some non-linearity into the circuit under transient
conditions (and due to leakage which will tend to discharge the capacitors)
so use with care. The diodes must be capable of passing the peak current
without damage.
* Connect two capacitors of twice the uF rating in series and bias the center
point from a positive or negative DC source greater than the maximum signal
expected for the circuit:
+12 V
o
|
/
\ 1K
/
- + | + -
o----------)|-----+-----|(-----------o
2,000 uF 2,000 uF
35 V 35 V
The resistor value should be high compared to the impedance of the driving
circuit but low compared to the leakage of the capacitors. Of course, the
voltage ratings of the capacitors need to be greater than the bias plus the
peak value of the signal in the opposite direction.
Last edited:
#### jcx
no its really better to buy a nonpolar/bipolar electrolytic with a full thickness oxide grown on both electrode foils
according to Bateman's "Capacitor Sound" series the 2 back-to-back electros gives much more distortion
#### jam
........yes, but only if it is available.
Jam
#### Fedess
Depending what do you need it for, maybe it's better to buy few (or lots!) of polyester, (or MKT, etc) caps and put them in parallel to get the value you need. I understand this way you get better sound quality...
Anyway I used two electros in series for a 1st order filter and they just worked well...
I have read on internet that using at least one polyester (or other kind of non polarized caps) in parallel with the resultant of the in-series electrolytics gives better sound quality... but I'm not sure of that...
#### pjp
Back to back electrolytics are sooo 1990's.
When you can buy 100µF Ceramic caps, I don't see any reason to use electrolytics.
#### PhantomBox
I'am recapping a old dbx160 compressor, and I need a 4.7uF/25V non polarized cap.
I could use 4x 4.7uF polarized caps and arrange them in 2 parallel pairs + to +.
What if I put 2x 10uF + to +, in order to safe space? Would the 0.6uF on each side in excess make a difference or is it neglectable?
I don't have any 4.7uF caps on hand but tons of 10uF/50V...
#### chip_mk
I'am recapping a old dbx160 compressor, and I need a 4.7uF/25V non polarized cap.
I could use 4x 4.7uF polarized caps and arrange them in 2 parallel pairs + to +.
What if I put 2x 10uF + to +, in order to safe space? Would the 0.6uF on each side in excess make a difference or is it neglectable?
I don't have any 4.7uF caps on hand but tons of 10uF/50V...
2x10uF should be OK, elcos tolerance is typically +-20% anyway. However, if space is not critical it's even better to go for 4.7uF MKT cap.
#### benb
...
* Connect two capacitors of twice the uF rating in series and bias the center
point from a positive or negative DC source greater than the maximum signal
expected for the circuit:
+12 V
o
|
/
\ 1K
/
- + | + -
o----------)|-----+-----|(-----------o
2,000 uF 2,000 uF
35 V 35 V
The resistor value should be high compared to the impedance of the driving
circuit but low compared to the leakage of the capacitors. Of course, the
voltage ratings of the capacitors need to be greater than the bias plus the
peak value of the signal in the opposite direction.
Charge Coupled Crossovers Article By Jeff Poth
http://www.diyaudio.com/forums/multi-way/13184-pre-bias-capacitors-crossover.html
Back to back electrolytics are sooo 1990's.
When you can buy 100µF Ceramic caps, I don't see any reason to use electrolytics.
For better or worse, I haven't seen a 100uF ceramic. I can imagine it used to drive a stylus in a record cutter.
Do they make ceramic capacitors big enough to be woofers?
#### indianajo
I've used 4.7 uf and 10 uf ceramic CPO caps in place of tantalums for amp input couplers. They sounded better than the tantalums I took out, which might have been rejects due to the popcorn noise from day one.
The secret to using ceramic for couplers and filters, use 50 v caps on 2 v signals. That linearizes the voltage/capacitance curve a lot. But 10 uf 50 v COG Aerovox caps are \$7 each when in stock. I usually can't buy those, got the originals on closeout at farnell for \$4.78. I would suspect a 100 uf ceramic cap of being counterfeit, made with something else, a silk screen, and a spray paint can.
10 uf polyester are huge since the smallest voltage stocked I can find from an authorized distributor is 63 V.
I've looked at building a 4700 uf 80 v cap out of NP caps for a speaker protector, but despite the cost, the assembly would be bigger and weigh as much as the amp. Two 10000 uf polar electrolytic caps back to back, without the diode, sounds funny. They produce top octave IM distortion (vibrato) on Steinway piano source. I had thought about charging up polar caps with one diode from the speaker drive. Thanks to Jam (5 years later) for making that crazy idea official.
Last edited:
#### davidsrsb
Info on X7R dielectric http://www.avx.com/docs/Catalogs/cdesc.pdf
Interestingly there is a burn in effect reducing capacitance with time.
These are ok for input/output blocking and amplifier negative feedback dc blocking capacitors but not for crossovers. You probably wouldn't use such high values in an active filter
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. | 1,897 | 7,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.910022 |
https://photo.stackexchange.com/questions/24826/is-there-a-formula-to-calculate-dof | 1,718,505,632,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00301.warc.gz | 406,261,338 | 45,530 | # Is there a formula to calculate DOF?
I am pretty clear about that DOF depends on:
1. Focal length
2. Aperture or f-stop
3. Distance from subject
4. Sensor size
and more (as pointed out in this comment).
But my question is: Is there any formula that relates all these factors with Depth of Field?
Given these values is it possible to accurately calculate the Depth of Field?
• There are two more things to consider: (5) the size of the final image; and (6) whether you are concerned with the "zone of acceptable sharpness" when the other five factors are taken into account, or with the "zone of sufficient blurriness".
– user2719
Commented Jul 1, 2012 at 11:15
Depth of field depends on two factors, magnification and f-number.
Focal length, subject distance, size and circle of confusion (the radius at which blur becomes visible) jointly determine the magnification.
Depth of field does not depend on lens or camera design other than the variables in the formula so there are indeed general formulas to calculate depth of field for all cameras and lenses. I don't have them all committed to memory so I'd only be copying and pasting from Wikipedia: Depth of field.
A better answer to your question would be to go through the derivation of the formulas from first principles, something I've been meaning to do for a while but haven't had time. If anyone wants to volunteer I'll give them an upvote ;)
• That's somewhat mixed up. Magnification is just determined by the ratio of focal length to subject distance. Magnification and f-number determine at which rate depth relative to an object's width grows into blur relative to the object's width. To determine actual depth of field, you need to define just what size of blur you still consider in-focus: that's essentially the circle of confusion. Commented Sep 26, 2020 at 12:54
You wanted the math, so here it goes:
You need to know the CoC of your camera, Canon APS-C sized sensors this number is 0.018, for Nikon APS-C 0.019, for full frame sensors and 35mm film the number is 0.029.
The formula is for completeness:
CoC (mm) = viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25
Anothe way of doing this is the Zeiss formula:
c = d/1730
Where d is the diagonal size of the sensor, and c is the maximum acceptable CoC. This yields slightly different numbers.
You need to calculate the hyperfocal distance first for your lens and camera (this formula is inaccurate with distances close to the focal length e.g. extreme macro):
HyperFocal[mm] = (FocalLength * FocalLength) / (Aperture * CoC)
e.g.:
50mm lens @ f/1.4 on a full frame: 61576mm (201.7 feet)
50mm lens @ f/2.8 on a full frame: 30788mm (101 feet)
50mm lens @ f/1.4 on a Canon APS frame: 99206mm (325.4 feet)
50mm lens @ f/2.8 on a Canon APS frame: 49600mm (162.7 feet)
Next you need to calculate the near point which is the closest distance that will be in focus given the distance between the camera and the subject:
NearPoint[mm] = (HyperFocal * distance) / (HyperFocal + (distance – focal))
e.g.:
50mm lens @ f/1.4 on a full frame with a subject at 1m distance: 0.984m (~16mm in front of target)
50mm lens @ f/1.4 on a full frame with a subject at 3m distance: 2.862m (~137mm in front of target)
50mm lens @ f/2.8 on a full frame with a subject at 1m distance: 0.970m (~30mm in front of target)
50mm lens @ f/2.8 on a full frame with a subject at 3m distance: 2.737m (~263mm in front of target)
50mm lens @ f/1.4 on a Canon APS frame with a subject at 1m distance: 0.990m (~10mm in front of target)
50mm lens @ f/1.4 on a Canon APS frame with a subject at 3m distance: 2.913m (~86mm in front of target)
50mm lens @ f/2.8 on a Canon APS frame with a subject at 1m distance: 0.981m (~19mm in front of target)
50mm lens @ f/2.8 on a Canon APS frame with a subject at 3m distance: 2.831m (~168mm in front of target)
Next you need to calculate the far point which is the furthest distance that will be in focus given the distance between the camera and the subject:
FarPoint[mm] = (HyperFocal * distance) / (HyperFocal – (distance – focal))
e.g.:
50mm lens @ f/1.4 on a full frame with a subject at 1m distance: 1.015m (~15mm behind of target)
50mm lens @ f/1.4 on a full frame with a subject at 3m distance: 3.150m (~150mm behind of target)
50mm lens @ f/2.8 on a full frame with a subject at 1m distance: 1.031m (~31mm behind of target)
50mm lens @ f/2.8 on a full frame with a subject at 3m distance: 3.317m (~317mm behind of target)
50mm lens @ f/1.4 on a Canon APS frame with a subject at 1m distance: 1.009m (~9mm behind of target)
50mm lens @ f/1.4 on a Canon APS frame with a subject at 3m distance: 3.091m (~91mm behind of target)
50mm lens @ f/2.8 on a Canon APS frame with a subject at 1m distance: 1.019m (~19mm behind of target)
50mm lens @ f/2.8 on a Canon APS frame with a subject at 3m distance: 3.189m (~189mm behind of target)
Now you can calculate the total focal distance:
TotalDoF = FarPoint - NearPoint
e.g.:
50mm lens @ f/1.4 on a full frame with a subject at 1m distance: 31mm
50mm lens @ f/1.4 on a full frame with a subject at 3m distance: 228mm
50mm lens @ f/2.8 on a full frame with a subject at 1m distance: 61mm
50mm lens @ f/2.8 on a full frame with a subject at 3m distance: 580mm
50mm lens @ f/1.4 on a Canon APS frame with a subject at 1m distance: 19mm
50mm lens @ f/1.4 on a Canon APS frame with a subject at 3m distance: 178mm
50mm lens @ f/2.8 on a Canon APS frame with a subject at 1m distance: 38mm
50mm lens @ f/2.8 on a Canon APS frame with a subject at 3m distance: 358mm
So the complete formula w/ CoC and HyperFocal precalculated:
TotalDoF[mm] = ((HyperFocal * distance) / (HyperFocal – (distance – focal))) -(HyperFocal * distance) / (HyperFocal + (distance – focal))
Or simplified:
TotalDoF[mm] = (2 * HyperFocal * distance * (distance - focal)) / (( HyperFocal + distance - focal) * (HyperFocal + focal - distance))
With CoC precalulated: I've made an attempt to simplify the following equations with the following substitutions: a = viewing distance (cm) b = desired final-image resolution (lp/mm) for a 25 cm viewing distance c = enlargement d = FocalLength e = Aperture f = distance X = CoC
TotalDoF = ((((d * d) / (e * X)) * f) / (((d * d) / (e * X)) – (f – d))) - ((((d * d) / (e * X)) * f) / (((d * d) / (e * X)) + (f – d)))
Simplified:
TotalDoF = (2*X*d^2*f*e(d-f))/((d^2 - X*d*e + X*f*e)*(d^2 + X*d*e - X*f*e))
Even further simplified with WolframAlpha:
TotalDoF = (2 * d^2 * e * (d - f) * f * X)/(d^4 - e^2 * (d - f)^2 * X^2)
Or if nothing is precalculated, you get get this monster, which is unusable:
TotalDoF = ((FocalLength * FocalLength) / (Aperture * (viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25)) * distance) / ((FocalLength * FocalLength) / (Aperture * (viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25)) – (distance – focal)) - ((FocalLength * FocalLength) / (Aperture * (viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25)) * distance) / ((FocalLength * FocalLength) / (Aperture * (viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25)) + (distance – focal))
Simplified:
(50*a*b*c*d^2*f*e*(d-f))/((25*b*c*d^2 - a*d*e + a*f*e)*(25*b*c*d^2 + a*d*e - a*f*e)
So basically use recalculated CoC and HyperFocal :)
If you want to see a practical implementation of the depth of field formulas you can check out this Online Depth of Field Calculator. The source of the linked HTML page has all the formulas implemented in Javascript.
Yes, there are formulas. One can be found at http://www.dofmaster.com/equations.html. These formulas are used on this calculator, it also explains depth of field in more detail. I have used this site several times and have found it to be reasonably accurate after doing practical tests myself.
Here's a simple DOF formula. Hope it helps.
DOF = 2 * (Lens_F_number) * (circle_of_confusion) * (subject_distance)^2 / (focal_length)^2
P = point focused upon
Pd = distant point sharply defined
Pn = near point sharply defined
D = diameter of circle of confusion
f = f-number
F = focal length
Pn = P ÷ (1+PDf÷F^2)
Pd = P ÷ (1-PDf÷F^2)
Industry standard to set D = 1/1000 of the focal length. For more precise work use 1/1500 of the focal length. Presume 100mm focal length then 1/1000 of 100mm = 0.1mm or 1/1500 = 0.6666mm | 2,519 | 8,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.944378 |
http://persistentstruggle.blogspot.com/2011_10_12_archive.html | 1,532,346,137,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00263.warc.gz | 286,065,131 | 26,031 | ## Wednesday, 12 October 2011
### Anatomy of a convert - Pioneers
J.J. Hill. His company issued the first convertible.
Meyer Weinstein. He was one of the first to hedge convertibles with other securities. Using heuristic techniques.
E.O. Thorp. He showed a mathematical approach to valuing a convertible as a bond converting into a warrant.
Fischer Black. He got a CAPM-friendly solution to the pricing of a call warrant and in return got a call on the Nobel prize for economic science, but it expired just out of the money.
Since then, absolutely nothing of real importance except for a melding of credit and volatility factors into pricing models.
A lot of basics of fixed income modelling already was in place as early as 1913. See the below book.
### Anatomy of a convert - crude terms
Just as a little break form the maths I'd like to consider the major clauses in a convertible and try to come up with memorable catchphrases which describe their effect or purpose. Some clauses in a convertible prospectus are more straight bond like (i.e. they're often seen in ordinary bonds), whereas others are more specific to the warrant/convertibility side. Also, since traders are generally a foul-mouthed lot, I'm trying to make them as textually authentic as I can.
First up, the bond like clauses.
The holder's put clause. "Take your shit and give me my fuckin' money". The bond holder, on certain named dates, can decide he's had enough and would like his money back.
The issuer's call clause. "Party's over, get out of my fuckin' house". The issuer, if its company stock has done particularly well, can decide in short notice to force holders to decide whether or not to convert or else get their money back.
And now a couple of convertibility clauses.
The conversion right. "First we loan you, now we own you". The initial investment constituted an income bearing loan to the company. The bond holder now wants to convert and own a piece of the company, since the share price now makes it desirable to own.
The reset clause. "We sin, you win". If the company stock price fails to perform, the bond holder is entitled to get a fatter slice of the company.
The exchangeable. "Pimpin' your daughter". The exchange property which the issuer refers to isn't the issuer itself, but a related entity - often a majority shareholding in an asset which they now deem non-strategic.
The contingent conversion clause. "We win, you're in". The holder's right to convert is itself contingent on the company stock having cleared a number of performance hurdles.
The mandatory structure. "A chattel forward". The holder starts off lending the company capital, gets a regular income from it, then ends up getting certain ownership of the entity. | 634 | 2,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-30 | latest | en | 0.950083 |
http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=2235971 | 1,508,792,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00158.warc.gz | 180,245,786 | 9,022 | ## #2235971
Solution for DSL_2_A: Range Minimum Query (RMQ) by Tsuta_J
Source Code Status Test Cases
Policy: public Reviewed: 63
00.11 sec 4144 KB 50 lines 1172 bytes 2017-03-29 16:00
```#include <bits/stdc++.h>
using namespace std;
int N, Q;
int const INF = INT_MAX;
struct SegmentTree {
private:
int n;
vector<int> node;
public:
SegmentTree(vector<int> v) {
int sz = v.size();
n = 1; while(n < sz) n *= 2;
node.resize(2*n-1, INF);
for(int i=0; i<sz; i++) node[i+n-1] = v[i];
for(int i=n-2; i>=0; i--) node[i] = min(node[2*i+1], node[2*i+2]);
}
void update(int x, int val) {
x += (n - 1);
node[x] = val;
while(x > 0) {
x = (x - 1) / 2;
node[x] = min(node[2*x+1], node[2*x+2]);
}
}
int getmin(int a, int b, int k=0, int l=0, int r=-1) {
if(r < 0) r = n;
if(r <= a || b <= l) return INF;
if(a <= l && r <= b) return node[k];
int vl = getmin(a, b, 2*k+1, l, (l+r)/2);
int vr = getmin(a, b, 2*k+2, (l+r)/2, r);
return min(vl, vr);
}
};
int main() {
cin >> N >> Q;
SegmentTree seg( vector<int>(N, INF) );
for(int i=0; i<Q; i++) {
int c, x, y; cin >> c >> x >> y;
if(c == 0) seg.update(x, y);
else cout << seg.getmin(x, y+1) << endl;
}
}
```
Compile Error Logs: You are not authorized to see the message.
Status
Judge: 20/20 C++ CPU: 00.11 sec Memory: 4144 KB Length: 1172 B 2017-03-29 16:00 2017-03-29 16:00
Results for testcases
Case # Verdict CPU Time Memory In Out Case Name Case #1: : Accepted 00.00 sec 3108 KB Case #2: : Accepted 00.00 sec 3104 KB Case #3: : Accepted 00.00 sec 3084 KB Case #4: : Accepted 00.00 sec 3008 KB Case #5: : Accepted 00.00 sec 3084 KB Case #6: : Accepted 00.00 sec 3096 KB Case #7: : Accepted 00.00 sec 2988 KB Case #8: : Accepted 00.00 sec 3004 KB Case #9: : Accepted 00.00 sec 3060 KB Case #10: : Accepted 00.00 sec 3080 KB Case #11: : Accepted 00.00 sec 3008 KB Case #12: : Accepted 00.09 sec 3592 KB Case #13: : Accepted 00.00 sec 3088 KB Case #14: : Accepted 00.00 sec 3032 KB Case #15: : Accepted 00.00 sec 3108 KB Case #16: : Accepted 00.00 sec 3116 KB Case #17: : Accepted 00.10 sec 4088 KB Case #18: : Accepted 00.11 sec 4036 KB Case #19: : Accepted 00.11 sec 4144 KB Case #20: : Accepted 00.11 sec 4052 KB
< prev | / | next >
Judge Input # ( | ) Judge Output # ( | ) | 868 | 2,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.327991 |
https://nrich.maths.org/public/leg.php?code=-334&cl=3&cldcmpid=2760 | 1,511,212,726,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00617.warc.gz | 670,368,123 | 9,814 | # Search by Topic
#### Resources tagged with Games similar to Diamond Mine:
Filter by: Content type:
Stage:
Challenge level:
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
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Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score.
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##### Stage: 2 and 3 Challenge Level:
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Collect as many diamonds as you can by drawing three straight lines.
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##### Stage: 1, 2 and 3
This article, the second in the series, looks at some different types of games and the sort of mathematical thinking they can develop.
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##### Stage: 1, 2 and 3
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##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
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##### Stage: 1, 2 and 3 Challenge Level:
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##### Stage: 1, 2, 3 and 4
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##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
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##### Stage: 3 and 4 Challenge Level:
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##### Stage: 3 Challenge Level:
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##### Stage: 3 and 4 Challenge Level:
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##### Stage: 3 and 4 Challenge Level:
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Advent Calendar 2010 - a mathematical game for every day during the run-up to Christmas.
### Twin Corresponding Sudoku
##### Stage: 3, 4 and 5 Challenge Level:
This sudoku requires you to have "double vision" - two Sudoku's for the price of one
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##### Stage: 3 and 4 Challenge Level:
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##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
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##### Stage: 4 Challenge Level:
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### Football World Cup Simulation
##### Stage: 2, 3 and 4 Challenge Level:
A maths-based Football World Cup simulation for teachers and students to use. | 2,271 | 9,490 | {"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} | 3.375 | 3 | CC-MAIN-2017-47 | latest | en | 0.86321 |
https://docs.google.com/document/d/1vIloU_JyLm5ClTn5qGa5lC2bizLG5AtIY6E9L_xSkvY/pub | 1,506,076,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688932.49/warc/CC-MAIN-20170922093346-20170922113346-00272.warc.gz | 638,583,218 | 15,418 | 15:04 - so we seem to have 8 problems, penalty time 1178. Thanks to everybody who was following our contest! I’m sorry we didn’t post much in the last couple hours - things got really intense and we were too drawn into solving problems. For example, in the last 10 minutes Jakub was working as a mutex by physically blocking either me or Bruce from using the keyboard.
15:01 - several switches between K and E, but couldn’t get either to work :( So we’re done with 8 problems. What’s our penalty time?
14:48 - Submitted I :) K or E? I accepted.
14:30 - Jakub proposed a much better solution for K - for every substring (set of letters that appear consecutively in all 3 strings) and a combination of three recursion methods, find the lex-smallest tree that is produced from that substring. Finishing E, then going to do this for K.
14:18 - now comes the hardest time - to choose what to do :( After we finish E, should we write I or optimize K?
14:13 - Sped up K a bit, finally submitted - time limit exceeded.
14:03 - we’ve read the statement for I, and it turns out to be not that difficult :) Apparently we can iterate over left and right boundaries and then the problem becomes 1-dimensional and simple.
13:52 - we’ve submitted B, Bruce is back to E. B is accepted!
13:41 - we seem to have got rid of numerical stability issues in B, the only question is to pick boundaries now. Going to try nested binary search.
13:33 - Bruce still coding E, Petr and Jakub are simplifying the formula in B. Getting close to a solution there!
13:16 - Petr tried to add some pruning, but it turns out to be not so easy. Bruce resumed E. While Petr was doing pruning, Bruce and Jakub could speed up the solution for E a bit :)
13:02 - Petr finished coding K, but it’s too slow without any pruning. Will need to add pruning at some point, right now Bruce started on E.
12:37 - we decided to write K first, Jakub will just look at his code in the meantime.
12:34 - Jakub found a bug in his code, now the hypothesis works for the sample input! We have to speed it up to work properly on “99.99 49.99” case, Jakub is working on that.
12:29 - The hypothesis doesn’t work out at the moment, so Petr is starting on K. Basically checking all possibilities, with some pruning.
12:23 - But ITMO already has 7 :) It will be hard to catch them. Now Jakub is testing a quick hypothesis in B.
12:22 - Petr got H accepted. The idea is somewhat standard - what is the smallest number of steps required to assemble each subsegment. The straightforward implementation is O(n^4) though, so one has to eliminate one n from the DP transition.
12:04 - J accepted as well! The bug was: to check if a circle arc is inside or outside the polygon, we should take the middle of the arc, not the middle of its base segment.
12:02 - C accepted!
12:00 - while Jakub was coding C, Petr and Bruce have re-read the solution for J and found one bug. Now submitted C and fixing the bug in J.
11:47 - got WA on J, Jakub is starting on C. It’s simply maximum flow.
11:42 - well, we can just try all trees - there should be only so many rooted trees with 26 vertices (with labeled left and right subtrees for each vertex).
11:36 - K looks approachable. There are only so many possible Pre/Post/In combinations, and then we need to somehow build the tree. The problem is, of course, is that we don’t know when we descend and when we ascend.
11:31 - Bruce has coded almost complete solution for J, now Jakub is typing a formula for the area of a “circle trapezoid” over his shoulder.
11:28 - meanwhile, Jakub is ready to code C and Petr is ready to code H. Now that we have a solution for all problems that have been solved by at least one team, let’s try the difficult ones :)
11:10 - now Bruce started writing a solution for G. It looks like it’s simply taking a hundred of linear constraints on a toroid... But how to solve two equations on a toroid? So Bruce started on J instead. There we simply need to intersect segments with a circle, and then take a signed sum of areas of trapezoids in the usual way.
11:09 - We got D accepted on our third try. Interestingly, the only optimization we added is memoization of DP states across test cases.
11:01 - A is accepted!
11:00 - The most difficult part was to notice that pieces can be reflected. Without that property, the problem is much more difficult. Submitted.
10:54 - Petr going to solve A. Just check if a graph has cycles.
10:49 - apparently Jakub didn’t factor in the fact that we have 1000 testcases. Optimizing now :)
10:44 - sent D, time limit exceeded :( And another TLE.
10:35 - Accepted! Hopefully we’ll get faster from now on :) Jakub still coding D.
10:33 - Petr found a bug: integer overflow :( Resubmitted.
10:27 - Jakub started by writing D, then thought his solution is too slow (O(num_divisors * log^2(n)), so Petr started solving F, got two Wrong Answers. Now Jakub is continuing to solve D, while Petr is trying to find bugs in F. Statement of B looks easy but somehow we can’t find a solution yet.
10:09 - we spent about 10 minutes looking for a room, didn’t find one, so now we’re solving in the hall. Jakub is solving D!
9:50 - John Clevenger explains the final technical aspects. Just 7 minutes to go.
9:44 - They’re using the big under-the-roof hockey scoreboard! Right now it shows a lot of Windows console java.net...Exceptions :)
9:40 - Roman Elizarov is trying to make teams nervous by emphasizing how great the competition is, using the microphone :)
9:39 - The scoreboard at http://myicpc2.icpcnews.com/ confirms 11 problems.
9:31 - There are special balloons for the first team to solve each problem, and one (heart-shaped :)) for the very first successful submission. So the very first successful submissions yields 3 balloons.
9:30 - According to the balloon color guide, there are 11 problems: A - Red, B - Black, C - Turquoise, D - Yellow, E - Dark Pink, F - Blue, G - Burgundy, H - Green, I - Orange, J - Pink, K - Teal.
9:24 - The link to submit problems after the contest starts: https://icpc.kattis.com/problems.
9:18 - the scoreboard still shows Dress Rehearsal results, and links to Dress Rehearsal problem statements.
9:14 - apparently the contest starts at 10, not 9:30. So about 45 more minutes before the official contestants see the problems, about an hour before we see the problems.
9:10 - a lot of people around have red badges - apparently those are press representatives. This makes it easier for teams to ignore them during the contest, I guess :)
9:04 - Egor’s blog at http://codeforces.ru/blog/Egor is awesome in many ways, but one of those ways is that he seems to be going to post a lot of pictures :) There won’t be too many pictures here, so don’t forget to check updates from Egor, too!
9:03 - You can ask questions at petr-mitrichev.blogspot.com/2013/07/acm-icpc-2013-world-finals-live.html, we’ll do our best to answer them. | 1,707 | 6,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-39 | longest | en | 0.952966 |
https://www.coursehero.com/file/122599/STA-301-Ch-08-pp-85-95/ | 1,529,350,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00533.warc.gz | 811,391,185 | 39,629 | STA 301 - Ch 08 - pp 85-95
# STA 301 - Ch 08 - pp 85-95 - Chapter 8 Fundamental Sampling...
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Chapter 8: Fundamental Sampling Distributions and Data Descriptions 85 CHAPTER 8 — FUNDAMENTAL SAMPLING DISTRIBUTIONS AND DATA DESCRIPTIONS Recall in Chapter 1 we introduced the idea of random sampling . In this chapter, we formalize the concept and introduce the idea of a sampling distribution . Definitions Population – “The totality of the observations with which we are concerned” o The population may be finite or infinite. o Each individual element of a population is a value of a random variable X having some probability distribution f ( x ). Parameter – A numerical characteristic of the population. o Typically, we think of the population mean and population variance, but any numerical characteristic of a population can be thought of as a parameter. Sample – A subset of a population o Generally, the sample is used for inference about some population parameter. o To avoid inadvertent bias in our inference, we prefer using random samples . Random Sample Let n X X X , , , 2 1 K be n independent random variables, each having the same probability distribution f ( x ). Then we define n X X X , , , 2 1 K to be a random sample of size n from the population f ( x ), and write its joint probability distribution as: ( ( ( ( n n x f x f x f x x x f L K 2 1 2 1 , , , = Statistic o A statistic is any function of the random variables that constitute a random sample. Note that since a statistic is a function of random variables, it is also a random variable.
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Chapter 8: Fundamental Sampling Distributions and Data Descriptions 86 Some Common Statistics Sample Mean If n X X X , , , 2 1 K represent a random sample of size n , then the sample mean is defined by the statistic: = = n i i X n X 1 1 . Sample Variance If n X X X , , , 2 1 K represent a random sample of size n , then the sample variance is defined by the statistic: ( 29 = - - = n i i X X n S 1 2 2 1 1 . Of course, this is computationally equivalent to: - - = = 2 1 2 2 1 1 X n X n S n i i . The sample standard deviation , S , is the positive square root of the sample variance.
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Jill Tulane University ‘16, Course Hero Intern | 787 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.802949 |
https://things-in-motion.blogspot.com/2018/12/how-to-estimate-torque-of-bldc-pmsm.html | 1,568,599,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572471.35/warc/CC-MAIN-20190916015552-20190916041552-00437.warc.gz | 707,601,175 | 25,087 | Tuesday, December 25, 2018
How to estimate the torque of a BLDC (PMSM) electric motor using only its Kv and current draw
There exists a fundamental relationship between an electric motors velocity constant (K_{V}), armature* current (I_{A}) and torque (\tau)This relationship is as follows:
\tau \approx \frac{8.3 \times I_{A} }{K_{V}}
were \tau is in N.m, I_{A} is in A and K_{V} is in RPM/V. This relationship is extremely useful since most 'hobby grade' BLDC/PMSM manufacturers do not publish the usual motor constants you would expect from an industrial product while K_{V} and peak I_{A} is normally specified on sites like hobbyking.com.
The above relationship works for two reasons:
1. A motors K_{V} and its torque constant (K_{\tau}) are fundamentally the same thing.
2. The torque generated by a motor for a given current is governed by K_{\tau}.
Of course, there are limitations to this approach. If a motor is close to saturation or if the current waveform supplied by a motor controller does not exactly match a motors back EMF waveform (i.e. the current is not exactly on the q-axis at all times) then your torque will be less than that suggested above. However, my own testing shows that this approximation is quite accurate when a 'hobby grade' PMSM motor is driven using field oriented control (FOC). Even when not using FOC or a PMSM this approach should still provide a good starting point.
Read on if you are interested in a more detailed understanding of why this relationship works and the limitations of this approach.
Fundamental torque production
Electric motors do useful work by producing torque and rotation. The amount of steady state torque produced by a well optimised and non-salient machine** with a specific volume (specific torque density) is ultimately dependent upon three factors:
• The average flux density acting upon the armature. For a BLDC/PMSM motor this 'flux' is provided by the rotor's permanent magnets.
• The average current that can be maintained by the armature without overheating**. For a BLDC/PMSM motor the armature consists of the copper windings in the stator.
• The total length of the armature windings which has 'flux' acting upon it. For a BLDC/PMSM motor this length represents the number of turns of wire in the stator armature which interact with the 'flux' provided by the rotor.
In other words, all the complexity of an electric motor ultimately boils down to the "BIL" Lorentz force law:
Force = Magnetic field \timesCurrent\times Conductor length
Increase one of the above terms without negatively impacting another and you have increased the torque output of your system. When discussing motor constants I find it useful to keep this simple relationship in mind.
Velocity constant = Torque constant
In the 'hobby' community most people seem comfortable with the concept of a motors velocity constant (K_{V}). K_{V} is easily measured and is given by half the peak to peak back EMF generated line to line (across any two motor leads) for a given mechanical frequency with a unit RPM/V.
What seems less well understood is that an electric motors torque constant (K_{\tau}) is equal to K_{V} so that
K_{\tau}^* = K_{V}^*
where here K_{\tau}^* is a motors per-phase torque constant and K_{V}^* is a motors per-phase velocity constant given by half the peak to peak back EMF generated line to neutral (from the centre tap point on a star wound motor) for a given electrical frequency (Elec. Freq. = Mech. Freq. / No. Pole Pairs) with the units \frac{V \cdot S}{rad}.
When first learning about electric motors it took me a while to accept that K_{\tau}^* does indeed equal to K_{V}^*. The mathematics is clear and their SI units are equivalent but something about it just didn't feel right. To overcome this I find it helps to keep in mind that all aspects of an electric motor which impact its K_{V}^* (e.g. flux gap size, magnet strength, winding turn number, rotor length etc.) also impact the amount of torque a motor can produce for a given current.
K_{V}^* is not especially useful since we are normally interested in a motors 'total torque constant' produced by all three phases and not that of a single phase. The 'total torque constant' of a 3 phase PMSM as estimated using its line-line K_{V} (RPM/V) is instead given by
K_{\tau} = \frac{3}{2} \times \frac{1}{\sqrt{3}}\times \frac{60}{2\pi} \times \frac{1}{K_{V}}
As is discussed by Oskar Weigl here, the factor \frac{3}{2} is derived from eq. 3.2 in this paper, the \frac{1}{\sqrt{3}} is for converting the line-line voltage (which is what is commonly used by hobby motor manufacturers in the determination of the K_{V}) to phase voltage (example here) and \frac{60}{2\p} is to convert from rpm to rad/s, which is needed to estimate the torque constant from the voltage constant, due to the voltage constant being specified in units of RPM/V.
Torque constant determines torque output for a given current
K_{\tau} is also defined as
K_{\tau} = \frac{\tau}{I_{A}}
where I_{A} was the armature current. Therefore the 'overall torque' output of a 3 phase PMSM is given by
\tau = \frac{3}{2} \times \frac{1}{\sqrt{3}}\times \frac{60}{2\pi} \times \frac{1}{K_{V}}\times I_{A}
This finally leads us to our 'conversion constant' of approximately 8.3 mentioned in the introduction
\tau \approx 8.3 \times \frac{1}{K_{V}} \times I_{A} \approx \frac{8.3 \times I_{A} }{K_{V}}
and so the 'conversion constant' of 8.3 is just a simplified approximation for converting K_{V} to K_{V}^*.
It is important to point out here that this relationship is true for any three PMSM. Star, delta, big, small, high K_{V}, low K_{V}in-runner, out-runner, cored or core-less, 'weak' or 'strong' magnets, it doesn't make any difference. Equally important, the torque referenced above is the 'electromagnetic' torque produced by the motor. This torque represents a 100% efficient conversion from electrical energy to mechanical energy. The 'shaft' torque, which is the usable output torque of the motor, will always be less than the electromagnetic torque due losses in the system. If you would like to go deeper into all the topics described above than this then I highly recommend James Mevey's Master's thesis from Kansas State which which was recommended and summarised by Shane Colton in his blog.
All this theory is great, but lets see if it actually works in practice.
Measuring the motor constants of some real motors
In order to put all this theory to the test I have measured the motor constants for the following collection of 'hobby grade' out-runner motors.
K_{V} was first measured using the method described here. I took a photo of my oscilloscope at around 500 RPM for each motor as shown below.
1000 Kv Racerstar BR2212
190 Kv Keda 6364
150 Kv Odrive 6374
280 kV Turnigy SK3 5055
270 Kv Odrive D6374
All of these motors are 12N14P and of a similar construction. Its clear that the shape of the back EMF is sinusoidal and not trapezoidal. Therefore, these motors are technically permanent magnet synchronous motors (PMSM) and not brushless DC motorsOverall, the measured K_{V} of each motor was quite close to their labelled values as will be shown in a table later.
To estimate K_{\tau} the output torque of each motor was measured with respect to the armature current. This was done by winding a string around the rotor of each motor, attaching that string to a lever arm that then pulled down onto a laboratory balance. This balance then measured the weight, and therefore force, produced by the motor. This method is only possible thanks to the use of a high resolution encoder (8192 counts per rotation) and an Odrive motor controller which uses field oriented control (FOC) to place all current on the motors q-axis for maximum torque, even when stationary.
The setup is crude and would have been much cleaner if I had just attach an arm to each motors rotor. However, by using a string I didn't need to make a new arm adaptor for each motor and could instead just consider the diameter of the rotor in the final calculations. This was enough to get the job done.
Using this setup and a simple script I slowly stepped up the commanded current for each motor while manually recording the weight on the scale. After some back calculations the torque output for each motor with commanded current was found.
No surprises so far with the bigger motors producing more torque per amp and the torque increasing linearly with current.
A summary of the motor parameters can be seen below and the raw data can be found here
Racerstar BR2212 Turnigy SK3 5055 Odrive D5065 Keda 6364 Odrive D6374
Rated kV rpm/V 1000 280 270 190 150
Measured kV rpm/V 1058 276 259 182 151
Phase Resistance Ohm 0.128 0.032 0.039 0.039 0.039
Phase Inductance H 1.84E-05 1.33E-05 2.02E-05 2.13E-05 2.81E-05
Weight kg 0.045 0.389 0.411 0.647 0.885
Price $USD 6 52 69 47 99 Torque constant (Kt) N·m/A 0.008 0.029 0.030 0.042 0.053 Kt/kg N·m/A/kg 0.172 0.073 0.073 0.065 0.060 Kt/$ USD N·m/A/\$ USD 0.00129 0.00055 0.00043 0.00090 0.00054
Motor constant (Km) N⋅m/sqrt(W) 0.02 0.16 0.15 0.21 0.27
Km/kg N⋅m/sqrt(W)/kg 0.485 0.417 0.369 0.329 0.308
Conversion constant 8.1 8.2 8.2 8.2 8.2
I have calculated the 'conversion constant' based on an average of a few different current vs torque measurements. The calculated 'conversion constant' is actually very close to the theoretical value, with measured values between 8.1 and 8.2. Any error is likely due to the friction in the system and my less than ideal measurement setup. Also note that these values are based on the manufacture rated K_{V} of the motors and they are little lower if my own K_{V} is used instead. I'm not sure why this is the case but it may have something to do with the 'fudge factor' each manufacture assumes when estimating K_{V}.
Also listed is K_{\tau} with respect to motor weight and motor cost. Surprisingly, the smallest motor (1000 K_{V} BR2212) came out on top by a considerable margin in both cases, with a K_{\tau} more than double that of the other motors. This suggests that the electrical loading (current density in the copper windings) is much higher in the smallest motor when compared to the others. This same result could be achieved for the other motors by re-winding them to have a lower K_{V}. However, since the 'torque efficiency' (torque produced per Watt) is the same no matter how a motor is wound provided the amount of copper remains the same, and that the mass fraction of copper is likely to be about the same for these motors, the smallest motor will produce no more torque per unit weight than the rest when thermally limited. This assumption is backed up by the fact that the motor constant (K_{M}) per weight is about the same for all motors tested. Its likely that the motor manufacture decided to wind the smallest motor this way so that its base speed matched that required by an appropriately sized prop and the typical battery voltage used on model aircraft and drones.
Limitations of this approach
Using K_{V} and a motors current draw to estimate its torque output only works provided that:
• A motor does not produce any useful reluctance torque (i.e. its a non-salient machine**). This is true for essentially all 'hobby grade' electric motors.
• A motors torque increase linearly with current. This is not true if your motor is close to saturation. However, most 'hobby grade' electric motors are designed to operate with a current limit well below that needed to saturate them and so this is generally a safe assumption for stead state use.
• A motor is operated below its base speed. Operating a motor above its base speed by field weakening effectively lowers a motors K_{V} in that region and so unless you know by how much K_{V} is reduced you can not calculate the torque output. However, since no 'hobby grade' motor controllers (ESC's) that I know of utilise field weakening this is also not an issue.
• The current waveform supplied by a motor controller exactly matches a motors back EMF waveform (i.e. the current is not exactly on the q-axis at all times). This is generally true when using field oriented control (FOC) on a PMSM motor that has inbuilt position sensors (hall effect, encoder etc.). Operating a motor with 'six step 120 degree' commutation at low speed without a position sensor will result in less torque being produced than that predicted using the 'conversion constant' while high speed operation should be pretty close.
Conclusion
The torque produced by brushless permanent magnet synchronous motor can be easily estimated so long as its K_{V} and armature current is known. This relationship works because a motors K_{V} is fundamentally the same thing as its motor torque constant provided the right units are used, which is where a 'conversion constant' of ~8.3 is required.
* The armature is considered the winding in which a rotational 'back emf' would be generated if the motor were used as a generator. In some motor designs the armature is on the rotor (e.g. brushed DC motor) or in the stator (e.g. brushless DC motor).
** A non-salient machine in this context is any motor which does not derive useful torque from reluctance torque. A motor can have salient poles on the rotor or stator and still be considered a non-salient machine with this definition.
Equations were produced in this post with the help of arachnoid.com. If you have noticed any errors in the above article then please let me know.
1. Trying to figure out this BLDC thing for a hobby project and your website has been of great help! Found it through the Odrive website. Thanks a lot!
2. Thank you for this very useful formula and explanation! But I do have one question... how can a BLDC motor ever achieve >87% efficiency when the torque is calculated this way?
Conversion from RPM and Newton-meters to Watts is RPM * Nm / 9.55, so if RPM = Volts x Kv and Nm = Amps x 8.3 / Kv, then Watts = Volts x Amps x 8.3 / 9.55, therefore output power = input power * 0.87. And since motors spin slower under load, that means the real efficiency will be even lower. And is this even accounting for resistive heat loss yet? So how is it that some motors like this http://nt-power.eu/motor_m.html claim 93% peak efficiency?
3. Hi Dekutree64
Thanks for your comment.
Electric motors can indeed have an efficiency >95%.
I think I can see the problem with your assumptions. Kv x Supply voltage = maximum no load speed that a motor can achieve and not the motor speed at a given power output. Refer to this post for more details: https://things-in-motion.blogspot.com/2019/05/understanding-bldc-pmsm-electric-motors.html
Assuming a power factor of 1, the electrical power supplied to a motor is given by the voltage drop across the motor multiplied by the current supplied to the motor (P = VI). The easiest way to measure this power is before the motor controller where the current is still DC and the voltage is near constant. However, keep in mind you will also be measuring the motor controller losses.
As you mentioned, the power output is the angular velocity multiplied by the torque and from that you can find the efficiency.
The source of electric motor inefficiencies is a deep topic but one which I plan on exploring in this blog in more detail soon.
1. Oh, I think I see where I went wrong. In my example I was using the voltage and current measured before the controller, but this torque formula uses the current after the controller. So this implies that the armature current must be 1.15x the input current, to cancel out that 0.87 factor.
I'm still a little fuzzy on exactly how armature current works, so that would be another very useful article if you ever feel like writing about it. | 3,981 | 15,769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-39 | latest | en | 0.863917 |
https://math.stackexchange.com/questions/4180650/prove-a-given-function-solves-the-transport-equation-pde | 1,719,026,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00685.warc.gz | 330,535,303 | 35,935 | # Prove a given function solves the transport equation PDE
From Evans' PDE, chapter 2.1.2
Prove $$u(x,t)=g(x-tb)+\int_0^t f(x+(s-t)b,s)\,ds, u\in C^1(\mathbb R^n\times[0,\infty))$$ solves $$\begin{cases} u_t+b\cdot Du=f &\text{in }\mathbb R^n\times(0,\infty),\\ u=g &\text{on }\Gamma:=\mathbb R^n\times\{t=0\}. \end{cases}$$ where $$b\in \mathbb R^n, g\in C^1(\Gamma), f\in C^0(\mathbb R^n\times(0,\infty))$$.
My attempt: Let $$P$$ be the primitive of $$f$$, and for brevity write $$g'$$ instead of $$g'(x-tb)$$ \begin{align} u_t &= g'\cdot(-b) +\frac{\partial}{\partial t}(P(x,t)-P(x-tb,0)) \\ &= -g'\cdot b+f(x,t)-f(x-tb,0)\cdot(-b)\\ Du &= g' +\frac{\partial}{\partial x}(P(x,t)-P(x-tb,0)) \\ &= g'+f(x,t)-f(x-tb,0)\\ \end{align} \begin{align}\require{cancel} u_t+b\cdot Du &= \cancel{-g'\cdot b}+f(x,t)\cancel{-f(x-tb,0)\cdot(-b)}+\cancel{b\cdot g'}+b\cdot f(x,t)\cancel{-b\cdot f(x-tb,0)}\\ &= f(x,t)+b\cdot f(x,t). \end{align} I was expecting $$u_t+b\cdot Du = f(x,t)$$, where is the error?
Remember the Leibniz integral rule. In the scalar case $$n=1$$, we have $$u_t(x,t) = -b g'(x-tb) + f(x,t) -b \int_0^t f_x(x+(s-t)b,s)\, ds$$ $$Du(x,t) = g'(x-tb) + \int_0^t f_x(x+(s-t)b,s)\, ds$$ i.e. $$u_t + b\cdot Du = f$$, and the boundary condition is satisfied as well. Now it remains to do the same for arbitrary dimension $$n$$.
• Many thanks! I have few questions, the integrand in first line is $f_x$ or $f_t$? Your computations are the solution of the problem or we have to write them for dimension $n$? (and if this is the case do you have any hint to start?) Thank you Commented Jun 23, 2021 at 12:31
• @soundwave It's $f_x$. According to the Leibniz rule, we must compute $$\partial_t f(X, s) = \frac{\partial X}{\partial t}\, f_X(X, s)$$ where $X=x+(s-t)b$ and $(x,t)\mapsto f_x(x,t)$ is a partial derivative of $f$. You can proceed the same way using the Leibniz rule in higher dimension $n>1$. Commented Jun 24, 2021 at 8:53 | 752 | 1,941 | {"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": 15, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.650799 |
https://www.tensorflow.org/api_docs/python/tf/contrib/optimizer_v2/AdamOptimizer | 1,566,480,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317130.77/warc/CC-MAIN-20190822130553-20190822152553-00401.warc.gz | 987,225,521 | 65,999 | TensorFlow 2.0 Beta is available
## Class AdamOptimizer
Optimizer that implements the Adam algorithm.
Inherits From: OptimizerV2
See Kingma et al., 2014 (pdf).
## __init__
View source
__init__(
learning_rate=0.001,
beta1=0.9,
beta2=0.999,
epsilon=1e-08,
use_locking=False,
)
Construct a new Adam optimizer.
#### Initialization:
$$m_0 := 0 \text{(Initialize initial 1st moment vector)}$$
$$v_0 := 0 \text{(Initialize initial 2nd moment vector)}$$
$$t := 0 \text{(Initialize timestep)}$$
The update rule for variable with gradient g uses an optimization described at the end of section2 of the paper:
$$t := t + 1$$
$$lr_t := \text{learning\_rate} * \sqrt{1 - beta_2^t} / (1 - beta_1^t)$$
$$m_t := beta_1 * m_{t-1} + (1 - beta_1) * g$$
$$v_t := beta_2 * v_{t-1} + (1 - beta_2) * g * g$$
$$variable := variable - lr_t * m_t / (\sqrt{v_t} + \epsilon)$$
The default value of 1e-8 for epsilon might not be a good default in general. For example, when training an Inception network on ImageNet a current good choice is 1.0 or 0.1. Note that since AdamOptimizer uses the formulation just before Section 2.1 of the Kingma and Ba paper rather than the formulation in Algorithm 1, the "epsilon" referred to here is "epsilon hat" in the paper.
The sparse implementation of this algorithm (used when the gradient is an IndexedSlices object, typically because of tf.gather or an embedding lookup in the forward pass) does apply momentum to variable slices even if they were not used in the forward pass (meaning they have a gradient equal to zero). Momentum decay (beta1) is also applied to the entire momentum accumulator. This means that the sparse behavior is equivalent to the dense behavior (in contrast to some momentum implementations which ignore momentum unless a variable slice was actually used).
Some of the args below are hyperparameters where a hyperparameter is defined as a scalar Tensor, a regular Python value or a callable (which will be evaluated when apply_gradients is called) returning a scalar Tensor or a Python value.
#### Args:
• learning_rate: A float hyperparameter. The learning rate.
• beta1: A float hyperparameter. The exponential decay rate for the 1st moment estimates.
• beta2: A float hyperparameter. The exponential decay rate for the 2nd moment estimates.
• epsilon: A float hyperparameter. This epsilon is "epsilon hat" in the Kingma and Ba paper (in the formula just before Section 2.1), not the epsilon in Algorithm 1 of the paper.
• use_locking: If True use locks for update operations.
• name: Optional name for the operations created when applying gradients. Defaults to "Adam".
## Methods
### apply_gradients
View source
apply_gradients(
global_step=None,
name=None
)
Apply gradients to variables.
This is the second part of minimize(). It returns an Operation that applies gradients.
#### Args:
• grads_and_vars: List of (gradient, variable) pairs as returned by compute_gradients().
• global_step: Optional Variable to increment by one after the variables have been updated.
• name: Optional name for the returned operation. Default to the name passed to the Optimizer constructor.
#### Returns:
An Operation that applies the specified gradients. If global_step was not None, that operation also increments global_step.
#### Raises:
• TypeError: If grads_and_vars is malformed.
• ValueError: If none of the variables have gradients.
### compute_gradients
View source
compute_gradients(
loss,
var_list=None,
aggregation_method=None,
scale_loss_by_num_replicas=False
)
Compute gradients of loss for the variables in var_list.
This is the first part of minimize(). It returns a list of (gradient, variable) pairs where "gradient" is the gradient for "variable". Note that "gradient" can be a Tensor, an IndexedSlices, or None if there is no gradient for the given variable.
#### Args:
• loss: A Tensor containing the value to minimize or a callable taking no arguments which returns the value to minimize. When eager execution is enabled it must be a callable.
• var_list: Optional list or tuple of tf.Variable to update to minimize loss. Defaults to the list of variables collected in the graph under the key GraphKeys.TRAINABLE_VARIABLES.
• gate_gradients: How to gate the computation of gradients. Can be GATE_NONE, GATE_OP, or GATE_GRAPH.
• aggregation_method: Specifies the method used to combine gradient terms. Valid values are defined in the class AggregationMethod.
• grad_loss: Optional. A Tensor holding the gradient computed for loss.
• stop_gradients: Optional. A Tensor or list of tensors not to differentiate through.
• scale_loss_by_num_replicas: Optional boolean. If true, scale the loss down by the number of replicas. DEPRECATED and generally no longer needed.
#### Returns:
A list of (gradient, variable) pairs. Variable is always present, but gradient can be None.
#### Raises:
• TypeError: If var_list contains anything else than Variable objects.
• ValueError: If some arguments are invalid.
• RuntimeError: If called with eager execution enabled and loss is not callable.
#### Eager Compatibility
When eager execution is enabled, gate_gradients, and aggregation_method are ignored.
### get_name
View source
get_name()
### get_slot
View source
get_slot(
var,
name
)
Return a slot named name created for var by the Optimizer.
Some Optimizer subclasses use additional variables. For example Momentum and Adagrad use variables to accumulate updates. This method gives access to these Variable objects if for some reason you need them.
Use get_slot_names() to get the list of slot names created by the Optimizer.
#### Args:
• var: A variable passed to minimize() or apply_gradients().
• name: A string.
#### Returns:
The Variable for the slot if it was created, None otherwise.
### get_slot_names
View source
get_slot_names()
Return a list of the names of slots created by the Optimizer.
See get_slot().
#### Returns:
A list of strings.
### minimize
View source
minimize(
loss,
global_step=None,
var_list=None,
aggregation_method=None,
name=None,
scale_loss_by_num_replicas=False
)
Add operations to minimize loss by updating var_list.
This method simply combines calls compute_gradients() and apply_gradients(). If you want to process the gradient before applying them call compute_gradients() and apply_gradients() explicitly instead of using this function.
#### Args:
• loss: A Tensor containing the value to minimize.
• global_step: Optional Variable to increment by one after the variables have been updated.
• var_list: Optional list or tuple of Variable objects to update to minimize loss. Defaults to the list of variables collected in the graph under the key GraphKeys.TRAINABLE_VARIABLES.
• gate_gradients: How to gate the computation of gradients. Can be GATE_NONE, GATE_OP, or GATE_GRAPH.
• aggregation_method: Specifies the method used to combine gradient terms. Valid values are defined in the class AggregationMethod.
• name: Optional name for the returned operation.
• grad_loss: Optional. A Tensor holding the gradient computed for loss.
• stop_gradients: Optional. A Tensor or list of tensors not to differentiate through.
• scale_loss_by_num_replicas: Optional boolean. If true, scale the loss down by the number of replicas. DEPRECATED and generally no longer needed.
#### Returns:
An Operation that updates the variables in var_list. If global_step was not None, that operation also increments global_step.
#### Raises:
• ValueError: If some of the variables are not Variable objects.
#### Eager Compatibility
When eager execution is enabled, loss should be a Python function that takes elements of var_list as arguments and computes the value to be minimized. If var_list is None, loss should take no arguments. Minimization (and gradient computation) is done with respect to the elements of var_list if not None, else with respect to any trainable variables created during the execution of the loss function. gate_gradients, aggregation_method, and grad_loss are ignored when eager execution is enabled.
### variables
View source
variables()
A list of variables which encode the current state of Optimizer.
Includes slot variables and additional global variables created by the optimizer in the current default graph.
#### Returns:
A list of variables.
## Class Members
• GATE_GRAPH = 2
• GATE_NONE = 0
• GATE_OP = 1 | 1,856 | 8,379 | {"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} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.756775 |
https://es.mathworks.com/matlabcentral/cody/problems/45474-sub-sequence-02/solutions/2231903 | 1,606,700,468,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00616.warc.gz | 285,596,147 | 19,116 | Cody
# Problem 45474. Sub-sequence - 02
Solution 2231903
Submitted on 24 Apr 2020 by bainhome
• Size: 61
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a=[1,1,1,1,1,2,3,1,4] b=[2,3,0,0,9,5,4,1] assert(isequal(longest_sub_common(a,b),3))
a = 1 1 1 1 1 2 3 1 4 b = 2 3 0 0 9 5 4 1
2 Pass
a=[1,1,1,1,1,2,3] b=[2,3,0,0,9,5,4,1] assert(isequal(longest_sub_common(a,b),2))
a = 1 1 1 1 1 2 3 b = 2 3 0 0 9 5 4 1
3 Pass
a=[1,1,1,1,1,2,3,1,4] b=zeros(1,500); assert(isequal(longest_sub_common(a,b),0))
a = 1 1 1 1 1 2 3 1 4
4 Pass
a=[1,1,1,1,1,2,3,1,4] b=[zeros(1,50),ones(1,200),ones(1,20)*3] assert(isequal(longest_sub_common(a,b),6))
a = 1 1 1 1 1 2 3 1 4 b = Columns 1 through 29 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 30 through 58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 Columns 59 through 87 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 88 through 116 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 117 through 145 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 146 through 174 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 175 through 203 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 204 through 232 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Columns 233 through 261 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 Columns 262 through 270 3 3 3 3 3 3 3 3 3
5 Pass
a='aaabbbcccxyz' b='abcyycbaabc' assert(isequal(longest_sub_common(a,b),5))
a = 'aaabbbcccxyz' b = 'abcyycbaabc'
6 Pass
a=[10 9 8 2 4 2 1 4 7 7 6 5 3 6 8 8 6 8 7 2 6 4 1 2 2 7 5 7 3 1 6 3 10 10 4 1 7 9 10 1 5 6 7 8 7 8 4 6 2 1 10 3 6 10 2 2 4 10 4 3 2 4 4 2 5 1 7 1 6 8 3 5 6 1 5 7 3 9 10 9 6 3 8 3 10 7 7 2 1 3 9 10 7 8 3 6 9 5 10 1 4 6 1 8 6 6 9 9 8 4 5 8 2 2 3 6 10 8 4 3 9 10 7 3 1 9 6 10 1 6 3 9 2 5 4 9 7 3 4 2 7 6 2 2 5 10 6 1 1 9 5 4 8 4 6 8 9 4 7 10 1 6 5 4 3 8 10 2 8 2 10 9 5 8 5 9 4 1 6 10 2 5 8 1 10 8 6 2 5 6 10 9 10 7 5 10 5 3 4 8 6 8 10 10 6 10 2 1 4 6 6 10 6 5 6 8 1 9 2 5 3 4 7 2 3 2 2 4 9 5 5 2 10 5 9 7 4 9 8 5 9 9 5 4 6 10 8 4 8 10 6 6 4 1 2 1 5 1 10 7 1 1 3 5 2 1 8 4 8 5 5 1 1 1 6 3 9 9 10 5 3 3 6 8 4 5 7 10 2 8 6 5 9 4 2 7 7 4 9 10 10 2 3 1 7 2 5 9 6 4 3 5 10 2 5 9 1 7 10 3 2 7 10 7 9 2 8 4 5 2 9 7 8 9 1 10 5 8 8 9 2 5 7 10 9 9 6 6 9 1 9 5 1 8 2 2 7 3 4 5 5 4 7 2 2 1 4 8 3 8 7 9 9 3 4 6 4 9 9 6 3 7 3 5 4 6 10 8 10 3 6 1 8 7 9 10 10 5 1 6 3 3 4 1 8 8 6 4 9 6 10 9 4 6 4 7 8 8 2 9 1 5 8 8 4 8 9 3 2 3 4 3 10 1 6 2 9 2 6 10 4 1 3 4 4 3 10 7 10 5 10 1 7 9 3 10 8 9 6 8 4 3 4 6 9 3 5 9 7 10 3 9 7 3 5 4 6 9 2 9 9 4 5 6 8 8 8 4 5 10 6 9 3 7 6 10 1 6 6 1] b=[14 14 7 12 3 10 4 7 13 3 5 8 6 12 15 3 4 11 6 15 15 10 13 7 10 15 9 15 11 8 10 14 3 6 15 7 10 14 15 10 2 1 10 9 15 12 10 8 4 15 9 1 11 8 1 14 5 4 2 5 4 10 1 5 5 14 7 12 10 12 2 15 13 1 7 5 10 4 9 10 9 7 1 8 7 2 7 7 9 13 11 14 1 4 7 15 12 7 6 1 12 8 3 7 3 12 6 15 1 13 10 9 10 11 2 14 1 5 3 14 2 9 10 10 13 1 13 8 11 4 9 11 15 7 2 1 10 12 11 6 15 8 15 2 4 12 14 12 5 3 13 12 5 4 5 13 13 9 9 5 11 12 7 7 7 5 11 14 14 12 4 11 2 2 3 3 9 2 13 11 14 8 10 14 9 5 15 1 5 15 6 5 2 14 3 5 14 8 10 9 11 1 8 1 13 6 13 4 9 15 1 1 1 11 9 2 12 10 2 2 3 12 2 4 4 2 13 11 12 10 8 5 10 2 3 1 15 15 2 8 10 5 12 9 7 5 12 14 11 7] assert(isequal(longest_sub_common(a,b),121))
a = Columns 1 through 29 10 9 8 2 4 2 1 4 7 7 6 5 3 6 8 8 6 8 7 2 6 4 1 2 2 7 5 7 3 Columns 30 through 58 1 6 3 10 10 4 1 7 9 10 1 5 6 7 8 7 8 4 6 2 1 10 3 6 10 2 2 4 10 Columns 59 through 87 4 3 2 4 4 2 5 1 7 1 6 8 3 5 6 1 5 7 3 9 10 9 6 3 8 3 10 7 7 Columns 88 through 116 2 1 3 9 10 7 8 3 6 9 5 10 1 4 6 1 8 6 6 9 9 8 4 5 8 2 2 3 6 Columns 117 through 145 10 8 4 3 9 10 7 3 1 9 6 10 1 6 3 9 2 5 4 9 7 3 4 2 7 6 2 2 5 Columns 146 through 174 10 6 1 1 9 5 4 8 4 6 8 9 4 7 10 1 6 5 4 3 8 10 2 8 2 10 9 5 8 Columns 175 through 203 5 9 4 1 6 10 2 5 8 1 10 8 6 2 5 6 10 9 10 7 5 10 5 3 4 8 6 8 10 Columns 204 through 232 10 6 10 2 1 4 6 6 10 6 5 6 8 1 9 2 5 3 4 7 2 3 2 2 4 9 5 5 2 Columns 233 through 261 10 5 9 7 4 9 8 5 9 9 5 4 6 10 8 4 8 10 6 6 4 1 2 1 5 1 10 7 1 Columns 262 through 290 1 3 5 2 1 8 4 8 5 5 1 1 1 6 3 9 9 10 5 3 3 6 8 4 5 7 10 2 8 Columns 291 through 319 6 5 9 4 2 7 7 4 9 10 10 2 3 1 7 2 5 9 6 4 3 5 10 2 5 9 1 7 10 Columns 320 through 348 3 2 7 10 7 9 2 8 4 5 2 9 7 8 9 1 10 5 8 8 9 2 5 7 10 9 9 6 6 Columns 349 through 377 9 1 9 5 1 8 2 2 7 3 4 5 5 4 7 2 2 1 4 8 3 8 7 9 9 3 4 6 4 Columns 378 through 406 9 9 6 3 7 3 5 4 6 10 8 10 3 6 1 8 7 9 10 10 5 1 6 3 3 4 1 8 8 Columns 407 through 435 6 4 9 6 10 9 4 6 4 7 8 8 2 9 1 5 8 8 4 8 9 3 2 3 4 3 10 1 6 Columns 436 through 464 2 9 2 6 10 4 1 3 4 4 3 10 7 10 5 10 1 7 9 3 10 8 9 6 8 4 3 4 6 Columns 465 through 493 9 3 5 9 7 10 3 9 7 3 5 4 6 9 2 9 9 4 5 6 8 8 8 4 5 10 6 9 3 Columns 494 through 500 7 6 10 1 6 6 1 b = Columns 1 through 29 14 14 7 12 3 10 4 7 13 3 5 8 6 12 15 3 4 11 6 15 15 10 13 7 10 15 9 15 11 Columns 30 through 58 8 10 14 3 6 15 7 10 14 15 10 2 1 10 9 15 12 10 8 4 15 9 1 11 8 1 14 5 4 Columns 59 through 87 2 5 4 10 1 5 5 14 7 12 10 12 2 15 13 1 7 5 10 4 9 10 9 7 1 8 7 2 7 Columns 88 through 116 7 9 13 11 14 1 4 7 15 12 7 6 1 12 8 3 7 3 12 6 15 1 13 10 9 10 11 2 14 Columns 117 through 145 1 5 3 14 2 9 10 10 13 1 13 8 11 4 9 11 15 7 2 1 10 12 11 6 15 8 15 2 4 Columns 146 through 174 12 14 12 5 3 13 12 5 4 5 13 13 9 9 5 11 12 7 7 7 5 11 14 14 12 4 11 2 2 Columns 175 through 203 3 3 9 2 13 11 14 8 10 14 9 5 15 1 5 15 6 5 2 14 3 5 14 8 10 9 11 1 8 Columns 204 through 232 1 13 6 13 4 9 15 1 1 1 11 9 2 12 10 2 2 3 12 2 4 4 2 13 11 12 10 8 5 Columns 233 through 250 10 2 3 1 15 15 2 8 10 5 12 9 7 5 12 14 11 7
7 Pass
a='aaabaaabaaabaaa' b='abababababa' assert(isequal(longest_sub_common(a,b),9))
a = 'aaabaaabaaabaaa' b = 'abababababa'
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 4,327 | 6,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-50 | latest | en | 0.451208 |
https://www.datameer.com/blog/data-wrangling-for-data-analysis-how-clean-data-can-improve-your-analysis-results/ | 1,679,353,997,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00446.warc.gz | 824,627,509 | 19,271 | # Data Wrangling for Data Analysis: How Clean Data Can Improve Your Analysis Results
• Ndz Anthony
• January 26, 2023
This article will discuss data wrangling, why it’s essential, and the prevalent differences between data wrangling and analysis.
I remember trying to solve Rubix puzzles in my childhood days. I had no idea how best to reintegrate this cryptic enigma and would sit attempting to for hours on end.
However, with some luck, I always figured it out, and when I did, it called for a ceremony of my gloating and a pat on the head.
What does my story have anything to do with understanding the data-wrangling process ?…
Stick around to find out ☺.
## What’s Data Wrangling?
Data wrangling goes by many names- data cleansing, data remediation, or data munging – all refer to a set of methods to convert raw data into more usable representations. The specific strategies vary based on the data you’re using and the aim you’re trying to achieve.
Data munging/wrangling requires a sophisticated understanding of the raw data, the analysis performed, and what information needs to be deleted.
### Why is it essential to wrangle data?
The amount of labor that goes into data wrangling is represented by the 80/20 rule in data science and data analysis – stats show that data scientists often spend 80% of their time ‘wrangling’ or preparing data and 20% of their time analyzing the data.
If you don’t run your analysis on good, clean data, your results will be skewed – often to the point where they’ll be more harmful than helpful. Essentially, the better the data you provide for upstream data activities, the more effective the output of your result sets.
Some may question whether the amount of work and time spent on data wrangling is worthwhile.
At this juncture, I’ll take you back to my Rubik’s story, if you don’t mind 😶
A Rubik’s cube only has uniformity, structure, and meaning when it comes together. It took me forever – yeah, no doubt. But the older I got, the more experience I had with this mystery cube, and as a result, the quicker I got.
Similarly, once the code, infrastructure, foundation, and best practices for handling data are in place, data wrangling becomes easier and upstream activities such as data visualization, data science, etc. – become achievable within shorter iterations.
Here are some of the essential characteristics of Data-wrangling:
1. #### Useable Data:
Data wrangling improves data usability by formatting data for the end user.
2. #### Aggregation:
It aids in integrating many forms of information and their sources, such as database catalogs, online services, files, etc.
3. #### Data preparation
Data prep is critical to attaining good outcomes from ML and deep learning initiatives, which is why data munging is vital.
4. #### Quick decision:
Wrangling the data allows the wrangler to make quick judgments while cleaning, enriching, and transforming the data into an excellent picture.
5. #### Automation:
Data wrangling techniques such as automated data integration tools clean and transform source data into a standard format that can be utilized repeatedly based on end requirements. This standardized data is used by businesses to undertake critical cross-data set analytics.
6. #### Saves time:
As stated earlier in this post, data analysts spend most of their time sourcing data from various sources and updating data sets rather than performing fundamental analysis. Data wrangling provides accurate data to analysts promptly.
### Difference between Data Wrangling and Data Cleaning
Despite their comparable approaches, data wrangling and data cleaning are independent operations. Upfront data cleansing ensures that downstream processes and analytics receive correct and consistent data, increasing consumer confidence in the information.
Here are some differences between data wrangling and data cleaning:
## Dedicated and Best Data Munging Tools For 2023
Knowing their data from front to back is one of the most valuable qualities an intelligent data wrangler can have. You’ll need to understand what is in your data sets, downstream data analysis tools, and end goals.
Historically, most data manipulation was done manually using spreadsheets such as Excel and Google Sheets. It is a viable choice when you have modest data sets that only require a little cleaning and enrichment.
More modern technology is generally preferable when dealing with enormous volumes of data; below, we summarized the Top 6 Best Data Wrangling tools for coders and non-coders:
• #### Numpy:
A fundamental package for scientific computing with Python, providing support for large, multi-dimensional arrays and matrices, along with an extensive library of mathematical functions to operate on these arrays.
• #### Datameer:
It is a data integration, preparation, and analysis platform that allows users to easily connect to and blend data from various sources and perform data transformations and analysis using a visual, spreadsheet-like interface. It also provides advanced features such as machine learning and predictive modeling capabilities, making it a powerful tool for data wrangling and preparation.
• #### OpenRefine:
A powerful tool for working with messy data, providing a wide range of data cleaning and transformation functionality, including support for regular expressions, clustering, and data reconciliation.
• #### Trifacta:
Allows users to easily clean, shape, and enrich data for analysis and visualization. It provides a user-friendly interface with a wide range of transformation options, including support for regular expressions, data parsing and pivoting, and machine learning-based suggestions for data cleaning.
• Dplyr: A powerful data manipulation library for R, providing a grammar of data manipulation for data in data frames, including filtering, aggregation, and transformation.
• Excel: A widely used spreadsheet software that provides essential data manipulation and analysis capabilities through its built-in functions and pivot table features.
### Transform your Data with ease using Datameer’s powerful Data-Wrangling capabilities
Datameer is a data integration, preparation, and analysis platform that helps streamline the data-munging process by providing various functionalities.
It allows users to easily connect to and blend data from various sources using a visual, spreadsheet-like interface. Users can perform data transformations, cleaning, and shaping using multiple options such as data parsing and pivoting, support for regular expressions, and machine learning-based suggestions for data cleaning.
It also includes advanced features like machine learning, predictive modeling capabilities, and other analytics functionalities, making it a one-stop solution for data wrangling and preparation.
Given below are the main three reasons why you should choose Datameer over any other tools available in the market:
• #### Easy-to-use interface:
Datameer provides a user-friendly interface that allows users to easily connect to and blend data from various sources using a visual, spreadsheet-like interface.
• #### Advanced data wrangling capabilities:
Offers a wide range of data munging functionalities, such as data parsing, pivoting, support for regular expressions, and machine learning-based suggestions for data cleaning.
• #### Comprehensive analytics capabilities:
Datameer is not just a data wrangling tool but also includes advanced analytics capabilities such as machine learning and predictive modeling, making it a one-stop solution for data wrangling and advanced analytics.
So what are you waiting for?
Stay away from struggling with that cube. Give yourself a backing of advanced technology tools – get started here!
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# Rational numbers have repeating decimal expansions
A couple of days ago a good friend of mine asked me for help on a more algebraic problem (I have studied more mathematical analysis), which I found cute, so I decided to write up proper proofs for it. The statement of the theorem is as follows:
A real number is rational if and only if its decimal expansion is repeating or terminating.
## Background
Here I introduce the basic objects that the proof uses, in an attempt to make it more accessible and for more people to appreciate it.
### Natural numbers and integers
When one begins their study of any type of mathematics, they often begin by counting. Counting uses the counting numbers; $$1$$, $$2$$, $$3$$, $$\dots$$. These are also known as natural numbers, often denoted by $$\mathbb{N}$$ in mathematics. We then tend to add zero, the count of nothing, and then start counting backwards, to get the set of whole numbers, or integers, denoted by $$\mathbb{Z}$$; the first couple are $$0$$, $$1$$, $$-1$$, $$2$$, $$-2$$, $$\dots$$.
### Rational numbers
We then continue to construct more numbers. What if we divide an object between a number of people? The next natural type of number is called a rational number, denoted by $$\mathbb{Q}$$. Although it is possible to enumerate them all (in the sense of listing them one-by-one), we have to choose an enumeration, and it is not completely obvious. Instead, we will give a general description of all rational numbers as the ratio of a whole number and a natural number. In other words, a number is rational if we can express it as $$p/q$$ where $$p$$ is a whole number, and $$q$$ is a natural number, in mathematical notation, $$p\in\mathbb{Z}$$ and $$q\in\mathbb{N}$$ (where the $$\in$$ means "in"). In other words, rational numbers are fractions. Examples of these are $$5/1$$, $$33/20$$, $$4/2$$, $$42/7$$, $$-99/17$$, and so forth.
So all natural numbers are whole numbers, and all whole numbers are rational numbers. But what comes after rational numbers?
### Real and irrational numbers
The next set of numbers is called the set of real numbers, these are numbers that lie anywhere on the number line, or can be written as any type of decimal (infinite, or finite in length). Examples of these are $$\pi$$; the ratio of the circumference of a circle to its diameter, $$e$$; the base of the natural logarithm, and $$\sqrt{2}$$. Real numbers are often denoted by $$\mathbb{R}$$, and the set of numbers that are real but not rational are called the set of irrational numbers, and are denoted by $$\mathbb{R}\setminus\mathbb{Q}$$ (where the $$\setminus$$ stands for "except for").
## The theorem
The content of the theorem is that any rational number, and only a rational number, has a repeating or terminating decimal expansion. A decimal expansion of the number, is if we write it in the decimal system, for instance $$2.365$$, these can also go forever, such as $$1.41421356\dots$$. This means that when we write any rational number such as $$1/3$$ as a decimal, we get $$1.3333\dots$$, where the decimal repeats with infinitely many threes, as opposed to a seemingly random sequence of numbers (as is the case with $$\pi$$, which is $$3.14159265\dots$$). The statement also says that any irrational number must have a non-repeating and non-terminating decimal expansion. From now on, I will refer to repeating or terminating decimal expansions as simply repeating decimals, as a terminating decimal can be thought of one that repeats with infinitely many zeroes.
## Proof that repeating decimals represent rational numbers
We first prove the backwards case, that if a decimal is repeating, then it represents a rational number.
### Intuition
Suppose we have a decimal such as $$1/3=0.33333\dots$$. There is a commonly known neat trick to convert this to a fraction. First we multiply it by $$10$$, or $$100$$, or with the power of ten corresponding to the number of repeating digits. We then subtract one copy of the original number, which gets rid of the repeating part, and simplify to get a fraction. For instance, if $$x=1.31313131\dots$$, we may proceed as:
\begin{aligned} x &= 1.31313131\dots \\ 100x - x &= 130 \\ 99x &= 130 \\ x &= 130/99. \end{aligned}
In fact, this works for any decimal which repeats on the right of the decimal point. But what if there are some non-repeating decimals on the right before it starts repeating, such as if $$z=4.21666666\dots$$, then we can simply multiply by a power of ten, and then perform the same trick, as in the example:
\begin{aligned} z &= 4.21666666\dots \\ 100x &= 421.66666666\dots \\ 10(100x) - (100x) &= 3795 \\ 1000x - 100x &= 3795 \\ 900x &= 3795 \\ x &= 3795/900. \end{aligned}
This trick now works for any repeating decimal. If the decimal is terminating, with $$n$$ digits, we may simply multiply by $$10^n$$ to get a whole number, and our number is this whole number divided by $$10^n$$. For instance, if $$w=5.44$$, then multiply by $$100$$ to get $$544$$, and we have $$w=544/100$$.
### Formal proof
Given a repeating decimal $$x=a.b\overline{c}$$ where $$a$$, $$b$$, and $$c$$ are groups of digits, we can separate it into the non-repeating, and repeating parts. First let $$n=\lceil{\log_{10}b\rceil}$$, the number of digits of $$b$$, and multiply by $$10^n$$. We are now left with something of the form $$10^nx=ab.\overline{c}$$. If the decimals terminate, then $$c=0$$, and the proof is complete. So now without loss of generality, assume $$c\neq0$$ and $$x=a.\overline{c}$$.
Now suppose $$c$$ has $$m$$ digits, then multiply by $$10^m-1$$ and we are left with an integer, that straight away gives a fraction. To see this, we may write it as a geometric series:
\begin{aligned} x &= a + c + 10^{-m}c + 10^{-2m}c + \cdots \end{aligned}
Now multiply by $$10^m-1$$ to get:
\begin{aligned} (10^m-1)x &= (10^m-1)a + (10^mc + c + 10^{-m}c + \cdots) \\ &\quad- (c + 10^{-m}c + 10^{-2m}c + \cdots) \\ &= (10^m-1)a + 10^mc. \end{aligned}
Since $$c$$ has $$m$$ digits, $$10^mc$$ is an integer. Therefore, $$x=((10^m-1)a+10^mc)/(10^m-1)$$, where $$(10^m-1)a+10^mc$$ is an integer and $$10^m-1$$ is a natural number, as required.
## Proof that rational numbers have repeating decimal expansions
This is the hard part. I originally came up with a proof working backwards with the above trick, but it is somewhat complicated, so I later produced a proof using another, simpler method.
### First proof
The first way of proving this uses the above trick backwards, in other words, given a rational number, we try to change its denominator to be of the form $$10^n$$ or $$10^n(10^m-1)$$.
To do this, suppose $$x=p/q$$ where $$p\in\mathbb{Z}$$ and $$q\in\mathbb{N}$$. Let the prime decomposition of $$q$$ be $$2^{l_1}3^{l_2}5^{l_3}p_4^{l_4}p_5^{l_5}\dots$$ where $$p_i$$ is the $$i$$-th prime and $$l_i$$ is a non-negative integer, the power of the $$i$$-th prime. Now let $$n=l_1\cdot l_3$$, these are the primes not co-prime to $$10$$. Furthermore, since $$3^{l_2}p_4^{l_4}p_5^{l_5}\cdots$$ is co-prime to $$10$$, Euler's theorem guarantees that $$10^{\varphi(3^{l_2}p_4^{l_4}p_5^{l_5}\cdots)-1}-1=0\pmod{3^{l_2}p_4^{l_4}p_5^{l_5}\cdots}$$, (where $$\varphi$$ is Euler's totient function) so we can let $$m=\varphi(3^{l_2}p_4^{l_4}p_5^{l_5}\cdots)-1$$. This is of the form discussed above, and hence proves the result.
### Second proof
This method manually computes the decimal places, and reasons why these must repeat. Suppose $$x=p/q$$ where $$p\in\mathbb{Z}$$ and $$q\in\mathbb{N}$$. To compute the parts on the left hand side of the decimal point, we simply compute $$p\ \text{div}\ q=0$$ (where $$\text{div}$$) denotes quotient). We can then compute the next digit (the one immediately on the right of the decimal point) by multiplying the remainder by $$10$$, and taking the quotient under $$q$$. To get the next digit, we again multiply the remainder of this by $$10$$ and take the quotient under $$q$$. Note now that we only ever use the remainder in the next step, but it is always between $$0$$ and $$q-1$$, with $$q$$ is fixed. Therefore, by the pigeonhole principle, we must eventually get back to a remainder we have already visited, and start looping, giving the required repeating decimal. This completes the proof.
The procedure described here is essentially long division. | 2,359 | 8,360 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-22 | latest | en | 0.936673 |
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