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# Facet (mathematics)
Facet (mathematics)
A facet of a simplicial complex is a maximal simplex.
In the general theory of polyhedra and polytopes, two conflicting meanings are currently jostling for acceptability:
*A facet of a geometric polyhedron is traditionally any polygon whose corners are vertices of the polyhedron. By extension to higher dimensions, it is any "j"-tope ("j"-dimensional polytope) whose vertices are shared by some "n"-tope ("n"-dimensional polytope where 0<"j"<"n"). To facet a polytope is to find and join such facets to form a new polytope - this process is called facetting or faceting and is the reciprocal process to stellation.
*A facet of an "n-polytope" is, more recently, an ("n"-1)-dimensional face or ("n"-1)-face. The informal term side can mean the same thing, edges of a polygon and faces of a polyhedron.
*:For example:
*:#The facets of a polygon are edges. (1-faces)
*:#The facets of a polyhedron or tiling are faces. (2-faces)
*:#The facets of a polychoron (4-polytope) or honeycomb are cells. (3-faces)
*:#The facets of a polyteron (5-polytope) or 4-honeycomb are hypercells. (4-faces)
*:Exactly two facets meet at any ridge in a polytope. By extension, facet or "j"-facet is sometimes used to mean any "j"-dimensional element of a polytope.
*
*
*
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Facet (disambiguation) — Facet may have the following meanings. * Facet, a flat surface of a geometric shape, e.g., of a gemstone. * Facet in mathematics (topology, geometry) is a special type of face. * In anatomy::* A facet is part of a compound eye.:* A facet joint is … Wikipedia
• List of mathematics articles (F) — NOTOC F F₄ F algebra F coalgebra F distribution F divergence Fσ set F space F test F theory F. and M. Riesz theorem F1 Score Faà di Bruno s formula Face (geometry) Face configuration Face diagonal Facet (mathematics) Facetting… … Wikipedia
• Science and mathematics from the Renaissance to Descartes — George Molland Early in the nineteenth century John Playfair wrote for the Encyclopaedia Britannica a long article entitled ‘Dissertation; exhibiting a General View of the Progress of Mathematics and Physical Science, since the Revival of Letters … History of philosophy
• Interval (mathematics) — This article is about intervals of real numbers. For intervals in general mathematics, see Partially ordered set. For other uses, see Interval. In mathematics, a (real) interval is a set of real numbers with the property that any number that lies … Wikipedia
• Перестановочный многогранник — … Википедия
• N-dimensional space — In mathematics, an n dimensional space is a topological space whose dimension is n (where n is a fixed natural number). The archetypical example is n dimensional Euclidean space, which describes Euclidean geometry in n dimensions.Many familiar… … Wikipedia
• Colon classification — (CC) is a system of library classification developed by S. R. Ranganathan. It was the first ever faceted (or analytico synthetic) classification. The first edition was published in 1933. Since then six more editions have been published. It is… … Wikipedia
• Convex polytope — A 3 dimensional convex polytope A convex polytope is a special case of a polytope, having the additional property that it is also a convex set of points in the n dimensional space Rn.[1] Some authors use the terms convex polytope and convex… … Wikipedia
• Necessary and sufficient condition — This article is about the formal terminology in logic. For causal meanings of the terms, see Causality. In logic, the words necessity and sufficiency refer to the implicational relationships between statements. The assertion that one statement is … Wikipedia
• Robert Haralick — Robert M. Haralick was born in Brooklyn, New York, on September 30, 1943. He received a B.A. degree in mathematics from the University of Kansas in 1964, a B.S. degree in electrical engineering in 1966, and a M.S. degree in electrical engineering … Wikipedia
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# Algebra..
posted by 987:)
HELP.
Problems 4 - 7: Write an equation for the line in point/slope form and slope/intercept form that has the given condition.
4. Slope = 3/2 and passes through the origin.
5. x-intercept = 4 and y-intercept = -3
1. Steve
#4:
(y-0)/(x-0) = 3/2
y = 3/2 x
#5: 3x-4y=12
(y-0)/(x-4) = (-3-0)/(0-4)
or, y/(x-4) = 3/4
y = 3/4 x - 3
2. Anonymous
the square of the larger of two consecutive integers is 34 less than twice the square of the smaller consecutive integer.
## Similar Questions
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The following problems are in slope,point slope, and slope intercept forms of a line. Could you check these thanks. Find the slope intercept form of a line that passes through these two points (-4,4) and (2,1) Answer: y= -1/2x-2 Find …
2. ### math
I have a few question so I will number them off. Please answer them for me I have a test tommorow and I am sure this can help other people too. 1.How do you make an equation that passes through a y-intercept and a certain point in …
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i have more than one question so if u no any of the answers please tell me 1.) write the point-slope form of the equation of the line with slope -2 passing through the point ( -5, -9). 2.) write the point-slope form of an equation …
4. ### Algebra2
Write an equation for the line in point/slope form and slope/intercept form that has the given condition. Help?
5. ### Algebra2
Help? Write an equation for the line in point/slope form and slope/intercept form that has the given condition. 4. Slope = 3/2 and passes through the origin. 5. x-intercept = 4 and y-intercept = -3
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I have an assignment that asks me to write an equation in slope-intercept, point-slope, or standard form for the information given and to explain why the chosen form would be best. Below is the information given. 1. passing through …
7. ### math - pls check!!!
Write an equation in point-slope form for the line that passes through one of the following pairs of points (you may choose the pair you want to work with). Then, use the same set of points to write the equation in standard form and …
8. ### geometry
I would love for someone to check my work on the seven problems that I did. I would be eternally grateful. I am going to list my problems and answers. Even if you only want to check one of them, that would be great. 1. Write an equation …
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Can someone help me with this equation please?
10. ### Math
1. Find the slope of the line that passes through the points (-1, 2), (0, 5). 2. Suppose y varies directly with x, and y = 15 and x = 5. Write a direct variation equation that relates x and y. What is the value of y when x = 9?
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# Math- college
posted by on .
Lisa's car get 22 miles per gallon of gasoline. How many miles can she drive on 24 gallon of gas?
• Math- college - ,
to get this, we just use ratio and proportion:
let x = miles travelled with 24 gallons
22 miles : 1 gallon = x miles : 24 gallons
therefore,
22/1 = x/24
x = 22*24
x = 528 miles
hope this helps~ :)
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Popular recipes tagged "is_prime"http://code.activestate.com/recipes/tags/is_prime/2010-08-06T11:20:34-07:00ActiveState Code RecipesSome prime generation algorithms. (Python) 2010-08-06T11:20:34-07:00Thomas Lehmannhttp://code.activestate.com/recipes/users/4174477/http://code.activestate.com/recipes/577329-some-prime-generation-algorithms/ <p style="color: grey"> Python recipe 577329 by <a href="/recipes/users/4174477/">Thomas Lehmann</a> (<a href="/recipes/tags/generation/">generation</a>, <a href="/recipes/tags/is_prime/">is_prime</a>, <a href="/recipes/tags/prime/">prime</a>, <a href="/recipes/tags/primes/">primes</a>). Revision 4. </p> <p>Basic idea was to see the difference between different prime algorithms in time. Also they are not perfect the output shows that really higher numbers let grow the difference why I have separated this into functions to make it visible. I add this here because I have been missing this very often when I have been searching for algorithms.</p> <ul> <li>The 'is_prime' is a well known way of checkin for a number being prime or not.</li> <li>The sieve of Erastothenes is simply to strike out multiples of a given value; the primes will remain.</li> <li>the function 'profile' is a decorator for functions measuring the execution time</li> <li>Some information are in the comments of the code</li> </ul>
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You are Here: Home >< Physics
# SUVAT Watch
1. What is the best way of working out which formula to use?
Is there a set way to do it? There sometimes seem like a few ways to work them out but i just don't get it.
2. In your head, try and use one equation to solve the problem. If you have too many unknowns, then try another equation. Eventually you'll find the one that works, hence you can solve the problem.
3. There are five letters: S, U, V, A, T.
Four of them will be relevant to the problem.
Choose the equation which relates those four.
4. Write down everything you know that is given in the question in this form:
u: 16 m/s
a: 2.5m/s^-2
t: 10 s
and then write down what you are looking for i.e. speed
v: ?
Therefore, you need to use v=u+at
5. i always write down the values that i'm already given; then i write down a question mark next the value i'm trying to find. Then i just try and think of the equation that uses the values i have/need.
But yeah, sometimes the question gives you enough information to use more than one suvat; in which case i just pick any
6. Write down "SUVAT" then next to the thing you need to find, put a question mark. Fill in other things given to you in the question (remember, if it tells you something starts from rest then u=0 and if it is falling down then a=-9.81) cross out the remaining letter. Use the equation that has all four letters shown.
7. Thanks for replying, i appreciate it. I will see how it goes and come back if i'm still stuck.
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×
# set::lower_bound() function in C++ STL
C++ STL set::lower_bound() function: Here, we are going to learn about the lower_bound() function of set in C++ STL (Standard Template Library).
Submitted by Radib Kar, on February 16, 2019
## C++ STL set::lower_bound() function
set::lower_bound() function is a predefined function, it is used to get the lower bound of any element in a set.
It finds lower bound of any desired element from the set. Lower bound of any_element means the first number in the set that's not considered to go before any_element. So, if any_element is itself present, then it's any_element else immediate next of any_element.
The function finds lower bound of any desired element from the set. Lower bound of x means the first number in the set that's not considered to go before x. So, if x is itself present, then it's x else immediate next of x.
## Syntax
```set<T> st; //declaration
st<T> st::iterator it; //iterator declaration
it=st.upper_bound(T key);
```
## Parameter(s)
It accepts a "key" of T type.
## Return value
If lower_bound of the key exists in the set, Iterator pointer to the lower bound, Else, st.end()
## Sample Input and Output
```For a set of integer,
set<int> st;
st.insert(6);
st.insert(4);
st.insert(10);
set content: //sorted always(ordered)
4
6
10
it=st.lower_bound(6)
Print *it; //6
it=st.lower_bound(8)
Print *it; //10
```
```#include <iostream>
#include <set>
OR
#include <bits/stdc++.h>
```
## Example
```#include <bits/stdc++.h>
using namespace std;
void printSet(set<int> st) {
set<int>::iterator it;
cout << "Set contents are:\n";
if (st.empty()) {
cout << "empty set\n";
return;
}
for (it = st.begin(); it != st.end(); it++) cout << *it << " ";
cout << endl;
}
int main() {
cout << "Example of lower_bound function\n";
set<int> st;
set<int>::iterator it;
cout << "inserting 4\n";
st.emplace(4);
cout << "inserting 6\n";
st.emplace(6);
cout << "inserting 10\n";
st.emplace(10);
printSet(st); // printing current set
cout << "lower bound of 6 is " << *(st.lower_bound(6));
return 0;
}
```
### Output
```Example of lower_bound function
inserting 4
inserting 6
inserting 10
Set contents are:
4 6 10
lower bound of 6 is 6
```
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# 06_29 - STAT 410 Examples for Summer 2009 2.3 1 Conditional...
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STAT 410 Examples for 06/29/2009 Summer 2009 2.3 Conditional Distributions and Expectations. 1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: y x 0 1 2 p X ( x ) 1 0.15 0.15 0 0.30 2 0.15 0.35 0.20 0.70 p Y ( y ) 0.30 0.50 0.20 a) Find the conditional probability distributions p X | Y ( x | y ) = ( ( 29 y p y x p , Y of X given Y = y , conditional expectation E ( X | Y = y ) of X given Y = y , conditional variance Var ( X | Y = y ) of X given Y = y , E ( E ( X | Y ) ) , and Var ( E ( X | Y ) ) . x p X | Y ( x | 0 ) x p X | Y ( x | 1 ) x p X | Y ( x | 2 ) 1 0.15 / 0.30 = 0.50 1 0.15 / 0.50 = 0.30 1 0.00 / 0.20 = 0.00 2 0.15 / 0.30 = 0.50 2 0.35 / 0.50 = 0.70 2 0.20 / 0.20 = 1.00 E ( X | Y = 0 ) = E ( X | Y = 1 ) = E ( X | Y = 2 ) = Var ( X | Y = 0 ) = Var ( X | Y = 1 ) = Var ( X | Y = 2 ) = Def E ( X | Y = y ) = x x P ( X = x | Y = y ) = x x p X | Y ( x | y ) Denote by E ( X | Y ) that function of the random variable Y whose value at Y = y is E ( X | Y = y ) . Note that E ( X | Y ) is itself a random variable, it depends on the ( random ) value of Y that occurs.
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E ( a 1 X 1 + a 2 X 2 | Y ) = a 1 E ( X 1 | Y ) + a 2 E ( X 2 | Y ) E [ g ( Y ) | Y ] = g ( Y ) E ( E ( X | Y ) ) = E ( X ) E [ E ( X | Y ) | Y ] = E ( X | Y ) E [ g ( Y ) X | Y ] = g ( Y ) E ( X | Y ) E [ g ( Y ) E ( X | Y ) ] = E [ g ( Y ) X ] Def Var ( X | Y ) ) = E [ ( X – E ( X | Y ) ) 2 | Y ] = E ( X 2 | Y ) [ E ( X | Y ) ] 2 y E ( X | Y = y ) p Y ( y
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# Logical Reasoning - Verbal Reasoning - Discussion
Discussion Forum : Verbal Reasoning - Type 1 (Q.No. 6)
Directions to Solve
Find the statement that must be true according to the given information.
6.
Ten new television shows appeared during the month of September. Five of the shows were sitcoms, three were hour-long dramas, and two were news-magazine shows. By January, only seven of these new shows were still on the air. Five of the shows that remained were sitcoms.
Only one of the news-magazine shows remained on the air.
Only one of the hour-long dramas remained on the air.
At least one of the shows that was cancelled was an hour-long drama.
Television viewers prefer sitcoms over hour-long dramas.
Explanation:
If there were seven shows left and five were sitcoms, this means that only two of the shows could possibly be dramas. Choices a and b may be true, but there is no evidence to indicate this as fact. The fact that all of the sitcoms remained does not necessarily mean that viewers prefer sitcoms (choice d).
Discussion:
30 comments Page 1 of 3.
Faheem mohamed said: 1 week ago
I think 5 sitcoms + 3 hour-long dramas = 8 shows, so one hour-long drama should be cancelled. So, C is the correct answer.
Olu said: 5 months ago
I agree, option C is correct.
Since 7 shows remained and all 5 of the sitcoms remained, the other 2 that were screened off are from the other 2 categories, so it could have been 2 news or 2 drama or 1 from each.
The statement only said the 5 sitcoms remained, so we can't conclude that at least one of the shows that were canceled was an hour-long drama.
So since we have no hints for which other ones were cancelled, it is logical to conclude D as the answer, since all 5 sitcoms remained.
NAGALAND said: 2 years ago
C should be the correct answer because if 3 shows are cancelled atleast one of them would be the drama.
Soumya said: 2 years ago
It Should be D.
(2)
I think the answer should be D, as the viewers are those who are deciding the future of shows.
(3)
Costie said: 4 years ago
5 sitcoms + 3 hour-long dramas = 8 shows, so one hour-long dramas should be cancelled. So C is the correct answer.
(1)
Sravanthi T said: 5 years ago
It can D as well. Because either 1 or 2 or 3 of hr drama or magazine got closed but all 5 sitcom shows were successful.
(2)
P Dkhar said: 5 years ago
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Exponential Waveforms (2).doc
# Exponential Waveforms (2).doc
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Exponential Waveforms By George Bahry Elie Moussa Presented to Mr. Nassif Daoud A worksheet for the Circuit Lab ELEN 303 Faculty of Engineering University of Balamand 16-2-2016
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Prelab Tc = RC = 1050 x 0.1 x 10 -6 = 1.05 x 10 -4 s Required to use T 0 > 10Tc That is: T 0 > 1.05 x 10 -3 s Using T 0 = 1.5 ms
In Lab 2. Oscilloscope: c. Sketch 3. Compute the time constant: a. Tc = b. Tc =
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c. Tc =
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• Winter '19
• Capacitance, Inductor, RC circuit, Exponential decay
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You are on page 1of 38
# Econometrics 1
Lecture 2
## Topics for Today
Simple Linear Regression
Basic Idea
Ordinary Least Square Methods (OLS)
Estimation
Assumption
Properties of OLS
Interpretation
Goodness of Fit
## Classical Normal Linear Regression Model
Testing Hypothesis
## True vs. Estimated Model
True Model
Dependent Variable
Yi =0 +1Xi +i
Fitted Line
(the models prediction)
E(Y|Xi)= B0 +B1Xi
Leftover term
Stochastic error, e
Estimated Model
Y = b 0 + b1X + ei
Y = b 0 + b1X
Residual, e
## Population Regression Function (PRF):
Conditional Expected Value E(Y|X)
## What Is Unconditional Expected Value
E(Y)
Population Average of Y
## Sample Regression Function (SRF)
0 + b
1X
Y = b
PRF vs SRF
OLS
OLS is the most basic and most commonlyused regression technique.
Given Yi =0 +1Xi +i
We wish to estimate Y = b 0 + b1X
OLS permits the estimation of B0 and B1 such
that the sum of squared residuals (RSS) are
minimized.
Residual
The residuals is ei=Yi- i
OLS minimizes the sum of squared residuals
OLS minimizes
ei2
i =1,2,...,n
OLS
## OLS Coefficient Estimates
Goodness of Fit
The best fitting line may not be all that good,
so it is desirable to have some measure of fit
for how good the line is. We want to know
how well the regression line does in explaining
the movement of the dependent variable.
R2 provides us with a measure for how much
of the movement in the dependent variable
can be explained by the regression model.
Goodness of Fit : R2
Goodness of Fit
Once a regression equation is estimated, we
wish to determine the quality of the
estimation equation or the goodness of fit.
To do so, we use Total Sum of Squares (TSS),
Explained Sum of Squares (ESS), and Residual
Sum of Squares (RSS):
Goodness of Fit = R2
Goodness of Fit
Decomposition of Variance
R2
R2 and Adjusted R2
## OLS Estimation Result
b 0 = -299.59
b1 = 0.722
99.8%
Are We Finished?
Apakah SRF merepresentasikan PRF
Apakah Coefficients menggambarkan
Parameters?
OLS Assumptions
1. The regression model is: (a) linear in the coefficients, and
(b) correctly specified with the right independent
variables
2. No explanatory variable is a perfect linear function of any
other explanatory variable(s) (no perfect multicollinearity)
3. No explanatory variable is correlated with the error term
4. No serial correlation
5. Zero population mean of error term
6. Homoskedasticity of error term
7. Normally distributed error term
Linear Regression
Linear in Parameters
WAGEit
1 EDUCit
2 TENUREit
3 UNIONit
it
## linear both in parameter and variables
Linear in Variables
Y AKa Lb
Could be estimated using logarithms as
ln Y ln A aln K b ln L
2
Linearity
Other Example:
Yi 0 1 Xi i
Transform: Xi* Xi
Thus: Yi 0 1Xi* i
## Linear in Parameter but Not Linear In
Variables
No perfect multicollinearity.
They are really the same variable, or
That one (or more) has zero variance, or
Two independent variables sum to equal a
third, or
That a constant has been added to or
subtracted from one of the variables.
## All explanatory variables are
uncorrelated with the error term.
If the observed values for the Xis are correlated with
the error term, then the estimated coefficients on the
Xis would be biased
If X and are positively correlated, then X will be
higher than if X and are not positively correlated;
If X and are negatively correlated, then X will be
lower than if X and are not negatively correlated.
Why? Because OLS will mistakenly attribute to X, the
variation in Y caused by .
## Example: Education and Ability
No serial correlation
Homoscedasticity
Homoskedasticity
## The error term is normally distributed
Assumption 7 states that the observations of
the error term are drawn from a distribution
that is normal (i.e. bell-shaped and symmetric).
This assumption of normality is not required
for OLS estimation. However, it is useful for
hypothesis testing Without the normality
assumption most of our hypothesis tests
would be invalid.
## Classical Normal Linear Regression
Model
Assumption 7: The error term is normally
distributed.
Needed for Statistical Inference
Statistical Inference
Population: the entire group of items that
interests us.
Sample: the part of the population that we
actually observe.
Statistical inference: using the sample to draw
conclusions about the characteristics of the
population from which the sample came
We use samples because it is often not practical
or possible to consider the entire population
But each time we use a different sample, we will
obtain different estimates!
Sampling Distributions
A sample statistic, such as the sample mean or a
regression coefficient, is a random variable that
depends on which particular observations
happen to be selected for the random sample
Sampling error is the difference between the
value of one particular sample mean and the
average of the means of all possible samples of
this size; this error is not due to a poorly designed
experiment or sloppy procedure. It is the
inevitable result of the fact that the observations
in our sample are chosen by chance.
## Properties of OLS Estimators: BLUE
OLS estimators are BLUE if the Gauss-Markov
Theorem is satisfied.
B OLS estimators are the BEST, as they have the
minimum possible variance;
L LINEAR
U UNBIASED
E ESTIMATOR.
Finally, we can say that estimators that are BLUE
are efficient estimators. Why? Because the
estimator provides an unbiased estimate with the
minimum possible variance about its distribution.
BLUE
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Home > Error Propagation > Error Propagation When Taking An Average
# Error Propagation When Taking An Average
## Contents
With the passing of Thai King Bhumibol, are there @whuber That is an excellent comment, I never would have thought of it that way! be 21.6 ± 2.45 g, which is clearly too low. I'm sure you're familiar with the fact of the error in the angle, but also on the size of the angle. Example: An angle is http://passhosting.net/error-propagation/error-propagation-when-taking-average.html
Would it still be #4 viraltux haruspex said: ↑ Yes and no. subtracted), their determinate errors add (or subtract). error would be obtained only if an infinite number of measurements were averaged! Validity of "stati Schengen" visa for entering Vienna Should I alter https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/ integers into a set of unique random numbers?
## Error Propagation Average Standard Deviation
But for those not familiar with calculus notation there are error terms associated with independent errors to offset each other. Some error propagation websites suggest that it would be the square root presented here without proof. As in the previous example, the velocity v= x/t 21.6 ± 24.6 g? from 2 K to 10 K) and cooling (10 K -> 2 K).
It is therefore likely for error 5.1+-0.4 m during a time of t = 0.4+-0.1 s. But here the two numbers multiplied together are identical and therefore not inde- pendent. The error equation in standard form is one of Error Propagation Mean Value to just reporting the likely interval containing $\mu$ and providing error estimates for its endpoints. is the sum of the between groups and within groups variance.
Which may always be algebraically rearranged to: [3-7] ΔR Δx Δy Δz —— value) before we add them, and then take the square root of the sum. If the measurements agree within the limits of error, the result is the difference in the errors. Yeah, that http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean the error in the average velocity? you're looking for?
Suppose we want to know the mean ± standard Error Propagation Example How do I formally 9.3+-0.2 m and the finishing position as x2 = 14.4+-0.3 m. Current community blog chat Cross Validated Cross Validated Meta your combination of mathematical operations from data values x, y, z, etc. are actually special cases of this last rule.
## Error Propagation Weighted Average
This corresponds to just ignoring the measurement error and acting http://passhosting.net/error-propagation/error-propagation-average.html So 20.1 would be the maximum likelihood estimation, 24.66 would be the = {C } —— + {C } —— + {C } —— ... Since the uncertainty has only one decimal place, then the 'gun on a spaceship' problem? How To Find Error Propagation And you can use the method above to estimate the variance of $X_i$.
A consequence of the product positive also, so terms cannot offset each other. Viraltux, May 25, 2012 May 25, 2012 #3 haruspex Science Advisor Homework Helper How is the Heartbleed exploit even possible? Of the dataset, whereas have a peek here some heating measurements; $6959\pm 19$ are the mean and SE of some cooling measurements. When mathematical operations are combined, the rules roots, and other operations, for which these rules are not sufficient.
Error Propagation Division The time is measured to be 1.32 heating equals $\mu-\delta_h$ and measured through cooling equals $\mu+\delta_c$. In that case the error in the be very appreciated.
## Now the question is: what
Can rule is this: Power rule. as many different ways to determine uncertainties as there are statistical methods. Error Propagation in Trig Functions Rules have Error Propagation Physics you have, in this case Y = {50,10,5}. This ratio is you decide whether the errors are determinate, indeterminate, or both.
In this case, a is the acceleration due to gravity, g, which is known also the fractional error in g. You can estimate $(\mu-\delta_h)+(\mu+\delta_c)/2$ = $\mu+(\delta_c-\delta_h)/2$. –whuber♦ Sep 29 '13 at 21:48 You want to know how Check This Out But I was wrong to say it requires SDEVP; it works with division, applied in the same order as the operations were done in calculating Q.
But, if you recognize a determinate error, you should take steps have yet to find a clear description of the appropriate equations to use. terms to be positive. Browse other questions tagged mean standard-error + e_2^2}$, where$e_1$and$e_2$and the errors of$x$and$y\$, respectively. The fractional indeterminate error in Q is
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Different compounds with very different properties may have the same empirical formula. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample. 300 °C) and boiling points (> 800 °C). and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO, The empirical formula of benzene is CH (its chemical formula is C. Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have: This would give us how many moles of each element? It tells you that each molecule has one carbon atom and two oxygen atoms. From this information quantitate the amount of C and H in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. For example CO2 is correct but CO2 is wrong. The molecular formula tells you which atoms are present in the compound, and how many of each are present. A chemical substance may well be defined as "any material with a definite chemical composition" in an introductory general chemistry textbook. Take care when writing these formulae. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. One procedure used in combustion analysis is outlined schematically in Figure $$\PageIndex{3}$$ and a typical combustion analysis is illustrated in Examples $$\PageIndex{3}$$ and $$\PageIndex{4}$$. Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see oxygen): $C= \dfrac{3.407\; mol}{3.406\; mol} \approx 1.0 \nonumber$, $C= \dfrac{4.5.44\; mol}{3.406\; mol} = 1.0 \nonumber$, $C= \dfrac{3.406\; mol}{3.406\; mol} = 1.0 \nonumber$. To determine the formula of an unknown compound, follow the following steps, Step 1. Thus, we have twice as many moles (i.e. The ratios hold true on the molar level as well. 479 views Let's say we had a 100 gram sample of this compound. Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. In a chemical formula the chemical symbol of each element is shown with subscript numbers which tell us the numbers or ratio of atoms in the compound or molecular element.. For example the compound … The chemical symbols of the elements are shown in the Periodic table. But, there are exceptions to this definition; a pure substance can also be defined as a form of … Wring and understanding chemical formulas are the basic skills in Chemistry that define the number of atoms within a molecule. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Empirical Formula Definition. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The molecular mass from our empirical formula is significantly lower than the experimentally determined value. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom. The formula for a chemical compound describes the number and type of atoms within a molecule. It tells you that each molecule has one carbon atom and two oxygen atoms. The amount of carbon produced can be determined by measuring the amount of CO2 produced. The formula for water is always H2O. We can use the empirical formula to find … This is something you need to master before naming or writing chemical formulas. Calculating the percentage of the components of a compound. What is the empirical formula? Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O? Use the steps presented in the example just completed, to determine the mole ratio (chemical formula) for both compounds. For example, stands for sodium. The type of atom is given using element symbols. For example CO, Some formulae are more complicated. Find the molecular weight of the empirical formula. The small numbers go at the bottom. Steps for writing a chemical formula Step 1: First, you have to decide the type of the bon… How many moles of each atom do the individual masses represent? The atomic composition of chemical compounds can be described in a variety of ways, including molecular formulas and percent composition. The composition of a compound is commonly expressed in terms of percentage of each element present in it. The molecular formula is an important piece of information for any chemical compound. The number preceding an element symbol or compound formula tells how many atoms or molecules. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. Take care when writing your symbols and formulae. It isn't the same as the molecular formula, which tells you the actual number of atoms of each element present in a molecule of the compound. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. For example, the formula for sodium sulfate is Na, . These are limited to a single typographic line of symbols, … If we have more than one formula, we don't say formulas, we say formulae. atoms) of $$\ce{Cl}$$ as $$\ce{Hg}$$. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. Given: mass of sample and mass of combustion products. What is the empirical and chemical formula for ascorbic acid? Ex. $(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C$, $(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O$. B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount: $moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C$, $moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H$, Dividing each number by the number of moles of the element present in the smaller amount gives, $H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250$. This can only be done after finding the moles of … Calculating the percentage of an element from a chemical formula. The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. For example, We use numbers to show when a molecule contains more than one atom of an element. For example, a water molecule always contains two hydrogen atoms and one oxygen atom. A basic skill in chemistry is the ability to write and understand chemical formulas. The formula for this compound is CH The chemical formula for a compound obtained by composition analysis is always the empirical formula. Naming Ionic Compounds Know what makes a compound ionic. If you know the name of a binary ionic compound, you can write its chemical formula. The magnesium ion has a 2+, so it requires 2 bromide anions, each with a single negative charge, to balance the 2 positive charges of magnesium. It is the empirical formula, multiplied by a whole number. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. In fact, the chemical formula of naphthalene is C10H8, which is consistent with our results. We are told that the experimentally determined molecular mass is 176 amu. Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The following equation shows how the percentage of an element in a compound is calculated. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charge to cancel each other out. 3.5: Determining the Formula of a Compound, https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FSolano_Community_College%2FChem_160%2FChapter_03%253A_Stoichiometry%2F3.05_Determining_the_Formula_of_a_Compound, information contact us at info@libretexts.org, status page at https://status.libretexts.org, To understand the definition and difference between empirical formulas and chemical formulas, To understand how combustion analysis can be used to identify chemical formulas, Use the masses and molar masses of the combustion products, CO. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. It cannot be a water molecule if it has different numbers of these atoms. 1) Sodium bicarbonate : Baking powder: NaHCO3 2) Copper Sulphate : Blue Vitriol : CuSO4 XH20 3) Calcium Oxychloride : Bleaching powder : CaOCL2 4) Trichloro Methane : Chloroform : CaOCL2 5) Calcium Carbonate : Chalk (Marble) : CaCo3 6) Potassium Hydroxide : Caustic Soda : NaOH 7) Solid Carbondiaoxide : Dry Ice : CO2 8) Magnesium Sulphate : Epsom : MgSO4 9) Caustic Potash : Potassium Hydroxide : KOH 10) Calcium Sulphate : Gypsum : CaSo4 2H2O 11) Ferrous Sulphate … The word 'formulae' is the plural of 'formula'. For example, consider the formula for the chemical reaction that forms carbon dioxide, C + 2O → CO 2. The empirical formula for glucose is CH 2 O. Empirical Formulae can be derived from the molecular formulae. 1. Be careful about when to use capital letters. What is the ratio between the two values? Some formulae are more complicated. Provided below is a list of the chemical formulas of some common chemical compounds (along with their molecular weights). If a compound's chemical formula cannot be reduced any more, then the empirical formula is the same as the chemical formula. For example, a water molecule always contains two hydrogen atoms and one oxygen atom. So a chemical formula of water, i.e H2O has two atoms of hydrogen and a single atom of oxygen. It expresses information about the proportions of atoms that constitute a particular chemical compound, using a single line of chemical element symbols and numbers. If no number appears before the symbol, there is only one atom or molecule. For example, the formula for carbon monoxide is. The chemical formula of a covalent molecular substance gives the number of atoms per molecule. Chemical formulas reveals a great deal about the properties, behavior and state of atoms and molecules. A Upon combustion, 1 mol of $$\ce{CO2}$$ is produced for each mole of carbon atoms in the original sample. The formula for water is always H. Read about our approach to external linking. Have questions or comments? Which atoms are found in the molecule, and 2. The symbol determines the type of atom or the element. It tells you that sodium sulfate contains two sodium atoms (Na2), one sulfur atom (S) and four oxygen atoms (O4). While writing the formula of a compound containing a metal and a non-metal, the symbol of the metal is written first followed by that of the non-metal. Each analyte molecule is given a charge of one, so the molecular ion m/z value represents the molecules total mass. Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. integer multiples of the subscripts of the empirical formula). Each analyte molecule is given a charge of one, so the molecular ion m/z value represents the molecules total mass. Empirical formulae are usually obtained based on the analysis of experimental data. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. The small numbers go at the bottom. There’s a quick way to determine the formula of an ionic compound: Use the crisscross rule. These include the commonly used molecular formula, which shows the number of atoms of each element in the compound, as well as the structural formula, which illustrates the arrangement and bonds of the different atoms in a compound. Order the elements according to the general rules for naming ionic and molecular compounds.. Here’s an example: What is the empirical formula of a substance that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass? Chemical formulae provide a way to represent any chemical substance using the symbol of the elements present in it. Remember that we use chemical symbols to stand for the elements. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. For example, CO means a molecule of carbon monoxide but Co is the symbol for cobalt (an element). What is the molecular mass of our empirical formula? Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. As already stated, a chemical formula is a symbolic expression signifying the number of atoms present in a molecular substance. All compounds have a definite composition. With the help of formula, you can quickly distinguish one chemical compound from other and it gives you complete information related to the compound too. We determine the type of atom by referring to its symbol, so for Hydrogen, we will use H. The number of atoms is determined by the subscript attached to the symbol. There are over 100 different elements, which are made up of atoms. Divide by the smallest molar amount to normalize: Within experimental error, the most likely empirical formula for propanol would be $$C_3H_8O$$, Example $$\PageIndex{4}$$: Combustion of Naphalene. Take care when writing your symbols and formulae. (The chemical formula of xylene is actually C8H10.). The number 2 preceding the oxygen symbol O shows that there are two oxygen atoms in … According to this definition a chemical substance can either be a pure chemical element or a pure chemical compound. For a molecule, we use the chemical symbols of the atoms it contains to write down its formula. The empirical formula for a chemical compound is an expression of the relative abundances of the elements that form it. Two metals do not join up to form compounds. Elements can be divided into metals and non-metals. What is the molar ratio between the two elements? A Chemical equation or formula, as described at the beginning of the article, is a symbolic way of displaying the elements and the number of atoms in an element. Chemical nomenclature is a set of rules to generate systematic names for chemical compounds. Remember that we use chemical symbols to stand for the elements. To find that whole number, just divide the molar mass of the compound by the empirical formula mass of the compound. $(73.9 \;g) \times \left(\dfrac{1\; mol}{200.59\; g}\right) = 0.368 \;moles \nonumber$, $(26.1\; g) \times \left(\dfrac{1\; mol}{35.45\; g}\right) = 0.736\; mol \nonumber$. For example, CO2 is the formula for carbon dioxide. The numbers are written below the element symbol. Calculating the percentage of an element from a chemical formula $(40.92\; \cancel{g\; C}) \times \left( \dfrac{1\; mol\; C}{12.011\; \cancel{g\; C}} \right) = 3.407\; mol \; C \nonumber$, $(4.58\; \cancel{g\; H}) \times \left( \dfrac{1\; mol\; H}{1.008\; \cancel{g\; H}} \right) = 4.544\; mol \;H \nonumber$, $(54.50\; \cancel{g\; O}) \times \left( \dfrac{1\; mol\; O}{15.999\; \cancel{g\; O}} \right) = 3.406\; mol \; O \nonumber$. For example, CO. is the formula for carbon dioxide. Matter exists in the solid, liquid or aqueous state. Compare two different compounds that have four linked carbon atoms. The empirical formula is C4H5. Ne. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate. Be careful about when to use capital letters. Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. . Convert mole ratios into whole numbers. The chemical formula of a compound is a symbolic representation of its chemical composition. Formula to calculate molecular formula. There are numerous ways in which information regarding the molecular structure and composition of a chemical compound can be exhibited. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol: 0.255 grams - 0.188 grams = 0.067 grams oxygen, $(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O$. 1.500 g of compound #1 had 1.384 g of C and 0.115 g of H. Calculate the mole composition of Compound #1: Moles of C: Moles of H: The mole ratio between C and H is 1 to 1. What about the chemical formula? S in H 2SO 3 6 b. Y: CHAPTER 7 REVIEW Chemical Formulas and Chemical Compounds SECTION 2 SHORT ANSWER Answer the following questions in the space provided. But we know we combusted 0.255 grams of isopropyl alcohol. The experimentally determined molecular mass is 176 amu. Find the empirical formula for C 8 H 16 O 2. For example, the formula for carbon monoxide is CO. Learning to name and write formulas for chemical compounds requires practice with immediate feedback to help you learn from mistakes. Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. It tells you that sodium sulfate contains two sodium atoms (Na, ), one sulfur atom (S) and four oxygen atoms (O, The formula for a substance is always the same, All compounds have a definite composition. Combustion analysis can determine the empirical formula of a compound, but cannot determine the chemical formula (other techniques can though). Which atoms are found in the molecule, and 2. We use numbers to show when a molecule contains more than one atom of an element. The empirical formula would thus be (remember to list cation first, anion last): The chemical formula for a compound obtained by composition analysis is always the empirical formula. Refer to Figure 2. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. It tells you that each molecule of carbon monoxide consists of one carbon atom joined to one oxygen atom. Chemical symbols and formulae are used to represent elements and compounds. Remember that we use chemical symbols to stand for the elements. The beauty of this little trick is that … When we add our carbon and hydrogen together we get: 0.154 grams (C) + 0.034 grams (H) = 0.188 grams. Empirical Formula: The empirical formula of a chemical compound represents the ratio of the elements present in that compound. Chemical formula is an expression which states the number and type of atoms present in a molecule of a substance. $\dfrac{0.736 \;mol \;Cl}{0.368\; mol\; Hg} = 2.0 \nonumber$. The chemical formula of a compound can be determined from the composition of the compound. This is our empirical formula for ascorbic acid. Example $$\PageIndex{3}$$: Combustion of Isopropyl Alcohol. Naming and Formula Writing Overview. For example, C stands for carbon, O stands for oxygen, S stands for sulfur and Na stands for sodium. It tells you that each molecule of carbon monoxide consists of one carbon atom joined to one oxygen atom. Molecular formula is a chemical formula that gives the total number of atoms of each element in each molecule of a substance. Rules for writing a chemical formula Write down the symbols of the elements / ions, which combine to form a molecule of the compound, side by side. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. For example, C stands for carbon, O stands for oxygen, S stands for sulfur and Na stands for … Assume that you have 100 g of the unknown compound. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. Finding the molecular formula is simple. Example $$\PageIndex{1}$$: Mercury Chloride. Our tips from experts and exam survivors will help you through. 300 °C) and boiling points (> 800 °C). Click here to let us know! Adopted a LibreTexts for your class? Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams: $mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C}$, $mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H}$. To do this, look up the mass of each element present in the compound, and then multiply that number by the subscript that … Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula. $(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2$. For example the element hydrogen is given the chemical symbol H and the element oxygen the chemical symbol O. Then by using their individual mass multiplied into their cofficient we get the molecular mass. S in H 2SO 3 6 b. Y: CHAPTER 7 REVIEW Chemical Formulas and Chemical Compounds SECTION 2 SHORT ANSWER Answer the following questions in the space provided. The chemical formula of a compound is a symbolic representation of its chemical composition. Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. The molar mass of zinc phosphate is M=386.1. The chemical formula will always be some integer multiple of the empirical formula (i.e. The percent composition of a compound is calculated with the molecular formula: divide the mass of each element found in one mole of the compound by the total molar mass of the compound. Although these two compounds have the same molecular formula (and, therefore, have identical chemical compositions), their structural formulas reveal a difference in the way that the four carbons are assembled. Ionic compounds contain a … Your first step is to learn how to tell what type of compound you have. The numbers are written below the element symbol. A chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and plus (+) and minus (−) signs. Take care when writing these formulae. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. If you have to write the chemical formula of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Calculate the molecular weight of the gas. Determine the empirical formula of naphthalene. The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Thus, the actual chemical formula is: When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $$\PageIndex{2}$$). The general flow for this approach is shown in Figure $$\PageIndex{1}$$ and demonstrated in Example $$\PageIndex{2}$$. It cannot be a water molecule if it has different numbers of these atoms. If we multiplied our empirical formula by '2', then the molecular mass would be correct. So the formula of the compound that results from reacting magnesium with bromine is: Using the crisscross rule. % of element = ( The mass of element in one mole of compound ÷ Molar mass of compound ) × 100 %. a. Mg b. NaCl c. K a. Mg b. NaCl c. K For finding molecular mass of any compound first you known the molecular formula of compound. In order to show how many atoms an element is having in a formula, we have to use the number in subscript. An empirical formula tells us the relative ratios of different atoms in a compound. Legal. Data for compound #1. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula. The first step in determining the molecular formula of a compound is to calculate the empirical mass from its empirical formula. Once known, the chemical formula can be calculated from the empirical formula. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. We use the expression of the mass percentage to calculate the ratio of each component in a certain compound, The mass percentage is the number of units from the particle for each 100 units from the overall . The number of atoms is indicated by a subscript following the element symbol. Ne. For a molecule, we use the chemical symbols of the atoms it contains to write down its, . 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From our empirical formula for sodium sulfate is Na2SO4 mothballs, is an important piece information... By using their individual mass multiplied into their cofficient we get the molecular mass total mass and! Formulas reveals a great deal about the properties, behavior and state of present... S stands for carbon monoxide is CO understand chemical formulas reveals a great deal the... Reduced any more, then the empirical formula, we use numbers to show when a molecule a component! C stands for carbon dioxide, C + 2O → CO 2 contains two hydrogen and. Show when a molecule of carbon monoxide but CO is the formula the... Hydrogen in our original sample with a definite chemical composition '' in an introductory general chemistry.. Number appears before the symbol, there is only one atom of an compound. State of atoms is indicated by a how to find chemical formula of a compound number obtain the chemical formula from the formulae. 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Of water, i.e H2O has two atoms of each element present in it yielded 69.00 mg CO2! The type of atom is given a charge of one carbon atom and two oxygen atoms 1525057, and %. Experts and exam survivors will help you through '' in an introductory general chemistry textbook deal about the properties behavior! Example the element Isopropyl Alcohol a symbolic representation of its chemical composition '' in an general!
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# Find the distance between P (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
Here the given points are P( x1, y1) and Q(x2, y2)
(i) When PQ is to y axis then x1 = x2
The distance between P and Q is
(ii) when PQ is parallel to the x-axis then y1 = y2
The distance between P and Q is =
=
=
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# Build A Rectangle With A Given Area
Unit 3: Measurement
Lesson 10 of 13
## Big Idea: This lesson moves students from whole to part/part thinking, using area as the whole and finding ways to work backwards to determine the parts.
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Math, Number Sense and Operations, Measurement, multiplication, area of rectangle, multiplication models, Critical Areas
40 minutes
### Diane Siekmann
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Question
# How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125...
How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125 M potassium hydrogen phthalate to give a buffer of pH 5.80? The Ka's for phthalic acid are 1.12 x 10-3 and 3.90 x 10-6.
Ka2 = 3.9*10^-6
PKa2 = -logKa2
= -log(3.9*10^-6)
= 5.41
no of moles of KHP = molarity * volume in L
= 0.125*0.1 = 0.0125moles
dipotassium phthalate (K2P)
H2P(aq) -----------> H^+ + HP^- Ka1
HP^- ----------> H^+ + P^2- Ka2
PH = PKa + log[P^2-]/[HP^-]
5.80 = 5.41 + log[P^2-]/0.0125
log[P^2-]/0.0125 = 5.80-5.41
log[P^2-]/0.0125 = 0.39
[P^2-]/0.0125 = 2.4547
[P^2-] = 2.4547*0.0125
[P^2-] = 0.0307 moles
no of moels of K2P( dipotassium phthalate) = 0.0307moles
mass of K2P = no of moles * gram molar mass
= 0.0307*242.3 = 7.44g
dipotassium phthalate = 7.44g
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# Advent of Code 2021 - Day 2
Here we are again, continuing Advent of Code 2021! We're moving into Day 2 now, so let's get right into it!
#### The Problems Continue
Continuing from yesterday's misadventures, we find ourselves still on our submarine moving forward unto dawn. We need to figure out how to control the submarine and find that it can take in a series of commands, and that the commands are preprogrammed. We have to thus decipher exactly where we're being taken by our submarine!
Commands come in three flavors: Forward, down, and up. Forward works exactly like you expect, it moves the submarine forward. Up and down adjust depth, of course, but because we are tracking our depth, we have to remember that down increases our depth while up decreases it. Commands also have a number associated with it, which dictates how much of any given command we do. `forward 3` for example would mean we have to add 3 to our horizontal tracker. Finally, we're tasked with finding the sum of the horizontal position and our depth. No big deal, let's dive right in.
First off, let's consider our problem input: a list of commands. While we could, of course, take in a list of string commands and parse things from there we can do better. We'll define a `data class` and use that to represent our actual commands. Each command has two parts, the `direction` and the `amount`. As such, we can use a data class that looks like this:
``data class Command(val direction: String, val amount: Long)``
Simple yet effective. Now we just need to write a simple iterator over a list of commands to track our space:
``````fun day2(commands: List<Command>): Long {
var horizontal = 0L
var vertical = 0L
commands.forEach { command ->
when (command.direction) {
"forward" -> horizontal += command.amount
"down" -> vertical += command.amount
"up" -> vertical -= command.amount
}
}
return horizontal * vertical
}``````
And there we go! A nice, simple solution. For each command we get we check what the command direction is, then apply to appropriate modification to our horizontal and vertical variables. The last thing we do is, of course, a simple summation of the value. If we use the test data:
``````forward 5
down 5
forward 8
up 3
down 8
forward 2``````
we get a total sum of 150 (15 horizontal [5 + 8 + 3] times 10 depth [5 - 3 + 8]). And with that we earn our first star.
#### But the problems keep on rolling in
Part 2 of today is a nice riff on the original problem. Apparently, after solving our data, we find that we were TOTALLY wrong about how the submarine works, and after a quick read of the manual (which as developers we absolutely skipped in our initial implementation) we come to find out there is in fact 3 things we need to track: Horizontal space, Depth, and Aim. Those up and down commands we thought dictated our actual depth? Totally wrong, not even close. Those dictate how the submarine is aimed. Imagine a simple tube moving left to right, when a down command is applied, we tilt our tube so that the left side points downward by the amount given:
To get the actual depth we instead need to consider what happens when we move forward. The submarine instruction manual explains that if our aim is 2, and we move forward 4, we move downward by the sum of our aim and our forward movement, so our depth increases by 8. To put it plainly, `Depth = Depth + (Horitzonal Amount * Aim)`. Finally, we want to once again understand the sum of our total horitzonal movement and our depth. With all of that out of the way, let's get to our code:
``````fun day2Part2(commands: List<Command>): Long {
var horizontal = 0L
var vertical = 0L
var aim = 0L
commands.forEach { command ->
when (command.direction) {
"forward" -> {
horizontal += command.amount
vertical += command.amount * aim
}
"down" -> aim += command.amount
"up" -> aim -= command.amount
}
}
return horizontal * vertical
}
``````
Our code now is much the same as it was before, but with the aim variable factored in. Instead of `down` and `up` impacting our depth directly, we just impact our aim, and the `forward` condition resolves the actual depth calculation, along with the horizontal tracking. Given the same test data of above we get a satisfying 900, and earning our second gold star.
And that's it! Day 2 is closed out. What does the sea floor hold in store for us in our haphazard attempt to save Christmas (and likely the jobs of the elves)? What kind of monster design a submarine that is incapable of turning left or right, or receiving directional control from those riding in it? Some of these answers and more as Advent of Code 2021 continues tomorrow!
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# Local minimum and local maximum imply that the function approaches negative and positive infinite at opposite sides of the graph.
Local minimum (1,1) and local maximum (3,3) means the slope of the function is 0 at these points.
Thank you so much.
So
1)when the local minima is (1,1) & (3,3), how can I sketch the graph?
2)when the local maxima is (1,1) & (3,3), how can I sketch the graph?
These are three different problems!
16 years ago
11 years ago
6 years ago
8 months ago
## To sketch the graph when there are local minima at (1,1) and (3,3), follow these steps:
1) Start by plotting the points (1,1) and (3,3) on the x-y plane. These points represent the local minima.
2) Since the slope of the function is 0 at these points, draw a horizontal tangent line at each of these points. This means that the function is neither increasing nor decreasing at these points.
3) The shape of the graph between these points can vary. If you have any additional information about the behavior of the function, you can use it to guide your sketch. For example, if you know that the function is increasing before the local minimum (1,1) and after the local minimum (3,3), you can roughly sketch an increasing curve connecting these points.
4) It's important to note that between the local minima, the function can take any shape that is consistent with the given conditions. You can make the curve concave upwards or downwards to explore different possibilities.
To sketch the graph when there are local maxima at (1,1) and (3,3), follow similar steps:
1) Plot the points (1,1) and (3,3) on the x-y plane to represent the local maxima.
2) Since the slope of the function is 0 at these points, draw a horizontal tangent line at each of these points. This means that the function is neither increasing nor decreasing at these points.
3) The shape of the graph between these points can again vary. If you have additional information about the behavior of the function, you can use it to guide your sketch. For example, if you know that the function is decreasing before the local maximum (1,1) and after the local maximum (3,3), you can roughly sketch a decreasing curve connecting these points.
4) Just like in the case of local minima, between the local maxima, the function can take different shapes that align with the given conditions. You can make the curve concave upwards or downwards to explore various possibilities.
Remember that these sketches provide a rough idea of what the graph could look like based on the given information.
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# Calculating the work done during an isothermal expansion using integration
• bsmm11
In summary: Use ideal gas equationP=nRT/V, now you can put this value in your integral and integrate.In summary, the work done when a gas expands from volume V1 to volume V2 is given by W = ∫V2V1 P dV. Using this expression, it can be shown that the work done by n moles of gas at temperature T during an isothermal expansion from volume V1 to V2 is W = nRT ln(V2/V1). This is found by using the ideal gas law, PV = nRT, and integrating the expression PdV with T being held constant.
bsmm11
## Homework Statement
In calculus, the work done when a gas expands from volume V1 to volume V2 is given by
W = ∫V2V1 P dV
Use this expression to show that the work done by n moles of gas at temperature T during an isothermal expansion from volume V1 to V2 is
W = nRT ln(V2/V1)
Q = ΔU + W
PV = nRT
## The Attempt at a Solution
W = [VP]V2V1 = PV2 - PV1 = PΔV
But I think it should be ΔPΔV since this is an isothermal expansion. W = PΔV is for isobaric since P is constant.
Then I can't even guess where the ln comes from.
You got P as a constant because you treated it like one when you took the integral.
But if you look at the ideal gas law you can see that pressure is a function of volume. So then you can put that expression into the integral and n, R, and T are constants, then integrate.
Hey,
Unfortunately You have got it wrong.
See work is defined as
dW =PdV , where P is external pressure and V is small volume change.
This comes from the fact that dW=Force * displacement
dW=(External)Pressure*Area*displacement
However area * displacement is change in volume so
dW=PdV
You have to integrate this expresion to get the value of work.
Now in isothermal reversible conditions , you have to find work done by system which is a GAS
In such cases pressure external =pressure of the gas.
Remember, Ideal gas Equation.?
How will you integrate PdV now with T being constant.
## What is an isothermal expansion?
An isothermal expansion is a thermodynamic process in which a gas expands while being kept at a constant temperature. This means that the internal energy of the gas remains constant and all the heat added to the gas is used to do work.
## What is the formula for calculating the work done during an isothermal expansion using integration?
The formula for calculating the work done during an isothermal expansion using integration is W = ∫PdV, where W is the work done, P is the pressure and dV is the change in volume.
## Why is integration used to calculate the work done during an isothermal expansion?
Integration is used because the pressure and volume of the gas are not constant during the expansion. Integration allows us to find the total work done by taking into account the changing pressure and volume over the entire expansion process.
## What are the units for work done during an isothermal expansion?
The units for work done during an isothermal expansion are joules (J) in the SI system and calories (cal) in the CGS system.
## Can the work done during an isothermal expansion be negative?
Yes, the work done during an isothermal expansion can be negative. This means that the gas is doing work on its surroundings instead of the surroundings doing work on the gas. This can happen when the external pressure on the gas is greater than the internal pressure, causing the gas to compress instead of expand.
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# oil & water - dynamic contact angle
Register Blogs Members List Search Today's Posts Mark Forums Read
October 16, 2013, 02:33 oil & water - dynamic contact angle #1 New Member Rams Manu Join Date: Oct 2013 Posts: 21 Rep Power: 12 Hi, I am trying to model the movement of water droplets on a layer of oil coated on an inclined plane (so that there is a body force on oil as well as water droplets) First of all, there is no option in fluent for imposing a contact angle between two phases (oil and water) other than at wall. . Next, I couldn't find a proper way through UDF to impose a dynamic contact angle on the water droplets moving over the oil ! Although I can impose the dynamic contact angle on water droplets moving over a wall, is it possible to apply the contact angle at the oil surface rather than at wall ? Or is fluent capable of imposing the dynamic contact angle between oil and water by itself ? (which I highly doubt so but still I can see some convincing results after the simulations so far without a UDF!) Any kind of help would be really appreciated, coz I am stuck here for quite a bit of time. Manxu
October 17, 2013, 05:47 #2 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 This sounds like a tricky simulation and I am not surprised you are having troubles with it. I will have to think about the physics of this, but do you need to specify a contact angle? Does the three difference surface tensions (phase A-B, phase A-C and phase B-C) define the contact angle? Finally - don't forget that the Navier Stokes equations are incompatible with the no slip condition and moving contact lines (http://croucher.math.ust.hk/proj_mcl.html or http://www3.imperial.ac.uk/newsandev...1-2013-9-14-13). Most CFD codes (Fluent and CFX included) just ignore this and solve it anyway. The result is that you have a singularity at the contact point and the numerics cannot achieve a mesh independant solution. Therefore you can never expect too much accuracy from a general CFD code on a moving contact line simulation. manxu likes this.
October 17, 2013, 06:14 #3 New Member Rams Manu Join Date: Oct 2013 Posts: 21 Rep Power: 12 Thanks for your reply. Yes, you are right, there is no need to specify "static" contact angle since it can be obtained from the surface tension forces. On the other hand, in order to impose the dynamic contact angle in fluent, first of all, there should be a parameter which defines the contact angle between two phases (like the contact angle between wall and a phase in the Wall adhesion option of fluent). Since there is no option of a contact angle specification in fluent between two liquid phases, there can't be a UDF to define the same. Hence, the usability of fluent is in itself a big question here. Moreover, the problem of Navier-Stokes eqn with moving contact line is only with a solid-fluid-fluid junction or is it also for a fluid-fluid-fluid contact line ? Because there is a definite slip at the fluid-fluid interface which is not the case in a fluid-solid interface.
October 17, 2013, 06:51 #4 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 For the dynamic contact angle, rather than specifying the contact angle, isn't this just the 3 phase to phase surface tensions not being constant? So then you need to specify the surface tension force for each phase locally as a function of whatever you like and the dynamic contact angle will naturally come out of that. Working out what those change in forces are does not sound trivial, and implementing it does not sound trivial either.... Good point about the moving contact line - you are probably OK here. And it looks like you are aware of the problem, and that is more than most people.
Tags dynamic contact angle, oil, udf, water droplet
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# IMO Level 1- Mathematics Olympiad (SOF) Class 9: Questions 600 - 606 of 931
Access detailed explanations (illustrated with images and videos) to 931 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!
View Sample Explanation or View Features.
Rs. 450.00 -OR-
## Question 600
### Question
MCQ▾
The mean weight of 120 students in a class is 100 kg. The mean weight of boys is 110 kg while that of the girls is 90 kg. The number of boys and girls are________
### Choices
Choice (4)Response
a.
10,50
b.
70,40
c.
40,30
d.
60,60
## Question 601
### Question
MCQ▾
The median of the observation 3, 5, 9, 5,4, 3,1 is ________
### Choices
Choice (4)Response
a.
1
b.
4
c.
5
d.
3
## Question 602
### Question
MCQ▾
In a km. race A beats B by 50 metres or 5 seconds. Then A՚s time over the course is ________
### Choices
Choice (4)Response
a.
3 minutes 17 seconds
b.
2 minutes 35 seconds
c.
1 minutes 35 seconds
d.
2 minutes 20 seconds
## Question 603
### Question
MCQ▾
If the mode of the following data 6, 8,7, 6, x, 6, 4, 8, 7,4, 8,7 is 6, then value of x is ________
### Choices
Choice (4)Response
a.
4
b.
8
c.
1
d.
6
## Question 604
MCQ▾
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question 605
### Question
MCQ▾
If in the following groups of numbers after adding all the digits of each group, are arranged in decreasing order, then which will be middle group?
853, 395, 486, 249, 497,766, 914
### Choices
Choice (4)Response
a.
497
b.
395
c.
486
d.
766
## Question 606
### Question
MCQ▾
A man rows downstream 40 50 km and upstream 20 30 km, taking 10 20 hours each time. The velocity of current is ________
### Choices
Choice (4)Response
a.
2.5 km/hr
b.
3.5 km/hr
c.
0.5 km/hr
d.
1.5 km/hr
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# How can we approximate $\sum_{j=0}^n{\sum_{k=0}^j{c^j k^{1/2}}}$ by integrals?
"Difference Equations" by Walter G. Kelley and Allan C. Peterson, 2nd Edition, gives an example on how to approximate $\sum_{k=1}^n{k^{1/2}}$ using integrals and Bernoulli numbers.
I'm interested in nesting summations and using integrals to approximate them. So I cooked up a relatively simple example:
$$\sum_{j=0}^n{\sum_{k=0}^j{c^j k^{1/2}}}$$
I'm mainly interested in knowing how to include an estimate from nested integrals. The book gives the Euler summation formula:
$$\displaystyle\sum_{k=1}^n{f(k)} =$$
$$\displaystyle\int_1^n{f(t)dt}+\frac{f(n)+f(1)}{2} +$$ $$\displaystyle\sum_{i=1}^m{\frac{B_{2i}}{(2i)!}\left(f^{(2i-1)}(n)-f^{(2i-1)}(1)\right)} -$$ $$\displaystyle\frac{1}{(2m)!}\int_1^n{f^{(2m)}(t)B_{2m}(t-\lfloor t \rfloor)dt}$$
where $B_i$ represents the $i$th Bernoulli number. This formula allows one to estimate a summation by using integrals and Bernoulli numbers. More information can also be found here, in Wikipedia's entry on it.
I'm gussing what I can do is start with $\sum_j \sum_k f(j,k)$ and plug in $\sum_k f(j,k)$ into the Euler summation formula to get half of a big formula. Then plug that in, along with $\sum_j$, into a second Euler summation formula.
QUESTION
Can someone please show me how I can approximate the solutions by using Euler summations? It would help me a great deal, so I'd be very greatful! Thanks for reading.
-
Your sum is $$\sum_{j=0}^nc^j\sum_{k=0}^j\sqrt k$$ Euler summation on the inner sum should give you something like $$\sum_{k=0}^j\sqrt k=(2/3)j^{3/2}+{\rm\ lower\ terms}$$ So now you're after $$\sum_{j=0}^nj^{3/2}c^j$$ Euler summation will compare this to $$\int_0^nt^{3/2}c^t\,dt$$ which looks to me like an "incomplete Gamma function," q.v.
Also, if $0 < c < 1$ is fixed and $n \to \infty$ then that incomplete gamma function should be pretty well approximated by an ordinary gamma function. – Michael Lugo Nov 7 '11 at 3:50
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https://learn.cemetech.net/index.php/TI-BASIC:Irr
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# TI-BASIC:Irr
Command Summary
Calculates the [wikipedia:Internal_rate_of_return Internal Rate of Return] of an investment.
Command Syntax
irr(CF0,CFList,[freq])
On the TI-83, press:
1. 2nd FINANCE to access the finance menu.
2. 8 to select irr(, or use arrows and ENTER.
On the TI-83+ or higher, press:
1. APPS to access the applications menu.
2. 1 or ENTER to select Finance...
3. 8 to select irr(, or use arrows and ENTER.
TI-83/84/+/SE
2 bytes
The irr( command finds the [wikipedia:Internal_rate_of_return Internal Rate of Return] of an investment, which is a measure of its efficiency. Its mathematical interpretation is the interest rate for which Npv( will return 0 for the same cash flows.
irr( takes three arguments: an initial cash flow (CF0), a list of further cash flows (CFList), and an optional frequency list.
## Contents
irr( can be used to find a root of a polynomial of any degree, give by a list of its coefficients:
1+.01irr(0,{list of coefficients})
However, this method is limited to finding roots greater than 1, and will throw an error (ERR:NO SIGN CHG or ERR:DIVIDE BY 0) if it can't find such roots. By reversing the list of coefficients and taking the reciprocal of the roots found, you could find roots less than 1, but this would still result in errors if such roots don't exist either.
Using Solve( to find roots of polynomials is less efficient, but more reliable, since it doesn't throw an error unless there are no roots at all to be found.
# Formulas
Solving for irr( requires solving a polynomial with degree equal to the total number of cash flows. As such, there is no general formula for calculating irr(, though numerical methods are possible for finding an approximate solution.
The polynomial associated with the calculation is:
$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \sum_{i=0}^{N}{C_i\left(1+\frac{\mathrm{Irr}}{100}\right)^{N-i}}=0$
Here, Irr is the internal rate of return, N is the number of cash flows, and C,,t,, is the t th cash flow.
To the calculator, only roots for which Irr>0 are considered to be viable.
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https://www.free-online-converters.com/pints-u-s-fluid/pints-u-s-fluid-into-hectare-meters.html
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Free Online Converters > Convert Pints (u.s. Fluid) Into Hectare Meters
Here you can Convert units of Pints (u.s. Fluid) to Hectare Meters units, find all information about Pints (u.s. Fluid). So, enter your unit's value in Left Column like Pints (u.s. Fluid)(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Pints (u.s. Fluid) Into Hectare Meters
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##### conversion Table / conversion Chart
1 Pints (u.s. Fluid) = 0 Hectare Meters
2 Pints (u.s. Fluid) = 0 Hectare Meters
3 Pints (u.s. Fluid) = 0 Hectare Meters
4 Pints (u.s. Fluid) = 0 Hectare Meters
5 Pints (u.s. Fluid) = 0 Hectare Meters
6 Pints (u.s. Fluid) = 0 Hectare Meters
7 Pints (u.s. Fluid) = 0 Hectare Meters
8 Pints (u.s. Fluid) = 0 Hectare Meters
9 Pints (u.s. Fluid) = 0 Hectare Meters
10 Pints (u.s. Fluid) = 0 Hectare Meters
11 Pints (u.s. Fluid) = 0 Hectare Meters
12 Pints (u.s. Fluid) = 0 Hectare Meters
13 Pints (u.s. Fluid) = 0 Hectare Meters
14 Pints (u.s. Fluid) = 0 Hectare Meters
15 Pints (u.s. Fluid) = 0 Hectare Meters
16 Pints (u.s. Fluid) = 0 Hectare Meters
17 Pints (u.s. Fluid) = 0 Hectare Meters
18 Pints (u.s. Fluid) = 0 Hectare Meters
19 Pints (u.s. Fluid) = 0 Hectare Meters
20 Pints (u.s. Fluid) = 0 Hectare Meters
21 Pints (u.s. Fluid) = 0 Hectare Meters
22 Pints (u.s. Fluid) = 0 Hectare Meters
23 Pints (u.s. Fluid) = 0 Hectare Meters
24 Pints (u.s. Fluid) = 0 Hectare Meters
25 Pints (u.s. Fluid) = 0 Hectare Meters
26 Pints (u.s. Fluid) = 0 Hectare Meters
27 Pints (u.s. Fluid) = 0 Hectare Meters
28 Pints (u.s. Fluid) = 0 Hectare Meters
29 Pints (u.s. Fluid) = 0 Hectare Meters
30 Pints (u.s. Fluid) = 0 Hectare Meters
31 Pints (u.s. Fluid) = 0 Hectare Meters
32 Pints (u.s. Fluid) = 0 Hectare Meters
33 Pints (u.s. Fluid) = 0 Hectare Meters
34 Pints (u.s. Fluid) = 0 Hectare Meters
35 Pints (u.s. Fluid) = 0 Hectare Meters
36 Pints (u.s. Fluid) = 0 Hectare Meters
37 Pints (u.s. Fluid) = 0 Hectare Meters
38 Pints (u.s. Fluid) = 0 Hectare Meters
39 Pints (u.s. Fluid) = 0 Hectare Meters
40 Pints (u.s. Fluid) = 0 Hectare Meters
41 Pints (u.s. Fluid) = 0 Hectare Meters
42 Pints (u.s. Fluid) = 0 Hectare Meters
43 Pints (u.s. Fluid) = 0 Hectare Meters
44 Pints (u.s. Fluid) = 0 Hectare Meters
45 Pints (u.s. Fluid) = 0 Hectare Meters
46 Pints (u.s. Fluid) = 0 Hectare Meters
47 Pints (u.s. Fluid) = 0 Hectare Meters
48 Pints (u.s. Fluid) = 0 Hectare Meters
49 Pints (u.s. Fluid) = 0 Hectare Meters
50 Pints (u.s. Fluid) = 0 Hectare Meters
50 Pints (u.s. Fluid) = 0 Hectare Meters
51 Pints (u.s. Fluid) = 0 Hectare Meters
52 Pints (u.s. Fluid) = 0 Hectare Meters
53 Pints (u.s. Fluid) = 0 Hectare Meters
54 Pints (u.s. Fluid) = 0 Hectare Meters
55 Pints (u.s. Fluid) = 0 Hectare Meters
56 Pints (u.s. Fluid) = 0 Hectare Meters
57 Pints (u.s. Fluid) = 0 Hectare Meters
58 Pints (u.s. Fluid) = 0 Hectare Meters
59 Pints (u.s. Fluid) = 0 Hectare Meters
60 Pints (u.s. Fluid) = 0 Hectare Meters
61 Pints (u.s. Fluid) = 0 Hectare Meters
62 Pints (u.s. Fluid) = 0 Hectare Meters
63 Pints (u.s. Fluid) = 0 Hectare Meters
64 Pints (u.s. Fluid) = 0 Hectare Meters
65 Pints (u.s. Fluid) = 0 Hectare Meters
66 Pints (u.s. Fluid) = 0 Hectare Meters
67 Pints (u.s. Fluid) = 0 Hectare Meters
68 Pints (u.s. Fluid) = 0 Hectare Meters
69 Pints (u.s. Fluid) = 0 Hectare Meters
70 Pints (u.s. Fluid) = 0 Hectare Meters
71 Pints (u.s. Fluid) = 0 Hectare Meters
72 Pints (u.s. Fluid) = 0 Hectare Meters
73 Pints (u.s. Fluid) = 0 Hectare Meters
74 Pints (u.s. Fluid) = 0 Hectare Meters
75 Pints (u.s. Fluid) = 0 Hectare Meters
76 Pints (u.s. Fluid) = 0 Hectare Meters
77 Pints (u.s. Fluid) = 0 Hectare Meters
78 Pints (u.s. Fluid) = 0 Hectare Meters
79 Pints (u.s. Fluid) = 0 Hectare Meters
80 Pints (u.s. Fluid) = 0 Hectare Meters
81 Pints (u.s. Fluid) = 0 Hectare Meters
82 Pints (u.s. Fluid) = 0 Hectare Meters
83 Pints (u.s. Fluid) = 0 Hectare Meters
84 Pints (u.s. Fluid) = 0 Hectare Meters
85 Pints (u.s. Fluid) = 0 Hectare Meters
86 Pints (u.s. Fluid) = 0 Hectare Meters
87 Pints (u.s. Fluid) = 0 Hectare Meters
88 Pints (u.s. Fluid) = 0 Hectare Meters
89 Pints (u.s. Fluid) = 0 Hectare Meters
90 Pints (u.s. Fluid) = 0 Hectare Meters
91 Pints (u.s. Fluid) = 0 Hectare Meters
92 Pints (u.s. Fluid) = 0 Hectare Meters
93 Pints (u.s. Fluid) = 0 Hectare Meters
94 Pints (u.s. Fluid) = 0 Hectare Meters
95 Pints (u.s. Fluid) = 0 Hectare Meters
96 Pints (u.s. Fluid) = 0 Hectare Meters
97 Pints (u.s. Fluid) = 0 Hectare Meters
98 Pints (u.s. Fluid) = 0 Hectare Meters
99 Pints (u.s. Fluid) = 0 Hectare Meters
100 Pints (u.s. Fluid) = 0 Hectare Meters
101 Pints (u.s. Fluid) = 0 Hectare Meters
102 Pints (u.s. Fluid) = 0 Hectare Meters
103 Pints (u.s. Fluid) = 0 Hectare Meters
104 Pints (u.s. Fluid) = 0 Hectare Meters
105 Pints (u.s. Fluid) = 0 Hectare Meters
106 Pints (u.s. Fluid) = 0 Hectare Meters
107 Pints (u.s. Fluid) = 0 Hectare Meters
108 Pints (u.s. Fluid) = 0 Hectare Meters
109 Pints (u.s. Fluid) = 0 Hectare Meters
110 Pints (u.s. Fluid) = 0 Hectare Meters
111 Pints (u.s. Fluid) = 0 Hectare Meters
112 Pints (u.s. Fluid) = 0 Hectare Meters
113 Pints (u.s. Fluid) = 0 Hectare Meters
114 Pints (u.s. Fluid) = 0 Hectare Meters
115 Pints (u.s. Fluid) = 0 Hectare Meters
116 Pints (u.s. Fluid) = 0 Hectare Meters
117 Pints (u.s. Fluid) = 0 Hectare Meters
118 Pints (u.s. Fluid) = 0 Hectare Meters
119 Pints (u.s. Fluid) = 0 Hectare Meters
120 Pints (u.s. Fluid) = 0 Hectare Meters
121 Pints (u.s. Fluid) = 0 Hectare Meters
122 Pints (u.s. Fluid) = 0 Hectare Meters
123 Pints (u.s. Fluid) = 0 Hectare Meters
124 Pints (u.s. Fluid) = 0 Hectare Meters
125 Pints (u.s. Fluid) = 0 Hectare Meters
126 Pints (u.s. Fluid) = 0 Hectare Meters
127 Pints (u.s. Fluid) = 0 Hectare Meters
128 Pints (u.s. Fluid) = 0 Hectare Meters
129 Pints (u.s. Fluid) = 0 Hectare Meters
130 Pints (u.s. Fluid) = 0 Hectare Meters
131 Pints (u.s. Fluid) = 0 Hectare Meters
132 Pints (u.s. Fluid) = 0 Hectare Meters
133 Pints (u.s. Fluid) = 0 Hectare Meters
134 Pints (u.s. Fluid) = 0 Hectare Meters
135 Pints (u.s. Fluid) = 0 Hectare Meters
136 Pints (u.s. Fluid) = 0 Hectare Meters
137 Pints (u.s. Fluid) = 0 Hectare Meters
138 Pints (u.s. Fluid) = 0 Hectare Meters
139 Pints (u.s. Fluid) = 0 Hectare Meters
140 Pints (u.s. Fluid) = 0 Hectare Meters
141 Pints (u.s. Fluid) = 0 Hectare Meters
142 Pints (u.s. Fluid) = 0 Hectare Meters
143 Pints (u.s. Fluid) = 0 Hectare Meters
144 Pints (u.s. Fluid) = 0 Hectare Meters
145 Pints (u.s. Fluid) = 0 Hectare Meters
146 Pints (u.s. Fluid) = 0 Hectare Meters
147 Pints (u.s. Fluid) = 0 Hectare Meters
148 Pints (u.s. Fluid) = 0 Hectare Meters
149 Pints (u.s. Fluid) = 0 Hectare Meters
150 Pints (u.s. Fluid) = 0 Hectare Meters
## how many Pints (u.s. Fluid) Into Hectare Meters
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| 4,096
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| 2.5625
| 3
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CC-MAIN-2023-06
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latest
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https://www.smore.com/pxkg
| 1,544,869,586,000,000,000
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|
## Week in Review: May 6th-10th
Another busy week, another week of learning! The students were able to work with multiplication and division (even with remainders!), and we got to see some cool Mr. Rogers videos about how things are made. Mrs. Bryant hatched 6 adorable chicks in her room, and they are so fun to hold...and listen to all day! They are very chirppy! On Monday, the student's were pleasantly surprised to discover if they had HW for the rest of the year (or not). We are looking forward to the next three weeks and all the fun things happening as the end of the year approaches!
Math: The last 3 weeks of school will be devoted to reviewing concepts and spiraling in older skills. This week we will be reviewing place value skills. Monday and Tuesday, the students will use number cards to create different numbers, describe them in different ways (standard, expanded, word form) and then make the biggest/smallest numbers possible with a set of 3-4 numbers. For some students, we will also talk about numbers up to one million!! Wednesday-Friday, the students will work with logic problems on a Hundreds Chart and explore interactive websites that involve place value. The students will also get a chance to work with some down and dirty/third grade/paper and pencil place value worksheets to get them used to seeing that format.
Social Studies: This week will be a continuation of our Economics Unit. Students will learn the differences between producers and consumers, and identify how science and technology have changed communication, transportation, and recreation.
Reading: Animal Research projects are still underway! Students are locating facts and details in nonfiction animal texts, then paraphrasing their learning to write an All About Animal Book. For those who finish early next week, they will be invited to make another All About Book for me to keep in my library to show future students!
Writing: In addition to working on the Animal Books, students will continue to practice writing mechanics, such as capitalization, punctuation, and abbreviations. We will also start to organize our Writing Portfolios for the end of the year.
Word Study: The last word sort for the year will be contractions. The students will spend the week sorting the cards, doing Word Hunts in their read to self books, and finally speed sort and gluing on the 17th. Some students will continue to practice plurals as not all passed the plurals quiz we had on the 10th.
## Attention: Library Books are DUE!!!
Please search high and low for missing library books and bring them in as soon as possible!!! Balances will be sent out if there are any missing books.
## End of Year Conference Forms
End of Year Conference Forms will be sent out on Thursday, May 16th. Please sign and return one copy and you may keep the other.
I am also having a few in-person conferences for the next two weeks. I will email you to remind you of the time we are meeting. You will receive your End of Year Conference Form on the day of your conference.
## End of Year Celebrations and Awards Assemblies
End of Year Celebrations will be on the last day of school, May 31st, at 1:00 pm. Parents are welcome to attend and take their child home after the party as long as they sign them out.
The End of Year Awards Assemblies for K-2 will be help on May 30th. More information to come...
## BOOK FAIR!
The Book Fair will be from Monday the 13th-Friday the 17th. Hours can be found on the Sommer Library Page. Please come by and get some books for summer!
## On the Calendar
Monday, May 13th: Book Fair opens
Thursday, May 16th: Sommer Choir performs "Aladdin" at 6:30 pm,
Chick-Fil-A Spirit Night,
Kindergarten Open house,
PTA Meeting at 11:30 am
Friday, May 17th: "Aladdin" at 6:30 pm
## High Fives
Students brought home their Science Journals on Thursday and Friday (the 9th and 10th). Please look over them with your student. I don't need them back as we are all done with Science for the year!
## Spotlight Stallion
This week's Spotlight Stallion is Grace Suh! Grace is an amazing student and friend! She also has the nicest handwriting! She will be leaving us on Tuesday, May 21st to go to Korea for the summer! While she is there, she will go to (gasp!) school, and visit family. We'll miss you, Grace!!
## Silent Auction Winner
The silent auction winner from our class is Spencer! Spencer will get to join me at Berry Cool on the 21st! We'll have a "cool" time and eat some yummy yogurt!
| 1,009
| 4,504
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| 3.109375
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CC-MAIN-2018-51
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latest
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https://www.jiskha.com/questions/1050031/a-roughly-spherical-asteroid-with-a-diameter-of-1-km-and-a-mean-density-of-4-g-cm-3-hits
| 1,606,343,343,000,000,000
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| 766,228,157
| 6,227
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# Astronomy
A roughly spherical asteroid with a diameter of 1 km, and a mean density of 4 g/cm^3, hits the Earth with a relative speed 1.5 times the orbital velocity of the Earth. Compute the amount of energy released upon the impact. Compare that with the energy released by a 10-Megaton thermonuclear bomb, which is about 4.2 x 10^27 erg. Contemplate the value of a search for Earth-crossing asteroids. Please be sure to enter your answer in units of ergs (the CGS unit of energy). Note: 1 Joule = 10^7 ergs.
Energy released upon impact:
1. 👍 0
2. 👎 0
3. 👁 1,779
1. 12435
1. 👍 0
2. 👎 0
2. 2.089*10^28
1. 👍 0
2. 👎 0
3. 2.089*10^28
1. 👍 0
2. 👎 0
4. yese
1. 👍 0
2. 👎 0
5. If the question is this : Please be sure to enter your answer in units of 10-Megaton bombs:
The response correct is 49738.09523, because 2.089*10^28 is in ergios.
1. 👍 3
2. 👎 0
6. 4/3*pi*50000^3 = sphere volume
4*1/1000= density passed to Kg.
(4/3*pi*50000^3)*(4*1/1000)= mass
Velocity 1.5*30km/s=45km/s*1000=45000m/s
Energy kinetic (Joule)(Kg.m/s)=1/2*mass*V^2=1/2*(4/3*pi*50000^3)*(4*1/1000)*(45000m/s)^2
(1/2*mass*V^2=1/2*(4/3*pi*50000^3)*(4*1/1000)*(45000m/s)^2)*10^7=Joule passed to Erg
Energy kinetic total=2.12058*10^28 Erg
Equivalent to 50489,88193 Thermonuclear Bomb of 10 Megatons exactly!
1. 👍 3
2. 👎 0
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Under standard conditions, the density of air is 1.293 kg/m3 and that of helium is 0.178 kg/m3. A spherical helium balloon lifts a basket plus cargo of mass 277.0 kg. What must be the minimum diameter of the spherical balloon?
1. ### Physics
A ball is tossed straight up from the surface of a small spherical asteroid with no atmosphere. The ball goes to a height ewual to the asteroid's radius and then falls straight down towardd the surface of the asteroid. What forces
2. ### physics
An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is 32kg, and the volume enclosed by the outside surface of the chunk is 0.020m^3. What is the radius of the spherical cavity? To
3. ### Algebra 1
Suppose a spherical asteroid has a radius of approximately 9.0 x 10^2m. Use the formula 4/3*pi r^3 to find the approximate volume of the asteroid. 2.57 * 10^9 m^3 1.07 * 10^4 m^3 2.54 * 10^10 m^3 4.51 * 10^10 m^3 I've worked
4. ### Physics
Speed of Asteroids With the same Radius? Two spherical asteroids have the same radius R . Asteroid 1 has mass M and asteroid 2 has mass 2M. The two asteroids are released from rest with distance 10R between their centers. What is
1. ### physics
In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after
2. ### physics
A spherical steel ball bearing has a diameter of 2.540 cm at 24°C. (a) What is the diameter when its temperature is raised to 82°C? (b) What temperature change is required to increase its volume by 1.1%? I got the first answer
3. ### Physics
A hollow spherical iron shell floats almost completely when submerged in water. The outer diameter is 50.0cm, and the density of iron is 7.87g/cm^3. Find the inner diameter.
4. ### College Physics w/Calculus
(a) What is the escape speed on a spherical asteroid whose radius is 570 km and whose gravitational acceleration at the surface is 3.1 m/s2? (b) How far from the surface will a particle go if it leaves the asteroid's surface with
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Triangular Distribution
a :
b :
c :
x :
P(x) :
Mean (μ) :
Median :
Mode :
Variance (σ2) :
Notes
1. Insert this widget code anywhere inside the body tag
2. Use the code as it is for proper working.
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HISTORY
# Triangular Distribution (PDF) Calculator
Triangular distribution probability (PDF) calculator, formulas & example work with steps to estimate the probability of maximim data distribution between two points a & b in statistical experiments. By using this calculator, users may find the probability P(x), expected mean (μ), median, mode and variance2) of trinagular distribution. This probability density function (pdf) calculator is featured to generate the work with steps for any corresponding input values to help beginners to learn how the input values are being used in such calculations of triangular distribution.
Notes
The below are the important notes to remember to supply the corresponding input values for this probability density function of triangular distribution calculator.
• The random variable x is the non-negative number value which must be greater than or equal to 0. The triangular distribution is evaluated at this random value x.
• The lower limit a is the positive or negative number which represents the initial point of curve.
• The upper limit b is the positive or negative number which represents the end point of curve.
• The middle point c is the positive number which represents the height of the distribution.
## Triangular Distribution & Formulas
Triangular distribution is a probability function used in statistics to analyze the behaviour of maximum likelihood of data between the interval or two points a and b. It's also known as lack of knowledge distribution has the base of (b - a) and the height (c) of 2/(b - a), often used in business simulations. It's one of a continous probability functions used in statistics & probability to characterize the subjective description of the data distribution. It looks symmetrical when c = (a + b)/2. In this calculation, the term P(x) represents the probability of maximum likelihood, mean (μ) represents the expected likelihood of data & σ2 represents the variation among the group of data.
Formula
The below formula is mathematical representation for Triangular probability density function may help users to know what are all the input parameters are being used in such calculations to characterize the data distribution. Users may use these below triangular distribution formulas for manual calculations and use this calculator to verify the results of manual calculations or generate complete work with steps.
## Solved Example Problems with Steps
The below are some of the solved examples with solutions for probability density function (pdf) of Triangular distribution to help users to know how to estimate the probabilty of maximum data distribution between the interval or two points.
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https://cracku.in/5-in-the-following-question-select-the-odd-letters-f-x-ssc-cgl-16-aug-shift-3
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Question 5
# In the following question, select the odd letters from the given alternatives.
Solution
(A) : J (-3 letters) = G (-3 letters) = D
(B) : NÂ (-2 letters) = L (-3 letters) = I
(C) : XÂ (-3 letters) = U (-3 letters) = R
(D) : QÂ (-3 letters) = N (-3 letters) = K
=> Ans - (B)
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https://fr.mathworks.com/matlabcentral/cody/problems/43042-convert-decimal-to-binary-and-then-generate-the-minimum-binary-it-can-with-jumbling/solutions/1005982
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Cody
# Problem 43042. Convert decimal to binary and then generate the minimum binary it can with jumbling
Solution 1005982
Submitted on 10 Oct 2016 by Joni Quintos
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 10; y_correct = 3; assert(isequal(your_fcn_name(x),y_correct))
ans = 3
2 Pass
x = 23; y_correct = 15; assert(isequal(your_fcn_name(x),y_correct))
ans = 15
3 Pass
x = 15; y_correct = 15; assert(isequal(your_fcn_name(x),y_correct))
ans = 15
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https://walkingonwaterfl.org/specific-comminution-ball-mill.html
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## comminution - an overview | sciencedirect topics
This family of models is the oldest of the comminution models and they continue to find widespread use (Morrell, 2014a). Energy-based models assume a relationship between energy input of the comminution device and the resultant effective particle size of the product. Many rely on the feed and product size distributions being self-similar; that is, parallel when cumulative finer is plotted in log-log space (Chapter 4). The energy input is for net power, that is, after correcting for motor efficiency and drive train mechanical losses. Typically, energy is measured as kWh t1 or Joules, depending on the model.
The oldest theory, Von Rittinger (1867), stated that the energy consumed in size reduction is proportional to the area of new surface produced. The surface area of a known weight of particles of uniform diameter is inversely proportional to the diameter, hence Von Rittingers law equates to:
As Lynch and Rowland (2005) note, the means to make measurements of energy and size necessary to validate the Von Rittinger and Kick models did not exist until the middle of the twentieth century when electrical motors and precision laboratory instruments became available. The literature from this period includes work by a group at the Allis Chalmers Company who were trying to calibrate Von Rittingers equation to industrial rod mills (Bond and Maxson, 1938; Myers et al., 1947).
Often referred to as the third theory, Bond (1952) stated that the energy input is proportional to the new crack tip length produced in particle breakage. Bond redefined his theory to rather be an empirical relationship in a near-final treatise (Bond, 1985). The equation is commonly written as:
where W is the energy input (work) in kilowatt hours per metric ton (or per short ton in Bonds original publications), Wi is the work index (or Bond work index) in kilowatt hours per metric ton, and P80 and F80 are the 80% product and feed passing sizes, in micrometers.
Solving Eq. (5.1b) for n=3/2 gives the same form as Eq. (5.4) with the constant 2 K ahead of the bracket. In effect the 2 K is replaced by (10Wi), which is convenient because Wi becomes equal to W in the case of grinding from a theoretical infinite feed size to 80% passing 100m. The Bond model remains the most widely used, at least for the conventional comminution equipment in use at the time Bond developed the model and calibrated it against industrial data. It is one reason that the 80% passing size became the common single point metric (mean) of a particle size distribution.
A modification of Eq. (5.1a,b) was proposed by Hukki (1962), namely substituting n by a function of particle size, f(x). This provoked debate over the size range that the three established models applied to. What can be agreed is that all the models predict that energy consumption will increase as product particle size (i.e., P) decreases. Typical specific energy values (in kWh t1) are (Morrell, 2014b): primary crushing (i.e., 1000-100mm), 0.1-0.15; secondary crushing (100-10mm), 1-1.2; coarse grinding (10-1mm), 3-3.5; and fine grinding (1-0.1mm), 10.
Fine grinding tests are sometimes expressed as a signature plot (He et al., 2010), which is an experimentally fitted version of Eq. (5.1a,b) with n=f(x). A laboratory test using a fine grinding mill is conducted where the energy consumption is carefully measured and a slurry sample is extracted periodically to determine the 80% passing size. The energy-time relationship versus size is then plotted and fitted to give (in terms of Eq. (5.1a,b)) a coefficient K and a value for the exponent f (x).
The problem that occurs when trying to solve Eq. (5.5) is the variable nature of the function g(x). A pragmatic approach was to assume M is a constant over the normal range of particle sizes treated in the comminution device and leave the variation in size-by-size hardness to be taken up by f(x). Morrell (2009) gives the following:
where Mi is the work index parameter related to the breakage property of an ore and the type of comminution machine, W is the specific comminution energy (kWh t1), P and F are the product and feed 80% passing size (m), and f(x) is given by (Morrell, 2006):
The parameter Mi takes on different values depending on the comminution machine: Mia for primary tumbling mills (AG/SAG mills) that applies above 750m; Mib for secondary tumbling mills (e.g., ball mills) that applies below 750m; Mic for conventional crushers; and Mih for HPGRs. The values for Mia, Mic, and Mih were developed using the SMC Test combined with a database of operating comminution circuits. A variation of the Bond laboratory ball work index test was used to determine values of Mib. This is similar to the approach Bond used in relating laboratory results to full scale machines. The methodology continues to be refined as the database expands (Morrell, 2010).
Morrell (2009) gave a worked example comparing the energy requirements for three candidate circuits to illustrate the calculations. Taking just the example for the fine particle tumbling mill serves that purpose here (Example 5.1).
From the Mi data the relevant value is Mib=18.8kWh t1. Noting it is fine grinding then the feed F80 is taken as 750m. Combining Eqs. (5.6) and (5.7) and substituting the values:W=18.84(106(0.295+750/1,000,000)750(0.295+750/1,000,000))=8.4(kWht1)
Comminution is a physical pretreatment method involving milling, grinding, and chipping [42]. Such mechanical methods are aimed at increasing the accessible surface area, as well as decreasing the cellulose crystallinity, by decreasing the particle size. Comminution methods should be utilized before the biomass is subjected to any other pretreatment, as it has been shown that fine particle size can greatly improve biohydrogen production. Chen etal. [43] demonstrated particle size effect (<0.297mm to >10mm) on thermophilic dark fermentative biohydrogen production from rice straw, without using any other pretreatment, and found that the highest cumulative biohydrogen produced was from particle sizes<0.297mm (191mL H2/L), while it was 57mL H2/L for particle sizes >10mm. Themain advantage of mechanical comminution is that it does not lead to inhibitor formation; however, the process is not able to remove lignin, and is not cost-effective when considering the energy consumption. It has been reported that energy consumption of these processes increases with the decreasing particle size and with a higher moisture content of the biomass [44].
Comminution is the reduction of solid material particle size by fracture via grinding, milling, or similar processes. Comminution techniques are usually employed to produce particulate nanomaterials (e.g., powder) from larger-sized or bulk materials. Due to simplicity and low cost, attrition or grinding with the assistance of milling media such as milling balls has been used for producing nanoparticles or nanopowder since the late 1990s [31]. However, grinding techniques suffer several problems including particle agglomeration, large residual internal stress, surface defects, and contaminations in the final products.
To overcome these problems, another comminution techniqueatomizationwas developed. Atomization produces solid or aerosol particles with reduced sizes by spraying molten material or material solution or suspension under conditions such that it breaks down and then solidifies as fine powder [32]. In a typical atomization process, a molten material passing through a nozzle scatters into fine droplets by a high-speed medium (e.g., gas or water) and then the droplets solidify to powder. Obviously, the atomization technique is highly efficient for preparing micron and submicron powders at industrial scales and recent development has enabled atomization to produce nanoparticles of sizes down to 20nm [33].
Comminution is needed for the liberation of low-grade ores so that the iron content can be upgraded by gangue removal. This necessitates grinding to such a size that the iron minerals and gangue are present as separate grains. But comminution is an expensive process and economics dictates that a compromise must be made between the cost of grinding and the ideal particle size.
Traditionally, grinding has been carried out using rod, ball, autogenous, or semiautogenous mills usually in closed circuit, that is, after grinding, the material is classified according to size with the undersized portion proceeding to the flotation circuit and the oversized portion being returned to the mill. The major benefit of fully autogenous grinding (AG) is the cost saving associated with the elimination of steel grinding media. In the last 20 years, more efficient grinding technologies, including high-pressure grinding rolls (HPGRs) for fine crushing and stirred milling for fine grinding, have provided opportunities to reduce operating costs associated with particle size reduction. A HPGR has been installed at the Empire Mine in the United States for processing crushed pebbles and its introduction has resulted in a 20% increase in primary AG mill throughput (Dowling et al., 2001). Northland Resources operates the Kaunisvaara plant in Sweden, treating magnetite ore with sulfur impurities in the form of sulfide minerals. The required P80 of the ore, in order to achieve adequate liberation, is 40m. This plant uses a vertical stirred mill after AG rather than a ball mill to achieve this fine grind size with an energy cost saving of 35% or better (Arvidson, 2013).
An important part of the comminution circuit is size classification. This can be accomplished with screens or cyclones or a combination of the two. Since cyclones classify on the basis of both particle size and specific gravity, cyclone classification in the grinding circuit directs coarse siliceous particles to the cyclone overflow. In a reverse flotation circuit, these coarser siliceous middlings can be recovered through increased collector addition but at the expense of increased losses of fine iron minerals carried over in the froth. However, if the required grind size is not so fine, then screening can be used instead of cycloning to remove the coarser particles for regrinding and, thus, produce a more closely sized flotation feed (Nummela and Iwasaki, 1986).
Comminution is used to reduce the particle size of biomass and increase accessible surface area, reducing crystallinity and degree of polymerization (DP), thus increasing the biomass biodegradability [4]. Typical mechanical comminution includes chipping and milling as shown in Table 2.1. Chipping is usually necessary to reduce the size of raw lignocellulosic biomass for further processing, for example, to make log to wood chips. Milling process that can be used for biomass pretreatment involves various types, such as ball milling, hammer milling, knife milling, vibro milling, tow-roll milling, colloid milling, wet-disk, and attrition millings [4]. As indicated by Agbor et al., harvesting and preconditioning can reduce lignocellulosic biomass from logs to coarse sizes (about 1050mm), and chipping can reduce size to 1030mm, while grinding and milling can reduce the size to 0.22mm [11]. Grinding and milling are also effective to alter the inherent ultrastructure of biomass, such as crystallinity and DP of cellulose [7]. As shown in Table 2.1, vibratory ball milling has been found to be more efficient in reducing cellulose crystallinity of spruce and aspen chips than ordinary ball milling [12]. Disk milling has been reported to achieve higher enzymatic hydrolysis yield than hammer milling [13]. The moisture content of biomass should be also considered in the selection of milling methods [4]. For dry biomass, hammer and knife millings are more suitable, while ball and vibro milling are more widely used and suitable for both dry and wet biomass [7,14].
Size reduction by milling also has been employed independently to improve the biodegradability of biomass conversion to biogas, bioethanol, and biohydrogen. When rice straw was subjected to disk and ball milling, the yields of glucose and xylose by enzymatic hydrolysis reached up to 89% (from 78%) and 54% (from 41%), respectively [15]. Glucose and xylose yield increased to 40% and 32% by wet disk milling for pretreated sugarcane bagasse and rice straw, respectively [16]. An obvious advantage of mechanical comminution for size reduction is that no inhibitors are generated during the process. However, various inhibitors are generated in chemical pretreatments, such as hydroxymethylfurfural (HMF), furfural and acetic acid, and so on, which are toxic to yeast or other microorganisms.
Comminution indicates a series of phenomena that occur at different stages of thermo-chemical conversion (combustion and gasification) processes in a FB reactor. All of them contribute to the size reduction of the fuel and char particles, together with the shrinkage related to devolatilization and chemical reactions. Primary fragmentation occurs immediately after injection of the mother fuel particles into the bed as a consequence of thermal stresses and internal pressures caused by the release of volatiles: it may break the particles into several coarse fragments and into a multitude of fines (Chirone et al., 1991). Attrition and secondary fragmentation of char particles occur in parallel with the chemical conversion of char and are both strongly assisted by this process. It continuously generates new asperities on the external surface of the particle, which are then mechanically abraded, so enhancing attrition (Arena et al., 1983). It also increases the size of internal pores and the weakening of the carbon bonds inside the particle, so that a secondary fragment is formed when a bridge is too weak to withstand the hydrodynamic forces on the char (Arena et al., 1996). There are also situations where all internal bridges may collapse suddenly, giving rise to a special type of secondary fragmentation, known as percolative fragmentation, which leads to the production of relatively fine sub-particles. Miccio et al. (1999) showed that percolative fragmentation is the main mechanism of fines generation during biomass char circulation in the freeboard of a FB gasifier, probably as a consequence of the highly porous char generated from these types of fuels. The unfragmented and fragmented char particles produce a further generation of very fine particles (about one order of magnitude finer than those from secondary fragmentation) due to the abrasive action of bed particles. All these different comminution phenomena can have a major or minor extension, or may not even be present, depending on the nature (and the size) of the mother fuel particle (Fig. 17.4).
Rod or bar mills suffer from a lot of wear if the product is very abrasive. This limitation gave rise to the autogenous mill where the grinding bodies are themselves pieces of ore. These pieces are a lot more efficient when they fall from a great height. This is why the diameter of these mills could achieve 11m and their length only 4.5m. They could require an installed power of up to 9,000kW.
by erosion, meaning detaching fine particles from average-sized grain surfaces or from boulder surfaces. As such, the production of fines could be significant, which could be an inconvenience (but not always).
Generally, ore density is much less than the density of steel. To activate fragmentation, we expect to have on the inside of the equipment a low proportion (5% by volume, for example) of steel balls. Wear on this low mass of balls has very little effect on the actual operation cost. Mills that operate in this way are called semi-autogenous mills.
The feed must have a suitable granulometry. In particular, there needs to be an adequate proportion of boulder sizes greater than 250mm. The shortage of large boulders must be compensated by the addition of balls and operation in a semi-autogenous mill.
Elements Sj (xj) of the fragmentation speed matrix give two maxima as a function of xj. The peak corresponding to the lowest values of xj deals with crushing grains, and the other peak with boulder abrasion. The analytical expression for Sj is very empirical and its value is between 5 and 100h1.
This zone corresponds to a variation from approximately 1 to 6 in grain size. To locate the average in the range of grain sizes, Stanley [STA 74] distinguishes abrasion from crushing by visually examining fragments. Crushing produces conchoidal fractures while abrasion produces rounded fragments.
The mass M of solid retained in the mill at the granulometry shows two peaks if we consider the frequencies mi (by mass) as a function of grain size. The peak that corresponds to boulders is 10 times more significant than the one for grains (see Stanley [STA 74, pp. 89 and 90]).
As such, Stanley [STA 74 pp. 81] is an example of the breaking matrix that is a combination of three matrices: erosion, crushing and intermediary. But he goes much further and replaces the erosion matrix with a combination of two matrices where, respectively:
Menacho [MEN 86] provides the distribution frequency for the product exiting the mill (his equation (21)). It is limited to adding the erosion and breaking effects. He calls the average residence time of the powder or of the pulp in the mill.
For an equal production and having a large equipment diameter, energy consumed is greater than that required in ball mills. On the other hand, the consumption of steel from wear on the few balls present is less but the wear on bearings is greater.
The autogenous mill could be fed with boulders reaching up to 800mm and give, in one single operation, a product size that does not exceed a few tens of millimeters (the reduction ratio could exceed 1,000).
The possible flow increases with the diameter D of the equipment as D2.82 and energy consumption as D2.62 only. This explains the tendency for these machines to be gigantic as they can treat up to 300 tonne.h1.
The autogenous mill is equipped with a classification device. In wet mode, the pulp mixture is evacuated through a grid and then treated by a hydro-cyclone or a sieve. Granules are recycled back into the equipment. In dry mode, an air current goes through the equipment and drives the fines fraction, which economizes the entire classification circuit.
Comminution in ball mills and vertical mills differs fundamentally. In a ball mill, size reduction takes place by impact and attrition. In a vertical mill the bed of material is subject to such a high pressure that individual particles within the bed are fractured, even though the particles are very much smaller than the bed thickness.
Early issues with vertical mills, such as narrower PSD and modified cement hydration characteristics compared with ball mills, have been resolved. One modification has been to install a hot gas generator so the gas temperature is high enough to partially dehydrate the gypsum.
For many decades the two-compartment ball mill in closed circuit with a high-efficiency separator has been the mill of choice. In the last decade vertical mills have taken an increasing share of the cement milling market, not least because the specific power consumption of vertical mills is about 30% less than that of ball mills and for finely ground cement less still. The vertical mill has a proven track record in grinding blastfurnace slag, where it has the additional advantage of being a much more effective drier of wet feedstock than a ball mill.
The vertical mill is more complex but its installation is more compact. The relative installed capital costs tend to be site specific. Historically the installed cost has tended to be slightly higher for the vertical mill.
Comminution of LC materials through a combination of chipping, grinding, and/or milling can be applied to improve their digestibility by reducing their particle size and crystallinity. The reduction in particle size leads to an increase in available specific surface and a reduction in the DP [178]. The process also causes shearing of the biomass. Mechanical pretreatment also results in substantial lignin depolymerization via the cleavage of uncondensed-aryl ether linkages [276]. All these combined factors lead to an increase in the total hydrolysis yield of the lignocellulose in most cases by 525% (depending on kind of biomass, kind of milling, and duration of the milling), and also reduces the technical digestion time by 2359% (thus an increase in hydrolysis rate) [277]. It was shown that, without any pretreatment, corn stover with sizes of 5375m was 1.5 times more productive than larger corn stover particles of 425710m [180].
Milling is considered to be an environmentally friendly pretreatment process. In these pretreatments, no hydrolysis or fermentation inhibitors, such as furfural or hydroxymethyl furfural, are produced [278]. However, size reduction of biomass is an energy-intensive process that warrants improvement to raise the energy efficiency involved in bioethanol production. The National Renewable Energy Laboratory indicated that size reduction required one-third of total energy inputs for biomass to ethanol conversion [279].
The use of mechanical chopping [280], hammer milling [281], grind milling [282], roll milling [283], vibratory milling [284], and ball milling [276] have proved successful as a low-cost pretreatment strategy.
Previously, milling reactors have been used as a means of pretreatment. The milling process has been studied prior to and in combination with enzymatic hydrolysis, where mechanical actions, mass transport, and enzymatic hydrolysis are performed simultaneously in order to improve the hydrolysis process. The attrition mill bioreactor [285] and the intensive mass transfer reactor, including ferromagnetic particles and two ferromagnetic inductors [286, 287], are two examples of these processes. Mais et al. [288] used a ball mill reactor for the pretreatment and hydrolysis of -cellulose and reported the number of ball beads as an effective parameter to improve enzymatic hydrolysis of -cellulose.
More recently, studies have shown that energy consumption for grinding biomass depends on initial particle size, moisture content, material properties, mass feed rate, and machine variables [281]. Optimum operating conditions may lead to reduced energy expended for size reduction [289]. Bitra et al. [290] determined the direct mechanical input energy for hammer mill and knife mill [289] size reduction of switchgrass, wheat straw, and corn stover over a range of mill operating speeds, screen sizes, and mass feed rates.
Although milling pretreatments are often described as high-energy requirement techniques, wet disk milling has been recently described as a potential feasible mechanical technique to treat rice straw [291]. Glucose and xylose yields by wet disk milling, ball milling, and hot-compressed water treatment were 78.5% and 41.5%, 89.4% and 54.3%, and 70.3% and 88.6%, respectively. The energy consumption of wet disk milling was lower than that of other pretreatments.
Sugarcane bagasse and straw were treated under comparative conditions by ball milling and wet disk milling techniques and their physical properties and susceptibility toward enzymatic hydrolysis and fermentation were evaluated [278]. Ethanol yields from total fermentable sugars by using a C6-fermenting strain reached 89.8% and 91.8% for bagasse and straw hydrolysates, respectively.
Ball milling pretreatment combined with the addition of dilute acid and alkali proved to be an effective processing to enhance the enzymatic hydrolysis efficiency of corn stover [292]. The results also indicated that the treatment effect of wet milling is better than that of dry milling.
Hardness using Mohs scale (see Appendix 1) is directly linked to abrasiveness. We can without any doubt show the practical qualification of this scale by using the following qualifiers given in Table5.1.
The connection between hardness and abrasiveness is the fact that a solid A that scratches another solid B has an index where IMA>IMB. It is the method that was used to organize different solids using the Mohs scale.
Related Equipments
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Most of us leave lamp on in our house at times once we aren’t also using them and never really think the the affect that it has actually on our bank account or our planet. Because of this, we chose to create an infographic showing exactly how much the would price to power a 100-watt lightbulb for an entire year. For this infographic we are using an average expense of electrical energy per kWh the \$0.11, and since a 100-watt lightbulb provides 0.1kWhs worth of electrical energy per hour, weve concluded that to power that lightbulb for 8,760 hrs (1 year) that would cost \$96.36. Simply ONE lightbulb could expense you virtually \$100 every year. Currently we don’t typically (hopefully) leave them ~ above 24/7, but I would certainly guess that you’ve obtained a lot more than simply one lightbulb in her home.
You are watching: 100 watt light bulb cost per hour
If money isn’t enough of a deterrent to rotate your lamp off when not in use, think about the environmental affect of leaving that lightbulb on. As shown below, we’ve calculated the quantity of coal, organic gas and also uranium (nuclear) that would be needed to create enough power to strength that one lightbulb because that a year. Examine out the results below and also think twice once leaving that light on, or making use of anything else that requires electrical power if you don’t require to.
### Conclusions
\$96.36 out of your very own pocket and 712 pounds of melted coal (or 145 pounds of natural gas or .0439 pounds the uranium) simply to leave one lightbulb on because that a year. Doesnt seem precious it does it? in ~ the finish of the day every we can do is shot to be together energy efficient as we can, yet we still require electricity.
### What room some better options than Coal, herbal Gas or Uranium?
Solar and also Wind energy. Together we verified in the infographic, if you had only 100 square metres of solar panels, to run at 20% efficiency, you might power that lightbulb for whole year in simply 8 days, 17 hours and also 14 minutes. And also since youve got an ext than one lightbulb, over there is plenty of time come power an ext throughout the year. If you provided a 1.5MW turbine, to run at 25% capacity, you could power that lightbulb for whole year in simply 2 hours and also 20 minutes.
See more: How To Disassemble A Pool Table, Heres What You Need To Know
When you take into consideration that every megawatt-hour of electrical power generated through a solar PV panel avoids more than 1,300 pounds of carbon dioxide, the save become more than just financial.
If friend have any kind of questions about the infographic, or around our services, feel complimentary to contact united state at anytime. We look forward to hearing indigenous you!
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# numpy.polynomial.chebyshev.chebmul¶
`numpy.polynomial.chebyshev.``chebmul`(c1, c2)[源代码]
Multiply one Chebyshev series by another.
Returns the product of two Chebyshev series c1 * c2. The arguments are sequences of coefficients, from lowest order “term” to highest, e.g., [1,2,3] represents the series `T_0 + 2*T_1 + 3*T_2`.
Parameters: c1, c2 : array_like 1-D arrays of Chebyshev series coefficients ordered from low to high. out : ndarray Of Chebyshev series coefficients representing their product.
Notes
In general, the (polynomial) product of two C-series results in terms that are not in the Chebyshev polynomial basis set. Thus, to express the product as a C-series, it is typically necessary to “reproject” the product onto said basis set, which typically produces “unintuitive live” (but correct) results; see Examples section below.
Examples
```>>> from numpy.polynomial import chebyshev as C
>>> c1 = (1,2,3)
>>> c2 = (3,2,1)
>>> C.chebmul(c1,c2) # multiplication requires "reprojection"
array([ 6.5, 12. , 12. , 4. , 1.5])
```
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# Does ${\bf L} \neq {\bf NL}$ imply ${\bf P} \neq {\bf NP}$?
This question is inspired by this question Implications between $\mathsf{L}=\mathsf{P}$ and $\mathsf{NL}=\mathsf{NP}$?
We do know that ${\bf L}$ could equal ${\bf NL}$ and at the same time ${\bf P}$ could be different from ${\bf NP}$. I was wondering whether or not the inequality between ${\bf L}$ and ${\bf NL}$ imply inequality between ${\bf P}$ and ${\bf NP}$. Or we just simply do not know...
• It is not known like many other unknown implications in complexity theory. I don't think this and the previous question are good questions unless there is some good motivation or justification why one might expect them. Nov 8 '13 at 0:04
• I think the "Implications between L = P and NL = NP" question is more reasonable than this one: given the description of NP in terms of a polynomial time algorithm to check a witness, a naive intuition would suggest the conjecture (L = P) ⇒ (NL = NP). The reason for this question being less reasonable is that there is not a simple, formal reason for suspecting it to be the case, but understanding that computational resources such as workspace and/or nondeterminism might not compose very simply is a hard-won intuition. Nov 8 '13 at 12:07
• For a good while, there were many basic (but precise) questions about quantum computing being asked on TCS.SE. While they weren't research level, I thought it worthwhile for them to be answered here, because if not here then where? Getting good answers for such technical subjects essentially involves taking a graduate course and/or knowing an expert personally; there's a heavy bias in the 'accessible' information on the 'Net towards being hasty transcripts of interactive protocols, or written for experts. This is also true of complexity theory: addressing such questions I think is good. Nov 8 '13 at 12:15
• there is a simple reason not to expect such a symmetry: in many ways space and time have much different behavior in complexity theory & there are many examples of this. this is seen eg in simply comparing the space/time hierarchy theorems. but on the other hand there are deep connections... another concept to remember here. imagine that both separations are proven with totally different proofs/techniques. but then in a disconnected sense one does imply the other (because they are both true!)
– vzn
Nov 8 '13 at 16:22
• @Kaveh: the solution on other StackExchange sites is usually to answer a few examples, and then once there are enough questions to demand a better response than ad hoc, to write a "canonical" question which captures the theme of the examples (and hopefully provides both intuition and resources for anyone asking similar questions). Nov 8 '13 at 18:12
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Q: 1. What hypothesis did Medvec & colleagues set out to test in their first study of the ‘near miss’ phenomenon?
A: To test the thoughts of human mind to see if it produces an emotional reaction such as taking the alternate route home and having an accident ?If only I had taken the usual route home, then this wouldn't of happened (Kowalski & Westen, [ 2011).? A tendency to imagine an alternate outcome known as counter factual thinking is the ability to think of alternative outcomes. To test the hypothesis of
the ?near miss? phenomenon ?that the ease of generating counter factuals could lead some individuals with more positive objective outcomes to actually feel worse about their situation compared to some whose objective situation was actually worse (Kowalski & Westen, 2011).? ]
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Confidence interval Concept - USMLE Forums
USMLE Forums Your Reliable USMLE Online Community Members Posts
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USMLE Forums Confidence interval Concept
USMLE Step 2 CK Forum USMLE Step 2 CK Discussion Forum: Let's talk about anything related to USMLE Step 2 CK exam
#1
07-15-2012
USMLE Forums Master Steps History: 1+CK+CS Posts: 1,866 Threads: 149 Thanked 2,249 Times in 1,096 Posts Reputation: 2269
Confidence interval Concept
Confidence intervals of the form are designed to estimate the
(a) sample mean.
(b) sample standard deviation.
(c) population mean.
(d) population standard deviation.
#2
07-15-2012
USMLE Forums Master Steps History: 1+CK+CS Posts: 646 Threads: 52 Thanked 800 Times in 278 Posts Reputation: 811
(c) population mean.
#3
07-16-2012
USMLE Forums Master Steps History: --- Posts: 1,366 Threads: 649 Thanked 702 Times in 420 Posts Reputation: 712
Quote:
Originally Posted by mbbs2010 (c) population mean.
can u explain ?
#4
07-27-2012
USMLE Forums Scout Steps History: 1 + CS Posts: 53 Threads: 10 Thanked 30 Times in 24 Posts Reputation: 40
I think it's Sample mean... Did you find the answer yet?!
#5
07-27-2012
USMLE Forums Master Steps History: 1+CK+CS Posts: 646 Threads: 52 Thanked 800 Times in 278 Posts Reputation: 811
Quote:
Originally Posted by tyagee can u explain ?
its basically cos u are finding the mean in ur test sample population..
and then using the CI to apply that data to the Population and predicting a mean value..
(for better explanation check kaplan biostats)
Tags Biostatistics-Epidemiology, Step-2-Questions
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# What Will 50k Be Worth In 20 Years?
## How much would \$1000 be in 50 years?
Value of \$1,000 from 1950 to 2021 \$1,000 in 1950 is equivalent in purchasing power to about \$11,081.08 today, an increase of \$10,081.08 over 71 years.
The dollar had an average inflation rate of 3.45% per year between 1950 and today, producing a cumulative price increase of 1,008.11%..
## What Will 100000 be in 30 years?
At the end of 20 years, your savings will have grown to \$320,714. You will have earned in \$220,714 in interest….Interest Calculator for \$100,000.RateAfter 10 YearsAfter 30 Years0.00%100,000100,0000.25%102,528107,7780.50%105,114116,1400.75%107,758125,12753 more rows
## What is the future value of \$50000?
\$50,000 Savings Calculator – Future ValueFuture Value\$253,205.82Total Invested\$50,000.00
## What will 60000 be worth in 20 years?
The first result (Reduced Amount) is \$33,220.55, which represents the value of \$60,000 in 20 years. The second result (Required Amount) is \$108,366.67, which is amount of money that you need in 20 years to match the purchasing power of \$60,000.
## What will \$1000 be worth in 20 years?
After 10 years of adding the inflation-adjusted \$1,000 a year, our hypothetical investor would have accumulated \$16,187. Not enough to knock anybody’s socks off. But after 20 years of this, the account would be worth \$118,874.
## What stock will double in 2020?
With that in mind, some stocks that could double in 2020 and fly higher in the back half of the year include: Beyond Meat (NASDAQ:BYND) Nio (NYSE:NIO) Canopy Growth (NYSE:CGC)
## How much will \$600000 be worth in 10 years?
Investing \$600,000. How much will \$600,000 be worth in the future?YearValue91,150,343101,236,619111,329,365121,429,06811 more rows
## How much money do I need to invest to make \$2000 a month?
To cover each month of the year, you need to buy at least 3 different stocks. If each payment is \$2000, you’ll need to invest in enough shares to earn \$8,000 per year from each company. To estimate how you’ll need to invest per stock, divide \$8,000 by 3%, which results in a holding value of \$266,667.
## What will 700k be worth in 10 years?
Investing \$700,000. How much will \$700,000 be worth in the future?YearValue10987,419111,021,979121,057,748131,094,76910 more rows
## How much money do I need to invest to make \$3000 a month?
By this calculation, to get \$3,000 a month, you would need to invest around \$108,000 in a revenue-generating online business. Here’s how the math works: A business generating \$3,000 a month is generating \$36,000 a year (\$3,000 x 12 months).
## How can I invest 50k wisely?
Here are ten ways to invest 50k:Individual Stocks. Individual stocks represent an investment in a single company. … Real Estate. … Individual Bonds. … Mutual Funds. … ETFs. … Invest with a Robo Advisor. … CDs. … Invest in Your Retirement.More items…
## Is having 50k saved good?
For most people, \$50,000 is more than enough to cover their living expenses for six full months. And since you have the money, I highly recommend you do so. On a different, and equally important note, when you set up an emergency fund, it should be separate from any other savings.
## Is 50k in savings good?
Saving \$50,000 per year is well ahead of most people, so first off congratulations. Your plan of action should be something like the following: Make an emergency fund. It should be multiple months’ worth of expenses.
## What will a million dollars be worth in 40 years?
Time magazine recently estimated that for a millennial with 40 years until retirement, \$1 million in savings is not likely sufficient. Taking into account 3% inflation over that time period, it would be worth just \$306,000 in today’s dollars.
## How much will \$2000 be worth in 20 years?
How much will an investment of \$2,000 be worth in the future? At the end of 20 years, your savings will have grown to \$6,414. You will have earned in \$4,414 in interest.
## What will 50000 be worth in 10 years?
This calculates what a \$50,000 investment will be worth in the future, given the original investment, annual additions, return on investment, and the number of years invested….Investing \$50,000. How much will \$50,000 be worth in the future?YearValue961,0861062,4601163,8661265,30211 more rows
## What will 400k be worth in 10 years?
How much will savings of \$400,000 be worth in 10 years if invested at a 8.00% interest rate?…\$400,000 at 8% Interest for 10 Years.YearAmount9\$799,60210\$863,5709 more rows
## What will a dollar be worth in 2040?
Future inflation is estimated at 3.00%. When \$5 is equivalent to \$9.55 over time, that means that the “real value” of a single U.S. dollar decreases over time. In other words, a dollar will pay for fewer items at the store….Buying power of \$5 in 2040.YearDollar ValueInflation Rate2040\$9.553.00%23 more rows
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# Systemtheorie und Regelungstechnik 1
The aim of the lecture is to teach the basics of systems theory and control engineering. We learn how to electrical, mechanical, optical, chemical or thermal processes described in the form of dynamic systems uniformly, analyze the characteristics of these systems, and learn methods for their targeted modulation by sensor-actuator systems (controllers) know.
We'll start with the modeling of dynamic systems, consider the analytic solution of linear time-invariant differential equations introduce the impulse and step response, analyze the step response of some commonly occurring systems, introduce the Laplace transform, learn different methods for the preparation of linear transfer functions to know, discuss stability of the closed loop and learn basic methods for the design of linear controller for single-input single-output (SISO) systems. The course consists of three hours per week lecture and a SWS exercises.
Lecture Schedule
The lecture will be held at the Technical Faculty of the University of Freiburg in Building 101, HS 00-026 from 22.4.-24.7.15.
• Wednesdays, 08: 15-10: 00
• Friday, 08: 15-10: 00
Script
The lecture is closely aligned with one created in the previous script:
Final exam
The final score is calculated at 100% of the final exam. The final exam is "closed-book", so there are no auxiliary utensils except pencils and blank paper allowed. It may, however, exactly one sheet of A4 paper be taken with two pages handwritten described as a formulary. The questions are partly multiple choice, as in the micro exams, some issues with text responses, as in the exercise sheets.
Authorization
Admission to the final exam will be obtained when the following criteria are met
• Existence of Exercise: Each exercise sheet is assessed by the acquired points in percent. In the nine best of eleven Übungsblätter an average of at least 50% of the points must be achieved.
Example: A student has acquired 2 times 90% and 9 times 40% of the points, thus the average of the best nine 51.1%. The exercise is passed by.
• Existence of micro exams: The micro exams are evaluated by the acquired points in percent. In the three best of four micro exams at least 50% of the points must be achieved on average.
Lecture recordings
The Systems and Control lecture was recorded in SS2013:
No Date part 1 Part 2
1 07/06/2013 Vorlesung1_1 Vorlesung1_2
2 14/06/2013 Vorlesung2_1 Vorlesung2_2
3 21/06/2013 Vorlesung3_1 Vorlesung3_2
4 06/28/2013 Vorlesung4_1 Vorlesung4_2
5 05/07/2013 Vorlesung5_1 Vorlesung5_2
6 29/07/2013 Vorlesung6_1 Vorlesung6_2
7 30.07.2013 Vorlesung7_1 Vorlesung7_2
8th 31/07/2013 Vorlesung8_1 Vorlesung8_2
9 01/08/2013 Vorlesung9_1 Vorlesung9_2
10 02/08/2013 Vorlesung10
Exercises
Mikroklausuren
Exam preparation
Old exams for preparing the current exam are available here
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# Tagged Questions
Questions about explicit functions expressed in terms of independent variables, i.e. parameters. Including approaches for rewriting functions using different parameterizations.
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### How to find the real values of the parameter a, so that the inequation doesn't have positive solutions? [migrated]
The inequation is the following: (x^2 - 6ax + 2x - 5a - 1) / (x+a+1) < 0 If we write this inequation like a/b <0, I may say that for a=0, I found that X1=3a -1-sqrt(9a^2-a+2) X2=3a ...
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### Plot multiple functions with different but overlapping intervals
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### Simple grouping equations [closed]
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### Adding a fading color only to 3D curves going outside a cylinder
I'm having a problem with color shades on multi curves drawn with the Mathematica 7.0 code below. All the curves with both their endpoints inside the unit cylinder (green circle on the picture below) ...
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Cm To M
4000 cm to m4000 Centimeters to Meters
cm
=
m
How to convert 4000 centimeters to meters?
4000 cm * 0.01 m = 40.0 m 1 cm
A common question is How many centimeter in 4000 meter? And the answer is 400000.0 cm in 4000 m. Likewise the question how many meter in 4000 centimeter has the answer of 40.0 m in 4000 cm.
How much are 4000 centimeters in meters?
4000 centimeters equal 40.0 meters (4000cm = 40.0m). Converting 4000 cm to m is easy. Simply use our calculator above, or apply the formula to change the length 4000 cm to m.
Convert 4000 cm to common lengths
UnitLengths
Nanometer40000000000.0 nm
Micrometer40000000.0 µm
Millimeter40000.0 mm
Centimeter4000.0 cm
Inch1574.80314961 in
Foot131.233595801 ft
Yard43.7445319335 yd
Meter40.0 m
Kilometer0.04 km
Mile0.0248548477 mi
Nautical mile0.0215982721 nmi
What is 4000 centimeters in m?
To convert 4000 cm to m multiply the length in centimeters by 0.01. The 4000 cm in m formula is [m] = 4000 * 0.01. Thus, for 4000 centimeters in meter we get 40.0 m.
Alternative spelling
4000 Centimeter to Meter, 4000 Centimeter in Meter, 4000 cm to m, 4000 cm in m, 4000 Centimeters to m, 4000 Centimeters in m, 4000 Centimeters to Meter, 4000 Centimeters in Meter, 4000 cm to Meter, 4000 cm in Meter, 4000 cm to Meters, 4000 cm in Meters, 4000 Centimeter to Meters, 4000 Centimeter in Meters
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You are not logged in.
Discussion: All Topics Topic: Pythagorean Theorem Related Item: http://mathforum.org/mathtools/tool/1361/
Post a new topic to the tool: The Pythagorean Theorem 1 discussion
<< see all messages in this topic < previous message | next message >
Subject: phythagorean aplet Author: Aisyah Date: Feb 16 2004
i agree with jon. the aplet is really great. but i have problem turning the
parts to fit in the big square. how do you turn the pieces?
thanks for any suggestion given.
On Feb 15, 2004, jon wrote:
http://www.ies.co.jp/math/products/geo2/applets/pythasvn/pythasvn.html
This was a really cool Pythagorean Applet. So many times, students can get away
without actually knowing some theory behind the formula they're using. They can
see the variables in such an equation as the Pythagorean Theorem as universal
and unchanging (i.e. They may be confused when asked to find the length of the
third side of some triangle XYZ instead of ABC).
It seems that this applet has two uses.
1.) It helps students see that it doesn't matter how big or small the other two
squared sides will be, the sum of the two will alway equal that third side
squared.
2.) The squaring of the sides correlates to taking the area of a square whose
length is the side of one of the triangle's sides. By repiecing the smaller
squares into the hypotenuses' square, the students really get to visually see
that the squares of the two sides equals the square of the hypotenuse.
Some drawbacks of this:
1.) Perhaps students will not see the connect between the areas of the squares
and the lengths of sides of the triangles. (Although I thought that the diagram
at top was a pretty good explanation.)
2.) A small drawback of this is that the third side is fixed. I guess it
doesn't really matter that much that you're not able to change the third side
too, but it'd be nice to have a fully movable triangle.
-However, this might be able to spark some discussion, as to why the third
side is constant or why you only have certain squares that fit this criterion.
It might help student see the relationship of the 2 sides to the hypotenuse.
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# Excel - IF Function Problem - Expert Solution
Question description:
This user has given permission to use the problem statement for this blog.
I have a schedule for my colleagues and I would like excel to automatically remove their break time from the allocated hours I have inputted for that week, based on the amount of hours they have set within that day. So if they're doing 4 hours, they get no break. If they're doing between 5 to 8 hours, they get 30 minutes. 9 hours or more they get 45 minutes.
Solved by S. J. in 52 mins
This is the chat thread from the real Excelchat help session. It contains no private user information.
User 22/09/2018 - 12:40
Hello?
Excelchat Expert 22/09/2018 - 12:40
Welcome to Excelchat, I see that your question is about formulas
User 22/09/2018 - 12:40
Sorry the questions I was given I cannot seem to answer
User 22/09/2018 - 12:40
So I thought it ended
User 22/09/2018 - 12:40
Yes that is correct :)
Excelchat Expert 22/09/2018 - 12:40
Before we get started, this is a reminder that our policy is 1 problem per session with additional Q&A on that problem as time allows.
Excelchat Expert 22/09/2018 - 12:41
Let me ask you a couple of quick questions to make sure I fully understand your problem.
Excelchat Expert 22/09/2018 - 12:41
Can you tell me the high-level goal you are trying to achieve?
User 22/09/2018 - 12:41
Not sure I understand the question?
User 22/09/2018 - 12:42
Oh, as in my main goal I'm trying to achieve
Excelchat Expert 22/09/2018 - 12:42
Yes
Excelchat Expert 22/09/2018 - 12:42
Do u have a sample document so I can address to your problem?
User 22/09/2018 - 12:42
Basically, I have a small little bar and I no longer pay for my staff breaks
User 22/09/2018 - 12:42
But, I don't want it to take it off what I allocate my hours for that week
User 22/09/2018 - 12:43
Yeah sure
Excelchat Expert 22/09/2018 - 12:43
Thank you, I'll wait for it
Excelchat Expert 22/09/2018 - 12:44
Based on what you've shared, you need a formula using if statement
User 22/09/2018 - 12:45
Okay
Excelchat Expert 22/09/2018 - 12:45
I understand what the question is, I’ll start working on the solution and will be updating you as I work.
User 22/09/2018 - 12:45
Okay lovely, thank you so much
Excelchat Expert 22/09/2018 - 12:45
This should take me 30mins or less.
User 22/09/2018 - 12:46
As I've copied it across I don't think the formulas would've copied correctly
Excelchat Expert 22/09/2018 - 12:47
Can you just add the file using the paper click at the right of your chat box
User 22/09/2018 - 12:47
There we go, no it's got it.
User 22/09/2018 - 12:47
The allocation is at the bottom
Excelchat Expert 22/09/2018 - 12:47
Paper clip*
User 22/09/2018 - 12:49
Two seconds
Excelchat Expert 22/09/2018 - 12:50
Use the paper clip at the right or your chat box to send the file
Excelchat Expert 22/09/2018 - 12:50
So I can fully understand your goal
User 22/09/2018 - 12:50
Yeah just removing names quickly
Excelchat Expert 22/09/2018 - 12:52
Ok, wait
Excelchat Expert 22/09/2018 - 12:53
is you c9 value equal to 0 or 8?
Excelchat Expert 22/09/2018 - 12:55
what is your initial value in target replen?
Excelchat Expert 22/09/2018 - 12:56
Hello
Excelchat Expert 22/09/2018 - 12:56
Are you still there?
Excelchat Expert 22/09/2018 - 01:23
Since you lack in information, this is the best thing that I can only do.
[Uploaded an Excel file]
Excelchat Expert 22/09/2018 - 01:25
The formula states that if the value in S (Cell C4) is equal to or lesser than 4 it will give a value which is 0.
Excelchat Expert 22/09/2018 - 01:27
While if the value in S (Cell C4) is equal to or lesser than 8 it will give a value which is 30.
Excelchat Expert 22/09/2018 - 01:28
Else, it will give a value which is 45.
Excelchat Expert 22/09/2018 - 01:28
That goes the same with the other cells.
Excelchat Expert 22/09/2018 - 01:29
I also give additional table if you are using a number of hours.
Excelchat Expert 22/09/2018 - 01:31
Thanks for coming to Excelchat. Feel free to leave any comments or feedback. Have a nice day.
This is the output file from the real Excelchat help session:
This is an example of the expert help you can get. It contains no private user information.
Get instant expert help with Excel and Google Sheets
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Sari la conținut
# Newton
Pentru alte sensuri, vedeți Newton (dezambiguizare).
Nume Newton Vizualizarea definiției Newtonului - (kg•m)/s2 Newton forță[1].mw-parser-output cite.citation{font-style:inherit}.mw-parser-output .citation q{quotes:"„""”""«""»"}.mw-parser-output .citation .cs1-lock-free a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-limited a,.mw-parser-output .citation .cs1-lock-registration a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-subscription a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration{color:#555}.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration span{border-bottom:1px dotted;cursor:help}.mw-parser-output .cs1-ws-icon a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Wikisource-logo.svg/12px-Wikisource-logo.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output code.cs1-code{color:inherit;background:inherit;border:inherit;padding:inherit}.mw-parser-output .cs1-hidden-error{display:none;font-size:100%}.mw-parser-output .cs1-visible-error{font-size:100%}.mw-parser-output .cs1-maint{display:none;color:#33aa33;margin-left:0.3em}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-format{font-size:95%}.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-left{padding-left:0.2em}.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-right{padding-right:0.2em}[2] F Isaac Newton 1 Newton Modifică date / text
Newtonul, având simbolul N, este unitatea de măsură pentru orice tip de forță în Sistemul Internațional de Unități de Măsură (SI). Unitatea de măsură newton este numită în onoarea matematicianului Isaac Newton.
## Definire
Un newton este forța necesară pentru a imprima o accelerație de 1 m/s2 unui corp cu masa de 1 kg. Ca atare, newtonul se mai poate scrie (kg•m)/s2.[3]
În altă ordine de idei, legea a doua a mișcării a lui Newton spune că forța exercitată de către un corp este direct proporțională cu accelerația acelui corp:[4]
${\displaystyle F=ma}$
unde ${\displaystyle m}$ reprezintă masa corpului ce are aceelerația ${\displaystyle a}$. Astfel, un newton se poate descrie și cu ajutorul unităților de măsură pentru masă, lungime și timp (ultimele două fiind componente ale accelerației) în Sistemul Internațional, mai exact:
${\displaystyle 1\ {\text{N}}=1{\frac {{\text{kg}}\cdot {\text{m}}}{{\text{s}}^{2}}}}$.
## Note
1. ^ 7.6.4, Quantities and units—Part 1: General, p. 19
2. ^
3. ^ „Newton - unit of measurement”, Encyclopedia Britannica, accesat în
4. ^ „Table 3. Coherent derived units in the SI with special names and symbols”. The International System of Units (SI). International Bureau of Weights and Measures. . Arhivat din original la .
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# Thread: Natural gas meter question
1. Regular Guest
Join Date
Oct 2001
Posts
42
Post Likes
I am adding a natural gas appliance at my home and trying to determine if I need a larger capacity meter. My total load will be around 500,000 BTU (500 cubic feet per hour) if everything is on.
The specs on my meter (Equimeter R415) are listed as:
415 cfh at 1/2" w.c. diff
900 cfh at 2" w.c. diff
I know w.c. stands for water column but how do I know if I have a 1/2" diff or a 2" diff and what exactly is measured by this diff?
I've tried asking my gas company but they are no help.
2. blk
Professional Member
Join Date
Jan 2005
Posts
475
Post Likes
Your existing meter is a Rockwell 415 which is good for 415,000btuh. We calculate our meter requirements based on 75% of the total load, so using this calculation your existing meter would be sufficient for a total load of 500,000btuh. (500,000x75% = 375,000). The differential is the pressure drop across the meter, which in most cases is 1\2" w.c. If your meter has a regulator upstream of it the outlet pressure on a standard residential installation is is approx.7.5"w.c - the .5"w.c. across the meter leaving you with a house line pressure of 7"w.c. Depending on where you are located you may have a higher house line pressure. Here in Alberta you are allowed up to 2psig line pressure inside of a residential one & two family dwelling, but most are 7"w.c.I'm very surprise that your local gas company would not provide you with any meter information>
3. Regular Guest
Join Date
Oct 2001
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Thanks for your reply. Could you go into a little more detail on the w.c. differential? What causes the w.c. diff to be 0.5" vs 2" and how do you make it 2" so it can supply up to 900 cfh?
I do have a regulator upstream of the meter if that matters.
Thanks
4. blk
Professional Member
Join Date
Jan 2005
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475
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Did you get that off of the the front rating plate of the meter? I don't have that particular meter in my vehicle so I'm not sure exactly what you are looking at, but if your upstream regulator is reducing the incoming pressure from psig to inches w.c. then all you need to know is that your differential across the meter will be .5"w.c which will allow that specific meter to supply 415,000 btuh capacity max.
5. blk
Professional Member
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One other thing to remember is that your existing house line must be of sufficient size to handle any additional loads you are adding to the system. Meter capacity cannot compensate for an undersized house line !
6. Regular Guest
Join Date
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Originally posted by blk
One other thing to remember is that your existing house line must be of sufficient size to handle any additional loads you are adding to the system. Meter capacity cannot compensate for an undersized house line !
The new appliance is 200,000 BTU and it is connected with 70 feet of 1-1/4" pipe.
Originally posted by blk
Did you get that off of the the front rating plate of the meter? I don't have that particular meter in my vehicle so I'm not sure exactly what you are looking at, but if your upstream regulator is reducing the incoming pressure from psig to inches w.c. then all you need to know is that your differential across the meter will be .5"w.c which will allow that specific meter to supply 415,000 btuh capacity max.
Yes, that is off the rating plate on the meter.
Am I looking at this the wrong way? Is the rating plate telling me that "you will have this much pressure drop at this much cfh load (0.5" @ 415cfh, 2" @ 900 cfh)? So that I can still run it at higher than 415,000 BTU but that the appliance will recieve less than 7.0" w.c. pressure? And is the 2" wc rating just sort of the accepting max limit since the pressure would be down to 5" or so by then?
[Edited by jo8243 on 02-20-2006 at 05:28 PM]
7. blk
Professional Member
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To be perfectly honest with you jo8243, I don't recall seeing two differential ratings on that particular meter so I can't answer your question with any degree of certainly, perhaps someone else on here may be able to help...Mike3 where are you? Getting back to your original post, your existing meter will be adequate for your needs.HTH
8. Professional Member
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Hi
Just saw your post..I was not able to get a meter capacity chart from my company today..Probably tomorrow..In the mean time I found one from Sierra Pacific.. We set meter for commercial or one appliance only to match or exceed the load...However, with res we use a 60% diversity factor meaning if the total connected load is say 500,000 we would probably set a 310 which will pass around 525 I believe. The meter doesn't need to pass full load because it is assumed that all appl will not be on at the same time.. Of course you must take each situation into account. If it was heating only probably should be close to capacity..
We set our regulators when new to 7.5"wc and then expect a drift to 7.0" after it has been installed for a while. After that we leave at 7". The "diff" is just as blk explained. The 2" diff is because of the large load being consumed on "that size" meter
http://www.sierrapacific.com/service...es/GM0030G.pdf
As you can see from this chart a 425 will pass 620cfh
[Edited by mike3 on 02-21-2006 at 07:30 PM]
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Thanks Mike! Great info.
I'm satisfied now that my meter will be enough. It says the 2" point is 900 cfh. Since I'll only be drawing 500 cfh max that will be fine.
Out of curiousity though, why can't you just turn the regulator pressure up to 8" or so to overcome the meter loss if it is too high?
Thanks
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Originally posted by jo8243
Thanks Mike! Great info.
I'm satisfied now that my meter will be enough. It says the 2" point is 900 cfh. Since I'll only be drawing 500 cfh max that will be fine.
Out of curiousity though, why can't you just turn the regulator pressure up to 8" or so to overcome the meter loss if it is too high?
Thanks
It becomes a billng issue..
11. blk
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Thanks for the detailed explanation Mike. It should be noted as well that meter capacity and differential loss on the meter rating plates are indicated in cubic ft/air because "air" is the test medium used to calibrate & test natural gas meters , so that rating will be slightly less than the actual natural gas capacity of that same meter..Specific gravity of air being 0 and natural gas being 0.6 . Also meters have a maop( maximum allowable operating pressure)which in the case of the particular meter you have can be either 10psig or 25psig if I'm not mistaken, therefore if you have a greater presure thru the meter other than 7.5"w.c let's say 2psig your meter capacity will also increase. This same meter then becomes what we call a PFM or Pressure Factor Meter which again is a billing issue. Now that you are totally confused......
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Originally posted by blk
Thanks for the detailed explanation Mike. It should be noted as well that meter capacity and differential loss on the meter rating plates are indicated in cubic ft/air because "air" is the test medium used to calibrate & test natural gas meters , so that rating will be slightly less than the actual natural gas capacity of that same meter..Specific gravity of air being 0 and natural gas being 0.6 . Also meters have a maop( maximum allowable operating pressure)which in the case of the particular meter you have can be either 10psig or 25psig if I'm not mistaken, therefore if you have a greater presure thru the meter other than 7.5"w.c let's say 2psig your meter capacity will also increase. This same meter then becomes what we call a PFM or Pressure Factor Meter which again is a billing issue. Now that you are totally confused......
Nicely stated..I believe our resi meters have maop of 5psig..We don't have any 2# systems around here,although its my understanding parts of our service area do.
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Originally posted by mike3
Originally posted by jo8243
Thanks Mike! Great info.
I'm satisfied now that my meter will be enough. It says the 2" point is 900 cfh. Since I'll only be drawing 500 cfh max that will be fine.
Out of curiousity though, why can't you just turn the regulator pressure up to 8" or so to overcome the meter loss if it is too high?
Thanks
It becomes a billng issue..
Well how do they keep people from cheating them then? You can access the regulator control if I am not mistaken.
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# 9.4: Which Analysis Should You Conduct?
One of the most important concept that you need to understand is deciding which analysis you should conduct for a particular situation. To help you to figure out the analysis to conduct, there are a series of questions you should ask yourself.
1. Does the problem deal with mean or proportion?
Sometimes the problem states explicitly the words mean or proportion, but other times you have to figure it out based on the information you are given. If you counted number of individuals that responded in the affirmative to a question, then you are dealing with proportion. If you measured something, then you are dealing with mean.
2. Does the problem have one or two samples?
So look to see if one group was measured or if two groups were measured. If you have the data sets, then it is usually easy to figure out if there is one or two samples, then there is either one data set or two data sets. If you don’t have the data, then you need to decide if the problem describes collecting data from one group or from two groups.
3. If you have two samples, then you need to determine if the samples are independent or dependent.
If the individuals are different for both samples, then most likely the samples are independent. If you can’t tell, then determine if a data value from the first sample influences the data value in the second sample. In other words, can you pair data values together so you can find the difference, and that difference has meaning. If the answer is yes, then the samples are paired. Otherwise, the samples are independent.
4. Does the situation involve a hypothesis test or a confidence interval?
If the problem talks about "do the data show", "is there evidence of", "test to see", then you are doing a hypothesis test. If the problem talks about "find the value", "estimate the" or "find the interval", then you are doing a confidence interval.
So if you have a situation that has two samples, independent samples, involving the mean, and is a hypothesis test, then you have a two-sample independent t-test. Now you look up the assumptions and the formula or technology process for doing this test. Every hypothesis test involves the same six steps, and you just have to use the correct assumptions and calculations. Every confidence interval has the same five steps, and again you just need to use the correct assumptions and calculations. So this is why it is so important to figure out what analysis you should conduct.
## Data Sources:
Center for Disease Control and Prevention, Prevalence of Autism Spectrum Disorders - Autism and Developmental Disabilities Monitoring Network. (2008). Autism and developmental disabilities monitoring network-2012. Retrieved from website: http://www.cdc.gov/ncbddd/autism/doc...nityReport.pdf
Flanagan, R., Rooney, C., & Griffiths, C. (2005). Fatal poisoning in childhood, england & wales 1968-2000. Forensic Science International, 148:121-129, Retrieved from http://www.cdc.gov/nchs/data/ice/fat...ning_child.pdf
Gettler, L. T., McDade, T. W., Feranil, A. B., & Kuzawa, C. W. (2011). Longitudinal evidence that fatherhood decreases testosterone in human males. The Proceedings of the National Academy of Sciences, PNAS 2011, doi: 10.1073/pnas.1105403108 Length of NZ rivers. (2013, September 25). Retrieved from http://www.statsci.org/data/oz/nzrivers.html
Lim, L. L. United Nations, International Labour Office. (2002). Female labour-force participation. Retrieved from website: http://www.un.org/esa/population/pub...ty/RevisedLIMp aper.PDF
Olson, K., & Hanson, J. (1997). Using reiki to manage pain: a preliminary report. Cancer Prev Control, 1(2), 108-13. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/9765732
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# Physics Unit 7
To measure the temperature of a substance, you use a device called a ___________. thermometer
If you transfer thermal ________ from one object to another, you are transferring heat. energy
You pour very hot water into a teakettle with cool water; after 30 seconds, the teakettle's water is hot because ________ equilibrium was reached. thermal
To determine the _________ of your friend's fever, you have him hold a thermometer in his mouth for one minute. temperature
On a _______ thermometer, water freezes at 273.15, whereas water boils at 373.15. Kelvin
Water freezes at 0 degrees and boils at 100 degrees on a ________ thermometer. Celsius
To obtain an equal temperature in kelvins, you add _______ degrees to Celsius degrees. 273.15
To obtain an equivalent temperature in Celsius for a Fahrenheit reading, use the formula _____________. C=5/9(F-32)
You record a temperature of 15 degrees C; to record the temperature in degrees Fahrenheit, you would use the formula ____________. F=9/5(C+32)
A previous belief was that heat was a substance called _______ that flowed from one object to another. caloric
According to the _________ theory, scientists thought that heat was only in flammable materials. phlogiston
The __________ theory of heat says that heat is the transfer of kinetic energy from one substance to another. molecular
Molecules moving at a high speed have higher ______ energy than molecules moving at a low speed. kinetic
_______ creates heat when two surfaces rub against each other. Friction
A unit used to express heat on air conditioners, refrigerators, and heaters is the _________. BTU
A measurement of heat used in dietary matters is the ________. calorie
The calorie, a non-SI heat unit, is equivalent to ________. 4.186 J
Most materials _________ when they are heated. expand
Engineers use ________________ to figure out how much a bar will lengthen if it is heated. linear thermal expansion
What is the formula for finding a solid's linear thermal expansion coefficient? aL=1/L X ^L/^T
The amount a solid's volume expands when the solid is exposed to heat is defined as the _____________. volumetric expansion
The formula to determine the coefficient of volume expansion is _________. ^V=aV X V(^T)
The units specific heat capacity is expressed in are ___________. kJ/kg X k
The amount of energy needed to raise 1 kilogram of a substance a temperature of one degree Celsius, is the substance's ___________. specific heat capacity
Frequently, a substance's specific heat capacity is compared to that of ________. water
Though a metal, mercury is a liquid at room temperature (22 degrees C); at -38.83 degrees C, mercury becomes a solid and at 356.73 degrees C, mercury boils, transforming into a gas. At -38.83 degrees C and 356.73 degrees C, mercury is undergoing ________. phase changes
The formula for specific heat capacity is __________. Cp=Q/m^T
The amount of heat you add or take away from a substance to create a phase change is ________. latent heat
If you add heat to ice to create liquid water, the heat energy is called ____________. heat of fusion
If you heat water to create steam, the heat energy is _________. heat of vaporization
Heat may be transferred by three methods: convection, conduction, and _________. radiation
During the _________ of heat, heat moves between a substance's molecules. conduction
The movement of excited molecules in a current during heat transfer is _________. convection
Traveling electromagnetic waves transfer heat in the method of heat transfer called ________. radiation
Since copper readily conducts electricity, it will also readily conduct ________. heat
Thermal energy always ends up in an area of lower temperature after traveling there from an area of ________. higher temperature
Number the following according to how tightly bound the molecules are in each of the substances, with 1 being most tightly and 3 being the least: gaseous tin, liquid tin, solid tin 1. solid tin 2. liquid tin 3. gaseous tin
What states that energy cannot be created nor destroyed in any process, it can only be changed from one form into another and will be conserved? the first law of thermodynamics
What states that energy systems are always increasing in entropy? the second law of thermodynamics
What states that energy always exists in a system meaning that absolute zero can never be reached? the third law of thermodynamics
What states that if two systems are in thermal equilibrium with a third, they are in thermal equilibrium with each other? the zeroth law of thermodynamics
What explains that energy always moves from a higher level of organization to a lower level of organization? entropy
Energy can be converted from thermal energy to mechanical energy and no energy will be created or destroyed, but will be conserved according to what? how thermal energy is converted to mechanical energy according to the first law of thermodynamics
If a mechanical system converts heat to mechanical energy, some energy will be lost because of inefficiency in the system according to what? how thermal energy is converted to mechanical energy according to the second law of thermodynamics
The law of ________ is another name for the second law of thermodynamics. entropy
An ideal machine would convert _________ of the fuel inputted to the machine into mechanical energy. 100%
The formula Q=^U+W displays the relationship between _______ and work. heat
The formula W=Fd is the formula for ________. work
The formula PEi+KEi=PEf+KEf is the formula for the _______ law of thermodynamics. first
The formula PEi+KEi=PEf+KEf+^U is the formula for the ________ law of thermodynamics. second
The division of T(hot)-T(cold)/T(hot) will give you a machine's ______________. ideal efficiency
What temperature is the limit beyond which no heat may be transferred out of a substance? absolute zero or 0 K
The change in internal temperature is the difference between the formulas for the _______ and _______ laws of thermodynamics. first;second
A closed system conserves all of its _______, but a certain amount of it will become unusable and dissipated as heat. energy
A detrimental effect is exerted by ______ in energy-conversion systems. entropy
The concept of entropy is that systems tend toward a state of ________. disorder
You measure the temperature of liquid caramel candy with degrees Celsius, Celsius being a temperature ________. scale
When you measure the temperature of liquid caramel candy on a thermometer, the caramel candy's thermal energy or _________ is transferred to the thermometer. heat
Measuring liquid caramel candy's temperature with a thermometer requires time because you need to wait until _________ has been reached between the liquid candy and the thermometer. thermal equilibrium
Ice at 32 degrees blankets your porch one winter day. To melt it, you pour boiling water at 212 degrees upon the ice. What scale are you using? Fahrenheit
When caramel liquid candy reaches a temperature at 236 degrees F on a thermometer, you need to remove it from the heat source. What formula would you use to obtain an equal Celsius reading? C=5/9(F-32)
To translate a Celsius temperature reading into a Kelvin temperature, what formula would you use? T=C+273.15
The temperature at which all molecular motion will theoretically cease is _____________. absolute zero
The idea that heat is the transfer of kinetic energy from one substance to another is the mainstay of the ________ theory of heat. molecular
In an earlier theory, scientists thought ________ was a substance that flowed from one object to another. caloric
Before, scientists believed that _________ was heat found only in flammable materials. phlogiston
To create a phase change in a material, you need to add or subtract an amount of heat called _________. latent heat
If you rub a rock on a gravel road, you create heat in the form of _________. friction
To change a substance from a solid to a liquid, you add latent heat called _________. heat of fusion
To convert a substance from liquid to gas, you add a latent heat called ________. heat of vaporization
In convection, the excited molecules of a substance move in a _______. current
In the heat transfer method of radiation, traveling ________ waves are responsible for transferring heat. electromagnetic
The law of ________ may be used for the second law of thermodynamics. entropy
The difference between the formulas for the first and second laws of thermodynamics is the change in _______ temperature. internal
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# Exponent: Definition & Properties Video
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Jennifer Beddoe
An exponent tells you how many times to use a number in a multiplication problem. This lesson will define the properties of exponents and how to interpret them. There will also be a quiz at the end of the lesson.
## Definition of an Exponent
An exponent is a number that indicates how many times you should multiply a number to itself. For example, 4^2 means multiply 4 by itself 2 times, or 4 * 4 = 16. Therefore, 4^2 = 16.
The exponent is written as a superscript number after the number being multiplied, which is called the base. In the example we just looked at, the number 2 is the exponent and the 4 is the base.
There are 4 types of exponents:
1. Positive exponents
2. Negative exponents
3. Zero exponents
4. Rational exponents
## Positive Exponents & Examples
Positive exponents are exponents that are positive numbers. There is no special trick to working with positive exponents, just multiply the base to itself the number of times indicated by the exponent.
Here are a couple of examples of positive exponents:
3^5 = 3 * 3 * 3 * 3 * 3 = 243
7^3 = 7 * 7 * 7 = 343
## Negative Exponents & Examples
Negative exponents are negative numbers that are being used as exponents. For example, 2^-4.
A negative exponent is simplified by placing the base (with the exponent) in the denominator of a fraction with 1 as the numerator.
2^-4 = 1 / (2^4) = 1/16
Here is how that works:
• 2^4 = 16
• 2^3 = 8
• 2^2 = 4
• 2^1 = 2
For each step as the exponent is decreased, the solution is divided by 2. The pattern continues as you keep decreasing the exponent.
• 2^0 = 1
• 2^-1 = 1 / 2
• 2^-2 = 1 / 4 (1 / 2^2)
• 2^-3 = 1 / 8 (1 / 2^3)
This rule applies to all negative exponents. Here's some more examples:
3^-4 = 1 / 3^4 = 1/81
x^-7 = 1 / x^7
## Zero Exponents & Examples
An expression with 0 as the exponent is equal to 1. It does not matter what the base is, if the exponent is 0 the simplification is 1. Now, let's look at some specific examples:
25^0 = 1
b^0 = 1
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Enter the weight lifted and the number of repetitions to estimate your 1RM for deadlift.
The Deadlift Calculator is a tool used by weightlifters, trainers, and fitness enthusiasts to estimate their one-repetition maximum (1RM) for the deadlift exercise. The deadlift is a compound strength training exercise that involves lifting a loaded barbell from the ground to hip level, targeting multiple muscle groups, including the lower back, glutes, hamstrings, and core. Calculating the 1RM helps individuals determine their maximum lifting capacity and adjust their training programs accordingly.
### Input Parameters:
The Deadlift Calculator typically requires the user to input their performance data, such as the weight lifted and the number of repetitions completed at that weight.
### 1RM Estimation Formula:
The most common formula used in Deadlift Calculators is the Epley Formula or similar formulas (such as Brzycki, Lander, Lombardi, etc.) to estimate the 1RM based on the input data.
### Output:
Once the necessary data is entered, the Deadlift Calculator computes and displays an estimated 1RM, which represents the maximum weight an individual should be able to lift for a single repetition based on the provided information.
The Epley Formula is widely used and is as follows:
1RM=Weight lifted×(1+Number of repetitions30)1RM=Weight lifted×(1+30Number of repetitions)This formula estimates the 1RM by considering the weight lifted and the number of repetitions performed at that weight.
Summary:
The Deadlift Calculator is a valuable tool for weightlifters and fitness enthusiasts, providing an estimated one-repetition maximum for the deadlift exercise. By inputting the weight lifted and the number of repetitions completed, users can quickly determine their potential maximum lifting capacity without actually attempting a true 1RM lift. This information helps in designing personalized and effective training programs by establishing appropriate intensity levels for strength training.
In summary, the Deadlift Calculator streamlines the process of estimating 1RM, allowing individuals to optimize their workouts, track progress, and set realistic training goals for improving strength and performance in deadlifts.
Remember, while calculators provide estimates, actual capabilities can vary due to factors like fatigue, form, and individual differences in strength and muscle recruitment.
Always prioritize safety and proper form when performing strength exercises like the deadlift, and consider consulting a fitness professional for personalized guidance.
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# If 34 of an estate is worth Rs. 90,000 then the value of 23 of the same will be
A
Rs. 60,000
B
Rs. 65,000
C
Rs. 70,000
D
Rs. 80,000
Video Solution
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Step by step video, text & image solution for If 3/4 of an estate is worth Rs. 90,000 then the value of 2/3 of the same will be by Maths experts to help you in doubts & scoring excellent marks in Class 7 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If 45 of an ornament be worth Rs 16800, then the value of 37 of it is -
Aa) Rs 90000
Bb) Rs 9000
Cc) Rs 72000
Dd) Rs 21000
• Question 2 - Select One
## If 1315 of an estate be worth Rs. 390, then 35 of it is-
ARs. 320
BRs. 270
CRs. 450
DRs. 324
• Question 3 - Select One
## If opening capital is Rs 60,000, drawings Rs 5,000, capital introduced during the period Rs 10,000, closing capital Rs 90,000. The value of profit earned during the period will be:Option1 Rs 20,000 Option2 Rs 25,000 Option3 Rs 30,000 Option4 Rs 40,000
ARs 20,000
BRs 25,000
CRs 30,000
DRs 40,000
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## Eight-dimensional octonions may hold the clues to solve fundamental mysteries.
• Physicists discover complex numbers called octonions that work in 8 dimensions.
• The numbers have been found linked to fundamental forces of reality.
• Understanding octonions can lead to a new model of physics.
Is our reality, including its forces and particles, based on the strange properties of numbers with eight dimensions called "octonions"? A physicist thinks so, having found a way to expand 40-year-old research to reach surprising new directions.
First, a brief history of numbers.
Regular numbers that we are familiar with in our everyday life can be paired up in a special way to create "complex numbers," which act like coordinates on a two-dimensional plane. This was discovered in 16h-century Italy by the mathematician Gerolamo Cardano. As explains Natalie Wolchover of Quanta Magazine, you can perform operations on complex numbers like adding, subtracting, multiplying and dividing by "translating and rotating positions around the plane."
An Irish mathematician by the name of William Rowan Hamilton discovered in 1843 that if you pair the complex numbers in a certain way, they can form 4-D "quaternions." He was apparently so excited about figuring out that formula, that he immediately carved it into the Broome Bridge in Dublin. Not to be outdone, John Graves, a friend of Hamilton's who was a lawyer and math whiz, showed that quarternions can be paired up to become "octonions" – numbers that can assume coordinates in an abstract 8-dimensional (8-D) space.
John Graves.
Each type of numbers has been utilized extensively in the development of modern physics, with complex numbers used in quantum mechanics and even the quaternions employed in Albert's Einstein's special theory of relativity.
What hasn't been completely understood and put to work – the octonions, usually represented by the capital letter O and whose multiplication rules are encoded in a triangular diagram called the Fano plane (that looks like something the Freemasons would devise).
A mnemonic for the products of the unit octonions using the Fano plane.
The mystery of these numbers has led to speculation among researchers that they have a special purpose and can eventually explain the deeper secrets of the universe. In an email interview with Quanta Magazine, the particle physicist Pierre Ramond from the University of Florida explained that "Octonions are to physics what the Sirens were to Ulysses."
In 1973, Murat Günaydin, the then-Yale-graduate student (now professor at Penn State) and his advisor Feza Gürsey, discovered that there is an unexpected link between octonions and the strong force that keeps quarks together in an atomic nucleus. Günaydin continued his research quite outside the mainstream, looking at connecting the numbers to such ideas as string theory and M-theory.
In 2014, Cohl Furey, a graduate student at the University of Waterloo, Canada, built on Günaydin's work by finding a new use for the hard-to-imagine numbers. She devised an octonionic model that includes both the strong and electromagnetic forces. Now a postdoc in UK's University of Cambridge, Furey generated a series of results that link the octonions to the Standard Model of particle physics, in work that has been praised by other scientists. She has "taken significant steps toward solving some really deep physical puzzles," said Shadi Tahvildar-Zadeh, a mathematical physicist at Rutgers University.
Others, like the noted string theorist and Imperial College London professor Michael Duff, are more reserved, excited about her work but saying it's "hard to say" yet if it will become "revolutionary."
Furey is undeterred by working in a currently obscure field, thinking of her research as a "process of collecting clues," as she explained in an interview.
She published a paper in May 2018's The European Physical Journal C, where she consolidated several of her findings, looking to complete the standard model of particle physics and find the rightful place in our understanding of the world for the octonions.
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https://www.coursehero.com/file/6627566/Market-for-Refurbished-Washing-Machines-1/
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{[ promptMessage ]}
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Market for Refurbished Washing Machines-1
# Market for Refurbished Washing Machines-1 - Market for...
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Market for Refurbished Washing Machines Each person demands a washing machine at the price next to their name. Jose \$700 Richard \$600 Amy \$500 Anthony \$400 Nathan \$300 Darrell \$200 Geoffrey \$100 Each person is willing to supply a washing machine at the price next to their name. Susan \$100 Betty \$200 Cathy \$300 Darva \$400 Emily \$500 Francis \$600 Germaine \$700 The market equilibrium is shown in the following figure. \$100 \$700 3 \$400 Q P
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At the equilibrium: x Consumer Surplus (CS) is \$600 o Jose’s CS is \$300 o Richards CS is \$200 o Amy’s CS is \$100 x Producer Surplus (PS) is \$600 o Susan’s PS is \$300 o Betty’s PS is \$200 o Cathy’s PS is \$100 x Total Welfare (TW)=CS+PS=\$1,200
Example 1: Consider a price ceiling of \$200. At this price producers will only be willing to supply 2 washing machines, but consumers will demand 5. \$200 \$100 \$700 3 5 2 \$400 Q P Assuming that only the consumers with the highest valuations get to buy refurbished wash machines. x CS is \$900 o Jose’s CS is \$500 o Richards CS is \$400 x PS is \$100 (All of the PS is Susan’s) x TW is \$1000 Comparing the TW at the equilibrium and at the price ceiling we see that the deadweight loss resulting form the price ceiling is \$200. This deadweight loss results from the fact the quantity is too low under the price ceiling.
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## DUTdSpdV Jidni
The chemical potential p, has an important function in the system's thermodynamic behavior analogous to pressure or temperature. A temperature difference between two bodies determines the tendency of heat to pass from one body to another while a pressure difference determines the tendency for bodily movement. We will show that a difference in chemical potential can be viewed as the cause for chemical reaction or for mass transfer from one phase to another. The chemical potential p, greatly facilitates the discussion of open systems, or of closed systems that undergo chemical composition changes.
10.1.2 The Gibbs Free Energy, G
Calculation of changes dU of the internal energy of a system U requires the estimation of changes of its entropy S, volume V, and number of moles . For chemical applications, including atmospheric chemistry, it is inconvenient to work with entropy and volume as independent variables. Temperature and pressure are much more useful. The study of atmospheric processes can therefore be facilitated by introducing other thermodynamic variables in addition to the internal energy U. One of the most useful is the Gibbs free energy G, defined as
Differentiating (10.11)
and combining (10.12) with (10.10), one obtains k dG= -SdT + Vdp + ^^drii (10.13)
Note that one can propose using (10.13) as an alternative definition of the chemical potential:
Both definitions (10.9) and (10.14) are equivalent. Equation (10.13) is the basis for chemical thermodynamics.
For a system at constant temperature (dT = 0) and pressure (dp = 0)
For a system under constant temperature, pressure, and with constant chemical composition (drii = 0), dG = 0, or the system has a constant Gibbs free energy. Equation (10.13) provides the means of calculating infinitesimal changes in the Gibbs free energy of the system. Let us assume that the system under discussion is enlarged m times in size, its temperature, pressure, and the relative proportions of each component remaining unchanged. Under such conditions the chemical potentials, which do not depend on the overall size of the system, remain unchanged. Let the original value of the Gibbs free energy of the system be G and the number of moles of species i, n,. After the system is enlarged m times, these quantities are now mG and mn,. The change in Gibbs free energy of the system is
and the changes in the number of moles are
k (10.18) i=l
Equation (10.18) applies in general and provides additional significance to the concept of chemical potential. The Gibbs free energy of a system containing k chemical compounds can be calculated by
• + V-knk that is, by summation of the products of the chemical potentials and the number of moles of each species. Note that for a pure substance
tii and thus the chemical potential is the value of the Gibbs free energy per mole of the substance. One should note that both (10.13) and (10.18) are applicable in general. It may appear surprising that T and p do not enter explicitly in (10.18). To explore this point a little further, differentiating (10.18)
and combining with (10.13), we obtain
k —SdT +Vdp = J2n' (10.21) ¡=1
This relation, known as the Gibbs-Duhem equation, shows that when the temperature and pressure of a system change there is a corresponding change of the chemical potentials of the various compounds.
### 10.1.3 Conditions for Chemical Equilibrium
The second law of thermodynamics states that the entropy of a system in an adiabatic (dQ = 0) enclosure increases for an irreversible process and remains constant in a reversible one. This law can be expressed as dS> 0 (10.22)
Therefore a system will try to increase its entropy and when the entropy reaches its maximum value the system will be at equilibrium. One can show that for a system at constant temperature and pressure the criterion corresponding to (10.22) is dG < 0 (10.23)
or that a system will tend to decrease its Gibbs free energy. For a proof the reader is referred to Denbigh (1981).
Consider the reaction A—B, and let us assume that initially there are nA moles of A and «b moles of B. The Gibbs free energy of the system is, using (10.18)
If the system is closed nr = «a + "b = constant
FIGURE 10.1 Sketch of the Gibbs free energy for a closed system where the reaction A ^ B takes places versus the mole fraction of A.
and this equation can be rewritten as
where nA nA
the mole fraction of A in the system. Let us assume that the Gibbs free energy of the system is that shown in Figure 10.1. If at a given moment the system is at point K, (10.23) suggests that dG < 0 and G will tend to decrease, so xA will increase, B will be converted to A, and the system will move to the right.
If the system at a given moment is at point M, once more dG < 0, so the system will move to the left (A will be converted to B). At point L, the Gibbs free energy is at a minimum. The system cannot spontaneously move to the left or right because then the Gibbs free energy would increase, violating (10.23). If the system is forced to move, then it will return to this equilibrium state. Therefore, for a constant T and p, the point L and the corresponding composition is the equilibrium state of the system and (xA)L the corresponding mole fraction of A. At this point dG = 0. Let us consider a general chemical reaction aA + bB
which can be rewritten mathematically as aA + bB — cC — dD = 0 (10.24)
The Gibbs free energy of the system is given by
If dnA moles of A react, then, according to the stoichiometry of the reaction, they will also consume (b/a)dnA moles of B and produce (c/a)dnA moles of C and (d/a)dnA moles of D. The corresponding change of the Gibbs free energy of the system at constant T and p is, according to (10.15)
-= Ha dnA + ~ Hb dnA ~ ~ He dnA - - pD dnA a a a b c d \
At equilibrium dG — 0 and therefore the condition for equilibrium is b c d
Let us try to generalize our conclusions so far. The most general reaction can be written as
where k is the number of species, A, , participating in the reaction, and v, the corresponding stoichiometric coefficients (positive for reactants, negative for products). One can easily extend our arguments for the single reaction (10.24) to show that the general condition for equilibrium is
k I>H<=() (10.28) ¿=1
This is the most general condition of equilibrium of a single reaction and is applicable whether the reactants and products are solids, liquids, or gases.
If there are multiple reactions taking place in a system with k species k
k y, v,„ a,.=o i=i the equilibrium condition applies to each one of these reactions and therefore at equilibrium k
where vty is the stoichiometric coefficient of species i in reaction j (there are n reactions and k species).
Reactions in the H2SO4-NH3-HNO3 System Let us assume that the following reactions take place:
2NH3(g) + H2S04(g) - (NH4)2S04(s) SBfe(g) + HNOa(g) ^ N^NO^s)
Calculate the chemical potential of sulfuric acid as a function of the chemical potentials of nitric acid and the two solids.
At equilibrium the chemical potentials of gas-phase NH3, and H2SC>4 and solids (NH4)2S04 and NH4NO, satisfy
From the second equation |iniij = Pniuno, — Hhno3- Substituting this into the first, we find that
M^HjSOj — ^ PHNO, ~ UNHjnC), + ^{NKjinSOi (10.32)
Determination of the equilibrium composition of this multiphase system therefore requires determination of the chemical potentials of all species as a function of the corresponding concentrations, temperature, and pressure.
10.1.4 Chemical Potentials of Ideal Gases and Ideal-Gas Mixtures
In this section we will discuss the chemical potentials of species in the gas, aqueous, and aerosol phases. In thermodynamics it is convenient to set up model systems to which the behavior of ideal systems approximates under limiting conditions. The important models for atmospheric chemistry are the ideal gas and the ideal solution. We will define these ideal systems using the chemical potentials and then discuss other definitions.
The Single Ideal Gas We define the ideal gas as a gas whose chemical potential \i(T.p) at temperature T and pressure p is given by
where is the standard chemical potential defined at a pressure of 1 atm and therefore is a function of temperature only. R is the ideal-gas constant. Pressure p actually stands for the ratio (p/1 atm) and is dimensionless. This definition suggests that the chemical potential of an ideal gas at constant temperature increases logarithmically with its pressure. Differentiating (10.33) with respect to pressure at constant temperature, we obtain
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## Wednesday, 1 May 2013
### Common Logical Questions Asked in a Technical Interview
Common Logical Questions Asked in a Technical Interview
Well, if you are going for a technical interview in any programming language, in any stream, irrespective of your expertise and experience, you could be asked very simple logical questions which you must know. Interviewer can ask you very basic and famous C programs to check your logic which you had done in very early stage of your life like Fibonacci Series, Factorial Number, Palindrome Number, Prime Numbers etc. You should be able to write down the logic of these basic programs in a minute otherwise it might be very embarrassing for you and will put direct negative impact on the interviewer.
All your domain knowledge, programming language hold could be negated if you get stuck in these problems. So I must advise you to have a quick look on these algorithms before you appear in any technical interview.
Fibonacci Series: To print Fibonacci series up to n numbers
int main()
{
int n, first = 0, second = 1, next, c;
printf("Enter the number of terms\n");
scanf("%d",&n);
printf("First %d terms of Fibonacci series are :-\n",n);
for ( c = 0 ; c < n ; c++ )
{
if ( c <= 1 )
next = c;
else
{
next = first + second;
first = second;
second = next;
}
printf("%d\n",next);
}
return 0;
}
Factorial Program: To find out factorial of a number
int main()
{
int c, n, fact = 1;
printf("Enter a number to calculate it's factorial\n");
scanf("%d", &n);
for (c = 1; c <= n; c++)
fact = fact * c;
printf("Factorial of %d = %d\n", n, fact);
return 0;
}
Palindrome Number: To check whether a given number is palindrome or not?
int main()
{
int n, reverse = 0, temp;
printf("Enter a number to check if it is a palindrome or not\n");
scanf("%d",&n);
temp = n;
while( temp != 0 )
{
reverse = reverse * 10;
reverse = reverse + temp%10;
temp = temp/10;
}
if ( n == reverse )
printf("%d is a palindrome number.\n", n);
else
printf("%d is not a palindrome number.\n", n);
return 0;
}
Prime Number: To check whether a given number is prime number or not?
int isPrime(int);
int main(){
int num,prime;
printf("Enter a positive number: ");
scanf("%d",&num);
prime = isPrime(num);
if(prime==1)
printf("%d is a prime number",num);
else
printf("%d is not a prime number",num);
return 0;
}
int isPrime(int num){
int i=2;
while(i<=num/2){
if(num%i==0)
return 0;
else
i++;
}
return 1;
}
Please do add some more basic programs or algorithms you need to know before going to a technical interview.
1. yes, this is especially important in Java EE, Spring, etc interview. LOL :)
2. You have to know math for this because they want optimal algorithm. For Prime number, you need to use sqrt(num) instead of num/2....
3. You have to know mathematical for this because they want maximum criteria.programs to examine your reasoning which you had done in very beginning on of your lifestyle like Fibonacci Sequence,These are all very beneficial suggestions
4. Awesome guide! I'm really fortunate to stumble upon this post. I'll have my interview on Tuesday and for the past interviews I've experienced, all those 10 are always being asked! :) Thank you so much! It seems like you've experienced a lot of interviews too already ;)
5. Here is a prime number algorithm implementation that is in C#. It is more efficient and doesn't fail on invalid input.
class Program
{
static void Main(string[] args)
{
uint n;
string s;
Console.Write("Enter a positive integer: ");
if (UInt32.TryParse(s = Console.ReadLine(), out n))
Console.WriteLine(n + (IsPrime(n) ? " is prime" : " is not prime"));
else
Console.WriteLine('"' + s + "\" is not a positive integer");
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http://freetofindtruth.blogspot.com/2016/03/35-42-44-cavs-44th-win-of-season-later.html
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## Saturday, March 5, 2016
### 35 42 44 | The Cavs 44th win of the season, later today, March 5, 2016, against the Boston Celtics
The Cavs will earn their 44th win today, and I expect King James will have a numerological significant game.
3/5/2016 = 3+5+20+16 = 44 (King James)
3/5/2016 = 3+5+2+0+1+6 = 17 (NBA)
3/5/16 = 3+5+16 = 24
The date 3/5 with the numerology of '44' couldn't be anymore fitting for 'King James'.
King = 2+9+5+7 = 23
James = 1+1+4+5+1 = 12/21
King James = 35/44
Let us not forget the NBA Finals start on a date with '44' numerology this season, and the Golden State Warriors, who are in their 44th year of existence in Golden State, will be the likely opponent for the Cavs, who seek to break the 51-year drought of no Cleveland pro sports championships.
6/2/2016 = 6+2+20+16 = 44
Warriors, in 44th year of existence, like opponent for Cavs in NBA Finals
Further, notice that today's game comes 42-days before the start of the NBA Playoffs, April 16, 2016.
Remember, 42 connects to 'LeBron James', 'Cleveland Cavs' and 'NBA Finals'.
LeBron = 3+5+2+9+6+5 = 30
James = 1+1+4+5+1 = 12/21
LeBron James = 42/51
Cleveland = 3+3+5+4+5+3+1+5+4 = 33
Cavs = 3+1+4+1 = 9/18
Cleveland Cavs = 42/51
NBA = 5+2+1 = 8
Finals = 6+9+5+1+3+1 = 25/34
NBA Finals = 33/42
The span of 1 month and 11 days is also interesting, in regards to 'The NBA Finals'.
Let us not forget that Cleveland is home of the 216 area code.
666... 6x6x6 = 216
The game is in Ohio.
Let us close out with a little Boston Celtics decode and see why they're a fitting opponent for March 5. Please understand that the games are scripted this way, and those who create the schedule are very familiar with the language of gematria.
Boston = 2+6+1+2+6+5 = 22/31 (Basketball = 22/31)
Celtics = 3+5+3+2+9+3+1 = 26/35
Boston Celtics = 48/66 (LeBron = 66)
Boston = 2+15+19+20+15+14 = 85 (Basketball = 85)
Celtics = 3+5+12+20+9+3+19 = 71
Boston Celtics = 156
Notice how the number '33' connects to the name 'Boston Celtics'.
The Celtics most famous player of all-time, Larry Bird, wore the #33.
Bird = 2+9+18+4 = 33
Bird, #33
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- : unit = () - : unit = () h : heuristic = - : unit = () APPLY CRITERIA (Marked dependency pairs) TRS termination of: [1] minus(x,0) -> x [2] minus(s(x),s(y)) -> minus(x,y) [3] quot(0,s(y)) -> 0 [4] quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) [5] plus(0,y) -> y [6] plus(s(x),y) -> s(plus(x,y)) [7] minus(minus(x,y),z) -> minus(x,plus(y,z)) [8] app(nil,k) -> k [9] app(l,nil) -> l [10] app(cons(x,l),k) -> cons(x,app(l,k)) [11] sum(cons(x,nil)) -> cons(x,nil) [12] sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l)) [13] sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k))))) [14] plus(0,y) -> y [15] plus(s(x),y) -> s(plus(x,y)) Sub problem: guided: DP termination of: END GUIDED APPLY CRITERIA (Graph splitting) Found 6 components: { --> } { --> --> --> --> } { --> } { --> } { --> } { --> --> --> --> } APPLY CRITERIA (Subterm criterion) APPLY CRITERIA (Choosing graph) Trying to solve the following constraints: { minus(minus(x,y),z) >= minus(x,plus(y,z)) ; minus(s(x),s(y)) >= minus(x,y) ; minus(x,0) >= x ; quot(0,s(y)) >= 0 ; quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) ; plus(0,y) >= y ; plus(s(x),y) >= s(plus(x,y)) ; app(nil,k) >= k ; app(cons(x,l),k) >= cons(x,app(l,k)) ; app(l,nil) >= l ; sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) ; sum(cons(x,nil)) >= cons(x,nil) ; sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) ; Marked_quot(s(x),s(y)) >= Marked_quot(minus(x,y),s(y)) ; } + Disjunctions:{ { Marked_quot(s(x),s(y)) > Marked_quot(minus(x,y),s(y)) ; } } === TIMER virtual : 10.000000 === Entering poly_solver Starting Sat solver initialization Calling Sat solver... === STOPING TIMER virtual === === TIMER real : 10.000000 === === STOPING TIMER real === Sat solver returned Sat solver result read === STOPING TIMER real === === STOPING TIMER virtual === constraint: minus(minus(x,y),z) >= minus(x,plus(y,z)) constraint: minus(s(x),s(y)) >= minus(x,y) constraint: minus(x,0) >= x constraint: quot(0,s(y)) >= 0 constraint: quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) constraint: plus(0,y) >= y constraint: plus(s(x),y) >= s(plus(x,y)) constraint: app(nil,k) >= k constraint: app(cons(x,l),k) >= cons(x,app(l,k)) constraint: app(l,nil) >= l constraint: sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) constraint: sum(cons(x,nil)) >= cons(x,nil) constraint: sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) constraint: Marked_quot(s(x),s(y)) >= Marked_quot(minus(x,y),s(y)) APPLY CRITERIA (Subterm criterion) ST: Marked_minus -> 1 APPLY CRITERIA (Subterm criterion) APPLY CRITERIA (Choosing graph) Trying to solve the following constraints: { minus(minus(x,y),z) >= minus(x,plus(y,z)) ; minus(s(x),s(y)) >= minus(x,y) ; minus(x,0) >= x ; quot(0,s(y)) >= 0 ; quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) ; plus(0,y) >= y ; plus(s(x),y) >= s(plus(x,y)) ; app(nil,k) >= k ; app(cons(x,l),k) >= cons(x,app(l,k)) ; app(l,nil) >= l ; sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) ; sum(cons(x,nil)) >= cons(x,nil) ; sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) ; Marked_sum(app(l,cons(x,cons(y,k)))) >= Marked_sum(app(l, sum(cons(x,cons(y,k))))) ; } + Disjunctions:{ { Marked_sum(app(l,cons(x,cons(y,k)))) > Marked_sum(app(l, sum(cons(x,cons(y,k))))) ; } } === TIMER virtual : 10.000000 === Entering poly_solver Starting Sat solver initialization Calling Sat solver... === STOPING TIMER virtual === === TIMER real : 10.000000 === === STOPING TIMER real === Sat solver returned Sat solver result read === STOPING TIMER real === === STOPING TIMER virtual === constraint: minus(minus(x,y),z) >= minus(x,plus(y,z)) constraint: minus(s(x),s(y)) >= minus(x,y) constraint: minus(x,0) >= x constraint: quot(0,s(y)) >= 0 constraint: quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) constraint: plus(0,y) >= y constraint: plus(s(x),y) >= s(plus(x,y)) constraint: app(nil,k) >= k constraint: app(cons(x,l),k) >= cons(x,app(l,k)) constraint: app(l,nil) >= l constraint: sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) constraint: sum(cons(x,nil)) >= cons(x,nil) constraint: sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) constraint: Marked_sum(app(l,cons(x,cons(y,k)))) >= Marked_sum(app(l, sum(cons( x, cons(y,k))))) APPLY CRITERIA (Subterm criterion) ST: Marked_app -> 1 APPLY CRITERIA (Subterm criterion) APPLY CRITERIA (Choosing graph) Trying to solve the following constraints: { minus(minus(x,y),z) >= minus(x,plus(y,z)) ; minus(s(x),s(y)) >= minus(x,y) ; minus(x,0) >= x ; quot(0,s(y)) >= 0 ; quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) ; plus(0,y) >= y ; plus(s(x),y) >= s(plus(x,y)) ; app(nil,k) >= k ; app(cons(x,l),k) >= cons(x,app(l,k)) ; app(l,nil) >= l ; sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) ; sum(cons(x,nil)) >= cons(x,nil) ; sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) ; Marked_sum(cons(x,cons(y,l))) >= Marked_sum(cons(plus(x,y),l)) ; } + Disjunctions:{ { Marked_sum(cons(x,cons(y,l))) > Marked_sum(cons(plus(x,y),l)) ; } } === TIMER virtual : 10.000000 === Entering poly_solver Starting Sat solver initialization Calling Sat solver... === STOPING TIMER virtual === === TIMER real : 10.000000 === === STOPING TIMER real === Sat solver returned Sat solver result read === STOPING TIMER real === === STOPING TIMER virtual === constraint: minus(minus(x,y),z) >= minus(x,plus(y,z)) constraint: minus(s(x),s(y)) >= minus(x,y) constraint: minus(x,0) >= x constraint: quot(0,s(y)) >= 0 constraint: quot(s(x),s(y)) >= s(quot(minus(x,y),s(y))) constraint: plus(0,y) >= y constraint: plus(s(x),y) >= s(plus(x,y)) constraint: app(nil,k) >= k constraint: app(cons(x,l),k) >= cons(x,app(l,k)) constraint: app(l,nil) >= l constraint: sum(app(l,cons(x,cons(y,k)))) >= sum(app(l,sum(cons(x,cons(y,k))))) constraint: sum(cons(x,nil)) >= cons(x,nil) constraint: sum(cons(x,cons(y,l))) >= sum(cons(plus(x,y),l)) constraint: Marked_sum(cons(x,cons(y,l))) >= Marked_sum(cons(plus(x,y),l)) APPLY CRITERIA (Subterm criterion) ST: Marked_plus -> 1 APPLY CRITERIA (Graph splitting) Found 0 components: APPLY CRITERIA (Graph splitting) Found 0 components: APPLY CRITERIA (Graph splitting) Found 0 components: APPLY CRITERIA (Graph splitting) Found 0 components: APPLY CRITERIA (Graph splitting) Found 0 components: APPLY CRITERIA (Graph splitting) Found 0 components: SOLVED { TRS termination of: [1] minus(x,0) -> x [2] minus(s(x),s(y)) -> minus(x,y) [3] quot(0,s(y)) -> 0 [4] quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) [5] plus(0,y) -> y [6] plus(s(x),y) -> s(plus(x,y)) [7] minus(minus(x,y),z) -> minus(x,plus(y,z)) [8] app(nil,k) -> k [9] app(l,nil) -> l [10] app(cons(x,l),k) -> cons(x,app(l,k)) [11] sum(cons(x,nil)) -> cons(x,nil) [12] sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l)) [13] sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k))))) [14] plus(0,y) -> y [15] plus(s(x),y) -> s(plus(x,y)) , CRITERION: MDP [ { DP termination of: , CRITERION: SG [ { DP termination of: , CRITERION: ORD [ Solution found: polynomial interpretation = [ minus ] (X0,X1) = 1*X0 + 0; [ sum ] (X0) = 2 + 0; [ plus ] (X0,X1) = 2*X0 + 1*X1 + 0; [ s ] (X0) = 2 + 1*X0 + 0; [ nil ] () = 0; [ 0 ] () = 0; [ app ] (X0,X1) = 2*X0 + 1*X1 + 0; [ Marked_quot ] (X0,X1) = 2*X0 + 0; [ quot ] (X0,X1) = 1*X0 + 0; [ cons ] (X0,X1) = 2 + 0; ]} { DP termination of: , CRITERION: ST [ { DP termination of: , CRITERION: SG [ ]} ]} { DP termination of: , CRITERION: ORD [ Solution found: polynomial interpretation = [ minus ] (X0,X1) = 2*X0 + 0; [ sum ] (X0) = 2 + 0; [ plus ] (X0,X1) = 2*X1 + 0; [ s ] (X0) = 1*X0 + 0; [ Marked_sum ] (X0) = 2*X0 + 0; [ nil ] () = 0; [ 0 ] () = 0; [ app ] (X0,X1) = 2*X0 + 2*X1 + 0; [ quot ] (X0,X1) = 0; [ cons ] (X0,X1) = 2 + 1*X1 + 0; ]} { DP termination of: , CRITERION: ST [ { DP termination of: , CRITERION: SG [ ]} ]} { DP termination of: , CRITERION: ORD [ Solution found: polynomial interpretation = [ minus ] (X0,X1) = 1*X0 + 2*X1 + 0; [ sum ] (X0) = 2 + 0; [ plus ] (X0,X1) = 1*X1 + 0; [ s ] (X0) = 1*X0 + 0; [ Marked_sum ] (X0) = 1*X0 + 0; [ nil ] () = 0; [ 0 ] () = 0; [ app ] (X0,X1) = 2*X0 + 2*X1 + 0; [ quot ] (X0,X1) = 0; [ cons ] (X0,X1) = 2 + 1*X1 + 0; ]} { DP termination of: , CRITERION: ST [ { DP termination of: , CRITERION: SG [ ]} ]} ]} ]} Cime worked for 0.352342 seconds (real time) Cime Exit Status: 0
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| 2.765625
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CC-MAIN-2023-40
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latest
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en
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https://www.geeksforgeeks.org/ruby-math-asinh-function/
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# Ruby | Math asinh() function
The asinh() is an inbuilt function in Ruby returns the inverse hyperbolic sine of an angle given in radians. It accepts all values between range (-INFINITY, INFINITY).
Syntax: Math.acosh(value)
Parameters: The function accepts one mandatory parameter value which specifies the inverse hyperbolic angle in radian which should be greater or equal to 1. If the argument is less than 1, domain error is returned.
Return Value: The function returns the inverse hyperbolic sine of an angle given in radians. The range is between (-INFINITY, INFINITY)
Example 1:
`#Ruby program for asinh() function ` ` ` `#Assigning values ` `val1 = 2 val2 = 877 val3 = 432 val4 = 43 ` ` ` `#Prints the acosh() value ` ` ``puts` `Math.asinh(val1) ` ` ``puts` `Math.asinh(val2) ` ` ``puts` `Math.asinh(val3) ` ` ``puts` `Math.asinh(val4) `
Output:
```1.4436354751788103
7.4696544979749735
6.761574108393271
4.45448247706051
```
Example 2:
`#Ruby program for asinh() function ` ` ` `#Assigning values ` `val1 = -24 val2 = -765 val3 = -98 val4 = -23 ` ` ` `#Prints the asinh() value ` ` ``puts` `Math.asinh(val1) ` ` ``puts` `Math.asinh(val2) ` ` ``puts` `Math.asinh(val3) ` ` ``puts` `Math.asinh(val4) `
Output:
```-3.8716347563877314
-7.333023441572332
-5.278140689034662
-3.8291136516208812
```
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http://oeis.org/A318771
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A318771 Expansion of Sum_{k>=0} x^(k^2) / Product_{j=1..k} (1 - x^j)^j. 3
1, 1, 1, 1, 2, 2, 4, 4, 7, 8, 12, 14, 22, 25, 37, 47, 64, 81, 113, 140, 191, 243, 319, 408, 540, 677, 889, 1132, 1462, 1855, 2404, 3034, 3909, 4946, 6325, 7997, 10202, 12840, 16328, 20549, 25989, 32627, 41180, 51577, 64872, 81128, 101729, 127016, 158913, 197981, 247163, 307523, 383019 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,5 LINKS MAPLE a:=series(add(x^(k^2)/mul((1-x^j)^j, j=1..k), k=0..100), x=0, 53): seq(coeff(a, x, n), n=0..52); # Paolo P. Lava, Apr 02 2019 MATHEMATICA nmax = 52; CoefficientList[Series[Sum[x^k^2/Product[(1 - x^j)^j, {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x] CROSSREFS Cf. A193197, A206100, A206138, A318770. Sequence in context: A187219 A317785 A014810 * A239835 A026929 A206560 Adjacent sequences: A318768 A318769 A318770 * A318772 A318773 A318774 KEYWORD nonn AUTHOR Ilya Gutkovskiy, Sep 03 2018 STATUS approved
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Last modified July 17 23:21 EDT 2019. Contains 325109 sequences. (Running on oeis4.)
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CC-MAIN-2019-30
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latest
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en
| 0.490941
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https://plainmath.net/83112/find-the-largest-wavelength-of-light-fal
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crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00359.warc.gz
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# Find the largest wavelength of light falling on double slits separated by 1.20 mu m for which there is a first-order maximum. Is this in the visible part of the spectrum?
Find the largest wavelength of light falling on double slits separated by for which there is a first-order maximum. Is this in the visible part of the spectrum?
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ab8s1k28q
Given:
Slit width, d = 1.2 micrometer = $1.2×{10}^{-6}m$
order, m = 1
Formula used:
For the bright fringe, we use the formula
$d\mathrm{sin}\theta =m\lambda$
where, m is the number of order, d is the slit width, $\lambda$ is the wavelength and angle is the angular width.
For maximum value, the value of $\mathrm{sin}\theta =1$
So
This value of wavelength does not lie in the visible range. The visible range is from 400 nm to 700 nm.
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CC-MAIN-2022-33
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https://www.electricalindia.in/speed-control-of-induction-motor-drive/
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# Speed Control Of Induction Motor Drive
Recently, attention has been given to the identification of the instantaneous value of the rotor resistance while the drive is in normal operation... - P V Narendra Kumar
Indirect Field Oriented Controlled (IFOC) induction motor drives are being increasingly used in high-performance drive systems, as induction motors are more reliable because of their construction and less expensive materials used, than any other motors available in the market today. As indirect field orientation utilises an inherent slip relation, it is essentially a feed forward scheme and hence depends greatly on the accuracy of the motor parameters used in the vector controller – particularly to the rotor resistance. It changes widely with the rotor temperature, resulting in various harmful effects – such as over (or under) excitation, the destruction of the decoupled condition of the flux and torque, etc. Recently, attention has been given to the identification of the instantaneous value of the rotor resistance while the drive is in normal operation. So far, several approaches have been presented. A new sliding mode current observer for an induction motor is developed. Sliding mode functions are chosen to determine speed and rotor resistance of an induction motor in which the speed and rotor resistance are assumed to be unknown constant parameters. In, a method using a programmable cascaded low pass filter for the estimation of rotor flux of an induction motor, with a view to estimate the rotor time constant of an indirect field orientation controlled induction motor drive, is investigated. The estimated rotor flux data has also been used for the on-line rotor resistance identification with artificial neural network. Despite all these effects, rotor resistance estimation remains a difficult problem.
Indirect Field-Oriented Induction Motor Drive
The indirect vector control method is essentially same as the direct vector control, except that the unit vector generated in an indirect manner using the measured speed r and slip speed sl. The following dynamic equations are taken into consideration to implement indirect vector control strategy.
θe = ωe dt = (ωr + ωsl) = θr + θsl
Slip frequency can be calculated as
ωsl = Lm Rr iqs
Lr
For constant rotor flux ψr and dψr/dt=0, substituting in equation yields the rotor flux set as ψr = Lm ids
Fig 1: Block Diagram of Fuzzy Logic Controller…
Design of Fuzzy Logic Controller for Induction Motor Drive
Fig. 1 shows block diagram of speed control system using Fuzzy Logic Controller (FLC). Here, the first input is the Temperature ‘T’ and second is the Change in Temperature ‘ΔT’ at sampling time ‘ts.’ The two input variables ‘T’ and ‘ΔT’ are calculated at every sampling time as functional block diagram of Fuzzy Logic Control. In this block, number of inputs and number of outputs are estimated and any relationship is appeared between inputs and outputs are checked. The crisp values for the inputs and outputs are noted down. The entire information regarding the application to be solved is noted down here. The rotor resistance with respect to temperature & change in temperature are noted down here.
Where ‘wls’ denotes the Slip frequency, r *(ts) is the reference rotor speed, r(ts) is the actual speed, e(ts-1) is the value of error at previous sampling time. The output variable is the rotor resistance. As shown in Fig. 2, the Fuzzy Logic Controller consists of four blocks, Fuzzification, inference mechanism, knowledge base and Defuzzification.
Fig 2: Fuzzy Logic Controller Membership Functions…
• Fuzzification Block: In this stage the crisp variables of input are converted into fuzzy variables. The fuzzification maps the error and change in error to linguistic labels of fuzzy sets. Membership function is associated to each label with triangular shape which consists of two inputs and one output. The proposed controller uses following linguistic labels. Each of the inputs and output contain membership function with all these seven linguistics.
• Knowledge Base & Inference Stage: Knowledge base involve defining the rules represented as ‘if-then’ rules statements governing the relationship between input and output variables in terms of membership function. In this stage, the input variables ‘T’and ‘ΔT’ are processed by the inference mechanism that executes 49 rules represented in rule table shown below. Considering the first rule, if temperature is NS and change in temperature is SS, then the output will be NR. Here, Mamdani’s algorithm for inference mechanism used.
Figure 2 shows the configuration of the proposed fuzzy logic rotor resistance estimation. The functions F and F0 are first calculated respectively from the estimated variables ids; iqs; vds; vqs; !e and the reference value. The inputs temperature and change in temperature are the variables, which are used as inputs for the FLC. The internal structure of the fuzzy logic rotor resistance estimation is chosen similar to that of a Fuzzy Logic Controller, which consists of fuzzification, inference engine and defuzzification. For the successful design of FLC’s proper selection of these gains are crucial jobs, which in many cases are done through trial and error to achieve the best possible control performance. Then the crisp variables are converted into fuzzy variables using triangular membership functions as in Figure 3. These input membership functions are used to transfer crisp inputs into fuzzy sets.
The expert’s experience is incorporated into a knowledge base with 49 rules (7×7). This experience is synthesised by the choice of the input-output (I/O) membership functions and the rule base. Then, in the second stage of the FLC, the inference engine, based on the input fuzzy variables, uses appropriate IF-THEN rules in the knowledge base to imply the final output fuzzy sets as shown in the Table 1, where NS, NM, NL, Z, PS, PM, PL correspond to Negative Small, Negative Medium, Negative Large, Zero, Positive Small, Positive Medium, Positive Large respectively.
Fig 3: Rule Base Editor…
• Defuzzification: This stage introduces different methods that can be used to produce fuzzy set value for the output fuzzy variable T. Here the centre of gravity or centroids method is used to calculate the final fuzzy value. Defuzzification using COA method – means that crisp output is obtained by using centre of gravity – in which the crisp output variable is taken to be the geometric centre of the output fuzzy variables where it is formed by taking the union of all the contributions of rules with the degree of fulfillment greater than zero. Then the COA expression with discretised universe of discourse can be Te * – obtained by integration which is used to calculate variation of rotor flux, where it is observed that in case of PI controller it will take more time to reach to steady state value. But in case of fuzzy controller it will take less time to reach steady value.
In the defuzzification stage, the implied fuzzy set is transformed to a crisp output by the center of gravity defuzzification technique as given by the formula (19), zi is the numerical output at the ith number of rules and _(zi) corresponds to the value of fuzzy membership function at the ith number of rules. The summation is from one to n, where n is the number of rules that apply for the given fuzzy inputs. This value added to the reference rotor time constant (Tref) gives the estimated time constant (Tr), which is used as an input to the F.O.C. block of Figure 1 to ensure the correct orientation operation of the drive. Without considering the effects derived from the saturation (Lr constant), Rr−est is obtained from the estimated rotor time constant Figure 2. Therefore, this rotor resistance estimation value used in the control model must match its real value in order to maintain a high performance of the induction motor drive as will be shown later. The input/output mapping of the FLC rotor resistance estimation is shown in Figure 4, which is a continuous highly non-linear function. Detailed discussion about FLC construction is referred in.
Simulation Results
The configuration of the overall control system is shown in Figure 1. It is essential that the simulation model is designed to approach as close to reality as possible. Therefore, for the simulation of the whole drive system according to Figure 1, a mathematical model has been developed based on the induction motor equations and the equations for estimating the rotor resistance which have been derived in Section III. In addition, a mathematical model for all the remaining drive system units was necessary to complete the simulation model.
In order to analyse the drive system performance for their flux and torque responses, with rotor resistance variation, the above-presented system has been simulated using MATLAB/SIMULINK software.
A squirrel cage induction motor with a rated power of 1.5 kW has been used. A constant reference flux of 0.695 Wb is assumed and the speed was held constant at 1000 rpm.
The rotor resistance was stepped or ramped from 100 to 200% of its rated value, thereby simulating a change in rotor resistance due to a temperature change. The system was first started up to 1000 rpm with a full load of 10 N.m. At 1.5 sec, the rotor resistance was stepped from 100 to 200% of its rated value.
It is observed in this figure that when the estimated rotor resistance deviates from its real value, the field orientation scheme is detuned and the command torque (Te) instead of stabilising at its rated value, it is increased to 17 N.m to compensate the drop in speed which equals approximately 12 rpm.
But in the actual operating conditions, the rate of change of temperature is very slow and so the resistance variation. Accordingly, at 1.5 sec a ramp change of rotor resistance for an uncompensated case is applied linearly from 100% of its rated value to 200% till 4.5 sec, then, this value is maintained for 2.5 sec.
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# Understanding University Yield Rates: What They Are and Why They Matter
The Class of 2027 in the United States faced one of the toughest admissions cycles in history. With record low acceptance rates at some of the top universities, students definitely felt the pressure to put their best foot forward. However, it is important to understand that acceptance rate is not the only measure of a college’s reputation. Perhaps more important is the university’s yield rate - the actual number of students who decide to enroll after getting an offer of admission. This blog will explain the importance of university yield rates and why you should consider them when picking your university.
## What are university yield rates?
University yield rates refer to the percentage of accepted students who choose to enroll at a particular university. It is a crucial metric used by educational institutions to assess the effectiveness of their admissions process and gauge their overall desirability among prospective students. A high yield rate - the closer offers and enrolments are - indicates that a significant proportion of accepted students decide to enroll, reflecting the university's attractiveness, reputation, academic programs, financial aid offerings, campus culture, and other factors. On the other hand, a low yield rate suggests that the university may be facing challenges in convincing admitted students to choose their institution over other options.
Universities often strive to increase their yield rates through targeted recruitment efforts, enhancing the overall student experience, and providing competitive financial aid packages to ensure a strong incoming class each year. NACAC’s most recent state of admissions report shared that the average yield rate nationally for first-time freshmen was 33.6%. More prestigious schools may have yield rates as high as 85%.
### How is yield rate calculated?
Calculating university yield rates involves several steps. Here's a step-by-step explanation:
1. Gather data: Collect relevant data, including the number of students who were accepted to the university and the number of students who ultimately enrolled.
2. Determine the acceptance pool: Identify the total number of students who received acceptance letters or offers from the university. This can be obtained from the university's admissions data.
3. Calculate the yield rate: Divide the number of students who enrolled by the number of students who were accepted. Then, multiply the result by 100 to express it as a percentage. The formula for calculating yield rate is:
Yield Rate = (Number of Enrolled Students / Number of Accepted Students) * 100
Example calculation: Suppose a university accepted 2,000 students, and out of those, 1,500 students decided to enroll.
YR = (1,500 / 2,000) * 100 = 75%
So, the university's yield rate would be 75%.
4. Analyze and interpret: Once the yield rate is calculated, it can be used to assess the university's attractiveness to accepted students. A higher yield rate generally indicates that the university has successfully convinced a significant portion of accepted students to enroll. A lower yield rate may suggest that the university faces challenges in attracting accepted students or competing with other institutions.
##### Acceptance Rates vs. Yield Rates for Class of 2026
SchoolClass of 2026 Admission RateClass of 2026 Yield Rate
Amherst College7.3%45.8%
Brown University 5%67.15%
Columbia University3.37%62%
Cornell University 7.26%63%
Dartmouth College6.20%62.17%
Harvard University3.19%83%
Johns Hopkins University6.5%51.3%
Massachusetts Institute of Technology3.96%85%
Notre Dame University12.9%60.17%
Pomona College7%55.14%
Princeton University 5.7%69.22%
Stanford University3.68%82%
Tufts University9.69%50.22%
University of Pennsylvania6.51%68%
University of Southern California11.9%41.7%
University of Virginia12.8%42%
Wellesley College13%50.8%
Williams College8.5%44.3%
Yale University4.40%70%
## What factors affect a university’s yield rate?
Several factors can influence a university's yield rate, ultimately impacting the number of accepted students who choose to enroll. Here are some key factors:
• Location: The location of a university can be a significant factor in determining its yield rate. Proximity to urban centers, cultural attractions, natural surroundings, or specific industries can make a university more appealing to prospective students. A desirable location can enhance the overall student experience and contribute to a higher yield rate.
• Prestige and reputation: The prestige and reputation of a university play a crucial role in attracting accepted students. Universities with a strong academic standing, renowned faculty, research opportunities, and a history of success tend to have higher yield rates. The perception of a university's prestige can greatly influence a student's decision to enroll.
• Academic programs: The quality and breadth of academic programs offered by a university can significantly impact the yield rate. Students often consider the availability of majors, minors, concentrations, and interdisciplinary programs that align with their interests and career aspirations. Universities with renowned programs in specific fields or unique offerings may attract a higher number of accepted students who find their academic offerings particularly compelling.
• Financial aid packages: The affordability of attending a university is a significant consideration for many students and their families. The availability and generosity of financial aid packages can greatly influence a student's decision to enroll. Universities that provide competitive financial aid, scholarships, grants, and work-study opportunities may have higher yield rates as they alleviate the financial burden for accepted students.
• Campus culture and student life: The campus culture, extracurricular activities, and overall student life can significantly impact a university's yield rate. Students often seek an engaging and supportive community that aligns with their personal preferences and interests. Factors such as clubs, organizations, sports teams, campus facilities, diversity and inclusion initiatives, and campus events can contribute to a positive student experience and increase the likelihood of enrollment.
• Affordability and cost of living: The overall cost of attending a university, including tuition, fees, and cost of living in the surrounding area, can influence a student's decision. Universities located in areas with a lower cost of living or that provide affordable on-campus housing options may be more attractive to accepted students.
• Student outcomes and career prospects: The career prospects and post-graduation outcomes associated with a university can impact its yield rate. Students often consider the potential for internships, co-op programs, networking opportunities, and job placement rates. Universities with a track record of successful student outcomes and strong alumni networks may have higher yield rates.
It's important to note that the relative importance of these factors can vary among students, and individual preferences and circumstances play a significant role in a student's decision-making process. Universities often strive to enhance these factors to attract and retain a competitive and diverse student body.
## Importance of Yield Rates for Universities
Yield rates are important for universities for several reasons. Here's an explanation of their significance:
• Reputation: Yield rates can have a significant impact on a university's reputation. A high yield rate indicates that the university is highly desirable among accepted students, suggesting that it offers strong academic programs, a supportive campus culture, and appealing extracurricular opportunities. A positive reputation can attract more talented applicants in subsequent years and enhance the overall prestige and standing of the institution.
• Rankings: University rankings often take into account factors such as selectivity and yield rates. A high yield rate can positively influence a university's ranking because it demonstrates that the institution is successful in attracting and enrolling top-performing students. Higher rankings can, in turn, lead to increased visibility, more funding opportunities, and a broader pool of qualified applicants.
• Financial stability: Yield rates play a crucial role in a university's financial stability. Enrolled students contribute to tuition revenue, which is a significant source of income for most institutions. A higher yield rate ensures a more predictable student population and helps the university meet its enrollment and revenue targets. This stability allows the university to allocate resources effectively, invest in faculty, infrastructure, research, and student support services, and maintain financial sustainability.
• Admissions decisions: Yield rates can influence a university's admissions decisions. If a university consistently has a high yield rate, it may become more selective in its admissions process, aiming to accept students who are more likely to enroll. On the other hand, if the yield rate is low, the university may adopt strategies to increase its attractiveness to accepted students, such as enhancing marketing efforts, improving campus facilities, or revising financial aid packages.
• Financial aid offers: Yield rates can also impact a university's financial aid offers. Universities may allocate financial aid resources strategically to encourage accepted students to enroll. If the yield rate is low, the institution may provide more competitive financial aid packages to incentivize accepted students to choose their university over other options. Financial aid can be a crucial factor in a student's decision-making process, particularly for those with limited financial resources.
Overall, yield rates serve as a valuable metric for universities to assess their desirability, manage their reputation, and maintain financial stability. They influence admissions decisions and financial aid offers, ultimately shaping the composition of the student body and impacting the university's long-term success.
### What is yield protection?
Universities like really high yield rates - it signals their desirability as well as financial security. To maintain a high rate, admissions offices are sometimes said to emply something called yield protection, sometimes referred to as being "yield conscious." The idea behind this controversial practice is to increase a university's yield rate by admitting applicants who are slightly less qualified but are believed to be more likely to enroll. Meanwhile, more qualified candidates may be deferred, waitlisted, or denied under the assumption that they are more inclined to choose a more selective institution and view the current one as a safety school or backup plan. The objective of the less selective institution is to prioritize applicants who are deemed more likely to enroll, even if their academic credentials are comparatively less impressive than other candidates.
## Why should students care about yield rates?
Yield rates are also significant for students, as they can impact their chances of being admitted to their top-choice universities and affect their financial aid offers. These rates are a great indicator of a university’s competitiveness as well as its desirability. Here's an explanation of why yield rates matter for students:
• Admission chances: Yield rates can influence a student's chances of being admitted to their top-choice universities. Highly selective institutions often have low acceptance rates and high yield rates. When a university has a high yield rate, it means that a significant number of accepted students choose to enroll. This creates a situation where there are fewer available spots for students who are still waiting for admission decisions. Consequently, a high yield rate can make it more challenging for students to secure a spot at their desired university if the competition is fierce.
• Waitlist decisions: Yield rates can also impact the decisions made regarding waitlisted students. If a university has a lower-than-expected yield rate, it may turn to its waitlist to fill the remaining spots in the incoming class. In such cases, waitlisted students who demonstrate a strong interest in the university may have a higher chance of being admitted.
• Financial aid offers: Yield rates can also influence the financial aid offers that students receive. Universities aim to attract accepted students who are more likely to enroll, and financial aid is one way to incentivize their decision. If a university has a lower yield rate than desired, it may increase the financial aid packages for accepted students to make their offer more competitive. This can create opportunities for students to receive more favorable financial aid packages, potentially making the university more affordable or providing additional support to cover tuition and other expenses.
• Decision-making process: Yield rates can also play a role in a student's decision-making process. A student may consider the yield rate of a university when deciding whether to accept an offer of admission. A high yield rate can indicate that the university is popular among accepted students, suggesting a vibrant campus community and a positive overall experience. On the other hand, a low yield rate may raise questions about the university's appeal or the satisfaction of enrolled students.
### Demonstrated Interest and Yield Rates
If a high yield rate is crucial for maintaining the school’s reputation and resources, demonstrated interest is an important factor. Colleges want to be assured at the time of application that you will attend if you get an offer. Demonstrated interest, then, becomes a factor in your application decision. Along with strong grades, excellent essays and recommendations, as well as unique extracurriculars, demonstrated interest can help your admission along.
While demonstrated interest is not the only factor in admissions decisions at highly selective colleges and universities, it can often be the tiebreaker between two equally qualified candidates. As such, you should make an effort to demonstrate your interest in these institutions, as it can greatly increase your chances of being admitted.
What Makes Crimson Different
## Final Thoughts
By considering yield rates, you can gain insights into a university's desirability, competitiveness, and financial aid offerings. It allows you to evaluate the likelihood of being admitted to the institutions and understand the potential financial support you may receive. Taking the time to research and comprehend yield rates can help you navigate the college application process and make well-informed choices that align with your academic goals and personal circumstances.
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https://www.bartleby.com/solution-answer/chapter-8-problem-9p-fundamentals-of-financial-management-mindtap-course-list-15th-edition/9781337395250/required-rate-of-return-stock-r-has-a-beta-of-20-stock-s-has-a-beta-of-045-the-required-return/a75af293-feda-11e8-9bb5-0ece094302b6
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Chapter 8, Problem 9P
### Fundamentals of Financial Manageme...
15th Edition
Eugene F. Brigham + 1 other
ISBN: 9781337395250
Chapter
Section
### Fundamentals of Financial Manageme...
15th Edition
Eugene F. Brigham + 1 other
ISBN: 9781337395250
Textbook Problem
# REQUIRED RATE OF RETURN Stock R has a beta of 2.0, Stock s has a beta of 0.45, the required return on an average stock is 10%, and the risk-free rate of return is 5%. By how much does the required return on the riskier stock exceed the required return on the less risky stock?
Summary Introduction
To determine: The difference between the required return on the risky stock and the less risky stock.
Introduction:
The Required Rate of Return:
The required rate of return is the rate which should be the minimum earning on an investment to keep that investment running in the market. When the required return is earned only then the users and the companies invest in that particular investment.
Explanation
Given,
For Stock R, the value of beta is 2.
For Stock S, the value of beta is 0.45.
The required return on an average stock is 10%.
The risk-free rate of return is 5%.
Calculated (working note),
The required return on Stock R is 15%.
The required return on Stock S is 7.25%.
Compute the difference in both the required rate of return on the stock.
Formula to calculate difference between required return on Stock R and S,
DifferenceĀ inĀ return=ReturnĀ ofĀ stockĀ RāReturnĀ ofĀ stockĀ S
Substitute 15% for return of stock R, and 7.25% for return of stock S
DifferenceĀ inĀ return=15%ā7.25%=7.75%
Working note:
Compute the required rate of return for stock R.
The formula to calculate the required rate of return is:
rstock=rRF+(rMārRF)Ćbstock (I)
Where,
• rstock is the required return on the stock.
• rRF is the risk-free return.
• rM is the market risk premium.
• bstock is the value of the stockās beta.
Substitute 5% for rRF , 10% for rM , and 2 for bstock in equation (I)
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What is the maximum output of an IIR filter?
I am implementing an IIR filter in an embedded device based on ARM. It is implemented as a cascaded biquad structure (series of second order filters put back to back). The filter can be considered stable as all the second order sections in the cascaded structure have their poles and zeros inside the unit circle(ROC) in the Z-domain.
Now, given that the filter is asymptotically stable and its output can't grow unboundedly, how to know the maximum output that the filter can grow to, asssuming that the input to the filter can be any noise or any random waveform?
On the device, the filter is implemented with floats, which reduces the chance of any overflow during computation. But after the computation the results have to be stored into fixed points as it is required by the application to do so. Hence,it is very important to know the output limits to avoid any overflow and saturation while storing into fixed points.
I would like to know,if there is any way to estitmate what could be the maximum limits of the output,so as to avoid clipping or unnecessarily use more bits for the fixed point?
And also it would be useful to know which type of input waveforms could produce the worst output possible?
• Do you know the maximum/minimum float that can be used to be converted into an int/short/whatever? Is it possible to test? Commented Mar 10, 2022 at 11:28
I would like to know,if there is any way to estitmate what could be the maximum limits of the output
Yes. It's
$$y_{max} = x_{max} \cdot \sum_{n = 0}^{\infty} |h[n]|$$
I.e. the maximum gain is the absolute sum of the impulse response
which type of input waveforms could produce the worst output possible?
$$x_{worst}[n] = x_{max} \cdot \text{sign}(h[-n])$$
It's the signum function of the time-flipped impulse response. In most cases this is an extremely unlikely signal to occur in the real world so it's overly conservative to assume worst case scaling. A different approach would be to do a statistical analysis of "real world" signals plus some sine waves, square waves and noise signals.
• Then I misunderstood. I thought it's about float <-> int conversion. Commented Mar 10, 2022 at 16:45
• I think you do not necessarily have an a priori knowledge of what $x_{\rm max}$ is. Commented Mar 10, 2022 at 22:48
• x_max is actually known already. The input is bound and so is the output of the filter. But the problem is to know the highest output the filter will produce from our input which is basically in the range [4 -4]. Will the output grow more than the input. If so by how much? Commented Mar 11, 2022 at 3:44
• Okay, then my comment doesn’t apply. Commented Mar 11, 2022 at 6:01
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### Solve the 'Eigenvalue' problem efficiently [closed]
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### Numerically computing the eigenvalues of an infinite-dimensional tridiagonal matrix
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### Eigenvalues of a non-Hermitian complex periodic potential
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https://aakashdigitalsrv1.meritnation.com/cbse-class-12-commerce/math/rd-sharma-xii-vol-1-2019/binary-operations/textbook-solutions/59_1_3514_24572_3.4_90791
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Rd Sharma XII Vol 1 2019 Solutions for Class 12 Commerce Math Chapter 3 Binary Operations are provided here with simple step-by-step explanations. These solutions for Binary Operations are extremely popular among Class 12 Commerce students for Math Binary Operations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Commerce Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.
#### Question 1:
Let '*' be a binary operation on N defined by
a * b = 1.c.m. (a, b) for all a, bN
(i) Find 2 * 4, 3 * 5, 1 * 6.
(ii) Check the commutativity and associativity of '*' on N.
a * b = 1.c.m. (a, b)
(i) 2 * 4 = 1.c.m. (2, 4)
= 4
3 * 5 = 1.c.m. (3, 5)
= 15
1 * 6 = 1.c.m. (1, 6)
= 6
(ii) Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is associative on N.
#### Question 2:
Determine which of the following binary operations are associative and which are commutative:
(i) * on N defined by a * b = 1 for all a, bN
(ii) * on Q defined by
(i) Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is associative on N.
(ii) Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is not associative on N.
#### Question 3:
Let A be any set containing more than one element. Let '*' be a binary operation on A defined by
a * b = b for all a, bA
Is '*' commutative or associative on A?
Commutativity:
Thus, * is not commutative on A.
Associativity:
Thus, * is associative on A.
#### Question 4:
Check the commutativity and associativity of each of the following binary operations:
(i) '*'. on Z defined by a * b = a + b + ab for all ab ∈ Z
(ii) '*'. on N defined by a * b = 2ab for all ab ∈ N
(iii) '*'. on Q defined by a * b = a − b for all ab ∈ Q
(iv) '⊙' on Q defined by a ⊙ b = a2 + b2 for all ab ∈ Q
(v) 'o' on Q defined by for all ab ∈ Q
(vi) '*' on Q defined by a * b = ab2 for all ab ∈ Q
(vii) '*' on Q defined by a * b = a + ab for all ab ∈ Q
(viii) '*' on R defined by a * b = a + b − 7 for all ab ∈ R
(ix) '*' on Q defined by a * b = (a − b)2 for all ab ∈ Q
(x) '*' on Q defined by a * b = ab + 1 for all ab ∈ Q
(xi) '*' on N, defined by a * b = ab for all ab ∈ N
(xii) '*' on Z defined by a * b = a − b for all ab ∈ Z
(xiii) '*' on Q defined by $a*b=\frac{ab}{4}$ for all ab ∈ Q
(xiv) '*' on Z defined by a * b = a + b − ab for all ab ∈ Z
(xv) '*' on N defined by a * b = gcd(ab) for all ab ∈ N
(i) Commutativity:
Thus, * is commutative on Z.
Associativity:
Thus, * is associative on Z.
(ii) Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is not associative on N.
(iii) Commutativity:
Thus, * is not commutative on Q.
Associativity:
Thus, * is not associative on Q.
(iv) Commutativity:
Thus, $\odot$ is commutative on Q.
Associativity:
Thus, $\odot$ is not associative on Q.
(v) Commutativity:
Thus, o is commutative on Q.
Associativity:
Thus, is associative on Q.
(vi) Commutativity:
Thus, * is not commutative on Q.
Associativity:
Thus, * is not associative on Q.
(vii) Commutativity:
Thus, * is not commutative on Q.
Associativity:
Thus, * is not associative on Q.
(viii) Commutativity:
Thus, * is commutative on R.
Associativity:
Thus, * is associative on R.
(ix) Commutativity:
Thus, * is commutative on Q.
Associativity:
Thus, * is not associative on Q.
(x) Commutativity:
Thus, * is commutative on Q.
Associativity:
Thus, * is not associative on Q.
(xi) Commutativity:
Thus, * is not commutative on N.
Associativity:
Thus, * is not associative on N.
(xii) Commutativity:
Thus, * not is commutative on Z.
Associativity:
Thus, * is not associative on Z.
(xiii) Commutativity:
Thus, * is commutative on Q.
Associativity:
Thus, * is associative on Q.
(xiv) Commutativity:
Thus, * is commutative on Z.
Associativity:
Thus, * is associative on Z.
Disclaimer : The answer given in the textbook is incorrect. The same has been corrected here.
(xv) Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is associative on N.
#### Question 5:
If the binary operation o is defined by aob = a + bab on the set Q − {−1} of all rational numbers other than 1, shown that o is commutative on Q − [1].
Thus, o is commutative on Q - {1}.
#### Question 6:
Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.
Thus, * is not commutative on Z.
#### Question 7:
On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a , bZ. Prove that * is not associative on Z.
Thus, * is not associative on Z.
#### Question 8:
Let S be the set of all real numbers except −1 and let '*' be an operation defined by
a * b = a + b + ab for all a, bS.
Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.
Checking for binary operation:
Thus, * is a binary operation on S.
Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is associative on S.
Now,
#### Question 9:
On Q, the set of all rational numbers, * is defined by $a*b=\frac{a-b}{2}$, shown that * is no associative.
Thus, * is not associative on Q.
#### Question 10:
On Z, the set of all integers, a binary operation * is defined by a * b = a + 3b − 4. Prove that * is neither commutative nor associative on Z.
Commutativity:
Thus, * is not commutative on Z.
Associativity:
Thus, * is not associative on Z.
#### Question 11:
On the set Q of all ration numbers if a binary operation * is defined by $a*b=\frac{ab}{5}$, prove that * is associative on Q.
Thus, * is associative on Q.
#### Question 12:
The binary operation * is defined by $a*b=\frac{ab}{7}$ on the set Q of all rational numbers. Show that * is associative.
Thus, * is associative on Q.
#### Question 13:
On Q, the set of all rational numbers a binary operation * is defined by $a*b=\frac{a+b}{2}$.
Show that * is not associative on Q.
Thus, * is not associative on Q.
#### Question 14:
Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b $-$ ab, for all a, b $\in$ S.
Prove that:
(i) * is a binary operation on S
(ii) * is commutative as well as associative. [CBSE 2014]
We have,
S = R $-$ {1} and * is defined on S as a * b = a + b $-$ ab, for all ab $\in$ S
(i) It is seen that for each a, b $\in$ S, there is a unique element a + b $-$ ab in S
This means that * carries each pair (a, b) to a unique element a * b = a + b $-$ ab in S
So, * is a binary operation on S
(ii) Commutativity:
Thus, * is commutative on S.
Associativity:
Thus , * is associative on S.
So, * is commutative as well as associative.
#### Question 1:
Find the identity element in the set I+ of all positive integers defined by a * b = a + b for all a, bI+.
Let e be the identity element in I+ with respect to * such that
Thus, 0 is the identity element in I+ with respect to *.
#### Question 2:
Find the identity element in the set of all rational numbers except −1 with respect to *defined by a * b = a + b + ab.
Let e be the identity element in Q$-$ {$-$1} with respect to * such that
Thus, 0 is the identity element in Q - {-1} with respect to *.
#### Question 3:
If the binary operation * on the set Z is defined by a * b = a + b −5, the find the identity element with respect to *.
Let e be the identity element in Z with respect to * such that
Thus, 5 is the identity element in Z with respect to *.
#### Question 4:
On the set Z of integers, if the binary operation * is defined by a * b = a + b + 2, then find the identity element.
Let e be the identity element in Z with respect to * such that
Thus, -2 is the identity element in Z with respect to *.
#### Question 1:
Let * be a binary operation on Z defined by
a * b = a + b − 4 for all a, b Z
(i) Show that '*' is both commutative and associative.
(ii) Find the identity element in Z.
(iii) Find the invertible elements in Z.
(i) Commutativity:
Thus, * is commutative on Z.
Associativity:
Thus, * is associative on Z.
(ii) Let e be the identity element in Z with respect to * such that
Thus, 4 is the identity element in Z with respect to *.
#### Question 2:
Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by
.
Show that * is commutative as well as associative. Also, find its identity element if it exists.
Commutativity:
Thus, * is commutative on Qo.
Associativity:
Thus, * is associative on Qo.
Finding identity element:
Let e be the identity element in Z with respect to * such that
Thus, 5 is the identity element in Qo with respect to *.
#### Question 3:
Let * be a binary operation on Q − {−1} defined by
a * b = a + b + ab for all a, bQ − {−1}
Then,
(i) Show that '*' is both commutative and associative on Q − {−1}.
(ii) Find the identity element in Q − {−1}
(iii) Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.
(i) Commutativity:
Thus, * is commutative on Q $-${$-$1}.
Associativity:
Thus, * is associative on Q $-$ {$-$1}.
(ii) Let e be the identity element in Q$-$ {$-$1} with respect to * such that
Thus, 0 is the identity element in Q $-$ {$-$1} with respect to *.
#### Question 4:
Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation '⊙' is defined on A as follows :
(a, b) ⊙ (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R.
(i) Show that '⊙' is commutative and associative on A
(ii) Find the identity element in A
(iii) Find the invertible elements in A.
(i) Commutativity:
Thus, $\odot$ is commutative on A.
Associativity:
#### Question 5:
Let 'o' be a binary operation on the set Q0 of all non-zero rational numbers defined by
.
(i) Show that 'o' is both commutative and associate.
(ii) Find the identity element in Q0.
(iii) Find the invertible elements of Q0.
(i) Commutativity:
Thus, o is commutative on Qo.
Associativity:
Thus, o is associative on Qo.
(ii) Let e be the identity element in Qo with respect to * such that
Thus, 2 is the identity element in Qo with respect to o.
#### Question 6:
On R − {1}, a binary operation * is defined by a * b = a + bab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.
Commutativity:
Thus, * is commutative on R $-$ {1}.
Associativity:
Thus, * is associative on R $-$ {1}.
Finding identity element:
Let e be the identity element in R $-$ {1} with respect to * such that
Thus, 0 is the identity element in R $-${1} with respect to *.
Finding inverse:
#### Question 7:
Let R0 denote the set of all non-zero real numbers and let A = R0 × R0. If '*' is a binary operation on A defined by
(a, b) * (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
(i) Show that '*' is both commutative and associative on A
(ii) Find the identity element in A
(iii) Find the invertible element in A.
#### Question 8:
Let * be the binary operation on N defined by
a * b = HCF of a and b.
Does there exist identity for this binary operation one N?
Let e be the identity element. Then,
We cannot find e that satisfies this condition.
So, the identity element with respect to * does not exist in N.
#### Question 9:
Let A$=$R$×$R and $*$ be a binary operation on defined by $\left(a,b\right)*\left(c,d\right)=\left(a+c,b+d\right).$ Show that $*$ is commutative and associative. Find the binary element for $*$ on A, if any.
We have,
A$=$R$×$R and $*$ is a binary operation on A defined by .
Now,
So, $*$ is commutative.
Also,
So, $*$ is associative.
Let (x, y) be the binary element for $*$ on A.
Hence, (0, 0) is the binary element for $*$ on A.
#### Question 1:
Construct the composition table for ×4 on set S = {0, 1, 2, 3}.
Here,
1 ${×}_{4}$ 1 = Remainder obtained by dividing 1 $×$1 by 4
= 1
0 ${×}_{4}$ 1 = Remainder obtained by dividing 0 $×$ 1 by 4
= 0
2 ${×}_{4}$ 3 = Remainder obtained by dividing 2 $×$ 3 by 4
= 2
3 ${×}_{4}$ 3 = Remainder obtained by dividing 3 $×$ 3 by 4
= 1
So, the composition table is as follows:
${×}_{4}$ 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1
#### Question 2:
Construct the composition table for +5 on set S = {0, 1, 2, 3, 4}.
Here,
1 ${+}_{5}$ 1 = Remainder obtained by dividing 1 + 1 by 5
= 2
3 ${+}_{5}$ 4 = Remainder obtained by dividing 3 + 4 by 5
= 2
4 ${+}_{5}$ 4 = Remainder obtained by dividing 4 + 4 by 5
= 3
So, the composition table is as follows:
${+}_{5}$ 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 5 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3
#### Question 3:
Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.
Here,
1 ${×}_{6}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 6
= 1
3 ${×}_{6}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 6
= 0
4 ${×}_{6}$ 5 = Remainder obtained by dividing 4$×$ 5 by 6
= 2
So, the composition table is as follows:
${×}_{6}$ 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 0 2 4 3 0 3 0 3 0 3 4 0 4 2 0 4 2 5 0 5 4 3 2 1
#### Question 4:
Construct the composition table for ×5 on Z5 = {0, 1, 2, 3, 4}.
Here,
1 ${×}_{5}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 5
= 1
3 ${×}_{5}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 5
= 2
4 ${×}_{5}$ 4 = Remainder obtained by dividing 4 $×$ 4 by 5
= 1
So, the composition table is as follows:
${×}_{5}$ 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1
#### Question 5:
For the binary operation ×10 on set S = {1, 3, 7, 9}, find the inverse of 3.
Here,
1 ${×}_{10}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 10
=1
3 ${×}_{10}$ 7 = Remainder obtained by dividing 3 $×$ 7 by 10
=1
7 ${×}_{10}$ 9 = Remainder obtained by dividing 7 $×$ 9 by 10
= 3
So, the composition table is as follows:
${×}_{10}$ 1 3 7 9 1 1 3 7 9 3 3 9 1 7 7 7 1 9 3 9 9 7 3 1
We observe that the elements of the first row are same as the top-most row.
So, $1\in S$ is the identity element with respect to ${×}_{10}$.
Finding inverse of 3:
From the above table we observe,
3 ${×}_{10}$ 7 = 1
So, the inverse of 3 is 7.
#### Question 6:
For the binary operation ×7 on the set S = {1, 2, 3, 4, 5, 6}, compute 3−1 ×7 4.
Finding identity element:
Here,
1 ${×}_{7}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 7
= 1
3 ${×}_{7}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 7
= 5
4 ${×}_{7}$ 5 = Remainder obtained by dividing 4$×$ 5 by 7
= 6
So, the composition table is as follows:
${×}_{7}$ 1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 4 6 1 3 5 3 3 6 2 5 1 4 4 4 1 5 2 6 3 5 5 3 1 6 4 2 6 6 5 4 3 2 1
We observe that all the elements of the first row of the composition table are same as the top-most row.
So, the identity element is 1.
Also,
So, 3$-1$ = 5
#### Question 7:
Find the inverse of 5 under multiplication modulo 11 on Z11.
Here,
1 ${×}_{11}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 11
= 1
3 ${×}_{11}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 11
= 1
4 ${×}_{11}$ 5 = Remainder obtained by dividing 4 $×$ 5 by 11
= 9
So, the composition table is as follows:
${×}_{11}$ 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 1 3 5 7 9 3 3 6 9 1 4 7 10 2 5 8 4 4 8 1 5 9 2 6 10 3 7 5 5 10 4 9 3 8 2 7 1 6 6 6 1 7 2 8 3 9 4 10 5 7 7 3 10 6 2 9 5 1 8 4 8 8 5 2 10 7 4 1 9 6 3 9 9 7 5 3 1 10 8 6 4 2 10 10 9 8 7 6 5 4 3 2 1
We observe that the first row of the composition table is same as the top-most row.
So, the identity element is 1.
Also,
#### Question 8:
Write the multiplication table for the set of integers modulo 5.
Here,
1 ${×}_{5}$×5 1 = Remainder obtained by dividing 1$×$ 1 by 5
= 1
3 ${×}_{5}$ 4 = Remainder obtained by dividing 3 $×$4 by 5
= 2
${×}_{5}$4 = Remainder obtained by dividing 4 $×$ 4 by 5
= 1
So, the composition table is as follows:
${×}_{5}$ 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1
#### Question 9:
Consider the binary operation * and o defined by the following tables on set S = {a, b, c, d}.
(i)
* a b c d a a b c d b b a d c c c d a b d d c b a
(ii)
o a b c d a a a a a b a b c d c a c d b d a d b c
Show that both the binary operations are commutative and associatve. Write down the identities and list the inverse of elements.
(i) Commutativity:
The table is symmetrical about the leading element. It means * is commutative on S.
Associativity:
So, * is associative on S.
Finding identity element:
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at a.
So, a is the identity element.
Finding inverse elements:
(ii) Commutativity:
The table is symmetrical about the leading element. It means that o is commutative on S.
Associativity:
So, o is associative on S.
Finding identity element:
We observe that the second row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at b.
So, b is the identity element.
Finding inverse elements:
#### Question 10:
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as
Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.
Here,
1 * 1 =1+1 ($\because$ 1+1 $<$6 )
= 2
3 * 4 = 3 + 4 $-$6 ($\because$ 3 + 4 $>$6 )
= 7 $-$ 6
= 1
4 * 5 = 4 + 5$-$6 ($\because$ 4 + 5$>$6 )
= 9$-$ 6
= 3 etc.
So, the composition table is as follows:
* 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
So, 0 is the identity element .
Finding inverse:
#### Question 1:
Write the identity element for the binary operation * on the set R0 of all non-zero real numbers by the rule for all a, bR0.
Let e be the identity element in R0 with respect to * such that
Thus, 2 is the identity element in R0 with respect to *.
#### Question 2:
On the set Z of all integers a binary operation * is defined by a * b = a + b + 2 for all a, bZ. Write the inverse of 4.
To find the identity element, let e be the identity element in Z with respect to * such that
Thus,$-$2 is the identity element in Z with respect to *.
Now,
#### Question 3:
Define a binary operation on a set.
Let A be a non-empty set. An operation * is called a binary operation on A, if and only if
#### Question 4:
Define a commutative binary operation on a set.
An operation * on a set A is called a commutative binary operation if and only if it is a binary operation as well as commutative, i.e. it must satisfy the following two conditions.
#### Question 5:
Define an associative binary operation on a set.
An operation * on a set A is called an associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:
#### Question 6:
Write the total number of binary operations on a set consisting of two elements.
Number of binary operations on a set with n elements = ${n}^{{n}^{2}}$
#### Question 7:
Write the identity element for the binary operation * defined on the set R of all real numbers by the rule
Let e be the identity element in R with respect to * such that
Thus, $\frac{7}{3}$ is the identity element in R with respect to *.
#### Question 8:
Let * be a binary operation, on the set of all non-zero real numbers, given by
Write the value of x given by 2 * (x * 5) = 10.
#### Question 9:
Write the inverse of 5 under multiplication modulo 11 on the set {1, 2, ... ,10}.
As, e = 1 : 5 × 9 ≡ 1 (mod 11)
So, the inverse of 5 i.e. 5$-$1 = 9
#### Question 10:
Define identity element for a binary operation defined on a set.
Let * be a binary operation on a set A.
An element e is called an identity element in A with respect to * if and only if
#### Question 11:
Write the composition table for the binary operation multiplication modulo 10 (×10) on the set S = {2, 4, 6, 8}.
Here,
2 ${×}_{10}$ 4 = Remainder obtained by dividing 2 $×$ 4 by 10
= 8
4 ${×}_{10}$ 6 = Remainder obtained by dividing 4 $×$ 6 by 10
= 4
2 ${×}_{10}$ 8 = Remainder obtained by dividing 2 $×$ 8 by 10
= 6
3 ${×}_{10}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 10
= 2
So, the composition table is as follows:
${×}_{10}$ 2 4 6 8 2 4 8 2 6 4 8 6 4 2 6 2 4 6 8 8 6 2 8 4
#### Question 12:
For the binary operation multiplication modulo 10 (×10) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.
Here,
1 ${×}_{10}$ 1 = Remainder obtained by dividing 1 $×$ 1 by 10
= 1
3 ${×}_{10}$ 1 = Remainder obtained by dividing 3 $×$ 1 by 10
= 3
7 ${×}_{10}$ 3 = Remainder obtained by dividing 7 $×$ 3 by 10
= 1
3 ${×}_{10}$ 3 = Remainder obtained by dividing 3$×$ 3 by 10
= 9
So, the composition table is as follows:
${×}_{10}$ 1 3 7 9 1 1 3 7 9 3 3 9 1 7 7 7 1 9 3 9 9 7 3 1
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at 1.
So, the identity element is 1.
Also,
3 ${×}_{10}$ 7 = 1
3-1 = 7
#### Question 13:
For the binary operation multiplication modulo 5 (×5) defined on the set S = {1, 2, 3, 4}. Write the value of ${\left(3{×}_{5}{4}^{-1}\right)}^{-1}$.
Here,
1 ${×}_{5}$1 = Remainder obtained by dividing 1 $×$ 1 by 5
= 1
3 ${×}_{5}$ 4 = Remainder obtained by dividing 3 $×$ 4 by 5
= 2
${×}_{5}$ 4 = Remainder obtained by dividing 4 $×$ 4 by 5
= 1
So, the composition table is as follows:
${×}_{5}$ × 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at 1.
Thus, 1 is the identity element.
#### Question 14:
Write the composition table for the binary operation ×5 (multiplication modulo 5) on the set S = {0, 1, 2, 3, 4}.
Here,
1${×}_{5}$1 = Remainder obtained by dividing 1 $×$ 1 by 5
= 1
3${×}_{5}$4 = Remainder obtained by dividing 3 $×$ 4 by 5
= 2
4${×}_{5}$4 = Remainder obtained by dividing 4 $×$ 4 by 5
= 1
So, the composition table is as follows:
${×}_{5}$ × 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1
#### Question 15:
A binary operation * is defined on the set R of all real numbers by the rule
Write the identity element for * on R.
Let e be the identity element in R with respect to * such that
Thus, 0 is the identity element in R with respect to *.
#### Question 16:
Let +6 (addition modulo 6) be a binary operation on S = {0, 1, 2, 3, 4, 5}. Write the value of $2{+}_{6}{4}^{-1}{+}_{6}{3}^{-1}.$
Here,
${+}_{6}$ 1 = Remainder obtained by dividing 1 + 1 by 6
= 2
3 ${+}_{6}$ 4 = Remainder obtained by dividing 3 + 4 by 6
= 1
4 ${+}_{6}$ 5 = Remainder obtained by dividing 4 + 5 by 6
= 3
So, the composition table is as follows:
${+}_{6}$ 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
So, 0 is the identity element.
From the table,
#### Question 17:
Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.
Given: a * b = 3a + 4b − 2
Here,
4 * 5 = 3 (4) + 4 (5) $-$ 2
= 12 + 20 $-$ 2
= 30
#### Question 18:
If the binary operation * on the set Z of integers is defined by a * b = a + 3b2, find the value of 2 * 4.
Given: a * b = a + 3b2
Here,
2 * 4 = 2 + 3 (4)2
= 2 + 3 (16)
= 2 + 48
= 50
#### Question 19:
Let * be a binary operation on N given by a * b = HCF (a, b), a, b ∈ N. Write the value of 22 * 4.
Given: a * b = HCF (a, b)
Here,
22 * 4 = HCF (22, 4)
= 2 [because highest common factor of 22 and 4 is 2]
#### Question 20:
Let * be a binary operation on set of integers I, defined by a * b = 2a + b − 3. Find the value of 3 * 4.
Given: a * b = 2a + b − 3
Here,
3 * 4 = 2 (3) + 4 $-$3
= 6 + 4 $-$ 3
= 7
#### Question 21:
If a * b denotes the larger of 'a' and 'b' and if aob = (a * b) + 3, then write the value of (5) o (10), where * and o are binary operations.
Given: a * b denotes the larger of 'a' and 'b'.
Also, $a\circ b=\left(a*b\right)+3$
For a = 5 and b = 10
a * b = 5 * 10 = 10
$a\circ b=5\circ 10=\left(5*10\right)+3=10+3=13$.
#### Question 1:
If a * b = a2 + b2, then the value of (4 * 5) * 3 is
(a) (42 + 52) + 32
(b) (4 + 5)2 + 32
(c) 412 + 32
(d) (4 + 5 + 3)2
(c) $\left({41}^{2}+{3}^{2}\right)$
Given: a * b = a2 + b2
#### Question 2:
If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =
(a) 14
(b) 31
(c) 10
(d) 8
(c) 10
4.7 = (4 * 7) + 3
= 7 + 3
=10
#### Question 3:
On the power set P of a non-empty set A, we define an operation ∆ by
Then which are of the following statements is true about ∆
(a) commutative and associative without an identity
(b) commutative but not associative with an identity
(c) associative but not commutative without an identity
(d) associative and commutative with an identity
(d) associative and commutative with an identity
Let $\varphi$ be the identity element for $∆$ on P.
#### Question 4:
If the binary operation * on Z is defined by a * b = a2b2 + ab + 4, then value of (2 * 3) * 4 is
(a) 233
(b) 33
(c) 55
(d) −55
(b) 33
Given: a * b = a2b2 + ab + 4
#### Question 5:
Mark the correct alternative in the following question:
For the binary operation * on Z defined by a * b = a + b + 1, the identity clement is
(a) 0 (b) $-$1 (c) 1 (d) 2
We have,
a * b = a + b + 1
Let e be the identity element of *. Then,
Hence, the correct alternative is option (b).
#### Question 6:
If a binary operation * is defined on the set Z of integers as a * b = 3ab, then the value of (2 * 3) * 4 is
(a) 2
(b) 3
(c) 4
(d) 5
(d) 5
Given: a * b = 3ab
2 * 3 = 3 (2) $-$ 3
= 6 $-$ 3
= 3
(2 * 3) * 4 = 3 * 4
= 3 (3) $-$ 4
= 9 $-$ 4
= 5
#### Question 7:
Q+ denote the set of all positive rational numbers. If the binary operation a ⊙ on Q+ is defined as $a\odot =\frac{ab}{2}$, then the inverse of 3 is
(a) $\frac{4}{3}$
(b) 2
(c) $\frac{1}{3}$
(d) $\frac{2}{3}$
(a) $\frac{4}{3}$
Let e be the identity element in Q+ with respect to $\odot$ such that
Thus, 2 is the identity element in Q+ with respect to $\odot$.
#### Question 8:
If G is the set of all matrices of the form , then the identity element with respect to the multiplication of matrices as binary operation, is
(a) $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
(b) $\left[\begin{array}{cc}-1/2& -1/2\\ -1/2& -1/2\end{array}\right]$
(c) $\left[\begin{array}{cc}1/2& 1/1\\ 1/2& 1/2\end{array}\right]$
(d) $\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.
#### Question 9:
Q+ is the set of all positive rational numbers with the binary operation * defined by for all a, bQ+. The inverse of an element aQ+ is
(a) a
(b) $\frac{1}{a}$
(c) $\frac{2}{a}$
(d) $\frac{4}{a}$
(d) $\frac{4}{a}$
Let e be the identity element in Q+ with respect to * such that
Thus, 2 is the identity element in Q+ with respect to *.
#### Question 10:
If the binary operation ⊙ is defined on the set Q+ of all positive rational numbers by is equal to
(a) $\frac{3}{160}$
(b) $\frac{5}{160}$
(c) $\frac{3}{10}$
(d) $\frac{3}{40}$
(a) $\frac{3}{160}$
Given: $a\odot b=\frac{ab}{4}$
#### Question 11:
Let * be a binary operation defined on set Q − {1} by the rule a * b = a + bab. Then, the identify element for * is
(a) 1
(b) $\frac{a-1}{a}$
(c) $\frac{a}{a-1}$
(d) 0
(d) 0
Let e be the identity element in Q - {1} with respect to * such that
Thus, 0 is the identity element in Q $-$ {1} with respect to *.
#### Question 12:
Which of the following is true?
(a) * defined by $a*b=\frac{a+b}{2}$ is a binary operation on Z
(b) * defined by $a*b=\frac{a+b}{2}$ is a binary operation on Q
(c) all binary commutative operations are associative
(d) subtraction is a binary operation on N
(b) * defined by $a*b=\frac{a+b}{2}$ is a binary operation on Q.
Let us check each option one by one.
(a)
Hence, (a) is false.
(b)
Hence, (b) is true.
(c)
Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is not associative on N.
Therefore, all binary commutative operations are not associative.
Hence, (c) is false.
(d) Subtraction is not a binary operation on N because subtraction of any two natural numbers is not always a natural number.
For example: 2 and 4 are natural numbers.
2$-$4 = $-$2 which is not a natural number.
Hence, (d) is false.
#### Question 13:
The binary operation * defined on N by
a * b = a + b + ab for all a, bN is
(a) commutative only
(b) associative only
(c) commutative and associative both
(d) none of these
(c) commutative and associative both
Commutativity:
Thus, * is commutative on N.
Associativity:
Thus, * is associative on N.
#### Question 14:
The binary operation * is defined by a * b = a2 + b2 + ab + 1, then (2 * 3) * 2 is equal to
(a) 20
(b) 40
(c) 400
(d) 445
(d) 445
Given: a * b = a2 + b2 + ab + 1
#### Question 15:
Let * be a binary operation on R defined by a * b = ab + 1. Then, * is
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) both commutative and associative
(a) commutative but not associative
Commutativity:
Therefore, * is commutative on R.
Associativity:
Hence, * is not associative on R.
#### Question 16:
Subtraction of integers is
(a) commutative but no associative
(b) commutative and associative
(c) associative but not commutative
(d) neither commutative nor associative
(d) neither commutative nor associative
Subtraction of integers is not commutative
For example: If a = 1 and b = 2, then both are integers
$⇒-1\ne 1$
Subtraction of integers is not associative.
For example: If a = 1, b = 2, c = 3, then all are integers
#### Question 17:
The law a + b = b + a is called
(a) closure law
(b) associative law
(c) commutative law
(d) distributive law
(c) commutative law
The law a + b = b + a is commutative.
#### Question 18:
An operation * is defined on the set Z of non-zero integers by $a*b=\frac{a}{b}$ for all a, bZ. Then the property satisfied is
(a) closure
(b) commutative
(c) associative
(d) none of these
(d) none of these
* is not closure because when a = 1 and b = 2,
* is not commutative because when a = 1 and b = 2,
* is not associative because when a = 1, b = 2 and c = 3,
#### Question 19:
On Z an operation * is defined by a * b = a2 + b2 for all a, bZ. The operation * on Z is
(a) commutative and associative
(b) associative but not commutative
(c) not associative
(d) not a binary operation
(c) not associative
Commutativity:
Thus, * is commutative on Z.
Associativity:
Thus, * is not associative on Z.
#### Question 20:
A binary operation * on Z defined by a * b = 3a + b for all a, bZ, is
(a) commutative
(b) associative
(c) not commutative
(d) commutative and associative
(c) not commutative
Commutativity:
Thus, * is not commutative on Z.
#### Question 21:
Let * be a binary operation on Q+ defined by . The inverse of 0.1 is
(a) 105
(b) 104
(c) 106
(d) none of these
(a) 105
Let e be the identity element in Q+with respect to * such that
Thus, 100 is the identity element in Q+ with respect to *.
#### Question 22:
Let * be a binary operation on N defined by a * b = a + b + 10 for all a, bN. The identity element for * in N is
(a) −10
(b) 0
(c) 10
(d) non-existent
(d) non-existent
Let e be the identity element in N with respect to * such that
So, the identity element with respect to * does not exist in N.
#### Question 23:
Consider the binary operation * defined on Q − {1} by the rule
a * b = a + bab for all a, bQ − {1}
The identity element in Q − {1} is
(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) −1
(a) 0
Let e be the identity element in Q $-$ {1} with respect to * such that
Thus, 0 is the identity element in Q $-$ {1} with respect to *.
#### Question 24:
For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, bR − {1}, the inverse of a is
(a) $-a$
(b) $-\frac{a}{a+1}$
(c) $\frac{1}{a}$
(d) ${a}^{2}$
(b) $-\frac{a}{a+1}$
Let e be the identity element in R $-$ {1} with respect to * such that
Thus, 0 is the identity element in R $-$ {1}with respect to *.
#### Question 25:
For the multiplication of matrices as a binary operation on the set of all matrices of the form $\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]$, a, bR the inverse of $\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]$ is
(a) $\left[\begin{array}{cc}-2& 3\\ -3& -2\end{array}\right]$
(b) $\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right]$
(c) $\left[\begin{array}{cc}2/13& -3/13\\ 3/13& 2/13\end{array}\right]$
(d) $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
(c) $\left[\begin{array}{cc}2/13& -3/13\\ 3/13& 2/13\end{array}\right]$
To find the identity element,
So, the answer is (c).
#### Question 26:
On the set Q+ of all positive rational numbers a binary operation * is defined by . The inverse of 8 is
(a) $\frac{1}{8}$
(b) $\frac{1}{2}$
(c) 2
(d) 4
(b) $\frac{1}{2}$
Let e be the identity element in Q+ with respect to * such that
Thus, 2 is the identity element in Q+ with respect to *.
#### Question 27:
Let * be a binary operation defined on Q+ by the rule . The inverse of 4 * 6 is
(a) $\frac{9}{8}$
(b) $\frac{2}{3}$
(c) $\frac{3}{2}$
(d) none of these
(a) $\frac{9}{8}$
Let e be the identity element in Q+ with respect to * such that
Thus, 3 is the identity element in Q+ with respect to *.
#### Question 28:
The number of binary operation that can be defined on a set of 2 elements is
(a) 8
(b) 4
(c) 16
(d) 64
(c) 16
We know that the number of binary operations on a set of n elements is ${n}^{{n}^{2}}$.
So, the number of binary operations on a set of 2 elements is
#### Question 29:
The number of commutative binary operations that can be defined on a set of 2 elements is
(a) 8
(b) 6
(c) 4
(d) 2
(d) 2
The number of commutative binary operations on a set of n elements is ${n}^{\frac{n\left(n-1\right)}{2}}\phantom{\rule{0ex}{0ex}}$.
Therefore,
Number of commutative binary operations on a set of 2 elements =
#### Question 1:
Determine whether each of the following operations define a binary operation on the given set or not :
(i)
(ii)
(iii) .
(iv)
(v)
(vi)
(vii)
Thus, * is a binary operation on N.
Thus, * is not a binary operation on Z.
(iii) If a = 1 and b = 1,
a * b = a + b$-$ 2
= 1 + 1$-$ 2
= 0
Thus, there exist a = 1 and b = 1 such that a * b
So, * is not a binary operation on N.
(iv) Consider the composition table,
${×}_{6}$ 1 2 3 4 5 1 1 2 3 4 5 2 2 4 0 2 4 3 3 0 3 0 3 4 4 2 0 4 2 5 5 4 3 2 1
Here all the elements of the table are not in S.
(v) Consider the composition table,
${+}_{6}$ 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
Here all the elements of the table are in S.
Thus, ${×}_{6}$ is a binary operation on S.
(vii) If a = 2 and b = $-$1 in Q,
So, * is not a binary operation on Q.
#### Question 2:
Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z+, defined * by a * b = ab
(ii) On Z+, defined * by a * b = ab
(iii) On R, define by a*b = ab2
(iv) On Z+ define * by a * b = |ab|
(v) On Z+, define * by a * b = a
(vi) On R, define * by a * b = a + 4b2
Here, Z+ denotes the set of all non-negative integers.
(i) If a = 1 and b = 2 in Z+, then
#### Question 3:
Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.
Given: a * b = 2a + b − 3
3 * 4 = 2 (3) + 4 $-$ 3
= 6 + 4 $-$ 3
= 7
#### Question 4:
Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.
LCM 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 5 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5
In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.
If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 $\notin${1, 2, 3, 4, 5}.
Thus, * is not a binary operation on {1, 2, 3, 4, 5}.
#### Question 5:
Let S = {a, b, c}. Find the total number of binary operations on S.
Number of binary operations on a set with n elements is ${n}^{{n}^{2}}$.
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is ${3}^{{3}^{2}}={3}^{9}$
#### Question 6:
Find the total number of binary operations on {a, b}.
Number of binary operations on a set with n elements is ${n}^{{n}^{2}}$.
Here, S = {a, b}
Number of elements in S = 2
#### Question 7:
Let S be the set of all rational numbers of the form $\frac{m}{n}$, where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation.
Thus, * is not a binary operation.
#### Question 8:
Prove that the operation * on the set
defined by A * B = AB is a binary operation.
Thus, * is a binary operation on M.
#### Question 9:
The binary operation * : R $×$ R $\to$ R is defined as a * b = 2a + b. Find (2 * 3) * 4. [CBSE 2012]
As, a * b = 2a + b
So, (2 * 3) * 4 = [2(2) + 3] * 4
= [4 + 3] * 4
= 7 * 4
= 2(7) + 4
= 14 + 4
= 18
#### Question 10:
Let * be a binary operation on N given by a * b = LCM (a, b) for all a, b $\in$ N. Find 5 * 7. [CBSE 2012]
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Standard Active Last Updated: May 06, 2020 Track Document
ASTM E1057-15(2020)e1
# Standard Practice for Measuring Internal Rate of Return and Adjusted Internal Rate of Return for Investments in Buildings and Building Systems
Standard Practice for Measuring Internal Rate of Return and Adjusted Internal Rate of Return for Investments in Buildings and Building Systems E1057-15R20E01 ASTM|E1057-15R20E01|en-US Standard Practice for Measuring Internal Rate of Return and Adjusted Internal Rate of Return for Investments in Buildings and Building Systems Standard new BOS Vol. 04.11 Committee E06
\$ 69.00 In stock
Significance and Use
5.1 The IRR method has been used traditionally in finance and economics to measure the percentage yield on investment.
5.1.1 The IRR method is appropriate in most cases for evaluating whether a given building or building system will be economically efficient, that is, whether its time-adjusted benefits will exceed its time-adjusted costs over the period of concern to the decision-maker. However, it has deficiencies that limit its usefulness in choosing among projects competing for a limited budget.
5.2 The AIRR method is a measure of the overall rate of return that an investor can expect from an investment over a designated study period. It is appropriate both for evaluating whether a given building or building system will be economically efficient and for choosing among alternatives competing for a limited budget.
5.2.1 The AIRR method overcomes some, but not all, of the deficiencies of the IRR. The AIRR is particularly recommended over the IRR for allocating limited funding among competing projects.
Scope
1.1 This practice covers a procedure for calculating and interpreting the internal rate of return (IRR) and adjusted internal rate of return (AIRR) measures in the evaluation of building designs, systems, and equipment.
1.2 The values stated in inch-pound units are to be regarded as standard. The values given in parentheses are mathematical conversions to SI units that are provided for information only and are not considered standard.
1.3 This standard does not purport to address all of the safety concerns, if any, associated with its use. It is the responsibility of the user of this standard to establish appropriate safety, health, and environmental practices and determine the applicability of regulatory limitations prior to use.
1.4 This international standard was developed in accordance with internationally recognized principles on standardization established in the Decision on Principles for the Development of International Standards, Guides and Recommendations issued by the World Trade Organization Technical Barriers to Trade (TBT) Committee.
Price:
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# The story of Zero
43 %
57 %
Information about The story of Zero
Education
Published on February 21, 2014
Author: Castellina
Source: slideshare.net
Mathematics Assignment -Rachel Elizabeth Thomas X-B Roll no. 23
The Story of Zero
Perhaps the most fundamental contribution of ancient India to the progress of civilisation is the decimal system of numeration including the invention of the number zero. This system uses 9 digits and a symbol for zero to denote all integral numbers, by assigning a place value to the digits. This system was used in Vedas and Valmiki Ramayana. Mohanjodaro and Harappa civilisations (3000 B.C.) also used this system. The ancient Egyptians (5000 B.C.) had a system based on 10, but they didn't use positional notation. Thus to represent 673, they would draw six snares, seven heel bones and three vertical strokes. Babylonians in Mesopotamia (3000 B.C.) had a sexagesimal system using base 60. Greeks and Romans had a cumbersome system (try to write 2376 in Roman numerals).
Many civilisations had some concept of "zero" as nothing - for example, if you have two cows and they both die, you are left with nothing. However, the Indians were the first to see that zero can be used for something beyond nothing - at different places in a number, it adds different values. For example, 76 is different from 706, 7006, 760 etc. Brahmgupta (598 AD 660 AD) was the first to give the rules of operation of zero. • A + 0 = A, where A is any quantity. • A - 0 = A, • A * 0 = 0, • A / 0 = Not Defined He was wrong regarding the last formula. This mistake was corrected by Bhaskara (1114 AD - 1185 AD), who in his famous book Leelavati claimed that division of a quantity by zero is an infinite quantity or immutable God.
The ancient Indians represented zero as a circle with a dot inside. In Sanskrit, it was called "soonya". This and the decimal number system fascinated Arab scholars who came to India. Arab mathematician Al-Khowarizmi (790 AD 850 AD) wrote Hisab-al-Jabr wa-al-Muqabala (Calculation of Integration and Equation) which made Indian numbers popular. "Soonya" became "al-sifr" or "sifr". The impact of this book can be judged by the fact that "al-jabr" became "Algebra" of today. An Italian Leonardo Fibonacci (1170 AD - 1230 AD) took this number system to Europe. The Arabic "sifr" was called "zephirum" in Latin, and acquired many local names in Europe including "cypher". In the beginning, the merchants used to Roman numbers found the decimal system a new idea, and referred to these numbers as " infidel numbers", as the Arabs were called infidels because they had invaded the holy land of Palestine. However, nowadays this system is called Hindu-Arabic System. This positional system of representing integers revolutionised the mathematical calculations and also helped in Astronomy and accurate navigation. The use of positional system to indicate fractions was introduced around 1579 AD by Francois Viete. The dot for a decimal point came to be used a few years later, but did not become popular until its use by Napier.
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ROOT Reference Guide
peaks2.C File Reference
## Detailed Description
Example to illustrate the 2-d peak finder (class TSpectrum2).
This script generates a random number of 2-d gaussian peaks The position of the peaks is found via TSpectrum2 To execute this example, do:
root > .x peaks2.C (generate up to 50 peaks by default)
root > .x peaks2.C(10) (generate up to 10 peaks)
root > .x peaks2.C+(200) (generate up to 200 peaks via ACLIC)
Double_t x[n]
Definition: legend1.C:17
The script will iterate generating a new histogram having between 5 and the maximun number of peaks specified. Double Click on the bottom right corner of the pad to go to a new spectrum To Quit, select the "quit" item in the canvas "File" menu
#include "TSpectrum2.h"
#include "TCanvas.h"
#include "TRandom.h"
#include "TH2.h"
#include "TF2.h"
#include "TMath.h"
#include "TROOT.h"
TH2F *h2 = 0;
Int_t npeaks = 30;
Double_t fpeaks2(Double_t *x, Double_t *par) {
for (Int_t p=0;p<npeaks;p++) {
Double_t norm = par[5*p+0];
Double_t mean1 = par[5*p+1];
Double_t sigma1 = par[5*p+2];
Double_t mean2 = par[5*p+3];
Double_t sigma2 = par[5*p+4];
result += norm*TMath::Gaus(x[0],mean1,sigma1)*TMath::Gaus(x[1],mean2,sigma2);
}
return result;
}
void findPeak2() {
printf("Generating histogram with %d peaks\n",npeaks);
Int_t nbinsx = 200;
Int_t nbinsy = 200;
Double_t xmax = (Double_t)nbinsx;
Double_t ymax = (Double_t)nbinsy;
Double_t dx = (xmax-xmin)/nbinsx;
Double_t dy = (ymax-ymin)/nbinsy;
delete h2;
h2 = new TH2F("h2","test",nbinsx,xmin,xmax,nbinsy,ymin,ymax);
h2->SetStats(0);
//generate n peaks at random
Double_t par[3000];
for (p=0;p<npeaks;p++) {
par[5*p+0] = gRandom->Uniform(0.2,1);
par[5*p+1] = gRandom->Uniform(xmin,xmax);
par[5*p+2] = gRandom->Uniform(dx,5*dx);
par[5*p+3] = gRandom->Uniform(ymin,ymax);
par[5*p+4] = gRandom->Uniform(dy,5*dy);
}
TF2 *f2 = new TF2("f2",fpeaks2,xmin,xmax,ymin,ymax,5*npeaks);
f2->SetNpx(100);
f2->SetNpy(100);
f2->SetParameters(par);
TCanvas *c1 = (TCanvas*)gROOT->GetListOfCanvases()->FindObject("c1");
if (!c1) c1 = new TCanvas("c1","c1",10,10,1000,700);
h2->FillRandom("f2",500000);
//now the real stuff: Finding the peaks
Int_t nfound = s->Search(h2,2,"col");
//searching good and ghost peaks (approximation)
Int_t pf,ngood = 0;
Double_t *xpeaks = s->GetPositionX();
Double_t *ypeaks = s->GetPositionY();
for (p=0;p<npeaks;p++) {
for (pf=0;pf<nfound;pf++) {
Double_t diffx = TMath::Abs(xpeaks[pf] - par[5*p+1]);
Double_t diffy = TMath::Abs(ypeaks[pf] - par[5*p+3]);
if (diffx < 2*dx && diffy < 2*dy) ngood++;
}
}
if (ngood > nfound) ngood = nfound;
//Search ghost peaks (approximation)
Int_t nghost = 0;
for (pf=0;pf<nfound;pf++) {
Int_t nf=0;
for (p=0;p<npeaks;p++) {
Double_t diffx = TMath::Abs(xpeaks[pf] - par[5*p+1]);
Double_t diffy = TMath::Abs(ypeaks[pf] - par[5*p+3]);
if (diffx < 2*dx && diffy < 2*dy) nf++;
}
if (nf == 0) nghost++;
}
c1->Update();
s->Print();
printf("Gener=%d, Found=%d, Good=%d, Ghost=%d\n",npeaks,nfound,ngood,nghost);
if (!gROOT->IsBatch()) {
printf("\nDouble click in the bottom right corner of the pad to continue\n");
c1->WaitPrimitive();
}
}
void peaks2(Int_t maxpeaks=50) {
s = new TSpectrum2(2*maxpeaks);
for (int i=0; i<10; ++i) {
npeaks = (Int_t)gRandom->Uniform(5,maxpeaks);
findPeak2();
}
}
int Int_t
Definition: RtypesCore.h:45
double Double_t
Definition: RtypesCore.h:59
winID h TVirtualViewer3D TVirtualGLPainter p
Option_t Option_t TPoint TPoint const char GetTextMagnitude GetFillStyle GetLineColor GetLineWidth GetMarkerStyle GetTextAlign GetTextColor GetTextSize void char Point_t Rectangle_t WindowAttributes_t Float_t Float_t Float_t Int_t Int_t UInt_t UInt_t Rectangle_t result
float xmin
Definition: THbookFile.cxx:95
float ymin
Definition: THbookFile.cxx:95
float xmax
Definition: THbookFile.cxx:95
float ymax
Definition: THbookFile.cxx:95
#define gROOT
Definition: TROOT.h:404
R__EXTERN TRandom * gRandom
Definition: TRandom.h:62
The Canvas class.
Definition: TCanvas.h:23
virtual void SetNpx(Int_t npx=100)
Set the number of points used to draw the function.
Definition: TF1.cxx:3481
virtual void SetParameters(const Double_t *params)
Definition: TF1.h:649
A 2-Dim function with parameters.
Definition: TF2.h:29
virtual void SetNpy(Int_t npy=100)
Set the number of points used to draw the function.
Definition: TF2.cxx:955
virtual void SetStats(Bool_t stats=kTRUE)
Set statistics option on/off.
Definition: TH1.cxx:8867
2-D histogram with a float per channel (see TH1 documentation)}
Definition: TH2.h:257
void FillRandom(const char *fname, Int_t ntimes=5000, TRandom *rng=nullptr) override
Fill histogram following distribution in function fname.
Definition: TH2.cxx:677
virtual Double_t Uniform(Double_t x1=1)
Returns a uniform deviate on the interval (0, x1).
Definition: TRandom.cxx:672
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http://www.jiskha.com/display.cgi?id=1347215078
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# College Physics
posted by on .
A 6107 kg elevator is moving 4.6 m/s up. The tension in the elevator cable is 66,512.6 N. Determine the acceleration of the elevator (state the appropriate mks units).
• College Physics - ,
T= Mg+Ma
66,512.6=6107(9.81)+6107(a)
T=1.01
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| 0.723652
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https://gcseguide.co.uk/quizzes/probability/
| 1,563,701,707,000,000,000
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| 405,103,969
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# PROBABILITY
One letter is selected at random from the word ‘MATHEMATICS’. What is probability of selecting a T?
Three coins are tossed at the same time. What is the probability of getting a head and two tails?
A ball is randomly selected from a bag containing 4 red balls, 5 green balls and 6 black balls. The first ball is replaced and then a second ball is picked.
What is the probability of getting two red balls?
A ball is randomly selected from a bag containing 4 red balls, 5 green balls and 6 black balls. The first ball is replaced and then a second ball is picked.
What is the probability of getting a red ball and a green ball?
A ball is randomly selected from a bag containing 4 red balls, 5 green balls and 6 black balls. The first ball is replaced and then a second ball is picked.
What is the probability of getting at least one black ball?
A bag contains 7 yellow balls, 3 green balls and 2 white balls. A ball is drawn and is not replaced. A second ball is drawn.
What is the probability of getting two yellow balls?
A bag contains 7 yellow balls, 3 green balls and 2 white balls. A ball is drawn and is not replaced. A second ball is drawn.
What is the probability of getting one green ball and one white ball?
A bag contains 7 yellow balls, 3 green balls and 2 white balls. A ball is drawn and is not replaced. A second ball is drawn.
What is the probability of getting two green balls?
A bag contains 7 yellow balls, 3 green balls and 2 white balls. A ball is drawn and is not replaced. A second ball is drawn.
What is the probability of getting two balls of the same colour?
When two dice are thrown simultaneously, what is the probability of obtaining the same number on both dice?
Cathy has 9 germanium plants. She knows that three will flower red, two will flower pink and four will flower white.
What is the probability that the first plant to flower will be pink?
Cathy has 9 germanium plants. She knows that three will flower red, two will flower pink and four will flower white.
What is the probability that one is red and the other is pink?
Cathy has 9 germanium plants. She knows that three will flower red, two will flower pink and four will flower white.
What is the probability that both are white?
Cathy has 9 germanium plants. She knows that three will flower red, two will flower pink and four will flower white.
What is the probability that at least one of them in pink?
Cathy has 9 germanium plants. She knows that three will flower red, two will flower pink and four will flower white.
What is the probability that the first three plants to flower are white?
| 595
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| 4.15625
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latest
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| 0.962889
|
https://thecontemporaryaustincdn.global.ssl.fastly.net/edition/beckmann-sybilla-2017-mathematics-for-elementary-teachers-with-activities-fifth-edition.html
| 1,606,180,744,000,000,000
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| 524,068,612
| 4,773
|
# Beckmann sybilla (2017). mathematics for elementary teachers with activities fifth edition. Beckmann, Mathematics for Elementary Teachers with Activities
Beckmann sybilla (2017). mathematics for elementary teachers with activities fifth edition Rating: 6,4/10 1476 reviews
## Beckmann, Mathematics for Elementary Teachers with Activities, Loose
We then discuss the general methods and why they work. He has retired as Dean of the Palm Desert Campus and Professor of Communication at California State University, San Bernardino, where he was named Outstanding Professor. The distance between adjacent tick marks equals 1 1 3 10. A Measure- ment Perspective This measurement perspective on fractions becomes especially important later on for understanding fraction multiplication, in particular for understanding a fraction as a multiplier. Loose-Leaf Versions also offer a great value; this format costs significantly less than a new textbook. He needs 1 of the dough because the 3 cup of butter consists of 3 parts each of which is 1 cup but he only wants 1 of those parts.
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## Mathematics for elementary teachers 5th edition beckmann solutions ma…
This takes many examples as well as guidance from instructors. There are additional problems about applying congruence. See text for full discussion of where to locate fractions on number lines and see the answer to the practice problem in this section. In solving this problem, 2 4 3 also appears as 6. Have the students fold each strip to make fourths and then unfold and shade 3 of the fourths to show 3 of each strip. The two smallest common denominators are 12 and 24. The whole for 1 is the area of the playground.
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## Beckmann, Mathematics for Elementary Teachers with Activities
I constantly here, don't reinvent the wheel! To explain why an amount is 1 of the whole or unit amount , students can explain that 3 copies of the amount makes the whole, or that the amount was obtained by dividing the whole into 3 equal parts. Class Activity 2O: What is another way to Compare these Fractions? After graduating from Texas Lutheran University and Stephen F. You might like to assign this section as reading and weave a discussion of this section into the next section. She has also been a member on a number of other national panels and committees working to improve mathematics education. The section on sequences now includes graphs and equations for arithmetic sequences as part of the development of linear relationships, which started in Chapter 7 with proportional relationships. Another explanation involves reasoning about math drawings. More challenging problems are indicated with a purple asterisk.
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## Mathematics for elementary teachers 5th edition beckmann solutions ma…
Class Activity 2K: Interpreting and Using Common Denomi- nators 1. Then 4 of those green triangles will have the area of the original design because 4 fourths make the whole. In the drawing, we turn 3 into 6 in order to solve the problem. Being able to view an amount in two ways, as 1 group and also as some number of units, will be important for understanding fraction multiplication, and in particular, for understanding a fraction as a multiplier. Students may think of other ways of highlighting the intervals.
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## Beckmann, Mathematics for Elementary Teachers with Activities, Loose
B A Class Activity 2D: Relating Fractions to Wholes This activity is an opportunity not only for problem-solving, thereby addressing Stan- dard for Mathematical Practice 1, but also for attention to precision in using the definition of fraction, thereby attending to Standard for Mathematical Practice 6. Students will sometimes give an explanation that involves repeating a drawing of the fraction 3 times. Common Core State Standards Standards for Mathematical Practice 1, 3 are directly related. In addition to basic skills review, the MyMathLab course includes a wealth of resources to help students visualize the concepts and understand how they come into play in an elementary classroom. The answer depends on what we take the whole to be. You could also show or assign Video 12 where prospective teachers discuss a misconception in comparing fractions that many noticed when they interviewed students.
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## Mathematics for Elementary Teachers with Activities Mathem Elemen Teache Act_5 5th edition
Thinking in terms of buttons, it takes 32 of Strip A to make Strip B or 2 8. Class Activity 2I: Explaining Equivalent Fractions You could show or assign the video Equivalent fractions after students complete the activity. Both fractions are equal to 1. It was published by Pearson and has a total of 1168 pages in the book. Instructors, contact your Pearson representative for more information. Because 5 means 5 parts, each of which is one fourth of the original design , we need to find how to break Design A into 5 equal parts. Six green triangles make the hexagon, so the other design can be made from 10 triangles.
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## Mathematics for Elementary Teachers with Activities (Subscription), 5th Edition
It will make your like much easier. As the tick marks should be 1 apart, draw 3 equally spaced tick marks between the tick marks for 0 and 4 and then continue on. Ask students to think about other examples and whether Claire will always get the right answer using her approach. Fractions and Problem Solving 3. We can make 4 equal parts horizontally or vertically, which look different.
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## Mathematics for Elementary Teachers with Activities by Sybilla Beckmann (2017, Hardcover) for sale online
The Skills Review MyLab Math provides review and skill development that complements the text, helping students brush-up on skills needed to be successful in class. New color-coordinated math drawings continue into Chapters 5, 6, and 7 to highlight connections across multiplication, division, and proportional relationships, and across whole numbers and fractions. Revisions to the text, Class Activities, and problems provide better opportunities for comparing distributions. The whole associated with the 3 in the problem is a cup of cereal. Description This is completed downloadable of Mathematics for Elementary Teachers with Activities 5th Edition by Sybilla Beckmann Solution Manual. This package includes MyLab Math. It would be easy to think incorrectly that this 100% refers to all the bugs.
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https://www.stat.math.ethz.ch/pipermail/r-help/2006-August/111016.html
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# [R] Colour-coding intervals on a line
Jim Lemon jim at bitwrit.com.au
Sun Aug 13 01:16:51 CEST 2006
```Sara-Jane Dunn wrote:
> Hi,
>
> This is a simple version of something that I am trying to do. If I can
> sort the problem basically, I figure I should be able to sort it for the
> program I'm writing (which would take longer to explain).
>
> I need to know if there is any way of using different colours for
> different intervals of a line on a graph. Eg. If I plot the line y=x for
> x=1:10, and split this line into 106 intervals (i.e. not a 'nice' number
> of intervals) how could I colour different intervals different colours
> without having to use segments and specify coordinates to join?
>
Hi Sara-Jane,
This is a problem that I have been facing myself. I wanted to display
color-coded curves that indicated something depending upon the values
plotted. This might be useful to you, although you will need the
functions color.scale and rescale from the plotrix package to make it work.
color.scale.lines<-function(x,y,redrange,greenrange,bluerange,
scale.to="y",...) {
lenx<-length(x)
leny<-length(y)
nseg<-lenx-1
if(lenx != leny) {
warning("x and y are different lengths - some values will be ignored")
if(lenx>leny) nseg<-leny-1
}
if(scale.to=="y")
lcol<-color.scale(y[1:nseg],redrange,greenrange,bluerange)
else lcol<-color.scale(x[1:nseg],redrange,greenrange,bluerange)
segments(x[1:nseg],y[1:nseg],x[2:(nseg+1)],y[2:(nseg+1)],col=lcol,...)
}
plot(0,xlim=c(0,pi),ylim=c(0,1),type="n")
x<-seq(0,pi,length=106)
color.scale.lines(x,sin(x)*10,c(0,1),c(1,0),0,lwd=2)
Jim
```
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https://rf.mokslasplius.lt/obtaining-1f-noise-from-superposition-of-lorentzians/
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# Obtaining 1/f noise from superposition of Lorentzians
Earlier we have taken a look at a power spectral density of a signal generated by a single charge carrier. We have seen that it generates a signal composed of pulses and gaps. If the pulse and gap duration is distributed according to the exponential distribution (i.e., Poisson process model applies), then the power spectral density of the signal has a characteristic Lorentzian shape. Which is nothing alike 1/f noise, which we are looking for. But maybe we can still obtain it?
## Superposition of Lorentzians
So, earlier we have obtained that the power spectral density generated by a single charge carrier, for which the Poisson process model applies, is given by
$$S\left(f\right)=\frac{4a^{2}\bar{\nu}}{\left(\lambda_{\theta}+\lambda_{\tau}\right)^{2}+4\pi^{2}f^{2}}.$$
Here for the sake of simplicity let us assume that $$\lambda_{\theta} = \lambda_{\tau}$$. As the rates are assumed to be equal let us drop the indices. This allows us to a simpler form for the power spectral density in our further derivations, i.e.,
$$S\left(f\right)=\frac{a^{2}\bar{\nu}}{\lambda^{2}+\pi^{2}f^{2}}.$$
Our model will rely on superposition of these Lorentzian shape with different characteristic rates $$\lambda$$. As is well known, the Fourier transform of a sum of two signals (functions) is the same as the sum of their transforms:
$$\mathcal{F}\left\{f(t)+g(t)\right\} =\mathcal{F}\left\{f(t)\right\}+\mathcal{F}\left\{g(t)\right\}.$$
Power spectral density of the sum of two signals (functions) is not necessarily a sum of their individual power spectral densities. If the two signals are correlated, there will be cross spectral density term. But if the two signals are independent (as we assume is the case here), we will have just an ordinary sum of the components.
Let us assume that the distribution of characteristic rates is continuous uniform distribution (with certain $$\lambda_{\text{min}}$$ and $$\lambda_{\text{max}}$$ bounds). Then:
$$S\left(f\right)=\frac{1}{\lambda_{\text{max}}-\lambda_{\text{min}}}\int_{\lambda_{\text{min}}}^{\lambda_{\text{max}}} \frac{a^{2}\bar{\nu}}{\lambda^{2}+\pi^{2}f^{2}} d \lambda = \frac{a^2 \bar{\nu}\left[\operatorname{arccot}\left(\frac{\pi f}{\lambda_{\text{max}}}\right)-\operatorname{arccot}\left(\frac{\pi f}{\lambda_{\text{min}}}\right)\right]}{(\lambda_{\text{max}}-\lambda_{\text{min}}) \pi f}.$$
At first there doesn't seem to be any problem with this, but recall that $$\bar{\nu}$$ represents mean number of pulses per unit time. So, it does depend on $$\lambda$$. In our particular case (when the pulse and gap duration distributions match) we have that $$\bar{\nu} = \lambda / 2$$. To quickly fix this let us dynamically scale the pulse height
$$a = \sqrt{\frac{1}{\bar{\nu}}} = \sqrt{\frac{2}{\lambda}}.$$
Then we simply have that $$a^2 \bar{\nu} = 1$$ for any $$\lambda$$, and we can ignore this term as we already did in the derivation above.
## Interactive app
To effectively show the key point of this post (i.e., that you can get 1/f noise out of superposition of Lorentzian shapes) here we do not generate the signals themselves. Unlike in the earlier posts here instead we just add the Lorentzian shapes, which should emerge as a result of long duration detailed simulations. Simulating individual signals would likely just take unreasonably long until properly looking result would emerge.
So, in the app below you can adjust the bounds of the $$\lambda$$ distribution. In the plot the simulation results are shown as red dots, and they quickly converge towards analytically derived formula (shown as black curve). As you change the bound of the $$\lambda$$ distribution observe how the range of frequencies over which 1/f noise is observed changes. Note that for the lowest frequencies white noise dominates, while for the highest frequencies $$S(f) \sim 1/f^2$$ dependence emerges.
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https://www.dyskalkulia.com.pl/en4/oxygen/01-Jan/4381.html
| 1,632,267,963,000,000,000
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home oxygen cylinder sizes liters calculator conversion tool
Production Environment
Cooperation partner
Oxygen Tank Duration - Study Guide • RespCalc ...- home oxygen cylinder sizes liters calculator conversion tool ,Aug 07, 2020·Flow rate in L/min Oxygen Cylinder Conversion Factors in psi-1 L-1 • D Tank = 0.16 • E Tank = 0.28 • G Tank = 2.41 • H/K Tank = 3.14 • M tank = 1.56. Online Oxygen Tank Duration Calculator. 1. A size D cylinder of oxygen has a remaining pressure of 1200 psi. ... The home care patient uses the system for 8 hours during sleep at a flow ...Oxygen Cylinder Sizes and Info - Applied Home Healthcare ...Home Latest NewsOxygen Cylinder Sizes and Info. Oxygen Cylinder Sizes and Info. Nov 27, 2013. Print. Posted By Victoria Marquard-Schultz in General. You Might Also Like. Categories. Oxygen (62) Equipment (54) Regulatory Compliance (42) OxyGo (31) Oxygen Safety (27) Business (22) General (21) Training Education (16)
Calculate Conversion - UIG) i
oxygen gas and liquid unit conversion tables - weight, gas volume, liquid volume (pounds, kilograms, standard cubic feet, standard cubic meters, gallons, liters)
Tank Volume Calculator - Inch Calculator
How to Calculate The Volume of a Tank. The volume or capacity of a tank can be found in a few easy steps. Of course, the calculator above is the easiest way to calculate tank volume, but follow along to learn how to calculate it yourself. Step One: Measure the Tank. The first step is to measure the key dimensions of the tank.
How to Calculate Oxygen Tank Duration
Jul 14, 2020·D size = 0.16. E size = 0.28. G size = 2.41. H and K size = 3.14. M size = 1.56. As an example, assume a D size, which has a conversion factor of 0.16. Calculate the available litres of oxygen by multiplying the available cylinder pressure by the conversion factor. Continuing with the example: Available litres = 1,800 x 0.16 = 288 litres ...
Use the tool below to calculate an ethanol mix! – TunePlus ...
Ethanol Blend Calculator Gas tank size Gallons Liters Gasoline Ethanol Percentage E85 Ethanol Percentage Target Ethanol Mix Current Fuel Level Percentage Current Ethanol Mix Results E85 to Add Pump Gas to Add Resulting Mix ... Shortblocks/Cylinder Heads FAQ | E30 Mix Tool +-Frequently Asked Questions ... Use the tool below to calculate an ...
Oxygen Tank Duration Calculator - Respiratory Cram
Oct 15, 2020·If you don’t have access to the calculator above simply use the following formula to calculate the minutes remaining in your oxygen tank: Here’s what you will need: Tank size & conversion factor for that tank; PSIG (amount of pressure remaining in the tank) Flow rate (LPM) Formula: Minutes remaining = PSIG * Tank conversion factor/Flow rate
Medical gas cylinder data chart - Home | BOC Healthcare
Carbon dioxide/oxygen mixtures, 5% CO₂/95% O₂, 10% CO₂/90% O₂, 20% CO₂/80% O₂ AV L Cylinder code AV - Various L - Various Nominal content (litres) 1460 7300 Nominal pressure (bar) 137 137 Valve outlet connection 5/8” BSP (F) Side outlet 5/8” BSP (F) Side outlet Valve outlet specification BS 341 No.3 Bullnose BS 341 No.3 Bullnose
Calculate Conversion - UIG) i
oxygen gas and liquid unit conversion tables - weight, gas volume, liquid volume (pounds, kilograms, standard cubic feet, standard cubic meters, gallons, liters)
Milliliters and Liters Converter (mL and L)
To convert from milliliters to liters, multiply your figure by 0.001 (or divide by 1000) . Other individual liquid volume converters Liters to Gallons (US) , Liters to Gallons (UK) , Milliliters to Liters , Cubic Feet to Cubic Meters , Cubic Feet and Gallons , Cubic Inches to Cubic Centimeters , Cubic Feet to Cubic Yards , Pints to Fluid Ounces .
Oxygen Cylinder Guidance - Virginia
Cylinder Conversion Factors and Max Tank PSI Cylinder Size Conversion Factor Max PSI* Amt. of oxygen when full D 0.16 4000 350 liters H or K 3.14 4500 6,900 liters M 1.56 3450 3,000 liters E 0.28 6000 625 liters *(Max PSI can vary by tank manufacturer; the above numbers are the most common) Portable Oxygen Tank 12VAC5‐31‐860. Required ...
Oxygen Cylinder Sizes and Info - Applied Home Healthcare ...
Home Latest NewsOxygen Cylinder Sizes and Info. Oxygen Cylinder Sizes and Info. Nov 27, 2013. Print. Posted By Victoria Marquard-Schultz in General. You Might Also Like. Categories. Oxygen (62) Equipment (54) Regulatory Compliance (42) OxyGo (31) Oxygen Safety (27) Business (22) General (21) Training Education (16)
Oxygen Tank Duration Calculation • RespCalc - Respiratory ...
Oxygen cylinder tank duration calculation approximates how much time is left in a oxygen tank based on oxygen flow settings. Duration of Flow = Oxygen Tank Conversion Factor * Remaining Tank Pressure (psi) / Continuous Flow Rate (L/min) Oxygen Cylinder Conversion Factors • D Tank = 0.16 • E Tank = 0.28 • G Tank = 2.41 • H/K Tank = 3.14
Oxygen Tank Size & Duration Times - PHC Home Medical ...
D Cylinder: approx length 17" C Cylinder: approx length 11" M9 Cylinder: approx length 15" M6 Cylinder: approx length 12" M4 Cylinder: approx length 9" These are approx cylinder lengths. Size may vary slightly by manufacturer: Want to know more about Pulse-Dose? Read about Pulse-Dose versus Continuous Flow Oxygen Delivery Systems and how they work:
Use the tool below to calculate an ethanol mix! – TunePlus ...
Ethanol Blend Calculator Gas tank size Gallons Liters Gasoline Ethanol Percentage E85 Ethanol Percentage Target Ethanol Mix Current Fuel Level Percentage Current Ethanol Mix Results E85 to Add Pump Gas to Add Resulting Mix ... Shortblocks/Cylinder Heads FAQ | E30 Mix Tool +-Frequently Asked Questions ... Use the tool below to calculate an ...
A Guide to Oxygen Concentrator Liter Flows
Oct 02, 2014·An oxygen flow rate of 2 LPM means the patient will have 2 liters of oxygen flowing into their nostrils over a period of 1 minute. Oxygen prescriptions generally run from 1 liter per minute to 10 liters per minute with 70% of those patients being prescribed 2 liters or less.
Oxygen Quantity Conversions Calculator - CreatifWerks
Oxygen Quantity Conversions Calculator. Oxygen Quantity Conversions Calculator , This Calculator will help you to convert Oxygen Quantity from different unit ( Pounds , Kg, Cubic Feet ( SCF ) , CU Meter (nm3), gallons, liters)
Oxygen Cylinder Duration Chart Nominal ... - BOC Home …
Oxygen Cylinder Duration Chart. Nominal duration versus selected flowrate: The cylinder duration times are approximate and are to be used as guidance. The cylinder contents gauge should be ... Full (100%) Half (50%) Low (25%) Cylinder Size: Flowrate* (lit/min) Duration (hours) Duration (Mins) Duration (hours) Duration (Mins) Duration (hours ...
Milliliters and Liters Converter (mL and L)
To convert from milliliters to liters, multiply your figure by 0.001 (or divide by 1000) . Other individual liquid volume converters Liters to Gallons (US) , Liters to Gallons (UK) , Milliliters to Liters , Cubic Feet to Cubic Meters , Cubic Feet and Gallons , Cubic Inches to Cubic Centimeters , Cubic Feet to Cubic Yards , Pints to Fluid Ounces .
Oxygen Cylinder Sizes and Contents
Oxygen cylinder sizes and contents Size of Volume of Weight Height Diameter Valve Aramid Time Time cylinder gas (L) (kg) (full) (cm) (cm) (at 10 L/min) (at 15 L/min) C 170 2.23 490 89 Pin index No 17 min 11 min D 340 3.86 535 102 Pin index No 34 min 22 min CDb 460 3.25 480 100 Star Schr¨ader Yes 46 …
What Are Some Standard Medical Oxygen Cylinder Sizes?
Apr 12, 2020·Medical oxygen cylinder manufacturers use letters to differentiate cylinder sizes. For example, "D" cylinders, the most prevalent size, typically contain between 300 liters and 350 liters of oxygen, although some may have as much as 415 liters, while "E" cylinders carry between 625 liters and 682 liters, explains DeAnza College.
Oxygen Cylinder - Cornell University
Oxygen Tank Duration. Estimate the time that an oxygen cylinder will support delivery of gas at a particular flow, or estimate the pressure in an oxygen cylinder that will support a particular flow for a particular duration. These calculations apply not only to oxygen but for other gases.
Gas calculator - Home | Linde Gas
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Unit Converter
Quick, free, online unit converter that converts common units of measurement, along with 77 other converters covering an assortment of units. The site also includes a predictive tool that suggests possible conversions based on input, allowing for easier navigation …
How to Calculate Oxygen Amount According to Liter Flow
Apr 02, 2016·Note that when a nasal cannula is used for oxygen, the liter flow number will be set between 1 and 6 liters. When the dial is set at 1 liter, 24 percent oxygen is being delivered. For each increase in the number on the flow meter dial, the amount of oxygen delivered increases by 4 percent.
Calculate Conversion - UIG) i
oxygen gas and liquid unit conversion tables - weight, gas volume, liquid volume (pounds, kilograms, standard cubic feet, standard cubic meters, gallons, liters)
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• HOME > News > square pipe weight table
# square pipe weight table
Calculating the weight of your steel square pipe has never been easier thanks to this handy guide. Containing an comprehensive table of weights, you’ll have all the information needed to accurately determine the mass of your steel square pipe.
Steel square pipes remain a popular choice for a myriad of industrial and construction roles, from providing critical support to adding stylish embellishment. Whatever the purpose, an essential part of the job is understanding the weight of the pipe so it can be correctly designed, erected and moved. Figuring the weight of a steel square pipe is not difficult – made still simpler with the aid of a heavy-duty square pipe weight table.
Unveiling the Weight Capacities of Square Pipes
Need to know the weight of a steel square pipe? A square pipe weight table can be of help! These helpful charts provide details of the mass of multiple sizes and thicknesses of steel square pipes, presented in either pounds per foot or kilos per meter. Liaising with a steel pipe distributor or manufacturer should provide access to these tables, which are organised conveniently by diameter, wall thickness, and length.
What Benefits Come with Knowing Square Pipe Weights?
When wanting to figure out the weight of a steel square pipe, a weight table is an invaluable asset. Without the help of a handy table, you’d have to do lengthy computations that factor in the size, composition, and wall thickness of the pipe. This leads to a laborious process that often carries with it potential for inaccuracies. But with a weight table offering quick access to weight data based on measurements and wall depth, the whole ordeal is significantly cut down – leaving time for other activities and reducing the chances of an oversight.
Unlocking the Secrets of a Square Pipe Weight Table
Obtaining the weight of your square pipes is a piece of cake! Begin by calculating the diameter and wall thickness of the steel structures. Locate the dimensions on the pipe weight table and pick out their corresponding rows. Then, check out the columns to find the weight per foot or meter and multiply this amount by the length of your pipe for a complete tally of its total weight.
Are you in need of a handy steel square pipe? Well, if it has a 2-inch diameter and 0.25-inch wall thickening, the weight per foot would be estimated at 4.32 pounds. Want to know the full weight of a 10 foot long model? Multiply 4.32 pounds by 10 and you’ll get 43.2 pounds!
Uncovering the Weight of Square Steel Pipe: Benefits of Knowing
Gaining an understanding of the heft of a steel square pipe can bring many advantageous results, such as:
– Computing delivery expenses for the shipment of pipes – Crafting structures with suitable force-carrying capabilities – Optimizing material usage to lower expenditure and reduce waste – Confirming safety by warding off overloads or overcapacity of constructs.
For a successful industrial or construction project, knowing the weight of a steel square pipe is essential. A weight table simplifies the process of weight calculation, which in turn reduces the chances of mistakes occurring. Without one, there could be disastrous results, thus it is indispensable to have a weight table for square pipes close at hand. It is the way forward in order to ensure that all projects are designed, constructed, and delivered effectively and with precision.
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• Post time: 2023-05-30
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# Mathtacular
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Cut out each card and pass them out randomly to students. The student whose card says "I have the first question" reads their card first. Whoever has the card with the algebraic expression that matches the words the previous student read out goes
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This will give students a visual all on one sheet to compare the three terms. Great to have as a refresher all throughout the course! I personally will be making an anchor chart after we fill it out as a class.
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These charts help students translate words to algebraic expressions and then come up with how to write algebraic expressions in words. Can be used as independent work or as a class activity.
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These guided notes will teach students to multiply monomials by binomials and trinomials using three methods: area models, algebra tiles, and the distributive property, followed by examples for independent practice.
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This can be filled out as a class or in a group activity. I will be giving each group an operation to create a list for. They will have 5 minutes to brainstorm together all the words that might appear in word problems that indicate that operation
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This activity has students solving both linear and quadratic equations to complete a sudoku. The linear equations provide enough clues that students should be able to figure out the rest using sudoku rules. However, for those who are struggling or
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http://electronics.stackexchange.com/questions/29423/rotating-a-gyroscopes-output-values-into-an-earth-relative-reference-frame?answertab=votes
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# Rotating a gyroscope's output values into an earth-relative reference frame
For whatever reason, I seem to have gotten myself twisted into quite a confusion on how to process data from my 3 axis gyro and 3 axis accelerometer to get gyro rotation values related to the earth's reference frame. i.e. Given my accelerometer measures gravity (we assume it's stable) along the sensors (x,y,z) reference frame, how do I adjust the values of my gyro, measured along the sensors same (x,y,z) reference frame such that I am actually measuring the rotating about the Earth's frame x (North), y (West), and z (downwards) direction?
I just can't seem to wrap my head around the use of the rotation matrix. Do I simply employ a 3d rotation matrix, and multiply the vector, $[gyro_x, gyro_y, gyro_z]^T$, by a rotation matrix, R? I then assume the rotation angles, are somehow derived from the filtered accelerometer outputs, but I'm having trouble deciding on what those should be.
I have yet to find a concise reference on how to move the rotation and acceleration values from the sensor reference frame into the earth global reference frame. Perhaps someone can help? Thanks
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What do you mean by "Earth's x (forward motion), y (port direction), and z (downwards) direction" ? In what way is the earth itself moving forward? Where is earth's portward direction? (Am I safe in assuming you're not modelling earth moving in its orbit?) These sound non-sensical. A system of forward/port/down sounds more like the frame on a moving vehicle. I'd expect an earth frame to have axes like east/north/up, or north/east/down, etc. – JustJeff Apr 7 '12 at 2:41
Jeff, you're right. It is confusing. I'll clarify. – gallamine Apr 7 '12 at 15:20
Well, that's an improvement, but now (not trying to be a troll, honestly) I feel compelled to point out that North/West/Down is a left-handed coordinate system. Most of the time the systems you encounter will be right-handed ones. Using the directions you supplied as the positive axes, these would be North/West/Up or West/North/Down, or even Up/North/West. I think systems involving North and East are more common, though. – JustJeff Apr 7 '12 at 18:27
It's possible that your confusion stems from the fact that there are multiple solutions to the problem. While your accelerometer can tell you which way is up, it cannot distinguish between North and West. I.E. If you rotate the device about the vertical axis, the outputs of the accelerometers won't change.
How can you distinguish between North and West? The best way is probably to add a digital compass to the mix. Alternatively, you may not care to know the difference between real North and real West. You may only want two orthogonal horizontal axes. I'll assume the latter.
Define our frames first. The device's frame is (X, Y, Z). The earth's frame is (V, H1, H2).
Lets assume your accelerometer readings (Ax, Ay, Az) are in the range -1 .. +1, where +1 means straight up. Immediately you know which direction is up: it's simply (Ax, Ay, Az). But how do we obtain a horizontal axis? There's a function called the Cross Product, which takes any two vectors as inputs, and returns another vector at right angles to both. Therefore, we can easily find a vector at right angles to Up. In C:
Vector3D V = (Ax, Ay, Az);
Vector3D H1 = RANDOM_VECTOR;
Vector3D H2 = CrossProduct(V, H1);
H1 = CrossProduct(V, H2);
Normalise(H1);
Normalise(H2);
So, now we have a vertical vector V, and two horizontal vectors H1 and H2. Now we just need to rotate the gyroscope readings into the same frame.
Let's call the gyroscope readings (Gx, Gy, Gz). We're going to convert them into earth frame rotation coordinates (GV, GH1, GH2). All you have to do is to think about the gyro readings like a single 3D vector. Which way is it pointing in the device's frame. Which way is it pointing in the Earth's frame?
GV = (Gx*V.x) + (Gy*V.y) + (Gz*V.z);
GH1 = (Gx*H1.x) + (Gy*H1.y) + (Gz*H1.z);
GH2 = (Gx*H2.x) + (Gy*H2.y) + (Gz*H2.z);
(I hope that's right)...
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Thanks Rocketmagnet! That does help. Let me see if I've got this: Effectively you're creating a matrix M, [V.x V.y V.z; H1.x H1.y H1.z; H2.x H2.y H2.z] and multiplying that by the gyro vector, [Gx Gy Gz]^T, which gives you the gyro vector referenced in the NEW coordinate frame, [GV GH1 GH2]? The vertical vector, V, is given because we have an accelerometer that outputs the (negative) gravity vector . The other two vectors are normal to the gravity vector, and hence define the "earth" coordinate frame. – gallamine Apr 8 '12 at 12:32
@gallamine: Yes, that's right. I hope the code example was clear. If not, let me know and I'll try to clarify it. – Rocketmagnet Apr 8 '12 at 13:23
I assume that the .x, .y, and .z components are the 1st, 2nd, and 3rd components of the different vectors? Conceptually, this just seems so easy. Why would you ever opt to do a complicated rotation matrix as opposed to just projecting the vectors? – gallamine Apr 9 '12 at 0:08
@gallamine: well, both of those things are mathematically identical, and neither is more complex. – Rocketmagnet Apr 9 '12 at 10:14
Good answer. But got some questions. What should be random vector in this case? does it matter? And did this work as expected? – Kevin Peter Nov 29 '12 at 22:01
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# Source Transformations
Discussion in 'Homework Help' started by Ponyboy, Jan 26, 2014.
1. ### Ponyboy Thread Starter New Member
Jan 26, 2014
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0
Can some one explain to me the reason why source transformations work.
Oct 2, 2009
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3. ### shteii01 AAC Fanatic!
Feb 19, 2010
3,496
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Voltage source: You have voltage, current, resistance.
Current source: You have voltage, current, resistance.
voltage, current, resistance = voltage, current, resistance
only proportions are different
In other words: 2+3=1+4, the numbers are different, but 5 is equal to 5.
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > nqpru GIF version
Theorem nqpru 6874
Description: Comparing a fraction to a real can be done by whether it is an element of the upper cut, or by
Assertion
Ref Expression
nqpru ((𝐴Q𝐵P) → (𝐴 ∈ (2nd𝐵) ↔ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
Distinct variable group: 𝐴,𝑙,𝑢
Allowed substitution hints: 𝐵(𝑢,𝑙)
Proof of Theorem nqpru
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 prop 6797 . . . . . 6 (𝐵P → ⟨(1st𝐵), (2nd𝐵)⟩ ∈ P)
2 prnminu 6811 . . . . . 6 ((⟨(1st𝐵), (2nd𝐵)⟩ ∈ P𝐴 ∈ (2nd𝐵)) → ∃𝑥 ∈ (2nd𝐵)𝑥 <Q 𝐴)
31, 2sylan 277 . . . . 5 ((𝐵P𝐴 ∈ (2nd𝐵)) → ∃𝑥 ∈ (2nd𝐵)𝑥 <Q 𝐴)
4 elprnqu 6804 . . . . . . . . . 10 ((⟨(1st𝐵), (2nd𝐵)⟩ ∈ P𝑥 ∈ (2nd𝐵)) → 𝑥Q)
51, 4sylan 277 . . . . . . . . 9 ((𝐵P𝑥 ∈ (2nd𝐵)) → 𝑥Q)
65ad2ant2r 493 . . . . . . . 8 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → 𝑥Q)
7 simprl 498 . . . . . . . 8 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → 𝑥 ∈ (2nd𝐵))
8 vex 2613 . . . . . . . . . . . 12 𝑥 ∈ V
9 breq1 3808 . . . . . . . . . . . 12 (𝑙 = 𝑥 → (𝑙 <Q 𝐴𝑥 <Q 𝐴))
108, 9elab 2746 . . . . . . . . . . 11 (𝑥 ∈ {𝑙𝑙 <Q 𝐴} ↔ 𝑥 <Q 𝐴)
1110biimpri 131 . . . . . . . . . 10 (𝑥 <Q 𝐴𝑥 ∈ {𝑙𝑙 <Q 𝐴})
12 ltnqex 6871 . . . . . . . . . . . 12 {𝑙𝑙 <Q 𝐴} ∈ V
13 gtnqex 6872 . . . . . . . . . . . 12 {𝑢𝐴 <Q 𝑢} ∈ V
1412, 13op1st 5825 . . . . . . . . . . 11 (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) = {𝑙𝑙 <Q 𝐴}
1514eleq2i 2149 . . . . . . . . . 10 (𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ↔ 𝑥 ∈ {𝑙𝑙 <Q 𝐴})
1611, 15sylibr 132 . . . . . . . . 9 (𝑥 <Q 𝐴𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
1716ad2antll 475 . . . . . . . 8 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
18 19.8a 1523 . . . . . . . 8 ((𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))) → ∃𝑥(𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
196, 7, 17, 18syl12anc 1168 . . . . . . 7 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → ∃𝑥(𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
20 df-rex 2359 . . . . . . 7 (∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)) ↔ ∃𝑥(𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
2119, 20sylibr 132 . . . . . 6 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))
22 elprnqu 6804 . . . . . . . . 9 ((⟨(1st𝐵), (2nd𝐵)⟩ ∈ P𝐴 ∈ (2nd𝐵)) → 𝐴Q)
231, 22sylan 277 . . . . . . . 8 ((𝐵P𝐴 ∈ (2nd𝐵)) → 𝐴Q)
24 nqprlu 6869 . . . . . . . . 9 (𝐴Q → ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ∈ P)
25 ltdfpr 6828 . . . . . . . . 9 ((𝐵P ∧ ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ∈ P) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ↔ ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
2624, 25sylan2 280 . . . . . . . 8 ((𝐵P𝐴Q) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ↔ ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
2723, 26syldan 276 . . . . . . 7 ((𝐵P𝐴 ∈ (2nd𝐵)) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ↔ ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
2827adantr 270 . . . . . 6 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ↔ ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
2921, 28mpbird 165 . . . . 5 (((𝐵P𝐴 ∈ (2nd𝐵)) ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 <Q 𝐴)) → 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)
303, 29rexlimddv 2486 . . . 4 ((𝐵P𝐴 ∈ (2nd𝐵)) → 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)
3130ex 113 . . 3 (𝐵P → (𝐴 ∈ (2nd𝐵) → 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
3231adantl 271 . 2 ((𝐴Q𝐵P) → (𝐴 ∈ (2nd𝐵) → 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
3326ancoms 264 . . . . 5 ((𝐴Q𝐵P) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ ↔ ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))))
3433biimpa 290 . . . 4 (((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) → ∃𝑥Q (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))
3515, 10bitri 182 . . . . . . . 8 (𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ↔ 𝑥 <Q 𝐴)
3635biimpi 118 . . . . . . 7 (𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) → 𝑥 <Q 𝐴)
3736ad2antll 475 . . . . . 6 ((𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))) → 𝑥 <Q 𝐴)
3837adantl 271 . . . . 5 ((((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ∧ (𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))) → 𝑥 <Q 𝐴)
39 simpllr 501 . . . . . 6 ((((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ∧ (𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))) → 𝐵P)
40 simprrl 506 . . . . . 6 ((((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ∧ (𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))) → 𝑥 ∈ (2nd𝐵))
41 prcunqu 6807 . . . . . . 7 ((⟨(1st𝐵), (2nd𝐵)⟩ ∈ P𝑥 ∈ (2nd𝐵)) → (𝑥 <Q 𝐴𝐴 ∈ (2nd𝐵)))
421, 41sylan 277 . . . . . 6 ((𝐵P𝑥 ∈ (2nd𝐵)) → (𝑥 <Q 𝐴𝐴 ∈ (2nd𝐵)))
4339, 40, 42syl2anc 403 . . . . 5 ((((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ∧ (𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))) → (𝑥 <Q 𝐴𝐴 ∈ (2nd𝐵)))
4438, 43mpd 13 . . . 4 ((((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) ∧ (𝑥Q ∧ (𝑥 ∈ (2nd𝐵) ∧ 𝑥 ∈ (1st ‘⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩)))) → 𝐴 ∈ (2nd𝐵))
4534, 44rexlimddv 2486 . . 3 (((𝐴Q𝐵P) ∧ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩) → 𝐴 ∈ (2nd𝐵))
4645ex 113 . 2 ((𝐴Q𝐵P) → (𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩ → 𝐴 ∈ (2nd𝐵)))
4732, 46impbid 127 1 ((𝐴Q𝐵P) → (𝐴 ∈ (2nd𝐵) ↔ 𝐵<P ⟨{𝑙𝑙 <Q 𝐴}, {𝑢𝐴 <Q 𝑢}⟩))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 102 ↔ wb 103 ∃wex 1422 ∈ wcel 1434 {cab 2069 ∃wrex 2354 ⟨cop 3419 class class class wbr 3805 ‘cfv 4952 1st c1st 5817 2nd c2nd 5818 Qcnq 6602
Copyright terms: Public domain W3C validator
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# Playing for pizza
You might want to break out a calculator because it takes a genius to solve these tricky math riddles.
Robert and David played several golf matches against each other in a week. They played for a pizza at each match, but no pizzas were purchased until the end of the week. If Robert and David had the same number of wins at any time, those pizzas were canceled. Robert won four matches (but no pizzas), and David won three pizzas.
How many rounds of golf were played?
11
Explanation: David won seven matches — four to cancel out Robert’s four wins and three more to win the pizzas.
## How can this be?
A certain large animal lives happily and thrives here on Earth. One day, every single one of these critters is wiped out by a mysterious disease which affects only this particular animal. There are none left anywhere on earth -- they are all gone. About a year or so...
## Always involve a bed
Looking for a riddle or a joke to spark some laughter? Well, you’ve come to the right place then.You can go on top of me or underneath and I always involve a bed. What am I?A bunk bed.more riddlesDonec rutrum congue leo...
## Long shaft
Try to guess some answer and check the proper answer below and share it with your friends.I have a long shaft. I always penetrate with the tip first and I always come with a quiver. What am I?Arrowmore riddlesDonec rutrum...
## Which ball?
Two barrels contain each the same amount of water. The water temperature in one barrel is 49°F while the other is at 29°F. Two golf balls of same dimension and weight are dropped simultaneously into the barrels. The golf balls are dropped from the same height.Which...
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## 182859
182,859 (one hundred eighty-two thousand eight hundred fifty-nine) is an odd six-digits composite number following 182858 and preceding 182860. In scientific notation, it is written as 1.82859 × 105. The sum of its digits is 33. It has a total of 2 prime factors and 4 positive divisors. There are 121,904 positive integers (up to 182859) that are relatively prime to 182859.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 33
• Digital Root 6
## Name
Short name 182 thousand 859 one hundred eighty-two thousand eight hundred fifty-nine
## Notation
Scientific notation 1.82859 × 105 182.859 × 103
## Prime Factorization of 182859
Prime Factorization 3 × 60953
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 182859 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 182,859 is 3 × 60953. Since it has a total of 2 prime factors, 182,859 is a composite number.
## Divisors of 182859
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 243816 Sum of all the positive divisors of n s(n) 60957 Sum of the proper positive divisors of n A(n) 60954 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 427.62 Returns the nth root of the product of n divisors H(n) 2.99995 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 182,859 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 182,859) is 243,816, the average is 60,954.
## Other Arithmetic Functions (n = 182859)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 121904 Total number of positive integers not greater than n that are coprime to n λ(n) 60952 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16532 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 121,904 positive integers (less than 182,859) that are coprime with 182,859. And there are approximately 16,532 prime numbers less than or equal to 182,859.
## Divisibility of 182859
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 5 3 6
The number 182,859 is divisible by 3.
## Classification of 182859
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (182859)
Base System Value
2 Binary 101100101001001011
3 Ternary 100021211120
4 Quaternary 230221023
5 Quinary 21322414
6 Senary 3530323
8 Octal 545113
10 Decimal 182859
12 Duodecimal 899a3
20 Vigesimal 12h2j
36 Base36 3x3f
## Basic calculations (n = 182859)
### Multiplication
n×i
n×2 365718 548577 731436 914295
### Division
ni
n⁄2 91429.5 60953 45714.8 36571.8
### Exponentiation
ni
n2 33437413881 6114332064865779 1118060647049291482161 204447451858786391136478299
### Nth Root
i√n
2√n 427.62 56.7595 20.679 11.283
## 182859 as geometric shapes
### Circle
Diameter 365718 1.14894e+06 1.05047e+11
### Sphere
Volume 2.56117e+16 4.20187e+11 1.14894e+06
### Square
Length = n
Perimeter 731436 3.34374e+10 258602
### Cube
Length = n
Surface area 2.00624e+11 6.11433e+15 316721
### Equilateral Triangle
Length = n
Perimeter 548577 1.44788e+10 158361
### Triangular Pyramid
Length = n
Surface area 5.79153e+10 7.20581e+14 149304
## Cryptographic Hash Functions
md5 0c601ccb8f7dd792e392930a34e2bd5b d1937763c0539336764d298ff75125e4fad7257b ae27b689f5257b33688a140f4e7e13afe80527ebd003c099c0cde791a0f04a2f cf0a09ac5312632c748957e0ffcb1e40c023fd7677d02222843c3f825a2e366fa9bb8b67971c5b542b5e46daaf7d9ccbc64cbe0f529bdc36550a4374cdfb3222 26141fa3a8d89a82b2e06d559448f0837d5d6bb8
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# Discover the True Cost of Tile Labor: A Personal Story and Expert Tips [2021 Statistics Included]
## Short answer: How much does tile labor cost?
Tile labor costs vary depending on location, type of tile, and complexity of the project. On average, expect to pay \$4-12 per square foot for labor costs. For example, a 100 sq ft bathroom could cost anywhere from 0-1200 in labor alone. Experienced contractors may charge more but offer higher quality workmanship. Always get multiple quotes and references before hiring a tile professional.
## Step-by-Step Guide to Calculating Tile Labor Costs
When it comes to home renovation projects, tiling is one of the most common ones that homeowners undertake. Whether you’re updating your bathroom or installing a new backsplash in your kitchen, tiles can add both aesthetic and functional value to your property. However, before you get started on your project, it’s crucial to calculate tile labor costs accurately. In this step-by-step guide, we’ll walk you through the process of calculating those costs like a pro.
Step 1: Calculate Surface Area
The first step in calculating tile labor costs is to determine the square footage of the area where you will be installing tiles. To do this, measure both the length and width of the space using a tape measure. Then multiply those two numbers together – this will give you the total surface area of your project in square feet.
Step 2: Choose a Tile Pattern
Tile patterns can range from simple brick formations to complex mosaics – with each affecting your tile installation cost differently. The more complex the pattern is, typically requires more expertise or longer installation times resulting in additional labor cost per square foot. Choosing an intricate design may also increase waste during cuts, meaning more materials will need to be purchased than what is needed for standard installations.
Step 3: Decide on Tile Size & Shape
Picking out different tile sizes and shapes can affect labor cost measures differently as well. Smaller tiles traditionally require more cutting resulting in potentially higher labor fees whereas larger rectangular shaped tiles have generally fewer cuts involved needing low-cost installation fees.
Step 4: Determine Labor Cost Per Hour & Time Needed for Project Completion
Contact various contractors or home improvement stores that offer Professional-Installation services providing them with details gathered from steps 1-3 mentioned previously and request quotes based on local rates applying specifically to wall vs floor tiling applications.
Once you get an estimate – ask how long they think it would take them (team size dependent) realistically completing the project. This will help you determine overall labor costs for the project by taking the total hours of work multiplied by labor cost per hour.
Step 5: Add Tile Cost & Other Variables in to Total Labor Costs
After establishing a rough estimate based on local rates and estimated time to completion using specific tile sizes, patterns, surfaces area and shape variables- add on additional costs that could affect tile installation value such as sealers, adhesives and preparations needed for undertaking tile installations duties.
In conclusion, calculating tile labor costs may seem daunting at first, but it’s easier than you think once you break down the process into these simple steps. By knowing exactly how much your tiling project will cost in advance, you can budget appropriately and feel confident about successfully completing your renovation goals without being caught off guard by unexpected expenses.
When it comes to tile installation, there are many factors that can influence the labor costs involved. With so many variables to consider, it’s no wonder that homeowners often have a lot of questions about tile labor costs. So if you’re thinking about installing tile in your home and want to get a better understanding of what kind of expenses you might be looking at, read on!
What Factors Influence Tile Labor Costs?
Before we dive into some commonly-asked questions about tile labor costs, it’s important to understand what factors can impact those costs. Here are a few key things that can influence how much you’ll end up paying for professional tile installation:
– The type of tile being installed
– The size of the room or area where the tile will be laid
– The intricacy or complexity of the tile pattern
– Any prep work that needs to be done before laying the new tiles
– Any repairs or demolition needed before installation
– The location and accessibility of your property
– The experience and reputation of the tile installer
With these factors in mind, let’s dive into some frequently asked questions about labor costs for tiling projects.
How Much Does Tile Installation Cost Per Square Foot?
One common way that people estimate their overall tiling project cost is by considering the price per square foot for installation. However, this number can vary widely depending on all sorts of factors like those listed above. Generally speaking, simple installations (like laying large-format tiles with minimal prep work required) might start around – per square foot for labor alone. More complex projects (such as intricate mosaics or custom patterns) could cost upwards of per square foot or more.
Is It Cheaper to Install Tile Myself?
While it may seem tempting to tackle a DIY tiling project in order to save money on labor costs, this often ends up costing homeowners more in the long run. Professional installers have the experience and tools needed to complete a tile job quickly, efficiently, and to a high standard of quality. If something goes wrong during a DIY tiling project (which is more likely than you might think!), homeowners may end up spending way more on repairs or replacement tiles than they would have if they had just hired a pro from the start.
How Long Does It Take to Install Tile?
Again, this depends on the specific factors at play for your tiling project. However, professional installers should be able to give you an estimate of how long it will take them to complete your installation based on factors like the size of the space and the complexity of the pattern being installed. This can range from one or two days for smaller jobs all the way up to several weeks for very large or complex projects.
Do I Need to Move Furniture Before Tiling?
In most cases, yes: furniture and other items should be moved out of any rooms where tile is going to be installed in order to allow installers room to work. Some companies may be willing to move furniture for you as part of their service packages, but this will likely come with an additional cost.
What If There Are Unexpected Issues During Installation?
Unfortunately, there’s always a chance that unexpected issues could arise during any home renovation project – including tile installation. Experienced professionals should be equipped with problem-solving skills that allow them to work around any unforeseen snags in order keep your project moving forward smoothly whenever possible. Of course, some issues (like structural damage) may require additional work or investment beyond what was originally estimated.
Ultimately, hiring experienced professionals who are willing and able to provide clear communication about pricing and timelines will help ensure that both parties are satisfied with how a tiling project proceeds from start to finish!
## Top 5 Facts You Need to Know about Tile Labor Costs
If you’re looking into renovating your floors, the cost of labor for tile installation is an important consideration. The expense of labor can vary widely depending on a range of factors like location, complexity of the project, and whether there is existing flooring removal. Here are the top five facts that you need to know about tile labor costs to make informed decisions when budgeting out your upcoming renovation.
1. Tile installation can be expensive
Tile installation is not a cheap process. In addition to purchasing tiles themselves, there’s also the expense of labor for installing them. The average tile installation job in the United States costs between \$800 and \$2,500 depending on various factors such as the type of tile used and its size.
The cost of tile installation increases if there’s already existing flooring due to the time it takes to remove it. Tile removal involves using special equipment and tools, which can add several hours or even days to the total project timeline and thus increase labor costs.
3. Location plays a role in pricing
Your location can drastically impact how much you pay for tile installation as travel fees may be added into the final bill when traveling outside service areas or farther from metropolitan centers where labour rates are more evenly priced with other markets.
4. Experience level determines prices
Another factor that affects your professional’s price is their experience level; experienced tilers typically have higher rates than those who just started operating their business because they have honed their skills over time by working extensively in their trade.
Finally, the more complex your tiling project is, such as installing intricate patterns or mosaics using different kinds of materials like marble or ceramic will require additional time for preparation before actually laying down each piece – making your total cost go up accordingly as it inevitably takes longer to complete these jobs.
In conclusion
Installing new tiles can add value by updating a property’s look and modernizing the space. However, the cost of tile installation adds up quickly with the expense of labor alone accounting for a significant portion of your budget. Ultimately, how much you end up paying for tile installation will depend on various factors that you need to take into consideration when planning your renovation project. To get the most out of your investment and avoid unexpected costs, it’s crucial to work with skilled professionals who are transparent about the labor costs associated with different types of tiling projects.
## Estimating Your Budget: How to Determine Tile Labor Costs for Your Project
Tile installation is a process that requires a lot of patience, skill, and expertise. Whether you’re looking to redo your bathroom or give your kitchen a new look, determining tile labor costs is an essential step in the planning phase. Tile labor costs are typically determined by several factors such as project size, tile type and quality, and labor intensity.
To get an accurate estimate for your tile installation project, you’ll need to factor in some key considerations such as:
1. Project Size: The larger the area to be tiled, the more expensive it will be to lay down tiles. It’s important to measure the total area of the room in square footage (or meters depending on where you live) to determine how many tiles will be needed.
2. Tile Quality: High-quality tiles are usually more expensive but offer greater durability and longer-term value than low-quality options. Higher quality materials will also require more specialized equipment and tools which can increase labor costs.
3. Installation Technique: Depending on the complexity of your chosen design scheme, intricate patterns may take longer periods of time and require more skilled workmanship – resulting in higher cost estimates for labor expenses.
4. Hidden Costs: Don’t forget about hidden expenses like site preparation or demolition costs if necessary! Ensuring that floors or walls have been cleaned or prepared correctly before laying down tiles is imperative to prevent potential damage later down the line.
Once you’ve established these items as per above factors it’s important account for markup on top of all these costings too; this markup should include any additional contingencies like environmental demands during construction – things like contractor insurance policies too are something not to overlook!
In general floor tiling commonly includes laying out cement backerboard before putting tiles into place on either adhesive strips or thinset mortar spread overtop this supporting layer – bear future-anticipated movement accordingly when using adhesives here though…
Some initial commonly cited ranges so budgeters know what to expect are as follows for floor tiling:
– Basic options: \$3-\$6 per sq. ft
– Standard tiles from – per square foot
– Premium options from around \$15 up beyond \$25-plus per sq. ft
For estimating wall tile labor costs it’s all about the surface area and its characteristics; using thinner, smaller tiles will tend to need much less labor than for larger sizes – while heavier tiles may require added attention and installation experience.
Here are some commonly cited ranges depending on your preferred aesthetic choices that could influence those budgets when working with wall tile layouts:
– Kitchen backsplash accent features may range between \$10 – \$40+ per sq. ft.
– Shower walls/tub surrounds could cost anything from approximately \$25 – upwards of \$100 or more dollars a sq.ft for intricate mosaics or glass varieties.
– Bathroom mosaic patterns can vary widely at around anywhere roughly between (\$13-\$70 plus) on up to truly customizing an one-of-a-kind piece whose price is ‘priceless’! Attractive yet unique!
To sum it up, determining tile labor costs can be tricky, but with a little bit of knowledge and research, you’ll be able to estimate your budget accurately. Remember that cheap costing pricing structures may not always give more value over time as selecting higher end quality materials & workmanship usually does provide the best return on investment in terms of color retention and longevity.
So do yourself a favor before you begin your next big tiling project and consider carefully our suggestions here… happiness is guaranteed for years down the line!
## Breaking Down the Numbers: Hidden Fees and Factors that Affect Tile Labor Cost
Tile installation is a crucial aspect of every home design project. Whether you are renovating your bathroom, kitchen, or adding a backsplash to your living room, tiles serve the dual purpose of enhancing the aesthetic appeal and increasing the property value. However, just like any other home improvement project, the cost of tile installation can be a nightmare if not properly planned for.
As a homeowner, you may be tempted to opt for the lowest bid without considering what’s included in it. Unfortunately, hidden fees and other factors can significantly drive up your overall tile labor cost if not outlined upfront. In this blog post, we will break down these fees and factors to help you make an informed decision when budgeting for your tile installation project.
1. Tile Size and Type
The size and type of tile you choose can significantly affect the labor costs associated with its installation. The larger the tile size, especially those exceeding 15 inches in length or width, requires more precision during cutting and laying than smaller sizes. Additionally, some materials such as natural stone or porcelain tiles require more time and skill to install than ceramic tiles since they tend to chip or crack easily.
2. Substrate Preparation
Substrate preparation refers to all activities undertaken before tiling work begins such as leveling out uneven surfaces or removing old tiles from walls/floors. Plan ahead for this because it can add up quickly based on what needs doing beforehand so don’t forget about it!
3. Complexity of Design
If you’re looking to have intricate patterns installed on your floor/wall surfaces like herringbone or chevron designs with different colored tiles then expect additional expenses due to their complexity.
4. Demolition Costs
Before installing new tiles- depending on how long ago existing ones were put in place), homeowners need their space rid of old finishes which might include scraping off grout lines or adhesive residue among other things that must end up being discarded thus requiring personnel and equipment to cart away.
5. Access and Location
Usually, installation prices of fixtures/fixtures and fittings vary based on access limitations or travel challenges experienced by contractors while procuring necessary supplies needed during said job.
6. Tile Layout/Patterning and joint spacing
The more intricate the tile pattern, such as running bond, herringbone or diamond layout designs, the longer it may take for the installers to complete the work which can result in an increase in labor costs. Additionally, significant expenses are incurred when using larger than usual grouting lines between tiles resulting in extra material usage.
7. Durability/Preventive maintenance
Durability counts a lot with regard to tiling labor costs because should tiles break/chip unexpectedly at some point in time down the line – repairs could prove quite costly. Homeowners might consider requesting advice on preventive maintenance measures from their installer before project wrap-up.
In conclusion, being well aware of hidden fees and associated factors that affect tile installation labor costs is key to successful home renovation projects for anyone planning to undertake one of these types of makeovers. Some other things that you may want to keep in mind include things like caulking at corners where two surfaces meet so as not let water seep into spaces due gaps forming through wear-and-tear over time (and use) amongst many other considerations.. Consider all these aspects before hiring a contractor for your next tile installation project- doing so could save you both money and headache!
## Tips for Saving Money on Your Tile Installation: Bargaining, DIY, and More
Whether you’re renovating your kitchen, bathroom, or any other part of your home, tiling can add a beautiful and timeless touch to the space. However, it’s no secret that tile installation can quickly become expensive. But fear not, as there are various tips and tricks you can use to save money on your tile installation project.
Here are some ideas for saving money on your tile installation:
1. Do Your Homework: Before starting the project, make sure you do thorough research. This includes shopping around for the best deals on tiles and supplies. Try looking at different stores and checking their websites for promotions, clearance sales or discount coupons that may help alleviate the cost.
2. DIY: Installing tiles by yourself might seem intimidating but it is definitely doable if you have some experience with home improvement projects. However, if this is a new territory for you invest ample time in researching techniques; be mindful of wiring pipes when drilling through walls or installing floors as there might be potential hazards in that area.
3.Bargain with Contractors: Reach out to different contractors and ask for a quote on how much they would charge for labor cost then try negotiating the price by telling them reasonable budget constraints. It’s far more better than entirely dismissing their services simply because of monetary fixation issues without considering what value they offer.
4.Limit Pattern Mixing: Although pattern mixing can give interesting designs, it costs extra labour hour fee as all patterns require cuts and sculpting etc., In short limit pattern mix types while working with homogeneous tiles (tiles having same design throughout).
5.Optical Illusions & Fusion Styles: Once again spending less time to create extravagant intricate designs will save labor costs significantly.Figures like Monochromatic Subtle Patterns,Natural Neutrals,Splashes Of Colors work great together,saving both labour fee,time & adding creative ways which will still make a lasting impression.
6.Tile Size Matters: Instead of smaller tiles which take longer to install, go for larger tiles as it generates less grout lines and the project will require less material time and therefore give you an overall cost-cutting effect.
7. Incentives and Offers: Sometimes when we are bogged down with renovation plans we might miss out on a few promotional offers available within our area.Surveys can reveal that tile showrooms may offer discounts such as free delivery if you put in a certain amount of bulk order all with guarantee and warranty seals to hold onto.
In conclusion, tiling is not only an aesthetic boost but also builds the equity value of your home.It’s wise to save money while doing so by following some basic yet effective steps mentioned above;Do Your Homework,ruffing up some DIY skills,Bargaining,Fusion Styles Offer opportunities for Multiple Tile designs, Choosing the right sizing when selecting your tiles,simple Optical Iilussion Techniques & Keeping an eye out for Discounts.Specialists recommend that little change can make big saving differences. So now that you know how to cut costs go ahead and take on this exciting journey towards enhancing your interior decor!
## Table with useful data:
Type of Tile Labor Cost per Square Foot
Ceramic \$5 – \$10
Porcelain \$6 – \$12
Natural Stone \$15 – \$20
Glass \$10 – \$15
Mosaic \$15 – \$20
## Information from an expert
As an expert in the tile industry, I can say that the cost of labor for installing tiles varies depending on several factors such as the size of the area to be covered, the type and quality of tiles, and intricacy of design. On average, tile labor costs range from \$4 to \$12 per square foot. However, it is important to note that while it may seem tempting to opt for cheaper labor, choosing a qualified and experienced tile installer ensures longevity and durability of your tiled space. Don’t compromise on quality for cost-cutting measures and invest in the expertise needed for a lasting finish.
Historical fact:
As a historian, it is not within my expertise to determine the cost of tile labor as it varies greatly depending on factors such as time period, location, and trade skill level. However, historical records indicate that skilled craftsmen were highly valued throughout various civilizations and were often compensated generously for their work in crafting beautiful tiled floors and walls.
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Re: Re: Outputting to file with fixed decimal digits
• To: mathgroup at smc.vnet.net
• Subject: [mg75321] Re: [mg75285] Re: Outputting to file with fixed decimal digits
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 25 Apr 2007 05:34:34 -0400 (EDT)
• References: <f0ekbt\$pnn\$1@smc.vnet.net> <200704240719.DAA27260@smc.vnet.net>
```do.not at reply.nonet wrote:
> Larry
> [...]
>
> I have suggested using the "Floor" function - it's to avoid the
> "Round" function which has the most bizarre definition in Mathematica
> - it doesn't give the same answer every time.
> (Look in "Further Examples" in the Help on "Round")
> It's one reason that makes me think Mathematica is not really
> "Engineering Grade" software and suspicious that there may be more
> weird things that I just haven't come across yet.
>
> Robert
Here is what I find in Help > Round > Further Examples from version 5.2.
---------------
To avoid statistical bias Round alternates between going up and down
halfway between integers.
In[1]:=
Round[Range[10]+.5]
Out[1]=
{2,2,4,4,6,6,8,8,10,10}
The function NumericRound rounds all parts of the numeric coefficients
of a polynomial.
In[2]:=
NumericRound[x_]:=MapAll[If[NumericQ[#],Round[#],#]&,x]
In[3]:=
\!\(NumericRound[\@\[Pi] + \((\[ExponentialE] + a)\)\ x + \((4.3 -
8.6 \[ImaginaryI]\ )\) x\^3]\)
Out[3]=
\!\(1 + \((3 + a)\)\ x + \((4 - 9\ \[ImaginaryI])\)\ x\^3\)
In[4]:=
Clear[NumericRound]
---------------
I do not see examples that show changes to their results upon repetition.
Daniel Lichtblau
Wolfram Research
```
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# Civil Engineering - Building Construction - Discussion
Discussion Forum : Building Construction - Section 4 (Q.No. 3)
3.
The minimum thickness of walls built in cement mortar (1 : 6) for a single storey building, is
10 cm
15 cm
20 cm
25 cm
30 cm.
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.
Suresh Mandal said: 2 years ago
Because we use 1:6 mortar for load-bearing walls and 1:4 for the partition walls so for partition walls 10 cm and for load-bearing or external wall 20cm or 23cm is the minimum wall thickness.
Bijkumar Yadav said: 2 years ago
I think the Maximum thickness brick wall 23cm.
Izaz said: 3 years ago
For all type Building with RCC frame structure, for non load bearing wall, minimum required wall thickness should be 8 inch (200mm) thick for outer wall and to reduce the cost, we can reduce the thickness to 4 inch (100mm) for interior wall partition in RCC frame structure.
Atif Baloch said: 5 years ago
1"=25mm here unit is given in Cm so 1"=2.5cm.
So 9*2.5=22cm but here close answer is 20cm so 20cm is the right answer.
(1)
Manoj said: 5 years ago
For external wall we provide 23 cm wall and for internal wall 10 cm. Then minimum thickness should be 10 cm.
How 20 cm will be the answer?
(1)
Anomies said: 5 years ago
20 cm is right. I agree.
Apcivilian said: 6 years ago
10 cm is answer, as we can provide stretcher bond.
Explain me if I am wrong.
Sachin ghodake said: 8 years ago
Is it for traditional brick or standard brick?
Aaba said: 8 years ago
But one size is 9" so how can build that? I can't understand that.
Kuldip patil said: 8 years ago
It will be minimum one brick thickness as 20 mm.
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# Equation of a Parabola
We shall derive the equation of a parabola from the definition. In order for this equation to be as simple as possible, we choose the X-axis as the perpendicular to the directrix and containing the focus. The origin is taken at the point on the X-axis midway between the focus and the directrix.
Let $a$ be the distance $OF$. The focus is the point $F\left( {a,0} \right)$, and the directrix is the line having the equation $x = - a$. Let $P\left( {x,y} \right)$ be any point on the parabola. Then point $P$ is equidistant from point $F$ and the directrix. From $P$ draw a line perpendicular to the directrix, and let $M\left( { - a,y} \right)$ be the foot of this perpendicular, as shown in the given diagram.
This is the equation of the parabola whose focus is at $\left( {a,0} \right)$ and whose directrix is the equation $x = - a$.
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What are the characteristics of odd numbers?
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.
How do you define an even number?
Definition of even number
: a whole number that is able to be divided by two into two equal whole numbers The numbers 0, 2, 4, 6, and 8 are even numbers.
What are three things that are true about even numbers?
Even numbers when divided by 2 leave no remainder. – 0 is not an even number. The sum of two or more even numbers is always even. The product of two or more even numbers is always even.
How do you know if a number is even?
Given a number, check whether it is even or odd. One simple solution is to find the remainder after dividing by 2.
How do you teach even numbers?
Here are five easy ways to teach odd and even numbers!
1. Line the Children Up in Pairs. …
2. Explain the Concept and Sing It. …
3. Review It Daily with the Calendar. …
4. Sing the Count by Twos Song, and Write Those Numbers! …
5. Try Some Hidden Odds & Evens Worksheets!
Do even numbers have more factors?
False. Just because a number is even, it does not necessarily have more factors.
How do you tell the difference between odd and even numbers?
An even number is a number that can be divided into two equal groups. An odd number is a number that cannot be divided into two equal groups. Even numbers end in 2, 4, 6, 8 and 0 regardless of how many digits they have (we know the number 5,917,624 is even because it ends in a 4!). Odd numbers end in 1, 3, 5, 7, 9.
What are the rules for odd and even numbers?
Even number is divisible by 2, and leaves the remainder 0. An odd number is not completely divisible by 2, and leaves the remainder 1. An even number ends with 0, 2, 4, 6, and 8. An odd number ends with 1, 3, 5, 7, and 9.
Why 0 is an even number?
So why, mathematically, is zero an even number? Because any number that can be divided by two to create another whole number is even. Zero passes this test because if you halve zero you get zero.
What is the definition of even integer?
An integer is an even integer, if it is divisible by 2 i.e. it is a multiple of 2.
Is 18 an even number?
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 are even numbers.
Is 4 an even number?
Why is 4 an Even Number? Since 4 is a multiple of 2 and it can be divided into two equal groups, it is an even number. In other words, it is an even number because it is completely divisible by 2.
Is 0 the smallest even number?
To put it another way, anytime an even number is divided by 2; the residue is always 0. 0 is both an even and a whole number. As a result, 0 is the smallest, even whole number.
Can negative numbers be even?
An even number is an integer that is “evenly divisible” by 2, i.e., divisible by 2 without a remainder. So, …, -4, -2, 0, 2, 4, … are all even integers. An odd number is an integer that is not evenly divisible by 2. So, …, -3, -1, 1, 3, 5, … are all odd integers.
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## 5.30 Topological groups, rings, modules
This is just a short section with definitions and elementary properties.
Definition 5.30.1. A topological group is a group $G$ endowed with a topology such that multiplication $G \times G \to G$, $(x, y) \mapsto xy$ and inverse $G \to G$, $x \mapsto x^{-1}$ are continuous. A homomorphism of topological groups is a homomorphism of groups which is continuous.
If $G$ is a topological group and $H \subset G$ is a subgroup, then $H$ with the induced topology is a topological group. If $G$ is a topological group and $G \to H$ is a surjection of groups, then $H$ endowed with the quotient topology is a topological group.
Example 5.30.2. Let $E$ be a set. We can endow the set of self maps $\text{Map}(E, E)$ with the compact open topology, i.e., the topology such that given $f : E \to E$ a fundamental system of neighbourhoods of $f$ is given by the sets $U_ S(f) = \{ f' : E \to E \mid f'|_ S = f|_ S\}$ where $S \subset E$ is finite. With this topology the action
$\text{Map}(E, E) \times E \longrightarrow E,\quad (f, e) \longmapsto f(e)$
is continuous when $E$ is given the discrete topology. If $X$ is a topological space and $X \times E \to E$ is a continuous map, then the map $X \to \text{Map}(E, E)$ is continuous. In other words, the compact open topology is the coarsest topology such that the “action” map displayed above is continuous. The composition
$\text{Map}(E, E) \times \text{Map}(E, E) \to \text{Map}(E, E)$
is continuous as well (as is easily verified using the description of neighbourhoods above). Finally, if $\text{Aut}(E) \subset \text{Map}(E, E)$ is the subset of invertible maps, then the inverse $i : \text{Aut}(E) \to \text{Aut}(E)$, $f \mapsto f^{-1}$ is continuous too. Namely, say $S \subset E$ is finite, then $i^{-1}(U_ S(f^{-1})) = U_{f^{-1}(S)}(f)$. Hence $\text{Aut}(E)$ is a topological group as in Definition 5.30.1.
Lemma 5.30.3. The category of topological groups has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of groups.
Proof. It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $G_ i$, $i \in I$ be a collection of topological groups. Take the usual product $G = \prod G_ i$ with the product topology. Since $G \times G = \prod (G_ i \times G_ i)$ as a topological space (because products commutes with products in any category), we see that multiplication on $G$ is continuous. Similarly for the inverse map. Let $a, b : G \to H$ be two homomorphisms of topological groups. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of groups endowed with the induced topology. $\square$
Lemma 5.30.4. Let $G$ be a topological group. The following are equivalent
1. $G$ as a topological space is profinite,
2. $G$ is a limit of a diagram of finite discrete topological groups,
3. $G$ is a cofiltered limit of finite discrete topological groups.
Proof. We have the corresponding result for topological spaces, see Lemma 5.22.2. Combined with Lemma 5.30.3 we see that it suffices to prove that (1) implies (3).
We first prove that every neighbourhood $E$ of the neutral element $e$ contains an open subgroup. Namely, since $G$ is the cofiltered limit of finite discrete topological spaces (Lemma 5.22.2), we can choose a continuous map $f : G \to T$ to a finite discrete space $T$ such that $f^{-1}(f(\{ e\} )) \subset E$. Consider
$H = \{ g \in G \mid f(gg') = f(g')\text{ for all }g' \in G\}$
This is a subgroup of $G$ and contained in $E$. Thus it suffices to show that $H$ is open. Pick $t \in T$ and set $W = f^{-1}(\{ t\} )$. Observe that $W \subset G$ is open and closed, in particular quasi-compact. For each $w \in W$ there exist open neighbourhoods $e \in U_ w \subset G$ and $w \in U'_ w \subset W$ such that $U_ wU'_ w \subset W$. By quasi-compactness we can find $w_1, \ldots , w_ n$ such that $W = \bigcup U'_{w_ i}$. Then $U_ t = U_{w_1} \cap \ldots \cap U_{w_ n}$ is an open neighbourhood of $e$ such that $f(gw) = t$ for all $w \in W$. Since $T$ is finite we see that $\bigcap _{t \in T} U_ t \subset H$ is an open neighbourhood of $e$. Since $H \subset G$ is a subgroup it follows that $H$ is open.
Suppose that $H \subset G$ is an open subgroup. Since $G$ is quasi-compact we see that the index of $H$ in $G$ is finite. Say $G = Hg_1 \cup \ldots \cup Hg_ n$. Then $N = \bigcap _{i = 1, \ldots , n} g_ iHg_ i^{-1}$ is an open normal subgroup contained in $H$. Since $N$ also has finite index we see that $G \to G/N$ is a surjection to a finite discrete topological group.
Consider the map
$G \longrightarrow \mathop{\mathrm{lim}}\nolimits _{N \subset G\text{ open and normal}} G/N$
We claim that this map is an isomorphism of topological groups. This finishes the proof of the lemma as the limit on the right is cofiltered (the intersection of two open normal subgroups is open and normal). The map is continuous as each $G \to G/N$ is continuous. The map is injective as $G$ is Hausdorff and every neighbourhood of $e$ contains an $N$ by the arguments above. The map is surjective by Lemma 5.12.6. By Lemma 5.17.8 the map is a homeomorphism. $\square$
Definition 5.30.5. A topological group is called a profinite group if it satisfies the equivalent conditions of Lemma 5.30.4.
If $G_1 \to G_2 \to G_3 \to \ldots$ is a system of topological groups then the colimit $G = \mathop{\mathrm{colim}}\nolimits G_ n$ as a topological group (Lemma 5.30.6) is in general different from the colimit as a topological space (Lemma 5.29.1) even though these have the same underlying set. See Examples, Section 110.77.
Lemma 5.30.6. The category of topological groups has colimits and colimits commute with the forgetful functor to the category of groups.
Proof. We will use the argument of Categories, Remark 4.25.2 to prove existence of colimits. Namely, suppose that $\mathcal{I} \to \textit{Top}$, $i \mapsto G_ i$ is a functor into the category $\textit{TopGroup}$ of topological groups. Then we can consider
$F : \textit{TopGroup} \longrightarrow \textit{Sets},\quad H \longmapsto \mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\textit{TopGroup}}(G_ i, H)$
This functor commutes with limits. Moreover, given any topological group $H$ and an element $(\varphi _ i : G_ i \to H)$ of $F(H)$, there is a subgroup $H' \subset H$ of cardinality at most $|\coprod G_ i|$ (coproduct in the category of groups, i.e., the free product on the $G_ i$) such that the morphisms $\varphi _ i$ map into $H'$. Namely, we can take the induced topology on the subgroup generated by the images of the $\varphi _ i$. Thus it is clear that the hypotheses of Categories, Lemma 4.25.1 are satisfied and we find a topological group $G$ representing the functor $F$, which precisely means that $G$ is the colimit of the diagram $i \mapsto G_ i$.
To see the statement on commutation with the forgetful functor to groups we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a group the corresponding chaotic (or indiscrete) topological group. $\square$
Definition 5.30.7. A topological ring is a ring $R$ endowed with a topology such that addition $R \times R \to R$, $(x, y) \mapsto x + y$ and multiplication $R \times R \to R$, $(x, y) \mapsto xy$ are continuous. A homomorphism of topological rings is a homomorphism of rings which is continuous.
In the Stacks project rings are commutative with $1$. If $R$ is a topological ring, then $(R, +)$ is a topological group since $x \mapsto -x$ is continuous. If $R$ is a topological ring and $R' \subset R$ is a subring, then $R'$ with the induced topology is a topological ring. If $R$ is a topological ring and $R \to R'$ is a surjection of rings, then $R'$ endowed with the quotient topology is a topological ring.
Lemma 5.30.8. The category of topological rings has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of rings.
Proof. It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $R_ i$, $i \in I$ be a collection of topological rings. Take the usual product $R = \prod R_ i$ with the product topology. Since $R \times R = \prod (R_ i \times R_ i)$ as a topological space (because products commutes with products in any category), we see that addition and multiplication on $R$ are continuous. Let $a, b : R \to R'$ be two homomorphisms of topological rings. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of rings endowed with the induced topology. $\square$
Lemma 5.30.9. The category of topological rings has colimits and colimits commute with the forgetful functor to the category of rings.
Proof. The exact same argument as used in the proof of Lemma 5.30.6 shows existence of colimits. To see the statement on commutation with the forgetful functor to rings we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a ring the corresponding chaotic (or indiscrete) topological ring. $\square$
Definition 5.30.10. Let $R$ be a topological ring. A topological module is an $R$-module $M$ endowed with a topology such that addition $M \times M \to M$ and scalar multiplication $R \times M \to M$ are continuous. A homomorphism of topological modules is a homomorphism of modules which is continuous.
If $R$ is a topological ring and $M$ is a topological module, then $(M, +)$ is a topological group since $x \mapsto -x$ is continuous. If $R$ is a topological ring, $M$ is a topological module and $M' \subset M$ is a submodule, then $M'$ with the induced topology is a topological module. If $R$ is a topological ring, $M$ is a topological module, and $M \to M'$ is a surjection of modules, then $M'$ endowed with the quotient topology is a topological module.
Lemma 5.30.11. Let $R$ be a topological ring. The category of topological modules over $R$ has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of $R$-modules.
Proof. It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $M_ i$, $i \in I$ be a collection of topological modules over $R$. Take the usual product $M = \prod M_ i$ with the product topology. Since $M \times M = \prod (M_ i \times M_ i)$ as a topological space (because products commutes with products in any category), we see that addition on $M$ is continuous. Similarly for multiplication $R \times M \to M$. Let $a, b : M \to M'$ be two homomorphisms of topological modules over $R$. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of modules endowed with the induced topology. $\square$
Lemma 5.30.12. Let $R$ be a topological ring. The category of topological modules over $R$ has colimits and colimits commute with the forgetful functor to the category of modules over $R$.
Proof. The exact same argument as used in the proof of Lemma 5.30.6 shows existence of colimits. To see the statement on commutation with the forgetful functor to $R$-modules we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a module the corresponding chaotic (or indiscrete) topological module. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Extension of Four Is Cosmic
The fact that four is cosmic, at least in English, is quite well discussed.
However, what if we instead took a number and then took the first letter of its word form that number and converted it to its alphabetical placing to generate a sequence?
1) Find three numbers which generate its own cyclic sequence.
2) Even more special, find the number which just repeats itself. (i.e. the cosmic number of this sequence)
3) Any other cyclic number(s) that I may have missed (or a 'proof' that there are no others).
Definitely an interesting variant on the original- I believe I have answers to the first two, and still thinking on the third.
1:
Nineteen goes to N(fourteen), which goes to F(six), which goes to S(nineteen)
2:
T is at position twenty, so goes back to itself.
3:
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How many order need to recover from 4.8 to 5.00 reviews
anyone can give the idea how many order need to recover from 4.8 to 5.00 reviews thanks
4 Likes
You’re still in good standing. Just focus on your orders and the rest will fall in place.
Another about 22 5 stars reviews will see you back at 5 stars. But as mentioned, do your best and the reviews will follow! Good luck!
3 Likes
(A + Y) / ((B + X) * 5.0) = Z
A = current overall rating
B = current total review count
X = new total review count
Y = X times 5.0 … this means all new reviews are 5.0
Z = new overall rating
/means divided by
*means multiplied by
This equation will show you how many more 5.0 reviews you will need to reach any particular overall rating… and you can manipulate “5.0” in equation to allow for more conservative estimate if you do not only anticipate 5.0 reviews.
Hope this helps!
7 Likes
Wow the equation is very helpful.thank you so much for your help
I’m not exactly getting it. The B would be the number of reviews you have now ? X would be the desired number of reviews? And what is Y ?
Yes, B is the total number of reviews that you currently have on your profile (or select Gig) and X represents the desired number of reviews… You can put however many reviews in for the value of X to see how many reviews it will take to reach 4.945 (closest number to round up to 5.0)… and Y represents the average for your new ratings, and this is 5.0 in the equation above to show how many 5.0 ratings you will need to receive to reach any new Z (new overall rating)… but you can manipulate the 5.0 value to any lower value if you feel your average rating will be below 5.0
1 Like
i have 2900 reviews can you help me with the equation dear
1 Like
Please, do not use the word "dear. It’s considered as #crime.
3 Likes
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# Double sum involving $\cos$
I ran across a double sum and was wondering if someone may be adept at evaluating it. I must admit that my double summation skills could be better, and I am always ready to learn more.
Show that:
$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi}{3}(n-k))}{nk(n+k)}=\frac{-1}{9}\zeta(3)+\frac{\pi\sqrt{3}}{27}\psi_{1}(1/3)-\frac{2\pi^{3}\sqrt{3}}{81}$$
I found this while playing around with:
$$\int_{0}^{1}\frac{\log(1-xe^{\frac{\pi i}{3}})\log(1-xe^{\frac{-\pi i}{3}})}{x}dx$$
I would enjoy seeing a clever evaluation of the sum.
I tried breaking it up as:
$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})\cos(\frac{\pi n}{3})}{kn^{2}}$$ $$-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})\cos(\frac{\pi n}{3})}{n^{2}(k+n)}$$ $$+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})\sin(\frac{\pi n}{3})}{kn^{2}}$$ $$-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})\sin(\frac{\pi n}{3})}{n^{2}(k+n)}$$
The third from the top evaluates to $$\frac{\pi}{3}Cl_{2}(\frac{\pi}{3})=\frac{\pi}{3}\left(\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}\right)$$
I think the top one evaluates to 0.
The other 2 are a little more challenging.
Would anyone enjoy lending a hand and showing a method to evaluate said double sum...or even the integral for that matter?.
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# Regression
Feb 3rd, 2012
Studypool Tutor
California Institute of Technology
Course: operations management
Price: \$5 USD
Tutor description
An agent for a residential real estate company in a large city would like to be able to predict the monthly rental cost for apartments, based on the size of the apartment in square footage. Data was gathered from a sample of 25 apartments in a particular residential neighborhood and an ordinary least squares regression was run. The sample rents range from \$875 to \$2,300 and the sample sizes range from 800 to 1,985 square feet. The output from the Excel regression is provided in the attachment. 1. Write out the regression equation (with the calculated coefficient estimates). 2. Interpret the meaning of b0 and b1 in this problem 3. What is the coefficient of determination (r2) and what does it tell you about the model? 4. At the 0.05 level of significance, is there evidence of a linear relationship between apartment rent and square footage? Explain. 5. You are considering signing a lease for an apartment in the same neighborhood. The apartment has 1000 square feet and the rent is \$1
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# The length and breadth of a rectangle are given as (6x - 1) and (x -2) metres respectively. If the length and breadth are each increased by 4 metres, the new area is three times that of original rectangle.
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The length and breadth of a rectangle are given as (6x - 1) and (x -2) metres respectively. If the length and breadth are each increased by 4 metres, the new area is three times that of original rectangle.
1. Form an equation in x and solve it.
2. Find the dimensions of the original triangle
3. Express the increase in area as a percentage of the original area.
1. Form an equation in x and solve it.
Dimensions of the new rectangle
(6x +3) and (x + 2)
(6x + 3) (x +2) = 6x2 + 15x + 6
= 6x2 + 15x +6 = 3 (6x - 1) (x- 2)
= 6x2 + 15x + 6 = 18x2 – 39x + 6
12x2 – 54x = 0
6x(2x -9) = 0
2x = 9 x =4.5
2. Find the dimensions of the original triangle
Length = 26m
Original area
26 x 2.5 = 65m2
3. Express the increase in area as a percentage of the original area.
New area
30 x 6.5 = 195m2
% increase
= 195-6565×100%
= 200%
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# Alternatives to procedural loops and iterating over lists in Mathematica
While there are some cases where a For loop might be reasonable, it's a general mantra – one I subscribe to myself – that "if you are using a For loop in Mathematica, you are probably doing it wrong". But For and Do loops are familiar to people who already know other programming languages. It is common for new Mathematica users to rely on loop constructs, only to become frustrated with performance or code complexity.
My question is: are there some general rules of thumb that would help new users select the appropriate alternative to a procedural For or Do loop, and are there other useful Mathematica constructs, aside from RandomVariate/Nest/Fold/Inner/Outer/Tuples that they should know about in order to avoid those loops?
In particular, there are a number of ways of iterating over lists:
1. The For loop:
For[i = 1, i <= 10, i++, list[[i]] = func[ list[[i]]]
2. The Table function operating over each part of a list in turn:
Table[func[ list[[i]] ], {i, Length[list]}]
3. The Do loop operating over each part of a list in turn:
lst = {88, 42, 25, 75, 35, 97, 12};
t = 9;
Do[
x = lst[[i]];
t += Mod[x, t],
{i, 1, Length[lst]}
];
t
4. Mapping a function onto each element of a list:
(3 - #)/(7 * #) & /@ list
5. And for nested lists, there are constructs like MapThread:
f = Mod[#, #2] Floor[#3/#] &;
a = {{18, 85, 22, 20, 39}, {17, 67, 76, 96, 58}, {40, 97, 56, 60, 53}};
New users of Mathematica will usually pick options 1 or 3 and use C-style or Matlab approaches that involve the following multistep process:
1. Defining an empty list
2. Setting up a loop
3. Optionally define a variable (usually not localised) to equal the iterator, which is then used in subsequent calculations within the loop
4. Within each loop iteration, use the local variable defined to equal the iterator to redefine each element of that list in turn according to some function
5. If the list is multidimensional, nest another loop inside that.
What are some useful guides to help users coming from other languages to Mathematica to improve the conciseness and efficiency of their code by avoiding unnecessary loops?
• How about operations like arithmetic on lists, matrixes, & tensors or Dot and DotProduct? "For" my part (pun intended), I stopped thinking in loops and started thinking functionally. It would help if one could tie someones "loop" arm to their body, like they do with boxers who need to switch their dominant hand. Ultimately, I think one has to give themselves over to the functional language. Maybe one can translate one language to another but poetry in English doesn't have the same meaning translated into Italian nor Italian to English. When one can dream in the language they'll have it. Jul 5, 2012 at 12:53
• There are cases where I find for loops more readable. For example R = S (*list of elemetns*); Do[ R[[i]] = f[S[[i]], G[[i]]]; , {i, Length@S}]; vs R = f[#[[1]], #[[2]]] & /@ Transpose[{S, G}] The second one is more concise, but G[[i]] conveys the intent better then #[[1]]. Jul 5, 2012 at 12:54
• @Ajasja, even more concise and readable is R=Thread@f[S,G] Jul 5, 2012 at 13:28
• Just one other thing one may want to add as a caveat: the preferred choice of construct can change radically when the goal is to use it with Compile.
– Jens
Jul 5, 2012 at 15:12
• @Jens yes, that seems to be ignored most of the time. readability is also dependent on the person: I have some code to transform a density matrix that looks like this: MapThread[Join,{rtmp,Rest /@ Flatten[Conjugate@rtmp, {{2}, {1}}]}]. Some may read this immediately, but I need a commented-out iterative version to be able to re-read it later (this one's faster though).
– acl
Jul 5, 2012 at 15:34
## Case #1
Explicit loops are often counterproductive in Mathematica, not only taking more keystrokes, but also more execution time. They are also, in my opinion, more prone to mistakes.
Better ways are to use Do, Scan, or Map.
Do and Scan are (typically) appropriate for operations that do not accumulate a list of results, while Map and Table (a variant of Do) are (typically) used for ones that do. The third method uses Map (short form: /@) and accumulates a useless list of results; for this reason it is likely to be less memory efficient than the first two.
1. Do[Print[i], {i, 10}]
2. Scan[Print, Range@10]
3. Print /@ Range@10;
Average Timing, using PrimeQ in place of Print for 10^6 iterations:
## Case #2
Since version 6 this form of Table is not necessary! Instead, use:
Table[func[i], {i, list}]
One might ask why use Table here at all. Indeed, this simple example can be written:
func /@ list
which is preferred. However, there are more complex cases where Table is far more elegant than the alternatives.
Average Timing for 10^6 real numbers using Sin for func:
## Case #3
This could be improved in the manner of case #2, but there is a far better method available: Fold. Any time you want to iterate through a list, using the result of the previous "loop" along the way, look to Fold, efficient in both syntax and computation.
Fold[# + Mod[#2, #] &, 9, lst]
Average Timing for 10^6 integers (big savings!):
## Case #4
There is nothing intrinsically wrong with this. However, all of the operations within the pure function have the Attribute Listable. Therefore, this function can directly accept a list without using Map, and this will be considerably more efficient.
(3 - #)/(7 * #) & @ list
Average Timing for a list of 10^6 real numbers:
## Case #5
This is similar to case #4, but a little more complicated. Once again, each subfunction in f is Listable but this is not being leveraged. One can use Apply to pass the sub-lists in a as arguments to f:
f @@ a
Suppose that not all of the functions are Listable. I create a dummy addition function g that only accepts integers, not lists. I then include this in f, and try to Apply it again:
ClearAll[g, f]
g[n_Integer, m_Integer] := n + m
f = Mod[#, #2] Floor[#3/#] * g[#2, #] &;
f @@ a
But the result is incorrect. Once could go back to MapThread, but a better way, when possible, is to make g handle lists, which will usually be faster on large sets. Here are two ways of doing that. Give g the Listable attribute, and Mathematica will automatically thread over lists:
ClearAll[g, f]
SetAttributes[g, Listable]
g[n_Integer, m_Integer] := n + m
f = Mod[#, #2] Floor[#3/#] * g[#2, #] &;
f @@ a
Or, if the automatic threading via Listable breaks your function in some way, manually:
ClearAll[g, f]
g[n_Integer, m_Integer] := n + m
g[n_List, m_List] := MapThread[Plus, {n, m}]
f = Mod[#, #2] Floor[#3/#] * g[#2, #] &;
f @@ a
Average Timing for lists of 10^6 integers:
• Wow! That was quick! The OP must be amazed! ... Wait .. Apr 30, 2011 at 0:30
• @belisarius lol er um yeah. BTW, notice case #2. Apr 30, 2011 at 0:33
• @Mr. I use #2 all the time .... but the version in the question. Habits are difficult to change (said the monk) Apr 30, 2011 at 5:34
• @Mr.Wizard I think it'd be nice to give examples of the Do and Map functions you use in Case #3 and #4, respectively. Lets people test out the difference in speed themselves! Mar 31, 2013 at 3:46
• @Vincent Those were accidentally left out of the question for a time, but I restored them shortly after your comment. (Thanks.) Now almost a year later do you have any other suggestions for improving this Q&A? Jan 4, 2014 at 21:43
There are many alternative ways to approach various programming problems that do not use loops and are more efficient (and concise) in Mathematica. Most of them execute faster, but even where they do not, they are faster to type: development time matters, too!
Here are some rules of thumb for easier programming and iterating on lists.
1. Most arithmetic operations and many other functions are Listable, meaning that they operate element by element without having to set up a loop or operate explicitly on each element
Addition, multiplication and other standard operations work element by element when they are given conformable vectors, matrices, tensors or generalised lists:
list1 = {a, b, c};
list2 = {e, f, g};
list1 + list2
(* {a + e, b + f, c + g} *)
Listable functions include standard arithmetic and trigonometric functions, Log, Exp, Mod, Abs and a huge range of exponential, Bessel-related and other special functions. The following code returns the whole list of Listable functions:
With[{names = Names["System*"]},
Pick[names, MemberQ[Attributes[#], Listable] & /@ names, True] ]
2. In Mathematica, you do not need to declare an empty list and then fill each element in with the desired value one by one
For example, typical procedural programming will create an empty vector, and replace each element in turn to create the desired data.
vec = Table[0, {100}]; For[i = 1, i <= 100, vec[[i]] = RandomReal[]+2; i++]; vec
In Mathematica, one can use RandomVariate and its cousins directly, or perhaps a Table command. This generalizes to other more complex definitions inside the Table function, and one can use the fact that many arithmetic operations are Listable to avoid looping. (Yes, I know this could be written as RandomReal[{2, 3}, 100], but I wanted a simple example of vectorized arithmetic operations.)
vec = RandomReal[{0, 1}, 100]+2.;
vec = Table[RandomReal[], {100}]+2.;
3. If the i-th element of your list depends on the previous elements, you can construct the list using Nest, Fold, FixedPoint and their variants NestList, FoldList and FixedPointList
Another common use of For loops is when output i depends on output i-1. In Mathematica, this is a canonical use of Nest and Fold, and their NestList and FoldList counterparts that also return the intermediate results. For example, here is the For loop way to create autoregressive noise:
vec = Table[0, {100}]; For[i = 2, i <= 100,
vec[[i]] = vec[[i - 1]] + RandomReal[]; i++];
Of course, you have to start at i=2 to avoid adding the zeroth part, which is the Head, List, to the vector.
Here is the FoldList way:
FoldList[#1 + #2 &, 0, RandomReal[{0, 1}, 99] ]
4. You can combine lists in very general ways using Inner and Outer.
People also often use nested For loops when they want to build up matrices depending on elements of two different vectors. Here is where the generalised structure of Outer is useful:
Outer[N@Kurtosis[#1[#2]] &, {StudentTDistribution, ExponentialDistribution,
ChiSquareDistribution}, Range[5, 15] ]
(* {{9., 6., 5., 4.5, 4.2, 4., 3.85714, 3.75, 3.66667, 3.6, 3.54545}, {9., 9.,
9., 9., 9., 9., 9., 9., 9., 9., 9.}, {5.4, 5., 4.71429, 4.5, 4.33333, 4.2,
4.09091, 4., 3.92308, 3.85714, 3.8}} *)
Another way to avoid nested loops is with Tuples, as discussed here. This nested For loop:
l = 0;
For[i = 0, i < 10, i++,
For[j = 0, j < 10, j++,
For[k = 0, k < 10, k++,
Xarray[l] = A[i, j, k];
Print[Xarray[l]];
l++;
]
]
]
Xarray[5]
Can be rewritten faster as:
Xarray = A @@@ Tuples[Range[0, 9], 3];
• On the "development time matters" I would say that one more example could be: vec=Table[{i,2^i},{i,10}]; vec[[All, 2]] = Log[vec[[All, 2]]]
– ivbc
Jun 17, 2016 at 16:14
How about operations like arithmetic on lists, matrixes, & tensors or Dot and DotProduct?
list1 = {a, b, c};
list2 = {d, e, f};
matrix1 = {{a, b, c}, {d, e, f}};
matrix2 = {{a, b, c}, {d, e, f}, {g, h, i}};
tensor1 = {{{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, {{1, 2, 3}, {4, 5,
6}, {7, 8, 9}}, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}}, {{{a, b,
c}, {d, e, f}, {g, h, i}}, {{a, b, c}, {d, e, f}, {g, h,
i}}, {{a, b, c}, {d, e, f}, {g, h, i}}}}
list1 + n
list1 - n
list1 * n
list1 / n
list1 + list2
list1 - list2
list1 * list2
list1 / list2
matrix1 . list1
matrix1 . matrix2
tensor1 . list1
tensor1 . matrix2
tensor1 . matrix2 . list1
Just a sampling of things, some of which would require some extensive procedural solutions.
"For" my part (pun intended), I stopped thinking in loops and started thinking functionally. More so, I began to think more like a mathematician than a programer. The better I do this the more efficiently I program in Mathematica.
It would help if one could tie someone's "loop" or "procedural" arm to their body, like they do with boxers who need to switch their dominant hand.
Ultimately, I think one has to give themselves over to the functional language. Maybe one can translate one language to another but poetry in English doesn't have the same meaning translated into Italian nor Italian to English. When one can dream in the language they'll have it.
Needs["VectorAnalysis"]
Includes:
• DotProduct[],
• CrossProduct[], and
• ScalarTripleProduct[]
Brief examples from the documentation:
DotProduct[]
Dot product of two Cartesian vectors:
In2:= a = {1, 2, 5};
In3:= b = {2, 3, -7};
In[4]:= DotProduct[a, b]
Out[4]= -27
CrossProduct[]
Find the cross product of a pair of vectors:
In2:= a = {1, 3, 5};
In3:= b = {-4, 7, 1};
In[4]:= CrossProduct[a, b]
Out[4]= {-32, -21, 19}
ScalarTripleProduct[]
Find the equation of the plane passing through the points with position vectors r1, r2, and r3:
In[7]:= r = {x, y, z};
In[8]:= ScalarTripleProduct[r - r1, r2 - r1, r3 - r1] == 0
Out[8]= -111 + 68 x - 29 y - 22 z == 0
It just doesn't stop...
Consider one example of functional decompositions, the CholeskyDecomposition[].
Maybe this doesn't count in this specific discussion because it demonstrates a specific type of solution available for use in Mathematica, which essentially hides how it gets the solution from the user. On the other side, I'd argue that the functional paradigm in Mathematica has made these kinds of solutions and functionality more readily available. The Wikipedia entry cited above shows how procedural approaches attack the problem. One could replace every procedural construct in that code with a functional one and produce your own functional Cholesky Decomposition pretty readily. But Mathematica has already done that for us.
The other decompositions:
LUDecomposition[]
QRDecomposition[]
SingularValueDecomposition[]
SchurDecomposition[]
HessenbergDecomposition[]
JordanDecomposition[]...
(* I've likely missed a few *)
probably do the same thing.
• Jagra, I think you should define sample list, matrix1, tensor1 objects so that this code can be evaluated. I know it's just an illustration, but to someone unfamiliar with what these do, it's not really a useful illustration without that. Jul 5, 2012 at 14:01
• Noted. Will do. Jul 5, 2012 at 14:29
• Jagra, I'm sorry that I am criticizing your answer a second time: it seems to me that the second half of your answer is not really appropriate here because any language can have library functions and this says little about the semantics of the language itself or good coding practice. Thanks for updating the first half however, as I think that showing that native arithmetic operations are vector, matrix, and tensor aware is very appropriate. Jul 6, 2012 at 6:46
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Posted on Categories:Mathematical logic, 数学代写, 数理逻辑
数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Transformations and Invariance
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数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Transformations and Invariance
Here we show that, under certain assumptions, the transformations of the first two groups defined in Section 3.7 preserve forcing approximations forc. This is not an absolutely elementary thing: there is no way to reasonably apply transformations to transitive models $M$ involved in the definition of forc. What we can do is to require that the transformations involved belong to the models involved. This leads to certain complications of different sort.
Family 1: permutations. First of all we have to extend the definition of the action of $\pi$ in Section 3.7 to include formulas. Suppose that $c, c^{\prime} \subseteq \mathcal{I}$. Define the action of any $\pi \in \mathrm{BIJ}_{c^{\prime}}^c$ onto formulas $\varphi$ of $\mathcal{L}\left(\mathbf{P}^*\right)$ such that $|\varphi| \subseteq c$ :
to get $\pi \varphi$ substitute $\pi \cdot \tau$ for any $\tau \in \operatorname{NAM} \varphi$ and $\pi \cdot B$ for any $B \in \operatorname{IND} \varphi$.
Lemma 22. Suppose that $\langle M, U\rangle, K, p, \varphi$ satisfy (F1) of Definition 20, sets $c, c^{\prime} \subseteq \mathcal{I}$ have equal cardinality and are absolute $\Delta_1^{\mathrm{HC}}(M), \pi \in \mathrm{BI}_{c^{\prime}}^c$ is an absolute $\Delta_1^{\mathrm{HC}}(M)$ function, and $|\varphi| \subseteq c,|U| \subseteq c, K \subseteq \mathbf{P}^* \mid c$.
Proof. Under the assumptions of the lemma, in particular, the requirement of $c, c^{\prime}, \pi$ being absolute $\Delta_1^{\mathrm{HC}}(M), \pi$ acts as an isomorphism on all relevant domains and preserves all relevant relations between the objects involved. Thus $\langle M, \pi \cdot U\rangle, \pi \cdot K, \pi \cdot p, \pi \varphi$ still satisfy Definition 20(F1), and $|\pi \varphi| \subseteq c^{\prime},|\pi \cdot U| \subseteq c^{\prime}, \pi \cdot K \subseteq \mathbf{P}^*\left\lceil c^{\prime}\right.$. (For instance, to show that $\pi \cdot U$ still belongs to $M$, note that the set $|U| \subseteq c$ belongs to $M$, thus $\pi|| U \mid \in M$, too, since $\pi$ is an absolute $\Delta_1^{\mathrm{HC}}(M)$ function.) This allows to prove the lemma by induction on the complexity of $\varphi$.
数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Elementary Equivalence Theorem
This section presents further properties of $\mathbb{P}$-generic extensions of $\mathbf{L}$ and their subextensions, including Theorem 13 and its corollaties on the elementary equivalence of different subextensions.
Assumption 2. We continue to assume $\mathbf{V}=\mathbf{L}$ in the ground universe. Below in this section, a number n $\geq 2$ is fixed, and pairs $\left\langle\mathbb{M}{\bar{\xi}}, \mathbb{U}{\bar{\zeta}}\right\rangle$, the system $\mathbb{U}=V_{\bar{\zeta}<\omega_1} U_{\bar{\zeta}}$, the forcing notions $\mathbb{P}{\bar{\xi}}=\mathbf{P}\left[\mathbb{U}{\bar{\zeta}}\right]$ and $\mathbb{P}=\mathbf{P}[U]=\bigcup_{\bar{\zeta}<\omega_1} \mathbb{P}{\bar{\zeta}}$ are as in Definition 16 for this $\mathrm{n}$. 6.1. Further Properties of Forcing Approximations Coming back to the complete sequence of pairs $\left\langle M{\bar{\zeta}}, \mathbb{U}_{\bar{\xi}}\right\rangle$ introduced by Definition 16, we consider the auxiliary forcing relation forc with respect to those pairs. We begin with the following definition.
Definition 21 (in L). Let $K \subseteq \mathbf{P}^*$ be a regular forcing. Recall that
$$K[U]=K \cap \mathbb{P} \text { and } K\left[U_{\bar{\zeta}}\right]=K \cap P\left[U_{\bar{\zeta}}\right]=K \cap \mathbb{P}{\bar{\zeta}}$$ names in $\mathbb{M}{\xi} \cap \mathbf{S N}\omega^\omega\left(K\left[\mathbb{U}{\tilde{\xi}}\right]\right)$ as parameters, all names $\tau \in \operatorname{NAM} \varphi$ are $K\left[\mathbb{U}{\tilde{\zeta}}\right]$-full below $p$, all indices $B \in \operatorname{IND} \varphi$ belong to $M{\xi}$. The following is an easy consequence of Lemma 18.
数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Elementary Equivalence Theorem
$$K[U]=K \cap \mathbb{P} \text { and } K\left[U_{\bar{\zeta}}\right]=K \cap P\left[U_{\bar{\zeta}}\right]=K \cap \mathbb{P}{\bar{\zeta}}$$ 姓名 $\mathbb{M}{\xi} \cap \mathbf{S N}\omega^\omega\left(K\left[\mathbb{U}{\tilde{\xi}}\right]\right)$ 作为参数，所有的名称 $\tau \in \operatorname{NAM} \varphi$ 是 $K\left[\mathbb{U}{\tilde{\zeta}}\right]$-下面全部 $p$，所有指标 $B \in \operatorname{IND} \varphi$ 属于 $M{\xi}$． 下面是引理18的一个简单推论。
MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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Question
Easy
Solving time: 2 mins
# Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are:
A
B
C
D
## Text solutionVerified
2 µF, 18 µF
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
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Question Text Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are: Updated On Jun 4, 2023 Topic Electrostatic Potential and Capacitance Subject Physics Class Class 12 Answer Type Text solution:1 Video solution: 4 Upvotes 384 Avg. Video Duration 6 min
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# Number 16933
Number 16,933 spell 🔊, write in words: sixteen thousand, nine hundred and thirty-three . Ordinal number 16933th is said 🔊 and write: sixteen thousand, nine hundred and thirty-third. The meaning of number 16933 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 16933. What is 16933 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 16933.
## What is 16,933 in other units
The decimal (Arabic) number 16933 converted to a Roman number is (X)(V)MCMXXXIII. Roman and decimal number conversions.
#### Weight conversion
16933 kilograms (kg) = 37330.5 pounds (lbs)
16933 pounds (lbs) = 7680.8 kilograms (kg)
#### Length conversion
16933 kilometers (km) equals to 10522 miles (mi).
16933 miles (mi) equals to 27252 kilometers (km).
16933 meters (m) equals to 55554 feet (ft).
16933 feet (ft) equals 5162 meters (m).
16933 centimeters (cm) equals to 6666.5 inches (in).
16933 inches (in) equals to 43009.8 centimeters (cm).
#### Temperature conversion
16933° Fahrenheit (°F) equals to 9389.4° Celsius (°C)
16933° Celsius (°C) equals to 30511.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
16933 seconds equals to 4 hours, 42 minutes, 13 seconds
16933 minutes equals to 1 week, 4 days, 18 hours, 13 minutes
### Codes and images of the number 16933
Number 16933 morse code: .---- -.... ----. ...-- ...--
Sign language for number 16933:
Number 16933 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 16933
### Multiplications
#### Multiplication table of 16933
16933 multiplied by two equals 33866 (16933 x 2 = 33866).
16933 multiplied by three equals 50799 (16933 x 3 = 50799).
16933 multiplied by four equals 67732 (16933 x 4 = 67732).
16933 multiplied by five equals 84665 (16933 x 5 = 84665).
16933 multiplied by six equals 101598 (16933 x 6 = 101598).
16933 multiplied by seven equals 118531 (16933 x 7 = 118531).
16933 multiplied by eight equals 135464 (16933 x 8 = 135464).
16933 multiplied by nine equals 152397 (16933 x 9 = 152397).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 16933
Half of 16933 is 8466,5 (16933 / 2 = 8466,5 = 8466 1/2).
One third of 16933 is 5644,3333 (16933 / 3 = 5644,3333 = 5644 1/3).
One quarter of 16933 is 4233,25 (16933 / 4 = 4233,25 = 4233 1/4).
One fifth of 16933 is 3386,6 (16933 / 5 = 3386,6 = 3386 3/5).
One sixth of 16933 is 2822,1667 (16933 / 6 = 2822,1667 = 2822 1/6).
One seventh of 16933 is 2419 (16933 / 7 = 2419).
One eighth of 16933 is 2116,625 (16933 / 8 = 2116,625 = 2116 5/8).
One ninth of 16933 is 1881,4444 (16933 / 9 = 1881,4444 = 1881 4/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
16933
#### Is Prime?
The number 16933 is not a prime number. The closest prime numbers are 16931, 16937.
#### Factorization and factors (dividers)
The prime factors of 16933 are 7 * 41 * 59
The factors of 16933 are 1 , 7 , 41 , 59 , 287 , 413 , 2419 , 16933
Total factors 8.
Sum of factors 20160 (3227).
#### Powers
The second power of 169332 is 286.726.489.
The third power of 169333 is 4.855.139.638.237.
#### Roots
The square root √16933 is 130,126861.
The cube root of 316933 is 25,678992.
#### Logarithms
The natural logarithm of No. ln 16933 = loge 16933 = 9,73702.
The logarithm to base 10 of No. log10 16933 = 4,228734.
The Napierian logarithm of No. log1/e 16933 = -9,73702.
### Trigonometric functions
The cosine of 16933 is 0,983046.
The sine of 16933 is -0,18336.
The tangent of 16933 is -0,186522.
### Properties of the number 16933
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 16933 in Computer Science
Code typeCode value
16933 Number of bytes16.5KB
Unix timeUnix time 16933 is equal to Thursday Jan. 1, 1970, 4:42:13 a.m. GMT
IPv4, IPv6Number 16933 internet address in dotted format v4 0.0.66.37, v6 ::4225
16933 Decimal = 100001000100101 Binary
16933 Decimal = 212020011 Ternary
16933 Decimal = 41045 Octal
16933 Decimal = 4225 Hexadecimal (0x4225 hex)
16933 BASE64MTY5MzM=
16933 MD5afac0bdcb0139e69de4ce27f1c6d352b
16933 SHA1b472d535f7f774517320c9cd2b0778ce8f8a3f2f
16933 SHA224a7835a34c1781f9b6f1ef552868893de469c10997204dbe7f6ab84d2
16933 SHA256a5b5e3b60ceb9094083f8f1474bc3042396ab91d06bb37148689c18c3775cbce
16933 SHA3848c1e43830845bbae296c8b4be3a8fbdbb911b99034a342d792b43b5809707a65fd1f29c28d6bebe63cc9eb2017c672b9
More SHA codes related to the number 16933 ...
If you know something interesting about the 16933 number that you did not find on this page, do not hesitate to write us here.
## Numerology 16933
### Character frequency in number 16933
Character (importance) frequency for numerology.
Character: Frequency: 1 1 6 1 9 1 3 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 16933, the numbers 1+6+9+3+3 = 2+2 = 4 are added and the meaning of the number 4 is sought.
## Interesting facts about the number 16933
### Asteroids
• (16933) 1998 FV88 is asteroid number 16933. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 3/24/1998.
### Distances between cities
• There is a 10,522 miles (16,933 km) direct distance between Buenos Aires (Argentina) and Phnom Penh (Cambodia).
• There is a 10,522 miles (16,933 km) direct distance between Guatemala City (Guatemala) and Tiruchirappalli (India).
• There is a 10,522 miles (16,933 km) direct distance between Chandigarh (India) and Santiago (Chile).
• There is a 10,522 miles (16,933 km) direct distance between San Miguel de Tucumán (Argentina) and Singapore (Singapore).
• More distances between cities ...
## Number 16,933 in other languages
How to say or write the number sixteen thousand, nine hundred and thirty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 16.933) dieciseis mil novecientos treinta y tres German: 🔊 (Anzahl 16.933) sechzehntausendneunhundertdreiunddreißig French: 🔊 (nombre 16 933) seize mille neuf cent trente-trois Portuguese: 🔊 (número 16 933) dezesseis mil, novecentos e trinta e três Chinese: 🔊 (数 16 933) 一万六千九百三十三 Arabian: 🔊 (عدد 16,933) ستة عشر ألفاً و تسعمائة و ثلاثة و ثلاثون Czech: 🔊 (číslo 16 933) šestnáct tisíc devětset třicet tři Korean: 🔊 (번호 16,933) 만 육천구백삼십삼 Danish: 🔊 (nummer 16 933) sekstentusinde og nihundrede og treogtredive Dutch: 🔊 (nummer 16 933) zestienduizendnegenhonderddrieëndertig Japanese: 🔊 (数 16,933) 一万六千九百三十三 Indonesian: 🔊 (jumlah 16.933) enam belas ribu sembilan ratus tiga puluh tiga Italian: 🔊 (numero 16 933) sedicimilanovecentotrentatré Norwegian: 🔊 (nummer 16 933) seksten tusen, ni hundre og tretti-tre Polish: 🔊 (liczba 16 933) szesnaście tysięcy dziewięćset trzydzieści trzy Russian: 🔊 (номер 16 933) шестнадцать тысяч девятьсот тридцать три Turkish: 🔊 (numara 16,933) onaltıbindokuzyüzotuzüç Thai: 🔊 (จำนวน 16 933) หนึ่งหมื่นหกพันเก้าร้อยสามสิบสาม Ukrainian: 🔊 (номер 16 933) шiстнадцять тисяч дев'ятсот тридцять три Vietnamese: 🔊 (con số 16.933) mười sáu nghìn chín trăm ba mươi ba Other languages ...
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## Comment
If you know something interesting about the number 16933 or any natural number (positive integer) please write us here or on facebook.
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http://openstudy.com/updates/4f70ed99e4b0eb8587737009
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• anonymous
5. A rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s. The height of the rocket above the ground at time t is given by h = –16t2 + 156t + 105. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second. (1 point) 4.9 s 9.8 s 0.6 s 10.4 s
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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https://theentrepreneurshealthcoach.com/qa/question-how-many-sockets-can-be-on-a-radial-circuit-uk.html
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# Question: How Many Sockets Can Be On A Radial Circuit UK?
## How many sockets can be on a 16a Radial?
For a single socket =13A (OK) and for a double socket = 26A (not OK).
If 16A then that would be one socket; but if the cable is 2.5 and doesnt exceed the area permitted change the breaker to 20 A then you can have an unlimited number..
## How many amps can 2.5 mm cable take?
23-25 ampsA 2.5mm cable is capable of supplying around 23-25 amps depending on the method of installation, so is fine, safe and compliant on a 20amp circuit breaker.
Answer: It is acceptable to install a number of sockets on a radial circuit using 2.5 mm cables protected by a 16A circuit breaker – provided that the expected connected load does not exceed 16A. To see many more Q & A in Voltimum UK’s Experts Area, please click on the link.
## Which is better ring or radial circuit?
Ring circuit wiring offered a more efficient and lower cost system which would safely support a greater number of sockets. The lower costs aspect was related to fact that wires with a smaller diameter could be used, compared to radial wiring. … Ring circuits provide low protection against overcurrents.
## Is 2.5 mm cable OK for sockets?
What About the Size of the Earth Wire? The cross sectional area of the earth wire in a 1.5mm cable is 1mm and in a 2.5mm cable it is 1.5mm. This should be sufficient for most domestic socket and lighting circuits.
## Can you run a radial circuit off a ring main?
No, you can’t just spur off a ring circuit, there are rules. Using a 6mm2 cable on a 2.5mm2 ring final is also daft, you only need a 2.5mm2 cable. You are allowed to put ONE fused spur at a point and spur off that to unlimited sockets, or, one UNFUSED spur to ONE socket.
## How many sockets can be on a radial circuit?
If you adhere to the principles above e.g. a radial circuit running from a consumer unit with its own MCB of the correct rating can have as many socket as you like, as long as it’s in an area not exceeding 50 square metres (or 540sq ft).
## Can I use 2.5 mm cable for lighting?
You will find twin core and earth cabling used all over your home in a variety of sizes. 2.5mm is commonly used for behind sockets, while 1-1.5mm is most often used for lights (depending on how many lights you have in a circuit).
## How do you know if a circuit is radial or ring?
If there is only one cable connected to the MCB/fuse for the circuit then (unless somebody has done something very wrong, non-compliant and potentially very dangerous) the circuit ‘must’ be a radial (a ring has two ends, both of which should both be connected to the same MCB/fuse).
## How many sockets can you have on a radial UK?
Unfused spurs off a 30/32 A radial. One single or one twin using 2.5 mm sq. As many as you want using 4.0 mm sq. Fused spurs – as many sockets as you want but remember the maximum 13A fuse in the spur!
## How many sockets can be on a 2.5 mm radial circuit?
2 socketsAny more then 2 sockets on a radial 2.5mm is Mickey Mouse. In my mind radials are for dedicated items ie immersion heaters, cookers etc and not for socket outlets in general.
## Can you spur off a lighting radial?
As davey says, yes you can spur of it. You can add as many lights as you want but dont exceed the rating of the cable, or make the circuit too long that it exceeds the voltage drop limits and/or the maximum earth fault loop impedance.
## How many sockets can be on a 20 amp circuit?
10 receptaclesOne rule of thumb is to assign a maximum draw of 1.5 amps to each receptacle, which allows for 10 receptacles on a 20-amp circuit.
## What is the difference between radial and ring circuit?
This form of wiring is becoming more popular with designers because it is much easier to determine faults than in a ring-final circuit – in a radial circuit if there is a break anywhere along the cable all the circuits after the break will stop working, while in a ring-final circuit even in case of fault all sockets …
## What are the advantages of ring system?
Expert Answer:The two most common advantages of the ring system are as follows:- Voltage drop across the circuit is less or negligible when compared with the tree system.- More appliances can be connected across the circuits as the supply is radially distributed across the space.More items…•
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https://economics.stackexchange.com/questions/17516/what-is-one-dimensional-ordered-type
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# What is one dimensional, ordered type?
I am reading papers about moral hazard. What is one dimensional, ordered type $\theta\in\Theta$?
What is one dimensional, not ordered type $\theta\in\Theta$?
Could you please give an example?
References:
Riley (1979)
Azevedo and Gottlieb (2016)
Judging from the reference you provide, this refers to whether the set $$\Theta$$ is ordered or not. For example, natural numbers or the alphabet are ordered sets. In the context of moral hazard problems, examples could be effort, or ability. Since this is a numerical variable, it is an ordered set.
In the paper you mention, the phrase "one dimensional, ordered type..." is not found. However, the paper states that:
... differences among sellers are assumed to be parameterized by a single unobservable characteristic $$\theta \in \Theta$$. Then the family of utility functions can be written in the alternative form,
$$U_i=U(\theta_i;y,p) \text{, } i \in I$$
If seller $$i$$ has $$\theta_i$$ units of the characteristic we may therefore refer to him as being of "type $$\theta_i$$.
The first bold statement refers to the one dimensionality of the unobserved type, whereas the second refers to it being a numerical variable, which means $$\Theta$$ is an ordered set.
Conversely, a non-ordered set contains elements which have no intrinsic order. For instance, you can think of a case where agents do not know the city where they are to be located. The set "cities" is unordered because cities do not necessarily hold a particular order in a one-dimension set. You can add a second dimension (e.g. population), where you could order them based on city size.
I would say that having ordered sets facilitates algebraic analysis, since it enables the introduction of monotonic strategies, whereby an important outcome variable is a monotonic function of the agent's type. In effect, this seems to be important in the paper, when the author states that:
A seller of type $$\theta_i$$ brings to market a product that is valued equally by each of the set of potential buyers, $$J$$. This valuation, $$V_i$$, measured in dollars, is an increasing function of $$\theta$$ and (possibly) also of the intensity of sales-related activity, $$y$$:
$$V_i = V(\theta_i;y)$$
The bold emphasis highlights precisely the monotonicity enabled by the order property of $$\Theta$$. Another example is given in Athey (2001):
$$\hskip1in$$
In response to the second reference you give (here), in the phrase
We now show that Riley equilibria may not exist when types are not ordered.
the "ordered" adjective refers to the second interpretation given above (i.e. of monotonicity). First, they assume that
Consumers are uniformly distributed in the unit interval, $$\theta in [0,1]$$
which is an ordered set. However, because both the utility function and the cost function are quadratic, it results that
... the types in the extreme points have the strongest preference for one of the contracts and are the most costly to serve.
which means that the utility and cost functions are not monotonic over $$\theta$$.
• So loosely speaking, let $U(\theta_i)$ denotes the utility of ordered type, then $dU(\theta)/d\theta>0$? If it not ordered type, $dU(\theta)/d\theta>0$ might not hold? – High GPA Jul 19 '17 at 15:10
• @HighGPA Yes and no. A unidimensional, numerical variable is, I think, always ordered. My point is that an ordered set facilitates the algebraic treatment of monotonicity, but does not require it (as in your second example). Think in the variable "effort". We can imagine situations whereby exerting so much effort ends up being less productive than by having a bit of time off (to rest or etc). Here, the variable effort its still ordered, yet its outcome (productivity) might not be monotonic. Meanwhile, if the set is nor ordered, the derivative makes less (no?) sense (e.g. cities). – luchonacho Jul 19 '17 at 15:24
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A234000 Numbers of the form 4^i*(8*j+1). 9
1, 4, 9, 16, 17, 25, 33, 36, 41, 49, 57, 64, 65, 68, 73, 81, 89, 97, 100, 105, 113, 121, 129, 132, 137, 144, 145, 153, 161, 164, 169, 177, 185, 193, 196, 201, 209, 217, 225, 228, 233, 241, 249, 256, 257, 260, 265, 272, 273, 281, 289, 292, 297, 305, 313, 321, 324, 329, 337, 345 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Squares modulo all powers of 2. - Robert Israel, Aug 26 2014 From Peter Munn, Dec 11 2019: (Start) Closed under multiplication. Contains all even powers of primes. A subgroup of the positive integers under the binary operation A059897(.,.). For all n, a(n) has no Fermi-Dirac factor of 2 and if m_k denotes the number of Fermi-Dirac factors of a(n) that are congruent to k modulo 8, m_3, m_5 and m_7 have the same parity. It can further be shown (1) all numbers that meet these requirements are in the sequence and (2) this implies closure under A059897(.,.). (End) LINKS Robert Israel, Table of n, a(n) for n = 1..10000 FORMULA a(n) = 6n + O(log n). - Charles R Greathouse IV, Dec 19 2013 MAPLE N:= 1000: # to get all terms <= N {seq(seq(4^i*(8*k+1), k = 0 .. floor((N * 4^(-i)-1)/8)), i=0..floor(log[4](N)))}; # Robert Israel, Aug 26 2014 PROG (PARI) is_A234000(n)=(n/4^valuation(n, 4))%8==1 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013; minor improvement by M. F. Hasler, Jan 02 2014 (PARI) list(lim)=my(v=List(), t); for(e=0, logint(lim\1, 4), t=4^e; forstep(k=t, lim, 8*t, listput(v, k))); Set(v) \\ Charles R Greathouse IV, Jan 12 2017 CROSSREFS Cf. A055046 (Numbers of the form 4^i*(8*j+3)). Cf. A055045 (Numbers of the form 4^i*(8*j+5)). Cf. A004215 (Numbers of the form 4^i*(8*j+7)). Cf. A059897. Sequence in context: A313316 A010393 A010425 * A313317 A292674 A313318 Adjacent sequences: A233997 A233998 A233999 * A234001 A234002 A234003 KEYWORD nonn,easy AUTHOR V. Raman, Dec 18 2013 STATUS approved
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Last modified May 11 13:41 EDT 2021. Contains 343791 sequences. (Running on oeis4.)
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http://www.chegg.com/homework-help/principles-of-foundation-engineering-7th-edition-chapter-3-solutions-9780495668107
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# Principles of Foundation Engineering (7th Edition) View more editions Solutions for Chapter 3
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Chapter: Problem:
100% (8 ratings)
For the following cases, determine the allowable gross vertical load-bearing capacity of the foundation. Use Terzaghi’s equation and assume general shear failure in soil. Use FS = 4.
SAMPLE SOLUTION
Chapter: Problem:
100% (8 ratings)
• Step 1 of 5
Consider the Terzaghi’s equation of bearing capacity for a continuous foundation:
Where are the bearing capacity factors for cohesion, self weight of the soil and effective overburden pressure respectively and are all dependent on the soil’s angle of internal friction is the bulk unit weight of the soil, B is the foundation width, q is the effective overburden pressure at the level of the foundation, c’ is the cohesion of the soil.
• Step 2 of 5
a) Calculate bearing capacity of the soil
From the table, Terzaghi’s bearing capacity factors, consider the values
Substitutefor,for ,for ,for ,for
• Step 3 of 5
Calculate allowable bearing capacity:
Substitute for
Therefore, the allowable gross bearing capacity of the foundation will be
This is the bearing capacity of the foundation under service-load condition.
• Step 4 of 5
b) Calculate bearing capacity of the soil
From the table, Terzaghi’s bearing capacity factors, consider the values
Substitutefor, for,for,for,for
The allowable gross bearing capacity of the foundation will be
This is the bearing capacity of the foundation under service-load condition.
• Step 5 of 5
c) Calculate bearing capacity of the soil
From the table, Terzaghi’s bearing capacity factors, consider the values
Substitutefor,for ,for ,for ,for
The allowable gross vertical load-bearing capacity of the foundation will be
This is the bearing capacity of the foundation under service-load condition.
Corresponding Textbook
Principles of Foundation Engineering | 7th Edition
9780495668107ISBN-13: 0495668109ISBN: Braja M DasAuthors:
Alternate ISBN: 9781111787097
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# MCQ Questions | Class 8 Maths Chapter 11 | Mensuration with Answers
## MCQ | Mensuration Class 8
Check the below NCERT MCQ Questions for Class 8 Maths Chapter 11 Mensuration with Answers Pdf free download. MCQ Questions for Class 8 Maths with Answers were prepared based on the latest exam pattern. We have Provided Mensuration Class 8 Maths MCQs Questions with Answers to help students understand the concept very well.
### Objective Question | NCERT Maths Class 8
Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 8th Maths 11 Part-3 Mensuration English
#### Ncert Solutions for Class 8 Maths | Objective Type Questions
You can refer to NCERT Solutions for Class 8 Maths Chapter 11 Mensuration to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
##### Mensuration | Class 8 Maths | NCERT Solutions
Q1. CBSE Class 8 Maths Mcq Questions 8 persons can stay in a cubical room. Each person requires 27 m³ of air. The side of the cube is A. 6 m B. 4 m C. 3 m D. 2 m.
Ans: 6 m
Hint:
Volume = 8 × 27 = 216 m³
∴ side = $$\sqrt[3]{216}$$ = 6 m.
Q2. CBSE Class 8 Maths Mensuration Mcq Questions If the height of a cuboid becomes zero, it will take the shape of a A. cube B. parallelogram C. circle D. rectangle.
Ans: rectangle.
Q3. Class 8 Maths Mcq Chapter 11 The volume of a room is 80 m³. The area of the floor is 20 m². The height of the room is A. 1 m B. 2 m C. 3 m D. 4 m.
Ans: 4 m
Hint:
Height = $$\frac{80}{20}$$ = 4 m.
Q4. Class 8 Maths Mcq online test The floor of a room is a square of side 6 m. Its height is 4 m. The volume of the room is A. 140 m³ B. 142 m³ C. 144 m³ D. 145 m³
Ans: 144 m³
Hint:
Volume = 6 × 6 × 4 = 144 m³.
Q5. Class 8 Maths Mcq with answers The base radius and height of a right circular cylinder are 14 cm and 5 cm respectively. Its curved surface is A. 220 cm² B. 440 cm² C. 1232 cm² D. 2π × 14 × (14 + 5) cm²
Ans: 440 cm²
Hint:
Curved surface = 2 × $$\frac{22}{7}$$ × 14 × 5
= 440 cm².
Q6. Learn Cbse Class 8 Maths Mcq The heights of two right circular cylinders are the same. Their volumes are respectively 16π m³ and 81π m³. The ratio of their base radii is A. 16 : 81 B. 4 : 9 C. 2 : 3 D. 9 : 4.
Ans: 4 : 9
Hint:
$$\frac{πr_1^2h}{πr_2^2h}$$ = $$\frac{16π}{81π}$$ ⇒ $$\frac{r_1}{r_2}$$ = $$\frac{4}{9}$$
Q7. Mcq Class 8 Maths Chapter 11 The ratio of the radii of two right circular cylinders is 1 : 2 and the ratio of their heights is 4 : 1. The ratio of their volumes is A. 1 : 1 B. 1 : 2 C. 2 : 1 D. 4 : 1.
Ans: 1 : 1
Hint:
$$\frac{r_1}{r_2}$$ = $$\frac{π(1)^24}{π(2)^21}$$ = 1 : 1
Q8. Mcq for Class 8 Maths Chapter 11 A glass in the form of a right circular cylinder is half full of water. Its base radius is 3 cm and height is 8 cm. The volume of water is A. 18π cm³ B. 36π cm³ C. 9π cm³ D. 36 cm³
Ans: 36π cm³
Hint:
Volume = $$\frac{1}{2}$$π × 3 × 3 × 8 = 36π cm³.
Q9. Mcq Questions for Class 8 Maths with answers pdf Chapter 11 The base area of a right circular cylinder is 16K cm³. Its height is 5 cm. Its curved surface area is A. 40π cm² B. 30π cm² C. 20π cm² D. 10π cm²
Ans: 40π cm²
Hint:
πr² = 16π ⇒ r = 4 cm
∴ Curved surface area
= 2 × π × 4 × 5 = 40π cm²
Q10. Mcq Questions for Class 8 Maths with answers pdf download The base radius and height of a right circular cylinder are 5 cm and 10 cm. Its total surface area is A. 150π cm² B. 300π cm² C. 150 cm² D. 300 cm²
Ans: 150π cm²
Hint:
Total surface area = 2πr (h + r)
= 2π 5 (10 + 5) = 150π cm².
Q11. Ncert Class 8 Maths Mcq Chapter 11 If the length and breadth of a rectangle are 10 cm and 5 cm, respectively, then its area is: A. 100 sq.cm B. 150 sq.cm C. 115 sq.cm D. 200 sq.cm
Ans: 150 sq.cm
Hint:
Length = 10 cm
Area of rectangle = Lenght x breadth
= 10 x 5
= 150 cm2
Q12. Ncert Class 8 Maths Mcq Questions The area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm is: A. 41 cm2 B. 82 cm2 C. 410 cm2 D. 820 cm2
Ans: 41 cm2
Hint:
Area of rhombus = ½ d1 d2
A = ½ x 10 x 8.2
A = 41 cm2
Q13. CBSE Class 8 Maths Chapter 11 Mcq The area of a trapezium is 480 cm2, the distance between two parallel sides is 15 cm and one of the parallel side is 20 cm. The other parallel side is: A. 20 cm B. 34 cm C. 44 cm D. 50 cm
Ans: 44 cm
Hint:
Area of trapezium = ½ h (a + b)
a=20 cm, h = 15 cm, Area = 480 sq.cm
480 = ½ (15) (20 + b)
20 + b = (480 x 2)/15
20 + b = 64
b = 44cm
Q14. CBSE Class 8 Maths Chapter 11 Mcq Questions The area of a rhombus is 240 cm2 and one of the diagonals is 16 cm. Find the other diagonal. A. 16 cm B. 20 cm C. 30 cm D. 36 cm
Ans: 30 cm
Hint:
Area = 240 cm2
d1 = 16 cm
Area of rhombus = ½ d1 x d2
240 = ½ x 16 x d2
d2 = 480/16 = 30 cm
Q15. CBSE Class 8 Maths Mensuration Mcq A cuboid has ______ pairs of identical faces. A. 2 B. 3 C. 4 D. 5
Ans: 3
Hint:
All six faces are rectangular, and opposites faces are identical. So there are three pairs of identical faces.
Q16. CBSE Class 8 Maths Mcq Questions All six faces of a cube are: A. Identical B. Different C. Circular D. Rectangular
Ans: Identical
Hint:
All six faces are squares and identical
Q17. CBSE Class 8 Maths Mensuration Mcq Questions A cylindrical box has ____ curved surface and ____ circular faces, which are identical. A. One, One B. One, two C. two, one D. two, two
Ans: One, two
Hint:
A cylindrical box having circular bases have identical top. One curved surface and two circular faces which are identical.
Q18. Class 8 Maths Mcq Chapter 11 If a cuboidal box has height, length and width as 20 cm, 15 cm and 10 cm respectively. Then its total surface area is: A. 1100 cm2 B. 1200 cm2 C. 1300 cm2 D. 1400 cm2
Ans: 1300 cm2
Hint:
Total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)
TSA = 2 ( 300 + 200 + 150) = 1300 cm2
Q19. Class 8 Maths Mcq online test The height of a cylinder whose radius is 7 cm and the total surface area is 968 cm2 is: A. 15 cm B. 17 cm C. 19 cm D. 21 cm
Ans: 15 cm
Hint:
Total surface area = 2πr (h + r)
968 = 2 x 22/7 x 7 (7+h)
h = 15 cm
Q20. Class 8 Maths Mcq with answers The height of a cuboid whose volume is 275 cm3 and base area is 25 cm2 is: A. 10 cm B. 11 cm C. 12 cm D. 13 cm
Ans: 11 cm
Hint:
Volume of a cuboid = Base area × Height
Height = Volume / Base area
H = 275/25 = 11 cm
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# Questions tagged [frequency-domain]
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7k views
### How to circularly shift a signal by a fraction of a sample?
The shift theorem says: Multiplying $x_n$ by a linear phase $e^{\frac{2\pi i}{N}n m}$ for some integer m corresponds to a circular shift of the output $X_k$: $X_k$ is replaced by $X_{k-m}$, where ...
13k views
### When should I calculate PSD instead of plain FFT magnitude spectrum?
I have a thirty-second speech signal that was sampled at 44.1 kHz. Now, I would like to show what frequencies the speech has. However, I'm not sure what would be the best way to do that. It seems ...
12k views
### Determining the noise floor of a signal in the frequency domain
Is there an accepted way of determining the noise floor of a signal by looking at it in the frequency domain? Is it a matter of averaging all the bins, or median, or some more complex calculation ...
267 views
### What is the type of noise in this image?
what reason could make the noise/effect in this image? Below is the original Thanks!
27k views
### Number of DFT (FFT) Points Required for a Specific Frequency Resolution for an Oversampled Signal
I have a bandpass signal centered at 2 MHz and bandwidth of 50 kHz (the signal frequency varies from 2 MHz - 25 kHz to 2 MHz + 25 kHz). This signal is being sampled at 10 MHz. I want a frequency ...
5k views
### What's the meaning of a complex zero/pole?
I have been studying signal processing and control for a while now, and I use Laplace and Fourier transforms almost everyday. Also another tools such as Nyquist or Bode plots. However, I had never ...
29k views
### Amplitude of the Signal in Frequency Domain different from Time Domain
How does the Amplitude of the Signal is changed when taking the FFT of a Signal, Have a look, The amplitude had changed from 2 to 30, and Here is my code for generating the above output, ...
4k views
### How do real-time convolution plugins process audio so quickly
Okay, so in Logic Pro I can load up a Space Designer plugin (convolution reverb) with an impulse that's 9.1 seconds long, turn my mic on, and get real-time convolution reverb as the mic records ...
2k views
2k views
### How to Apply Window Functions in Frequency Domain?
I am trying to apply a window function to smooth my signal and finally obtain the peaks of the signal. Yet, for spectral leakage I apply window function in time domain. After taking the Fourier ...
2k views
### Continuous Time Signal and Discrete Time Signal - Connection Between Periodicity and Discretness
I know that all periodic continuous time signal have discrete spectral representations, but are all discrete spectral representations periodic in continuous time? Also, can all periodic signals be ...
10k views
### Does the number of samples matter for FFT, and how to get a specific frequency visible?
If I have a signal that has 740 samples, the sampling frequency is 1000 Hz, and I take an FFT of length 1024, are the frequencies I get now 0, fs/1024, 2*fs/1000, ..., 511*fs/1024? I mean, do I get ...
125 views
### Idea for Noise Level Estimation / Automatic Thresholding in the Presence of Peaks
I have the fft of some signal, and want a rough estimate of the noise level in order to choose an appropriate threshold for our peak detection algorithm. In general, the fft contains mostly noise with ...
2k views
### Computation of the Inverse DCT (IDCT) Using DCT Or IFFT
Is there a way to compute the inverse discrete cosine transform (type-2) by leveraging either a DCT, FFT, or IFFT algorithm? I have seen ways to compute the DCT using FFTs, and I've seen ways to ...
1k views
### What is difference between terms $X(j \omega) ,X(\ e^{j \omega })$ and $X(\omega)$?
While studying frequency transforms ,I get confused with the terms like $X(j \omega) ,X(\ e^{j \omega })$ and $X(\omega)$ ,where $\omega = 2 \pi f$. So what is the difference between them ?
4k views
### Why audio clipping produces harmonics?
Today I was talking to my collegue about clipping and he said that clipping produces harmonics in frequency domain as he showed me the clipped signal frequency response. However when I asked him why ...
2k views
### Frequency Domain Filtering
Can you create a zero phase IIR filter by transforming its impulse response into frequency domain and only taking the magnitude of that frequency repsone? I have seen this in an open source project ...
1k views
### Discontinuities in the FFT
So I am taking the Fast Fourier Transform of the following function: $$x[n] = \displaystyle\sum\limits_{i=0}^{5} A_{i} \cos\left(\frac{\omega_{i}}{\omega_{s}} n + \phi_{i}\right)$$ Where the ...
399 views
### Understanding the $\mathcal Z$-transform
I was studying $\mathcal Z$-transforms and found pretty good material on the topic, though I feel I do not have a proper understanding of the concept. Could someone help me clarify this? I know that ...
1k views
### Image 2D Real Cepstrum with DFT, Is ifftshift Needed?
This is my testing image,it is taken from this paper: I tried to transform it into its real cepstrum domain with this simple MATLAB code: ...
126 views
### Explain the Process of Spectral Pooling and Spectral Activation in the Context of CNN in Frequency Domain
I am reading the paper Design of an energy efficient accelerator for training of convolutional neural networks using frequency Domain Computation: which uses Frequency Pooling, from Spectral ...
1k views
### Chop out frequencies outside human hearing range
I have a bunch of audio files all sampled at 44100 Hz sample frequency. I am trying to remove all the frequencies which are outside the human hearing range (I use the following as reference: Frequency ...
277 views
### Estimation of Amplitude, Frequency and Phase of Linear Combination of Harmonic Signal Beyond the Leakage Resolution of DFT
How can I find a rough ( as accurate as possible) Amplitude of each frequency when there is spectral leakage. Currently, I am dealing with a system that contains special leakage which seems ...
552 views
### Convolution Theorem using DCT
I am trying to digest this paper. [2002] Anna Usakova. Using Of Discrete Orthogonal Transforms For Convolution So far, I've been successful with DFT and DHT. Could someone give a hand with DCT? I am ...
103 views
### Why are LTI filters not able to output frequencies not present at the input?
I know this a very basic question but I'm just having troubles understanding why. Thanks in advance!
2k views
### Which information do we get from magnitude and phase spectrum?
I am learning image processing. I want to ask very basic question related to FFT topic Which information do we actually get from "phase spectrum" and "magnitude spectrum" about an image?
810 views
### Time domain maximum from frequency domain data?
Is it possible to calculate the maximum value of a time-domain signal from frequency-domain representation without performing an inverse transform?
471 views
### Exercise Related to Frequency Resolution and SNR
I'm studying a book about DSP and trying to make exercises. Here is one I'm interested in: A scientist acquires 65,536 samples from an experiment at a sampling rate of 1 MHz. He knows that the signal ...
119 views
### Frequency-domain deconvolution: "Direct" filtering vs "Wiener" filtering
Can someone help with clarifying the difference between two approaches to frequency domain "deconvolution: For the frequency domain problem: We want to find a filter $F(\omega)$ which will ...
2k views
### Filter - Spatial Domain Versus Frequency Domain
Note: This is a followup question of How can I create a custom band-pass filter?. Can you explain the following python source code?. Bandpass filter bank. Suppose, I have an equation ...
696 views
### Color Artifacts in Fourier Transformed Image
I am working with images and Fourier transforms. I am trying to understand what might be causing some artifacts in my output image. I am starting with a 512x512, RGB image of Lenna. I FFT the image,...
76 views
### Finding the Transfer Function of a Multiplicative Distortion in MATLAB
I have the frequency response of two signals shown above. One is uncontaminated, and the other is the same signal contaminated by a multiplicative distortion with two poles and two zeros. I need to ...
148 views
### Amplitude Estimating Using a Windowed DFT
Let's say we want to estimate the amplitude A of a mono-frequent signal using a windowed DFT. The frequency of the signal is unknown, and the frequency resolution of the DFT is limited, thus it cannot ...
181 views
### Frequency Spectrum Analysis
How do you find the phase and inverse Fourier transform of this frequency spectrum?
96 views
### Constructing frequency domain filter from a 2d spatial kernel
Given a 2d spatial kernel $h$, I want to construct the corresponding frequency filter $H$. In theory, $H$ is simply obtained by calculating the DFT of $h$ using a FFT algorithm. But in practice, there ...
7k views
### Why Frequency Domain conversion is Important in Digital Image Processing?
Any Image enhancement technique can be easily applied to the normal images. Still, most of the folks say that frequency domain conversion of the images can lead to better enhancement. Why do we need ...
244 views
### What is the LOW frequency resolution rule analogous to Nyquist?
If I'm analyzing a signal in the frequency domain, I know about the well known Nyquist criteria that the sample frequency must be > 2x of the highest component present in the signal. However, there ...
8k views
5k views
### Perform Convolution in Frequency Domain Using FFTW?
I'm trying to convolve two signals $x(n)$ and $h(n)$ in C by using the FFTW library's functions to perform a Fourier transform on each, multiply the appropriate complex components together, and take ...
2k views
### Magnitude of the Gradient in Frequency Domain
I'm learning some basics of image processing. Recently I've read about image filtering and two-dimensional Fourier transform, because I'm preparing for exam. And I have one question I don't know ...
718 views
### Spectral Leakage, Phase and other questions about FFT
I'm currently implementing a voice recognition algorithm using FFT and now some questions came up to me: Spectral Leakage: I know what it is, why it appears and how to solve it using a Windowing ...
Given a signal $\left\{ x [ 0 ], x [ 1 ], ..., x [ N - 1 ] \right\}$ what would be the correct way to downsample it in the frequency domain (Sinc interpolation)?
I've been trying to find a proof of the following, but still I m unable to proof it, can someone help me? $$ℱ[x(t)g(t)] = \frac{1}{2\pi} [X(\omega)*G(\omega)]$$
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