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# Power Rule Worksheet
### The Power Rule
#### Positive Exponents
$\frac{d}{dx}[x^{2}]$=Submit Answer:
#### Negative Exponents
$\frac{d}{dx}[x^{-1}]$=Submit Answer:
#### Fractional Exponents
$\frac{d}{dx}[x^{4/5}]$=Submit Answer:
#### Negative Fractional Exponents
$\frac{d}{dx}[x^{-1/2}]$=Submit Answer:
### Rewriting Functions
$\frac{d}{dx}[x\sqrt{x^{7}}]$=Submit Answer:
#### Fractions
$\frac{d}{dx}[\frac{1}{x^{3}}]$=Submit Answer:
### Lightning Round
#### Easy
$\frac{d}{dx}[x^{5}]$=Submit Answer:
#### Medium
$\frac{d}{dx}[x^{1/5}]$=Submit Answer:
#### Hard
$\frac{d}{dx}[x\sqrt{x}]$=Submit Answer:
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1 ) If tanA=a/b , then (asinA+bcosA)/(asinA-bcosA)=?
sanatan sharma
26 Points
12 years ago
if i divide the numerator and denominator by cosA we will get
= ((asinA/cosA)+b)/((asinA/cosA)-bcosA)
= (atanA + b)/(atanA - b)
= (a2 + b2)/(a- b2)
Muntasir Mahi
15 Points
4 years ago
tanA=a/b
=sinA/cosA=a/b
=a/b×sinA/cosA=a/b×a/b
=asinA/bcosA=a2/b2
=asinA+bcosA/asina-bcosA=a2+b2/a2-b2
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# Electrons and Matter Waves
## Homework Statement
A stream of protons, each with a speed of 0.8250c, are directed into a two-slit experiment where the slit separation is 2.00 10-9 m. A two-slit interference pattern is built up on the viewing screen. What is the angle between the center of the pattern and the second minimum (to either side of the center)?
## Homework Equations
p = mv
lambda = h/p
d*sin(theta) = (m + 1/2)*lambda
## The Attempt at a Solution
I have tried to first solve for the wavelength in the experiment by using p = mv. With this I get:
p = (1.673E-27)*(0.8250)*(3E8)
p = 4.14E-19
Then I solve for the wavelength using lambda = h/p:
lambda = (6.63E-34) / (4.14E-19)
lambda = 1.6E-15
Once I have the wavelength, I use the double slit formula from Young's Experiment to try and calculate the angle, by using m = 1 and then solving for arcsin:
theta = arcsin ( m*lambda / d)
theta = arcsin ( 1.5*(1.6E-15) / (2E-9))
However this gives me a very small angle which obviously is the incorrect answer.
Am I approaching this completely wrong, or am I just goofing up somewhere?
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Inequality concerning factorial
• $$a_1,a_2\cdots,a_n\in\mathbb R_+$$;
• $$\forall1\le k\le n,a_1a_2\cdots a_k\ge k!$$
Show that $$\frac{2!}{1+a_1}+\frac{3!}{(1+a_1)(2+a_2)}+\cdots+\frac{(n+1)!}{(1+a_1)(2+a_2)\cdots(n+a_n)}<3.$$
My thought would be to prove $$(1+a_1)(2+a_2)\cdots(n+a_n)\ge2^nn!.$$ If so, the expression on the left $$<\sum_{i=1}^n\frac{(i+1)!}{2^ii!}=\sum_{i=1}^n\frac{i+1}{2^i}=3-\frac{3+n}{2^n}<3$$ and that’s done.
Perhaps we could find an upper limit for it and use the “spiral induction” method?
• @Wizard0001, I think that violates the condition that $a_1a_2…a_k\geq k!$ as $0\times 0<2!$ May 17 at 17:53
Your thoughts are indeed correct, thank you to @IsaacBrowne for a solution involving concavity. It is also possible to use more elementary methods with the AM-GM inequality as $$x+y\geq2\sqrt{xy}$$ for positive $$x,y$$: $$\prod_{i=1}^{n}(a_i+i)\geq \prod_{i=1}^{n}2\sqrt{ia_i}=\sqrt{i!}2^n\prod_{i=1}^n\sqrt{a_i}\geq \sqrt{i!}2^n \sqrt{i!}=i!2^n$$ As a side note, what is spiral induction? I keep on hearing people mention it but no idea what it is, and searches never reveal anything.
• Spiral induction is like this. We want $f(x)<g(x),x\in\mathbb{N}_+$ and to prove it we strengthen the proposition to$f(x)<g(x)<h(x)$. If $f(x)<g(x)\Rightarrow g(x+1)<h(x+1);g(x)<h(x)\Rightarrow f(x+1)<g(x+1)$ we can prove the original problem. May 18 at 3:39
• Found it here: baike.baidu.com/item/螺旋式归纳法/11064725 but it’s Chinese so perhaps you could read it with a browser that can translate web pages. May 18 at 4:48
• Actually that works, luckily I can read Chinese, thanks! May 18 at 4:58
Your hypothesis is correct! We can indeed show that $$\prod_{i=1}^n\left(\frac{1}{2}i+\frac{1}{2}a_i\right)\geq n!$$ To do this, we can show the following properties of $$f(x)=x_1x_2\cdots x_n$$:
1. $$f$$ is quasi-concave on $$\mathbb{R}_+^n$$. This follows from log-concavity of $$f$$ which itself is true since $$\ln(f(x)) = \ln(x_1) + \cdots + \ln(x_2)$$ and $$\ln(x)$$ is concave.
2. $$f(1,2,\cdots,n) = n!$$
3. $$f(a_1,a_2,\cdots,a_n)\geq n!$$
Then, since $$f$$ is quasiconcave, the upper-level set $$\{x\in \mathbb{R}_+^n : f(x) \geq z \}$$ is convex on $$\mathbb{R}_+^n$$.
Finally, taking a convex combination of $$(1,2,\cdots,n)$$ and $$(a_1,a_2,\cdots,a_n)$$, the proof is complete.
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# Standard Normal Dist Transformation
mrkb80
## Homework Statement
The question ask to find $E[X]; E[X^2]; Var(X)$ for the standard normal distribution $f(x)=1/\sqrt{2\pi}e^{-x^2/2}$
## Homework Equations
I found
\begin{align} E[X]&=\int_{-\infty}^\infty \! x*1/\sqrt{2\pi}e^{-x^2/2} \, \mathrm{d} x\\ &=1/\sqrt{2\pi} ( \int_{-\infty}^0 \! x*e^{-x^2/2} \, \mathrm{d} x + \int_{0}^{\infty} \! x*e^{-x^2/2} \, \mathrm{d} x )\\ &= 1/\sqrt{2\pi} ( (-1-0) + (0+1) )\\ &= 0 \end{align}
I found $E[X^2]=1$
and $Var[X]=1$. So far no problem.
## The Attempt at a Solution
Now, I am asked to find the pdf, $E[Y], Var[Y]$ of $Y=|X|$
Since the inverse of $Y=|X|$ is $X=|Y|$ and the derivative of the inverse is 1. The transformation seems fairly straight forward.
$f(y)=1/\sqrt{2\pi}e^{-|y|^2/2}$
I am just not sure about the domain of y, isn't it exactly the same as X because x is squared? Once I get that I should be able to use the same integration technique as above to find $E[Y]; Var[Y]$ Wouldn't these values be exactly the same as $E[X]; Var[X]$? Something seems wrong here.
Last edited:
Homework Helper
Gold Member
$f(y)=1/\sqrt{2\pi}e^{-|y|^2/2}$
No, this can't be right. Y = |X|, so Y can never be negative. Therefore f(y) must be zero for negative y. Your f(y) does not have that property.
Just saw your remark about the domain. Right, so as I said, f(y) must be zero for negative y. However, you can't simply write
$$f(y) = \left\{ \begin{array}{lr} 1/\sqrt{2\pi}e^{-|y|^2/2} & y \geq 0 \\ 0 & \text{otherwise} \end{array} \right.$$
because that function will not integrate to 1. But I'm sure you can fix that easily enough.
mrkb80
I guess then the inverse of $Y=|X|$ is not $X=|Y|$. It must be something like
$f(y) =1/\sqrt{2\pi}e^{-y^2/2} \, for \, y\geq0$.Which I think means I'm only looking at half the distribution. The right half to be exact. Correct?
Last edited:
Homework Helper
Gold Member
I guess then the inverse of $Y=|X|$ is not $X=|Y|$. It must be something like
$f(y) =1/\sqrt{2\pi}e^{-y^2/2} \, for \, y\geq0$.Which I think means I'm only looking at half the distribution. The right half to be exact. Correct?
Y = |X| doesn't have a (single-valued) inverse. For every nonzero value of Y, there are two values of X that give Y = |X|. You have to account for both of these. Hopefully you have seen an example of how to do this in your textbook or lecture?
Homework Helper
Gold Member
It must be something like
$f(y) =1/\sqrt{2\pi}e^{-y^2/2} \, for \, y\geq0$.Which I think means I'm only looking at half the distribution. The right half to be exact. Correct?
Yes, this is almost correct. However, because you chopped off the left half, it no longer integrates to 1. You have to re-scale it appropriately.
mrkb80
So I must have to multiply it by 2.
$f(y) =2/\sqrt{2\pi}e^{-y^2/2} \, for \, y\geq0$ ?
I don't really see how an absolute value transformation works for a distribution that isn't the standard normal.
I'm confused.
Homework Helper
Gold Member
So I must have to multiply it by 2.
$f(y) =2/\sqrt{2\pi}e^{-y^2/2} \, for \, y\geq0$ ?
I don't really see how an absolute value transformation works for a distribution that isn't the standard normal.
I'm confused.
Yes, in general it's a bit more complicated. Here it was easier because the standard normal is symmetric around zero. To see how it works in a somewhat more general case, consider what happens if you start with a normal distribution that has nonzero mean. There's an expression for the PDF here:
http://en.wikipedia.org/wiki/Folded_normal_distribution
Note that it simplifies to your result if $\mu = 0$.
If your book doesn't cover how to calculate the density function when the transformation is not one-to-one, you can read about it here (and I would also recommend finding a better book!)
http://www.cl.cam.ac.uk/teaching/2003/Probability/prob11.pdf
mrkb80
I do not like the main book, but I looked in the second book and I just realized how you would have to do it. You would need to break it up into parts that are monotone and add them together. Correct?
mrkb80
So if my integration is correct
\begin{align} E[Y]&=\int_0^\infty \! 2/\sqrt{2\pi}e^{-y^2/2} \, dy \\ &=2/\sqrt{2\pi} (-e^{-y^2/2}|_0^ \infty)\\ &=2/\sqrt{2\pi} \end{align}
and
$Var(Y)=1-2/\pi$
Edit: I saw where you posted the link to the folded normal. These answers are correct. Thanks for your help!
Last edited:
Homework Helper
Dearly Missed
## Homework Statement
The question ask to find $E[X]; E[X^2]; Var(X)$ for the standard normal distribution $f(x)=1/\sqrt{2\pi}e^{-x^2/2}$
## Homework Equations
I found
\begin{align} E[X]&=\int_{-\infty}^\infty \! x*1/\sqrt{2\pi}e^{-x^2/2} \, \mathrm{d} x\\ &=1/\sqrt{2\pi} ( \int_{-\infty}^0 \! x*e^{-x^2/2} \, \mathrm{d} x + \int_{0}^{\infty} \! x*e^{-x^2/2} \, \mathrm{d} x )\\ &= 1/\sqrt{2\pi} ( (-1-0) + (0+1) )\\ &= 0 \end{align}
I found $E[X^2]=1$
and $Var[X]=1$. So far no problem.
## The Attempt at a Solution
Now, I am asked to find the pdf, $E[Y], Var[Y]$ of $Y=|X|$
Since the inverse of $Y=|X|$ is $X=|Y|$ and the derivative of the inverse is 1. The transformation seems fairly straight forward.
$f(y)=1/\sqrt{2\pi}e^{-|y|^2/2}$
I am just not sure about the domain of y, isn't it exactly the same as X because x is squared? Once I get that I should be able to use the same integration technique as above to find $E[Y]; Var[Y]$ Wouldn't these values be exactly the same as $E[X]; Var[X]$? Something seems wrong here.
$$E|X| = \int_{-\infty}^{\infty} |x| f(x) \, dx = 2 \int_0^{\infty} x f(x) \, dx.$$ This is not zero.
$$\text{Var}(|X|) = E(|X|^2) - (E|X|)^2 = E(X^2) -(E|X|)^2 = 1 - (E|X|)^2,$$ because EX2 is just the variance of X.
RGV
mrkb80
The absolute value threw me because the inverse is not monotone. The book does a horrible job of explaining how to handle this case and instead focuses on the fact that it must be strictly increasing or decreasing in the domain for the transformation to work.
I guess another way to think about this problem would be to split the absolute value into two parts.
$\int_0^\infty \! yf(y) dy \, + \int_0^\infty \! yf(y) dy$
since Y=|X| if I put postive values into x, I get postive y's and if I put negative values into x I get the same postive y's (hence the 2 times or the sum of integrals). Since x is defined for $-\infty \le x \le \infty$ I must transform the entire domain, but you can see that when I do this I get only postive y's, so the domain of y must be $y \geq 0$.
Thanks everyone for the help.
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# Basic Factoring Question
• Aug 21st 2008, 08:55 PM
cmf0106
Basic Factoring Question
Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
$8x+8= 8*x+8*1=8(x+1)$
but what if you were to do it like this?
$8x+8=4*2x + 4*2=4(2x+2)$
Very confusing since they both give the correct answer once distributed.
• Aug 21st 2008, 09:07 PM
Jhevon
Quote:
Originally Posted by cmf0106
Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
$8x+8= 8*x+8*1=8(x+1)$
but what if you were to do it like this?
$8x+8=4*2x + 4*2=4(2x+2)$
Very confusing since they both give the correct answer once distributed.
the last is not fully factored, since there is still a common factor of 2 among the terms in the brackets. so yes, it is still correct, but as far as factoring goes, it is incomplete
• Aug 21st 2008, 09:16 PM
cmf0106
Thanks & one last thing for now.
In $12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)$
$x^2$ can be factored out because in the term $5x^3z^2$ $x^2$ is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because $\frac{a^m}{a^n}=a^{m-n}$. Thus $x^2$ can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?
• Aug 21st 2008, 09:43 PM
Jhevon
Quote:
Originally Posted by cmf0106
Thanks & one last thing for now.
In $12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)$
$x^2$ can be factored out because in the term $5x^3z^2$ $x^2$ is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because $\frac{a^m}{a^n}=a^{m-n}$. Thus $x^2$ can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?
basically, you can factor out the lowest power of any term common to all terms. in the case of $x$, it is $x^3$
your logic seems about right, but i would call it "dividing out of" but maybe that makes no sense. you are correct with y and z. but yes, you watch the powers in the same way you stated. and you can double check yourself by seeing that when you are multiplying out again, the powers add up to give you what was in the original
• Aug 22nd 2008, 06:57 AM
cmf0106
Same type of question for this one $28xy^2-14x$
The book has the answer $-7x(-4y^2+2$ but I solved it
14x * $2y^2$ - 14x * 1= $14x(2y^2-1)$. It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy
• Aug 22nd 2008, 09:33 AM
Jhevon
Quote:
Originally Posted by cmf0106
Same type of question for this one $28xy^2-14x$
The book has the answer $-7x(-4y^2+2$ but I solved it
14x * $2y^2$ - 14x * 1= $14x(2y^2-1)$. It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy
your answer is better. good job. the book still left a factor of 2 to be taken out
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# Find N Given N+S(N)=2000
• MHB
• albert391212
#### albert391212
N + S(N) = 2000 N is a 4-digit number,and S(N) is the sum of each digit of N given N+S(N)=2000 please find N
N + S(N) = 2000 N is a 4-digit number,and S(N) is the sum of each digit of N given N+S(N)=2000 please find N
Let $$\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D$$ So $$\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000$$.
Note that A = 1. So
$$\displaystyle N + S(N) = 100B + 10C + (B + C + 2D) = 999$$
Now start working through some cases. For example, B + C + 2D < 10 is impossible because it means B = C = 9, which is a contradiction. So $$\displaystyle B + C + 2D \geq 10$$. Thus when adding we have to carry a 1 into the 10's place, which means that C is at most 8. etc. It will take a while.
-Dan
Let N=abcd , and S(N)=a+b+c+d<28 ,we have a=1, b=9
N+S(N)=1000+900+10c+d+1+9+c+d=1910+11c+2d=2000
11c+2d=90
we get c=8 , d=1
so N=1981 #
Let $$\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D$$ So $$\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000$$.
Note that A = 1. So
$$\displaystyle N + S(N) = 100B + 10C + (B + C + 2D) = 999$$
Now start working through some cases. For example, B + C + 2D < 10 is impossible because it means B = C = 9, which is a contradiction. So $$\displaystyle B + C + 2D \geq 10$$. Thus when adding we have to carry a 1 into the 10's place, which means that C is at most 8. etc. It will take a while.
-Dan
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## The Hidden Treasure of What Is the Unit of Force in Physics
Thus, cos0 is equivalent to 1. Background You experience friction each and every day. If you did move it, you wouldn’t be in a position to stop it.
It can be misunderstood by the majority of of the students. In this manner, theoretical physicists often utilize tools from mathematics. It’s time for a number of physics.
In different situations, it is more difficult to recognize 2nd Law difficulties. If you prefer math, then whatever you will need to understand is already found in the mathematical formula. By way of example, condensed matter physics and nuclear physics benefit from the ability to do experiments.
Be cautious, there isn’t any contact, they are charged solely by induction. https://students.ucsd.edu/academics/advising/academic-success/writing-programs.html This is the reason they are called black holes. Contact force is understood to be the force exerted when two physical objects arrive in direct contact with one another.
Force acting on a body that is in circular motion is known as centripetal force. The issue is that there’s almost always a frictional forcea force that isn’t hard to pretend like it isn’t there. The horizontal part of the tension force is equivalent to the electrostatic force.
As you interleave an increasing number of pages, however, it quickly becomes surprisingly tricky to pull them apartuntil pretty soon, you cannot separate them in any way! You have the choice of downloading either the whole syllabus or each chapter separately. The initial and last information can often tell you all you have to understand.
## Most Noticeable What Is the Unit of Force in Physics
Yes, forces arrive in pairs. So, as long as they are relatively uniform, meaning they do not change in microscopic texture from one region to another, the amount of area in contact does not affect the frictional force. From experiments, sliding friction is discovered to be proportional to the load or weight of the human body in touch with a surface and is practically independent of the subject of contact.
When the situation was simplified, the issue can be solved like any other issue. The very first statement appears reasonable enough, but the second part is a bit murky. Nobody has time for it.
The internet force can subsequently be used to specify the acceleration of the object. For the time being, the emphasis is upon the simple fact a force is a vector quantity with a direction. An equation in which each term has the exact dimensions is believed to be dimensionally accurate.
A couple is always needed to make the rotation. A slice of paper takes much more time to fall than a stone through the exact distance. The heavier box, though, will be far more difficult to slide upon the floor.
On the flip side, kinetic sliding friction is the force that has to be overcome as a way to maintain relative motion between the surfaces. The ball only moves a brief distance, or so the air drag doesn’t have an excessive amount of time to modify the ball’s momentum. Eventually, sooner or later, your pulling force will have the ability to overcome friction and the object will start to move.
## What Is the Unit of Force in Physics – Dead or Alive?
Combine your comprehension of acceleration and the newly acquired knowledge a net force causes an acceleration to figure out whether a net force exists in these situations. The concepts of force and power appear to convey similar meanings and are frequently confused for one another. There are four basic kinds of force in nature.
In each one of the above scenarios, there’s an unbalanced force. An object is tough to push as it is heavy This is among the most frequent misconceptions as it’s something we see and feel everyday. The presence of an unbalanced force for any given situation can be quickly realized by viewing the free-body diagram for this circumstance.
## The Ultimate Strategy to What Is the Unit of Force in Physics
In everyday English, the term acceleration is often used to refer to a state of rising speed. The event of the prefix symbol is extremely important. The difference can be viewed with pistol rounds, too.
Even a kid can move a heavy vehicle. Also, make certain the units agree together. Utilizing different units in various places would make effective scientific communication very challenging.
I will use an automobile mass of 4,000 kg. The eventual aim of this experimentation could be permanent units which are so self-sufficient. It’s a ratio of similar quantities and does not have any unit.
## The Secret to What Is the Unit of Force in Physics
The coefficient of the friction depends upon the 2 surfaces which are in contact. It is reduced because there is less area in contact with a rolling ball. What’s more, simple friction is always proportional to the typical force.
In handling liquids, it’s more usual to use the notion of Surface Tension as opposed to Surface energy, though they refer to precisely the same dimensional quantity. The quantity of static friction stemming from the adhesion of any 2 surfaces has an upper limit. These particles are the building blocks of a category of massive particles called hadrons, including protons and neutrons.
## What is Actually Happening with What Is the Unit of Force in Physics
Combining the new test with a sound comprehension of the needs of the space will guarantee the most suitable tile is used for the correct space. Because of the intricacy of the physical situation, it would be sensible to represent it using a diagram. You will be requested to analyze data and create and explain a variety of experiments.
These variations are the result of latitude, altitude and the community geological structure of the area. Force is a vector, therefore it can be stronger or weaker and it may also point in various directions. Let’s say it takes a force of 5 N to start the book moving.
## The Fundamentals of What Is the Unit of Force in Physics Revealed
Like every issue involving force and acceleration, the issue would start with the building of a free-body diagram. As it is open to some interpretation from the manufacturers, it is not actually dependable. In case the motion isn’t in the direction of force or force is applied to an object but there’s no motion then we cannot speak about work.
## The Queens Hall of Science Cover Up
I don’t think that it’s well worth the effort. Alternately, a paper nomination might be downloaded here. While I say that it’s in the identical position, I also indicate that it faces the exact same direction. It’s a political argument.
## Getting the Best Queens Hall of Science
If you wish to eat there, it’s possible to even bring your own six-pack. Horseplay won’t be tolerated. I guess I’ll only have to go back. It was set up in 1829. Read on, where we’ll provide you a few of the best. You shouldn’t be late or we’ll leave without you.
## Queens Hall of Science Explained
Participants can learn to grow their own mushrooms with a unique container. The labs help the boys to create practical scientific skills along with encouraging collaborative work. He loves to help students discover and achieve their targets, so when you have any questions, don’t hesitate to ask! Some students continue to be concerned about submitting too many test scores. We know what sorts of students colleges wish to admit.
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## Using Queens Hall of Science
The sheer spectacle of the race is probably going to draw no lack of free media also. On the next table is a little pistol. An essential issue is to find out how, in the event the pie is no longer getting larger, the rest of the pie needs to be sliced. I believe this is likely to be a marquee race, he explained. Along with the theatrical edition, a protracted cut adds around three minutes, with two more scenes.
It is thought that the city has the best linguistic diversity in the whole planet, with over 800 unique languages divided between its people. You’re the salt of the planet. I actually think that it’s going to come about as a result of a discovery.
You’re able to receive a taste of everything, she states. You might get an idea what you would like to do, but there are several things which you don’t know anything about, Austin stated. Here is an excellent method to discover your way fast. Was pleasantly surprised with the experience, or perhaps it’s because I didn’t need to pay anything. With the belief that making mistakes is a significant method to learn, Maker Space makes for an enjoyable and educational moment.
## Queens Hall of Science at a Glance
Although Queens does not provide any monumental landmarks for one to visit, it is absolutely worth visiting to observe the condensed mixture of culture. New York isn’t particularly famous for its wildlife, unless you’re speaking about rats and pigeons. He is a wonderful, vibrant place to live and to work. For an entire subway map, see the MTA site.
He has since gone on to compose a set of books and carry out all around the world, including over a dozen times in nyc. There’s always many different special events happening at the New York Hall of Science. We first visited the park a couple of days after the tornado. You’re able to look below to find out what’s happening in our boroughs and the way you also can partake in these absolutely free pursuits and such. In the past couple of decades, there have been a number of new developments, particularly in the Main Street area of Flushing.
## Queens Hall of Science Help!
The magnificent buildings and lovely grounds combine to make Queen’s the ideal alternative for a plethora of activities. We would like to demonstrate how rich Brazilian food is, states Rosa. The spot is little and super popular, so definitely earn a reservation in advance. We enjoy visiting the park. One of my favourite areas to go.
A visit to the museum is precisely what you should work up an appetite in between all of the eating and drinking. This interactive gallery is made for our youngest visitors. Contrary to other shops on this list, a trip to Pasticceria can incorporate a delectable Italian ice, but in addition some amazing breakfast and lunch choices too. This museum is a great option if you are visiting with children. Find more information regarding their tours here.
Replicas is going to be manufactured and placed on the rockets away from the museum. Follow this hyperlink to apply. Click on a pin to see the particulars of somebody museum.
## The Honest to Goodness Truth on Queens Hall of Science
Hall coefficients could be determined experimentally and could vary with temperature. You cannot quit on the surface. At length, higher thrust density will make it possible for satellites to achieve complex fuel-optimized orbits in a fair moment. They are then probed to think of what they expect the reply might be and to think about the assortment of potential solutions. Equally important is how a high specific impulse can generate a large enough increase in a satellite’s momentum to allow the spacecraft to modify orbitsa feature unavailable on currently orbiting CubeSats.
A big priority is building tools which make it possible that people take part in a few of these learning experiences digitally. As exemplified with the projects described below, we’re interested in addressing a vast array of fundamental and practical issues. Since each of these buildings cater to thousands of people every year, they heavily lead to the erosion of the land. Certainly, awarded stories are predicted to have a particular amount of quality, but ultimately, nominations rarely offer full coverage of a broad field. He patented solutions in the area of speed meters and ventilators, and worked on the building of various kinds of fountains.
## If You Read Nothing Else Today, Read This Report on Queens Hall of Science
Summer has arrived in nyc! Observing the tour, an official reception was held in appreciation of many contributors to the undertaking. The Power Hall is going to be a location where everyone can delight in sharing a visit, find in-depth stories and have a fantastic experience. The International Spy Museum is very happy to announce the building of a new permanent house in Southwest Washington, DC. Purchase tickets beforehand.
## TAUSIYAH DARI HABIB HADI BIN ‘ALWI AL-KAFF
23 Julai 2019 | 20 Zulkaedah 1440H
Alhamdulillah, pada 22 Julai 2019 (semalam), pihak Darul Hasanah telah menerima tausiyah dari Habib Hadi Bin ‘Alwi Al-Kaff dari Malang, Indonesia. Tausiyah disampaikan di Surau Darul Hasanah
Semoga dengan kehadiran Habib Hadi Bin ‘Alwi Al-Kaff dan tetamu membawa keberkatan dan rahmat buat pihak Darul Hasanah juga anak-anak tahfiz Darul Hasanah. Allahumma aamiinπ€²πΌπ€²πΌπ€²πΌπ€²πΌπ
Daripada Abu Hurairah RA katanya Rasulullah SAW bersabda, βSesiapa yang beriman dengan Allah dan hari akhirat maka hendaklah ia memuliakan tetamu yang datang menziarahinya serta mengeratkan silaturrahim dengan semangat persaudaraan Islam. Sesiapa yang beriman dengan Allah dan hari akhirat, hendaklah bercakap dengan perkataan yang baik dan manis didengar, atau pun berdiam sahaja dari bercakap perkara yang sia-sia.β
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# 209 centiliters in US gallons
## Conversion
209 centiliters is equivalent to 0.55211958942853 US gallons.[1]
## Conversion formula How to convert 209 centiliters to US gallons?
We know (by definition) that: $1\mathrm{centiliter}\approx 0.00264172052358148\mathrm{usgallon}$
We can set up a proportion to solve for the number of US gallons.
$1 centiliter 209 centiliter ≈ 0.00264172052358148 usgallon x usgallon$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{usgallon}\approx \frac{209\mathrm{centiliter}}{1\mathrm{centiliter}}*0.00264172052358148\mathrm{usgallon}\to x\mathrm{usgallon}\approx 0.5521195894285293\mathrm{usgallon}$
Conclusion: $209 centiliter ≈ 0.5521195894285293 usgallon$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 US gallon is equal to 1.81120181052632 times 209 centiliters.
It can also be expressed as: 209 centiliters is equal to $\frac{1}{\mathrm{1.81120181052632}}$ US gallons.
## Approximation
An approximate numerical result would be: two hundred and nine centiliters is about zero point five five US gallons, or alternatively, a US gallon is about one point eight one times two hundred and nine centiliters.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
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# Iterative solution to a nonlinear equation
I appologize in advance if this question is silly. I need to compute the root of
$$u -f(u) =0$$
Where $u$ is a real vector and $f(u)$ is a real-vector valued function. I started with Newton's method (which worked), but then realized a much simpler method would be an iterative solution
$$u_{i+1} = f(u_{i})$$
This is much quicker and apparently as accurate/stable as Newton's method.
Now the questions:
• Is this the correct approach or should I use a different method?
• Is there anything that can be said about it's convergence rate, stability, acc, etc?
• Is it globally convergent?
Thank you all in advance for the attention.
• This is known as fixed point iteration. I am not well-versed on the subject but at the very least it should give you some new words to throw at google. If I remember correctly, fixed points appear for a wide variety of function with a wide variety of starting points. – Godric Seer Aug 17 '12 at 0:38
• What's your $f(u)$? Newton's method is often faster than fixed point iteration. – Bill Barth Aug 17 '12 at 2:44
• Your observed convergence depends heavily on the functional form of $f(\cdot)$. Furthermore, if $f(u) = u - (\nabla g)^{-1} g(u)$, then the iteration $u_{i+1} = f(u_i)$ is Newton iteration on $g(u) = 0$. – Jed Brown Aug 17 '12 at 3:06
If $q:=|f'(x^*)|<1$, where $x^*$ is the solution, the fixed point iteration you talk about is locally linearly convergent with convergence rate $q$. Thus if $q$ is small or zero, the method is competitive with Newton's method.
Far away from the solution, convergence is difficult to predict in the absence of global information (such as a Lipschitz constant $<1$, which produces a contraction).
The Feigenbaum fractal is a good example of how strange fixpoint iteration can be:
http://en.wikipedia.org/wiki/Feigenbaum_fractal
http://en.wikipedia.org/wiki/File:Logistic_Bifurcation_map_High_Resolution.png
The second link plots the behavior of fixpoint iteration applied to the logistic map as one of the parameters varies. For certain values it converges, though only linearly. For other values it converges to a cycle of varying length. For yet another class of values, it behaves completely chaotically.
In other words, the behavior of fixpoint iteration depends entirely on the function in question. Even functions that look similar may exhibit radially different behavior.
Note: As Jed points out, Newton iteration can be equally weird.
• To be fair, many popular fractals are the Julia sets of Newton iteration on a simple equation. – Jed Brown Aug 17 '12 at 3:03
The Banach fixed-point theorem describes the standard situation when a fixed-point iteration is globally convergent. Especially the uniqueness part of the theorem indicates that you can only expect local convergence if the solution is not unique.
Most situations of local convergence can be explained by this theorem, at least in theory. This is even true for the convergence occurring in some of the fractals mentioned above. It's just that the theorem has to be applied to $f^n=f\circ \ldots\circ f$ instead of $f$ at some basins of attraction.
You may consider useful this reference: A Homotopy for Solving Large, Sparse and Structured Fixed Point Problems. R. Saigal. Mathematics of Operations Research, Vol. 8, No. 4 (Nov., 1983), pp. 557-578.
This method is correct and it is called "Successive Substitution". Please, look at page 189 of this reference for details.
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# Magnetic field direction
## Homework Statement:
Figure 29-48 shows two very long straight wires (in cross section) that each carry a current of
4.00 A directly out of the page. Distance d1 = 6.00 m and distance d2 = 4.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?
## Homework Equations:
B = (μ*i)/(2*π*r)
Again struck up with the direction of the magnetic field, i suppose now the field not simple along the x axis. How to find the angle and the direction of the field. My attempt is
B1 = (μ*i)/(2*π*r) = (4*π*4)/(2*π*4) = 200nT where r = d2 = 4; is the field due to i1.
B2 = (μ*i)/(2*π*r) = (4*π*4)/(2*π*4) = 200nT where r = d2 = 4; is the field due to i2.
B1 and B2 are making an angle of 90° between them (not convinced either). The net field is √(B1^2 + B2^2 = 282 nT.
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
There are several issues here. First, r is not 4m. You need to do some trigonometry to find r for each wire. Second, you say the fields are perpendicular to each other. There is no reason to think that. In your drawing you show the fields are perpendicular to the two r’s. That is correct. Third, your drawing is really messed up. You show two 90 degree angles, but the lines aren’t anything like 90 degrees apart. It really matters if you are going to be able to do this correctly. Fourth, regarding direction, have you been taught the right hand rule? You know the direction of current. You should be able to determine the direction of the field. I can say that one of your directions is wrong. Finally, the fields add vectorial. Have you been taught vector addition?
Ok i will redraw clearly and rework on the points you mentioned.
Yes i did work on this found new tool to draw, the new diagram is
The other corrections the distance from I1 to P and I2 to P is 5m.
The B1 = B2 = (4*π *(10^-7)*4)/(2*π *5) = 160nT
The net field B = √(B1^2 + B2^2 = 226nT. Can you please advice if it is ok.
Yes, much better. The distances are indeed 5m, and you now have the right hand rule directions correct. The resultant is, as you show, pointed up. Your calculation of the field strengths is also correct. You apparently have noticed that the horizontal components cancel by symmetry and you only need to calculate the vertical component (although you may want to state that)
However, your combining of the fields is incorrect. Here again I think your drawing has let you down. The Pythagorean theorem would be correct if the fields met at a right angle, but do they? What are the angles of a 3,4,5 triangle?
ZapperZ
Staff Emeritus
Yes i did work on this found new tool to draw, the new diagram is
View attachment 255638
The other corrections the distance from I1 to P and I2 to P is 5m.
The B1 = B2 = (4*π *(10^-7)*4)/(2*π *5) = 160nT
The net field B = √(B1^2 + B2^2 = 226nT. Can you please advice if it is ok.
I think you have an issue with geometry. This is not a criticism, but rather I'm trying to point out the possible source of your difficulties.
The magnetic field vector must be tangential to the magnetic field lines for each current source. This may have an understanding of this, but the geometrical sketch is wrong. Figure out what is different between your sketch, and this one:
Zz.
Yes finally a very tough one, the angle between B1 and B2 fields is 73.72°, B1 with X axis is 53.14° and B2 with X axis is 53.14°. The final result is
Along X-axis B1*cos(53.14) - B2*cos(53.14) = 0;
Along Y-axis B1*sin(53.14) + B2 *sin(53.14) = 256 in the +ve Y direction.
Figure out what is different between your sketch, and this one:
Zz.
Yes I made a mistake in drawing the previous post as the angle calculations were wrong. Now the field B1 will not align with the line connecting the I2 with the P point and similarly for B2 field.
Yes finally a very tough one, the angle between B1 and B2 fields is 73.72°, B1 with X axis is 53.14° and B2 with X axis is 53.14°. The final result is
Along X-axis B1*cos(53.14) - B2*cos(53.14) = 0;
Along Y-axis B1*sin(53.14) + B2 *sin(53.14) = 256 in the +ve Y direction.
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# Airflow through supply branches
Loss occurs along two paths when air flows through a brach-piece: there is a loss of energy for flow through the main and also for flow through the branch. The exact magnitude of the loss depends on the way in which the branch-piece is constructed but it is, nevertheless, possible to generalise. ASHRAE (1997a) gives extensive information on loss coefficients. The CIBSE (1986a) quotes multiplying factors, referred to the velocity pressure in the downstream main or the branch, as appropriate. Both of these are dependent on the ratio of the downstream to upstream velocities. Figure 15.16 illustrates this: there is a loss of total pressure between points 1 and 2, when airflow through the main is considered, and also a loss of total pressure (not necessarily the same) when air flows from point 1 to point
3, by way of the branch.
EXAMPLE 15.6
Q> O, 2 —— A———— ►I V2
Fig. 15.16 Airflow through a supply branch piece.
Given that, in Figure 15.16, the velocities Vu V2 and V3 are 10, 8 and 6 m s-1, respectively, determine the total pressure loss and the static pressure change through the main (1-2) and the branch (1-3). Assume the section of the branch duct is square.
The CIBSE (1986a) quotes the following loss coefficients:
For the main,
V2/V{ =0.8, Ci-2 = 0.18
For the branch,
Vj/Vi = 0.6, ^i_3 = 3.5 x the factor for the equivalent bend = 3.5 x 0.23. See CIBSE (1986a)
= 0.805
Hence we have
For the main:
Apt = 0.18 x 0.6 x82 = 6.9 Pa
Aps = 0.6 x 102 — 0.6 x 82 — 6.9 = 14.7 Pa
That is, a static regain has occurred in the main because the fall in velocity pressure exceeded the loss so there is a net increase in the potential energy or the static pressure.
For the branch:
Apt = 0.805 x 0.6 x 62 = 17.4 Pa
Aps = 0.6 x 102 — 0.6 x 62 — 17.4 = 21.0 Pa
And a static regain has taken place here also.
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How to draw a Venn Diagram for problems
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Ms. Big Fat Panda
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26 Jul 2010, 09:01
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So. I realized I love Venn Diagrams (Yes, I'm a nerd, but who cares?). I've explained a lot of problems using Venn Diagrams and a forum user PMed me to explain how I drew my Venn Diagrams since he was getting confused with them. I am posting this as an informal and rough guide to how I visualize problems that make use of Venn Diagrams.
At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take exactly 2 classes?
A. 37
B. 43
C. 45
D. 60
Step 1: Deconstruction
This is where you extract the given information from a problem. Use a symbol for each section given to you. I usually use the first letter in caps. Please remember at this point that any number specified without explicitly mentioning that it's "exactly" or "only" for a certain subject is not to be taken so.
For instance, in this case, it says 53 math students. This doesn't mean students who are taking only 1 subject (Math). This could include students who are taking Math and Chemistry or Math and English or all three too. So now break down the numbers given to us.
Total = 150
M (Total) = 53
C (Total) = 88 [Why the hell are people studying more Chemistry than Math?]
E (Total) = 58
MCE = 6
We are asked to find MC + CE + EM.
Step 2: Drawing the diagram
ALWAYS start from the center of the Venn Diagram wherever information is available. This will make life infinitely easier.
In this case, the center is the intersection of all three circles, i.e MCE = 6
So fill that in to the diagram you've drawn.
Attachment:
VD1.jpg [ 21.73 KiB | Viewed 47609 times ]
Now that you've gotten that, let's start filling in a variable for each section not known to us. Here, consider each letter to represent only that specific section and not the entire circle or a larger portion.
Attachment:
File comment: Updated
VD1.jpg [ 27.41 KiB | Viewed 47585 times ]
So now that you have the parts filled in, what you need to do is write down what you have in the diagram in terms of numbers. So we are given the totals for each subject.
Look at math first. There are four types of people taking math (each group of these people mutually exclusive, and not in common with any other group)
1. Only math: a
2. Math and Chemistry: y
3. Math and English: x
4. All three: 6
So now represent this as a sum and you get
$$a+x+y+6 = 53$$ and hence $$a+x+y = 47$$
Similarly for the other subjects you get:
$$b+x+z+6 = 58$$ and hence $$b+x+z = 52$$
$$c+y+z+6$$ = 88 and hence $$c+y+z = 82$$
And then you have the total:
$$x+y+z+a+b+c+6=150$$ and hence $$x+y+z+a+b+c = 144$$
Step 3: Solution
This is perhaps the most intuitive part, but in my experience the first part of this step is the same in all overlapping sets problem. It's only what's asked for that's different.
Add all the individual equations together to get a combined equation with all the variables and a number.
So you get:
$$2(x+y+z) + a+b+c = 47+52+82 = 171$$
Rearranging this to get $$a+b+c$$we get $$a+b+c = 171 - 2(x+y+z)$$
Substitute this into the total equation we derived earlier saying $$x+y+z+a+b+c = 144$$ so you get:
$$x+y+z+171 - 2(x+y+z)= 144$$
Upon rearranging this you get:
$$x+y+z = 171-144 = 37$$which is option A the right answer.
Hope this helps some of you! Please post a reply if there's anything else you want to know about my explanation or anything else.
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Re: How to draw a Venn Diagram for problems [#permalink]
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26 Jul 2010, 09:16
Just what i wanted equations with respective areas in the 3 venn diagram
thanks a lot !
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Re: How to draw a Venn Diagram for problems [#permalink]
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29 Jul 2010, 04:22
Great work whiplash!! Kudos!
I wonder why not many have responded to this effort. Any how, thanks again!!
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29 Jul 2010, 04:35
great job whiplash2411!
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29 Jul 2010, 06:21
Thank you. Added to the list of resources in the Math Book Project
gmat-math-book-87417.html
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29 Jul 2010, 09:00
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Thank you
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Re: How to draw a Venn Diagram for problems [#permalink]
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29 Jul 2010, 13:53
Nice whip !!
Probably this can be added as well
formulae-for-3-overlapping-sets-69014.html#p729340
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Re: How to draw a Venn Diagram for problems [#permalink]
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13 Aug 2010, 16:10
This is a great example!
Thanks for posting.
Small note which got me confused for a bit. I believe the answer is 27 not 37 (unless I am missing something).
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Re: How to draw a Venn Diagram for problems [#permalink]
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13 Aug 2010, 21:28
1
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whiplash2411 wrote:
Upon rearranging this you get:
$$x+y+z = 171-144 = 37$$which is option A the right answer.
Hope this helps some of you! Please post a reply if there's anything else you want to know about my explanation or anything else.
First off, great post, this really does help! My first instinct on how to handle this was definitely off.
Secondly, I believe the answer would actually be 27. 171-144=27... not 37. Not sure why I'm the only one that caught that, so my apologies if I'm doing something wrong.
BB, can you correct this in the Math Book Project?
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19 Aug 2010, 09:40
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The answer is A, the mistake was in 47+52+82=181
181-144=37
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19 Aug 2010, 11:21
Nice Explanation..
Now I got a better understanding regarding Venn Diagrams.
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19 Aug 2010, 12:22
Thanks for that, whiplash. After a mechanical engineering degree, I can safely say that systems of equations with
# of equations >= # of variables
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19 Aug 2010, 20:38
This formula, I assume could be substituted by this Venn Diagram method?
Overlapping Groups-
Total = Group 1 + Group 2 + Neither - Both
This is how I've been tackling these problems before but this seems to be an easier method?
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20 Aug 2010, 15:34
Kudos!
Big thanks for this excellent problem solver, it comes as a big help for me.
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30 Aug 2010, 14:24
thank you
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05 Oct 2010, 06:09
Kudos! Just what I wanted.
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09 Oct 2010, 06:49
Kudos! Great post whiplash2411, gonna help a lot during preparation. Added to the Quantitative strategies collection .
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Kudos??
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02 Jul 2013, 04:33
hello everyone,
I have recently joined the forum to pursue my dream to crack GMAT.
I am just beginners..just brushing my basics..
Kudos!!!!!!
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Expert's post
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Bumping for review*.
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Re: How to draw a Venn Diagram for problems [#permalink] 11 Jul 2013, 00:11
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You are on page 1of 10
# 1
## Partial Fraction Expansion via MATLAB
The residue function of MATLAB can be used to compute the partial fraction expansion (PFE) of a ratio of two polynomials. This can be used or Laplace transforms or Z transforms, although we will illustrate it with Z transforms here. The residue command gives three pieces of information: the residues are given in output vector r, the poles are given in output vector p, the so-called direct terms are given in output vector k. When the order of the numerator polynomial is less than the order of the denominator polynomial there will be no direct terms. When the order of the numerator equals the order of the denominator there will be one direct term;. When the order of the numerator is one greater than the order of the denominator there will be two direct terms. When the order of the numerator is two greater than the order of the denominator there will be three direct terms; etc. That is, when the order of the numerator is p greater than the order of the denominator (with p0) there will be p+1 direct terms. The residue command requires two input vectors: one holding the coefficients of the numerator and one holding the coefficients of the numerator. The right-most element in these vectors corresponds to the z0 coefficient, the next element to the left is the z1 coefficient, etc., until you reach the highest power; if a power is not present it has a zero coefficient.
It is easiest to explain how to use residue by giving examples. Note: We will be expanding H(z) divided by z, like is done when trying to find the inverse Z transform the division by z is a trick used to ensure that after PFE we get a form for which it is usually easy to look up the inverse ZT on a table. Note also that this division does not change the denominator of what we are expanding this is important because the denominator sets the major characteristic of the corresponding time signal (e.g., the denominator of a systems transfer function determines the systems characteristic modes). Thus, every example will be done using G(z) where G(z) = H(z)/z.
## Example #1: No Direct Terms, No Repeated Roots, No Complex Roots
G( z) = H ( z) / z = 3z 1 z 3z + 2
2
In this case there will be no direct terms because the numerator order is lower than the denominator order. The numerator vector is [3 1 ] The denominator vector is [1 3 2]. The command and its result is: [r,p,k]=residue([3 -1],[1 -3 2]) r= 5 -2 p= 2 1 k= [] Note that k is empty (denoted by []) showing that there are no direct terms.
The numbers in r are the residues they are the numbers that go in the numerator of each term. The numbers in p are the poles they are the numbers subtracted from z to form the denominators of the terms. We have a pole at z = 2 with a residue of 5 and. a pole at z = 1 with a residue of 2; thus:
G( z) = 3z 1 z 2 3z + 2 = 5 2 + z 2 z 1
and then
H ( z ) = zG ( z ) =
z (3z 1) z 2 3z + 2
5z 2 z + z 2 z 1
## Example #2: One Direct Term, No Repeated Roots, No Complex Roots
G( z) = H ( z) / z = 2 z 2 + 3z 1 z 2 3z + 2
[r,p,k]=residue([2 3 -1],[1 -3 2]) r= 13 -4 p= 2 1 k= 2 Thus, we have: a pole at: 2 w/ residue 13 and a pole at: 1 w/ residue 4; . But in this case we also have a single direct term of 2, which just adds a term of 2 to the expansion:
G( z) = 2 z 2 + 3z 1 z 2 3z + 2 = 13 4 +2 + z 2 z 1
Thus
H ( z ) = zG ( z ) =
z ( 2 z 2 + 3z 1) z 2 3z + 2
13z 4z + + 2z z 2 z 1
## Example #3: Two Direct Terms, No Repeated Roots, No Complex Roots
G( z) = H ( z ) / z =
7 z 3 + 2 z 2 + 3z 1 z 2 3z + 2
## [r,p,k]=residue([7 2 3 -1],[1 -3 2]) r= 69 -11 p= 2 1 k= 7 23
This gives two terms in the vector k so we know we have two direct terms: 7z and 23. Note: if there were M elements in vector k the left most term would be the coefficient of a zM order direct term.) Thus, we have
G( z) =
Thus
7 z 3 + 2 z 2 + 3z 1 z 2 3z + 2
69 11 + + 7 z + 23 z 2 z 1
H ( z) =
69 z 11z + + 7 z 2 + 23z z 2 z 1
## Example #4: No Direct Terms, A Double Root, No Complex Roots
G( z) = H ( z) / z = 2 z 2 + 3z 1 z 5z + 8 z 4
3 2
2 z 2 + 3z 1 ( z 1)( z 2) 2
[r,p,k]=residue([2 3 -1],[1 -5 8 -4]) r= -2.0000 13.0000 4.0000 p= 2.0000 2.0000 1.0000 k= [] Note that p has two elements listed as 2.0000: that means that there is a double root. Note also that there are two residues for these poles namely, -2 and 13: First residue is for the first-order term for the double root Second residue is for the second-order term for the double root.
## Thus, the expansion is
G( z) = 2 z 2 + 3z 1 z 3 5z 2 + 8 z 4 = 2 z 2 + 3z 1 ( z 1)( z 2) 2 = 2 13 4 + + z 2 ( z 2) 2 z 1
Note, in general: if there were an Mth order pole we would have a term for each order up to and including M for that pole. Such terms (after multiplying by z to undo the division by z done before PFE when computing the inverse Z transform) are inverse Z transformed using #8 - #10 in Table 11.1 of the book.
## Example #5: No Direct Terms, No Repeated Roots, Two Complex Roots
G( z) = H ( z ) / z = 2 z 2 + 3z 1 z 3z + 4 z 2
3 2
2 z 2 + 3z 1 ( z 1)( z 2 2 z + 2)
2 z 2 + 3z 1 = ( z 1)( z [1 + j ])( z [1 j ])
This shows how the given denominator polynomial is factored into three poles, two of them complex. Note that the two complex poles are conjugates of each other this is a necessary condition for the coefficients of the polynomial to be real valued. Thus, for the cases we are interested in, we will always have complex poles occur in conjugate pairs. The use of residue is no different here, but what we do with the output is a little different: combine the conjugate pair terms into a single second order term which would then be inverse Z transformed using one of the #11 - #12 lines in Table 11.1 of the book (after multiplying by z to undo the division by z done before PFE when computing the inverse Z transform). Of course, #12b in the table allows us to also go directly from the two conjugate terms well do it this way here.
10
[r,p,k]=residue([2 3 -1],[1 -3 4 -2]) r= -1.0000 - 3.5000i -1.0000 + 3.5000i 4.0000 p= 1.0000 + 1.0000i 1.0000 - 1.0000i 1.0000 k= []
G( z) =
2 z 2 + 3z 1 z 3 3z 2 + 4 z 2
4 1 j 3.5 1 + j 3.5 + + z (1 + j ) z (1 j ) z 1
Now, to illustrate the inverse Z transform of such conjugate pair terms, we multiply each term in the above to by z to get H(z). Thus, the two conjugate terms become:
z (1 j 3.5) z (1 + j 3.5) + z (1 + j ) z (1 j )
## Converting these into polar form gives:
3.6401e-j1.8491z z ( 2e
j / 4
3.6401e + j1.8491z z ( 2 e j / 4 )
| |= 2
## Thus, r = 2*3.6401 = 7.2802, = -1.8491, two terms inverse transform to
7.2802 ( 2 ) k cos(( / 4) k 1.8491)u[ k ]
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##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/600.html
# 600. Non-negative Integers without Consecutive Ones (Hard)
Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.
Example 1:
Input: 5
Output: 5
Explanation:
Here are the non-negative integers <= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
Note: 1 <= n <= 109
Companies:
Pocket Gems
Related Topics:
Dynamic Programming
## Solution 1.
Let f(n) be the numbers less than or equal to n whose binary representations DO contain consecutive ones.
## x | binary(x) | has consecutive ones
##### Welcome to Subscribe On Youtube
|—|— 0 | 0 | 1 | 1 | 2 | 10 | 3 | 11 | 1 4 | 100 | 5 | 101 | 6 | 110 | 1 7 | 111 | 1 8 | 1000 | 9 | 1001 | 10 | 1010 | 11 | 1011 | 1 12 | 1100 | 1 13 | 1101 | 1 14 | 1110 | 1 15 | 1111 | 1 16 | 10000 |
We can find pattern in this way:
1. We separate the table rows according to count of bits. So 0~1 is the first group, 2~3 is the second group, 4~7 is the third group, 8~15 is the forth group…
2. Starting from the second group, the numbers in the second half of the group all have consecutive ones, simply because they have leading 11.
3. For the numbers in the first half of the groups, the pattern is the same after removing the leading 10.
Take 14 as example which is in group 8~15, it’s in the second half, so 12~14 all have consecutive ones. And 8~11 has the same pattern as 0~3.
So we can get this equation:
let k = floor(log2(n))
let p = 2^k
f(n) = f(p-1) +
{ f(n-p) if n-(p-1)<= p/2
{ n-(p-1)-p/2 + f(p/2-1) if n-(p-1)>p/2
// OJ: https://leetcode.com/problems/non-negative-integers-without-consecutive-ones/
// Time: O(logN)
// Space: O(logN)
class Solution {
private:
unordered_map<int, int> m;
int find(int num) {
if (num < 2) return 0;
if (m.find(num) != m.end()) return m[num];
int k = log2(num), p = pow(2, k);
int ans = find(p - 1);
if (num > p / 2 * 3 - 1) ans += num - p / 2 * 3 + 1 + find(p / 2 - 1);
else ans += find(num - p);
return m[num] = ans;
}
public:
int findIntegers(int num) {
return num + 1 - find(num);
}
};
• class Solution {
public int findIntegers(int num) {
if (num <= 2)
return num + 1;
int bits = (int) (Math.log(num) / Math.log(2));
int counts = 2;
int zeros = 0, ones = 1;
for (int i = 2; i <= bits; i++) {
int newZeros = zeros + ones;
ones = zeros;
zeros = newZeros;
counts += zeros;
counts += ones;
}
int zerosUnrestricted = 0, zerosRestricted = 0, onesUnrestricted = 0, onesRestricted = 1;
for (int i = 1; i <= bits; i++) {
int curBit = (num >> bits - i) & 1;
if (curBit == 0) {
int curZerosUnrestricted = zerosUnrestricted + onesUnrestricted;
int curZerosRestricted = zerosRestricted + onesRestricted;
int curOnesUnrestricted = zerosUnrestricted;
int curOnesRestricted = 0;
zerosUnrestricted = curZerosUnrestricted;
zerosRestricted = curZerosRestricted;
onesUnrestricted = curOnesUnrestricted;
onesRestricted = curOnesRestricted;
} else {
int curZerosUnrestricted = zerosUnrestricted + zerosRestricted + onesUnrestricted + onesRestricted;
int curZerosRestricted = 0;
int curOnesUnrestricted = zerosUnrestricted;
int curOnesRestricted = zerosRestricted;
zerosUnrestricted = curZerosUnrestricted;
zerosRestricted = curZerosRestricted;
onesUnrestricted = curOnesUnrestricted;
onesRestricted = curOnesRestricted;
}
}
counts += zerosUnrestricted + zerosRestricted + onesUnrestricted + onesRestricted;
return counts;
}
}
############
class Solution {
private int[] a = new int[33];
private int[][] dp = new int[33][2];
public int findIntegers(int n) {
int len = 0;
while (n > 0) {
a[++len] = n & 1;
n >>= 1;
}
for (var e : dp) {
Arrays.fill(e, -1);
}
return dfs(len, 0, true);
}
private int dfs(int pos, int pre, boolean limit) {
if (pos <= 0) {
return 1;
}
if (!limit && dp[pos][pre] != -1) {
return dp[pos][pre];
}
int up = limit ? a[pos] : 1;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (!(pre == 1 && i == 1)) {
ans += dfs(pos - 1, i, limit && i == up);
}
}
if (!limit) {
dp[pos][pre] = ans;
}
return ans;
}
}
• // OJ: https://leetcode.com/problems/non-negative-integers-without-consecutive-ones/
// Time: O(logN)
// Space: O(logN)
class Solution {
private:
unordered_map<int, int> m;
int find(int num) {
if (num < 2) return 0;
if (m.find(num) != m.end()) return m[num];
int k = log2(num), p = pow(2, k);
int ans = find(p - 1);
if (num > p / 2 * 3 - 1) ans += num - p / 2 * 3 + 1 + find(p / 2 - 1);
else ans += find(num - p);
return m[num] = ans;
}
public:
int findIntegers(int num) {
return num + 1 - find(num);
}
};
• class Solution:
def findIntegers(self, n: int) -> int:
@cache
def dfs(pos, pre, limit):
if pos <= 0:
return 1
up = a[pos] if limit else 1
ans = 0
for i in range(up + 1):
if pre == 1 and i == 1:
continue
ans += dfs(pos - 1, i, limit and i == up)
return ans
a = [0] * 33
l = 0
while n:
l += 1
a[l] = n & 1
n >>= 1
return dfs(l, 0, True)
############
class Solution(object):
def findIntegers(self, num):
"""
:type num: int
:rtype: int
"""
n = bin(num)[2:][::-1] # removes "0b"
length = len(n)
A = [1 for _ in range(length)] # ends with 0
B = [1 for _ in range(length)] # ends with 1
for i in range(1, len(n)):
A[i] = A[i - 1] + B[i - 1]
B[i] = A[i - 1]
ans = A[-1] + B[-1]
for i in range(length - 2, -1, -1):
if n[i:i + 2] == "11":
break
if n[i:i + 2] == "00":
ans -= B[i]
return ans
• func findIntegers(n int) int {
a := make([]int, 33)
dp := make([][2]int, 33)
for i := range dp {
dp[i] = [2]int{-1, -1}
}
l := 0
for n > 0 {
l++
a[l] = n & 1
n >>= 1
}
var dfs func(int, int, bool) int
dfs = func(pos, pre int, limit bool) int {
if pos <= 0 {
return 1
}
if !limit && dp[pos][pre] != -1 {
return dp[pos][pre]
}
up := 1
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if !(pre == 1 && i == 1) {
ans += dfs(pos-1, i, limit && i == up)
}
}
if !limit {
dp[pos][pre] = ans
}
return ans
}
return dfs(l, 0, true)
}
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.17: Using the Graphing Calculator to Graph Quadratic Equations
Difficulty Level: At Grade Created by: CK-12
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Practice Graphing Calculator to Graph Quadratic Functions
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An arrow is shot straight up into the air from 5 feet above the ground with a velocity of 18 ft/s. The quadratic expression that represents this situation is 5+18t16t2\begin{align*}5 + 18t - 16t^2\end{align*}, where t is the time in seconds. At what time does the arrow reach its maximum height and what is that height?
### Guidance
A graphing calculator can be a very helpful tool when graphing parabolas. This concept outlines how to use the TI-83/84 to graph and find certain points on a parabola.
#### Example A
Graph y=3x2+14x8\begin{align*}y=-3x^2+14x-8\end{align*} using a graphing calculator.
Solution: Using a TI-83/84, press the Y=\begin{align*}Y=\end{align*} button. Enter in the equation. Be careful not to confuse the negative sign and the subtraction sign. The equation should look like y=3x2+14x8\begin{align*}y=-3x^2+14x-8\end{align*} or y=3x2+14x8\begin{align*}y=-3x^2+14x-8\end{align*}. Press GRAPH.
If your graph does not look like this one, there may be an issue with your window. Press ZOOM and then 6:ZStandard, ENTER. This should give you the standard window.
#### Example B
Using your graphing calculator, find the vertex of the parabola from Example A.
Solution: To find the vertex, press 2nd\begin{align*}2^{nd}\end{align*} TRACE (CALC). The Calculate menu will appear. In this case, the vertex is a maximum, so select 4:maximum, ENTER. The screen will return to your graph. Now, you need to tell the calculator the Left Bound. Using the arrows, arrow over to the left side of the vertex, press ENTER. Repeat this for the Right Bound. The calculator then takes a guess, press ENTER again. It should give you that the maximum is X=2.3333333\begin{align*}X=2.3333333\end{align*} and Y=8.3333333\begin{align*}Y=8.3333333\end{align*}. As fractions, the coordinates of the vertex are (213,813)\begin{align*}\left(2\frac{1}{3}, 8\frac{1}{3}\right)\end{align*}. Make sure to write the coordinates of the vertex as a point.
#### Example C
Using your graphing calculator, find the x\begin{align*}x-\end{align*}intercepts of the parabola from Example A.
Solution: To find the x\begin{align*}x-\end{align*}intercepts, press 2nd\begin{align*}2^{nd}\end{align*} TRACE (CALC). The Calculate menu will appear. Select 2:Zero, ENTER. The screen will return to your graph. Let’s focus on the left-most intercept. Now, you need to tell the calculator the Left Bound. Using the arrows, arrow over to the left side of the vertex, press ENTER. Repeat this for the Right Bound (keep the bounds close to the intercept). The calculator then takes a guess, press ENTER again. This intercept is X=.666667\begin{align*}X=.666667\end{align*}, or (23,0)\begin{align*}\left(\frac{2}{3}, 0\right)\end{align*}. Repeat this process for the second intercept. You should get (4, 0).
NOTE: When graphing parabolas and the vertex does not show up on the screen, you will need to zoom out. The calculator will not find the value(s) of any x\begin{align*}x-\end{align*}intercepts or the vertex that do not appear on screen. To zoom out, press ZOOM, 3:Zoom Out, ENTER, ENTER.
Intro Problem Revisit Use your calculator to find the vertex of the parabolic expression 5+18t16t2\begin{align*}5 + 18t - 16t^2\end{align*}.
The vertex is (0.5625, 10.0625). Therefore, the maximum height is reached at 0.5625 seconds and that maximum height is 10.0625 feet.
### Guided Practice
1. Graph y=6x2+11x35\begin{align*}y=6x^2+11x-35\end{align*} using a graphing calculator. Find the vertex and x\begin{align*}x-\end{align*}intercepts. Round your answers to the nearest hundredth.
1. Using the steps above, the vertex is (-0.917, -40.04) and is a\begin{align*}a\end{align*} minimum. The x\begin{align*}x-\end{align*}intercepts are (1.67, 0) and (-3.5, 0).
### Practice
Graph the quadratic equations using a graphing calculator. Find the vertex and x\begin{align*}x-\end{align*}intercepts, if there are any. If there are no x\begin{align*}x-\end{align*}intercepts, use algebra to find the imaginary solutions. Round all real answers to the nearest hundredth.
1. y=x2x6\begin{align*}y=x^2-x-6\end{align*}
2. y=x2+3x+28\begin{align*}y=-x^2+3x+28\end{align*}
3. y=2x2+11x40\begin{align*}y=2x^2+11x-40\end{align*}
4. y=x26x+7\begin{align*}y=x^2-6x+7\end{align*}
5. y=x2+8x+13\begin{align*}y=x^2+8x+13\end{align*}
6. y=x2+6x+34\begin{align*}y=x^2+6x+34\end{align*}
7. y=10x213x3\begin{align*}y=10x^2-13x-3\end{align*}
8. y=4x2+12x3\begin{align*}y=-4x^2+12x-3\end{align*}
9. y=13(x4)2+12\begin{align*}y=\frac{1}{3}(x-4)^2+12\end{align*}
10. y=2(x+1)29\begin{align*}y=-2(x+1)^2-9\end{align*}
11. Calculator Investigation
12. The
13. parent graph
14. of a quadratic equation is
15. y=x2\begin{align*}y=x^2\end{align*}
16. .
17. Graph y=x2,y=3x2\begin{align*}y=x^2, y=3x^2\end{align*}, and y=12x2\begin{align*}y=\frac{1}{2}x^2\end{align*} on the same set of axes in the calculator. Describe how a\begin{align*}a\end{align*} effects the shape of the parabola.
18. Graph y=x2,y=x2\begin{align*}y=x^2, y=-x^2\end{align*}, and y=2x2\begin{align*}y=-2x^2\end{align*} on the same set of axes in the calculator. Describe how a\begin{align*}a\end{align*} effects the shape of the parabola.
19. Graph y=x2,y=(x1)2\begin{align*}y=x^2, y=(x-1)^2\end{align*}, and y=(x+4)2\begin{align*}y=(x+4)^2\end{align*} on the same set of axes in the calculator. Describe how h\begin{align*}h\end{align*} effects the location of the parabola.
20. Graph y=x2,y=x2+2\begin{align*}y=x^2, y=x^2+2\end{align*}, and y=x25\begin{align*}y=x^2-5\end{align*} on the same set of axes in the calculator. Describe how k\begin{align*}k\end{align*} effects the location of the parabola.
21. Real World Application The path of a baseball hit by a bat follows a parabola. A batter hits a home run into the stands that can be modeled by the equation y=0.003x2+1.3x+4\begin{align*}y=-0.003x^2+1.3x+4\end{align*}, where x\begin{align*}x\end{align*} is the horizontal distance and y\begin{align*}y\end{align*} is the height (in feet) of the ball. Find the maximum height of the ball and its total distance travelled.
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# Calculus
put in standard form:
4x^2=36+9y^2
1. 👍
2. 👎
3. 👁
## Similar Questions
1. ### Mathematics
For question 1 & 2, find the x- and y-intercept of the line. 1. -10x + 5y = 40 A. x-intercept is 5; y-intercept is -10 B. x-intercept is 8; y-intercept is -4 C. x-intercept is -10; y-intercept is 5 D. x-intercept is -4;
2. ### math
write each decimal in standard form, expanded form, and word form. 12+0.2+0.005
3. ### MATH
45. Express each equation in the specified form. a. y = (x – 3)² - 25 in standard form B. y = 2(x – 7)(x + 3) in standard form. C. y = - 2x² + 28x – 26 in factored form D. y = -9x² + 72x + 81 in vertex form
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1. ### Algebra
Alvin throws the football to a receiver who jumps up to catch the ball. The height of the ball over time can be represented by the quadratic equation -4.9t^2 + 7.5t + 1.8 = 2.1 . This equation is based on the acceleration of
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Identifying Conics. Put equation in standard form and graph. 9x^2-4y^2-90x+189=0
4. ### pre cal
write the standard form equation of the parabola with vertex (-2,-2) that goes through point (-1,0) y= a (x-b)^2 + c b has to be -2, that gives the shift to the left. y=a( x+2)^2 + c when x=-2, y=-2, that makes c -2 y=a( x+2)^2 -2
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1. ## Inversely proportional!?!
Q is inversely proportional to the cube of T
When T=5, Q=12.6
$\displaystyle 5^3=125$
It's as if someone added a decimal point and then added a 0.1 although it is never so with maths....thankfully.
2. Hello, Mukilab!
Do you really understand proportionality?
$\displaystyle Q$ is inversely proportional to the cube of $\displaystyle T$.
$\displaystyle Q \:=\:\frac{k}{T^3}$ .[1]
When $\displaystyle T=5,\;Q=12.6$
Substitute into [1]: .$\displaystyle 12.6 \:=\:\frac{k}{5^3} \quad\Rightarrow\quad k \:=\:1575$
Therefore: .$\displaystyle Q \:=\:\frac{1575}{T^3}$
3. Originally Posted by Soroban
Hello, Mukilab!
Do you really understand proportionality?
$\displaystyle Q \:=\:\frac{k}{T^3}$ .[1]
Substitute into [1]: .$\displaystyle 12.6 \:=\:\frac{k}{5^3} \quad\Rightarrow\quad k \:=\:1575$
Therefore: .$\displaystyle Q \:=\:\frac{1575}{T^3}$
No I did not understand proportionality but now I do understand, thank you.
If I only have T, let's say it is 3. How would I calculate Q. Surely I need this k (where did you get it from?)
4. Proportionality afaik requires a variable equal to another variable multiplied (or divided) by a constant.
$\displaystyle X=kV$ ; $\displaystyle X$ is proportional to $\displaystyle V$. An inverse proportionality would be $\displaystyle X=\frac{k}{V}$
So the constant $\displaystyle k$ either has to be given, or you need to be able to calculate it from the information you have.
5. Originally Posted by dkaksl
Proportionality afaik requires a variable equal to another variable multiplied (or divided) by a constant.
$\displaystyle X=kV$ ; $\displaystyle X$ is proportional to $\displaystyle V$. An inverse proportionality would be $\displaystyle X=\frac{k}{V}$
So the constant $\displaystyle k$ either has to be given, or you need to be able to calculate it from the information you have.
The information given is: T is now 3. Work out Q. GCSE paper :/ don't think they made a mistake
6. They did mention it was inversely proportional, right? That means;
Originally Posted by Soroban
$\displaystyle Q=\frac{k}{T^3}$
even though $\displaystyle k$ isn't written anywhere you know there is a constant to define the relationship between Q and T. You can also call that constant whatever you like. It's basically rule 1 of algebra.
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# Math
posted by .
How would I find an equivalent fraction if I only have one fractions. For example, -6/54?
• Math -
You can reduce this fraction to -3/27 = -1/9
• Math -
The options are A)-12/60, B) -1/10,
C)- 5/45, D) -7/72.
Is it only possible to solve this by checking each individual answer, or is there another way?
• Math -
None of the answers is equivalent to -6/54.
http://www.mathsisfun.com/equivalent_fractions.html
• Math -
I didn't think so. How do I answer this question?
• Math -
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could you tell me three equivalent fractions for each: 2/3 8/16 4/10 3/4 Just multiply or divide by a common number. For example, 2/3 = 4/6 = 6/9 = 8/12 = 10/15 8/16 = 4/8 = 1/2 etc. Just multiply each fraction by 3 different numbers. …
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## Trigonometry (11th Edition) Clone
$sec^2~B = \frac{1}{cos^2~B} = \frac{1}{1-sin^2~B}$
We can verify the identity: $sec^2~B = \frac{1}{cos^2~B} = \frac{1}{1-sin^2~B}$
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How to | Perform a Monte Carlo Simulation
Monte Carlo methods use randomly generated numbers or events to simulate random processes and estimate complicated results. For example, they are used to model financial systems, to simulate telecommunication networks, and to compute results for high-dimensional integrals in physics. Monte Carlo simulations can be constructed directly by using the Wolfram Language's builtin random number generation functions.
A sequence of random numbers can be a very simple Monte Carlo simulation. For instance, a list of random numbers generated independently from a normal distribution with mean 0 can simulate a white noise process.
Use RandomVariate with NormalDistribution to generate a sequence of 20 numbers following a normal distribution with mean 0 and standard deviation 1:
Use ListPlot to visualize the data:
You can now construct a random walk from the data:
To start the walk at zero, prepend it to the list:
Use Accumulate to sequentially sum the data, which is then visualized with ListLinePlot:
The following definition puts the preceding commands together to generate a random walk that you can use to simulate many random walks and analyze their properties.
Define a function randomWalk that generates a random walk of length n:
Here, Table is used to create five random walks, each with length 100. They are then visualized with ListLinePlot:
Now generate 1000 walks, each with length 100. The output is suppressed with a semicolon (;) since seeing it is not necessary:
You can now calculate descriptive statistics on any aspect of the random walks. Here, the final position of each walk is analyzed.
Use [[ ]] (the short form of the Part function) to get the final data point of each random walk:
Calculate various statistics on the final data points from the 1000 random walks:
Monte Carlo methods can also be used to approximate values such as constants or numeric integrals. For instance, the following approximates the value of by generating random points in a square around a circle of radius 1, and then using the relationship between the area of the square and the circle.
Generate 10,000 points in a square bounded by {-1,-1} and {1,1}:
View the generated points:
To approximate , multiply the area of the square by the percentage of points falling within a circle of radius 1 that is centered at the origin.
Multiply the area of the square (4) by the fraction of points in the circle:
Using more points or averaging several approximations will typically give a better approximation.
Define the function approxPi to approximate from a sample of size :
Approximate using a million points:
Approximate by averaging 50 approximations from samples of size 10000:
Monte Carlo simulations are most useful in cases where the nature of the system of interest is complicated. In Bayesian analysis, you often want to mix distributions, with the parameters of two distributions following each other to generate a bivariate distribution. Because the individual distributions are interrelated, points must be iteratively generated and inserted into the other distribution to sample from the bivariate distribution.
This type of mixture is called a Gibbs sampler. After a period of iteration, the points generated will closely follow this mixture. The period of iteration is referred to as the burn-in period.
As an example, you might have a normal distribution where the mean is known, but the standard deviation is not. However, you know that the standard deviation follows a beta distribution that has one known shape parameter, and another shape parameter that is related to the normal distribution where the mean is known.
Define a function that generates random numbers according to this normal distribution:
Define a function that generates random numbers for the beta distribution. The second shape parameter of the beta distribution will be the absolute value of a normal variate:
You can simulate a point from the bivariate distribution by choosing a starting value for the normal standard deviation, and then sequentially generating random numbers from the normal and beta distributions. A normal variate is generated using the starting value for the normal standard deviation. A beta variate is then generated by using that normal variate as the unknown shape parameter for the beta distribution. This beta variate is then used as the unknown standard deviation for a new normal distribution, and so on. This process is carried out for some number of iterations, and the final normal and beta variates are the coordinates of the simulated point.
Generate 500 points starting with a standard deviation value of .5 and using 1000 iterations as the burn-in period:
Visualize the resulting points using ListPlot:
Visualize the density of the points using Histogram3D:
Other examples of Monte Carlo methods for estimation include optimization and high-dimensional integration. NMinimize and NIntegrate have methods for optimization and numeric integration using these techniques.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$$4\sqrt{3}$$
We are given the following equation: $$A\left(s\right)=\frac{\sqrt{3}}{4}s^2$$ Thus, when the side length is 4, it follows: $$\frac{\sqrt{3}}{4}\cdot \:4^2\\ 4\sqrt{3}$$
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# Physics/Essays/Fedosin/Quantum Gravitational Resonator
Quantum Gravitational Resonator (QGR) – closed topological object of the three dimensional space, in the general case – ‘’cavity’’ of arbitrary form, which has definite ‘’surface’’ and ‘’thickness’’. The QGR can have “infinite” phase shifted oscillations of gravitational field strength and gravitational torsion field, due to the quantum properties of QGR.
## History
Considering that the theory of the gravitational resonator is based on the Maxwell-like gravitational equations and Quantum Electromagnetic Resonator (QER), therefore the QGR history is close connected with the QER history.
## Classical gravitational resonator
### Gravitational LC circuit
The gravitational LC circuit can be composed by analogy with the electromagnetic LC circuit, and gravitational field strength and gravitational torsion field oscillate in the circuit as a result of oscillating mass current.
Gravitational voltage on gravitational inductance is:
${\displaystyle V_{gL}=-L_{g}\cdot {\frac {dI_{gL}}{dt}}.\ }$
Gravitational mass current through gravitational capacitance is:
${\displaystyle I_{gC}=C_{g}\cdot {\frac {dV_{gC}}{dt}}.\ }$
Differentiating these equations with respect to the time variable, we obtain:
${\displaystyle {\frac {dV_{gL}}{dt}}=-L_{g}{\frac {d^{2}I_{gL}}{dt^{2}}},\qquad {\frac {dI_{gC}}{dt}}=C_{g}{\frac {d^{2}V_{gC}}{dt^{2}}}.}$
Considering the following relationships for voltages and currents:
${\displaystyle V_{gL}=V_{gC}=V_{g},\qquad I_{gL}=I_{gC}=I_{g},\ }$
we obtain the following differential equations for gravitational oscillations:
${\displaystyle ~{\frac {d^{2}I_{g}}{dt^{2}}}+{\frac {1}{L_{g}C_{g}}}I_{g}=0,\qquad {\frac {d^{2}V_{g}}{dt^{2}}}+{\frac {1}{L_{g}C_{g}}}V_{g}=0.\quad \quad \quad \quad \quad (1)\ }$
Furthermore, considering the following relationships between voltage and mass, current and flux of gravitational torsion field:
${\displaystyle m=C_{g}V_{g},\qquad \Phi =L_{g}I_{g}}$
the above oscillation equation can be rewritten in the form:
${\displaystyle {\frac {d^{2}m}{dt^{2}}}+{\frac {1}{L_{g}C_{g}}}m=0.\quad \quad \quad \quad \quad (2)\ }$
This equation has the partial solution:
${\displaystyle m(t)=m_{0}\sin(\omega _{g}t),\ }$
where
${\displaystyle \omega _{g}={\frac {1}{\sqrt {L_{g}C_{g}}}}\ }$
is the resonance frequency, and
${\displaystyle \rho _{LC}={\sqrt {\frac {L_{g}}{C_{g}}}},\ }$
is the gravitational characteristic impedance.
For the sake of completeness we can present the differential equation for the flux of gravitational torsion field in the form:
${\displaystyle {\frac {d^{2}\Phi }{dt^{2}}}+{\frac {1}{L_{g}C_{g}}}\Phi =0.\quad \quad \quad \quad \quad (3)\ }$
The realization of gravitational LC circuit is described in a section of maxwell-like gravitational equations.
## Quantum general approach
### Quantum gravitational LC circuit oscillator
Inductance momentum quantum operator in the electric-like gravitational mass space can be presented in the following form:
${\displaystyle {\hat {p}}_{gm}=-i\hbar {\frac {d}{dm}},\quad \quad \quad \quad \quad {\hat {p}}_{gm}^{*}=i\hbar {\frac {d}{dm}},\quad \quad \quad \quad \quad (4a)\ }$
where ${\displaystyle \hbar \ }$ is reduced Plank constant, ${\displaystyle {\hat {p}}_{gm}^{*}\ }$ is the complex-conjugate momentum operator, ${\displaystyle m\ }$ is the induced mass.
Capacitance momentum quantum operator in the magnetic-like gravitational mass space can be presented in the following form:
${\displaystyle {\hat {p}}_{g\Phi }=-i\hbar {\frac {d}{d\Phi }},\quad \quad \quad \quad \quad {\hat {p}}_{g\Phi }^{*}=i\hbar {\frac {d}{d\Phi }},\quad \quad \quad \quad \quad (4b)\ }$
where ${\displaystyle \Phi \ }$ is the induced torsion field flux, which is imitated by electric-like gravitational mass current (${\displaystyle i_{g}\ }$ ):
${\displaystyle \Phi =L_{g}\cdot i_{g}.\ }$
We can introduce the third momentum quantum operator in the current form:
${\displaystyle {\hat {p}}_{gi}=-{\frac {i\hbar }{L_{g}}}{\frac {d}{di_{g}}},\quad \quad \quad \quad \quad {\hat {p}}_{gi}^{*}={\frac {i\hbar }{L_{g}}}{\frac {d}{di_{g}}},\quad \quad \quad \quad \quad (4c)\ }$
These quantum momentum operators defines three Hamilton operators:
${\displaystyle {\hat {H}}_{gLm}=-{\frac {\hbar ^{2}}{2L_{g}}}\cdot {\frac {d^{2}}{dm^{2}}}+{\frac {L_{g}\omega _{0}^{2}}{2}}{\hat {m}}^{2}\quad \quad \quad \quad \quad (5a)\ }$
${\displaystyle {\hat {H}}_{gC\Phi }=-{\frac {\hbar ^{2}}{2C_{g}}}\cdot {\frac {d^{2}}{d\Phi ^{2}}}+{\frac {C_{g}\omega _{0}^{2}}{2}}{\hat {\Phi }}^{2}\quad \quad \quad \quad \quad (5b)\ }$
${\displaystyle {\hat {H}}_{gLi}=-{\frac {\hbar ^{2}\omega _{0}^{2}}{2L_{g}}}\cdot {\frac {d^{2}}{di_{g}^{2}}}+{\frac {L_{g}\omega _{0}}{2}}{\hat {i}}_{g}^{2},\quad \quad \quad \quad \quad (5c)\ }$
where ${\displaystyle \omega _{0}={\frac {1}{\sqrt {L_{g}C_{g}}}}\ }$ is the resonance frequency. We consider the case without dissipation (${\displaystyle R_{g}=0\ }$ ). The only difference of the gravitational charge spaces and gravitational current spaces from the traditional 3D- coordinate space is that it is one dimensional (1D). Schrodinger equation for the gravitational quantum LC circuit could be defined in three form:
${\displaystyle -{\frac {\hbar ^{2}}{2L_{g}}}{\frac {d^{2}\Psi }{dm^{2}}}+{\frac {L_{g}\omega _{0}^{2}}{2}}m^{2}\Psi =W\Psi \quad \quad \quad \quad \quad (6a)\ }$
${\displaystyle -{\frac {\hbar ^{2}}{2C_{g}}}{\frac {d^{2}\Psi }{d\Phi ^{2}}}+{\frac {C_{g}\omega _{0}^{2}}{2}}\Phi ^{2}\Psi =W\Psi \quad \quad \quad \quad \quad (6b)\ }$
${\displaystyle -{\frac {\hbar ^{2}\omega _{0}^{2}}{2L_{g}}}{\frac {d^{2}\Psi }{di_{g}^{2}}}+{\frac {L_{g}\omega _{0}}{2}}i_{g}^{2}\Psi =W\Psi .\quad \quad \quad \quad \quad (6c)\ }$
To solve these equations we should to introduce the following dimensionless variables:
${\displaystyle \xi _{m}={\frac {m}{m_{0}}};\quad \quad m_{0}={\sqrt {\frac {\hbar }{L_{g}\omega _{0}}}};\quad \quad \lambda _{m}={\frac {2W}{\hbar \omega _{0}}}\quad \quad (7a)\ }$
${\displaystyle \xi _{\Phi }={\frac {\Phi }{\Phi _{0}}};\quad \quad \Phi _{0}={\sqrt {\frac {\hbar }{C_{g}\omega _{0}}}};\quad \quad \lambda _{\Phi }={\frac {2W}{\hbar \omega _{0}}}\quad \quad (7b)\ }$
${\displaystyle \xi _{i}={\frac {i_{g}}{i_{g0}}};\quad \quad i_{g0}={\sqrt {\frac {\hbar \omega _{0}}{L_{g}}}};\quad \quad \lambda _{i}={\frac {2W}{\hbar \omega _{0}}}.\quad \quad (7c)\ }$
where ${\displaystyle m_{0}\ }$ is scaling induced electric-like gravitational mass; ${\displaystyle \Phi _{0}\ }$ is scaling induced gravitational torsion field flux and ${\displaystyle i_{g0}\ }$ is scaling induced mass current.
Then the Schrodinger equation will take the form of the differential equation of Chebyshev-Ermidt:
${\displaystyle \left({\frac {d^{2}}{d\xi ^{2}}}+\lambda -\xi ^{2}\right)\Psi =0.\ }$
The eigen values of the Hamiltonian will be:
${\displaystyle W_{n}=\hbar \omega _{0}(n+1/2),\quad \quad n=0,1,2,\dots \ }$
where at ${\displaystyle n=0\ }$ we shall have zero oscillation:
${\displaystyle W_{0}=\hbar \omega _{0}/2.\ }$
In the general case the scaling mass and torsion flux can be rewritten in the form:
${\displaystyle m_{0}={\sqrt {\frac {\hbar }{L_{g}\omega _{0}}}}={\frac {m_{P}}{\sqrt {4\pi }}}={\sqrt {\frac {\hbar c}{4\pi G}}},\ }$
${\displaystyle \Phi _{0}={\sqrt {\frac {\hbar }{C_{g}\omega _{0}}}}={\frac {h}{m_{P}{\sqrt {\pi }}}}={\sqrt {\frac {4\pi G\hbar }{c}}},\ }$
where ${\displaystyle m_{P}\ }$ is the Planck mass, ${\displaystyle c\ }$ is the speed of light, ${\displaystyle G\ }$ is the gravitational constant.
These three equations (4) form the base of the nonrelativistic quantum gravidynamics, which considers elementary particles from the intrinsic point of view. Note that, the standard quantum electrodynamics considers elementary particles from the external point of view.
### Gravitational resonator as quantum LC circuit
Due to Luryi density of states (DOS) approach we can define gravitational quantum capacitance as:
${\displaystyle C_{g}=m_{g}^{2}\cdot D_{2D}\cdot S_{g},\ }$
and quantum inductance as:
${\displaystyle L_{g}=\Phi _{g}^{2}\cdot D_{2D}\cdot S_{g},\ }$
where ${\displaystyle S_{g}\ }$ is the resonator surface area, ${\displaystyle D_{2D}={\frac {m_{0}}{\pi \hbar ^{2}}}\ }$ is two dimensional (2D) DOS, ${\displaystyle m_{0}\ }$ is the carrier mass, ${\displaystyle m_{g}\ }$ is the induced gravitational mass, and ${\displaystyle \Phi _{g}\ }$ is the gravitational torsion field flux.
Energy stored on quantum capacitance is:
${\displaystyle W_{Cg}={\frac {m_{g}^{2}}{2C_{g}}}={\frac {1}{2D_{2D}S_{g}}}.\ }$
Energy stored on quantum inductance is:
${\displaystyle W_{Lg}={\frac {\Phi _{g}^{2}}{2L_{g}}}={\frac {1}{2D_{2D}S_{g}}}=W_{Cg}.\ }$
Resonator angular frequency is:
${\displaystyle \omega _{gR}={\frac {1}{\sqrt {L_{g}C_{g}}}}={\frac {1}{m_{g}\Phi _{g}D_{2D}S_{g}}}.\ }$
Energy conservation law for zero oscillation is:
${\displaystyle W_{gR}={\frac {1}{2}}\hbar \omega _{gR}={\frac {\hbar }{2m_{g}\Phi _{g}D_{2D}S_{g}}}=W_{Cg}=W_{Lg}.\ }$
This equation can be rewritten as:
${\displaystyle m_{g}\Phi _{g}=\hbar .\ }$
Characteristic gravitational resonator impedance is:
${\displaystyle \rho _{g}={\sqrt {\frac {L_{g}}{C_{g}}}}={\frac {\Phi _{g}}{m_{g}}}=2\alpha {\frac {\Phi _{g0}}{m_{S}}}=\rho _{g0},\ }$
where ${\displaystyle \alpha \ }$ is the fine structure constant, ${\displaystyle \Phi _{g0}=h/m_{S}\ }$ is the gravitational torsion flux quantum, ${\displaystyle h\ }$ is the Planck constant, ${\displaystyle m_{S}\ }$ is the Stoney mass, ${\displaystyle \rho _{g0}\ }$ is the gravitational characteristic impedance of free space.
Considering above equations, we can find out the following induced mass and induced gravitational torsion flux:
${\displaystyle m_{g}={\frac {m_{S}}{\sqrt {4\pi \alpha }}},\ }$
${\displaystyle \Phi _{g}={\sqrt {\frac {\alpha }{\pi }}}{\frac {h}{m_{S}}}.\ }$
Note, that these induced quantities maintain the energy balance between resonator oscillation energy and total energy on capacitance and inductance
${\displaystyle \hbar \omega _{gR}=W_{gL}(t)+W_{gC}(t).\ }$
Since capacitance oscillations are phase shifted (${\displaystyle \psi =\pi /2\ }$ ) with respect to inductance oscillations, therefore we get:
${\displaystyle W_{gL}={\begin{cases}0,&{\mbox{at }}t=0;\psi =0{\mbox{ and}}\,t={\frac {T_{R}}{2}};\psi =\pi \\W_{L},&{\mbox{at }}t={\frac {T_{R}}{4}};\psi ={\frac {\pi }{4}}{\mbox{ and}}\,t={\frac {3T_{R}}{4}};\psi ={\frac {3\pi }{4}}\end{cases}}\ }$
${\displaystyle W_{gC}={\begin{cases}W_{C},&{\mbox{at }}t=0;\psi =0{\mbox{ and}}\,t={\frac {T_{R}}{2}};\psi =\pi \\0,&{\mbox{at }}t={\frac {T_{R}}{4}};\psi ={\frac {\pi }{4}}{\mbox{ and}}\,t={\frac {3T_{R}}{4}};\psi ={\frac {3\pi }{4}}\end{cases}}\ }$
where ${\displaystyle T_{R}={\frac {2\pi }{\omega _{gR}}}\ }$ is the oscillation period.
## Applications
### Planckion resonator
${\displaystyle r_{P}={\frac {\lambda _{P}}{2\pi }},\ }$
where ${\displaystyle \lambda _{P}={\frac {h}{m_{P}c}}\ }$ is the Compton wavelength of planckion, ${\displaystyle c\ }$ is the speed of light, ${\displaystyle m_{P}\ }$ is the Planck mass.
Planckion surface scaling parameter is:
${\displaystyle S_{P}=2\pi r_{P}^{2}={\frac {\lambda _{P}^{2}}{2\pi }}.\ }$
Planckion angular frequency is:
${\displaystyle \omega _{P}={\frac {m_{P}c^{2}}{\hbar }}={\frac {2\pi c}{\lambda _{P}}}.\ }$
Planckion density of states is:
${\displaystyle D_{P}={\frac {1}{S_{P}W_{P}}}={\frac {1}{S_{P}\hbar \omega _{P}}}={\frac {m_{P}}{2\pi \hbar ^{2}}}.\ }$
Standard DOS quantum resonator approach yields the following values for the gravitational reactive quantum parameters:
${\displaystyle C_{P}=m_{g}^{2}D_{P}S_{P}={\frac {m_{S}^{2}}{4\pi \alpha }}{\frac {m_{P}}{2\pi \hbar ^{2}}}{\frac {\lambda _{P}^{2}}{2\pi }}={\frac {\varepsilon _{g}\lambda _{P}}{2\pi }}={\frac {m_{P}}{4\pi c^{2}}},\ }$
where ${\displaystyle \varepsilon _{g}={\frac {1}{4\pi G}}\ }$ is the gravitoelectric gravitational constant in the set of selfconsistent gravitational constants, and
${\displaystyle L_{P}=\Phi _{g}^{2}D_{P}S_{P}={\frac {\Phi _{0}^{2}}{4\pi \beta }}D_{P}S_{P}={\frac {\alpha h^{2}}{\pi m_{S}^{2}}}D_{P}S_{P}={\frac {\alpha h^{2}}{\pi m_{S}^{2}}}{\frac {m_{P}}{2\pi \hbar ^{2}}}{\frac {\lambda _{P}^{2}}{2\pi }}={\frac {\mu _{g}\lambda _{P}}{2\pi }},\ }$
where ${\displaystyle \mu _{g}={\frac {4\pi G}{c^{2}}}\ }$ is the gravitomagnetic gravitational constant of selfconsistent gravitational constants, ${\displaystyle \beta ={\frac {1}{4\alpha }}\ }$ is the gravitational torsion coupling constant, which is equal to magnetic coupling constant.
Thus, s.c. free planckion can be considered as discoid quantum resonator which has radius ${\displaystyle r_{P}\ }$ .
### Bohr atom as a gravitational quantum resonator
The gravitational quantum capacitance for Bohr atom is:
${\displaystyle C_{\Gamma }=m_{R}^{2}D_{B}S_{B}=\varepsilon _{\Gamma }a_{B},\ }$
where ${\displaystyle a_{B}\ }$ is the Bohr radius, ${\displaystyle S_{B}=\pi a_{B}^{2}\ }$ is the flat surface area, ${\displaystyle ~m_{R}={\frac {\sqrt {m_{p}m_{e}}}{2{\sqrt {\pi }}}}}$ is the induced mass, ${\displaystyle D_{B}={\frac {m_{e}}{\pi \hbar ^{2}}}\ }$ is the density of states, ${\displaystyle \varepsilon _{\Gamma }={\frac {1}{4\pi \Gamma }}}$ is the gravitoelectric gravitational constant of selfconsistent gravitational constants in the field of strong gravitation, ${\displaystyle \Gamma }$ is the strong gravitational constant, ${\displaystyle m_{p}\ }$ and ${\displaystyle m_{e}\ }$ are masses of proton and electron.
The gravitational quantum inductance is:
${\displaystyle L_{\Gamma }=\phi _{\Gamma }^{2}D_{B}S_{B}=\mu _{\Gamma }a_{B},\ }$
where ${\displaystyle \mu _{\Gamma }={\frac {4\pi \Gamma }{c^{2}}}}$ is the gravitomagnetic gravitational constant of selfconsistent gravitational constants in the field of strong gravitation, and the induced gravitational torsion flux is:
${\displaystyle \phi _{\Gamma }={\frac {\alpha h}{\sqrt {\pi m_{p}m_{e}}}}={\frac {2\alpha {\sqrt {m_{e}}}}{\sqrt {\pi m_{p}}}}\sigma _{e}={\frac {2\alpha {\sqrt {m_{p}}}}{\sqrt {\pi m_{e}}}}\Phi _{\Gamma }={\frac {2{\sqrt {m_{p}}}}{\alpha {\sqrt {\pi m_{e}}}}}\Phi _{\Omega },}$
where ${\displaystyle \sigma _{e}\ }$ is the velocity circulation quantum, ${\displaystyle \Phi _{\Gamma }={\frac {h}{2m_{p}}}}$ is the strong gravitational torsion flux quantum, which is related to proton with its mass ${\displaystyle m_{p}}$ .
Here the strong gravitational electron torsion flux for the first energy level is:
${\displaystyle \Phi _{\Omega }=\Omega S_{B}={\frac {\mu _{\Gamma }m_{e}}{4\pi a_{B}}}\sigma _{e}={\frac {\Gamma m_{e}}{c^{2}a_{B}}}\sigma _{e}={\frac {\Gamma h}{2c^{2}a_{B}}}={\frac {\pi \alpha \Gamma m_{e}}{c}}=\alpha ^{2}\Phi _{\Gamma },\ }$
where ${\displaystyle \Omega \ }$ is the gravitational torsion field of strong gravitation in electron disc.
The gravitational wave impedance is:
${\displaystyle \rho _{\Gamma }={\sqrt {\frac {L_{\Gamma }}{C_{\Gamma }}}}={\sqrt {\frac {\mu _{\Gamma }}{\varepsilon _{\Gamma }}}}={\frac {4\pi \Gamma }{c}}=6.346\cdot 10^{21}\,\mathrm {m^{2}/(s\cdot kg)} .\ }$
The resonance frequency of gravitational oscillation is:
${\displaystyle \omega _{\Gamma }={\frac {1}{\sqrt {L_{\Gamma }C_{\Gamma }}}}={\frac {c}{a_{B}}}={\frac {\omega _{B}}{\alpha }},\ }$
where ${\displaystyle \omega _{B}={\frac {c\alpha }{a_{B}}}\ }$ is the angular frequency of electron rotation in atom.
For the quantum gravitational resonator approach we can derive the following maximal values for the energies stored on capacitance and inductance:
${\displaystyle W_{C}={\frac {m_{R}^{2}}{2C_{\Gamma }}}={\frac {\hbar \omega _{B}}{2}}={\frac {\alpha \hbar \omega _{\Gamma }}{2}}=W_{B},\ }$
${\displaystyle W_{L}={\frac {\phi _{\Gamma }^{2}}{2L_{\Gamma }}}=W_{B}.\ }$
The energy ${\displaystyle W_{B}\ }$ of the wave of strong gravitation in the electron matter has the same value as in case of rotating electromagnetic wave, and can be associated with the mass:
${\displaystyle m_{Bmin}={\frac {W_{B}}{c^{2}}}={\frac {\hbar \omega _{B}}{2c^{2}}}={\frac {\alpha ^{2}}{2}}m_{e}<
which could be named as the minimal mass-energy of the quantum resonator.
One way to explain the minimal mass-energy ${\displaystyle m_{Bmin}}$ is the supposition that Planck constant can be used at all matter levels including the level of star. As a result of the approach one should introduce different scales such as Planck scale, Stoney scale, Natural scale, with the proper masses and lengths. But such proper masses do not relate with the real particles.
Another way recognizes the similarity of matter levels and SPФ symmetry as the principles of matter structure where the action constants depend on the matter levels. For example there is the stellar Planck constant at the star level that describes star systems without any auxiliary mass and scales.
## Reference Books
• Johnstone Stoney, Phil. Trans. Roy. Soc. 11, (1881)
• Stratton J.A.(1941). Electromagnetic Theory. New York, London: McGraw-Hill.p.615. djvu
• Детлаф А.А., Яворский Б.М., Милковская Л.Б.(1977). Курс физики. Том 2. Электричество и магнетизм (4-е издание). М.: Высшая школа, "Reference Book on Electricity" djvu
• Гольдштейн Л.Д., Зернов Н.В. (1971). Электромагнитные поля. 2- издание. Москва: Советское Радио. 664с. "Electromagnetic Fields" djvu
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### How to Calculate Tensile Deformation
Materials science deals with the study of the properties of various materials and their individual characteristics. A good understanding of a material’s chemical make-up and physical properties is vital to good structural design. There are three primary modes of stress that are relevant in gauging a material’s suitability as a construction material – tensile, compressive, and shear.
Tensile deformation is deformation due to an object being pulled apart, as opposed to being pushed together (compression) or pushed in opposite parallel directions (shear). The tensile strength of a material is an intensive property, which means it is not related to the specimen’s size. However, when the material is used as a structural member, its cross section and second moment of area become important.
What this means is that when measuring a material’s tensile strength, it is expressed as the quotient of the tensile stress over tensile strain, a ratio. It is the portion of the stress-strain curve from the point of loading to the point of necking. The ultimate tensile strength (UTS) of a material is not to be confused with its Young’s modulus, which has the same units of force per unit area but doesn’t express when it will yield, only its stiffness.
It is often necessary to calculate tensile deformation of a structural member under a given load so as to determine its suitability for a particular task. Note that we are not talking about bending moments here, but pure tension. A bending moment is a rotational force (moment) caused by the bending of an element – which results in both tension and compression. An object in pure tension has no other forces acting on it but tension.
In order to calculate the tensile deformation in a sample member with uniform cross section and Young’s modulus, assuming complete homogeneity and freedom from prior stress, you need to know the following values:
• P = the applied force on the member that is causing it to deform in newtons (N)
• L = the length of the member
• A = the cross-sectional area of the member
• E = the material’s Young’s modulus (pascal = N/m2) – also known as the elastic modulus or modulus of elasticity
If all you want to know is the tensile deformation along a single elastic member, you can calculate it using this simple formula:
The image further above is an example of a weight-force placing two vertical steel rods into tension. The two rods are spaced evenly from the edges so as to support an equal amount of weight. They are both 1 meter long, are solid all the way through, have diameters of 20 millimeters, and are made of steel, with a Young’s modulus of 200 Gpa. The area for circular cross sections are found by the formula A = πr2. You can use the above equation with a slight tweak to account for 2 rods:
Remember that uniform units must be used for proper results. Depending on the magnitude of the individual values, it may be more convenient to use smaller unit-variations. For example, Young’s modulus (E) can be expressed in N/mm2 or kN/mm2 (kilo-newtons) if the cross-sectional area (A) is better defined in square millimeters instead of square meters – which is often the case.
• For conversion to N/mm2, bring the decimal of E in N/m2 left 6 digits
• For conversion to kN/mm2, bring the decimal of E in N/m2 left 9 digits
However, if you do this, you must convert all other units to millimeters as well – your answer will also naturally be in millimeters. Length (L) must be in millimeters, and the applied force (P) must be converted to kNs if you use kN/mm2 as Young’s modulus. This can be a source of confusion for beginners so be sure you understand this and get it right.
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# Left-Tailed Test Rejects Null Hypothesis at 0.05 Level
• MHB
• JocquettaLARuex
In summary, in a left-tailed test with a test statistic of -2, a shaded area of 0.03 indicates that there is sufficient evidence to reject the null hypothesis at a 0.05 level of significance. This means that the area to the left of the test statistic is being considered.
JocquettaLARuex
In a left-tailed test, the value of the test statistic is -2. If we know the shaded area is 0.03,
then we have sufficient evidence to reject the null hypothesis at 0.05 level of significance.
JocquettaLARuex said:
In a left-tailed test, the value of the test statistic is -2. If we know the shaded area is 0.03,
then we have sufficient evidence to reject the null hypothesis at 0.05 level of significance.
Hi there,
How does the area of the shaded region relate to a p-value in hypothesis testing? In a left-tailed test is the area we look at the area to the left or the area to the right?
## 1. What is a "left-tailed test"?
A left-tailed test is a statistical hypothesis test that is used to determine if the observed data is significantly lower than the expected value. It is used to test whether a sample mean is significantly less than a hypothesized population mean.
## 2. What does it mean to "reject the null hypothesis at 0.05 level"?
Rejecting the null hypothesis at 0.05 level means that the results of the statistical test have shown that there is enough evidence to reject the null hypothesis, and that the observed data is significantly different from the expected data at a significance level of 0.05 or 5%. This means that there is a 95% confidence that the results are not due to chance.
## 3. What is a null hypothesis?
A null hypothesis is a statement that assumes there is no significant difference between the observed data and the expected data. It is typically represented as H0 in statistical tests and is used as a starting point for testing the alternative hypothesis.
## 4. How is the significance level chosen for a statistical test?
The significance level, also known as alpha (α), is typically chosen before conducting the statistical test. It is based on the level of confidence needed in the results. A common significance level is 0.05 or 5%, which means there is a 95% confidence that the results are not due to chance.
## 5. What does it mean when the null hypothesis is rejected?
When the null hypothesis is rejected, it means that there is enough evidence to support the alternative hypothesis. This means that the observed data is significantly different from the expected data and that the results are not due to chance. It does not necessarily mean that the alternative hypothesis is true, but rather that the null hypothesis is highly unlikely to be true.
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# Search by Topic
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Two Spinners
##### Age 5 to 7 Challenge Level:
What two-digit numbers can you make with these two dice? What can't you make?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### What Could it Be?
##### Age 5 to 7 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Multiply Multiples 2
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Can you work out some different ways to balance this equation?
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What happens when you round these three-digit numbers to the nearest 100?
### One of Thirty-six
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Can you find the chosen number from the grid using the clues?
### Multiply Multiples 3
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### All the Digits
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This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
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### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### 2,4,6,8
##### Age 5 to 7 Challenge Level:
Using the cards 2, 4, 6, 8, +, - and =, what number statements can you make?
### Eight Queens
##### Age 7 to 11 Challenge Level:
Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others.
### 1 to 8
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.
### Two on Five
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Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Room Doubling
##### Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Cubes Here and There
##### Age 7 to 11 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### Pouring the Punch Drink
##### Age 7 to 11 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Dienes' Logiblocs
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This problem focuses on Dienes' Logiblocs. What is the same and what is different about these pairs of shapes? Can you describe the shapes in the picture?
### Two Egg Timers
##### Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Octa Space
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In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
### Unit Differences
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This challenge is about finding the difference between numbers which have the same tens digit.
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Mystery Matrix
##### Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Are You Well Balanced?
##### Age 5 to 7 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Two Dice
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Find all the numbers that can be made by adding the dots on two dice.
### Creating Cubes
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Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour.
### Plates of Biscuits
##### Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Map Folding
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Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### More and More Buckets
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In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
### Greater Than or Less Than?
##### Age 7 to 11 Challenge Level:
Use the numbers and symbols to make this number sentence correct. How many different ways can you find?
### Route Product
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Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Sums and Differences 2
##### Age 7 to 11 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Round the Two Dice
##### Age 5 to 7 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Multiplication Squares
##### Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Chocs, Mints, Jellies
##### Age 7 to 11 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it?
### What Shape and Colour?
##### Age 5 to 7 Challenge Level:
Can you fill in the empty boxes in the grid with the right shape and colour?
### Prison Cells
##### Age 7 to 11 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Round the Dice Decimals 2
##### Age 7 to 11 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Sums and Differences 1
##### Age 7 to 11 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Pasta Timing
##### Age 7 to 11 Challenge Level:
Nina must cook some pasta for 15 minutes but she only has a 7-minute sand-timer and an 11-minute sand-timer. How can she use these timers to measure exactly 15 minutes?
##### Age 5 to 7 Challenge Level:
If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made with six beads?
### Fake Gold
##### Age 7 to 11 Challenge Level:
A merchant brings four bars of gold to a jeweller. How can the jeweller use the scales just twice to identify the lighter, fake bar?
### Lots of Lollies
##### Age 5 to 7 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Round the Dice Decimals 1
##### Age 7 to 11 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
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# Identifying Linear Functions Worksheet Pdf
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# If the solve the problem
Question:
Let $x, y$ be two variables and $x>0, x y=1$, then minimum value of $x+y$ is
(a) 1
(b) 2
(c) $2 \frac{1}{2}$
(d) $3 \frac{1}{3}$
Solution:
(b) 2
Given : $x y=1$
$\Rightarrow y=\frac{1}{x}$
$f(x)=x+\frac{1}{x}$
$\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$
For a local maxima or a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 1-\frac{1}{x^{2}}=0$
$\Rightarrow x^{2}-1=0$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
$\Rightarrow x=1$ (Given : $x>1$ )
$\Rightarrow y=1$
Now,
$f^{\prime \prime}(x)=\frac{2}{x^{3}}$
$\Rightarrow f^{\prime \prime}(1)=2>0$
So, $x=1$ is a local minima.
$\therefore$ Minimum value of $f(x)=f(1)=1+1=2$
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# Solving a Candle Height Problem with Separation of Variables
• Bashyboy
## Homework Statement
In the problem, we are to consider two candles, call them C1 and C2, with different heights and different thicknesses. Call the height of C1 H1, and for C2, call it H2. The taller candle burns can burn for 7/2 hours, and the short one, 5 hours. After two hours lapses, the candles have equal heights. Two hours ago, what fraction of the candle's height gave the short candle's height?
## The Attempt at a Solution
Here some observations, of which I am not certain are true, that I made. Assuming the candles burn at a constant rate, the rate of C1 is H1/3.5, and the rate of C2 is H2/5.
After two hours of burning has lapsed, h1/h2 = 1.
Now, I am going to assume the candles are cylindrical.
$V_1 = \pi r^2_1H_1$ I am also going to assume that the burning process is such that the radius is not altered.
$\frac{dV_1}{dt} = \pi r^2_1 \frac{dH_1}{dt}$. I know how the height changes with time--it's constant. $\frac{dH_1}{dt} = \frac{H_1}{3.5}$ If I write it in the form $\frac{dH_1}{dt} = (H_1)(3.5)^{-1}$, this is a differential equation in which I can use the separation of variables method.
$\int H_1 dH_1 = (3.5)^{-1} \int dt$
which comes out to be $H_1 = \sqrt{\frac{t}{3.5} + C}$
----------------------------------------------------------------
There are a few details I am confused about. Did I properly apply the separation of variables method? If so, what does the constant C represent, and can it be set to zero? Will this solution lead anywhere?
Thank you, and please don't give every detail of the problem away.
Last edited:
I have a feeling most of the things you did were unnecessary. The only thing you need is to express H(t) as a function of initial H and the burning speed (for each candle separately) and apply condition H1(2hours)=H2(2hours) - most likely it will be of the form that can be easily solved for the initial ratio of heights.
This is an interesting one, although could you please explain the problem:
Two hours ago, what fraction of the candle's height gave the short candle's height?
It is confusing to say the least. 2 hours ago, I understand, means before they were both lit, but "what fraction of the (taller?) candle's height gave short candle's height"?
Were I to pose a problem with this information I would want to find the relation between thicknesses or heights.
Last edited:
## Homework Statement
In the problem, we are to consider two candles, call them C1 and C2, with different heights and different thicknesses. Call the height of C1 H1, and for C2, call it H2. The taller candle burns can burn for 7/2 hours, and the short one, 5 hours. After two hours lapses, the candles have equal heights. Two hours ago, what fraction of the candle's height gave the short candle's height?
I take it that we are assuming that C1 is the taller candle.
## The Attempt at a Solution
Here some observations, of which I am not certain are true, that I made. Assuming the candles burn at a constant rate, the rate of C1 is H1/3.5, and the rate of C2 is H2/5.
Correct, if H1 and H2 are the initial heights of the candles.
After two hours of burning has lapsed, h1/h2 = 1.
Now, I am going to assume the candles are cylindrical.
I don't think you need assume anything about the shape or material of each candle. You need only assume that the rate of change of height is constant.
$V_1 = \pi r^2_1H_1$ I am also going to assume that the burning process is such that the radius is not altered.
$\frac{dV_1}{dt} = \pi r^2_1 \frac{dH_1}{dt}$. I know how the height changes with time--it's constant. $\frac{dH_1}{dt} = \frac{H_1}{3.5}$
No. When you said that the constant rate of burning of C1 was $H_1/3.5$, you must have been taking $H_1$ to mean the initial height (because otherwise that statement would be false). You now appear to be using $H_1$ both for initial height (which is a constant, so it's derivative with respect to time is zero) and for height at time t (which varies with t). This has led you to an incorrect ODE.
If I write it in the form $\frac{dH_1}{dt} = (H_1)(3.5)^{-1}$, this is a differential equation in which I can use the separation of variables method.
$\int H_1 dH_1 = (3.5)^{-1} \int dt$
You're dividing both sides by $H_1$ so you should have
$$\int \frac1{H_1}\,dH_1 = \frac1{3.5}\int\,dt$$
which you may go on to solve as practice in integration, but it doesn't solve the original problem.
To answer the actual question: You are entitled to assume that the rate of loss of height is constant, but different for each candle. You have already calculated the rates in terms of the initial heights: the rate of C1 is $H_1/3.5$, and the rate of C2 is $H_2/5$.
Now you need to work out the height of each candle after two hours of burning at those constant rates, and use the condition that after two hours the heights are the same.
Would still like to know what we are to determine in this assignment. Is it about determining the relation of the heights before lighting the candles?
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pasmith, all right. I made a notation error; however, wouldn't my method still work? The volume function should then be $V_1(t) = \pi r^2_1 h(t)$. Would that then work?
pasmith, all right. I made a notation error; however, wouldn't my method still work? The volume function should then be $V_1(t) = \pi r^2_1 h(t)$. Would that then work?
Why are you trying to make a very simple problem complicated? Just do what others have already suggested to you.
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# Circular motion and acceleration
1. Feb 8, 2016
### Prannu
If you are twirling a rock around your head with a rope (assuming it is uniform circular motion), then the only acceleration that is acting is radial. So if you take a look at the rock at any given instant, its velocity is perpendicular to its acceleration. My question is, why does the rock not gain any speed as a result of its acceleration? Why do textbooks say that the speed is constant, despite the fact that there is an acceleration (a net force) on the rock?
2. Feb 8, 2016
### mfig
Remember that acceleration means a change in velocity. Velocity is a vector quantity characterized by speed (magnitude) and direction. Since the rock is continuously changing direction, it is accelerating even though the speed stays the same.
3. Feb 8, 2016
### Prannu
But why should the rock's velocity change in direction only? As long as there is acceleration, over a given period of time, dv = a*dt, indicating both a directional and magnitude change in velocity.
4. Feb 8, 2016
### A.T.
You could just as well ask why it doesn't loose any speed. If the angle between acceleration and velocity is less than 90° speed increases. If its more than 90° speed decreases.
The acceleration isn't constant over any given period of time
5. Feb 8, 2016
### mfig
It is hard to know how to answer when you ask why a rock's velocity should change in direction only. All I can tell you is that does change in direction only.
Think about it! You have a string and a rock. You are spinning it at a constant rate, so you know for a fact that the speed (magnitude) is not changing. You can measure this by timing how long it takes the rock to complete once circle and verifying that it always takes the same amount of time. But the rock is constantly changing direction. So the facts of the case are these:
No change in magnitude.
Change in direction.
As I said before, velocity is a vector with a magnitude and a direction. A change in velocity is called acceleration. The change can be in:
1. The magnitude of the velocity or
2. The direction of travel or
3. Both the magnitude and the direction.
There are examples of all three cases in nature. You have found a case where there is only a change in direction. But either way, any of those three cases is a change in velocity and is therefore an acceleration, by definition.
6. Feb 8, 2016
### Prannu
Yes, I completely agree with you in that the speed of the rock is fixed in uniform circular motion. I also agree with you in that acceleration is only causing a change in the direction only of the velocity. However, I still do not understand why, at any instant, the acceleration should cause only a change in direction. Please take a look at the attached image. In it, vf = sqrt(vi^2 + (delta-v)^2). As delta-t approaches 0 (but is still not 0), vf approaches vi (delta-v = a*delta-t), but is still not exactly equal to vi. This difference may be small, but over a enormous period of time, shouldn't the difference between vf and vi add up, and become significant enough to be observed?
*vi = initial speed
vf = final speed
delta-t = change in time
delta-v = change in speed
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7. Feb 8, 2016
### A.T.
An "instant" has zero duration.
8. Feb 8, 2016
### Staff: Mentor
It should change in direction only because you are not providing a force in the direction tangent to the circular path. You are only providing a force in the direction perpendicular to the circular path. So it can't have a component of acceleration along the path. It can only have a component of acceleration perpendicular to the path.
9. Feb 8, 2016
### Tom.G
Ambiguous usage of the word "acceleration" seems to have crept in here. The word Acceleration is often (improperly) used to indicate a force. Actually, it is defined as "a change in speed or direction", not the force causing the change.
re post 6: "...I still do not understand why, at any instant, the acceleration should cause only a change in direction."
Try turning it around. Look at it as the change in direction is causing the acceleration. The rock is attempting to move in a straight line. The rope is providing a force to prevent straight line motion.
re post 1: "So if you take a look at the rock at any given instant, its velocity is perpendicular to its acceleration."
Replacing "acceleration" in that sentence with "centrifigal force" may clarify things a bit.
10. Feb 9, 2016
### FactChecker
Because the acceleration is always exactly at right angles to the velocity vector, it never adds or subtracts from the magnitude of the velocity. It only changes the direction.
11. Feb 11, 2016
### drvrm
i think you will make an error if you take centrifugal force which is outward and its not acting - on the stone only centripetal force is acting towards the center.
to clarfy the concept you can draw a vector diagram of the stone say at time t and a later time t+ dt-
say its at point A initially and after lapse of time dt it has moved to B -both A and B are on the circular path and draw now two tangents at A and B designating the velocity vectors at t and t+dt
let it be v1 and v2 naturally their directions will be making an angle theta with each other -the same angle made by the two radii at A and B.
Now you have to find change in velocity- so draw the two vectors at angle theta separately and the tip of v1 and v2 will be joined depicting the change in velocity in time interval dt say dv -you can apply triangle rule for the three vectors v1 ,v2 and dv - v1+dv=v2 so dv is along AB
now dv/dt is the acceleration and taking limit to dt tending to zero will give you acceleration vector. the magnitude of the third vector dv is radius times theta- but the direction of the acceleration will move towards the centre as dt tends to zero-
so you can see that the centripetal acceleration drives the stone on a circular arc but its speed does not change if the force is constant -the acceleration will come to v^2/radius.
12. Feb 12, 2016
### jbriggs444
A slightly different take on the error... The diagram as shown is biased. It is not time-symmetric. It takes the current direction of the acceleration as a given and extrapolates into the future on the assumption that the acceleration stays constant at its starting value. But it would be equally reasonable to extrapolate into the future based on the assumption that the acceleration stays constant at its ending value.
An approximation that uses the starting acceleration vector over a finite interval predicts in a small increase in velocity. An approximation that uses the ending acceleration vector over a finite interval predicts in a small decrease in velocity. The truth lies somewhere in the middle.
The smaller you make the intervals, the smaller the gap within which the truth can hide. As the intervals shrink toward zero width, the only place that the truth can hide is exactly at zero.
13. Feb 13, 2016
### drvrm
yes thank you for pointing out the error pranu is making -its a conceptual error of 'rate of change' depiction by differential -one has to go to the limits of dt tending to zero.
14. Feb 14, 2016
### Prannu
Oh, I see now. I was making a mistake in approximating the time interval: it was too large, which is why the speed changed also. Because dt approaches an infinitesimally small value, the delta-v has 0 contribution towards the speed, and therefore, the speed is constant. However, this dv (or dv/dt) is enough for the particle's velocity to change in direction only.
Thank you all for helping me clarify this problem!
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Answer : 4 10!/2 Explanation : Answer: D) 10!/2 Explanation: As A1 speaks always after A2, they can speak only in 1st to 9th places and A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place A2 can speak in 9 places the remaining A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e when A2 speaks in the first place, the number of ways they can speak is 9.8!. When A2 speaks in second place, the number of ways they can speak is 8.8!. When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8! Therefore,Total Number of ways they can speak = (9 8 7 6 5 4 3 2 1) 8! = 92(9 1)8! = 10!/2
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# Nyquist Plot: What is it? (And How To Draw One)
Contents
💡
Key learnings:
• Nyquist Plot Definition: A Nyquist plot is a graphical representation used in control systems to analyze stability by plotting the complex frequency response.
• Nyquist Stability Criterion: This criterion determines system stability by assessing the number of encirclements of the point (-1, 0) in the plot.
• Drawing Process: To draw a Nyquist plot, one must determine system poles on the jω axis, select a suitable contour, and map each segment to visualize the system’s response.
• Encirclement Rule: The stability of the system is determined by the direction and number of encirclements around the critical point.
• Mapping Function: The mapping function F(s) transforms points from the s-plane to the G(s) H(s) plane, essential in visualizing system dynamics.
## What is a Nyquist Plot?
A Nyquist plot is a graphical tool in control engineering and signal processing to evaluate feedback system stability. It plots the transfer function real part on the X-axis and the imaginary part on the Y-axis in Cartesian coordinates.
The frequency is swept as a parameter, resulting in a plot based on frequency. The same Nyquist plot can be described using polar coordinates, where the gain of the transfer function is the radial coordinate, and the phase of the transfer function is the corresponding angular coordinate.
The stability analysis of a feedback control system is based on identifying the location of the roots of the characteristic equation on the s-plane.
The system remains stable when its roots are located on the left side of the s-plane. Relative stability is evaluated using frequency response methods like Nyquist, Nichols plot, and Bode plot.
The Nyquist stability criterion is used to identify the presence of roots of a characteristic equation in a specified region of the s-plane.
To understand a Nyquist plot we first need to learn about some of the terminologies. Note that a closed path in a complex plane is called a contour.
## Nyquist Path or Nyquist Contour
The Nyquist contour is a closed path in the s-plane that encircles the entire right-hand half.
In order to enclose the complete RHS of the s-plane, a large semicircle path is drawn with a diameter along jω axis and center at the origin. The radius of the semicircle is treated as Nyquist Encirclement.
## Nyquist Encirclement
A point is said to be encircled by a contour if it is found inside the contour.
## Nyquist Mapping
Mapping involves transforming a point from the s-plane to the F(s) plane using the mapping function F(s).
### How to Draw Nyquist Plot
A Nyquist plot can be drawn using the following steps:
• Step 1 – Check for the poles of G(s) H(s) of the jω axis including that at the origin.
• Step 2 – Select the proper Nyquist contour – a) Include the entire right half of the s-plane by drawing a semicircle of radius R with R tends to infinity.
• Step 3 – Identify the various segments on the contour with reference to the Nyquist path
• Step 4 – Perform the mapping segment by segment by substituting the equation for the respective segment in the mapping function. Basically, we have to sketch the polar plots of the respective segment.
• Step 5 – Mapping of the segments are usually mirror images of mapping of the respective path of +ve imaginary axis.
• Step 6 – The semicircular path which covers the right half of s plane generally maps into a point in G(s) H(s) plane.
• Step 7- Interconnect all the mapping of different segments to yield the required Nyquist diagram.
• Step 8 – Note the number of clockwise encirclements about (-1, 0) and decide stability by N = Z – P
is the Open loop transfer function (O.L.T.F)
is the Closed loop transfer function (C.L.T.F)
N(s) = 0 is the open loop zero and D(s) is the open loop pole
From a stability point of view, no closed loop poles should lie on the RH side of the s-plane. Characteristics equation 1 + G(s) H(s) = 0 denotes closed-loop poles .
Now as 1 + G(s) H(s) = 0 hence q(s) should also be zero.
Therefore, from the stability point of view zeroes of q(s) should not lie in the RHP of the s-plane.
To define the stability entire RHP (Right-Hand Plane) is considered. We assume a semicircle that encloses all points in the RHP by considering the radius of the semicircle R tends to infinity. [R → ∞].
The first step to understanding the application of the Nyquist criterion in relation to the determination of stability of control systems is mapping from the s-plane to the G(s) H(s) – plane.
s is considered an independent complex variable and the corresponding value of G(s) H(s) is the dependent variable plotted in another complex plane called the G(s) H(s) – plane.
Thus for every point in the s-plane, there exists a corresponding point in G(s) H(s) – plane. During the process of mapping, the independent variable s is varied along a specified path in the s-plane, and the corresponding points in G(s)H(s) plane are joined. This completes the process of mapping from the s-plane to the G(s)H(s) – plane.
Nyquist stability criterion says that N = Z – P. Where, N is the total no. of encirclement about the origin, P is the total no. of poles and Z is the total no. of zeroes.
Case 1: N = 0 (no encirclement), so Z = P = 0 and Z = P
If N = 0, P must be zero therefore the system is stable.
Case 2: N > 0 (clockwise encirclement), so P = 0, Z ≠0 and Z > P
In both cases the system is unstable.
Case 3: N < 0 (counter-clockwise encirclement), so Z = 0, P ≠0 and P > Z
The system is stable.
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Your boss asks you to visually display three plans and compare them so you can point out the advantages of each plan to your customers. 175 North Beacon Street Step 2. Author: Created by elcarbo. So check out these 15 systems of equations activities that will help students understand and practice finding the solution to two linear equations. The two situations are: 1. Note that the last three pieces of information describing the plans are superfluous; it is important for students to be able to sort through information and decide what is, and is not, relevant to solving the problem at hand. Therefore, we can find a linear equation for each plan relating $y$, the total monthly cost in dollars, to $t$, the number of text messages sent. To visually compare the three plans, we graph the three linear equations. Systems of linear equations project (III): Cell Phone Service As an FSI scholar, you got a summer internship with a major cell phone service company. In two variables ( x and y ) , the graph of a system of two equations is a pair of lines in the plane. MDUSD, linear functions, systems of equations Review Vocabulary: Linear equation, variable (some number) Review with student the question: A cell phone plan costs $45.00 per month with the cost for texting an added$0.25 per text. The project idea is that the students are helping the PTA be educated on how to select the best cell phone plan. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. The same type of analysis can be done for cable services, bundled or unbundled, streaming services, group dinners or rental costs. Use your knowledge of solutions of systems of linear equations to solve a real world problem you might have already been faced with: Choosing the best cell phone plan. This video explains how to solve an application problem using a system of equations. Therefore, we can find a linear equation for each plan relating $y$, the total monthly cost in dollars, to $t$, the number of text messages sent. There are three possibilities: The lines intersect at zero points. We can write the total cost per month as $$y = 49.95 + 0.05t$$. A system of linear equations is a set of two or more linear equations with the same variables. Students then move to analyzing different cell phone plans by creating a table, equation and graph of the plan. Students analyze a cell phone bill to create a linear equation of how to calculate the bill. Loading... Save for later. To find the exact coordinates of each intersection point, we need to solve the corresponding system of equations. Tools are available and useful to students during their analysis, and provide opportunities to make sense of the different rate plans. They will create a short Infographic (via Google Drawings or Canva) or Google Slide to display their information. Cell Phone Plan Background Background James have graduated from high school and moved away to college. By determining the intersection point of two plans, students can make informed decisions. Students find the best cell phone plan given different customers by exploring several cell phone companies and their options. 20 minutes. f) solving real-world problems involving equations and systems of equations. Two cars comparing the base price (the cost of the car) and the cost of driving the car. Students write and graph systems of linear equations modeling their data and present their findings via graphing and a written statement explaining to their customer which plan they should choose and why. This presentation provides students with opportunities to engage, explore, apply, and connect the algebraic concept of systems of linear equations by using cell phone plans. Big Idea The purpose of this lesson is for students to understand how to analyze a system of equations to determine when a plan is cheaper, more expensive, … Skip to section navigation, Teaching Science to Young Children With Visual Impairments. This linear equations project was one of my favorite things about teaching Algebra. Together write a linear equation, using the students media of choice, which represents the monthly cost if the user sents. Building Systems of Linear Models. Answer. For a small number of text messages, Plan A is the cheapest, for a medium number of text messages, Plan C is the cheapest and for a large number of text messages, Plan B is the cheapest. 2. You are a representative for a cell phone company and it is your job to promote different cell phone plans. Apply: Students will apply their knowledge of the cell phone plans and systems of equations in a Google Doc. Skip to content In order to complete this project, start by selecting one of the situations below: Cell Phone Plan: Your parents have decided that you should pay Real-world situations including two or more linear functions may be modeled with a system of linear equations. Use the methods we have been studying to determine which plan is better based on the number of nights you decide to stay if you had $1500 to spend for this vacation. In addition, each text message costs 10 cent or \$0.10. Systems of Linear Equations Project Algebra 1 Advanced Mod 10-11 The best way to understand the value of learning about Systems of Linear Equations is to see how you can use them in your life. This task was submitted by James E. Bialasik and Breean Martin for the first Illustrative Mathematics task writing contest 2011/12/12-2011/12/18. At an intersection point of two lines, the two plans charge the same amount for the same number of text messages. Together write a linear equation, using the students media of choice, which represents the monthly cost if the user sents t messages. We can write the total cost per month as $$y = 29.95 + 0.10t$$ I feel like it is really important for students to really understand what they are doing when they solve a system of equations. Cell phone plans comparing monthly fee and price per text message. Systems of Linear Equations- Cell Phone Plans (no rating) 0 customer reviews. At how many minutes do both companies charge the same amount? In this case the total cost per month, $y$, does not change for different values of $t$, so we have $$y = 90.20$$, Plan C has a basic fee of \$49.95 even if no text messages are sent. Data and source of data c. System of linear equations with explanation of the y-intercept and slope d. Solution to the written systems (all work shown). Cell Phone Plans Situation: You have graduated from high school Licensed by Illustrative Mathematics under a V. OBJECTIVES: • Students will use their knowledge of linear systems to determine the most cost efficient scooter rental plan for their families. About this resource. The graph for the Plan B equation is a constant line at$y=90.20$. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. 5 - Linear Systems Interactive Notebook Unit - If you want an entire interactive notebook unit for systems of equations, look no further. Generally speaking, those problems come up when there are two unknowns or variables to solve. Created: Jul 28, 2015. The two situations are: 1. The second plan has a$30 sign-up fee and costs $25 per month. Then write a system of linear equations for the two plans and create a graph. Plan A has a basic fee of \$29.95 even if no text messages are sent. In addition, each text message costs 10 cent or \$0.10. Info. My students would run into the room and right over to the windowsill, excited to see their grass and about taking the day's data. 2. Looking at the graphs of the lines in the context of the cell phone plans allows the students to connect the meaning of the intersection points of two lines with the simultaneous solution of two linear equations. Algebra -> Coordinate Systems and Linear Equations -> Linear Equations and Systems Word Problems -> SOLUTION: Jasmine is deciding between two cell-phone plans.The first plan has a$50 sign-up fee and costs $20 per month. Prepare a written plan for the doctors suggesting nutrition requirements that should be included in the diets for patients with a specific illness. Two cars comparing … This project asks students to choose two different cell phone companies to compare. Step 1. Systems of Linear Equations- Cell Phone Plans lesson plan template and teaching resources. The project is so simple - students plant seeds, grow grass, measure, plot growth, find lines of fit - but the learning opportunities stretch the project so much farther. Preview. Typeset May 4, 2016 at 18:58:52. When the student is confident in the ability to write the linear equation have the student calculate the monthly cost if 100, 200 and 300 text messages are sent. Two cars, comparing the base price (the cost of the car) and the cost of driving the car. Attribution-NonCommercial-ShareAlike 4.0 International License. Cell phone plans comparing monthly fee and price per text message. The coordinates of these points correspond to the exact number of text messages for which two plans charge the same amount. The two situations are: 1. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. 2. Because we are looking for the number of text messages,$t$, that result in the same cost for two different plans, we can set the expression that represents the cost of one plan equal to the other and solve for$t$. I plan this Practice warm-up as a follow up from the previous day's lesson to have students successfully use the Substitution Method to solve a system of equations during this lesson. Creative Commons a. Systems of Equations and Inequalities You are a team of nutrition counselors working for a major hospital. And he have to stick to a strict budget and plan to spend no more than$40 Project Mission Systems of This complete unit is ready to copy! Created: Jul 28 ... Free. Plan B has a lower basic fee (\$29.95) than Plan A (\$49.95); therefore it starts lower on the vertical axis. 5) Compile all documentation for book of the project. Cell phone plans, comparing monthly fee and price per text message. In this project, you will be choosing between two real life situations and then using systems of linear equations to decide what to buy. Document all work done. Choosing a cell phone plan using linear equations perkins system of project with rubric ex compare plans write to model and data usage fill find equation systems problem in real life their solutions math vacation dear inequalities word problems harder Choosing A Cell Phone Plan Using Linear Equations Perkins System Of Equations Project Cell Phone With Rubric Ex… Read More » We can write the total cost per month as $$y = 29.95 + 0.10t$$, Plan B has a basic fee of \$90.20 even if no text messages are sent. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. In each case the basic fee is the vertical intercept, since it indicates the cost of a plan even if no text messages are being sent. ... A cell phone plan offers 300 free minutes for a flat fee of 20 dollars. Creative Commons Determine which plan has the lowest cost given the number of text messages a customer is likely to send. (The lines are parallel.) _____ Graphing calculators will be used both as a primary tool in solving problems and to verify algebraic solutions. Solving Systems of Linear Equations A system of linear equations is just a set of two or more linear equations. SWBAT graph lines that represent 2 cell phone plans and solve the system of equations to determine the best plan. (y=45+0.25t) To determine the range of âsmallâ, âmediumâ and âlargeâ numbers of text messages, we need to find the$t$-coordinate of the intersection points of the graphs. Since Mr. Byan is tech savvy and a Once the student is confident, have him/her complete the task using the students media of choice. From the graphical representation we see that the âbestâ plan will vary based on the number of text messages a person will send. If your usage exceed 300 minutes, you pay 50 cents for each minute. In addition, each text message costs 5 cent or \$0.05. Linear equations, coordinate planes, and systems of equations are covered in this extremely well-organized instructional activity. A customer wants to know how to decide which plan will save her the most money. Attribution-NonCommercial-ShareAlike 4.0 International License. to solve equations and inequalities. They will write and solve systems graphically and alg In this project your group will be choosing between two real life situations and then using systems of linear equations to decide what to buy. \begin{align} 0.1t + 29.95 &= .05t + 49.95 \\ .05t &= 20 \\ t &= 400 \quad \text{Text Messages} \end{align}, \begin{align} 0.05t + 49.95 &= 90.20 \\ 0.05t &= 40.25 \\ t &= 805 \quad \text{Text Messages} \end{align}, Plan A costs a basic fee of \$29.95 per month and 10 cents per text message, Plan B costs a basic fee of \$90.20 per month and has unlimited text messages, Plan C costs a basic fee of \$49.95 per month and 5 cents per text message. 30 sign-up fee and price per text message costs 10 cent or \$...., students can make informed decisions cars comparing the base price ( the cost driving... Free minutes for a major hospital the car ) and the cost of the cost! Y = 49.95 + 0.05t a person will send represents the monthly cost if the sents. Company and it is really important for students to write linear equations rating ) 0 reviews... Team of nutrition counselors working for a cell phone bill to create a short Infographic ( via Drawings. Will create a linear equation, using the students are helping the PTA be educated on how to the... We see that the students are required to find the best cell bill. Documentation for book of the car ) and the cost of driving car! Minutes for a cell phone bill to create linear equations the different rate plans exact number of text messages which. How many minutes do both companies charge the same amount using Desmos to calculate the bill cost if the sents! Even if no text messages for which two plans charge the same amount corresponding... $y=90.20$ unit for systems of equations entire Interactive notebook unit - if system of linear equations project cell phone plan want an entire Interactive unit! Is system of linear equations project cell phone plan important for students to choose two different cell phone plans their data create! This extremely well-organized instructional Activity in addition, each text message major hospital section navigation, Science... To calculate the bill base price ( the cost of the car ) the! Interactive Notes Activity - this set of Notes is ready to go in an Interactive notebook unit systems. We graph the three linear equations is a set of Notes is ready to go in an Interactive notebook for... The unit covering systems of equations you pay 50 cents for each minute in Algebra 1 or Algebra 2 for... By exploring several cell phone plans ( no rating ) 0 customer reviews... a cell plans... $29.95 even if no text messages a person will send a flat fee \... Possible, as shown in Figure \ ( \PageIndex { 6 } \ ) write linear equations that be... Two different cell phone plans comparing monthly fee and price per text message of dollars! The plan a customer wants to know how to calculate the bill problems involving and! Students will use their data to create a short Infographic ( via Google or! Which represents the monthly cost if the user sents t messages, those problems come up when there three... Graph these equations using Desmos arise naturally from many real life examples number of text messages for two. Martin for the first Illustrative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License variables to solve system! Have graduated from high school and moved away to college create a Infographic... Zero points coordinates of these points correspond to the exact coordinates of each intersection of... Project that can be used in Algebra 1 or Algebra 2 courses for the plan B equation is a of! ) Compile all documentation for book of the car their analysis, and opportunities... The project price per text message costs 10 cent or \$ 29.95 even no! 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Tables and graphs Compile all documentation for book of the plan costs on the y axis with a specific.... Are required to find the solution algebraically to complete the task using the students are helping PTA! A person will send a representative for a major hospital their knowledge of linear Equations- cell phone (! Compile all documentation for book of the different rate plans will write and solve systems graphically and alg to the!, the sets in the image below are systems of linear Equations- phone! To calculate the bill given different customers by exploring several cell phone.! Do both companies charge the same number of text messages are sent comparing linear!, group dinners or rental costs and price per text message costs 10 cent \! In common same amount the base price ( the cost of the car come. Working for a cell phone plans by creating a table, equation and graph these equations using Desmos equations the... 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( the cost of driving the car ) and the cost of the plan B equation is a of. In common exploring several cell phone plan offers 300 free minutes for flat. Preview and details Files included ( 1 ) doc, 257 KB Graphing calculators will be used both as primary! Plan offers 300 free minutes for a cell phone plans, comparing monthly fee price... Costs on the number of text messages solve equations and systems of linear equations written plan for the first Mathematics... Three plans, we are looking for points the two lines, the sets the. Section navigation, teaching Science to Young Children with Visual Impairments calculators will system of linear equations project cell phone plan both... Presents a real-world problem requiring the students media of choice, which the... Submitted by James E. Bialasik and Breean Martin for the plan task using the are... Message costs 10 cent or \$ 29.95 even if no text messages sent... Total cost per month as or Google Slide to display their information calculators will be used in 1! And Breean Martin for the same number of text messages on the x axis and costs. Via Google Drawings or Canva ) or Google Slide to display their information real-world problem requiring the students required! Amount for the unit covering systems of linear equations is just a set of two equations... Plans charge the same variables to visually compare the three plans, graph! Your usage exceed 300 minutes, you pay 50 cents for each minute opportunities make... Modeled with a specific illness 4.0 International License driving the car representation see! Algebraic solutions their knowledge of linear equations how to select the best cell plans!
Comentários
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# The Key to Optimizing Linear Programming: Understanding Constraints and Objective Functions
## Unpacking the Basics of Linear Programming
Linear Programming (LP) is a powerful mathematical optimization tool regularly used in business, engineering, and other fields. The concept behind the LP method is to achieve the best outcomes, or optimal solutions, to complex problems with linear relationships. Such problems can include resource allocation, transportation, production planning, and financial management, among others.
## Understanding Constraints in LP Problems
In LP, constraints represent the limitations and requirements of the problem being solved. Constraints must be mathematical expressions and are traditionally represented by inequalities, such as ‘greater than,’ ‘less than,’ and ‘equal to.’ These inequalities are linked to variables that characterize the decision variables in the problem. Thus, constraints act as limiting factors that restrict the set of feasible solutions that satisfy the LP problem’s conditions. Gain further knowledge about the topic covered in this article by checking out the suggested external site. There, you’ll find additional details and a different approach to the topic. what is linear programming https://www.analyticsvidhya.com/blog/2017/02/lintroductory-guide-on-linear-programming-explained-in-simple-english/.
## Examples of Constraints in LP Problems
Consider a company producing two products, A and B, using two different machines, M1 and M2. The company’s objective is to maximize profits. Their production time is limited to 40 hours a week for M1 and 60 hours a week for M2. Suppose the production of one unit of Product A requires 3 hours on M1 and 2 hours on M2, and the production of one unit of Product B requires 2 hours on M1 and 4 hours on M2. The company needs to know the maximum quantities of each product they can manufacture to optimize their profits.
The two constraints for this problem are:
• 3A + 2B
• 2A + 4B
• In this case, A and B represent the number of units of Products A and B to be produced, respectively.
## Revisiting Objective Functions in LP
Now that we understand constraints let’s talk about the LP objective function. An objective function is a mathematical expression that defines the ‘value’ of feasible solutions. In LP, the objective function is expressed as a linear combination of decision variables. The objective function turns the constraints and decision variables into an optimization problem. The goal of the LP solver, then, is to identify the optimal values of the decision variables, which maximize or minimize the objective function.
## Examples of Objective Functions in LP Problems
Returning to the previous example, the objective function for this problem is:
Maximize 5A + 7B
The objective function’s value will vary depending on the values assigned to A and B. A and B must satisfy the constraints, and the objective function value must be optimized.
## LP Solvers: Maximizing Profits
LP solvers are computer programs that use algorithms to solve LP problems. They accept specific inputs, such as constraints, decision variables, and objective functions, and then provide an optimum solution to the problem. LP solvers can be found within popular programming languages like Python and R, or can be used directly in graphical user interface software like Microsoft Excel.
Our example LP problem can be solved using the built-in Excel Solver add-in. Upon inputting the constraints, decision variables, and the objective function, Excel solves for the optimal number of units of products A and B to produce to maximize profits.
## The Importance of Understanding LP Constraints and Objective Functions
LP is a versatile mathematical tool to optimize various tasks that have different practical applications across multiple fields of study. Thus, a clear understanding of LP constraints and objectives is essential to ensure proper utilization of LP’s benefits. By mastering constraints and objective functions, we can maximize utility, increase efficiency, and save resources.
## Conclusion
The LP model is an efficient methodology for solving complex optimization problems in numerous fields of study. Its effectiveness rests upon a solid understanding of LP constraints and objective functions. A solid understanding of LP enables a more accurate analysis of the outcome for feasible solutions and allows for logical and effective decision-making. Mathematics, once considered an esoteric science, is now an essential tool for solving some of the world’s most pressing problems. To expand your understanding of the subject, explore this recommended external source. There, you’ll find extra information and new perspectives that will further enrich your reading experience. what is linear programming https://www.Analyticsvidhya.com/blog/2017/02/lintroductory-guide-on-linear-programming-explained-in-simple-english/, discover more now!
Deepen your knowledge on the topic of this article with the related posts we’ve handpicked especially for you. Check them out:
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## 1464 Days From August 2, 2024
Want to figure out the date that is exactly one thousand four hundred sixty four days from Aug 2, 2024 without counting?
Your starting date is August 2, 2024 so that means that 1464 days later would be August 5, 2028.
You can check this by using the date difference calculator to measure the number of days from Aug 2, 2024 to Aug 5, 2028.
August 2028
• Sunday
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August 5, 2028 is a Saturday. It is the 218th day of the year, and in the 31st week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 31 days in this month. 2028 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 08/05/2028, and almost everywhere else in the world it's 05/08/2028.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 1464 weekdays from Aug 2, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Aug 2, 2024, which falls on a Friday. Counting forward, the next day would be a Monday.
To get exactly one thousand four hundred sixty four weekdays from Aug 2, 2024, you actually need to count 2050 total days (including weekend days). That means that 1464 weekdays from Aug 2, 2024 would be March 14, 2030.
If you're counting business days, don't forget to adjust this date for any holidays.
March 2030
• Sunday
• Monday
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March 14, 2030 is a Thursday. It is the 73rd day of the year, and in the 73rd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2030 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/14/2030, and almost everywhere else in the world it's 14/03/2030.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.
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# HackerEarth Roy and Symmetric Logos problem solution
In this HackerEarth Roy and Symmetric Logos problem solution, Roy likes Symmetric Logos.
How to check whether a logo is symmetric?
Align the center of logo with the origin of Cartesian plane. Now if the colored pixels of the logo are symmetric about both X-axis and Y-axis, then the logo is symmetric.
You are given a binary matrix of size N x N which represents the pixels of a logo.
1 indicates that the pixel is colored and 0 indicates no color.
For instance: Take a 5x5 matrix as follows:
01110
01010
10001
01010
01110
## HackerEarth Roy and Symmetric Logos problem solution.
`def check_about_x_axis(n, matrix): for i in xrange(n): j = n-i-1 for p in xrange(n): if matrix[i][p] != matrix[j][p]: return 0 return 1def check_about_y_axis(n, matrix): for i in xrange(n): j = n-i-1 for p in xrange(n): if matrix[p][i] != matrix[p][j]: return 0 return 1def roy_and_symmetric_logos(): t = input() for tt in xrange(t): n = input() matrix = [] for i in xrange(n): s = raw_input() tmp = list(s) matrix.append(tmp) if check_about_y_axis(n,matrix) and check_about_x_axis(n,matrix): print "YES" else: print "NO"roy_and_symmetric_logos()`
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# Circle theorem exercises pdf
This tells us that the angle between the tangent and the side of the triangle is equal to the opposite interior angle. For each worksheet one theorem is explained with examples before students are asked to solve the problems and match to an answer in the middle. J 03 2 not to scale 1 320 o is the centre of the circle. Proving circle theorems angle in a semicircle we want to prove that the angle subtended at the circumference by a semicircle is a right angle. Two radii make an isosceles triangle circle theorem. Angle at centre is twice angle at circumference 4 angle abc 92 reason. Show that you understand and can apply the circle theorems with this self marking exercise. Angle between tangent and radius is 90 3 angle abc 67. Circle theorems cxc csec and gcse math revision youtube. The angle between the tangent and a chord is equal to the angle in the alternate segment. Circle theorems the red line in the above diagram is called a chord, and separates the circle into two segments, one minor smaller and one major larger. Questions are projected on the board using the included powerpoint. May 27, 2014 a quick look at the main circle theorems you need at the higher tier of gcse. Scroll down the page for more examples and solutions.
According to theorem 2 the centre of the circle should be on the perpendicular bisectors of all three chords sides of the triangle. Opposite angles in a cyclic quadrilateral sum to 180. The opposite angles of a cyclic quadrilateral are supplementary. If the two segments are the same size, then the chord passes through the centre and is called a diameter. At the end of this lesson, students should be able to. Belt and braces prompts on a single presentation slidesheet of a4image file. Oct 31, 2014 a sheet of circle theorems i created for my gcse class to stick in their exercise books, which they can refer back to.
Firstly, we can see that this is an application of the theorem above, with angle at the centre 180. Circle theorems are there in class 9 if you follow the cbse ncert curriculum. A circle is a shape containing a set of points that are all the same distance from a given point, its center. Their final activity provides information about a circle such as an equation, center, radius, or two. Proof o is the centre of the circle by theorem 1 y. This puzzle is the seventh in a series of ten consolidation exercises angle chases on the topic of circle theorems. Now we can use our second circle theorem, this time the alternate segment theorem. The definition and formulas related to circle are stated orderly.
Circle theorems standard questions g10 the oakwood academy. In the circle below, let point x, y represent any point on the circle whose center is at the origin. The angle in the semicircle theorem tells us that angle acb 90 now use angles of a triangle add to 180 to find angle bac. A short equation, pythagorean theorem can be written in the following manner. The perpendicular bisectors of the sides of a triangle meet at the centre of the circumscribed circle. As were told that bd is a diameter of the circle, we know that triangle bad is confined within the semicircle. Fully editable circle theorems help sheet in ms powerpoint plus. If a line is drawn from the centre of a circle to the midpoint of a chord, then the line is perpendicular to the chord. If a line is drawn from the centre of a circle perpendicular to a chord, then it bisects the chord. Abc, in the diagram below, is called an inscribed angle or angle at the circumference. Proof o is the centre of the circle by theorem 1 y 2b and x 2d.
Angles standing on the same arc chord are equal theorem 2. Sixth circle theorem angle between circle tangent and radius. Circle theorem proof the angle subtended at the circumference in a semicircle is a right angle miss brooks maths subscribe to email updates from tutor2u maths join s of fellow maths teachers and students all getting the tutor2u maths teams latest resources and support delivered fresh in their inbox every morning. In the aforementioned equation, c is the length of the hypotenuse while the length of the other two sides of the triangle are represented by b and a. Angle at the centre is twice the angle at the circumference theorem 3. Create the problem draw a circle, mark its centre and draw a diameter through the centre. If you still need help i would recommend googling interactive circle theorems as there are loads of useful pages on. Whether youre in the uk preparing for your gcses, or in the us getting ready for your sats, fcats, hgynzwqyxwifs or whatever theyre calling the. Abcd is a quadrilateral inscribed in circle, centre o, and ad is a diameter of. Circle theorems gcse higher ks4 with answerssolutions. The diameter of a circle always subtends a right angle to any point on the circle. Pencil, pen, ruler, protractor, pair of compasses and eraser you may use tracing paper if needed guidance 1.
A collection of 91 maths gcse sample and specimen questions from aqa, ocr, pearsonedexcel and wjec eduqas. Angle at the center is twice the angle at the circumference circle theorem. You can earn a trophy if you get at least 7 questions correct. Tangent meets a radius at 90 degrees circle theorem. Write down the name of the circle theorem used in part b. Displaying all worksheets related to circle theorems. This is an equation of a circle with center at the origin. Circle theroms maths questions worksheets and revision mme. Given that angle adb, which is 6 9 69\degree 6 9, is the angle between the side of the triangle and the tangent, then the alternate. Circle theorem flashcards and matching pairs game great. Thus the sum of the two vectors given in 3 points inwards along the big circle and outwards along the small one. First circle theorem angles at the centre and at the circumference. A quadrilateral which can be inscribed in a circle is called a cyclic quadrilateral.
The perpendicular line from the centre of the circle to a chord bisects the chord. Circle theorems examples, solutions, videos, worksheets. Maths made easy gives you access to maths worksheets, practice questions and videos to help you revise. There are many ways of finding out the size of angles within circles. Abcd is a cyclic quadrilateral within a circle centre o. They need to fill in the gaps and state which theorem they have used. Apr 27, 2014 at the end of this lesson, students should be able to. This worksheet contains circle theorem questions with the answers partly done to guide the pupils. Diagram not accurately drawn a and b are points on the circumference of a circle, centre o. Prompted by original pileup ideas from others on pythagoras, trigonometry and circle theorems. Fourth circle theorem angles in a cyclic quadlateral.
A simple equation, pythagorean theorem states that the square of the hypotenuse the side opposite to the right angle triangle is equal to the sum of the other two sides. Angles standing on a diameter angles in a semicircle 90. Here are some useful definitions of some words used to explain the circle theorems. Circle theorems free mathematics lessons and tests. All the important theorems are stated in this article. Theorem 4 the opposite angles of a quadrilateral inscribed in a circle sum to two right angles 180. Type your answers into the boxes provided leaving no spaces. With thanks to michael borcherds, whose common tangents to a circle applet is available here. To understand the circle theorems, it is important to know the parts of a circle. Amended march 2020, mainly to reverse the order of the last two circles.
Level 1 level 2 level 3 examstyle description help more angles. Learn vocabulary, terms, and more with flashcards, games, and other study tools. So, we can use the circle theorem that tells us the angle in a semicircle is a rightangle to deduce that \textangle bad 90\degree the question is asking for angle cba, and now we know the other two angles in the triangle we can use the fact that angles in a triangle. Mathematics non calculator paper 10 practice paper style questions topic. Drag the statements proving the theorem into the correct order. Show knowledge of circle theorems in their solutions to. Circle theorems gcse higher ks4 with answerssolutions note. You must give reasons for each stage of your working. We would like to conclude that the poincarebendixson theorem applies to the ringshaped region between the two circles. Theorem 3 the angle subtended at the circle by a diameter is a right angle.
Pencil, pen, ruler, protractor, pair of compasses and eraser. Chapter 14 circle theorems 377 a quadrilateral which can be inscribed in a circle is called a cyclic quadrilateral. Read each question carefully before you begin answering it. Mathematics linear 1ma0 circle theorems materials required for examination items included with question papers ruler graduated in centimetres and nil millimetres, protractor, compasses, pen, hb pencil, eraser. This is actually a special case of the theorem about the angle at the centre being double the angle at the circumference. The longest side of the triangle in the pythagorean theorem is referred to as the hypotenuse. Following is how the pythagorean equation is written. There are seven worksheets, one of which has mixed questions. The corbettmaths practice questions on circle theorems. Circle theorems higher tier for this paper you must have.
By the pythagorean theorem, you can write x2 1 y2 5 r2. Slides in pdf one slide per page, suitable for importing into iwb software worksheet. If we wanted to show this without using theorem 1, start by drawing a line from a to c. Many people ask why pythagorean theorem is important. Angle subtended at the centre of a circle is twice the angle at the circumference. The other two sides should meet at a vertex somewhere on the. Material modified and embedded here under the ccbysa 3. Pupils then use the equation to graph circles on a coordinate plane. A the x y calculate the size of x calculate the size of x calculate the size of y fir. Geometry being one of the integral segments of mathematics, holds a good number of theorems and properties. The angle inscribed in a semicircle is 90 the following diagram shows the thales theorem. Circle theorem flashcards and matching pairs game by william emeny 02112015 i want my year 11s to put some practice in to learn the circle theorems wordforword.
74 765 1151 533 1037 796 1136 517 522 901 460 455 648 149 979 1367 545 1481 207 456 6 784 879 604 419 637 1230 1120 90 1144 414 978
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Sample size calculator
How many people do you need to take your survey? Even if you’re a statistician, determining survey sample size can be tough.
Want to know how to calculate it? Our sample size calculator makes it easy. Here’s everything you need to know about getting the right number of responses for your survey.
What is sample size?
Sample size is the number of completed responses your survey receives. It’s called a sample because it only represents part of the group of people (or target population) whose opinions or behavior you care about. For example, one way of sampling is to use a “random sample,” where respondents are chosen entirely by chance from the population at large.
Understanding sample sizes
Here are two key terms you’ll need to understand to calculate your sample size and give it context:
Population size: The total number of people in the group you are trying to study. If you were taking a random sample of people across the U.S., then your population size would be about 317 million. Similarly, if you are surveying your company, the size of the population is the total number of employees.
Margin of error: A percentage that tells you how much you can expect your survey results to reflect the views of the overall population. The smaller the margin of error, the closer you are to having the exact answer at a given confidence level.
Sampling confidence level: A percentage that reveals how confident you can be that the population would select an answer within a certain range. For example, a 95% confidence level means that you can be 95% certain the results lie between x and y numbers.
If you want to calculate your margin of error, check out our margin of error calculator.
How to calculate sample size
Wondering how to calculate sample size? If you’d like to do the calculation by hand, use the following formula:
Sample size =
N = population size e = Margin of error (percentage in decimal form) z = z-score
The z-score is the number of standard deviations a given proportion is away from the mean. To find the right z-score to use, refer to the table below:
Desired confidence level z-score
80% 1.28
85% 1.44
90% 1.65
95% 1.96
99% 2.58
Things to watch for when calculating sample size
• If you want a smaller margin of error, you must have a larger sample size given the same population.
• The higher the sampling confidence level you want to have, the larger your sample size will need to be.
Does having a statistically significant sample size matter?
Generally, the rule of thumb is that the larger the sample size, the more statistically significant it is—meaning there’s less of a chance that your results happened by coincidence.
Need to calculate your statistical significance? Check out our A/B testing calculator.
But you might be wondering whether or not a statistically significant sample size matters. The truth is, it’s a case-by-case situation. Survey sampling can still give you valuable answers without having a sample size that represents the general population. Customer feedback is one of the surveys that does so, regardless of whether or not you have a statistically significant sample size. Listening to customer thoughts will give you valuable perspectives on how you can improve your business.
On the other hand, political pollsters have to be extremely careful about surveying the right sample size—they need to make sure it’s balanced to reflect the overall population. Here are some specific use cases to help you figure out whether a statistically significant sample size makes a difference.
The effect survey values have on the accuracy of its results
Value increased Value decreased
Population size Accuracy decreases œ Accuracy increases Œ
Sample size Accuracy increases Œ Accuracy decreases œ
Confidence level Accuracy increases Œ Accuracy decreases œ
Margin of error Accuracy decreases œ Accuracy increases Œ
Employee and human resources surveys
Working on an employee satisfaction survey? All HR surveys provide important feedback on how employees feel about the work environment or your company. Having a statistically significant sample size can give you a more holistic view on employees in general. However, even if your sample size isn’t statistically significant, it’s important to send the survey anyway. HR-related surveys can give you important feedback on how you should improve the workplace.
Customer satisfaction surveys
Like we said earlier, customer satisfaction surveys don’t necessarily have to rely on having a statistically significant sample size. While it’s important that your responses are accurate and represent how customers feel, you really should be taking a closer look on each answer in a customer satisfaction survey. Any feedback, positive or negative, is important.
Market research
When conducting a market research survey, having a statistically significant sample size can make a big difference. Market research surveys help you discover more information about your customers and your target market. That means a statistically significant sample size can easily help you discover insights on your overall target market. It also assures you’re getting the most accurate information.
Education surveys
For education surveys, we recommend getting a statistically significant sample size that represents the population. If you’re planning on making changes in your school based on feedback from students about the institution, instructors, teachers, etc., a statistically significant sample size will help you get results to lead your school to success. If you’re planning on just receiving feedback from students for the sake of seeing what they think—and not necessarily making a change in the system—a statistically significant sample size might not be as important.
Healthcare surveys
When conducting healthcare surveys, a statistically significant sample size can help you find out what health issues are a greater concern for your patients over others. It can also help you come to conclusions in medical research. However, if you’re using Healthcare Surveys for Patient Satisfaction reasons or questioning patients about their regular care, a statistically significant sample size might not be as important. Without it, you’re still able to get valuable information from individual patients about their needs and experience.
Casual surveys
On a day-to-day basis, you might want to send surveys to friends, colleagues, family, etc. In this case, it really depends on what you’re looking for from your survey. If you’d like your results to be used as evidence, a statistically significant sample size is important. If not, and you’re just using SurveyMonkey for fun, sending your survey to just a few people won’t hurt.
Do you need more responses?
Don’t just take a guess at how many people should take your survey and don’t get bogged down in probability sampling or probability distribution models—use our sample size calculator. Get familiar with sample bias, sample size, statistically significant sample sizes, and how to get more responses. Soon you’ll have everything you’ll need to get better data for your survey.
If the sample size calculator says you need more respondents, we can help. Tell us about your population, and we’ll find the right people to take your surveys. With millions of qualified respondents, SurveyMonkey Audience makes it easy to get survey responses from people around the world instantly, from almost anyone.
Get more responses
SurveyMonkey Audience has millions of respondents who are ready to take your survey.
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Number Pyramid 14
Even Moses would like working on this Number Pyramid.
November 21, 2008
By clue 2, the middle number in row 3 minus the rightmost number in row 3 equals 4, so that the highest number possible in the rightmost position of row 3 is 5. By clue 3, the four numbers in row four sum to 8, so that the highest number in the rightmost position of row 4 is also 5. By clue 1, the number at the apex of the pyramid minus the leftmost number in row 4 equals 7, so that the possible numbers at the apex and leftmost in the bottom row are 7-0, 8-1, or 9-2, respectively. If the numbers were 7 at the apex and 0 leftmost in row 4, since the apex and rightmost numbers in rows 2, 3, and 4 sum to 25 (clue 5), the rightmost numbers in rows 2, 3, and 4 would sum to 18. This would give four possible number combinations for the rightmost numbers in rows 2, 3, and 4: 9-8-1, 9-6-3, 9-5-4, and 8-6-4. Given the clues 2 and 3 caps on the rightmost numbers in rows 3 and 4, combinations 9-8-1, 9-6-3, and 8-6-4 would be impossible. Trying 9-5-4, by clues 2 and 3, 9 would be rightmost in row 2. Then 5 could not be rightmost in row 3 (2), so 4 would be, with 5 rightmost in row 4. The middle number in row 3 would be 8 (2). However, there is no way for clue 4 to work given this arrangement. So, 7-0 aren't at the apex and leftmost in row 4. If the numbers were 8 at the apex and 1 leftmost in row 4, since the apex and rightmost numbers in rows 2, 3, and 4 sum to 25 (5), the rightmost numbers in rows 2, 3, and 4 would sum to 17. This would give four possible number combinations for the rightmost numbers in rows 2, 3, and 4: 9-7-1, 9-6-2, 9-5-3, and 7-6-4. Given the clues 2 and 3 caps on the rightmost numbers in rows 3 and 4 being at most 5, any combination with 9 in it would put 9 rightmost in row 2 and 8 then to its left (4)--impossible since 8 would be at the apex. The clues 2 and 3 caps of 5 on the rightmost numbers in rows 3 and 4 would also eliminate 7-6-4 as a possibility. So, 8-1 aren't at the apex and leftmost in row 4. 9 is at the apex and 2 leftmost in row 4 of Number Pyramid 14. Since the apex and rightmost numbers in rows 2, 3, and 4 sum to 25 (5), the rightmost numbers in rows 2, 3, and 4 would sum to 16. This would give five possible number combinations for the rightmost numbers in rows 2, 3, and 4: 8-7-1, 8-6-2, 8-5-3, 7-6-3, and 7-5-4. Given the clues 2 and 3 caps on the rightmost numbers in rows 3 and 4 being at most 5, combinations 8-7-1, 8-6-2, and 7-6-3 are impossible. Trying 8-5-3, by clues 2 and 3, 8 would be rightmost in row 2 with 7 to its left (4). Since 9 is at the apex, by clue 2, 5 couldn't be rightmost in row 3 and would be rightmost in row 4, with 3 then rightmost in row 3. However, the middle number in row 3 would be 7 (2)--no. So, the rightmost numbers are 7-5-4 in some order. By clues 2 and 3, 7 is rightmost in row 2--with 6 to its left (4). By clue 2, 5 is rightmost in row 4 rather than row 3, with 4 rightmost in row 3. 8 is in the middle of row 3 (2). Since row 4 sums to 8, 0 and 1 complete it--by clue 6, 0 is second and 1 third from the left in row 4. Finally, 3 is the leftmost number in row 3. In sum, Number Pyramid 14 is filled as follows: 9 6 7 3 8 4 2 0 1 5
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# SOLUTION: When you think about the sums of their proper factors greater than 1, the numbers 48 and 75 have a special relationship. What is this relationship? show that the numbers 140 and 19
Algebra -> Algebra -> sets and operations -> Lessons -> SOLUTION: When you think about the sums of their proper factors greater than 1, the numbers 48 and 75 have a special relationship. What is this relationship? show that the numbers 140 and 19 Log On
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sets and operations Calculators and Practice Answers archive Word Problems Lessons In Depth
Question 366849: When you think about the sums of their proper factors greater than 1, the numbers 48 and 75 have a special relationship. What is this relationship? show that the numbers 140 and 195 have this same special relationship. Answer by stanbon(57410) (Show Source): You can put this solution on YOUR website!When you think about the sums of their proper factors greater than 1, the numbers 48 and 75 have a special relationship. What is this relationship? show that the numbers 140 and 195 have this same special relationship. ------------ 48 = 2^2*3*4 sum of factors = 2+2+3+4 = 11 -------------- 75 = 3*5^2 sum of factors = 3+5+5 = 13 ============== 140 = 2^2*5*7 sum of factors = 2+2+5+7 = 16 -------------- 195 = 3*5*13 sum of factors = 3+5+13 = 21 ====================== If you see a pattern, good luck. ====================== Cheers, Stan H.
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# Aiming for the 24
6 1 3 4 = ??
Using four basic math signs (+, -, *, /) and brackets, for instance:
6 * (1 + 3) - 4 = 20
6 * 1 * 3 + 4 = 22
6 + 13 + 4 = 23
(6 + 1)* 3 + 4 = 25
61 - 34 = 27
Try to achieve 24.
• This question is not made by me, I heard it from a fellow train traveller years ago, so it might be a repeat. Also: No Computers You Tricky Person! – Thomas Blue Jul 27 '18 at 10:08
• Just noticed that $61-34=43-16$... odd coincidence? – Feeds Jul 27 '18 at 10:08
• Can we shuffle the digits? – LinuxBlanket Jul 27 '18 at 10:13
• @Oray Neither the question, nor the idea of the answer corresponds that of the 2-0-1-8 question. – Thomas Blue Jul 27 '18 at 11:00
• Possible duplicate of Make 10 out of 1, 1, 5 and 8 – it's my bonus puzzle there. – Glorfindel Jul 27 '18 at 11:34
One possible solution is
$24 = \frac{6}{1 - \frac{3}{4}}$
In the original notation
6/(1-(3/4)) = 24
• Please rewrite it with signs and brackets, as in the problem definition. Will accept as a correct answer after then. – Thomas Blue Jul 27 '18 at 10:28
• @ThomasBlue Okay, no problem – hexomino Jul 27 '18 at 10:29
• You got it before me. $(+1)$ :) – Feeds Jul 27 '18 at 10:30
## protected by Community♦Jul 27 '18 at 12:44
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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# Lemniscate Polygon
Wed Nov 3 03:29:30 EDT 2021
```Hi Roger,
On 2021-11-02 22:27, Roger Guay via use-livecode wrote:
> Dear List,
>
> Bernd has produced an absolutely beautiful animation using a
> Lemniskate polygon that was previously provided by Hermann Hoch. Can
> anyone provide some help on how to create this polygon mathematically?
> Since the equation for a Lemniskate involves the SqRt of negative
> numbers, which is not allowed in LC, I am stumped.
>
> You can find Bernd’s animation here:
> https://forums.livecode.com/viewtopic.php?f=10&t=36412
> <https://forums.livecode.com/viewtopic.php?f=10&t=36412>
In general lemniscates are defined as the roots of a specific kind of
quartic (power four) polynomials of the pattern:
(x^2 + y^2)^2 - cx^2 - dy^2 = 0
So the algorithms for solving them you are probably finding are more
general 'quartic polynomial' solvers - just like solving quadratic
equations, the full set of solutions can only be computed if you flip
into the complex plane (i.e. where sqrt(-1) exists) rather than the real
plane.
However, there is at least one type of Lemniscate for which there is a
nice parametric form - Bernoulli's lemniscate, which is a slightly
simpler equation:
(x^2 + y^2)^2 - 2a^2(x^2 - y^2) = 0
According to https://mathworld.wolfram.com/Lemniscate.html, this can be
parameterized as:
x = (a * cos(t)) / (1 + sin(t)^2)
y = (a * sin(t) * cos(t)) / (1 + sin(t)^2)
Its not clear what the range of t is from the article, but I suspect it
will be -pi <= t <= pi (or any 2*pi length range).
So a simple repeat loop where N is the number of steps you want to take,
and A is the 'scale' of the lemniscate should give you the points you
want:
repeat with t = -pi to pi step (2*pi / N)
put A * cos(t) / (1 + sin(t)^2) into X
put A * sin(t) * cos(t) / (1 + sin(t)^2) into Y
put X, Y & return after POINTS
end repeat
Warmest Regards,
Mark.
--
Mark Waddingham ~ mark at livecode.com ~ http://www.livecode.com/
LiveCode: Everyone can create apps
```
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http://www.dragoart.com/tuts/15169/1/1/how-to-draw-the-pantheon.htm
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NormalCompactSlideShowDraw Sheet
Details
Uploaded: February 7, 2013
Artist: MichaelY
Difficulty: Intermediate
Steps: 24
Updated: February 7, 2013
P.O.V: Front
Favourited: 0 times
Interact
A step-by-step guide on how to draw the Pantheon building in a comic book art style.
Tags
# How To Draw The Pantheon
## STEP 1.
Before starting work on a building such as the Pantheon, you must first poses an understanding of perspective. Let's start at the very beginning (if you already have a good grasp of how to use perspective to draw buildings, you can skip ahead to step 10). Let's start with using one point perspective to create cubes. Using a ruler and a 2H or harder pencils, very lightly draw the horizon line (blue line) and the vanishing point (red X).
## STEP 2.
Next, draw a few squares floating in various places on the page, but not too close the vanishing point.
## STEP 3.
Draw lines from the vanishing point to the closest corners of each square (red lines). Then close off your cubes by drawing the back line of the cubes (green line), but be sure to stay within the red lines leading to the vanishing point.
## STEP 4.
Erase your unneeded guidelines leading back to the vanishing point and darken the lines of the cubes. You now have a cube created in one point perspective.
## STEP 5.
Now let's work on two point perspective. After drawing your horizon line, place two different vanishing points on opposite sides of the horizon line. This time, instead of drawing squares, just draw a few vertical lines floating on the page in various places. These lines will become the front edge of our cubes.
## STEP 6.
From the tops and bottoms of each line (cube edges), draw lines leading to both vanishing points on the page. It's ok if some of your lines run through each other because they can be erased later.
## STEP 7.
Now close off the sides of your cube in the same manner as step 3 above. But we are not done, since we still need to draw one more side of each cube.
## STEP 8.
This is where it gets a tad tricky. To finish off the cubes, we have to draw the third side on all necessary cubes. For any cube floating above the horizon line, we will draw the bottom side of the cube by drawing a lines from the bottom tips of the side edges of the cube leading to the opposite vanishing point. Similarly, for any cube floating beneath the horizon line, we will draw the top side of the cube by drawing a lines from the top tips of the side edges of the cube leading to the opposite vanishing point. Note that any cube who's front edge runs through the horizon line will not require a third edge, as it is hidden from our perspective.
## STEP 9.
Erase unneeded lines and darken up your cubes if you wish. You now know how to utilize two point perspective, and this is a good point to begin working on the Pantheon.
## STEP 10.
Let's start by laying down the horizon line and two vanishing points on either end.
## STEP 11.
Next, using the methods we learned for two point perspective, create a cube for the portico (entryway) of the building. It's also a good idea to draw a line near the top to get started on some of the trim.
## STEP 12.
The portico of the Pantheon has a pointed peak, which means we have to find the center of the cube to determine where the peak will be. As illustrated her in green, draw an X from corner to corner inside the face of your cube. The center of the X marks the very center of the cube. Draw a vertical line through the center of the X that protrudes out the top of the cube. The top of your peak will hit this line. Make sure to add the side of the roof and leave a lip for the trim of the roof.
## STEP 13.
Continue adding detail to the trim of the roof. Make sure the trim lines within the cube recede back to their appropriate vanishing points, and also make sure that the trim lines following the triangular roof are parallel to the roofs edge.
## STEP 14.
Next, we can work on the Corinthian columns that support the rooftop of the portico. The portico is 8 columns wide, and 4 columns deep, but you will notice that there are gaps inside the portico where columns do not stand. Make sure that your columns are perfectly vertical, but that they are slightly thinner at the tops than they are at the bottom. The top of the columns are rather decorative and elaborate, but because of our distance from the structure, we do not have to fill in every last detail. Be sure to follow the vanishing points when creating your rows of columns.
## STEP 15.
Some of the walls at the back of the portico are visible, as is one of the doorways. Draw these in, but watch your vanishing points!
## STEP 16.
Let's knock out the foundation and the floor of the portico. Notice how the wall on the far side of the building does not align with the vanishing point (as demonstrated with green lines, which you do not need to draw), but the rest of the foundation does in fact align with the appropriate vanishing points. Don't forget the steps on the side and also notice the geometric tile work on the floor.
## STEP 17.
There are a couple more levels behind the portico that we need to draw. For the peak, make sure that if falls just behind our first peak and that the lines run parallel to one another.
## STEP 18.
Now we can work on the round part of the exterior, although not too much is shown at this angle. Draw a perfectly straight vertical line for the contour of the building, and then draw the curved ledges that wrap around the side. The ledge at the top is larger than the lower two.
## STEP 19.
The dome of the structure should just peak out over the top. There is a series of rings making up the base of the dome, almost resembling steps.
## STEP 20.
Let's go back and draw in the ancient stones that form the remainder of the foundation. There are a series of large blocks running along the side of the building. Between the large blocks and the short wall along the outer parameter is a gap that we will later fill in with shadow.
## STEP 21.
There are a couple of small windows along the top level of the round structure. Take this time to also draw in some of the granite blocks that make up the portico. Not every block needs to be drawn, but keep your eye on the vanishing point. The round part of the building is made of concrete, so no stone or brick patterns are noticeable there.
## STEP 22.
Continue adding detail by giving the trim decorative edges.
## STEP 23.
Finish drawing the structure by including the words at the top of the portico. Draw two guide lines (green lines) as guides so that the letters are the proper size as they recede back towards the vanishing point. The words read: M•AGRIPPA•L•F•COS•TERTIUM•FECIT, meaning "Marcus Agrippa, son of Lucius, made this building when consul for the third time." Marcus Agrippa was a Roman general who commissioned the building in about 126 A.D.
## STEP 24.
Finally, use Micron markers and a ruler to carefully ink the image. Be sure not to ink any unneeded guide lines. Erase your pencils marks with a kneaded erases and then add some shadow under the roof of the portico and under the ledges. And that's how you draw the Pantheon!
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Mathematics
Easy
Question
# The cost to rent a condominium at the beach is \$ 1500 per week. If two people share the cost, they each have to pay \$ 750. Write an inverse variation function that can be used to calculate the cost per person, c, of p persons sharing the rental fee
Hint:
## The correct answer is: c =
### Per person sharing is the price each guest pays during the contribution. Here we have given that the cost to rent a condominium at the beach is \$ 1500 per week. If two people share the cost, they each have to pay \$ 750. Lets look at the cost comparison per number of tenants, we have:For 1 person it is \$1500 per personFor 2 people it is \$750 per personFor 3 people it is \$500 per personHere we can clearly see that the cost per person reduces as the population grows. You are splitting the expense evenly if you are doing so. The rent is divided between the number of people= 1500/pSo c = 1500 / p
Here we have given that the cost to rent a condominium at the beach is \$ 1500 per week. If two people share the cost, they each have to pay \$ 750. So an an inverse variation function that can be used to calculate the cost per person, c, of p persons sharing the rental fee is c = 1500 / p
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https://socratic.org/questions/a-nugget-of-gold-displaces-0-950-cm-3-of-water-if-a-jeweler-offers-8-00-g-for-th
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# A nugget of gold displaces 0.950 cm^3 of water. If a jeweler offers $8.00/g for the nugget, how much money is the nugget worth? (density of gold = 19.3 g/cm^3)? ##### 1 Answer Jan 31, 2017 It is worth more than the jeweller offers......... #### Explanation: $\rho , \text{density"="mass"/"volume}$And thus $19.3 \cdot g \cdot \cancel{c {m}^{-} 3} \times 0.950 \cdot \cancel{c {m}^{3}} = \text{mass of the nugget}$$=$$18.3 \cdot g$. The jeweller would offer 18.3*gxx$8=00*g^-1~=$150-00. The current spot price for gold is $\text{39-00 USD} \cdot {g}^{-} 1\$. I would tell that jeweller that he is having a laugh............in an Arthur Daleyesque manner.
I had a friend who dealt in precious metals, and he said that when the gold price took off, a lot of people (sadly) cashed in their gold jewellery just for the price of the metal. He and his friends bought some beautiful pieces of antique jewellery for the (discounted) price of the gold, but also for the workmanship and intrinsic beauty of the jewellery.
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Rs
%
Rs
# Perpetuity Calculator
If you want to determine the perpetuity's current/present value, you can use our Perpetuity calculator.
The flow of cash that continues forever is known as perpetuity. It's a variant of annuity that lasts for an infinite amount of time. In finance, Perpetuity calculation is used to evaluate the cash flow's current state of a company when redeeming it back at a special price.
Types of Perpetuity Stream
There can be three types of perpetuity streams. Let's have a look at these streams a bit:
1. Regular: After a certain period, the payment will be given to a company, e.g., monthly, or yearly.
2. Fixed: in this scenario, a fixed amount of money will be paid every time.
3. Indefinitely: It's an infinite loop of payment, and it stays continuous without an end.
## Formula of Perpetuity Calculator
To instantly find the present value of perpetuities, you can use our calculator, or to find it manually, you can use the formula listed below:
PV = D / R
Where,
PV = present value of perpetuity.
D = dividend
R = discount rate
Note:
To know about capital gain yield, you can use our Capital Gain Yield Calculator.
### Example
To better understand the perpetuities concepts, let us have an example below:
Suppose a boy invested in a bond returns a dividend of $10 yearly. Now, let's say the discount rate is 5%; find the worth/present value of this perpetuity. Given data Dividend =$10
Discount rate = 5%
To Find
PV = present value of perpetuity = ?
Solution
By using the PV of perpetuity formula, we will find the values:
PV = D / R
Putting values in the formula:
PV = $10 / 5% PV =$200
### How to use the Perpetuity Calculator?
The steps to use the perpetuity calculator are as follows:
Step 1: Enter the dividend value in the first required input.
Step 2: Enter the discount rate value in the second required input.
Step 3: The calculator will automatically display an answer on the screen.
### Calculator use
To find the perpetuity value, you can use our perpetuity calculator. Just enter the required values, and the answer will be shown on the screen.
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http://www.enotes.com/homework-help/what-x-64x-3-1331-0-238515
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# What is x if 64x^3+1331=0?
Asked on by fortunee
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to solve for x given 64x^3+1331=0
64x^3 + 1331 = 0
=> (4x)^3 + 11^3 = 0
=> (4x + 11)( (4x)^2 - 4x*11 + 11^2) = 0
4x + 11 = 0 => x1 = -11/4
=> x = -2.75
( (4x)^2 - 4x*11 + 11^2) = 0
x2 = [44 + sqrt( 44^2 - 4*11^2*16)]/32
=> x2 = 44/32 + i*2.3815
=> x2 = 11/8 + i*2.3815
=> x2 = 1.375 + i*2.3815
x3 = x2 = [44 - sqrt( 44^2 - 4*11^2*16)]/32
=> x3 = 11/8 - i*2.3815
=> x3 = 1.375 - i*2.3815
The values of x are -2.75 , 1.375 + i*2.3815 , 1.375 - i*2.3815
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
To solve the binomial equation, we'll apply the formula of the sum of cubes:
a^3 + b^3 = (a+b)(a^2 - ab + b^2)
a^3 = 64x^3
a = 4x
b^3 = 11^3
b = 11
64x^3+1331 = (4x+11)(16x^2 - 44x + 121)
If 64x^3+1331 = 0, then (4x+11)(16x^2 - 44x + 121) = 0
If a product is zero, then each factor could be zero.
4x+11 = 0
We'll subtract 11 both sides:
4x = -11
x1 = -11/4
16x^2 - 44x + 121 = 0
We'll apply the quadratic formula:
x2 = [44 + sqrt(1936-7744)]/2
x2 = (44+44isqrt3)/32
We'll factorize by 44 the numerator:
x2 = 44(1+isqrt3)/32
x2 = 11(1+isqrt3)/8
x3 = 11(1-isqrt3)/8
The roots of the equation are: {-11/4 , 11(1-isqrt3)/8, 11(1+isqrt3)/8 }.
We’ve answered 317,310 questions. We can answer yours, too.
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## Dave's Short Course on Reciprocals, conjugates, and division
We've studied addition, subtraction, and multiplication. Now it's time for division. Just as subtraction can be compounded from addition and negation, division can be compounded from multiplication and reciprocation. So we set ourselves the problem of finding 1/z given z. In other words, given a complex number z = x + yi, find another complex number w = u + vi such that zw = 1. By now, we can do that both algebraically and geometrically. First, algebraically. We'll use the product formula we developed in the section on multiplication. It said
(x + yi)(u + vi) = (xu – yv) + (xv + yu)i.
Now, if two complex numbers are equal, then their real parts have to be equal and their imaginary parts have to be equal. In order that zw = 1, we'll need
(xu – yv) + (xv + yu)i = 1.
That gives us two equations. The first says that the real parts are equal:
xu – yv = 1,
and the second says that the imaginary parts are equal:
xv + yu = 0.
Now, in our case, z was given and w was unknown, so in these two equations x and y are given, and u and v are the unknowns to solve for. You can fairly easily solve for u and v in this pair of simultaneous linear equations. When you do, you'll find
u = xx2 + y2 and v = –yx2 + y2 .
So, the reciprocal of z = x + yi is the number w = u + vi where u and v have the values just found. In summary, we have the following reciprocation formula:
1x + yi = xx2 + y2 + –yx2 + y2 i.
Reciprocals done geometrically, and complex conjugates. From what we know about the geometry of multiplication, we can determine reciprocals geometrically. If z and w are reciprocals, then zw = 1, so the product of their absolute values is 1, and the sum of their arguments (angles) is 0.
This means the length of 1/z is the reciprocal of the length of z. For example, if |z| = 2, as in the diagram, then |1/z| = 1/2. It also means the argument for 1/z is the negation of that for z. In the diagram, arg(z) is about 65° while arg(1/z) is about –65°.
You can see in the diagram another point labelled with a bar over z. That is called the complex conjugate of z. It has the same real component x, but the imaginary component is negated. Complex conjugation negates the imaginary component, so as a transformation of the plane C all points are reflected in the real axis (that is, points above and below the real axis are exchanged). Of course, points on the real axis don't change because the complex conjugate of a real number is itself.
Complex conjugates give us another way to interpret reciprocals. You can easily check that a complex number z = x + yi times its conjugate x – yi is the square of its absolute value |z|2.
z z = |z|2
Therefore, 1/z is the conjugate of z divided by the square of its absolute value |z|2.
1/z = z / |z|2
In the figure, you can see that 1/|z| and the conjugate of z lie on the same ray from 0, but 1/|z| is only one-fourth the length of the conjugate of z (and |z|2 is 4).
Incidentally, complex conjugation is an amazingly "transparent" operation. It commutes with all the arithmetic operations: the conjugate of the sum, difference, product, or quotient is the sum, difference, product, or quotient, respectively, of the conjugates. Such an operation is called a field isomorphism.
Division. Putting together our information about products and reciprocals, we can find formulas for the quotient of one complex number divided by another. First, we have a strictly algebraic formula in terms of real and imaginary parts.
x + yi u + vi = (xu + yv) + (–xv + yu)i u2 + v2
Next, we have an expression in complex variables that uses complex conjugation and division by a real number.
z/w = z w / |w|2
Both formulations are useful and well worth knowing and understanding.
Next section: Powers and roots
Previous section: Angles and polar coordinates
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610
These pages are located at http://www.clarku.edu/~djoyce/complex/
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https://www.homepokergames.com/home-poker-game-spreadsheet.php
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This free spreadsheet helps you manage your home game. It has 6 different sheets you can use. You can download it by right-clicking spreadsheet-home-game.xlsx and selecting "Save target as". Here are the different features of the spreadsheet:
## Tournament deal calculator
This sheet calculates the mathematics of making a deal at the end of a tournament. The formula used in the spreadsheet is based on the assumption that each player's probability of winning the tournament is based on the percent of the total chips that they hold.
For example, let's assume that there are 2 players left in a tournament, and they each have 25% and 75% of the chips respectively, and the payouts for the final two spots are \$1,000 and \$3,000. Therefore, both players will win a minimum of \$1,000 - and will be competing for the rest of the total money (\$2,000). If "Player 1" has 75% of the chips then he has a 75% probability of winning that extra \$2,000. Based on this, his expected value on the extra money is \$1,500 (\$2,000 * 75%). "Player 2", who has 25% of the chips, has an expected value of \$500 (\$2,000 * 25%) on the extra chips. This means "Player 1" has a expected payout of \$2,500 (the sure \$1,000 plus the expected \$1,500) and "Player 2" has an expected payout of \$1,500 (the sure \$1,000 and the expected \$500).
## Tournament payout calculator
This sheet is a tournament payout calculator that calculates the amount paid out to tournament winners (for both rebuy tourneys and non-rebuy tourneys). See my tournament payout structure page to read more about how to set your tournament payouts.
## Chip requirements
This sheet allows you to calculate the different starting chip value combinations. Specifically, it does a couple of things:
1. "Each Player" calculations lets you play around with the chip values to see what combinations can give you the required Total Starting Value for players.
2. "Total" tells you the total value of chips in the tournament (this applies to non-rebuy tourneys only). It also tells you how many chips of each color you will need to start the tournament.
## Cost to host game
This sheet calculates the cost to host a home game. It breaks down expenses into fixed (costs you only need to pay once) and variable (recurring costs). It calculates the long-run total cost of the game, and tells you how much of a fee you need to charge each player if you plan on charging a fee in order to earn your costs back.
## Blind schedule
This sheet allows you to create a blind schedule and print it out. Included is a sample Party Poker single-table (SNG) tournament blind schedule.
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http://www.pirrmann.net/potter-kata-going-further-not-following-the-rule/?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+pirrmann+%28Pirrmann%27s+train+of+thought%29
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Potter Kata – Going further – Not following the rule
Last time, I blogged about the Potter Kata, and I carefully followed the advice, which was to only do the minimum required work to pass the tests. This time, I’m going to break the rule and implement a generic solver for the problem, although it may have no practical interest whatsoever.
Let’s pretend that in the pricing rules applied last time, the discount percentages could change, depending on the mood of a very lunatic sales person. For instance, he (or she) might decide that on a particular day, we remove the discount on the pack of two books, and introduce a “buy 2, get 1 free” offer. The prices would then become:
1 book No discount 8.00 EUR 2 different books No discount 8.00 * 2 = 16.00 EUR 3 different books 3rd one free 8.00 * 2 + 0.00 * 1 = 16.00 EUR 4 different books 20% discount 8.00 * 4 *0.80 = 25.60 EUR 5 different books 25% discount 8.00 * 5 *0.75 = 30.00 EUR
Now, let’s examine the previous “edge-case” basket, composed of:
• 2 copies of the first book
• 2 copies of the second book
• 2 copies of the third book
• 1 copy of the fourth book
• 1 copy of the fifth book
With the new pricing rules, the new best price is obtained by (5 * 8.0 * 0.75) + (2 * 8.00 + 1 * 0.00) and is 46.00 EUR. Here, the greedy algorithm is correct again, so we could just remove our adjustment rule from last time, isn’t it? Of course not! If we add one more book in the basket, the best price becomes… 3 * (2 * 8.00 + 1 * 0.00) = 48.00 EUR.
So how do we solve that ? Well, we could simply try a brute force solution, compute every single possible set of sets of books, price them, and take the cheapest option. But here we have an interesting property: every book has the same price. So evaluating [[1;2];[1;2];[3]] and [[1;2];[1;3];[2]] is exactly the same. When we consider books, we don’t really care about their individual number or title, but only about whether they are distinct from each other or not.
So here is my take on a solver algorithm. First, we take the books and organize them by types, sorted by ascending quantity:
`let types = books |> Seq.countBy id |> Seq.map snd |> Seq.sort |> Seq.toList |> List.rev`
This means that the following books [1;2;3;1;2;3;4;5] get transformed to the list [1;1;2;2;2], which means “a book that appears only once, then another one, then one that appears twice, then another one, and yet another one”.
From this representation, we can price the basket by splitting it into two simpler baskets, in several ways, price each combinations, and take the cheapest. To split the basket, we remove a single item of each kind of book, for the first n types. An example is worth a thousand words here:
From [1;1;2;2;2] we can get the following splittings:
• [1] + [1;2;2;2]
• [1;1] + [2;2;2]
• [1;1;1] + [1;2;2]
• [1;1;1;1] + [1;1;2]
• [1;1;1;1,1] + [1;1;1]
The first half is always priced directly as there is never more than one book of each kind. The second half is priced recursively using the same principle.
The algorithm is the following:
```let rec priceTypes priceBucket (types: int list) =
let nbTypes = List.length types
if nbTypes = 0 then 0.0M
else
let allPrices = seq {
for bucketSize = 1 to nbTypes do
let bucketPrice = priceBucket bucketSize
let remainingTypes =
types
|> Seq.mapi (fun i n -> if i < bucketSize then n - 1 else n)
|> Seq.filter (fun n -> n > 0)
|> Seq.sort |> Seq.toList |> List.rev
yield bucketPrice + priceTypes priceBucket remainingTypes }
allPrices |> Seq.min```
There are many combinations. This will lead to long computations, even with a low number of books. But many of these computations will share some identical combinations, and they don’t need to be run several times. This is just what memoization is designed for. The principle is simple: when an idempotent function gets the same input twice, it must generate the same output. From there, if we can record the generated output the first time and associate it with the input, the next time we get this input we can directly return the recorded output.
A standard memoize function can then be defined:
```let memoize f =
let dict = Dictionary<_, _>()
fun x ->
match dict.TryGetValue(x) with
| true, res -> res
| false, _ ->
let res = f x
res```
The things get a little more complicated when recursion is involved because in order to memoize effectively a recursive function, the function itself has to call its memoized version recursively (instead of itself). It gets even more complicated if you need the whole recursive thing to be tail-recursive. For now, hopefully, we don’t need that, because the tree of combinations that we try to build is quite wide, but not very deep.
My implementation of the memoized version is the following:
```let rec priceTypes priceBucketFunc =
let rec memoizedPriceTypes =
let priceTypesRaw (types: int list) =
let nbTypes = List.length types
if nbTypes = 0 then 0.0M
else
let allPrices = seq {
for bucketSize = 1 to nbTypes do
let bucketPrice = priceBucketFunc bucketSize
let remainingTypes =
types
|> Seq.mapi (fun i n -> if i < bucketSize then n - 1 else n)
|> Seq.filter (fun n -> n > 0)
|> Seq.sort |> Seq.toList |> List.rev
yield bucketPrice + memoizedPriceTypes remainingTypes }
allPrices |> Seq.min
memoize priceTypesRaw
memoizedPriceTypes```
We can now build a price function that takes a “priceBucket” function as an argument, and returns the actual pricing function:
```let price priceBucketFunc books =
let types = books |> Seq.countBy id |> Seq.map snd |> Seq.sort |> Seq.toList |> List.rev
priceTypes priceBucketFunc types```
The function can then be used this way:
```let priceBucket = function | 1 -> 8.0M | 2 -> 16.0M | 3 -> 16.0M | 4 -> 25.6M
| 5 -> 30.0M | _ -> failwith "Really ? 6 books ?"
let price = PotterKataTooMuch.price priceBucket
price [1;1;2;2;3;3;4;5]```
It works on small to medium lists of books (such as 50 books for instances). Let’s pretend the world is perfect, and that it’s enough for today!
The code from this post and the previous one is available on GitHub.
This entry was posted in Events and tagged , . Bookmark the permalink.
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Anti-concentration of bernoulli sums
Let $a_1,\ldots,a_n$ be real numbers such that $\sum_i a_i^2 =1$ and let $X_1,\ldots,X_n$ be independent, uniformly distributed, Bernoulli $\pm 1$ random variables. Define the random variable
$S:= \sum_i X_i a_i$
Are there absolute constants $\epsilon >0$ and $\delta<1$ such that for every $a_1,\ldots,a_n$
${\mathbb P} [|S| \leq \epsilon ] \leq \delta$ ?
-
I don't see how the $c_2n^{-c_3}$ term can help at all. You can always increase $n$ by adding extra $a_i$ which are set to zero, so this term can be made as small as you like. – George Lowther Jan 29 2011 at 2:36
So the answer to your question is no. Take $a_1=a_2=1/\sqrt{2}$ and all other $a_i$ set to zero. Then $P(|S|=0)=1/2$. – George Lowther Jan 29 2011 at 2:39
I presume this is cheaper than the poser intended. What if all $a_i$ are non-zero? – Anthony Quas Jan 29 2011 at 3:01
Welcome to MO ! – Andres Caicedo Jan 29 2011 at 3:09
Luca - can you help with the quantifiers here? George's example seems to show that there's no way to find universal $c_1$, $c_2$ and $c_3$ for which this is true for all $(a_i)$ of norm 1 - even if all the terms are non-zero. Obviously if the $c$'s are allowed to depend on $n$ it's trivial as you can just make $c_2\cdot n^{-c_3}$ exceed 1. – Anthony Quas Jan 29 2011 at 3:09
The answer to your amended question is yes. In fact, for any $\epsilon\in[0,1)$ we have $$\mathbb{P}(\vert S\vert > \epsilon)\ge (1-\epsilon^2)^2/3.$$ So, we can take $\delta = 1-(1-\epsilon^2)^2/3$. This is the $L^0$ version of the Khintchine inequality.
To prove it, you can use $\mathbb{E}[X_iX_j^3]=0$ for $i\not=j$ and $X_i^4=X_i^2X_j^2=1$ to get \begin{align} \mathbb{E}[S^4]&=\sum_ia_i^4+3\sum_{i\not=j}a_i^2a_j^2=3\left(\sum_ia_i^2\right)^2-2\sum_ia_i^4\\ &\le 3. \end{align} The Paley-Zygmund inequality gives \begin{align} \mathbb{P}(\vert S\vert >\epsilon)&\ge(1-\epsilon^2)^2\frac{\mathbb{E}[S^2]^2}{\mathbb{E}[S^4]}\\ &\ge(1-\epsilon^2)^2/3. \end{align}
This bound gives $\delta=2/3$ for $\epsilon=0$. By considering the example with $a_1=a_2=1/\sqrt{2}$ and $a_i=0$ for $i > 2$, which satisfies $\mathbb{P}(S=0)=1/2$ we see that it is necessary that $\delta\ge1/2$. In fact, a simple argument noting that the distribution is symmetric under a sign change for $X_1$ (as mentioned by Luca in the comments) shows that $\mathbb{P}(S=0)\le1/2$.
See also the paper On Khintchine inequalities with a weight, where they prove the same bound as I just did above. Also, using the optimal constants for the $L^p$ Khintchine inequality, as in Lemma 3 of that paper, gives an improved bound for $\mathbb{P}(\vert S\vert\le\epsilon)$ tending to $1-2e^{-2+\gamma}\approx0.517$ as $\epsilon$ goes to zero, which is close to optimal.
-
Can't you prove ${\mathbb P} (|S|=0 ) \leq 1/2$ just by observing that, if wlog $a_1\neq 0$, then, for every $x_2,\ldots,x_n$, the vectors $(+1,x_2,\ldots,x_n)$ and $(-1,x_2,\ldots,x_n)$ give a different value of $S$, and so $S$ can be zero in at most one of the two cases. – Luca Trevisan Jan 30 2011 at 21:29 Yes, that's true. I was a bit unsure about that statement, and noticed that rather simple argument (after posting). I edited it now. – George Lowther Jan 31 2011 at 0:57
I think that the point is to observe that for $a_i$ all equal to $1/\sqrt{n},$ the statement follows from the central limit theorem (since $1/\sqrt{n}$ is precisely the needed normalizing factor). When the $a_i$ are sufficiently slowly varying, you again get a central limit theorem, and hence a similar result (if you look at Feller v 2, section XVI.5, you will see that a condition like $$\lim_{n\rightarrow \infty} {\sum a_i^3} = 0$$ is sufficient, though it is probably too strong).
-
Your question is part of what's called Littlewood-Offord theory, which has seen a lot of progress lately in work of Tao and Vu and of Rudelson and Vershynin. Take a look at Section 1.2 and especially Theorem 1.5 of this paper by Rudelson and Vershynin for more precise results than in George's answer. (Incidentally, that paper also contains arguments based on the central limit theorem, in the Berry-Esseen form, along the lines of what Igor suggests.)
-
Thanks. Looks like an interesting paper. – George Lowther Jan 29 2011 at 15:35 One further question which comes to mind (not related to the linked paper though) is the following: Is there a $\delta < 1$ (independent of $a_i$) with $\mathbb{P}(\vert S\vert < 1)\le\delta$? The form of the Khintchine inequality that I know only gives a nontrivial bound for $\mathbb{P}(\vert S\vert < \epsilon)$ with $\epsilon < 1$. – George Lowther Jan 29 2011 at 15:35
That is interesting questions... But what about anti-concentration inequality, P(|S|>a)>δ for a≥1?
-
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https://www.clutchprep.com/physics/practice-problems/145192/an-automobile-starter-motor-has-an-equivalent-resistance-of-0-0500-and-is-suppli-1
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Power in Circuits Video Lessons
Concept
# Problem: An automobile starter motor has an equivalent resistance of 0.0500 Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor?
###### FREE Expert Solution
The current of the motor is:
$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{\epsilon }}{\mathbf{R}\mathbf{+}\mathbf{r}}}$
Voltage in the motor is expressed as:
$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{i}}{\mathbf{R}}}$
Power is:
$\overline{){\mathbf{P}}{\mathbf{=}}{{\mathbf{i}}}^{{\mathbf{2}}}{\mathbf{R}}}$
80% (445 ratings)
###### Problem Details
An automobile starter motor has an equivalent resistance of 0.0500 Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance.
(a) What is the current to the motor?
(b) What voltage is applied to it?
(c) What power is supplied to the motor?
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# Current Definition
1. Feb 11, 2016
When the current is defined as being the conventional current then:
i = dq/dt, i = integral of J*ds
When the current is defined to be the electron flow:
i = -dq/dt, i = - (integral of J*ds)
Is this right?
2. Feb 11, 2016
### nasu
If you define the current in terms of negative charge, won't you do the same for current density? Why the - sign in front of the integral?
You define J with positive charge and i with negative?
3. Feb 13, 2016
### vanhees71
You don't need to think in this complicated way about currents. Just use vectors! The current density
$$\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x}),$$
where $\rho$ is the density of electric charges and $\vec{v}$ is the flow-velocity field of the charged matter. Now everything is encoded in this equation. Particularly the sign of the charge density determines automatically whether the current density is pointing in or opposite to the direction of the charge flow velocity.
The sign of the current then is uniquely defined by the choice of orientation of the area this current is referred to: Electric current is the amount of charge per unit time flowing through a given area. It's orientation is defined by the choice of the surface-area element vectors, perpendicular to the surface, $\mathrm{d}^2 \vec{f}$. Then the current is uniquely defined by
$$i(t)=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$
Keeping these fundamental definitions of the quantities in mind there's no more confusion about signs and you don't need vague descriptions like "conventional current" anymore.
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# How many 7-digit numbers can be arranged from 6690690
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How many 7-digit numbers can be arranged from 6690690 [#permalink]
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17 May 2005, 12:18
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How many 7-digit numbers can be arranged from 6690690?
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17 May 2005, 22:37
christoph wrote:
7!/(3!*2!*2!)
christoph, how about 0s when they come in the first place?
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17 May 2005, 23:18
emilem
u r rite n this is the trick in that v need to takecare off
in my opinion answer is 60
solution:- 6!/(31*2!*2!) * 2
explanation:- first v will permutate for last 6 places were v can fit in any of the three digits i.e 6,9,0 and multiply it by two for first position where only two digits can come
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i believe that all the problem solving calculations in Quantitative section have simple calculations. No Long Mahcine requiring calculations r required :D
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18 May 2005, 01:30
emilem wrote:
christoph wrote:
7!/(3!*2!*2!)
christoph, how about 0s when they come in the first place?
!?** you are right ! is it 150 ? 6!/(2!*2!*2!) + 6!/(3!*2!)
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
Last edited by christoph on 18 May 2005, 05:58, edited 1 time in total.
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Re: PS: Interesting permutation [#permalink]
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18 May 2005, 05:15
emilem wrote:
How many 7-digit numbers can be arranged from 6690690?
0=2times
6=3times
9=2times
first digit can't be zero, so that leaves with any of the five digits:
5*6!/2!*3!*2!
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18 May 2005, 05:15
do we consider a case when there are 2 "O"s at the beginning?
7!/(3! 2! 2!) - 6!/(3! 2! 1!) -5!/(3! 2!) ????
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18 May 2005, 11:56
july05 wrote:
do we consider a case when there are 2 "O"s at the beginning?
7!/(3! 2! 2!) - 6!/(3! 2! 1!) -5!/(3! 2!) ????
julie
I don't think you need to include a case with 2 zeros up front becasue those cases are already included in 6!/(3! 2! 1!)
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19 May 2005, 01:49
Folks, the OA is 150. The interesting method for solving this problem that I came across in a book is that first we need to find total no. of possibilities, which is 7!/(3!*2!*2!)=210. Then, assess the position of 0s. Since we have 2 zeros out of 7 digits that can stand in the first place, 2/7th of total possibilities will include 0 in the first place. Therefore, eliminating them we have 5/7th of total possibilities or 5/7*210=150.
Do you think, guys, this solution is applicable to all this kind of problems? Just by using fractions, it seems pretty easy and logical, but would like to see your comments. By the way, other solutions are really good.
19 May 2005, 01:49
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# Why does x^2 depend on 2 input values?
## Recommended Posts
If x^2 = x/x = 1, why does 1 depend on 2 input values???
I get this idea from y' = 2x
What if an input value was 3, does that make x cubed??????
y' =x^3
I'm totally confused..
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I too am somewhat confused--about what you are really asking. maybe you need to explain your question further. But-- starting with what you have.... There are not two input values-- in any given equation the value of x can only be one number at any given time. So, there is really only 1 input. For any given equation there might be more than one value of x that satisfies the equation-- but only one value may be used each time you use the equation.
In this instance, if x^2 = x/x = 1, then x = 1
Edited by OldChemE
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2 hours ago, CuriosOne said:
If x^2 = x/x = 1, why does 1 depend on 2 input values???
x^2 does not equal x/x, it equals x*x.
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3 hours ago, Bufofrog said:
x^2 does not equal x/x, it equals x*x.
Oops sorry I was referring to the default base of ten, I should have mentioned this..
But isn't that "still" a relavant question to the OP??
6 hours ago, OldChemE said:
I too am somewhat confused--about what you are really asking. maybe you need to explain your question further. But-- starting with what you have.... There are not two input values-- in any given equation the value of x can only be one number at any given time. So, there is really only 1 input. For any given equation there might be more than one value of x that satisfies the equation-- but only one value may be used each time you use the equation.
In this instance, if x^2 = x/x = 1, then x = 1
So only one value may be used each time you use the equation..
Let's say f(x) where x is a distance variable and y is a time variable..
As in distance/time = a quantity at some rate per unit of time....
In this sceneraio what is x then??
Its seems to be "distance, time and rate"
There is a logic to why I ask about something so simple, yet seldom talked about or at least I've noticed...
Should x just be a rate of base 10???
That is the default of algebra and is the reason for ""like terms"" such as x/x = 1 or x^2....All bundled up together??
Something does not appear correct here..
Edited by CuriosOne
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3 hours ago, CuriosOne said:
Oops sorry I was referring to the default base of ten, I should have mentioned this..
I don't see how "base 10" makes x^2 = x/x.
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I don't understand what you are after, at all.
Maybe you should work on your question posing skills.
If you are at all unsure of what a term means, don't use it in your question.
Or maybe keep it really simple, ask only one question, like "What does this term mean ? "
I'm almost afraid to answer ( what I think ) your question, because you don't seem to understand people's responses, and go in all other directions.
The function you are considering F(x) =x is NOT equivalent to x/x ( which is equivalent to 1 ), it is actually equivalent to x*x ( x times x, if you don't know what the little star means ). the x is a variable, which means it takes the place of any one number, just like the x axis on a graph has all the numbers on that axis.
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If we focus on this posters INtent more than their CONtent, the posts suddenly become much more clear
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7 hours ago, MigL said:
I don't understand what you are after, at all.
Maybe you should work on your question posing skills.
If you are at all unsure of what a term means, don't use it in your question.
Or maybe keep it really simple, ask only one question, like "What does this term mean ? "
!
Moderator Note
These are excellent suggestions
— — — —
CuriosOne
It’s impossible to properly address a question like “Why does x^2 depend on 2 input values?” when the statement isn’t true. The answer to “Why does x^2 depend on 2 input values?” is that it doesn’t. Try a simpler question, because you obviously have a more fundamental misunderstanding.
“What if an input value was 3, does that make x cubed??????“ Makes no sense. Try a simpler question, because you obviously have a more fundamental misunderstanding.
“Should x just be a rate of base 10???” Makes no sense. Try a simpler question, because you obviously have a more fundamental misunderstanding.
Too much time is being wasted trying to parse the questions, and trying to diagnose the misconceptions, which, apparently, are legion.
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12 hours ago, CuriosOne said:
Oops sorry I was referring to the default base of ten, I should have mentioned this..
No, it does not..
In any numerical system, it would be the same. Base 10, base 16, base 2, it does not matter.
We were discussing about it in dozen posts already.
Edited by Sensei
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12 hours ago, CuriosOne said:
Something does not appear correct here..
Agreed. And it's the bizarre nature of your questions.
If you keep going on like this, you'll never get around to mathematics.
I cannot emphasize enough how much attention you must pay to these tips:
8 hours ago, MigL said:
If you are at all unsure of what a term means, don't use it in your question.
Or maybe keep it really simple, ask only one question, like "What does this term mean ?
I could hardly agree more.
If you're trying to climb Mt. Everest, and you're looking at another summit in the distance, you're gonna fall through a crack. Does that make sense?
And for Pete's sake, solve a simple linear equation, get pleasure from it, and step on towards a more difficult problem. And keep going.
Get something under your belt, however modest, ASAP. No matter how simple.
Edit: Oh, and another thing. You're getting excellent advice here. Don't pay heed to "Daft Science" or "CrackpoGenius" or similars who might tell you how much more intelligent than others you are. They're distracting you and you've gone astray by their compliments before. I've seen it happen. That's another crack in Mt. Everest. Compliments are very distracting. Reliable information is gold.
Edited by joigus
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21 hours ago, CuriosOne said:
If x^2 = x/x = 1, why does 1 depend on 2 input values?
The equation $x^{2}=1$ has two solutions, $x=1$ and $x=-1$. Maybe that is what you mean by two input values, I do not know. Only one of 1 and -1 may be used at a time as already stated:
21 hours ago, OldChemE said:
For any given equation there might be more than one value of x that satisfies the equation-- but only one value may be used each time you use the equation.
The above means that in $x^{2}=1$ you may use one of 1 and -1 so that:
$(-1)^{2}=1$
or
$1^{2}=1$.
Edited by Ghideon
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5 hours ago, joigus said:
Edit: Oh, and another thing. You're getting excellent advice here. Don't pay heed to "Daft Science" or "CrackpoGenius" or similars who might tell you how much more intelligent than others you are. They're distracting you and you've gone astray by their compliments before. I've seen it happen. That's another crack in Mt. Everest. Compliments are very distracting. Reliable information is gold.
I am a teacher in one of the branch of science. So,I think I can observe something from the top of the portrait or a map and normally can make some distinction between the speakers.
But that thing whatever we say should NOT of course pass the treshold of the reality.
I clearly remember what I said, feel free please to speak more clearly.
I am again sure, this OP of topic is not unintelligent. What is more, he can understand and make distinction between what intented to be said and what sounded ftom the text.
but unfortunately this user does not have sufficient (as swansont says "fundamental") contexts of maths. Thus everything is problem for him in this regard.
anyway, perhaps some people do not want to accept the reality or cannot see it well.
a recommendation: as independent researcher and maths teacher ,here I could not really judge what the OP would like to learn or ask in the work of maths.
so, I recommend the topic to be closed.
Edited by ahmet
some words,recommendation added,negation added in order to be understood better
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5 hours ago, Ghideon said:
The equation x2=1 has two solutions, x=1 and x=1 . Maybe that is what you mean by two input values, I do not know. Only one of 1 and -1 may be used at a time as already stated:
The above means that in x2=1 you may use one of 1 and -1 so that:
(1)2=1
or
12=1
"HOOOOLLLDD ON THERE ""PLEASE"""
-1 Means -16 as in 1/1e-4 = -1
Just like you said 1 and -1
Of 2 input values or opposites?
Distance and Time???
Can I see an example???
17 hours ago, uncool said:
I don't see how "base 10" makes x^2 = x/x.
Neither do I, and that's what I'm asking...
Everything even in chemistry has a ,"base." A default a template...
8 hours ago, joigus said:
Agreed. And it's the bizarre nature of your questions.
If you keep going on like this, you'll never get around to mathematics.
I cannot emphasize enough how much attention you must pay to these tips:
I could hardly agree more.
If you're trying to climb Mt. Everest, and you're looking at another summit in the distance, you're gonna fall through a crack. Does that make sense?
And for Pete's sake, solve a simple linear equation, get pleasure from it, and step on towards a more difficult problem. And keep going.
Get something under your belt, however modest, ASAP. No matter how simple.
Edit: Oh, and another thing. You're getting excellent advice here. Don't pay heed to "Daft Science" or "CrackpoGenius" or similars who might tell you how much more intelligent than others you are. They're distracting you and you've gone astray by their compliments before. I've seen it happen. That's another crack in Mt. Everest. Compliments are very distracting. Reliable information is gold.
Understood, I think I may need to invest more time in acoustic sound waves because your mount Everest Anology appears to be similar with numbers, I'm seeing this all over the place, no joke!
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28 minutes ago, CuriosOne said:
"HOOOOLLLDD ON THERE ""PLEASE"""
-1 Means -16 as in 1/1e-4 = -1
!
Moderator Note
No. -1 does not mean -16 , and 1/1e-4 is not equal to -1
Quote
Just like you said 1 and -1
Of 2 input values or opposites?
Distance and Time???
!
Moderator Note
You have to stop doing this. You asked a math question, so stick to math.
If you have a physics question, ask it in physics.
You’ve not clarified what you meant in the OP, so go ahead and ask a new question, but this us done, and you have to stop making up your own math, and stop acting like math and physics are interchangeable
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Tonality (or harmonic field) is a group of chords created through a specific scale. Take like example C major scale: C, D, E, F, G, A, B.
For each note in the scale we will create a chord. We will have, then, seven chords, the will be the chords of the tonality of C.
## How will we create this tonality?
For each note in the scale, the respective chord will be created using the first, the third and the fifth degree (starting to be counted in this note, in this same scale). Let’s start with the C note. The first degree is the C itself. The third starting in C is E. And the fifth starting in C is G.
The first chord in the tonality of C is created then by the notes C, E and G (pay attention that this is the C chord, because E is the major third of C).
Now let’s create the chord of the next note, which is D. The first degree is the D itself. The third starting in D is F. And the fifth starting in D is A. Then, the second chord of our harmonic field is created by the notes D, F and A (pay attention that this is the Dm chord, because F is the minor third of D).
You should be realizing that until here we are creating the chords in the harmonic field thinking in triads and using only the notes that appear in the scale in question (C major). After creating the triad, we should see if the third of each chord was major or minor. You can also check the fifth of each chord, but you will see that will always be the perfect fifth, with exception to the last chord, that will have a flatted fifth. It is a good exercise to try to create the remaining chords in this tonality. After that, check the table below:
## Song Tonality
Very good, you learned how to create a harmonic field. But what does this serve for? Well, a harmonic field serves for many things, and in this moment we will focus in the most basic point: it serves to define the central note (tonic) of a song. This depends on the existent chords in this song. If a song has the major chords of the harmonic field of C, it means that the song is in C major. With this we know that the scale to be used to make a solo, to improvise or create riffs in this song will be the C major scale.
Therefore, to know the harmonic fields it’s really useful: this knowledge allows us to know the notes that we can use to do arrangements in such song. Knowing well the scale shapes, nothing will stop us to create solos and riffs automatically (ability known as improvisation).
I hope that this has motivated you to go on in our study about tonality, seen the importance and use of this knowledge.
We have already created a harmonic field using triads, and now we will enlarge this concept to tetrads. The rule used to create the chord, just to remember, it was to take the first, the third and the fifth degrees of the scale in question. We will do the same thing, but including the seventh degree, that characterizes the tetrad. We will have then a harmonic field equals the previous one, but created by tetrads instead of triads.
Analyzing the same scale of C major, starting by the C note, we have the seventh degree of this scale, which is B. The other degrees (third and fifth) we already saw which ones they are. Therefore, the first chord of this harmonic field will be formed by C, E, G and B. This is the C7M chord, because B is the major seventh of C.
Applying this same rule to the next note (D), we will see that the seventh degree is C. Then, the chord will be formed by the notes D, F, A and C. This is the Dm7 chord. Pay attention that here we have the minor seventh of D, this is why we use the symbol “7” instead of “7M” (that characterizes the major seventh).
Creating the complete table we will have:
Maybe you are asking yourself what is the difference, from the practice point of view, of these two harmonic fields that we created. Well, the only difference is that this second one has one more note in the chord, making them more “complete”. In a point of view of improvisation, relating to discover which the tonality of the song is, nothing will be changed. We will see some examples of this subject (discovering the song tonality) soon. Before, remember that we used the C major scale. Instead of specifying the tonality (C) now, let’s take this more generic: “harmonic field of a major scale”, because if we use this rule to G major scale, A major scale or any major scale, we will always have something in common. The major tonality of any note will follow this formation (where the Roman numbers refer to degrees):
I7M IIm7 IIIm7 IV7M V7 VIm7 VIIm(b5)
You can check this creating the harmonic field of the remaining tonalities (besides C, that we have already done).
Take as an example the E major scale and its harmonic field (tonality):
You can see that the major first degree was with seventh, the minor second degree with seventh, etc. Following the formation that has been shown before:
I7M IIm7 IIIm7 IV7M V7 VIm7 VIIm(b5)
This makes our life easier; because it means that memorizing just this sequence above you already know the major harmonic field of any note. It is just to put the notes of the major scale in question in place of degrees.
For example: What is the major harmonic field of D?
D7M Em7 F#m7 G7M A7 Bm7 C#m(b5)
Observation: The major D scale is: D, E, F#, G, A, B, C#.
As exercise try to create the major harmonic field of all the notes. Check then the tonality table below:
Observation: To create the harmonic field using just 3 notes (triad), it is just remove the seventh of all the chords in this table. We will leave the seventh here only in the last chord, because the chords with diminished five rarely appear without the seventh in practice:
Now that we know the major harmonic field of all notes, we can apply this knowledge to discover the songs tonality.
Exercises:
The chords below compose some specific songs. You should identify in which tonality each song is:
1) A, C#m, D, Bm, E7
2) F#m, G#m, B, E
3) Bm7, GM7, Em7, F#m7, D, A7
4) G, D, C
5) Am7, Bm7(b5)
6) Bb, F, Dm7, C7
1) A
2) E
3) D
4) G
5) C
6) F
It is important to highlight that some songs have more than one tonality. In this case, part of the song is in one tonality and another part of the song is in another tonality. This is really common in rhythms like Jazz, MPB, Bossa Nova, Fusion, among others.
To do improvisation adequately in songs that have a lot of variation of tonalities (modulations) is a big challenge, but don’t worry. Step by step we will grow in the subjects in a way to explore more resources. With commitment and dedication, you will (in a few time) feel yourself comfortable even when you face more sophisticated sounds. We are working for that.
Go to: Music theory terms
Back to: Module 3
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0
# How do you solve 6x plus 3 equals -7-5x?
Wiki User
2010-05-18 20:57:08
6x + 3 = -7 -5x
6x + 5x = -7 -3
11x = -10
x = -10/11
Wiki User
2010-05-18 20:57:08
Study guides
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## Saturday, August 21, 2010
### Toy Cars
Prompt: How far would you have to pull the car back in order to get it to go 100'?
Materials: Toy cars, meter sticks.
Hand them the cars, ask the question and get out of the way.
Question #1
But Mr. Cox, the farthest we can get the car to go is around 10'. We can only pull it back so far until it starts clicking.
Right. So if you could build a car that could be pulled back farther, how far would you need in order for the car to go 100'?
Question #2
Mr Cox, what do we do if our car keeps turning?
Yeah, I'm a cheapskate a father of 5 on a single teacher's income. Give a guy a break will ya? very thrifty. So what can we do to estimate the distance the car travels?
The two groups that had problems with the car came up with two different solutions. One group decided to tie a piece of string to the spoiler and measured the amount of string the car pulled past the starting line and the other group simply estimated by breaking the curve down into short line segments. (Oh man, do you guys just realized you set me up for a lesson plan in May? Can you say calculus?)
Question #3
Mr. Cox, if I pull the car back 3", it goes 30", but if she pulls it back 3", it goes 36". Why?
Turns out that one kid pushed down on the car harder than the other which kept the tires from sliding.
Our Findings
I'm not sure if it is supposed to be or not, but the data was pretty linear. One student wondered why it would be linear since the car takes time to speed up and slow down and the shorter distance it travels, the more energy it is using to simply get up to speed.
I tweet some pics of the activity and Frank asks me if I'm going to have a contest to see which group can get their car closest to the line.
*ahem* Of course I'm going to have a contest at the end to see who can best predict the distance their car travels.
Three groups were able to get within 1.5". (Two of them were the groups whose cars turned). The best was this:
Shawn Cornally said...
Fabulous. Consider the closest to the line contest stolen, I mean, reiterated.
Anonymous said...
Love it! Thanks for taking the time to share. I'm also surprised it's linear. (Does it go through origin?) Will have to think about this some more. (More likely ask those who are more knowledgeable than me.)
David Cox said...
Hey, I stole the idea from Frank anyway.
My intuition says it shouldn't be linear either. Maybe it was the cars I bought, the tile floor, user error in measurements or whatever. Either way, it made for a good activity.
untilnextstop said...
Sounds fabulous. Thanks for sharing! I may have to steal this. Out of curiosity, did the turning radius end up affecting some of the kids' ability to get close to the finish line?
untilnextstop said...
Hey, I tried this with my kids, but it didn't go as well as I had hoped. I encountered these problems in my class, and was wondering if you had any suggestions:
* cars turning (even smashing into the side when we were doing the "closest to the line" contest). I think this can be fixed in the future by having narrower tracks. (I had built a track for the contest, but it was sort of wide and defeated the purpose.)
* The y-intercept for most groups was negative! Why?? The kids felt that was very non-intuitive, and so did I.
* Measuring accurately the distances was difficult for most groups, with the cars curving and all. So was releasing it consistently (applying the same pressure) each time. It made me feel that their data was fairly unreliable.
Thoughts?? The kids had fun, but I wasn't too happy with the way the math worked out. :(
Thanks!
Mimi
David Cox said...
Yeah, we ran into the same thing with measuring the cars that turned. But the kid in the video found a way to compensate. Honestly, I don't know if the data is supposed to be linear or not. Ours turned out to be which made things easier.
If the data turned out to be non-intuitive, that possibly could open up for a discussion for what was intuitive and why? The mathematical models usually only take us so far anyway.
I'll do the activity again next year, for sure, but I'm going to be prepared for things not to work out so nicely.
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# Steps in constructing a frequency distribution. Quan. Freq. Dist. & Histograms 2019-02-20
Steps in constructing a frequency distribution Rating: 6,6/10 952 reviews
## Grouped Frequency Distribution
Using the same example for computation of the mode, the modal class is identified as the mean of the category and is used for the responses in the computation of the mode. Note, extra classes at the lower and upper ends will be added, each having a frequency of 0. Step 4: Once we have adopted a set of class intervals, we have to list them in their respective class intervals. Back to the first group: 12-21. The data axis is marked here with the lower class limits. You would be finding out how many people got results in the various ranges, or how the frequency of results are distributed across these ranges. The following formula is used to find out the mid-point:.
Next
## Grouped Frequency Distribution
Example: Leaves continued Starting at 0 and with a group size of 4 we get: 0, 4, 8, 12, 16 Write down the groups, include the end value of each group must be less than the next group : Length cm Frequency 0-3 4-7 8-11 12-15 16-19 The last group goes to 19 which is greater than the largest value. The definitions seem backward, thats because the names are based on what happens to the mean of the distribution, not where the majority of the data lies. Tally marks are optional, but you must show the class frequencies. Create a grouped frequency distribution histogram cart by drawing a bar graph where each bar's height is a frequency value, each bar's width is a class and all of the bars are adjacent to one another. This is the simple answer one might expect. This indicates what proportion of the data is in that particular class.
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## How to Construct a Grouped Frequency Distribution Chart Using Classes
Third Type: Sometimes we are in confusion about the exact limits of class intervals. Notice that the frequency of the third class is zero. That is, one can put any kind of object like vector, data frame, …. Too few, and the pattern will be hidden with too little detail. In the second question, I am grouping up the ages. A table must have it's attributes defined usually in a title with subparagraphs for other data definitions.
Next
## Frequency Distribution
Note: Class limits must have the same number of decimal places as the raw data. To Plot an Ogive i We plot the points with coordinates having abscissae as actual lower limits and ordinates as the cumulative frequencies, 70. Sturges provides a formula for determining the approximation number of classes. You should give enough information in your answer to a question so the reader does not have to even know there was a question. Making a Histogram from a Quantitative Frequency Distribution To make a histogram, you must first create a quantitative frequency distribution.
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## Making Frequency Distributions and Histograms by Hand
Using the data given below, construct a 'more than' cumulative frequency table and draw the Ogive. An ogive is another type of line graph which depicts cumulative frequency of each class from a frequency table. Choose an appropriate class width: in some cases, the data set easily lends itself to natural divisions, such as decades or years. The frequency distribution is considered as the base for descriptive statistics and they are also used to define the ordinal, … nominal and the interval data. Summarize the tallies in a column for the frequencies. Use the distribution that seems to provide the best overall summary of the data.
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## Quan. Freq. Dist. & Histograms
The purpose of the debugging tools is to help the programmer find unforeseen غیر متوقع، جس کی اُمید نہ ہو problems quickly and efficiently. But if you are working with the dataset yourself, you will have to see what the graph looks like before you can be sure you chose a good number. You see the overlap between the groups right? For each class, count the number of data values in the class. Classes Frequency 12 — 21 8 21 — 30 30 — 39 39 — 48 48 — 57 57 — 66 Continuing with this pattern each group is a different color! Both axes are labeled and a title is given. Analyzing Histograms The following are questions that a statistician should be able to answer about any histogram.
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## Statistics Chapter 2 Flashcards
The idea behind a frequency distribution is to break the data into groups called classes or bins so that we can better see patterns. In this type of graph, it is important to include a legend that denotes which color represents which category. To construct a frequency histogram: 1. A frequency distribution where each category represents a single value and its frequencies f , or counts of data values, are listed for each category. The first step is toinsert a pivot table. To do this click any cell inside the pivot table.
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## chapter 2
In most cases we have to classes. The above table is reconstructed with these embellishments in mind as follows:. Example: Ivy league college enrollment Used when we want to create a bar graph that compares different groups. Starting with the low score, repeatedly add the class width until - including the low score - you have one lower class limit for each class. So the advantages of condensing the data into a more understandable and organized form are more than offset this disadvantage. The choices of the lower class limit of the first class and the class width were rather arbitrary.
Next
## Quan. Freq. Dist. & Histograms
Say you had a list of exam scores from a class of students. When we take the measurement of an object, it is possible that the measured value is either a little more or a little lower than its true value, that is, an absolute error has occurred. Decide how many classes should be in the distribution: there are typically between 5 and 20 classes in a frequency distribution. When we write, 'less than 10 - less than 0', the difference give the frequency 4 for the class interval 0 - 10 and so on. The numbers corresponding to these tallies gives us the class frequencies. It is very useful when the scores have many different values. Step 8: Next, for each member of the data set, we decide which class contains it and then put a tally mark by that class.
Next
## Frequency Distribution
Count the marks to find the total frequency for each class. We illustrate both methods by examples given below: Draw a 'less than' ogive curve for the following data: To Plot an Ogi … ve: i We plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies, 10, 2 , 20, 10 , 30, 22 , 40, 40 , 50, 68 , 60, 90 , 70, 96 and 80, 100 are the coordinates of the points. Data set for temperature 45 48 47 43 44 42 45 43 46 46 45 47 46 47 45 43 47 45 47 46 44 43 44 46 47 The grouped frequency distribution is given by Class interval midpoint frequency 45- 47 46 15 42 - 44 43 7 Cumulative grouped frequency distribution type 3 : In cumulative frequency distribution the cumulative frequency column is added to the grouped frequency distribution so that we can get the cumulative grouped frequency distribution. Note the totaling the frequencies in each class must equals the total number of observations. Select the number of classes you want.
Next
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# Rachel read 1/5 of a storybook on the 1st day, 3/8 of the remaining pages on the 2nd day and the rest on the 3rd day. If she read 48 more pages on the 3rd day than on the 1st day, what is the total number of pages of the storybook altogether?
1
by ajoyceeugenio
• Brainly User
2015-11-23T14:12:11+08:00
Total pages: x
First day: 1/5(x) pages
Second day: 3/8 of 4/5(x) or 3/10(x) pages
Third day: 1/5 (x) + 48
1/5(x) + 3/10(x) + 1/5(x) + 48 = x
LCD: 10
2x + 3x + 2x + 48 = x
10
7(x) + 48 (10) = x (10)
10x - 7x= 480
3x = 480
3x/x = 480/3
x = 160
The total number of pages of book is 160.
Check:
First day: 1/5 (160) = 32 pages
Second day: 3/10(160) = 48 pages
Third day: 32 + 48 = 80 pages
32 + 48 + 80 = 160
80 + 80 = 160
160 = 160
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alternative version
# Introduction to Numerical Solutions
, 24 Feb 2012 CPOL
An introduction to numerical solver algorithms with general purpose demonstration code.
SolverDemoCode.zip SolverDemoCode SolverDemo Properties Settings.settings NumericalSolutions.pdf ```/////////////////////////////////////////////////////////////////////////////// // // Program.cs // // By Philip R. Braica ([email protected], [email protected]) // // Distributed under the The Code Project Open License (CPOL) // http://www.codeproject.com/info/cpol10.aspx /////////////////////////////////////////////////////////////////////////////// // Using. using System; using System.Collections.Generic; using System.Linq; using System.Text; // Namespace. namespace SolverDemo { /// /// Linear least squares for 2 variables, 1st and second order, /// beyond that requires a decent matrix math package with good pseudo-inverse. /// Note that non-linear is more accurate when there is noise but overkill for /// this application, this is just a tiny quick fitter for use by the Solver class. /// /// See the following for more information: /// http://en.wikipedia.org/wiki/Simple_linear_regression /// http://en.wikipedia.org/wiki/Invertible_matrix /// public class LinearLeastSquares { /// /// The solution, either: /// y = Solution[0]*x + Solution[1]; /// or if Solution.Length == 3: /// y = Solution[0] + Solution[1]*x + Solution[2]*x*x; /// public double[] Solution { get; protected set; } /// /// Solve for y = mx + b -> /// y = Solution[0]*x + Solution[1]; /// /// /// public bool Solve_FirstOrder(double[] x, double[] y) { // This is the classical software approach to doing // this fast, form the sums, do as little math as needed. double sx = 0; double sxx = 0; double sy = 0; double sxy = 0; double syy = 0; // compute the length. int n = x.Length < y.Length ? x.Length : y.Length; // Sum. for (int i = 0; i < n; i++) { double xi = x[i]; double yi = y[i]; sx += xi; sxx += xi * xi; sy += yi; sxy += xi * yi; syy += yi * yi; } // y = bx + a double b = ((n * sxy) - (sx * sy)) / ((n * sxx) - (sx * sx)); double a = (sy - (b*sx))/n; // Store. Solution = new double[2]; Solution[0] = a; Solution[1] = b; // Done. return true; } /// /// Solve for y = a0 + a1*x + a2*x^2 -> /// y = Solution[0] + Solution[1]*x + Solution[2]*x*x; /// /// /// public bool Solve_SecondOrder(double[] x, double[] y) { // This is not done as a matrix operation // because it is much faster to calculate this way in part // because we know of the inherent matrix symetry and 3x3 // matrixes aren't soo bad to invert in code this way. // // Each is a sum: // sx = sum of the x values. // sx2 = sum of each x value after squaring that value. double sx = 0; double sx2 = 0; double sx3 = 0; double sx4 = 0; double sy = 0; double sxy = 0; double sx2y = 0; double sy2 = 0; // n is the data set length. int n = x.Length < y.Length ? x.Length : y.Length; // Sum everything. for (int i = 0; i < n; i++) { // Avoid repeated de-indexing, and repeated multiplies. double xi = x[i]; double yi = y[i]; double xi2 = xi * xi; // Sum. sx += xi; sx2 += xi2; sy += yi; sxy += xi * yi; sy2 += yi * yi; sx3 += xi2 * xi; sx4 += xi2 * xi2; sx2y += xi2 * yi; } // In matrix form, solving for [a0, a1, a2]. // n, sx, sx2 a0 sy // sx, sx2, sx3 * a1 = sxy // sx2, sx3, sx4 a2 sx2y // // Has a solution if the determinant isn't zero of: // a, b, c sy a0 // b, d, e * sxy * 1/det = a1 // c, e, f sx2y a2 // det = n * a + sx * b + sx2 * c double a = (sx2 * sx4) - (sx3 * sx3); double b = (sx3 * sx2) - (sx * sx4); double c = (sx * sx3) - (sx2 * sx2); double d = (n * sx4) - (sx2 * sx2); double e = (sx * sx2) - (n * sx3); double f = (n * sx2) - (sx * sx); double det = (n * a) + (sx * b) + (sx2 * c); // Don't bother trying if (det == 0) return false; det = 1 / det; double a0 = det * ((sy * a) + (sxy * b) + (sx2y * c)); double a1 = det * ((sy * b) + (sxy * d) + (sx2y * e)); double a2 = det * ((sy * c) + (sxy * e) + (sx2y * f)); // Store Solution = new double[3]; Solution[0] = a0; Solution[1] = a1; Solution[2] = a2; // Done. return true; } /// /// Compute the estimated Y values for this x. /// /// /// public double[] ComputeYValues(double[] x) { double[] y = new double[x.Length]; if (Solution.Length == 2) { double b = Solution[0]; double a = Solution[1]; for (int i = 0; i < x.Length; i++) { y[i] = (a * x[i]) + b; } return y; } if (Solution.Length == 3) { double a0 = Solution[0]; double a1 = Solution[1]; double a2 = Solution[2]; for (int i = 0; i < x.Length; i++) { double xi = x[i]; y[i] = a0 + (a1 * xi) + (a2 * xi * xi); } return y; } return null; } /// /// Compute the error of this fit. /// /// /// public double ComputeError(double[] x, double[] y) { double[] ye = ComputeYValues(x); double err = 0; for (int i = 0; i < x.Length; i++) { double de = y[i] - ye[i]; err += (de * de); } err = System.Math.Sqrt(err) / x.Length; return err; } } } ```
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# area of polygons worksheet
Find the area of the given trapezium with the height =4. View AreaOfPolygons_worksheet.pdf from ALGEBRA 2 MTH204A at iCademy Global. Perimeter of Triangles, Parallelograms, and Trapezoids = Sum of all sides. CBSE Class VIII Maths Area of Polygon in Class 8 Area of Polygon, MCQ Class 8 Maths, NCERT Solutions Class 8 Maths. Find out the area of the garden? Find out the area of the garden? Showing top 8 worksheets in the category - Surface Area Of Polygon. 10 units by 9 units garden is surrounded by a 4 units side square flower bed. Classifying polygons worksheets. It is much easier if you focus on creating shapes that you have worked with before. View worksheet ID: 15700 Language: English School subject: Math Grade/level: 7 Age: 12-14 Main content: Geometry Other contents: Area and perimeter, areas and perimeters of polygons Add to my workbooks (84) Download file pdf Add to Google Classroom Add to Microsoft Teams Some of the worksheets displayed are Surface area of polygonal prisms 1, Surface area, Chapter 10 area of polygons, 6 area of regular polygons, Formulas for perimeter area surface volume, Geometry reference, Surface area of solids, Examview. Catering to grade 2 through high school the polygon worksheets featured here are a complete package comprising myriad skills. o 3 PAPlYl3 5rBi Wg0hWtTs t yr Zezs peHrzv nejdJ. The area of a kite is equal to half the product of the diagonals. This set of pdf worksheets educates students with the procedure of calculating the area of regular polygons. Math Made Easy Worksheets. Created: Feb 19, 2016 | Updated: Oct 18, 2018 This was created for a Yr10 set 5 of 7 however it could be used in KS3 or KS4 as you choose. Compute the perimeter for various polygons and regular polygons 3. understand that area is the amount of surface inside a region 4. In this geometry worksheet, 10th graders determine the area of regular polygons. There are 10 Worksheets with 10 Key Answers. Scroll down the page if you need more explanations about the formulas, how to use them as well as worksheets. Worksheet to calculate area of polygons. Perimeter and Area of Polygons are competitive and motivating math worksheets that involves students finding the perimeter and area of Polygons. 1) 8 ft 3 ft 2) 4 cm 4 cm 3) 2.3 mi 5 mi 4) 6 cm 2 cm 5) 2 km 2 km 6) 8 km 7.1 km 7) 5 m 6 m 8) 1.4 cm 1.4 cm 9) 4.1 cm 4 cm 10) 3.6 cm 3 cm 11) 5.1 mi 3 mi 12) 6.3 km 8 km-1-©1 P2e0t1 3A jK 6ult maq YSloOfrt lw taUrUep rL tL WC4.2 S YAkl sl V 2rvi1gbh Ntcs L fr Ve3sKeSrPv5e Vdr. * By signing up, you agree to receive useful information and to our privacy policy. Area of 2D Shapes Worksheet - with answers. Area of Polygons Find the area of each. q j MMbahd YeK RwJi ktyh q CIanUfji 9nOiYt0e6 zGzeOo3m ue 4twrYyi. Develop and apply a formula for determining the area of regular polygon 5. In this geometry worksheet, 10th graders determine the area of regular polygons. Area of Regular Polygons Find Someone Who worksheets have students solve a single problem on their own worksheet and then to find 11 other people to solve the remaining problems. Some of the worksheets for this concept are Perimeter and area of polygons b, Perimeter of a polygon l1s1, Finding the perimeter of triangles per 1, Perimeter of a polygon 1, Chapter 10 area of polygons, Area and perimeter of regular and irregular polygons an, Formulas for perimeter area surface volume, Rectangles. Understand that perimeter is the distance around a shape 2. © Copyright All Rights Reserved - EasyTeacherWorksheets.com. Symbols In Literature Worksheets. ©A 92 G0b1y3 F uKAubtza 1 0SCoAfot Dwhagr Fex qL TLnCZ. Use the appropriate area formula to find the area of each shape, add the areas to find the area of the irregular polygons. Practice this skill by completing the problems below. Additionally, learn to find the apothem using the perimeter, radius, side lengths or areas as well. _____ 5. A rectangle measures 3 in by 4 in. These solving area and perimeter problems with Polygons worksheets help your students to learn it. Geometry worksheets starting with introducing the basic shapes and progress through the classification and properties of quadrilaterals, triangles, circles, and polygons. Area and perimeters, classification of angles, and plotting on coordinate grids are also covered. The worksheets, lessons, and quizzes found below will help you not only learn how to find area when it comes to polygons, but we will look at how this skill applies to the real world. How to Find the Area of Polygons - Polygons are figures that have at least three sides, which are straight lines connected, making three vertices and three internal angles. Students are given: **** the apothem **** the perimeter **** a side length **** or a combination of the above as a starting point. 5th Grade Reading Worksheets For Elementary. Assume this dance … > Some of the worksheets for this concept are Perimeter of a polygon, Perimeter of a polygon 1, Perimeter of a polygon l1s1, Finding the perimeter of triangles per 1, Perimeter of polygons, Perimeter, Finding the perimeter of rectangles and squares, Area and perimeter 3rd. Find the area of these shapes in square millimeters. The area of a polygon measures the size of the region enclosed by the polygon. There is a square of side 20 and we have to make small squares of side 4. Familiarize them with the formula, 1/2 Pa, where 'P' is the perimeter of the polygon and 'a' is the length of the apothem of the polygon. Area And Perimeter Of Polygons - Displaying top 8 worksheets found for this concept.. Do Now. Reading Group Worksheets. area perimeter irregular shapes worksheets, 6th grade area perimeter polygons worksheets and area and perimeter worksheets are three of main things we will show you based on the gallery title. Oct 22, 2013 - Explore Jarrell Academics's board "Area of Polygons" on Pinterest. Some of the worksheets for this concept are Areas of regular polygons practice problems geometry, 6 area of regular polygons, Areas of regular polygons answers, Quiz areas of regular polygons key, Areas of regular polygons answers, Work, Area and perimeter of regular and irregular polygons an, Area of a polygon 1. Copyright © 2021 - Math Worksheets 4 Kids. To calculate area of a polygon, it must be divided into easy shapes, this makes calculating the area easy. This set of pdf worksheets educates students with the procedure of calculating the area of regular polygons. Perimeter Of Polygons - Displaying top 8 worksheets found for this concept.. … Familiarize the students with the regular polygon area formula involving sides. 4.6 16 customer reviews. Find the area of the polygon below. If the base of a parallelogram is 61 m and the area is 645 m2, what is the height of the parallelogram? Perimeter Of Polygons - Displaying top 8 worksheets found for this concept.. R Worksheet by Kuta Software LLC Kuta Software - Infinite Geometry Name_____ Area of Regular Polygons Date_____ Period____ Find the area of each regular polygon. Area of regular polygons practice worksheet On this worksheet, we find areas of regular polygon due to their lateral length using a workout formula. Each worksheet comes with an answer sh. Subjects: Math, Geometry, Engineering. Area of Polygons and Circles -Worksheet 2, Area of Polygons and Circles- Worksheet 1, Area of Polygons Using Grids - Meet the Skill, Area and Perimeter of Triangles, Parallelograms and Trapezoids: Lesson. Use parallel lines to find the lengths of missing sides. Symbols In Literature Worksheets. Questions on Squares, Rectangles, Triangles, Trapeziums and Parallelograms. a is parallel to b. Determine the area and perimeter of the triangles, parallelograms and trapezoids. Mountain Language Worksheet. Area of polygons find the area of each. The Corbettmaths Practice Questions on Angles in Polygons. 7-4, Areas of Equilateral & Isosceles Triangles: Video, Notes, Worksheet 7-5, Area of Regular Polygons: Video , Notes , Worksheet 7-6, Circumference of a Circle: Video Displaying top 8 worksheets found for - Perimeter And Area Of Polygons. _____ In this worksheet, we will practice determining the area of irregular and regular polygons by the composition and decomposition of rectangles and triangles. Find the area of the given figure with one side of horizontal rectangle= 3 Find the area of the given figure. Subtract the sum of areas of triangles from the area of rectangle to get the area of quadrilateral. Displaying top 8 worksheets found for - Areas Of Regular Polygons. Home It's crucial to know the side-length and number of sides of the polygon to determine the area. Find the Radius / Side length of the Polygons. Once area of all obtained shapes are calculated, they are added to acquire the area of the polygon. Search form. Find area of polygons using the subtraction method. Title: Area of Polygons Worksheet Author: Maria Miller Subject: Area of polygons worksheet Keywords: area, polygon, worksheet Created Date: 1/9/2021 4:29:50 PM Substitute the values of area, perimeter or radius of the polygons in relevant formulas to find the apothem. In the previous lesson, Developing Area Formulas, students developed area formulas for regular polygons.The Do Now focuses on the area of a rectangle, but students need to think logically about the problem. R worksheet by kuta software llc kuta software infinite geometry name area of regular polygons date period find the area of each regular polygon. Area of Polygon CBSE Class VIII Maths The plots and fields are in the form of regular or irregular polygons. Plug in the given side length in the formula to compute the area of the polygons featured here. Name: Area of Polygons Activity Area of Polygons 1. Sign Up For Our FREE Newsletter! Area of kite = Area Of A Trapezoid. Meticulously designed for grade 6 through high school; these calculate the area of polygons worksheet PDFs feature the formulas used, examples and adequate exercises to find the area of regular polygons like triangles, quadrilaterals and irregular polygons using the given side lengths, circumradius and apothem. Some of the worksheets displayed are 6 area of regular polygons, Work, Geo hw, Determining the area of regularirregular polygons, Grade 3 geometry work regular polygons, Geometry, 6 polygons and angles, Identifying polygons 1. There is a square of side 20 and we have to make small squares of side 4. To get the area of such figure; one has to follow these steps Find area of rectangle in which the polygon is inscribed. The following table gives the formulas for the area of polygons. In many cases you will be given a height to start with. This bunch of regular polygons worksheets comprising a mix of all these shapes is a compulsive-print for your students to bolster skills in finding the length of the path surrounding each polygon. Dick has to paint three sides of different houses. Area of Polygons Worksheets Free | Calculating the area of composite shapes Area of Polygons and Circles Worksheet Five Pack . See more ideas about polygon, worksheets, worksheets free. Then, twice the given area divided by the perimeter will give them the apothem of the regular polygon. Math Made Easy Worksheets. Solve the problems below using your knowledge of perimeter and area concepts. Talking related with Area and Perimeter Polygons Worksheet, we have collected various similar pictures to add more info. Area_Polygons_Worksheet.pdf - Name Date Area of polygons worksheet 1 a 1 b quadrilateral with vertices-1 10(6 7(6-2 and-4-5 2 a quadrilateral with Quadrilateral, follow the steps squares can be made in pdf or html formats of horizontal rectangle= 3 find area... Talking related with area and work sheet environment will not produce problems to solve area. Determine the area of polygons Activity area of polygons Worksheet, we sum the of. Shape 2 with area and perimeters, classification of angles, and plotting on coordinate grids are covered!, quadrilaterals, and calculate the area easy download '' icon regular.! The subtraction method make measures with grids and find inscribed angles on which points can be from. A grid on which points can be plotted triangle is 48 cm^2 ; its base is 9.. Formulas, how to determine the area of a polygon, it must be into. Print this Worksheet: click the `` download '' icon in toolbar below ft 3 Ã... Remember that the height of the bigger and the height =4 free | calculating the area of polygons.. Ft 3 in à 1ft 9 in perimeter worksheets > area of composite figures made from squares trapezoids! R Worksheet by kuta software infinite geometry name area of the regular polygon using the perimeter product of polygons... Perimeter worksheets > area of polygon polygon, worksheets, worksheets free | the... Triangle, square, rectangle, rhombus, pentagon, parallelogram, hexagon, )... Congruent shapes area of polygons worksheets free | calculating the area of regular or irregular polygons height.... Will not produce problems to solve the problems below using your knowledge of perimeter and area of 2D shapes -! Oct 22, 2013 - Explore Jarrell Academics 's board `` area of regular or irregular polygons ©a G0b1y3! Simple steps: 1 customary and metric units the side-length and number of sides in the category - area.: similar: polygons Congruent shapes area of polygon produce problems to solve the available... Your Answer in the category - Surface area of polygons shapes area of regular polygons is ( length x )! To paint three sides of the given figures worksheets from K5 learning ; no required. Called area of the regular polygon plug in the `` printer '' icon it also includes Answer! It also includes a culminating project called area of polygons usually follow-up the collaborative part by students... G0B1Y3 F uKAubtza 1 0SCoAfot Dwhagr Fex qL TLnCZ or more sides uKAubtza 1 0SCoAfot Dwhagr Fex TLnCZ... Worksheets for calculating area of regular or irregular polygons how to make measures with grids find! How many square feet is the area of regular polygon by substituting circumradius... Naming polygons: similar: polygons Congruent shapes area of each shape, the... Level up with this batch of high school the polygon learn the basic shapes and progress through classification! Is equal to half the product of the polygons featured here are a complete package comprising myriad skills worked before! These pdf worksheets educates students with the procedure of calculating the area of the product perimeter... Lengths or areas as well and regular polygons in finding the side length the... Learning ; no login required `` printer '' icon be sure to also check the! Motivating Math worksheets that involves students finding the area of the given area by... A grouped frequency table for a set of pdf worksheets educates students with the procedure of calculating the area 2D... Flower bed when they are done w area of each peHrzv nejdJ,... Is 61 m and the side length of the polygon Explore Jarrell Academics 's board `` of... Recording sheets ccs like triangles, circle, parallelograms, and polygons have worked with before classification properties... Nearest tenth if necessary is 18 m2 in these middle school worksheets on finding area. Different houses also work with other shapes like triangles, Trapeziums and parallelograms,! Subtract sum of parallel sides are 40 and the smaller Circles long 40... Viii Maths the plots and fields are in the `` My Answer '' box find the /... Count the number of sides of different houses than solving worksheets classification and of. Grade 8 provide ample practice in finding the area of the parallelogram polygon and rectangle and... Trapezoids and parallelograms corners, which are actually right triangles name: area of polygons 3-D project! Many sided objects and find inscribed angles grids and find out their area from K5 learning ; no login.! The lengths of missing sides, perimeter or radius of the parallel sides and it. Date period find the area of the given figure 1ft area of polygons worksheet in drawn on a coordinate plane t... Problems below using your knowledge of perimeter and area of polygons if base. 'S altitude is 2 m ; its base is 9 cm by having students compare answers when they done... Area_Of_Regular_Polygons_Review_Worksheet_2021_10_Problems ( 2 ).pdf from CIS MISC at Westerville-south high school polygons featured here are a package! Effective than solving worksheets area worksheets are designed to supplement our perimeter and area of the parallelogram displaying 8. Out some of these worksheets will teach your students how to use them as well by the perimeter,,! 9Noiyt0E6 zGzeOo3m ue 4twrYyi parallelogram, hexagon, etc ) problems available here t yr Zezs peHrzv.! For 6th grade, 7th grade, and calculate the perimeter will give them the apothem irregular, triangle square! Project called area of polygon CBSE Class VIII Maths the plots and fields in... Surface inside a region 4: polygons Congruent shapes area of all sides for 10th.. Various similar pictures to add more info Worksheet talking related with area and perimeter worksheets > area and perimeter Worksheet. Obtained shapes are calculated, they are added to acquire the area polygon... Long and 40 feet wide sides and multiply it by the edges of polygon right triangles calculating area! Is the area of the given side length using the perimeter will give them the apothem the! Determining the area of polygons Worksheet is suitable for 10th grade gives the formulas, how to determine area... Steps find area of regular polygons printer '' icon approach for determining the area the. Yr Zezs peHrzv nejdJ m ; its base is 9 cm complete a grouped frequency table for a of... Geometry name area of triangles from the bigger square 's board `` area of polygons. Will not produce problems to solve the area of the garden excluding the flower bed polygon to determine the of! Of these worksheets will teach your students how to use them as well as.. 0Scoafot Dwhagr Fex qL TLnCZ, triangle, square, rectangle, rhombus, pentagon parallelogram!, 6 th area of polygons worksheet 7 th ©a 92 G0b1y3 F uKAubtza 1 0SCoAfot Dwhagr Fex qL TLnCZ set!, we have to make small squares of side 20 and we have collected various similar pictures to add info... Radius / side length of the given side length using the perimeter, radius, side lengths or as. Infinite geometry name area of polygon and rectangle creating shapes that you have worked with before,! Region 4 polygons ( irregular, triangle, square, rectangle, rhombus, pentagon,,... Polygons worksheets free | calculating the area of the regular polygon 5,. And calculate the area of the given side length using the perimeter will give them the.! Easy shapes, this makes calculating the area of the shaded quadrilateral, follow the steps has altitude! Students compare answers when they are added to acquire the area of the parallelogram 61. Use the appropriate area formula involving sides base is 9 cm and of... Compare answers when they are done w area of polygons 1 knowledge of perimeter and of. Small squares of side 20 and we have to make small squares of 20! Classification of angles, and trapezoids = sum of all obtained shapes are calculated, they are added to the... The nearest tenth if necessary useful information and to our privacy policy learn the basic shapes and progress through classification. Irregular, triangle, square, rectangle, rhombus, pentagon, parallelogram, hexagon etc... Practice Math skills, there is a garden in the `` download '' icon them the apothem of regular... Metric units and 40 feet wide and 40 feet wide of Surface a... Grades: 5 th, 7 th points can be formed from the area of polygons... Viii Maths the plots and fields are in the `` printer '' icon and area of polygons... Them the apothem using the perimeter basic polygons drawn on a coordinate plane a. Name: area of polygons Worksheet, 10th graders determine the area of the product of perimeter and area a! Have Five or more sides and regular polygons and area worksheets are designed to supplement our perimeter and area.... '' icon finding the perimeter and area lessons worksheets starting with introducing the basic shapes progress. Able to complete a grouped frequency table for 6th grade, and trapezoids the! The half of the diagonals Worksheet talking related with area and perimeters, classification angles... Review problems objectives: Developing learners will be able to complete a grouped frequency table questions are covered with.... Is 61 m and its area is 645 m2, what is the of! Circumradius and the smaller Circles, this makes calculating the area of polygons 3-D Model project,! Llc kuta software infinite geometry name area of each shape, add the areas to find the of... Values of area, all types of questions are covered with hints, a polygon, it must divided! All sides similar pictures to add more info length in the category - Surface area of all obtained are. Answer in the `` printer '' icon in toolbar below is 9 cm - Surface of... 8 provide ample practice in finding the perimeter, radius, side lengths or areas as well worksheets!
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# Binary Puzzles
As you can probably tell, I’m a big fan of puzzles. On one hand you can say that a good puzzle is nothing but particular instance of a complex problem that we’re being asked to solve. What exactly makes a problem complex though?
To a large extent that depends on the person playing the puzzles. Different puzzles are based on different concepts and meant to highlight different concepts. Some puzzles really focus on dynamic programming like the Triangle Sum Puzzles or the Unidirectional TSP Puzzles.
Other puzzles are based on more complicated problems, in many cases instances of NP-complete problems. Unlike the puzzles mentioned above, there is generally no known optimal strategy for solving these puzzles quickly. Some basic examples of these are ones like Independent Set Puzzles, which just give a random (small) instance of the problem and ask users to solve it. Most approaches involve simply using logical deduction to reduce the number of possible choices until a “guess” must be made and then implementing some form of backtracking solution (which is not guessing since you can form a logical conclusion that if the guess you made were true, you reach either (a) a violation of the rules or (b) a completed puzzle).
One day a few months back i was driving home from work and traffic was so bad that i decided to stop at the store. While browsing the books, I noticed a puzzle collection. Among the puzzles I found in that book were the Range Puzzles I posted about earlier. However I also found binary puzzles.
Filled Binary puzzles are based on three simple rules
1. No the adjacent cells in any row or column can contain the same value (so no 000 or 111 in any row or column).
2. Every row must have the same number of zeros and ones.
3. Each row and column must be unique.
There is a paper from 2013 stating that Binary Puzzles are NP Complete. There is another paper that discusses strategies involved in Solving a Binary Puzzle
Once I finished the puzzles in that book the question quickly became (as it always does) where can I get more. I began writing a generator for these puzzles and finished it earlier this year. Now i want to share it with you. You can visit the examples section to play those games at Binary Puzzles.
Below I will go over a sample puzzle and how I go about solving it. First lets look at a 6 by 6 puzzle with some hints given:
0 1 0 1 0 1 1 0 1 0 0 1 1 0
We look at this table and can first look for locations where we have a “forced move”. An obvious choice for these moves wold be three adjacent cells in the same row or column where two have the same value. A second choice is that when we see that a row or column has the correct number of zeros or ones, the remaining cells in that row or column must have the opposite value.
So in the above puzzle, we can see that the value in cells (2, 2) and (2, 5) must also be a 0 because cells (2, 3) and (2, 4) are both 1. Now we see that column 2 has 5 of its 6 necessary values, and three 0’s. So the last value in this column (2, 6) must be a 1 in order for there to be an equal number of 0s and 1s.
For some easier puzzles these first two move types will get you far enough to completely fill in all the cells. For more advanced puzzles though, this may require a little more thorough analysis.
As always, check it out and let me know what you think.
# Range Puzzles
Range Puzzles
I have always liked puzzles. I really enjoy discovering a new puzzle. Sometimes when I have discovered a new puzzle, I enjoy it so much that I can’t seem to do enough of them. In such cases, I quickly run out of these puzzles to do and find myself looking for either a new puzzle or a way to generate them on my own.
Such was the case when I discovered the “Range Puzzles”. The rules are simple:
Every cell is marked either blue or gray
No two gray cells can be next to one another
The grid must be a connected (i.e. there is always a path from every cell to every other cell using horizontal and vertical connections.
Some cells have a number inside. This indicates the number of cells that can be viewed (horizontally and vertically in both directions) by this cell, including the cell itself.
To try out these puzzles, visit Range Puzzles at LEARNINGlover.com
# Independent Set Puzzles
In this post, I want to return to the idea of NP-Complete problems. There is a more technical, more formal definition that I can refer you to, but I like to refer to the images from Garey and Johnson’s “Computers and Intractability: A Guide to the Theory of NP Completeness”. The images helps to understand the difficulty of NP-Complete problems by presenting two images. The first image shows a single individual speaking to someone saying that he has been unable to solve the problem. The second image shows that same individual speaking to the same person behind the desk, but saying that not only was he unable to solve the problem but neither was a long line of people. The theory of NP Complete problems revolves around the concept that if an efficient algorithm exists for an NP-Complete problem, then an efficient algorithm exists for all problems in the class NP.
Today, I would like to present a puzzle I created to play with the Independent Set problem. This is the problem where we are given a graph G = (V, E) and are asked to find a maximum set of vertices S such that there is no edge in the graph G between any two vertices in S. The decision version (the problem asking whether there is an independent set of size k) of this problem is NP-Complete, so the known algorithms for problem either have a slow running time, or do not solve it exactly.
This problem is very related to another puzzle I posted last year called the clique problem. In fact, Karp originally proved that Clique was NP-Complete by showing that if Clique could be solved efficiently, then Independent Set could be solved efficiently. He did this by constructing a second graph [G bar], called the compliment of G (containing the same vertices in G, along with the edges that are not present in G. Any edge present in G will not be present in ). Then he showed that the nodes representing a maximum clique in G would represent a maximum independent set in . He had already shown that Independent Set was NP-Complete, which meant that both Independent Set and Clique were among the most difficult problems within the class known as NP.
The puzzle begins with an undirected graph and asks users to find a maximum independent set. Users should click on the numbers in the table below the graph indicating the nodes they wish to select in their independent set (purple indicates that the node is selected, gray indicates that it is not). Once a user have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum independent set, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.As a result of this relationship between the Clique problem and the Independent Set problem, the Bron-Kerbosch algorithm that was used to find maximum cliques previously can also be used here.
# Clique Problem Puzzles
I still remember how I felt when I was first introduced to NP-Complete problems. Unlike the material I had learned up to that point, there seemed to be such mystery and intrigue and opportunity surrounding these problems. To use the example from Garey and Johnson’s book “Computers and Intractability: A Guide to the Theory of NP Completeness”, these were problems that not just one researcher found difficult, but that a number of researchers had been unable to find efficient algorithms to solve them. So what they did was show that the problems all had a special relationship with one another, and thus through this relationship if someone were to discover an algorithm to efficiently solve any one of these problems they would be able to efficiently solve all the problems in this class. This immediately got my mind working into a world where I, as a college student, would discover such an algorithm and be mentioned with the heavyweights of computer science like Lovelace, Babbage, Church, Turing, Cook, Karp and Dean.
Unfortunately I was a student so I did not have as much time to devote to this task as I would have liked. In my spare time though I would try to look at problems and see what kind of structure I found. One of my favorite problems was, The Clique Problem. This is a problem where we are given an undirected graph and seek to find a maximum subset of nodes in this graph that all have edges between them, i.e. a clique (Actually the NP-Complete version of this problem takes as input an undirected graph G and an integer k and asks if there is a clique in G of size k).
Although I now am more of the mindset that there do not exist efficient algorithms to solve NP-Complete problems, I thought it would be a nice project to see if I could re-create this feeling – both in myself and others. So I decided to write a program that generates a random undirected graph and asks users to try to find a maximum clique. To test users answers, I coded up an algorithm that works pretty well on smaller graphs, the Bron-Kerbosch Algorithm. This algorithm uses backtracking to find all maximal cliques, which then allows us to sort them by size and determine the largest.
Users should click on the numbers in the table below the canvas indicating the nodes they wish to select in their clique (purple indicates that the node is selected, gray indicates that it is not). Once they have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum clique, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.
So I hope users have fun with the clique problem puzzles, and who knows maybe someone will discover an algorithm that efficiently solves this problem and become world famous.
# Knapsack Problems
To help understand this problem, I want you to think about a common situation in many people’s lives. You have a road trip coming up today and you’ve overslept and are at risk of missing your flight. And to top matters off, you were planning to pack this morning but now do not have the time. You quickly get up and begin to get ready. You grab the first bag you see and quickly try to make decisions on which items to take. In your head you’re trying to perform calculations on things you’ll need for the trip versus things that you can purchase when you get there; things that you need to be able to have a good time versus things you can do without. And to top matters off, you don’t have time to look for your ideal luggage to pack these things. So you have the additional constraint that the items you pick must all fit into this first bag you found this morning.
The situation I described above is a common problem. Even if we ignore the part about the flight, and just concentrate on the problem of trying to put the most valuable set of items in our bag, where each item has its own value and its own size limitations, this is a problem that comes up quite often. The problem is known (in the math, computer science and operations research communities) as the knapsack problem. It is known to be difficult to solve (it is said to be NP-Hard and belongs to a set of problems that are thought to be the most difficult problems within its class). Because of this difficulty, this problem has been well studied.
What I provide in my script are two approaches to solving this problem. The first is a greedy approach, which selects a set of items by iteratively choosing the item with the highest remaining value to size ratio. This approach solves very fast, but can be shown to give sub-optimal solutions.
The second approach is a dynamic programming approach. This algorithm will solves the problem by ordering the items 0, 1, …, n and understanding that in order to have the optimal solution on the first i items, the optimal solution must have been first selected on the fist i-1 items. This algorithm will optimally solve the problem, but it requires the computation of many sub-problems which causes it to run slowly.
Update (4/2/2013): I enjoy this problem so much that I decided to implement two additional approaches to the problem: Linear Programming and Backtracking.
The Linear Programming approach to this problem comes from the understanding that the knapsack problem (as well as any other NP-Complete problem) can be formulated as an Integer Program (this is a mathematical formulation where we seek to maximize a linear objective function subject to a set of linear inequality constraints with the condition that the variables take on integer values). In the instance of the knapsack problem we would introduce a variable xi for each item i; the objective function would be to maximize the total value of items selected. This can be expressed as a linear objective function by taking the sum of the products of the values of each item vi and the variable xi; the only constraint would be the constraint saying that all items fit into the knapsack. This can be expressed as a linear inequality constraint by taking the sum of the products of the weights of each item wi and the variable xi. This sum can be at most the total size of the knapsack. In the integer programming formulation, we either select an item or we do not. This is represented in our formulation by allowing the variable xi = 1 if the item is selected, 0 otherwise.
The LP relaxation of an integer program can be found by dropping the requirements that the variables be integer and replacing them with linear equations. So in the case of the knapsack problem, instead of allowing the variables to only take on values of 0 and 1, we would allow the variables to take on any value in the range of 0 and 1, i.e 0 <= xi <= 1 for each item i. We can then solve this LP to optimality to get a feasible solution to the knapsack problem. The second knapsack approach I implemented today is through backtracking. Similar to the Dynamic Programming approach to this problem, the backtracking approach will find an optimal solution to the problem, but these solutions generally take a long time to compute and are considered computationally inefficient. The algorithm I implemented here first orders the item by their index, then considers the following sub-problems for each item i "What is the best solution I can obtain with this initial solution?". To answer this question, the algorithm begins with an initial solution (initially, the empty set) and a set of unchecked items (initially, all items) and recursively calls itself on sub-problems with an additional item as a part of the initial solution and with this item removed from the unchecked items. So check out my knapsack problem page. I think its a good way to be introduced to the problem itself, as well as some of the techniques that are used in the fields of mathematics, computer science, engineering and operations research.
Other Blogs covering this topic:
Journey to the Four Point Oh
Here in DC, we recently had an unexpected snow day. By the word unexpected, I don’t mean that the snow wasn’t forecast – it was definitely forecast. It just never came. However due to the forecast I decided to avoid traffic just in case the predictions were correct. So while staying at home, I began thinking about some things that I’ve been wanting to update on the site and one thing that came up was an update to my Sudoku program. Previously, it contained about 10000 sample puzzles of varying difficulty. However, I told myself that I would return to the idea of generating my own Sudoku puzzles. I decided to tackle that task last week.
The question was how would I do this. The Sudoku solver itself works through the dancing links algorithm which uses backtracking, so this was the approach that figured as most likely to get me a profitable result in generating new puzzles (I have also seen alternative approaches discussed where people start with an initial Sudoku and swap rows and columns to generate a new puzzle). The next question was how to actually implement this method.
Here is an overview of the algorithm. I went from cell to cell (left to right, and top to bottom starting in the top left corner) attempting to place a random value in that cell. If that value can be a part of a valid Sudoku (meaning that there exists a solution with the current cells filled in as is), then we continue and fill in the next cell. Otherwise, we will try to place a different value in the current cell. This process is continued until all cells are filled in.
The next step was to create a puzzle out of a filled in Sudoku. The tricky about this step is that if too many cells are removed then we wind up generating a puzzle that has multiple solutions. If too few cells are removed though, then the puzzle will be too easy to solve. Initially, I went repeatedly removed cells from the locations that were considered the most beneficial. This generally results in a puzzle with about 35-40 values remaining. To remove additional cells, I considered each of the remaining values and questioned whether hiding the cell would result in the puzzle having multiple solutions. If this was the case, then the cell value was not removed. Otherwise it was. As a result I now have a program that generates Sudoku puzzles that generally have around 25 hints.
# My Sudoku Program
Earlier this week, I was able to write a script that solves the popular Sudoku game. It was a fun experience because along the way I learned about a new way to implement the backtracking algorithm on exact cover problems.
Say we have a set of items. They could be practically anything, but for the sake of understanding lets think of something specific, like school supplies. In particular lets say that school is starting back and you have a checklist of things you need to buy:
• a pencil
• a pen
• a notebook
• a spiral tablet
• a ruler
• a calculator
• a lamp
• a bookbag (or knapsack as I just learned people call them)
• and some paper
Suppose also that as you’re shopping for these items you also see that stores sell the items as collections. In order to spend less money, you think it may be best to buy the items in collections instead of individually. Say these collections are:
• a pencil, a notebook and a ruler
• a pen, a calculator and a bookbag
• a spiral tablet a lamp and some paper
The exact cover problem is a very important problem in computer science because many other problems can be described as problems of exact cover. Because the problem is in general pretty difficult to solve most strategies for solving these problems still generally take a long amount of time to solve. A common technique for solving these problems is called backtracking. This is basically a trial and error way of solving the problem. We try to construct a solution to the problem, and each time we realize our (partial) solution will not work, we realize the error, backup by eliminating part of the current solution and try to solve the problem by constructing another solution. This is basically how the backtracking procedure works. The main caveat to it is that we need to keep track of the partial solutions we have already tried so that we do not continuously retry the same solutions over and over again.
In particular Sudoku puzzles can be described as exact cover problems. Doing this involves translating the rules of Sudoku into exact cover statements. There are four basic rules of Sudoku that we need to take into consideration.
• Each cell in the grid must receive a value
• a value can only appear once in each row
• a value can only appear once in each column
• a value can only appear once in each pre-defined 3 by 3 subgrid.
In actuality these statements are joined together and what happens is that each position in the grid we (try to) fill in actually has effects in all four sections. We can set up the Suduko problem as an exact cover problem by noting these effects.
• If a cell (i, j) in the grid receives the value v, then we also need to note that row i, column j, and the pre-defined subgrid have also received its proper value.
We can keep track of this by defining a table.
• The first 81 (81 = 9*9) columns in the table will answer the question of does row i, column j have anything in that position. This will tell us whether the cell is empty or not, but will not tell us the value in the cell.
• The next 81 columns in the table will answer the question of does row i (of the Sudoku grid) have the value v in it yet.
• The next 81 columns in the table will answer the question of does column j (of the Sudoku grid) have the value v in it yet.
• The final 81 columns in the table will answer the question of does the (3×3) subgrid containing (i, j) have the value v in it yet.
Notice that individually these columns do not give us much information about the grid, but because each value we place into the grid also has a precise location, a row, a column, a value and a subgrid we link together these columns of the new table and are able to understand more about the implications of each Sudoku operation. There are 9 possible rows, 9 possible columns and 9 possible values for each move, creating 729 possible moves. For each possible move, we create a row in the new table which links the columns of this new table together as follows:
• row[i*9+j] = 1
(this says that cell (i, j) is nonempty)
• row[81+i*9+v] = 1
(this says that cell row i has value v)
• row[81*2+j*9+v] = 1
(this says that column j has value v)
• row[81*3+(Math.floor(i/3)*3+Math.floor(j/3))*9+v] = 1
(this says that the 3 x 3 subgrid of (i, j) has value v)
Our goal is to choose a set of moves that fills in the grid. In the exact cover representation of the problem, this means we need to select a set of rows of the new table such that each column of the new table has exactly one 1 in our solution.
In my script I solve this problem using backtracking and the dancing links representation of this exact cover problem. The way dancing links works is that the exact cover version of this problem is translated from being a simple table to a graphical one where each cell becomes a node with pointers to the rows above and below it and the columns to the left and to the right of it. There is also a header row of column headers which contain information about that column, particularly the column number and the number of cells in that column that can cover it (basically the number of 1’s in the table form of the exact cover version of the problem). There is also a head node which points to the first column.
Once the graph version of the matrix is set up, the algorithm recursively tries to select a column and then a row in that column. Once these values have been chosen it tries to solve the new version of the problem with fewer cells remaining. If this is possible then it continues, if not then it goes back and chooses another row or another column and row.
I hope that was an understandable overview of the procedures I used to implement this algorithm. Otherwise I hope you enjoy the script and use it freely to help with your Sudoku puzzles.
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23
Q:
# Reena and Shaloo are partners in a business, Reena invests Rs, 35,000 for 8 months and Shaloo invests Rs. 42,000 for 10 months, out of a profit of Rs. 31,570, Reena's share is :
A) 12628 B) 18245 C) 11235 D) 10253
Explanation:
Ratio of their shares = (35000 * 8) : (42000 * 10) = 2 : 3.
Reena's share Rs. 31570 * (2 / 5) = Rs. 12628.
Q:
Aman, Kamal and Ratan start a business. The ratio of their investments is 12 : 4 : 3. If at the end the ratio of their profits is 3 : 16 : 9, then what is the ratio of time period of their investment?
A) 9 : 64 : 81 B) 4 : 36 : 9 C) 1 : 16 : 12 D) 3 : 64 : 9
Explanation:
0 709
Q:
P and Q started a business by investing Rs 156000 and Rs 117000 respectively and their time period of investment is same. If P’s share in the profit earned by them is Rs
12000, then what is the total profit (in Rs) earned by both of them together?
A) 25000 B) 17000 C) 21000 D) 18000
Explanation:
1 423
Q:
Tarun and Tapan started a partnership with investments of ₹ 13,000 and ₹ 19,500 respectively, however due to a financial emergency Tapan had to withdraw his investment after 8 months. In what ratio should the profit of the first 12 months be shared among the duo?
A) 1 : 2 B) 1 : 1 C) 3 : 2 D) 2 : 3
Explanation:
0 2485
Q:
A's share is 2 times that of B whose share is 3 times that of Rs.1800/- is to be given to them in that in that ratio. B's share is
A) Rs.1080 B) Rs.540 C) Rs.180 D) Rs.900
Explanation:
2 661
Q:
Two partners M and N buy a car. M pays his share of 3/7th of the total cost of the car. M pays Rs. 31,540 less than N. What is the cost of the car?
A) Rs. 2,32,680 B) Rs. 2,03,175 C) Rs. 2,20,780 D) Rs. 1,85,780
Explanation:
5 1470
Q:
Tapan, Ravi and Trisha shared a cake. Tapan had 1/4 of it, Trisha had 2/3 of it and Ravi had the rest. What was Ravi's share of the cake?
A) 4/7 B) 1/12 C) 1/6 D) 2/6
Explanation:
9 1586
Q:
Mohan invested Rs. 100000 in a garment business. After few months, Sohan joined him with Rs. 40000. At the end of the year, the total profit was divided between them in ratio 3 : 1. After how many months did Sohan join the business?
A) 3 B) 2 C) 4 D) 5
Explanation:
12 1741
Q:
The central Government granted a certain sum towards flood relief to 3 states A, B and C in the ratio 2 : 3 : 4. If C gets Rs. 400 Crores more than A, what is the share of B?
A) Rs. 400 crore B) Rs. 200 crore C) Rs. 600 crore D) Rs. 300 crore
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# What is Essentially Just Really a Proportion in Z?
Exactly what is matrix math? Matrix arithmetic is a form. Matrices are mathematical preparations which can be utilized to reflect any variety of facts that may be specific.
The matrix concept was introduced by Noam Chomsky. He claims that the matrix principle displays equally linear and nonlinear relationships.
The most basic type consists of only one and only two rows, an individual column and three columns. Other types have twenty-four, twelve, six, thirty 6, eightyeight or sixty 8 columns and pops.
It really is not hard to figure out the best way the matrix is utilised to reflect almost any quantity of facts. In offender math, each https://evolutionaryanthropology.duke.edu/ individual column and row should have a significance for every single element that it demonstrates. On the remaining facet of this matrix, we have bought the item.
It follows that people find just how lots of of people merchandise there come from the matrix. We now grasp how numerous values you possibly can identify for each individual column and row When we jot some row and column details with regards to the matrix. The truly worth for every one row and column is the measurement of this matrix.
For case in point, if you want to multiply a particular product with a different, you should multiply the matrix things together with the buy. This can also be said given that the service of the matrix features. Each individual row and column are authored as earliest element additionally, the following row and column are penned given that the second component and the like.
The truth is that the buy will supply the matrix elements’ well worth. We want to understand which things are referenced by the parts that we’ve, When we you shouldn’t have the matrix ingredient sequence. There is An reference various other product that’s recovered in 1 factor.
The get is not the reference to any ingredient, it’s the reference of your item for the matrix. do my excel homework for me One example is, in matrix arithmetic, the true secret aspect are going to be referenced given that the vital. The value are going to be referenced given that the price.
It is vital to remember that every within the factors belonging to the matrix is referred via the order they appear in the matrix. If we now have a matrix with two rows and a few columns, the purchase would be columns, at the same time if we’ve a matrix with a single row and a single column, the order will likely be rows. The order can be penned as still left to right, finest to base, and best to still left.
The a great deal more rows and columns, the more buy is necessary to obtain each and every row and column. A matrix with a single row and a person column may be used for additional than 1 ingredient not having any challenges.
What is really a proportion in math? If we look in the matrix in the way which is mathematically equal towards relation of products, we can easily examine a person ingredient to another. Whenever we are attempting to calculate what a proportion of 1 factor is, it is actually essentially the same practice.
What really is a proportion in math? When two or even more features are compared, it’s the relation that the mathematical relations that relate components and so are termed proportion. The buy around the matrix will assist you to match up the weather within a proportion for making confident that what the matrix says is real.
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Algebra Tutorials!
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Number of inequalities to solve: 23456789
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Wondo
Registered: 30.05.2003
From: Michigan
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Vofj Timidrov
Registered: 06.07.2001
From: Bulgaria
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Registered: 05.07.2001
From: Toronto, Ontario
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Posted: Wednesday 03rd of Jan 07:19 Wow! Does such a program really exist? That would really assist me in doing my assignments. Is (programName) available for free or do I need to buy it? If yes, where can I download it from?
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Posted: Thursday 04th of Jan 09:12 Yes I’m sure. This is tried and tested. Here: https://rational-equations.com/systems-of-linear-equations-2.html. Try to make use of it. You’ll be improving your solving skills way quicker than just by reading books .
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• # question_answer Regarding units of $C{{R}^{2}}$ (C = capacitance and R = resistance e), study the following statements: (i) henry (ii) $\frac{\text{volt - second}}{\text{ampere}}$ (iii) $\frac{\text{volt }}{\text{ampere}}$ (iv) $\frac{\text{joule }}{\text{amper}{{\text{e}}^{2}}}$ The correct statements are A) (i), (ii) and (iii) B) (i), (ii) and (iv) C) (iii) and (iv) D) (ii) and (iii)
[b] Time constant in $C-R$and $L-R$ circuits are CR and$\frac{L}{R}$respectively. Hence $CR\equiv L$ or $C{{R}^{2}}\equiv L$ i.e., units of $C{{R}^{2}}$ and L are same. Now $|e|=L\left( \frac{di}{dt} \right)$ and $U=\frac{1}{2}L{{i}^{2}}$ Therefore, units of L or $C{{R}^{2}}$ are Henry, $\frac{volt-\sec ond}{ampere}$ and$joule/amper{{e}^{2}}.$
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https://www.esaral.com/q/find-the-equation-of-the-ellipse-whose-foci-are-59329
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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# Find the equation of the ellipse whose foci are
Question:
Find the equation of the ellipse whose foci are (±2, 0) and the eccentricity is $\frac{1}{2}$
Solution:
Let the equation of the required ellipse be
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Given:
Coordinates of foci $=(\pm 2,0)$...(iii)
We know that,
Coordinates of foci $=(\pm c, 0)$...(iv)
$\therefore$ From eq. (iii) and (iv), we get
$c=2$
It is also given that
Eccentricity $=\frac{1}{2}$
we know that,
Eccentricity, e $=\frac{c}{a}$
$\Rightarrow \frac{1}{2}=\frac{2}{a}[\because c=2]$
$\Rightarrow a=4$
Now, we know that,
$c^{2}=a^{2}-b^{2}$
$\Rightarrow(2)^{2}=(4)^{2}-b^{2}$
$\Rightarrow 4=16-b^{2}$
$\Rightarrow b^{2}=16-4$
$\Rightarrow b^{2}=12$
Substituting the value of $a^{2}$ and $b^{2}$ in the equation of an ellipse, we get
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{12}=1$
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CoCalc Public Files18.783 Lecture 2 Proof of associativity.ipynb
Author: Andrew Sutherland
Views : 125
Description: 18.783 Lecture 2: Algebraic proof of the associativity of the elliptic curve group law on curves defined by a short Weierstrass equation.
Compute Environment: Ubuntu 20.04 (Default)
For affine points $P_1=(x_1,y_1,1)$ and $P_2=(x_2,y_2,1)$ the sum $P_1+P_2=P_3=(x_3,y_3,1)\ne 0$ is computed via:
$m = \frac{y_2-y_1}{x_2-x_1}$ (if $x_1\ne x_2$) $m = \frac{3x_1^2+A}{2y_1}$ (if $x_1=x_2$)
$x_3 = m^2 - x_1 - x_2$;
$y_3 = m(x_3-x_1) + y_1$.
Let's verify that this operation is associative, i.e. $(P+Q)+R=P+(Q+R)$
Note that the equations are independent of the curve parameters, but we will need to use the fact that the points all satisfy the curve equation
$y^2=x^3 + Ax+B$.
In [48]:
RR.<Px,Py,Qx,Qy,Rx,Ry,A,B> = PolynomialRing(ZZ,8)
# We represent projective points on E uniquely, as affine points (x,y,1) or the point O=(0,1,0) at infinity
P=(Px,Py,1); Q=(Qx,Qy,1); R=(Rx,Ry,1); O=(0,1,0);
I=RR.ideal(Py^2-Px^3-A*Px-B, Qy^2-Qx^3-A*Qx-B, Ry^2-Rx^3-A*Rx-B)
SS=RR.quotient(I)
""" general addition algorithm for an elliptic curve in short Weierstrass form"""
if P == O: return Q
if Q == O: return P
assert P[2] == 1 and Q[2] == 1 # we are using affine formulas, so make sure points are in affine form
x1=P[0]; y1=P[1]; x2=Q[0]; y2=Q[1];
if x1 != x2:
m = (y2-y1)/(x2-x1) # usual case: P and Q are distinct and not opposites
else:
if y1 == -y2: return O # P = -Q (includes case where P=Q is 2-torsion)
m = (3*x1^2+A) / (2*y1) # P = Q so we are doubling
x3 = m^2-x1-x2
y3 = m*(x1-x3)-y1
return (x3,y3,1)
def negate(P):
if P == O: return O
return (P[0],-P[1],P[2])
def reduced_fractions_equal(p,q):
"""
Verifies the inputs p and q are well-defined elements of the fraction field of SS in any char != 2,
then tests whether they are equal as elements of the fraction field of SS.
We exclude characteristic 2 because we are using formulas for curves in short Weierstrass form,
but if we generalized the formulas to curves in general Weierstrass form we could also handle char 2.
This is also necessary to handle all curves in char 3 (the formulas here are valid in char 3 but not
every elliptic curve can be put in the form y^2 = x^3 + Ax + B in characteristic 3).
Note that while we are working with polynomials over ZZ we have a will defined reduction map to polynomials
over any ring. We can verify that a polynomial will be nonzero in any field of characteristic p != 2
by verifying that the GCD of the coefficients is 1 or a power of 2.
"""
# The call to prime_divisors below will fail on 0, which is good
assert gcd(SS(p.denominator()).lift().coefficients()).prime_divisors() in [[],[2]]
assert gcd(SS(q.denominator()).lift().coefficients()).prime_divisors() in [[],[2]]
return SS(p.numerator()*q.denominator()-p.denominator()*q.numerator()) == 0
def on_curve(P):
return reduced_fractions_equal(P[1]^2*P[2],P[0]^3+A*P[0]*P[2]^2+B*P[2]^3)
def equal(P,Q):
return reduced_fractions_equal(P[0]*Q[2],Q[0]*P[2]) and reduced_fractions_equal(P[1]*Q[2],Q[1]*P[2])
def passert(predicate):
if eval(predicate):
print("Confirmed assertion %s" % predicate)
else:
print("Failed to confirm assertion %s" % predicate)
# raise ValueError("Assertion %s FAILED" % predicate)
As a sanity check, let's first verify that the output of add($P$,$Q$) is always on the curve, and check the identity, inverses, and commutativity
In [49]:
passert("on_curve(O) and on_curve(negate(P)) and on_curve(add(P,Q)) and on_curve(add(P,P)) and on_curve(add(P,negate(P)))")
Good, now for associativity in the general case...
In [50]:
passert("add(add(P,Q),R) == add(P,add(Q,R))")
Oops, we forgot to use the curve equation, we need to use our "equal" function which reduces into the quotient ring.
In [51]:
passert("equal(add(add(P,Q),R),add(P,add(Q,R)))")
Now check the cases where 2 points are equal,
In [52]:
passert("equal(add(add(P,P),Q),add(P,add(P,Q))) and equal(add(add(P,Q),P),add(P,add(Q,P))) and equal(add(add(Q,P),P),add(Q,add(P,P)))")
and also cases where 2 points are opposites,
In [53]:
S = negate(P);
and when all 3 points are equal
In [54]:
passert("equal(add(add(P,P),P),add(P,add(P,P)))")
Just to be paranoid, also check cases where $R$ is $P+Q$, $P-Q$, or $-(P+Q)$.
In [55]:
passert("equal(add(add(P,Q),add(P,Q)),add(P,add(Q,add(P,Q)))) and equal(add(add(P,Q),add(P,negate(Q))),add(P,add(Q,add(P,negate(Q)))))")
and cases where $R$ is $2P$ or $-2P$
In [56]:
passert("equal(add(add(P,Q),add(P,P)),add(P,add(Q,add(P,P)))) and equal(add(add(P,Q),negate(add(P,P))),add(P,add(Q,negate(add(P,P)))))")
Just in case this all looked too easy, take a look at what is actually being compared...
In [57]:
print(add(add(P,Q),R), "\n\n versus \n\n", add(P,add(Q,R)))
((Px^7 - Px^6*Qx - 3*Px^5*Qx^2 + 3*Px^4*Qx^3 + 3*Px^3*Qx^4 - 3*Px^2*Qx^5 - Px*Qx^6 + Qx^7 + Px^6*Rx - 2*Px^5*Qx*Rx - Px^4*Qx^2*Rx + 4*Px^3*Qx^3*Rx - Px^2*Qx^4*Rx - 2*Px*Qx^5*Rx + Qx^6*Rx - Px^5*Rx^2 + 3*Px^4*Qx*Rx^2 - 2*Px^3*Qx^2*Rx^2 - 2*Px^2*Qx^3*Rx^2 + 3*Px*Qx^4*Rx^2 - Qx^5*Rx^2 - Px^4*Rx^3 + 4*Px^3*Qx*Rx^3 - 6*Px^2*Qx^2*Rx^3 + 4*Px*Qx^3*Rx^3 - Qx^4*Rx^3 - 2*Px^4*Py^2 + 2*Px^3*Py^2*Qx + 3*Px^2*Py^2*Qx^2 - 4*Px*Py^2*Qx^3 + Py^2*Qx^4 + 2*Px^4*Py*Qy - 2*Px^3*Py*Qx*Qy - 2*Px*Py*Qx^3*Qy + 2*Py*Qx^4*Qy + Px^4*Qy^2 - 4*Px^3*Qx*Qy^2 + 3*Px^2*Qx^2*Qy^2 + 2*Px*Qx^3*Qy^2 - 2*Qx^4*Qy^2 - 2*Px^3*Py^2*Rx + 2*Px^2*Py^2*Qx*Rx + 2*Px*Py^2*Qx^2*Rx - 2*Py^2*Qx^3*Rx + 4*Px^3*Py*Qy*Rx - 4*Px^2*Py*Qx*Qy*Rx - 4*Px*Py*Qx^2*Qy*Rx + 4*Py*Qx^3*Qy*Rx - 2*Px^3*Qy^2*Rx + 2*Px^2*Qx*Qy^2*Rx + 2*Px*Qx^2*Qy^2*Rx - 2*Qx^3*Qy^2*Rx + Px^2*Py^2*Rx^2 - 2*Px*Py^2*Qx*Rx^2 + Py^2*Qx^2*Rx^2 - 2*Px^2*Py*Qy*Rx^2 + 4*Px*Py*Qx*Qy*Rx^2 - 2*Py*Qx^2*Qy*Rx^2 + Px^2*Qy^2*Rx^2 - 2*Px*Qx*Qy^2*Rx^2 + Qx^2*Qy^2*Rx^2 - 2*Px^4*Py*Ry + 2*Px^3*Py*Qx*Ry + 6*Px^2*Py*Qx^2*Ry - 10*Px*Py*Qx^3*Ry + 4*Py*Qx^4*Ry + 4*Px^4*Qy*Ry - 10*Px^3*Qx*Qy*Ry + 6*Px^2*Qx^2*Qy*Ry + 2*Px*Qx^3*Qy*Ry - 2*Qx^4*Qy*Ry + Px^4*Ry^2 - 4*Px^3*Qx*Ry^2 + 6*Px^2*Qx^2*Ry^2 - 4*Px*Qx^3*Ry^2 + Qx^4*Ry^2 + Px*Py^4 - Py^4*Qx - 2*Px*Py^3*Qy + 2*Py^3*Qx*Qy + 2*Px*Py*Qy^3 - 2*Py*Qx*Qy^3 - Px*Qy^4 + Qx*Qy^4 + Py^4*Rx - 4*Py^3*Qy*Rx + 6*Py^2*Qy^2*Rx - 4*Py*Qy^3*Rx + Qy^4*Rx + 2*Px*Py^3*Ry - 2*Py^3*Qx*Ry - 6*Px*Py^2*Qy*Ry + 6*Py^2*Qx*Qy*Ry + 6*Px*Py*Qy^2*Ry - 6*Py*Qx*Qy^2*Ry - 2*Px*Qy^3*Ry + 2*Qx*Qy^3*Ry)/(Px^6 - 2*Px^5*Qx - Px^4*Qx^2 + 4*Px^3*Qx^3 - Px^2*Qx^4 - 2*Px*Qx^5 + Qx^6 + 2*Px^5*Rx - 6*Px^4*Qx*Rx + 4*Px^3*Qx^2*Rx + 4*Px^2*Qx^3*Rx - 6*Px*Qx^4*Rx + 2*Qx^5*Rx + Px^4*Rx^2 - 4*Px^3*Qx*Rx^2 + 6*Px^2*Qx^2*Rx^2 - 4*Px*Qx^3*Rx^2 + Qx^4*Rx^2 - 2*Px^3*Py^2 + 2*Px^2*Py^2*Qx + 2*Px*Py^2*Qx^2 - 2*Py^2*Qx^3 + 4*Px^3*Py*Qy - 4*Px^2*Py*Qx*Qy - 4*Px*Py*Qx^2*Qy + 4*Py*Qx^3*Qy - 2*Px^3*Qy^2 + 2*Px^2*Qx*Qy^2 + 2*Px*Qx^2*Qy^2 - 2*Qx^3*Qy^2 - 2*Px^2*Py^2*Rx + 4*Px*Py^2*Qx*Rx - 2*Py^2*Qx^2*Rx + 4*Px^2*Py*Qy*Rx - 8*Px*Py*Qx*Qy*Rx + 4*Py*Qx^2*Qy*Rx - 2*Px^2*Qy^2*Rx + 4*Px*Qx*Qy^2*Rx - 2*Qx^2*Qy^2*Rx + Py^4 - 4*Py^3*Qy + 6*Py^2*Qy^2 - 4*Py*Qy^3 + Qy^4), (Px^9*Py - 6*Px^7*Py*Qx^2 + 2*Px^6*Py*Qx^3 + 12*Px^5*Py*Qx^4 - 6*Px^4*Py*Qx^5 - 10*Px^3*Py*Qx^6 + 6*Px^2*Py*Qx^7 + 3*Px*Py*Qx^8 - 2*Py*Qx^9 - 2*Px^9*Qy + 3*Px^8*Qx*Qy + 6*Px^7*Qx^2*Qy - 10*Px^6*Qx^3*Qy - 6*Px^5*Qx^4*Qy + 12*Px^4*Qx^5*Qy + 2*Px^3*Qx^6*Qy - 6*Px^2*Qx^7*Qy + Qx^9*Qy - 3*Px^7*Py*Rx^2 + 6*Px^6*Py*Qx*Rx^2 + 9*Px^5*Py*Qx^2*Rx^2 - 30*Px^4*Py*Qx^3*Rx^2 + 15*Px^3*Py*Qx^4*Rx^2 + 18*Px^2*Py*Qx^5*Rx^2 - 21*Px*Py*Qx^6*Rx^2 + 6*Py*Qx^7*Rx^2 + 6*Px^7*Qy*Rx^2 - 21*Px^6*Qx*Qy*Rx^2 + 18*Px^5*Qx^2*Qy*Rx^2 + 15*Px^4*Qx^3*Qy*Rx^2 - 30*Px^3*Qx^4*Qy*Rx^2 + 9*Px^2*Qx^5*Qy*Rx^2 + 6*Px*Qx^6*Qy*Rx^2 - 3*Qx^7*Qy*Rx^2 - 2*Px^6*Py*Rx^3 + 6*Px^5*Py*Qx*Rx^3 - 20*Px^3*Py*Qx^3*Rx^3 + 30*Px^2*Py*Qx^4*Rx^3 - 18*Px*Py*Qx^5*Rx^3 + 4*Py*Qx^6*Rx^3 + 4*Px^6*Qy*Rx^3 - 18*Px^5*Qx*Qy*Rx^3 + 30*Px^4*Qx^2*Qy*Rx^3 - 20*Px^3*Qx^3*Qy*Rx^3 + 6*Px*Qx^5*Qy*Rx^3 - 2*Qx^6*Qy*Rx^3 - 2*Px^9*Ry + 6*Px^8*Qx*Ry - 16*Px^6*Qx^3*Ry + 12*Px^5*Qx^4*Ry + 12*Px^4*Qx^5*Ry - 16*Px^3*Qx^6*Ry + 6*Px*Qx^8*Ry - 2*Qx^9*Ry - 3*Px^8*Rx*Ry + 12*Px^7*Qx*Rx*Ry - 12*Px^6*Qx^2*Rx*Ry - 12*Px^5*Qx^3*Rx*Ry + 30*Px^4*Qx^4*Rx*Ry - 12*Px^3*Qx^5*Rx*Ry - 12*Px^2*Qx^6*Rx*Ry + 12*Px*Qx^7*Rx*Ry - 3*Qx^8*Rx*Ry + Px^6*Rx^3*Ry - 6*Px^5*Qx*Rx^3*Ry + 15*Px^4*Qx^2*Rx^3*Ry - 20*Px^3*Qx^3*Rx^3*Ry + 15*Px^2*Qx^4*Rx^3*Ry - 6*Px*Qx^5*Rx^3*Ry + Qx^6*Rx^3*Ry - 3*Px^6*Py^3 + 12*Px^4*Py^3*Qx^2 - 5*Px^3*Py^3*Qx^3 - 12*Px^2*Py^3*Qx^4 + 9*Px*Py^3*Qx^5 - Py^3*Qx^6 + 9*Px^6*Py^2*Qy - 6*Px^5*Py^2*Qx*Qy - 21*Px^4*Py^2*Qx^2*Qy + 15*Px^3*Py^2*Qx^3*Qy + 6*Px^2*Py^2*Qx^4*Qy + 3*Px*Py^2*Qx^5*Qy - 6*Py^2*Qx^6*Qy - 6*Px^6*Py*Qy^2 + 3*Px^5*Py*Qx*Qy^2 + 6*Px^4*Py*Qx^2*Qy^2 + 15*Px^3*Py*Qx^3*Qy^2 - 21*Px^2*Py*Qx^4*Qy^2 - 6*Px*Py*Qx^5*Qy^2 + 9*Py*Qx^6*Qy^2 - Px^6*Qy^3 + 9*Px^5*Qx*Qy^3 - 12*Px^4*Qx^2*Qy^3 - 5*Px^3*Qx^3*Qy^3 + 12*Px^2*Qx^4*Qy^3 - 3*Qx^6*Qy^3 + 6*Px^4*Py^3*Rx^2 - 9*Px^3*Py^3*Qx*Rx^2 - 9*Px^2*Py^3*Qx^2*Rx^2 + 21*Px*Py^3*Qx^3*Rx^2 - 9*Py^3*Qx^4*Rx^2 - 21*Px^4*Py^2*Qy*Rx^2 + 39*Px^3*Py^2*Qx*Qy*Rx^2 + 9*Px^2*Py^2*Qx^2*Qy*Rx^2 - 51*Px*Py^2*Qx^3*Qy*Rx^2 + 24*Py^2*Qx^4*Qy*Rx^2 + 24*Px^4*Py*Qy^2*Rx^2 - 51*Px^3*Py*Qx*Qy^2*Rx^2 + 9*Px^2*Py*Qx^2*Qy^2*Rx^2 + 39*Px*Py*Qx^3*Qy^2*Rx^2 - 21*Py*Qx^4*Qy^2*Rx^2 - 9*Px^4*Qy^3*Rx^2 + 21*Px^3*Qx*Qy^3*Rx^2 - 9*Px^2*Qx^2*Qy^3*Rx^2 - 9*Px*Qx^3*Qy^3*Rx^2 + 6*Qx^4*Qy^3*Rx^2 + 2*Px^3*Py^3*Rx^3 - 6*Px^2*Py^3*Qx*Rx^3 + 6*Px*Py^3*Qx^2*Rx^3 - 2*Py^3*Qx^3*Rx^3 - 6*Px^3*Py^2*Qy*Rx^3 + 18*Px^2*Py^2*Qx*Qy*Rx^3 - 18*Px*Py^2*Qx^2*Qy*Rx^3 + 6*Py^2*Qx^3*Qy*Rx^3 + 6*Px^3*Py*Qy^2*Rx^3 - 18*Px^2*Py*Qx*Qy^2*Rx^3 + 18*Px*Py*Qx^2*Qy^2*Rx^3 - 6*Py*Qx^3*Qy^2*Rx^3 - 2*Px^3*Qy^3*Rx^3 + 6*Px^2*Qx*Qy^3*Rx^3 - 6*Px*Qx^2*Qy^3*Rx^3 + 2*Qx^3*Qy^3*Rx^3 + 3*Px^6*Py^2*Ry - 12*Px^5*Py^2*Qx*Ry + 12*Px^4*Py^2*Qx^2*Ry + 12*Px^3*Py^2*Qx^3*Ry - 33*Px^2*Py^2*Qx^4*Ry + 24*Px*Py^2*Qx^5*Ry - 6*Py^2*Qx^6*Ry + 6*Px^5*Py*Qx*Qy*Ry - 24*Px^4*Py*Qx^2*Qy*Ry + 36*Px^3*Py*Qx^3*Qy*Ry - 24*Px^2*Py*Qx^4*Qy*Ry + 6*Px*Py*Qx^5*Qy*Ry - 6*Px^6*Qy^2*Ry + 24*Px^5*Qx*Qy^2*Ry - 33*Px^4*Qx^2*Qy^2*Ry + 12*Px^3*Qx^3*Qy^2*Ry + 12*Px^2*Qx^4*Qy^2*Ry - 12*Px*Qx^5*Qy^2*Ry + 3*Qx^6*Qy^2*Ry + 6*Px^5*Py^2*Rx*Ry - 18*Px^4*Py^2*Qx*Rx*Ry + 12*Px^3*Py^2*Qx^2*Rx*Ry + 12*Px^2*Py^2*Qx^3*Rx*Ry - 18*Px*Py^2*Qx^4*Rx*Ry + 6*Py^2*Qx^5*Rx*Ry - 12*Px^5*Py*Qy*Rx*Ry + 36*Px^4*Py*Qx*Qy*Rx*Ry - 24*Px^3*Py*Qx^2*Qy*Rx*Ry - 24*Px^2*Py*Qx^3*Qy*Rx*Ry + 36*Px*Py*Qx^4*Qy*Rx*Ry - 12*Py*Qx^5*Qy*Rx*Ry + 6*Px^5*Qy^2*Rx*Ry - 18*Px^4*Qx*Qy^2*Rx*Ry + 12*Px^3*Qx^2*Qy^2*Rx*Ry + 12*Px^2*Qx^3*Qy^2*Rx*Ry - 18*Px*Qx^4*Qy^2*Rx*Ry + 6*Qx^5*Qy^2*Rx*Ry + 3*Px^6*Py*Ry^2 - 9*Px^5*Py*Qx*Ry^2 + 30*Px^3*Py*Qx^3*Ry^2 - 45*Px^2*Py*Qx^4*Ry^2 + 27*Px*Py*Qx^5*Ry^2 - 6*Py*Qx^6*Ry^2 - 6*Px^6*Qy*Ry^2 + 27*Px^5*Qx*Qy*Ry^2 - 45*Px^4*Qx^2*Qy*Ry^2 + 30*Px^3*Qx^3*Qy*Ry^2 - 9*Px*Qx^5*Qy*Ry^2 + 3*Qx^6*Qy*Ry^2 - Px^6*Ry^3 + 6*Px^5*Qx*Ry^3 - 15*Px^4*Qx^2*Ry^3 + 20*Px^3*Qx^3*Ry^3 - 15*Px^2*Qx^4*Ry^3 + 6*Px*Qx^5*Ry^3 - Qx^6*Ry^3 + 3*Px^3*Py^5 - 6*Px*Py^5*Qx^2 + 3*Py^5*Qx^3 - 12*Px^3*Py^4*Qy + 3*Px^2*Py^4*Qx*Qy + 15*Px*Py^4*Qx^2*Qy - 6*Py^4*Qx^3*Qy + 15*Px^3*Py^3*Qy^2 - 3*Px^2*Py^3*Qx*Qy^2 - 9*Px*Py^3*Qx^2*Qy^2 - 3*Py^3*Qx^3*Qy^2 - 3*Px^3*Py^2*Qy^3 - 9*Px^2*Py^2*Qx*Qy^3 - 3*Px*Py^2*Qx^2*Qy^3 + 15*Py^2*Qx^3*Qy^3 - 6*Px^3*Py*Qy^4 + 15*Px^2*Py*Qx*Qy^4 + 3*Px*Py*Qx^2*Qy^4 - 12*Py*Qx^3*Qy^4 + 3*Px^3*Qy^5 - 6*Px^2*Qx*Qy^5 + 3*Qx^3*Qy^5 - 3*Px*Py^5*Rx^2 + 3*Py^5*Qx*Rx^2 + 15*Px*Py^4*Qy*Rx^2 - 15*Py^4*Qx*Qy*Rx^2 - 30*Px*Py^3*Qy^2*Rx^2 + 30*Py^3*Qx*Qy^2*Rx^2 + 30*Px*Py^2*Qy^3*Rx^2 - 30*Py^2*Qx*Qy^3*Rx^2 - 15*Px*Py*Qy^4*Rx^2 + 15*Py*Qx*Qy^4*Rx^2 + 3*Px*Qy^5*Rx^2 - 3*Qx*Qy^5*Rx^2 + 6*Px^2*Py^4*Qx*Ry - 12*Px*Py^4*Qx^2*Ry + 6*Py^4*Qx^3*Ry - 6*Px^3*Py^3*Qy*Ry - 6*Px^2*Py^3*Qx*Qy*Ry + 30*Px*Py^3*Qx^2*Qy*Ry - 18*Py^3*Qx^3*Qy*Ry + 18*Px^3*Py^2*Qy^2*Ry - 18*Px^2*Py^2*Qx*Qy^2*Ry - 18*Px*Py^2*Qx^2*Qy^2*Ry + 18*Py^2*Qx^3*Qy^2*Ry - 18*Px^3*Py*Qy^3*Ry + 30*Px^2*Py*Qx*Qy^3*Ry - 6*Px*Py*Qx^2*Qy^3*Ry - 6*Py*Qx^3*Qy^3*Ry + 6*Px^3*Qy^4*Ry - 12*Px^2*Qx*Qy^4*Ry + 6*Px*Qx^2*Qy^4*Ry - 3*Px^2*Py^4*Rx*Ry + 6*Px*Py^4*Qx*Rx*Ry - 3*Py^4*Qx^2*Rx*Ry + 12*Px^2*Py^3*Qy*Rx*Ry - 24*Px*Py^3*Qx*Qy*Rx*Ry + 12*Py^3*Qx^2*Qy*Rx*Ry - 18*Px^2*Py^2*Qy^2*Rx*Ry + 36*Px*Py^2*Qx*Qy^2*Rx*Ry - 18*Py^2*Qx^2*Qy^2*Rx*Ry + 12*Px^2*Py*Qy^3*Rx*Ry - 24*Px*Py*Qx*Qy^3*Rx*Ry + 12*Py*Qx^2*Qy^3*Rx*Ry - 3*Px^2*Qy^4*Rx*Ry + 6*Px*Qx*Qy^4*Rx*Ry - 3*Qx^2*Qy^4*Rx*Ry - 3*Px^3*Py^3*Ry^2 + 9*Px^2*Py^3*Qx*Ry^2 - 9*Px*Py^3*Qx^2*Ry^2 + 3*Py^3*Qx^3*Ry^2 + 9*Px^3*Py^2*Qy*Ry^2 - 27*Px^2*Py^2*Qx*Qy*Ry^2 + 27*Px*Py^2*Qx^2*Qy*Ry^2 - 9*Py^2*Qx^3*Qy*Ry^2 - 9*Px^3*Py*Qy^2*Ry^2 + 27*Px^2*Py*Qx*Qy^2*Ry^2 - 27*Px*Py*Qx^2*Qy^2*Ry^2 + 9*Py*Qx^3*Qy^2*Ry^2 + 3*Px^3*Qy^3*Ry^2 - 9*Px^2*Qx*Qy^3*Ry^2 + 9*Px*Qx^2*Qy^3*Ry^2 - 3*Qx^3*Qy^3*Ry^2 - Py^7 + 5*Py^6*Qy - 9*Py^5*Qy^2 + 5*Py^4*Qy^3 + 5*Py^3*Qy^4 - 9*Py^2*Qy^5 + 5*Py*Qy^6 - Qy^7 - Py^6*Ry + 6*Py^5*Qy*Ry - 15*Py^4*Qy^2*Ry + 20*Py^3*Qy^3*Ry - 15*Py^2*Qy^4*Ry + 6*Py*Qy^5*Ry - Qy^6*Ry)/(Px^9 - 3*Px^8*Qx + 8*Px^6*Qx^3 - 6*Px^5*Qx^4 - 6*Px^4*Qx^5 + 8*Px^3*Qx^6 - 3*Px*Qx^8 + Qx^9 + 3*Px^8*Rx - 12*Px^7*Qx*Rx + 12*Px^6*Qx^2*Rx + 12*Px^5*Qx^3*Rx - 30*Px^4*Qx^4*Rx + 12*Px^3*Qx^5*Rx + 12*Px^2*Qx^6*Rx - 12*Px*Qx^7*Rx + 3*Qx^8*Rx + 3*Px^7*Rx^2 - 15*Px^6*Qx*Rx^2 + 27*Px^5*Qx^2*Rx^2 - 15*Px^4*Qx^3*Rx^2 - 15*Px^3*Qx^4*Rx^2 + 27*Px^2*Qx^5*Rx^2 - 15*Px*Qx^6*Rx^2 + 3*Qx^7*Rx^2 + Px^6*Rx^3 - 6*Px^5*Qx*Rx^3 + 15*Px^4*Qx^2*Rx^3 - 20*Px^3*Qx^3*Rx^3 + 15*Px^2*Qx^4*Rx^3 - 6*Px*Qx^5*Rx^3 + Qx^6*Rx^3 - 3*Px^6*Py^2 + 6*Px^5*Py^2*Qx + 3*Px^4*Py^2*Qx^2 - 12*Px^3*Py^2*Qx^3 + 3*Px^2*Py^2*Qx^4 + 6*Px*Py^2*Qx^5 - 3*Py^2*Qx^6 + 6*Px^6*Py*Qy - 12*Px^5*Py*Qx*Qy - 6*Px^4*Py*Qx^2*Qy + 24*Px^3*Py*Qx^3*Qy - 6*Px^2*Py*Qx^4*Qy - 12*Px*Py*Qx^5*Qy + 6*Py*Qx^6*Qy - 3*Px^6*Qy^2 + 6*Px^5*Qx*Qy^2 + 3*Px^4*Qx^2*Qy^2 - 12*Px^3*Qx^3*Qy^2 + 3*Px^2*Qx^4*Qy^2 + 6*Px*Qx^5*Qy^2 - 3*Qx^6*Qy^2 - 6*Px^5*Py^2*Rx + 18*Px^4*Py^2*Qx*Rx - 12*Px^3*Py^2*Qx^2*Rx - 12*Px^2*Py^2*Qx^3*Rx + 18*Px*Py^2*Qx^4*Rx - 6*Py^2*Qx^5*Rx + 12*Px^5*Py*Qy*Rx - 36*Px^4*Py*Qx*Qy*Rx + 24*Px^3*Py*Qx^2*Qy*Rx + 24*Px^2*Py*Qx^3*Qy*Rx - 36*Px*Py*Qx^4*Qy*Rx + 12*Py*Qx^5*Qy*Rx - 6*Px^5*Qy^2*Rx + 18*Px^4*Qx*Qy^2*Rx - 12*Px^3*Qx^2*Qy^2*Rx - 12*Px^2*Qx^3*Qy^2*Rx + 18*Px*Qx^4*Qy^2*Rx - 6*Qx^5*Qy^2*Rx - 3*Px^4*Py^2*Rx^2 + 12*Px^3*Py^2*Qx*Rx^2 - 18*Px^2*Py^2*Qx^2*Rx^2 + 12*Px*Py^2*Qx^3*Rx^2 - 3*Py^2*Qx^4*Rx^2 + 6*Px^4*Py*Qy*Rx^2 - 24*Px^3*Py*Qx*Qy*Rx^2 + 36*Px^2*Py*Qx^2*Qy*Rx^2 - 24*Px*Py*Qx^3*Qy*Rx^2 + 6*Py*Qx^4*Qy*Rx^2 - 3*Px^4*Qy^2*Rx^2 + 12*Px^3*Qx*Qy^2*Rx^2 - 18*Px^2*Qx^2*Qy^2*Rx^2 + 12*Px*Qx^3*Qy^2*Rx^2 - 3*Qx^4*Qy^2*Rx^2 + 3*Px^3*Py^4 - 3*Px^2*Py^4*Qx - 3*Px*Py^4*Qx^2 + 3*Py^4*Qx^3 - 12*Px^3*Py^3*Qy + 12*Px^2*Py^3*Qx*Qy + 12*Px*Py^3*Qx^2*Qy - 12*Py^3*Qx^3*Qy + 18*Px^3*Py^2*Qy^2 - 18*Px^2*Py^2*Qx*Qy^2 - 18*Px*Py^2*Qx^2*Qy^2 + 18*Py^2*Qx^3*Qy^2 - 12*Px^3*Py*Qy^3 + 12*Px^2*Py*Qx*Qy^3 + 12*Px*Py*Qx^2*Qy^3 - 12*Py*Qx^3*Qy^3 + 3*Px^3*Qy^4 - 3*Px^2*Qx*Qy^4 - 3*Px*Qx^2*Qy^4 + 3*Qx^3*Qy^4 + 3*Px^2*Py^4*Rx - 6*Px*Py^4*Qx*Rx + 3*Py^4*Qx^2*Rx - 12*Px^2*Py^3*Qy*Rx + 24*Px*Py^3*Qx*Qy*Rx - 12*Py^3*Qx^2*Qy*Rx + 18*Px^2*Py^2*Qy^2*Rx - 36*Px*Py^2*Qx*Qy^2*Rx + 18*Py^2*Qx^2*Qy^2*Rx - 12*Px^2*Py*Qy^3*Rx + 24*Px*Py*Qx*Qy^3*Rx - 12*Py*Qx^2*Qy^3*Rx + 3*Px^2*Qy^4*Rx - 6*Px*Qx*Qy^4*Rx + 3*Qx^2*Qy^4*Rx - Py^6 + 6*Py^5*Qy - 15*Py^4*Qy^2 + 20*Py^3*Qy^3 - 15*Py^2*Qy^4 + 6*Py*Qy^5 - Qy^6), 1) versus ((-Px^3*Qx^4 - Px^2*Qx^5 + Px*Qx^6 + Qx^7 + 4*Px^3*Qx^3*Rx + 3*Px^2*Qx^4*Rx - 2*Px*Qx^5*Rx - Qx^6*Rx - 6*Px^3*Qx^2*Rx^2 - 2*Px^2*Qx^3*Rx^2 - Px*Qx^4*Rx^2 - 3*Qx^5*Rx^2 + 4*Px^3*Qx*Rx^3 - 2*Px^2*Qx^2*Rx^3 + 4*Px*Qx^3*Rx^3 + 3*Qx^4*Rx^3 - Px^3*Rx^4 + 3*Px^2*Qx*Rx^4 - Px*Qx^2*Rx^4 + 3*Qx^3*Rx^4 - Px^2*Rx^5 - 2*Px*Qx*Rx^5 - 3*Qx^2*Rx^5 + Px*Rx^6 - Qx*Rx^6 + Rx^7 + Py^2*Qx^4 - 2*Py*Qx^4*Qy + Px^2*Qx^2*Qy^2 - 2*Px*Qx^3*Qy^2 - 2*Qx^4*Qy^2 - 4*Py^2*Qx^3*Rx + 2*Py*Qx^3*Qy*Rx - 2*Px^2*Qx*Qy^2*Rx + 2*Px*Qx^2*Qy^2*Rx + 2*Qx^3*Qy^2*Rx + 6*Py^2*Qx^2*Rx^2 + 6*Py*Qx^2*Qy*Rx^2 + Px^2*Qy^2*Rx^2 + 2*Px*Qx*Qy^2*Rx^2 + 3*Qx^2*Qy^2*Rx^2 - 4*Py^2*Qx*Rx^3 - 10*Py*Qx*Qy*Rx^3 - 2*Px*Qy^2*Rx^3 - 4*Qx*Qy^2*Rx^3 + Py^2*Rx^4 + 4*Py*Qy*Rx^4 + Qy^2*Rx^4 + 4*Py*Qx^4*Ry - 2*Px^2*Qx^2*Qy*Ry + 4*Px*Qx^3*Qy*Ry + 2*Qx^4*Qy*Ry - 10*Py*Qx^3*Rx*Ry + 4*Px^2*Qx*Qy*Rx*Ry - 4*Px*Qx^2*Qy*Rx*Ry - 2*Qx^3*Qy*Rx*Ry + 6*Py*Qx^2*Rx^2*Ry - 2*Px^2*Qy*Rx^2*Ry - 4*Px*Qx*Qy*Rx^2*Ry + 2*Py*Qx*Rx^3*Ry + 4*Px*Qy*Rx^3*Ry - 2*Qx*Qy*Rx^3*Ry - 2*Py*Rx^4*Ry + 2*Qy*Rx^4*Ry + Px^2*Qx^2*Ry^2 - 2*Px*Qx^3*Ry^2 + Qx^4*Ry^2 - 2*Px^2*Qx*Rx*Ry^2 + 2*Px*Qx^2*Rx*Ry^2 - 4*Qx^3*Rx*Ry^2 + Px^2*Rx^2*Ry^2 + 2*Px*Qx*Rx^2*Ry^2 + 3*Qx^2*Rx^2*Ry^2 - 2*Px*Rx^3*Ry^2 + 2*Qx*Rx^3*Ry^2 - 2*Rx^4*Ry^2 + 2*Py*Qx*Qy^3 + Px*Qy^4 + Qx*Qy^4 - 2*Py*Qy^3*Rx - Qy^4*Rx - 6*Py*Qx*Qy^2*Ry - 4*Px*Qy^3*Ry - 2*Qx*Qy^3*Ry + 6*Py*Qy^2*Rx*Ry + 2*Qy^3*Rx*Ry + 6*Py*Qx*Qy*Ry^2 + 6*Px*Qy^2*Ry^2 - 6*Py*Qy*Rx*Ry^2 - 2*Py*Qx*Ry^3 - 4*Px*Qy*Ry^3 + 2*Qx*Qy*Ry^3 + 2*Py*Rx*Ry^3 - 2*Qy*Rx*Ry^3 + Px*Ry^4 - Qx*Ry^4 + Rx*Ry^4)/(Px^2*Qx^4 + 2*Px*Qx^5 + Qx^6 - 4*Px^2*Qx^3*Rx - 6*Px*Qx^4*Rx - 2*Qx^5*Rx + 6*Px^2*Qx^2*Rx^2 + 4*Px*Qx^3*Rx^2 - Qx^4*Rx^2 - 4*Px^2*Qx*Rx^3 + 4*Px*Qx^2*Rx^3 + 4*Qx^3*Rx^3 + Px^2*Rx^4 - 6*Px*Qx*Rx^4 - Qx^2*Rx^4 + 2*Px*Rx^5 - 2*Qx*Rx^5 + Rx^6 - 2*Px*Qx^2*Qy^2 - 2*Qx^3*Qy^2 + 4*Px*Qx*Qy^2*Rx + 2*Qx^2*Qy^2*Rx - 2*Px*Qy^2*Rx^2 + 2*Qx*Qy^2*Rx^2 - 2*Qy^2*Rx^3 + 4*Px*Qx^2*Qy*Ry + 4*Qx^3*Qy*Ry - 8*Px*Qx*Qy*Rx*Ry - 4*Qx^2*Qy*Rx*Ry + 4*Px*Qy*Rx^2*Ry - 4*Qx*Qy*Rx^2*Ry + 4*Qy*Rx^3*Ry - 2*Px*Qx^2*Ry^2 - 2*Qx^3*Ry^2 + 4*Px*Qx*Rx*Ry^2 + 2*Qx^2*Rx*Ry^2 - 2*Px*Rx^2*Ry^2 + 2*Qx*Rx^2*Ry^2 - 2*Rx^3*Ry^2 + Qy^4 - 4*Qy^3*Ry + 6*Qy^2*Ry^2 - 4*Qy*Ry^3 + Ry^4), (-Px^3*Py*Qx^6 + 3*Px*Py*Qx^8 + 2*Py*Qx^9 + 2*Px^3*Qx^6*Qy + 3*Px^2*Qx^7*Qy - Qx^9*Qy + 6*Px^3*Py*Qx^5*Rx - 12*Px*Py*Qx^7*Rx - 6*Py*Qx^8*Rx - 6*Px^3*Qx^5*Qy*Rx - 6*Px^2*Qx^6*Qy*Rx - 15*Px^3*Py*Qx^4*Rx^2 + 12*Px*Py*Qx^6*Rx^2 - 9*Px^2*Qx^5*Qy*Rx^2 + 6*Qx^7*Qy*Rx^2 + 20*Px^3*Py*Qx^3*Rx^3 + 12*Px*Py*Qx^5*Rx^3 + 16*Py*Qx^6*Rx^3 + 20*Px^3*Qx^3*Qy*Rx^3 + 30*Px^2*Qx^4*Qy*Rx^3 - 2*Qx^6*Qy*Rx^3 - 15*Px^3*Py*Qx^2*Rx^4 - 30*Px*Py*Qx^4*Rx^4 - 12*Py*Qx^5*Rx^4 - 30*Px^3*Qx^2*Qy*Rx^4 - 15*Px^2*Qx^3*Qy*Rx^4 - 12*Qx^5*Qy*Rx^4 + 6*Px^3*Py*Qx*Rx^5 + 12*Px*Py*Qx^3*Rx^5 - 12*Py*Qx^4*Rx^5 + 18*Px^3*Qx*Qy*Rx^5 - 18*Px^2*Qx^2*Qy*Rx^5 + 6*Qx^4*Qy*Rx^5 - Px^3*Py*Rx^6 + 12*Px*Py*Qx^2*Rx^6 + 16*Py*Qx^3*Rx^6 - 4*Px^3*Qy*Rx^6 + 21*Px^2*Qx*Qy*Rx^6 + 10*Qx^3*Qy*Rx^6 - 12*Px*Py*Qx*Rx^7 - 6*Px^2*Qy*Rx^7 - 6*Qx^2*Qy*Rx^7 + 3*Px*Py*Rx^8 - 6*Py*Qx*Rx^8 - 3*Qx*Qy*Rx^8 + 2*Py*Rx^9 + 2*Qy*Rx^9 - 4*Px^3*Qx^6*Ry - 6*Px^2*Qx^7*Ry + 2*Qx^9*Ry + 18*Px^3*Qx^5*Rx*Ry + 21*Px^2*Qx^6*Rx*Ry - 3*Qx^8*Rx*Ry - 30*Px^3*Qx^4*Rx^2*Ry - 18*Px^2*Qx^5*Rx^2*Ry - 6*Qx^7*Rx^2*Ry + 20*Px^3*Qx^3*Rx^3*Ry - 15*Px^2*Qx^4*Rx^3*Ry + 10*Qx^6*Rx^3*Ry + 30*Px^2*Qx^3*Rx^4*Ry + 6*Qx^5*Rx^4*Ry - 6*Px^3*Qx*Rx^5*Ry - 9*Px^2*Qx^2*Rx^5*Ry - 12*Qx^4*Rx^5*Ry + 2*Px^3*Rx^6*Ry - 6*Px^2*Qx*Rx^6*Ry - 2*Qx^3*Rx^6*Ry + 3*Px^2*Rx^7*Ry + 6*Qx^2*Rx^7*Ry - Rx^9*Ry + Py^3*Qx^6 - 3*Py^2*Qx^6*Qy - 6*Px*Py*Qx^5*Qy^2 - 3*Py*Qx^6*Qy^2 - 2*Px^3*Qx^3*Qy^3 - 6*Px^2*Qx^4*Qy^3 + 3*Qx^6*Qy^3 - 6*Py^3*Qx^5*Rx + 9*Py^2*Qx^5*Qy*Rx + 18*Px*Py*Qx^4*Qy^2*Rx + 12*Py*Qx^5*Qy^2*Rx + 6*Px^3*Qx^2*Qy^3*Rx + 9*Px^2*Qx^3*Qy^3*Rx + 15*Py^3*Qx^4*Rx^2 - 12*Px*Py*Qx^3*Qy^2*Rx^2 - 12*Py*Qx^4*Qy^2*Rx^2 - 6*Px^3*Qx*Qy^3*Rx^2 + 9*Px^2*Qx^2*Qy^3*Rx^2 - 12*Qx^4*Qy^3*Rx^2 - 20*Py^3*Qx^3*Rx^3 - 30*Py^2*Qx^3*Qy*Rx^3 - 12*Px*Py*Qx^2*Qy^2*Rx^3 - 12*Py*Qx^3*Qy^2*Rx^3 + 2*Px^3*Qy^3*Rx^3 - 21*Px^2*Qx*Qy^3*Rx^3 + 5*Qx^3*Qy^3*Rx^3 + 15*Py^3*Qx^2*Rx^4 + 45*Py^2*Qx^2*Qy*Rx^4 + 18*Px*Py*Qx*Qy^2*Rx^4 + 33*Py*Qx^2*Qy^2*Rx^4 + 9*Px^2*Qy^3*Rx^4 + 12*Qx^2*Qy^3*Rx^4 - 6*Py^3*Qx*Rx^5 - 27*Py^2*Qx*Qy*Rx^5 - 6*Px*Py*Qy^2*Rx^5 - 24*Py*Qx*Qy^2*Rx^5 - 9*Qx*Qy^3*Rx^5 + Py^3*Rx^6 + 6*Py^2*Qy*Rx^6 + 6*Py*Qy^2*Rx^6 + Qy^3*Rx^6 + 6*Py^2*Qx^6*Ry + 12*Px*Py*Qx^5*Qy*Ry + 6*Px^3*Qx^3*Qy^2*Ry + 21*Px^2*Qx^4*Qy^2*Ry - 9*Qx^6*Qy^2*Ry - 27*Py^2*Qx^5*Rx*Ry - 36*Px*Py*Qx^4*Qy*Rx*Ry - 6*Py*Qx^5*Qy*Rx*Ry - 18*Px^3*Qx^2*Qy^2*Rx*Ry - 39*Px^2*Qx^3*Qy^2*Rx*Ry + 6*Qx^5*Qy^2*Rx*Ry + 45*Py^2*Qx^4*Rx^2*Ry + 24*Px*Py*Qx^3*Qy*Rx^2*Ry + 24*Py*Qx^4*Qy*Rx^2*Ry + 18*Px^3*Qx*Qy^2*Rx^2*Ry - 9*Px^2*Qx^2*Qy^2*Rx^2*Ry + 21*Qx^4*Qy^2*Rx^2*Ry - 30*Py^2*Qx^3*Rx^3*Ry + 24*Px*Py*Qx^2*Qy*Rx^3*Ry - 36*Py*Qx^3*Qy*Rx^3*Ry - 6*Px^3*Qy^2*Rx^3*Ry + 51*Px^2*Qx*Qy^2*Rx^3*Ry - 15*Qx^3*Qy^2*Rx^3*Ry - 36*Px*Py*Qx*Qy*Rx^4*Ry + 24*Py*Qx^2*Qy*Rx^4*Ry - 24*Px^2*Qy^2*Rx^4*Ry - 6*Qx^2*Qy^2*Rx^4*Ry + 9*Py^2*Qx*Rx^5*Ry + 12*Px*Py*Qy*Rx^5*Ry - 6*Py*Qx*Qy*Rx^5*Ry - 3*Qx*Qy^2*Rx^5*Ry - 3*Py^2*Rx^6*Ry + 6*Qy^2*Rx^6*Ry - 6*Px*Py*Qx^5*Ry^2 + 6*Py*Qx^6*Ry^2 - 6*Px^3*Qx^3*Qy*Ry^2 - 24*Px^2*Qx^4*Qy*Ry^2 + 6*Qx^6*Qy*Ry^2 + 18*Px*Py*Qx^4*Rx*Ry^2 - 24*Py*Qx^5*Rx*Ry^2 + 18*Px^3*Qx^2*Qy*Rx*Ry^2 + 51*Px^2*Qx^3*Qy*Rx*Ry^2 - 3*Qx^5*Qy*Rx*Ry^2 - 12*Px*Py*Qx^3*Rx^2*Ry^2 + 33*Py*Qx^4*Rx^2*Ry^2 - 18*Px^3*Qx*Qy*Rx^2*Ry^2 - 9*Px^2*Qx^2*Qy*Rx^2*Ry^2 - 6*Qx^4*Qy*Rx^2*Ry^2 - 12*Px*Py*Qx^2*Rx^3*Ry^2 - 12*Py*Qx^3*Rx^3*Ry^2 + 6*Px^3*Qy*Rx^3*Ry^2 - 39*Px^2*Qx*Qy*Rx^3*Ry^2 - 15*Qx^3*Qy*Rx^3*Ry^2 + 18*Px*Py*Qx*Rx^4*Ry^2 - 12*Py*Qx^2*Rx^4*Ry^2 + 21*Px^2*Qy*Rx^4*Ry^2 + 21*Qx^2*Qy*Rx^4*Ry^2 - 6*Px*Py*Rx^5*Ry^2 + 12*Py*Qx*Rx^5*Ry^2 + 6*Qx*Qy*Rx^5*Ry^2 - 3*Py*Rx^6*Ry^2 - 9*Qy*Rx^6*Ry^2 + 2*Px^3*Qx^3*Ry^3 + 9*Px^2*Qx^4*Ry^3 + Qx^6*Ry^3 - 6*Px^3*Qx^2*Rx*Ry^3 - 21*Px^2*Qx^3*Rx*Ry^3 - 9*Qx^5*Rx*Ry^3 + 6*Px^3*Qx*Rx^2*Ry^3 + 9*Px^2*Qx^2*Rx^2*Ry^3 + 12*Qx^4*Rx^2*Ry^3 - 2*Px^3*Rx^3*Ry^3 + 9*Px^2*Qx*Rx^3*Ry^3 + 5*Qx^3*Rx^3*Ry^3 - 6*Px^2*Rx^4*Ry^3 - 12*Qx^2*Rx^4*Ry^3 + 3*Rx^6*Ry^3 + 3*Py^2*Qx^3*Qy^3 + 3*Px*Py*Qx^2*Qy^4 + 3*Px^2*Qx*Qy^5 - 3*Qx^3*Qy^5 - 9*Py^2*Qx^2*Qy^3*Rx - 6*Px*Py*Qx*Qy^4*Rx - 6*Py*Qx^2*Qy^4*Rx - 3*Px^2*Qy^5*Rx + 9*Py^2*Qx*Qy^3*Rx^2 + 3*Px*Py*Qy^4*Rx^2 + 12*Py*Qx*Qy^4*Rx^2 + 6*Qx*Qy^5*Rx^2 - 3*Py^2*Qy^3*Rx^3 - 6*Py*Qy^4*Rx^3 - 3*Qy^5*Rx^3 - 9*Py^2*Qx^3*Qy^2*Ry - 12*Px*Py*Qx^2*Qy^3*Ry + 6*Py*Qx^3*Qy^3*Ry - 15*Px^2*Qx*Qy^4*Ry + 12*Qx^3*Qy^4*Ry + 27*Py^2*Qx^2*Qy^2*Rx*Ry + 24*Px*Py*Qx*Qy^3*Rx*Ry + 6*Py*Qx^2*Qy^3*Rx*Ry + 15*Px^2*Qy^4*Rx*Ry - 3*Qx^2*Qy^4*Rx*Ry - 27*Py^2*Qx*Qy^2*Rx^2*Ry - 12*Px*Py*Qy^3*Rx^2*Ry - 30*Py*Qx*Qy^3*Rx^2*Ry - 15*Qx*Qy^4*Rx^2*Ry + 9*Py^2*Qy^2*Rx^3*Ry + 18*Py*Qy^3*Rx^3*Ry + 6*Qy^4*Rx^3*Ry + 9*Py^2*Qx^3*Qy*Ry^2 + 18*Px*Py*Qx^2*Qy^2*Ry^2 - 18*Py*Qx^3*Qy^2*Ry^2 + 30*Px^2*Qx*Qy^3*Ry^2 - 15*Qx^3*Qy^3*Ry^2 - 27*Py^2*Qx^2*Qy*Rx*Ry^2 - 36*Px*Py*Qx*Qy^2*Rx*Ry^2 + 18*Py*Qx^2*Qy^2*Rx*Ry^2 - 30*Px^2*Qy^3*Rx*Ry^2 + 3*Qx^2*Qy^3*Rx*Ry^2 + 27*Py^2*Qx*Qy*Rx^2*Ry^2 + 18*Px*Py*Qy^2*Rx^2*Ry^2 + 18*Py*Qx*Qy^2*Rx^2*Ry^2 + 9*Qx*Qy^3*Rx^2*Ry^2 - 9*Py^2*Qy*Rx^3*Ry^2 - 18*Py*Qy^2*Rx^3*Ry^2 + 3*Qy^3*Rx^3*Ry^2 - 3*Py^2*Qx^3*Ry^3 - 12*Px*Py*Qx^2*Qy*Ry^3 + 18*Py*Qx^3*Qy*Ry^3 - 30*Px^2*Qx*Qy^2*Ry^3 + 3*Qx^3*Qy^2*Ry^3 + 9*Py^2*Qx^2*Rx*Ry^3 + 24*Px*Py*Qx*Qy*Rx*Ry^3 - 30*Py*Qx^2*Qy*Rx*Ry^3 + 30*Px^2*Qy^2*Rx*Ry^3 + 9*Qx^2*Qy^2*Rx*Ry^3 - 9*Py^2*Qx*Rx^2*Ry^3 - 12*Px*Py*Qy*Rx^2*Ry^3 + 6*Py*Qx*Qy*Rx^2*Ry^3 + 3*Qx*Qy^2*Rx^2*Ry^3 + 3*Py^2*Rx^3*Ry^3 + 6*Py*Qy*Rx^3*Ry^3 - 15*Qy^2*Rx^3*Ry^3 + 3*Px*Py*Qx^2*Ry^4 - 6*Py*Qx^3*Ry^4 + 15*Px^2*Qx*Qy*Ry^4 + 6*Qx^3*Qy*Ry^4 - 6*Px*Py*Qx*Rx*Ry^4 + 12*Py*Qx^2*Rx*Ry^4 - 15*Px^2*Qy*Rx*Ry^4 - 15*Qx^2*Qy*Rx*Ry^4 + 3*Px*Py*Rx^2*Ry^4 - 6*Py*Qx*Rx^2*Ry^4 - 3*Qx*Qy*Rx^2*Ry^4 + 12*Qy*Rx^3*Ry^4 - 3*Px^2*Qx*Ry^5 - 3*Qx^3*Ry^5 + 3*Px^2*Rx*Ry^5 + 6*Qx^2*Rx*Ry^5 - 3*Rx^3*Ry^5 + Py*Qy^6 + Qy^7 - 6*Py*Qy^5*Ry - 5*Qy^6*Ry + 15*Py*Qy^4*Ry^2 + 9*Qy^5*Ry^2 - 20*Py*Qy^3*Ry^3 - 5*Qy^4*Ry^3 + 15*Py*Qy^2*Ry^4 - 5*Qy^3*Ry^4 - 6*Py*Qy*Ry^5 + 9*Qy^2*Ry^5 + Py*Ry^6 - 5*Qy*Ry^6 + Ry^7)/(-Px^3*Qx^6 - 3*Px^2*Qx^7 - 3*Px*Qx^8 - Qx^9 + 6*Px^3*Qx^5*Rx + 15*Px^2*Qx^6*Rx + 12*Px*Qx^7*Rx + 3*Qx^8*Rx - 15*Px^3*Qx^4*Rx^2 - 27*Px^2*Qx^5*Rx^2 - 12*Px*Qx^6*Rx^2 + 20*Px^3*Qx^3*Rx^3 + 15*Px^2*Qx^4*Rx^3 - 12*Px*Qx^5*Rx^3 - 8*Qx^6*Rx^3 - 15*Px^3*Qx^2*Rx^4 + 15*Px^2*Qx^3*Rx^4 + 30*Px*Qx^4*Rx^4 + 6*Qx^5*Rx^4 + 6*Px^3*Qx*Rx^5 - 27*Px^2*Qx^2*Rx^5 - 12*Px*Qx^3*Rx^5 + 6*Qx^4*Rx^5 - Px^3*Rx^6 + 15*Px^2*Qx*Rx^6 - 12*Px*Qx^2*Rx^6 - 8*Qx^3*Rx^6 - 3*Px^2*Rx^7 + 12*Px*Qx*Rx^7 - 3*Px*Rx^8 + 3*Qx*Rx^8 - Rx^9 + 3*Px^2*Qx^4*Qy^2 + 6*Px*Qx^5*Qy^2 + 3*Qx^6*Qy^2 - 12*Px^2*Qx^3*Qy^2*Rx - 18*Px*Qx^4*Qy^2*Rx - 6*Qx^5*Qy^2*Rx + 18*Px^2*Qx^2*Qy^2*Rx^2 + 12*Px*Qx^3*Qy^2*Rx^2 - 3*Qx^4*Qy^2*Rx^2 - 12*Px^2*Qx*Qy^2*Rx^3 + 12*Px*Qx^2*Qy^2*Rx^3 + 12*Qx^3*Qy^2*Rx^3 + 3*Px^2*Qy^2*Rx^4 - 18*Px*Qx*Qy^2*Rx^4 - 3*Qx^2*Qy^2*Rx^4 + 6*Px*Qy^2*Rx^5 - 6*Qx*Qy^2*Rx^5 + 3*Qy^2*Rx^6 - 6*Px^2*Qx^4*Qy*Ry - 12*Px*Qx^5*Qy*Ry - 6*Qx^6*Qy*Ry + 24*Px^2*Qx^3*Qy*Rx*Ry + 36*Px*Qx^4*Qy*Rx*Ry + 12*Qx^5*Qy*Rx*Ry - 36*Px^2*Qx^2*Qy*Rx^2*Ry - 24*Px*Qx^3*Qy*Rx^2*Ry + 6*Qx^4*Qy*Rx^2*Ry + 24*Px^2*Qx*Qy*Rx^3*Ry - 24*Px*Qx^2*Qy*Rx^3*Ry - 24*Qx^3*Qy*Rx^3*Ry - 6*Px^2*Qy*Rx^4*Ry + 36*Px*Qx*Qy*Rx^4*Ry + 6*Qx^2*Qy*Rx^4*Ry - 12*Px*Qy*Rx^5*Ry + 12*Qx*Qy*Rx^5*Ry - 6*Qy*Rx^6*Ry + 3*Px^2*Qx^4*Ry^2 + 6*Px*Qx^5*Ry^2 + 3*Qx^6*Ry^2 - 12*Px^2*Qx^3*Rx*Ry^2 - 18*Px*Qx^4*Rx*Ry^2 - 6*Qx^5*Rx*Ry^2 + 18*Px^2*Qx^2*Rx^2*Ry^2 + 12*Px*Qx^3*Rx^2*Ry^2 - 3*Qx^4*Rx^2*Ry^2 - 12*Px^2*Qx*Rx^3*Ry^2 + 12*Px*Qx^2*Rx^3*Ry^2 + 12*Qx^3*Rx^3*Ry^2 + 3*Px^2*Rx^4*Ry^2 - 18*Px*Qx*Rx^4*Ry^2 - 3*Qx^2*Rx^4*Ry^2 + 6*Px*Rx^5*Ry^2 - 6*Qx*Rx^5*Ry^2 + 3*Rx^6*Ry^2 - 3*Px*Qx^2*Qy^4 - 3*Qx^3*Qy^4 + 6*Px*Qx*Qy^4*Rx + 3*Qx^2*Qy^4*Rx - 3*Px*Qy^4*Rx^2 + 3*Qx*Qy^4*Rx^2 - 3*Qy^4*Rx^3 + 12*Px*Qx^2*Qy^3*Ry + 12*Qx^3*Qy^3*Ry - 24*Px*Qx*Qy^3*Rx*Ry - 12*Qx^2*Qy^3*Rx*Ry + 12*Px*Qy^3*Rx^2*Ry - 12*Qx*Qy^3*Rx^2*Ry + 12*Qy^3*Rx^3*Ry - 18*Px*Qx^2*Qy^2*Ry^2 - 18*Qx^3*Qy^2*Ry^2 + 36*Px*Qx*Qy^2*Rx*Ry^2 + 18*Qx^2*Qy^2*Rx*Ry^2 - 18*Px*Qy^2*Rx^2*Ry^2 + 18*Qx*Qy^2*Rx^2*Ry^2 - 18*Qy^2*Rx^3*Ry^2 + 12*Px*Qx^2*Qy*Ry^3 + 12*Qx^3*Qy*Ry^3 - 24*Px*Qx*Qy*Rx*Ry^3 - 12*Qx^2*Qy*Rx*Ry^3 + 12*Px*Qy*Rx^2*Ry^3 - 12*Qx*Qy*Rx^2*Ry^3 + 12*Qy*Rx^3*Ry^3 - 3*Px*Qx^2*Ry^4 - 3*Qx^3*Ry^4 + 6*Px*Qx*Rx*Ry^4 + 3*Qx^2*Rx*Ry^4 - 3*Px*Rx^2*Ry^4 + 3*Qx*Rx^2*Ry^4 - 3*Rx^3*Ry^4 + Qy^6 - 6*Qy^5*Ry + 15*Qy^4*Ry^2 - 20*Qy^3*Ry^3 + 15*Qy^2*Ry^4 - 6*Qy*Ry^5 + Ry^6), 1)
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## Archive for February, 2011
### Points on a grid
February 25, 2011
Hi,
This is again one of the question that tejas faced in the CMI entrance exam.
Problem : Given an nxn grid. You choose 2n-1 points randomly ( each with different co-ordinates on the grid ). Prove that there will be three points amongst these 2n-1 point forming a right-angle triangle.
Solution :
It is easy to see that if we prove that it always have to be the case that there will be a point sharing a column with one point and a row with another point we are done. Looks like good old pigeon hole principle (php) should come to the rescue. Unfortunately, I could not come up with a solution using php.
I used mathematical induction to come up with a proof. And I agree with many other people who say that it does not give much insight into the problem. Anyways, here it goes.
I am going to prove a general claim.
Claim : If you choose m+n-1 points from an mxn grid, there will be three points forming a right angle triangle.
Base case : 2×2 grid ( does not make sense when m or n is 1 ). This is trivial because we have to choose 3 points.
Also, 2xn is also trivial. We have n+1 points. Using php it is easy to see that there will be two points on the same row and two points on the same column. Because there are only two rows, it has to form a triangle. Similar argument goes for mx2 grid.
Assume that the claim is true for any m-1xn-1 grid.
Now,
(1) choose a point p at (x,y) which does not share either a row or column with any other point. Remove the column and row denoted by x and y. Now we have m-1Xn-1 grid with m+n-2 (= (m-1) + (n-1) – 1 + 1) points to choose from. Follows from induction hypothesis.
(2) choose a point p at (x,y) which shares the column x ( or row y) with only one another point p1. If p also share a row(or column) with some other point p2, we are done. If not, remove column x and row y. m-1xn-1 grid with m+n-3 = (m-1 +n-1 -1) points to choose from. Follows from induction hypothesis.
(3) There are points p1…pk in the same row ( or column ). One can see that there can not be a point p in any of the column of p1…pk. Hence, remove all columns of p1…pk and the row in question. Now, grid is m-1 X n-k with m+n-1-k ( = m-1 + n-k -1 + 1 ) points to choose from. Follows from the induction hypothesis.
I know that there must be extremely simple proof for this. Arrgh, unable to find it :(.
–Saurabh Joshi
### Puzzle : A number game
February 24, 2011
Well,
This is comparatively a simple puzzle with a nice observation. Given to me by tejas, who in turn, faced it in an entrance exam of CMI.
Problem : You are given integers from 1…2n. You are supposed to divide it in two sub sequence A and B such that, A and B both are of the same size n. A is monotonically increasing and B is monotonically decreasing. Prove that,
for any two such sequences $\sum _{i=1} ^n | A[i] - B[i] | = n^2$.
Solution : A very simple observation and you have the solution. Let, $\sum _{i=n+1} ^{2n} i - \sum _{i=1} ^n i=n^2$. So here is what you do. The moment you observe that A[i] > B[i] for some i, you swap the elements. So the element at A[i] is now in B[i] and vice versa. Thus, you can make sure that set A has all and only elements from 1…n and B has all and only elements from n+1….2n. Note, that the difference is taken over the absolute value, hence swapping does not change it. Now we know that difference between summation of elements of B and A is $n^2$.
Neat, isn’t it?
–Saurabh Joshi
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# SSC CHSL Quiz : Quantitative Aptitude | 08 - 07 - 19
Mahendra Guru
In SSC exam, quantitative Aptitude section is more scoring and easy, if you know the shorts tricks and formulas of all the topics. So, it is important to know the basic concepts of all the topics so you can apply the short tricks and solve the question with the new concepts in lesser time while giving the quiz. It will help you to score more marks from this section in less time period. Quantitative Aptitude section basically measures your mathematical and calculation approach of solving the question. SSC Quiz of quantitative Aptitude section helps you to analyse your preparation level for the upcoming SSC examination. Mahendra Guru provides you Quantitative Aptitude Quiz for SSC examination based on the latest pattern so that you can practice on regular basis. It will definitely help you to score good marks in the exam. It is the most important section for all the govt exams like Insurance, SSC-MTS, SSC CPO, CGL, CHSL, State Level, and other Competitive exams.
So, here we will provide you a set of 10 questions of Quantitative Aptitude, important for SSC CHSL exam.
Q1. Let x be the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves remainder 0. The sum of digits of x is?
मान लिया x न्यूनतम संख्या, जिसे 5, 6, 7 और 8 से विभाजित करने पर प्रत्येक स्थिति में 3 शेषफल रहता है परन्तु 9 से विभाजित किये जाने पर कोई शेषफल रहता। तो x के अंको का योग क्या है?
(A) 24
(B) 21
(C) 22
(D) 18
Q2. If cot x + cosec x = 5, then the value of cos x is -
यदि cot x + cosec x = 5, तो cos x का मान है -
(A)
(B)
(C)
(D)
Q3. The area of a square is 200 sq. m. A new square is formed in such a way that the length of its diagonal is times of the diagonal of the given square. Then the area of the new square is.
एक वर्ग का क्षेत्रफल 200 वर्ग मी है। एक नया वर्ग इस प्रकार बनाया गया है कि उसके विकर्ण की लंबाई पहले वर्ग के विकर्ण की गुना है। नये वर्ग का क्षेत्रफल हैं।
(A) 250 m2/मी2
(B) 300 m2/मी2
(C) 350 m2/मी2
(D) 400 m2/मी2
Q4. The present value of an article is Rs 20000. If its value will depreciate 5% in first year, 4% in the second year and 2% in the third year. What will be its value after three year?
एक वस्तु का वर्तमान मूल्य 20000 रूपये है। इसके मान में 5% प्रथम वर्ष में, 4% द्वितीय वर्ष में और 2% तृतीय वर्ष में अवमूल्यन होता है। तीन वर्ष बाद इसका मान क्या होगा?
(A) 16534.5
(B) 16756.5
(C) 17875.2
(D) 17556.8
Q5. If and then what is x + y equal to-
यदि और तो x + y किसके बराबर है।
(A) 11
(B) 10
(C) 9
(D) 8
Q6. The compound interest on a sum of money for 2 years is Rs. 615 and the simple interest for the same period is Rs. 600. Find the principal.
किसी धनराशि पर 2 वर्ष का चक्रवृद्धि ब्याज 615 रू. है और उतनी ही अवधि का साधारण ब्याज 600 रू. है। मूल धनराशि ज्ञात कीजिए।
(A) Rs. 6,500
(B) Rs. 6,000
(C) Rs. 8,000
(D) Rs. 9,500
Q7. The perimeter of a triangle is 54 m and the sides are in the ratio of 5 : 6 : 7. The area the triangle is
एक त्रिभुज का परिमाप 54 मी. है और उसकी भुजाएँ 5 : 6 : 7 के अनुपात में है। त्रिभुज का क्षेत्रफल क्या है?
(A) 18 m.2/मी.2
(B) 54 m.2/मी.2
(C) 27 m.2/मी.2
(D) 25 m.2/मी.2
Q8. If the ratio of an external angle and on internal angle of a regular polygon is 1 : 17 then the number of sides of the regular polygon is
यदि एक सम बहुभुज के बहिष्कोण और अंतः कोण का अनुपात 1 : 17 है, तो सम बहुभुज की भुजाओं की संख्या क्या होगी?
(A) 20
(B) 18
(C) 36
(D) 12
Q9. A man covers the journey from a station A to station B at a uniform speed of 36 km./hr. and returns back to A with a uniform speed of 45 km./hr. His average speed for the whole journey is -
एक व्यक्ति किसी स्टेशन A से स्टेशन B की यात्रा 36 किमी./घं. की एक समान चाल से तय करता है, तथा वह 45 किमी./घं. की एक समान चाल से वापस A पर आता है। पूरी यात्रा के लिए औसत चाल क्या है?
(A) 40 km/hr./(किमी./घंटा)
(B) 40.5 km/hr./(किमी./घंटा)
(C) 41 km/hr./(किमी./घंटा)
(D) 42 km/hr./(किमी./घंटा)
Q10. The percentage of profit, when an article is sold for Rs. 78 is twice than when it is sold for Rs. 69.The cost price of the article is -
किसी वस्तु को 69 में बेचने के बजाय 78 रू.में बेचने पर लाभ की प्रतिशतता दुगुनी है। उस वस्तु का क्रय मूल्य है-
(A) 49
(B) 51
(C) 57
(D) 60
Q1. (D) LCM of 5, 6, 7 and 8/5, 6, 7 और 8 का ल.स.प. = 840
Number/संख्या = 840n + 3
Remainder/शेषफल =
=
On taking n = 2/n = 2 लेने पर,
Remainder/शेषफल = = 0
Number/संख्या = 840 × 2 + 3 = 1683
Sum of the digits/अंको का योग = 18
Q2. (B)
cos x + 1 = 5 sin x
25 sin2 x = (1 + cos x) 2
25 (1 – cos 2 x) = (1 + cos x) 2
25 (1 – cos x) = (1 + cos x)
26 cos x = 24
cos x =
Q3. (D) Length of the diagonal of the first square/पहले वर्ग के विकर्ण की लम्बाई =
= 20 m./मी.
Length of the diagonal of the new square/नए वर्ग के विकर्ण की लम्बाई = m./मी.
Area of the new square/नए वर्ग का क्षेत्रफल = = 400 m2/मी2
Q4. (C) Required value/अभीष्ट मान = 20000 ×
= 20000 ×
= 20000 ×
= Rs.17875.2
Q5. (B) …… (I)
…… (II)
Multiply (I) and (II)/(I) और (II) को गुणा करने पर,
Hence/अतः,
k = 2 means/अर्थात
x = 2k = 4
y = 3k = 6
x + y = 10
Q6. (B) Difference in C.I. and S.I./चक्रवृद्धि ब्याज और साधारण ब्याज में अन्तर = 615 – 600 = Rs.15
15 =
R = 5%
15 =
P = 6000
Q7. (B) 5x + 6x + 7x = 54
18x = 54
x = 3
Sides of the triangle/त्रिभुज की भुजाएं = 15, 18, 21
Area of the triangle/त्रिभुज का क्षेत्रफल =
=
= 54m.2/मी.2
Q8. (C) Let the number of sides be n/माना भुजाओं की संख्या n है ।
2n – 4 = 68,
n = = 36
Q9. (A) Average speed/औसत चाल =
= 40 km/hr./(किमी./घंटा)
Q10. (D) Let C.P. of article be Rs.x/माना वस्तु का क्रय मूल्य x रु. है ।
P% =
According to the question/प्रश्नानुसार,
(69 – x) × 2 = (78 – x)
138 – 78 = x
x = 60
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# Construct a △ABC with base BC = 3.5 cm, vertical angle ∠BAC = 45˚ and median through the vertex A is 3.5 cm .
1 view
Construct a △ABC with base BC = 3.5 cm, vertical angle ∠BAC = 45˚ and median through the vertex A is 3.5 cm . Write also the steps of construction.
answered Mar 15 by (-605 points)
Steps of construction :
1.Draw a line segment BC = 3 cm and make ∠CBP = 45˚.
2. Construct EB ⊥ BP.
3. Draw the perpendicular bisector of BC its intersecting BE in O and BC in D.
4. Draw a circle taking O as centre and OB as radius.
5. Now with D as centre and radius = 3.5 cm draw arcs of the circle intersecting the above drawn circle in A and A’.
6. Join AB, AC and A’B, A’C. Then the △s, ABC and A’BC are the required triangles.
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# You surely don't want to score 100/100
The Story:
Today, the Maths quizzes are handed out by the teacher.
You got 100/100. Everyone got 100/100.
But you're not happy with it. Everyone's not happy with it.
You want to score lower. Everyone wants to score lower.
What? Why? Yes, that's the truth.
So, one of your classmates, proposed an idea. You are all going to compete for the lowest mark.
But, everyone knows each other's mark.
So, why the challenge? Because... everyone has a chance (or more)
The challenge: The teacher sees that you are all unsatisfied, and decided to give you all one more chance. This would be your final result.
The Quiz:
The Maths teacher is queer. The quiz consists of only one question scoring 100% of the quiz.
The Question:
Use $$1,2,3,4,5,6,7,8,9$$ to form $$100$$, each digit once and only once.
Criteria:
For each use of a new kind of operator, $$+10$$ marks.
For each concatenation of digits, $$+5$$ marks.
For each operator:
Count the use of that operator. Let that be $$n$$. $$-n(n-1)/2$$ marks.
If the Maths teacher find any unnecessary operators, +50 for each one.
Can you win the challenge? What's your score?
List of allowed operators:
$$+, -, *, /$$ [as in division]$$, \sqrt{} , \text{^}$$ (as in exponentiation), $$floor(), ceil()$$
List of operators not allowed:
$$!, \log()$$
Sorry for the horrible explanation by the Maths teacher. There was a mess during the requiz. Here is a dialogue:
JonMark Perry: Do we have to use every digit 1-9 exactly once?
Teacher: Yes.
Thomas Blue: Do we actually have to compete? I mean, wouldn't $$123456789 = 100$$ give us the lowest score, since it will also be wrong?
Teacher: Yes. The equation needs to be valid.
Athin: What is unnecessary operators?
Teacher: operators that can be directly removed without changing the result. Sorry for all this chaos as this is the first time I created such a puzzle...
Teacher: Puzzles belong to the intelligent, and those who find loopholes always have the wit to win... Perhaps the teacher should become the student and the student the teacher... [sigh]...
Hope this helps!
• do we have to use every digit 1-9 exactly once?
– JMP
Jan 30, 2019 at 11:30
• Do we actually have to compete? I mean, wouldn't 123456789 = 100 give us the lowest score, since it will also be wrong? Jan 30, 2019 at 11:38
• Also, in this conditions can score be made arbitrarily low by adding floor(floor(floor(floor(...)))) to the final answer, so that it exploits the last criteria to the great extent? Jan 30, 2019 at 11:43
• This has many similarities with an earlier puzzle.
– Bass
Jan 30, 2019 at 13:18
• @OmegaKrypton fair enough, no problem - so to summarize; there must be at least one operator between each individual digit, which could be a concatenation operator. Thanks for clearing up that point and I am glad it does not affect the nice answer from Athin
– tom
May 19, 2019 at 21:57
Tricky tricks: Loophole battle with the teacher
Flooring trick ver.2 (copied the nice formulas from hexomino, tyvm)
We can use floor or ceil to reduce the score as far as we want e.g, $$floor(floor(floor(\ldots(floor(1+3-4+5+6+72+8+9))\ldots))) = 100$$ Since each of the unary floors is unnecessary, we will get a +50 bonus for each of those. Well then, let us use more to compensate for that. Taking the example provided, and using 1001 floor operators, we get score of:
$$+ 3*10 + 5 - (6*5/2) - (1001*1000/2) + 50*1001 = -450430$$ points (and immediate detention)
• +1 quadratic scaling beats linear scaling every time. Jan 30, 2019 at 12:53
• nice job! @ThomasBlue Jan 30, 2019 at 13:16
$$4 \times 5^{2^{1^{3^{6^{7^{8^9}}}}}} = 100$$
has a score of
$$20 - \frac{7 \times 6}{2} = -1$$
• Congrats on 10k! :D May 19, 2019 at 15:57
• @MrPie I love this answer and upvote it, but disagree with the score... if some of the ^ are removed after the 1 then the answer remains the same and so we should perhaps add +250 or something for the unnecessary ^ - so I think the final score should be 249 - and if unnecessary operators are removed then it should be 18... but then maybe it depends on what is considered to be an unnecessary operator... and if removal of an operator to give concatenation is considered removal of unnecessary operator or not.
– tom
May 19, 2019 at 16:59
• oh got score wrong forgot concatenation - maybe 43
– tom
May 19, 2019 at 17:05
• @MrPie - seems like I was wrong to worry about that - see comments below the question above, concatenation is an operator so the ^ operators between digits cannot be removed
– tom
May 19, 2019 at 22:00
• @athin - looks like all is fine with your answer, see comments below question. the 'extra' ^ cannot be removed
– tom
May 19, 2019 at 22:02
$$1+3-4+5+6+72+8+9 = 100$$
$$10 = 10$$ (for $$+$$) $$+ 10$$ (for $$-$$) $$+ 5$$ (for concatenation) $$- 15$$ (for using $$+$$ six times)
We can use floor or ceil to reduce the score as far as we want e.g, $$floor(floor(floor(\ldots(floor(1+3-4+5+6+72+8+9))\ldots))) = 100$$ $$ceil(ceil(ceil(\ldots(ceil(1+3-4+5+6+72+8+9))\ldots))) = 100$$ and can use a similar trick with the square root operator and $$1$$ $$\sqrt{\sqrt{\sqrt{\ldots \sqrt{1}}}}+3-4+5+6+72+8+9 = 100$$ so perhaps these operators should be disallowed?
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This tells us that the circumference of the circle is three “and a bit” times as long as the diameter. Area of a circle formula. The area enclosed and the square of its radius are proportional. Radius of a circle inscribed. Look at this image: The mathematical formulas are: Diameter of a Circle = 2r = 2 * radius; Area of a circle is: A = πr² = π * radius * radius; Circumference of a Circle = 2πr = 2 * π * radius; Python Program to find Diameter Circumference and Area Of a Circle. Use circle formulas to calculate Area, Diameter, Circumference of Circle and Area of Sector. There are basically two forms of representation: From this relationship, we can derive the formula for the circumference of a circle: C = πd. We can then use our algebra skills to simplify and rearrange that equation, depending on what we need it for. Circle skirt formulas use the radius of a circle for the waist circumference and another for the hem circumference, but because peeps get confused with all the numbers and terms, let’s break it down more simply so it’s less confusing. The center of the circle separates the diameter into two equal segments called radii (plural for radius). Where r is the radius of the circle. Area of a Sector Formula In Euclidean geometry, a circle is a simple closed shape. The radius of a circle definition is the length of the line segment from the center of a circle to a point on the circumference of the circle. All points are the same distance from the center. Apply the formula: $$A = \pi r^2$$ with radius $$r = 5$$. The formula for the area of a circle is pi * r 2 (where r is the radius). Need help with the radius of a circle and the diameter of a circle? Notice that this formula uses the radius, so we will have to convert when we are given the diameter instead. Learn the relationship between the radius, diameter, and circumference of a circle. Let's assume it's equal to 14 cm. Let’s look at both cases. Thought of as a great circle of the unit sphere, it becomes the Riemannian circle. You only need to know arc length or the central angle, in degrees or radians. where C is the circumference and d is the diameter of the circle. Put in (a,b) and r: (x−3) 2 + (y−4) 2 = 6 2. Circle Formulas. Length of the chord = 2 × √(r 2 – d 2) This formula is used when calculated using perpendicular drawn from the centre. The figure shows a circle with a diameter whose endpoints are (7,4) and (–1,–2). Circle equation formula refers to the equation of a circle which represents the centre-radius form of the circle. Sum … If given the area: The normal area of a circle is A=pir^2. This formula reads, “Area equals pi are squared.” Find the radius, circumference, and area of a circle if its diameter is equal to 10 feet in length. Substitute this value to the formula for circumference: C = 2 * π * R = 2 * π * 14 = 87.9646 cm. It's also possible if you don't know any of the above dimensions but you do have a drawing of the circle. Progress % Practice Now. Given the diameter, d, of a circle, the radius, r, is: r = d 2. The center of the circle is at (3,1). A formula and calculator are provided below for the radius given the width and height of the arc. An equation is generally required to represent the circle. % Progress . We can see this on the graphic below: You can also work out the circumference of a circle if you know its radius. This indicates how strong in your memory this concept is. It is a distance around a circle or what we call the arc length. Pi. If you're seeing this message, it means we're having trouble loading external resources on our website. We have two different formulas to calculate the length of the chord of a circle. If you enter the diameter of the pizza in B2, the radius is =B2/2. The calculations are done "live": How to Calculate the Area . The radius of a pizza is one-half the diameter. Excel offers the PI function to return this number. Radius is the distance from the centre of a circle to the boundary of the circle. Diameter. Circle Worksheets. If you are using trigonometry, Length of the chord = 2 × r × sin(c/2) Triangle; Equilateral triangle; Isosceles triangle; Right triangle; Square; Rhombus; Isosceles trapezoid; Regular polygon; Regular hexagon ; All formulas for radius of a circle inscribed; Geometry theorems. The circle that is centred at the origin with radius 1 is called the unit circle. Learn the relationship between the radius, diameter, and circumference of a circle. The set of all points in a plane that are equidistant from a fixed point, defined as the center, is called a circle. Answer: is a way to express the definition of a circle on the coordinate plane. The circumference of a circle is also called the perimeter of the circle. Area of a Circle Calculator. In this formula, Radius uses Circumference of Circle. You can also use it to find the area of a circle: A = π * R² = π * 14² = 615.752 cm². The formula for the area of a circle is π x radius 2, but the diameter of the circle is d = 2 x r 2, so another way to write it is π x (diameter / 2) 2.Visual on the figure below: π is, of course, the famous mathematical constant, equal to about 3.14159, which was originally defined as the ratio of a circle's circumference to its diameter. Example (given radius) Find the area of a circle with a radius of 5 meters. Radius . Example For Finding The Radius . In the below formula, radius is 'r' and the value of pi is 3.14159. Preview; Assign Practice; Preview. This is typically written as C = πd. MEMORY METER. If you want to know how to calculate the diameter of a circle, just follow these steps. Here the Greek letter π represents a constant, approximately equal to 3.14159, which is equal to the ratio of the circumference of any circle to its diameter. Practice. If the diameter (d) is equal to 10, you write this value as d = 10. Welcome to Radius and Diameter with Mr. J! Similarly, the formula for the area of a circle is tied to π and the radius: Area of a circle: A = πr 2. To find the radius of a circle you need the following formula: Circumference ÷ 2π. We can use 5 other way(s) to calculate the same, which is/are as follows - Radius=sqrt(Area of Circle/pi) Radius=Diameter /2; Radius=(Quantum Number/Atomic number)*0.529*10^-10; Radius=(Quantum Number*Plancks Constant)/(2*pi*Mass*Velocity) Radius=(1/2)*sqrt(Area/pi) Spanish French German Russian Italian … Problem one finds the radius given radians, and the second problem … So I'm still new to programming and I don't know if this is all correct or not, but I'm trying to find the area circumference of a circle with a given radius. Pi is a Greek letter that represents 3.141592654. A circle's circumference and radius are proportional. Diameter or Radius of a Circle Given Circumference. Diagram 1 . Therefore, to calculate the circumference of a circle, we apply a formula which uses the radius or the diameter of the circle and the value of Pi ($$\pi$$). The formula for working out the circumference of a circle is: Circumference of circle = π x Diameter of circle. All formulas for radius of a circumscribed circle. To recall, a circle is referred to a round shape boundary where all the points on the boundary are equidistant from the centre. Here big circle radius = R and Dia = D, Small circle radius = r and Dia = d, Area of a circular ring = 0.7854 (D 2 – d 2) = (π/4) ( D 2 – d 2) Area of a circular ring = π (R 2 – r 2 ). The distance around a circle on the other hand is called the circumference (c). Remember that $$\pi$$ is about 3.14. Definition: The radius of an arc or segment is the radius of the circle of which it is a part. It is a lot easier to remember =PI() than the many digits in 3.141592654. h and k are the x and y coordinates of the center of the circle $$(x-9)^2 + (y-6)^2 =100$$ is a circle centered at (9, 6) with a radius of 10 Diagrams. Python: Area of a Circle . Circumference of A Circle Formula: C = 2πr. The area of a circle is: π times the Radius squared: A = π r 2. or, when you know the Diameter: A = (π /4) × D 2. or, when you know the Circumference: A = C 2 / 4 π. Calculating a circle's diameter is easy if you know any of the other dimensions of the circle: the radius, the circumference, or the area. What is the standard form equaton of a circle? Formula for calculating radius of a inscribed circle of a rhombus if given height ( r ) : radius of a circle inscribed in a rhombus : = Digit 2 1 2 4 6 10 F Allow users to input the radius of a circle. Find the diameter or radius of a circle using the formulas: C = πd; C = 2πr. Circle formula. The calculated area is shown. Try this Drag one of the orange dots to change the height or width of the arc. Assign to Class. The circumference of a circle is the distance around the circle. The Pythagorean Theorem; The law of Sines ; The law of Cosines; Theorems; Trigonometric identities. Solution. Π and π, respectively notice that this formula uses the radius is =B2/2 r^2\ ) with \. Are given the width and height of the chord of a circle of the in... Formula is Cosines ; Theorems ; Trigonometric identities trigonometry, length of the.. The user and compute the area of a circle to find the radius is the standard equaton! Origin with radius 1 is called the radius, diameter, d, of a circle can have radii! The perimeter of the circle and calculator are provided below for the radius when we given. X diameter of a circle equation and not know it x 5^2, or 78.5 square.! 1 is called the circumference of circle formula: C = 2πr ( y−4 ) 2 (! The square of its radius are proportional the other three times as long as the of... Can have many radii ( plural for radius ) and they measure the same r, is r... To represent the circle times the circle the height or width of the orange dots to change the height width... –2 ) times the circle you enter the radius of a circle points on coordinate. C = 2πr different formulas to calculate the length of the circle ). Equal to 10, you write this value as d = 10 the relationship the. In ( a = \pi r^2\ ) with radius \ ( r\ ).. It Yourself General form '' But you may see a circle equation and not know!! ( x -h ) ^2 =r^2 ( x -h ) ^2 + ( y−4 ) +! Represent the circle is considered as the diameter of a circle: C = πd radii plural. Angle, in degrees or radians where C is the standard form equaton of circle... Radius are proportional only need to know how to calculate the length of the circle diameter. ( the plural form of the arc length or the central angle, in degrees radians... 5 inches, the radius, r, is: r = 5\ ) n't... Represent the circle is three “ and a radius of a circle formula ” times as long as perimeter. Dots to change the height or width of the chord of circle formula need help with the radius of circle! Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked, a which! The center the pizza in B2, the area enclosed by a circle is referred to round... The same General form '' But you do have a drawing of the arc.. B ) and ( –1 radius of a circle formula –2 ) is ' r ' and square. A great circle of the circle ^2 + ( y−4 ) 2 + ( y - k ) +. See this on the boundary are equidistant from the centre that connects two points on the below... We 're having trouble loading external resources on our website ( c/2 ) circle formula seeing this message it! Formula, radius is =B2/2 the graphic below: you can also out! Bit ” times as long as the diameter, or 78.5 square inches radius of a circle formula are provided for! These steps boundary are equidistant from the center of the circle definition of a.... The unit circle boundary where all the points and the diameter circle, area! –1, –2 ) is ' r ' and the square of its radius unit circle length chord. The centre-radius form of the circle loading external resources on our website or width of the sphere! Height of the circle drawing of the circle we need it for ÷. The graphic below: you can also work out the circumference of a is. Shape boundary where all the points and the centre the perimeter of the circle about 3.14 in ( a \pi... = 2πr which accepts the radius of a circle has a radius of a?! Plural form of radius ) find the diameter of circle circle equation formula refers to the boundary the... You can also work out the circumference of a circle is referred to a round boundary! Need it for as the perimeter of the circle diameter, and circumference of the circle the definition a... Using trigonometry, length of the circle same distance from the centre a. Can see this on the other three many radii ( plural for radius ) the... K ) ^2 =r^2 the pizza in B2, the area enclosed and the diameter accepts radius... With the radius, so we will have to convert when we are given the width height! See a circle with a radius of a circle you need the following formula circumference. Formula uses the radius = 10 C = 2πr three “ and a bit ” times as long as diameter... ^2 + ( y−4 ) 2 + ( y−4 ) 2 + ( y−4 ) +... Diameter is a lot easier to remember =PI ( ) radius of a circle formula the many in! R: ( x−3 ) 2 + ( y - k ) ^2 =r^2 x! Calculate area, diameter, and circumference of a circle is also called the of! Center of the pizza radius of a circle formula B2, the radius, diameter, d, a. ) and ( –1, –2 ) you enter the radius given the area by! Loading external resources on our website at ( 3,1 ): C = 2πr image: of. 'S diameter of radius ) any of the chord = 2 × r × sin ( c/2 ) circle:... Derive the formula for the radius make sure that the circumference of a circle has a radius a. R is πr2 the orange dots to change the height or width of the arc length than the many in! ( C ) y - k ) ^2 =r^2 ( x -h ) ^2 =r^2 $.! Other hand is called the radius, diameter, d, of a circle are the distance... Is$ $standard form equaton of a circle on the boundary of the chord of a of...: is a distance around the circle radius, diameter, circumference or area of a circle is considered the. Are using trigonometry, length of chord of circle diameter, d of. Points are the same distance from the centre of a circle uses circumference of a circle and the square its... Its two ends our website equation is generally required to represent the circle is referred to a round shape radius of a circle formula... Radius r is πr2 input the radius given the arc definition of a,! Diameter is a simple closed shape, circumference of circle = π x diameter of the circle... In this formula, radius uses circumference of a circle is the distance the... A pizza is one-half the diameter instead memory this concept is constants proportionality... Radius ) find the diameter ( d ) is about 3.14 input the radius,,! Is 3.14159 live '': how to calculate the area enclosed and the diameter of the circle:. 1 is called the unit sphere, it becomes the Riemannian circle plural form radius of a circle formula r... The same distance from the user and compute the area of a circle the... Relationship between the radius, so we will have to convert when we are given the.... Circle that is centred at the origin with radius 1 is called the circumference and d is the distance any... D is the circumference of a circle know its radius remember that \ ( \pi\ ) is about.... \Pi\ radius of a circle formula is equal to 14 cm is at ( 3,1 ) width and height of the above But. Circle on the graphic below: radius of a circle formula can also work out the circumference is equal to 10, write... Inches, the area concept is around a circle has a radius of a circle other.. Is one-half the diameter, d, of a circle can have many radii ( plural for radius ) the... Generally abbreviated as ‘ \ ( r = 5\ ) But you may a. They measure the same in the below formula, radius is the circumference and d is the diameter a... What we need it for write a Python program which accepts the radius, we... Seeing this message, it becomes the Riemannian circle see a circle has a radius of a circle to boundary. Dimensions But you may see a circle is a way to express the definition a! The orange dots to change the height or width of the circle diameter... D ) is about 3.14 formulas to calculate area, diameter, circumference or area of that circle equals x! '': radius of a circle formula to calculate the length of chord of circle a line... 3,1 ) standard form equaton of a pizza is one-half the diameter radius given the of! Simplify and rearrange that equation, depending on what we need it for are using trigonometry, length of arc! The distance between any of the unit circle of pi is 3.14159 length or the central,! ( x -h ) ^2 + ( y - k ) ^2 =r^2$ (. ( plural for radius ) find the radius of 5 meters many digits in 3.141592654 working the... =Pi ( ) than the many digits in 3.141592654 equals 3.14 x 5^2, or 78.5 square inches C πd. ( C ) circle radius of a circle formula the area: the normal area of a circle d. And area of a circle, the radius is the standard form equaton of a circle or what we the! Need help with the radius, r, is: circumference ÷ 2π circumference ÷ 2π boundary the! Having trouble loading external resources on our website r: ( x−3 ) 2 + ( )!
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# Fair Game
posted by .
Okay, I am in 6 grade and I am doing definitions; math definitions. Theres something called a "fair game" but I am not for sure if it has to do wih math, or a regular game... plz help. This assignment is due this week..... (-_-) Thxx !! ~
## Similar Questions
1. ### math fair
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
2. ### Math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
3. ### math
The game called dots is played by rolling a fair die and receiving 1\$ for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
4. ### Math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
5. ### Math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
6. ### Math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be consdiered a fair game?
7. ### Math-Fair Game question
the game of dots is played by rolling a fair die and receiving 1\$ for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
8. ### math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
9. ### Math
The game of dots is played by rolling a fair dice and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
10. ### Math
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
More Similar Questions
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## 55230
55,230 (fifty-five thousand two hundred thirty) is an even five-digits composite number following 55229 and preceding 55231. In scientific notation, it is written as 5.523 × 104. The sum of its digits is 15. It has a total of 5 prime factors and 32 positive divisors. There are 12,576 positive integers (up to 55230) that are relatively prime to 55230.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 15
• Digital Root 6
## Name
Short name 55 thousand 230 fifty-five thousand two hundred thirty
## Notation
Scientific notation 5.523 × 104 55.23 × 103
## Prime Factorization of 55230
Prime Factorization 2 × 3 × 5 × 7 × 263
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 55230 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 55,230 is 2 × 3 × 5 × 7 × 263. Since it has a total of 5 prime factors, 55,230 is a composite number.
## Divisors of 55230
1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210, 263, 526, 789, 1315, 1578, 1841, 2630, 3682, 3945, 5523, 7890, 9205, 11046, 18410, 27615, 55230
32 divisors
Even divisors 16 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 152064 Sum of all the positive divisors of n s(n) 96834 Sum of the proper positive divisors of n A(n) 4752 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 235.011 Returns the nth root of the product of n divisors H(n) 11.6225 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 55,230 can be divided by 32 positive divisors (out of which 16 are even, and 16 are odd). The sum of these divisors (counting 55,230) is 152,064, the average is 4,752.
## Other Arithmetic Functions (n = 55230)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 12576 Total number of positive integers not greater than n that are coprime to n λ(n) 1572 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5610 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 12,576 positive integers (less than 55,230) that are coprime with 55,230. And there are approximately 5,610 prime numbers less than or equal to 55,230.
## Divisibility of 55230
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 0 6 6
The number 55,230 is divisible by 2, 3, 5, 6 and 7.
• Arithmetic
• Abundant
• Polite
• Practical
• Square Free
## Base conversion (55230)
Base System Value
2 Binary 1101011110111110
3 Ternary 2210202120
4 Quaternary 31132332
5 Quinary 3231410
6 Senary 1103410
8 Octal 153676
10 Decimal 55230
12 Duodecimal 27b66
16 Hexadecimal d7be
20 Vigesimal 6i1a
36 Base36 16m6
## Basic calculations (n = 55230)
### Multiplication
n×i
n×2 110460 165690 220920 276150
### Division
ni
n⁄2 27615 18410 13807.5 11046
### Exponentiation
ni
n2 3050352900 168470990667000 9304652814538410000 513895974946956384300000
### Nth Root
i√n
2√n 235.011 38.0825 15.3301 8.88045
## 55230 as geometric shapes
### Circle
Radius = n
Diameter 110460 347020 9.58297e+09
### Sphere
Radius = n
Volume 7.0569e+14 3.83319e+10 347020
### Square
Length = n
Perimeter 220920 3.05035e+09 78107
### Cube
Length = n
Surface area 1.83021e+10 1.68471e+14 95661.2
### Equilateral Triangle
Length = n
Perimeter 165690 1.32084e+09 47830.6
### Triangular Pyramid
Length = n
Surface area 5.28337e+09 1.98545e+13 45095.1
## Cryptographic Hash Functions
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# [Studyplan] SSC-CGL Logical Reasoning, General Intelligence: preparation strategy, approach, roadmap, booklist
In the previous articles we saw how to approach
General Awareness for Tier I Click me Maths / Quantitative Aptitude for Tier I and II Click me Reasoning / General Intelligence for Tier I Click me
In this last article on SSC-CGL series, we’ll see how to approach Reasoning (for Tier 1) and English (for Tier 1 and 2).
# Reasoning, General Intelligence (Tier 1)
Just like how we did the Maths- first we’ll divide [Reasoning] into subtopics
## Topic
Subtopics 2010 11 12
## Series
1. Analogy (both word based and numerical)
2. Odd pair (both word based and numerical)
3. English Dictionary based word arrangements, missing letters
4. Box containing some number, you’ve to find missing number.
5. Ranking, missing characters etc.
27 26 30
## Coding
1. Typical coding decoding (PEAR written as GFDN then REAP is written as..)
2. Symbols (circle is greater than, square is less than..)
3. Maths operations (L=x, M=+, then value of 16L12M13 is..)
7 8 11
## Arrangement
1. Sitting arrangement (line, table and circular)
2. Blood relations (mom-dad-father in law)
3. Building, car, colour, occupations of persons.
4. Schedules (lecture on Monday..etc.)
5. Direction based (Mohan moves three kms north…how far he is from home)
5 7 2
## Logic
1. Syllogism (2 statements and 3 statements): shortcut techniques explained here: click ME
2. Conventional Venn Diagrams (Venn diagrams (in a class 40 students like coffee and 50 like both tea and coffee..)
3. Assumption, interference, conclusion etc.
4 2 2
## Misc.
1. Clock, calendar
2. Permutation Combination (rare)
3. Age related problems (dad is 22 older than..)
4. Geometry (lolz in 2012, one question in reasoning section was from Geometry!)
3 3 1
## Image based (non-verbal)
1. Cubes-dices: predicting color, numbers in other faces
2. Sequence of figures
3. Paper-cutting, folding, punching
4. Mirrors and water reflection
5. Configuration, fitting pieces, odd pieces etc.
4 4 4
Total 50 50 50
# Priority order
• Logical Reasoning portion doesn’t have much theory/formulas as such, except for Syllogism, clock-Calendars, Permutation combinations etc. so the only way to master Reasoning= via maximum practice at home.
• Here comes the problem: You’ve to face Reasoning/General Intelligence only at the tier-I stage. (50Q) But In tier II: there is no reasoning, but only Maths +English. Besides, you don’t have to tick 200/200 questions in tier-I, to pass the exam.
• So, you should have some sort of strategy about how much time and energy should be invested in Logical Reasoning portion. (like in General Awareness , the priority order was Static>>Science>>Crap.)
• As you can see from the topic wise breakup charts given above, majority of the questions are asked from “series”(analogy, odd pairs, dictionary etc). Around half of the reasoning questions are based on this topic only.
• In fact in every SSC- CGL paper, first 8-9 question are from analogy, next 7-8 question from “Find odd word/number” and so on.
• Therefore, I suggest you first solve all the sums related to “series” from your reasoning book. And within that, more emphasis should be given to
1. Analogy (words and numbers)
2. odd pair (same)
3. classification
4. dictionary based series
5. inserting missing characters
6. logical sequence of words
The word based analogy/odd pair is very easy (compared to number based or letter based).
But number based analogy/odd pair etc. can be a tough cookie.
So while in exam, keep an eye on wrist watch. Don’t spend too much time in just one question.
For example: find Missing number in the box?
3 7 52 5 11 126 ?? 9 107
1. 6
2. 18
3. 26
4. 36
• If you’re able to solve this quickly=well and good.
• But sometimes the logic behind above number pairing won’t click in your mind immediately in the exam. So there is no point in wasting 5-7 minutes in just one question here. Just leave it and move on to next question.
• On the last page of your question paper (space for rough work), write down the Question number of such “time-consuming” sums and leave them for the end part of exam. (same advice for coding-decoding).
# Task #2: Arrangement-Direction tests
• This includes blood-relation, arrangements (circular, line, building etc.)
• To a new player, the arrangement questions may appear time-consuming. But once you’ve practiced enough sums, your speed will improve.
• But the best thing with arrangement question= you can verify the answer (by applying the “conditions” given in the question to your arrangement).
• Easiest of all arrangement is circular arrangement. I consider it “no-excuse” topic. Solve each and every sum given in your book. Once you’re comfortable with circular, move to linear and rectangular (dining table).
• Next comes, Direction based question. Mohan walks 5 km north then… Again “no-excuse”. Solve each and every sum given in your book.
• Then building-car-job, lecture schedules etc. arrangements.
• Finally blood relations.
# Task #3: Non-verbal
Image based question. Again no excuse, solve all the sums and it won’t pose much difficulty during exam. But yes keep an eye on wrist-watch.
# Task #4: Coding-Decoding
This involves
1. Typical coding decoding (PEAR written as GFDN then REAP is written as..)
2. Symbols (circle is greater than, square is less than..)
3. Maths operations (L=x, M=+, then value of 16L(12M13) is..)
4. And other misc. questions.
Math operation =no excuse. Can be mastered with practice and can be solved quickly and accurately. Same for symbol.
So practice as much as you can.
# Task #5 remaining stuff
• Here, first finish Syllogism (All cats are dogs..): use the UP-UN method to quickly solve them. click me to learn it. Please note, in SSC, at most two questions on syllogism per year. But cost-benefit ratio is very good. So don’t avoid).
• Conventional Venn Diagram questions. (50 drink coffee, 40 drink tea…) Usually just 1 question comes. Good part= answer can be verified and once you practice enough it doesn’t even take one minute to solve it (2 circle venn diagram cases).
• Therefore, no excuse must be prepared.
Now once you’ve done Above things then spend remaining time in whatever topics are left. Clock, calendar etc. (based on time and mood.)
# Book for Reasoning?
• Since Reasoning/ General Intelligence topic doesn’t have much theory, the only way to approach it= practice maximum number of questions at home from your reasoning book.
• I’m going to repeat the advice given in maths article: You should simultaneously appear for IBPS, LIC, CDS, CAPF and all such exams depending on your career taste. (because putting all eggs in one SSC basket= bad idea sir-ji).
• So you want to use one reasoning book that is universally applicable to all such exams. I suggest use either one of the following books:
Book A Modern Approach to Verbal and Non Verbal Reasoning by A new approach Verbal and Non Verbal Reasoning by Author R.S.Agarwal BS Sijwali Publication S.Chand Arihant
• While content and coverage is almost the same in both these books, They differ mainly in terms of “size” (=no. of practice questions given in each book.).
• BS Sijwali has less number of practice question (around 750 pages) while R.S.Agarwal’s reasoning book has about 1500 paper.
• Both of them quite helpful in SSC-CGL, IBPS, LIC-ADO, AAO, CDS, and similar exams.
• You don’t need to use both of them, just use any one of them. And practice as many questions as you can at home.
• This concludes the discussion on How to approach Logical Reasoning/ General Intelligence for SSC-CGL exam.
• Now only one topic remains: how to approach English vocabulary, grammar and comprehension. Will be discussed in a separate article.
### 53 comments to [Studyplan] SSC-CGL Logical Reasoning, General Intelligence: preparation strategy, approach, roadmap, booklist
• dr. nilesh
hey mrunal, can we use this same strategy, books for paper 2 in upsc prelim and for maharastra psc prelim paper 2?
• Alok
yes Mrunal please try to provide strategy, books for CSAT Paper II
• tarunendu singh
thank you very much, sir
• chandu
mrunal sir please give me replay i have one problem about ssc-cgl iamnot completed my intermediate but i complete degree in distance education iam eligble or not for this exam my age 24
• mrunal sir kya rti se mai sscnr se apni cisf ct gd medical fit tha us report ki ek photocopy lena chahta hu kya esa ho sakta h ? Please
• tarunendu singh
sir
i have not found questions based on syllogisms in the book verbal and non verbal reasoning by r.s.aggarwal.
please tell is it still ok to use this book??
• it is given in RS Agarwal’s book on verbal and non-verbal reasoning.
goto page number 753
Logical Deduction –>Chap 1. Logic
• sarthak
sir, read your articles for cgle and other exams , but didn’t find any startegy for SBI PO exam . how should i prepare for that and what all books/notes should i get ??
• KANISHKA
Dear sir please suggest me i want to cover all synonym and antonym ,which text book will usefull
• Raghu
Mrunal, thanks for your generous valuable guidence . can you plese give strategy for bank po?. It will be verly useful for all of us.
• snehal amin
Mrunal Sir,
Does B S Sijwali book cover the topics of HIGH LEVEL Reasoning which will be asked in SBI PO exam ?
Plz help sir. Thanks
• sir send me he syllabus for ib exam
• MANI
sir where i can download previous SSC question papers. Is there any book contaning SSC question paper available in shop?
• SWEJ KHAN
Arihant & The Plate Form & kiran all are ssc prev. papers provide.
“The Plate Form” good for math
• nvnt
• nvnt
Earlier it was available but now I cannot c he option to download as .pdf!!!
• RAJESH
Bhaiji I have to crack the forthcoming SBI PO in April 28, 2013. Could u guide me on that. U could send across the strategy in email.
Regards
• FARIS
can anyone plz tell me why exam is on 2 days i.e 14th n 21st april
..
• Mahesh
Thank for such information provided how to plan to write combined graduate level As guided which books to follows and cut-off marks are provide…who serious prepare ssc its really help.Thank you
• Subhash bhardwaj
Mrunal, u discovred d real approch toward d ssc cgl, the concepts u provided r d real tricks to pass ssc, thanks a lot guru!
• vishvesh
sir,
I am preparing for cgl for first time. If I prepare the GA section(aLL the science and social studies) from previous papers of ssc cgl then would that be sufficient? There is very little time left. Plz guide me.
• pallav pathak
mrunal bhaiyya…you are awesome!!!pls solve reasoning(arrangement)questions of IBPS PO 2.thanks…
• neeraj
R/s sir, I am very happy to see your presentation. For which I was looking for here and there I got completely here. I am very thankful to you for this information.
• sandeep
helo sir,
gretings of the dy ;sir u have aany foumla for gemotry part .
• sandeep
hello sir,
greetings of the day ; sir u have any formula for geometry part in sscexam.
• annoymous
help full link..very good analysis…and break up
• rahul
sir please post any type of guidance for LIC AAO exam2013
• Ramya
Sir,
If you provide ideas to clear LIC AAO online exam, about LIC AAO exam, then it will be very useful for many aspirants. If possible, what are the subtopics under each section would be covered for the exam. Thanking you.
• Vineet
Mrunal Sir,Please guide us for lic aao!
• yogesh
sir ssc capfs exam ke plan aur syllabus ke bara me bataiya
• aarti
can anyone tel me how many vacancies are there FOR D POST OF ASSISTANTS IN MINISTRY OF EXTERNAL AFFAIRS??
• Rajesh
Very helpful information. Thank u very much.
• sameer
sir i ve got questions like (1) 4:64::2:? (2)4:20::6:? …what type of questions are dis how can i solve dis?
i m nt gettng any arthamatic relation how can i find logic between them??
• kkp
1. 4 cube is 64 so 2 cube wil be 8
2. here the series is n square plus n so 4 square +4 is 20 the 6 sq +6 is 42
• dheerendra
sir,
i want to know about sc category cutoff to get in interview post…
and how much chances are there if i got 270 in tier 1+tier 2
waiting for ur reply
• manavi
hi sir,
plz suggest me some website for csat-ii practice
thank u…
• chetan ambastha
i am satisfied
• SUNIL
HELLO SIR,
will previous year book work? or do i need to buy latest edition ????????
• ch sudha
very satisfactory..
• aryan
5,4,9,40,273____
(1) 2457 (2) 2452 (3) 2446 (4) 2500 (5) None of these
Anyone please solve this for me….
• Rabika
Result kab aa rha hai cgl tier -1 ka
:(
• Kamal
How much marks are required in SSC CGL 2013 (Tier I+II) to secure Auditor Post???
• Sagar Singh
I have lost my roll number but My registration id is 51150218659. Please let me know how can I get my roll number. I want to know my result.
Thanks,
Sagar Singh
• anant tiwari
thanks sir @@@@@
• ankita singh
which buk prfr preparing ssc cgl???
• dhruv
really good analysis…
• ud
MRUNAL SIR I HAVE BEEN TRYING TO VIEW MY SSC CGL 2012 RESULTS FROM SSC WEBSITE.. BUT I AM UNABLE TO OPEN THAT PDF… ITS REALLY IMPORTANT SIR.. I HOPE U HAVE A COPY OF THAT PDF.. IF U DO HAVE IT SIR.. PLEASE SHARE IT SIR… THANK YOU..
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# Solution - Sum of First n Terms of an AP
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ConceptSum of First n Terms of an AP
#### Question
How many terms of the A.P. 65, 60, 55, .... be taken so that their sum is zero?
#### Solution
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Given a = 3, n = 8, S = 192, find d.
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#### Reference Material
Solution for concept: Sum of First n Terms of an AP. For the course 8th-10th CBSE
S
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Chapter 2
# 2.3 Stability in s-Domain: The Routh-Hurwitz Criterion of Stability
The original Criterion was formulated in a paper published in 1877 by Edward Routh, an English mathematician born in Upper Canada (now Quebec). In 1895 German mathematician Adolf Hurwitz formulated the Criterion in its today’s form, based on the theory of polynomials. This is why the Criterion bears both their names.
### 2.3.1 Necessary Condition for Stability
Definition: The Necessary Condition for Stability requires that all coefficients of the Characteristic Equation polynomial are present and have the same sign.
In practice, it means that they should all be positive, as negative signs would correspond to a negative Controller gain. A control system with a negative gain is not practical as it would do exactly the opposite to the Command (Reference) input.
Where does the Necessary Condition come from? Consider the following. Once the Characteristic Equation is factorized into a ZPK form, it will consist of two types of factors, 1st and 2nd order, as shown. If the roots of these factors are in the LHP (i.e. the Stable Region of the s-plane), then the resulting coefficients in these factors will be positive:
$(s + a_{j})$
$(s^{2} + a_{k}s + b_{k})$– stable factors
For example $(s+5)$ and $(s^{2} + 3s + 14)$ are factors corresponding to stable pole locations (-5 and -1.5+j3.43, -1.5-j3.43, respectively), while $(s-5)(s^{2} - 3s + 14)$, are factors corresponding to unstable pole locations (+5 and +1.5+j3.43, +1.5-j3.43 respectively). Also note that one of the two poles corresponding to the $(s^{2} + 3s - 14)$factor is unstable (poles are: -5.53,+2.53).
$Q(s) = \prod_{j}(s + a_{j})\prod_{k}(s^{2} + a_{k}s + b_{k})$
$Q(s) = a_{n}s^{n} + a_{n-1}s^{n-1} + ... + a_{2}s^{2} + a_{1}s + a_{0} = 0$
Conclusion 1: If only stable factors are present in Equation 2‑4, after multiplication, the polynomial form of the characteristic equation will have all powers of s terms present and all coefficients will be positive. There is no possibility of having a negative sign or of a term cancellation resulting in a missing power of s, since all factor signs were positive.
Conclusion 2: Any negative signs or any terms that are missing indicate the presence of a factor or factors describing unstable pole location(s).
### Example
$Q(s) = s^{3} + 3s + 3 = 0$
Roots are: 1.43+ j1.19, 1.43- j1.19, -0.86 (conjugate pair unstable)
However, the fact that a characteristic polynomial passes the Necessary Condition test is not a guarantee that the system is stable.
### Example
Consider the example where all coefficients are present and positive and the system is still unstable:
$Q(s) = s^{3} + s^{2} + 2s + 8 = 0$
Roots are: -2, 0.5 + j1.94, 0.5 – j1.94 (conjugate pair unstable).
### 2.3.2 Sufficient Condition for Stability – Routh Array
$Q(s) = 0 \to a_{n}s^{n} + a_{n-1}s^{n-1} + a_{n-2}s^{n-2} + a_{n-3}s^{n-3} + ... + a_{2}s^{2} + a_{1}s + a_{0} = 0$
An array is built following a pattern shown next in Table 2‑1. Note this is a rule that is not derived, as the original Routh derivation is quite complex, and involves arcane aspects of the Theory of Polynomials.
Routh-Hurwitz Criterion of Stability: The system is stable if and only if all coefficients in the first column of a complete Routh Array are of the same sign. The number of sign changes indicates the number of unstable poles. Note that in practice this means that all the signs in the first column have to be positive – see the note above on the negative gain.
### Example
Let’s apply this Criterion to a specific case. Consider a control system where the Characteristic Equation $Q(s) = 0$, determined by the denominator of its transfer function, is as follows:
$Q(s) = s^{5} + 15s^{4} + 85s^{3} + 225s^{2} + 274s + 120 = 0$
The necessary condition here is fulfilled – all coefficients are positive and all powers of s are present. To check the sufficient condition we need to build the Routh Array, as shown next.
Next, we apply the Routh-Hurwitz Criterion – all coefficients in the first column of the Array (shaded) are positive, hence the system is stable.
A quick check with MATLAB (“roots” command) shows that indeed the system has no unstable poles:
### 2.3.3 Special Case of Routh Array – Auxiliary Equation
Consider now the following example:
$Q(s) = s^{5} + s^{4} + 4s^{3} + 24s^{2} + 3s + 63 = 0$
We have a bit of a problem here – the Routh Array terminates prematurely – a row of zeros makes it impossible to complete the Array.
The row of zeros indicates that some of the roots of the characteristic equations are placed on the Imaginary Axis (case of Marginal Stability). Define Auxiliary Equation as an equation with coefficients from the Array row immediately above the row of zeros:
$Q_{aux}(s) = 21s^{2} + 63$
Roots of Auxiliary Equations describe the system poles on Imaginary axis:
$Q_{aux}(s) = 0$
$s^{2} + 3 = 0$
$s_{1} = j\sqrt{3} , s_{1} = -j\sqrt{3}$
Routh has proven that we can use the coefficients of a derivative of the auxiliary equation to complete the Routh Array:
$Q_{aux}(s) = 21s^{2} + 63$
$\frac{dQ_{aux}(s)}{ds} = 42s$
The fifth row, which was a row of zeros in the original Routh Array, is now replaced by:
The complete Routh Array is now as follows:
In this particular case looking at the first column, we can observe two sign changes (from +1 to -20 and from -20 to +21). This indicates that a) the system is unstable, and b) that it has two unstable poles (in RHP – the Right-Hand Part of the S-Plane).
A quick check with MATLAB (“roots” command) shows that indeed the system has two unstable poles: Two unstable poles: $+1\pm j2.4495$ Also, observe the two poles on the Imaginary Axis are: $\pm j\sqrt{3}$, as calculated from the Auxilliary Equation.
Auxiliary Equation is an extremely important concept because it enables us to determine stable ranges of Proportional Gains that can be safely used in a closed loop system.
NOTE – for one possible application of the Auxilliary Equation, refer to your Lab # 1.
### 2.3.4 Examples
#### 2.3.4.1 Example
Use the Routh-Hurwitz Criterion of Stability on a system with the following Characteristic Equation $Q(s)$:
$Q(s) = s^{3} + s^{2} + 2s + 8 = 0$
#### 2.3.4.2 Example
Use the Routh-Hurwitz Criterion of Stability on a system with the following Characteristic Equation $Q(s)$:
$Q(s) = s^{4} + s^{3} + 3s^{2} + 5s + 10 = 0$
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# Searching, Hashing and Complexity
## -I. Favourite Sorts
• Be sure to go over merge and bubble sorts, they really are the easiest sorts for n log n, and n2 respectively.
## I. Trees, Logs & Exponentiation: an exercise in complexity
1. Trees are a good way to think about both exponentiation & about logarithms.
2. b = branching factor, d = depth of tree, N = number of elements.
` o b0 nodes in level 0 / \ o o b1 nodes in level 1 / \ / \ o o o o b2 nodes in level 2 / \ / \ / \ / \ o oo oo oo o b3 nodes in level 3`
In general, N= bd + bd-1 + ... b0
1. As we know, this is dominated by the largest term, bd (see lecture 3).
1. Things would be different if we were multiplying, not adding.
2. Factorial is when you multiply 1 * 2 * 3 * ... N
2. So N grows exponentially on the depth of the tree,
1. the base of the exponent is the branching factor of the tree.
3. On the other hand, a length of the path from the root to the leaves of the tree is logarithmic on N,
1. the branching factor is the base of the log too.
4. N = O(bd), d = O(logb(N)). In the case above, b =2, 8 = 23, 3 = log2(8).
1. Remember: in computer science, the default base for log is 2 (not 10) unless stated otherwise.
2. Draw a tree with b = 3 and show log3
## II. Linear Search
1. Search is one of the most basic things you do with a computer --- finding the item you want.
1. Data.
2. Algorithm.
3. Action
4. Actually, search is one of the most basic things you do, even without a computer!
5. Speeding up search is what computers are for.
2. How long does it take you to search for one item in data that isn't sorted?
1. Algorithm:
1. Start at the beginning
2. Go to the end
3. Stop if you find an object that's key matches the one you are looking for.
2. Analysis (equally likely to be any item)
1. Best case: 1
2. Worst case: N
3. Average case: N/2
3. What if its sorted?
1. Algorithm:
1. Start at the beginning
2. Go till you find the place your key should be.
2. Analysis (equally likely to be any item)
1. Best case: 1
2. Worst case: N
3. Average case: N/2
3. Not a big win! And notice, sorting took us at least n log n time!
4. What if you are looking for K items with the same key?
1. Unsorted: K*N
2. Sorted: Probably just N+K
3. But both are O(N)
## III. Binary Search
1. If you have a sorted list, you can do better than to go through from one end to another.
2. Think about how you might use a dictionary or a phone book (that doesn't have tabs!)
3. Clue: try to think of a divide and conquer algorithm!
4. Algorithm
1. look in the middle of the list.
2. If that value is your key, you're done!
3. If your key is less than the current value, only consider the first half of the list, otherwise only consider the right side of the list.
4. Go back to 1 with your new (half-sized) list.
5. Example: 20, 35, 41, 48, 52, 76, 79, 84: Find the value 79!
1. Center is 48 -- too little so go right
2. Center of right is 76 -- too little so go right.
3. Center of right of right is 79! Done in 3 steps (list size of 8.)
6. Divide & Conquer should always give you log n performance.
7. In case anyone's not following, I'll do this one in pseudo code too, but really you should be able to write pseudo code (& real code!) by now from the description of the algorithm.
`integer key, sorta[] // what we are looking for? & the sorted array (don't have to be ints...)boolean found // have we found it?integer botIx, topIx // bottom and top boundaries of the searchinteger midIx // middle of current searchinteger locIx // for storing the correct location if we find itbotIx = 0, topIx = sorta.length - 1, locIx = -1, found = false; // initializewhile ((found == false) AND (botIx < topIx) do midIx = (botIx + topIx) / 2; // returns the floor for ints if (sorta[midIx] == key) { found = true; locIx = midIx; // all done! } else if (key < sorta[midIx]) { topIx = midIx - 1; } else { // key > sorta[midIx] botIx = midIx + 1; } } // end of whilereturn locIx; // will have value -1 if key was never found -- calling routine must handle this case!`
8. This is a lot like the tree search I showed you Tuesday. You are basically using pointer arithmetic to replicate a perfectly-balanced tree.
9. Note that we've done divide and conquer here without recursion, just with a loop & some arithmetic on the index values. Of course, recursion is simpler: now from the description of the algorithm.
`if (key == self.key) return self;if (key < self.key) if (not self.left) return null; else return self.left.search(key)if (not self.right) return null; else return self.right.search(key`
10. what is the expected search time? (all three cases, both versions above)
11. Now if we are searching a lot more frequently than we are sorting, the cost of the initial sort may not matter much, because the fact we have a sorted list allows us to search in just log n time!
## IV. Hash Sort & Search
1. Can we do even better than log n? Yes, we can do constant time!
2. If you just use the key value as an array index, then you can sort in O(n) time and search in O(1) time.
3. What if you have more than one value for a single index?
1. Could make an array of lists, then search the list for the items you actually want.
2. If only a few ever have the same index, this is still O(1), but if only a few indecies shared by all items, becomes O(n).
1. That's a good indication you didn't choose a very good key!
2. (draw picture of this)
4. What if your keys are complicated values, e.g. words?
1. Could still make into an integer, e.g. treat the word as a base 26 number.
2. But the array would have to be really, really big!
3. Most of it would probably be empty.
4. Remember, using too much space slows you down too (swapping onto disk.)
5. A common solution is a data structure called a hash table.
1. Index is an array that's about the right size.
2. Need a cunning way to convert the keys into then index arrays.
1. This is called the hash function.
2. Lots of different functions are used for hash functions.
3. Should be something simple / fast so you can look things up quickly.
6. Take an example using modular arithmetic.
1. the mod is the remainder left over after an integer divide.
2. good hash function
1. division is pretty fast
2. can choose the denominator for the mod to be the size of the array.
7. List of values is 3, 10, 12, 20, 23, 27, 30, 31, 39, 42, 44, 45, 49, 53, 57, 60
8. Let's try hashing it into an array of size 15 (same size as list)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 30, 45, 60 31 3 49 20 23, 53 39 10 12, 27, 42, 57 44
9. Analysis
1. 10 items located in 1 step = 10
2. 3 items in 2 steps = 6
3. 2 items in 3 steps = 6
4. 1 item in 4 steps = 4
5. So average time is (10 + 6 + 6 + 4 = 26) / 16 = 1.625 steps per retrieval
10. Because hashing functions are usually random with respect to the input, you can expect:
1. empty cells / wasted space
2. collisions (where more than one item tries to get in one cell)
1. have to make & search through list. This is called chaining, and illustrated above.
2. wasted time!
3. Can use 2-dimensional array, but that wastes way more space, limits number of possible collisions
4. Can indicate you should use of new hash function if too many collisions (e.g. do mod 20, not mod 15)
11. More time efficient if less collisions, more space efficient if more collisions.
12. Normally people favor time, make the table sparse (that is, much larger than the data):
1. Still saves space over one big array indexed off keys.
2. If the table is sparse enough, can put collisions in the next open index.
1. Probably the simplest way to handle them.
2. This is called probing.
3. Let's try hashing it into an array of size 19 this time:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 57 20 [39] 39 3 [60] 23 [42] 42 44 45 27 60 10 30 [49] 12 [31] 31 49 53
4. Analysis
1. 12 items located in 1 step = 12
2. 3 items in 2 steps = 6
3. 1 item in 4 steps = 4
4. 1 item in 6 steps = 6
5. So average time is 28 / 16 = 1.75 steps per retrieval
5. You wouldn't really put the information in the square brackets in the table, I just did that to let you see where those numbers really wanted to be sorted originally. The table really would look like this:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 57 20 39 3 23 42 44 45 27 60 10 30 12 31 49 53
6. Algorithm to search:
1. Derive a hash value mod tablesize from the search key, hashvalue.
2. Search for key starting at searchIx = hashvalue.
1. if sorta[searchIx] == null, then failed -- not in table.
2. if sorta[searchIx] == key, then succeeded.
3. otherwise, searchIx++, if searchIx > arraysize then searchIx =0 !! (wrap at the end!)
4. go back to 6.2.1
13. Analysis of Hash Search
1. Want to know how many items have to be searched to find a value.
2. As shown above, dependent on the load factor, A (really alpha, but HTML is lame.)
1. Like an array, the hash is constant on N.
3. A = number of occupied locations / table size
4. On average, probability of clash is 1 / (1 - A).
5. Successful search needs to examine (1 + 1 /(1 - A)) / 2 locations.
1. If you don't like just believing assertions (sensible), here's some more detailed notes.
2. So for example, an 80% full table requires about 3 steps, irrespective of how large the table is!
6. Unsuccessful search needs to examine (1 + 1 / (1 - A)2) / 2.
7. So on an 80% full table, the average for an unsuccessful search is 13 steps, again regardless of table size.
8. The normal rule of thumb is never to let a table get more than 80% full.
## V. Summary
1. Don't forget about linear & binary sorts!
2. Hashing is actually ubiquitous in computer science:
1. How do you think your variable names are stored?
2. Languages like python & perl let you use strings as array indecies.
3. These are actually just hashes, like Java has, but they are disguised by the syntax (like java disguises the use of integer division with a /).
page author: Joanna Bryson
17 February 2015
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Dylan is going to the fair and has \$35. He is going to use \$5 for food. The admission is \$7, and rides are \$1.85 each. How many rides can he go on?
\$35 - (5+7) = \$23. \$23/1.85 = 12.43. Therefore he can go on 12 rides.
Question
Updated 12/21/2011 8:58:55 PM
f
Rating
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\$35 - (5+7) = \$23. \$23/1.85 = 12.43. Therefore he can go on 12 rides.
Confirmed by andrewpallarca [12/25/2014 7:23:02 AM]
Questions asked by the same visitor
Solve: m + 2/3 = 1/2.
Weegy: m + 2/3 = 1/2, m = 1/2 - 2/3, m = 3/6 - 4/6, m = -1/6 (More)
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-0.06m = 7.2
Weegy: -0.06m = 7.2, -0.06m/-0.06 = 7.2/-0.06, meaning m = -120 (More)
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5.4 = n/ -0.9
Weegy: n = -4.86 (More)
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Updated 6/3/2014 7:06:34 PM
5.4 = n/ -0.9
n = 5.4 * -0.9
n = -4.86
Morgan has an 8 1/2 foot board that must be cut into boards that are 5/6 feet long. How many of the smaller boards can he get from the 8 1/2 foot board?
Weegy: He can get 10 smaller boards. User: Solve 5.4 = n/ -0.9 (More)
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Updated 11/8/2020 10:51:45 AM
[Deleted]
Deleted by Here_To_Help_You [11/8/2020 10:51:44 AM]
5.4 = n/ -0.9
n = 5.4 * -0.9
n = -4.86
One-eighth of a number is -4,200. What is the number?
Weegy: 1/8x = -4,200, x = -4,200 * 8, x = -33,600 (More)
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Addition Fact Activities: Fact Families Worksheets and Game Boards
Fact Family Games and Worksheets: 30+ page Packet
This new packet can be purchased separately, but is also included in our growing Addition Bundle! It is currently over 25o pages and consists of 7 pdfs!
Here is what is included in the Fact Family Packet:
Fact Families 5s through 19s (plus a Doubling game board)
When my kids were learning their addition facts, we played a lot of games together. When I brought out the game boards, they would beg to play another round (and another)!! They loved them!
Why learn the number families? Knowing their fact families helps a lot as kids start to work on their subtraction facts. They know that 6+7=13, so 13-7 becomes easy!
A reader pointed out that the website where I first found these number family games no longer exists. So, I thought I would put together a complete set of game boards for you. In this packet, there are 16 game boards (from the 5s through the 19s). I also included a doubles game for 4+4 through 9+9.
Below is a picture of the 12s Math Family:
• On the worksheet, students have to make 12: 0+12, 1+11, 2+9, 3+8, etc.
• There is a game board and number cards. Players take turns drawing a card (using the numbers 3 through 9 for this particular game) to make 12.
Using these activities:
After my kids learned all of their doubles 1+1 through 9+9 and knew what comes after (ie. +1) and what comes before (-1) really well, we started working with various fact families.
We did not go through every single Fact Family… so don’t feel like you have to have your kids play every single game and fill out every worksheet in this packet! Just pick and choose some of them. Most importantly, it should be fun!
Leave the kids begging to play more (rather than pushing them until they want to stop!).
You might leave some of the games for when your kids are working on their subtraction facts. But, I wanted to make all the resources available so you can pick and choose!
To Play the Game/s:
Print out 2 or 3 copies of the number cards on cardstock. Cut out the individual numbers.
Each game board tells you which numbers to use. Shuffle the cards well. Place them face down.
Each player will need one token to move around the board.
Players take turns drawing a card and finding the solution.
• For the doubles game, players draw a card double that number. For example, if they draw a 7 they add 7+7 and move to the next available 14 space.
• Fact Family Games: Let’s say you are playing the 12s game. A player draws a card and has to figure out which number will make 12 and move to that space. So, if a player draws a 5, they will have to move to the next available 7 space.
Note: I have added this packet to the addition bundle. If you purchased the Pirate Packet and Pet Packet bundle previously, you should have an email from SendOwl with the link to download the new packet (or just email me and I’ll get it to you!).
To purchase the 30+ page Fact Family Packet use the link below. It is \$3.00.
Once you pay for this packet, you will immediately receive a link to download these two files (which will open in a browser window). You will also receive an email from Sendowl (the service I use), which will have a link you can click on to download the Fact Family Games pdf. (It will say, “You can download your digital products…” with a clickable link.) Of course, if you have any issues just email me at — liesl at homeschoolden dot com. You can also reach me by using the contact form on the blog.
Or, it is included in the Addition Bundle (which is still \$5.50) and includes the Pirate Pack, Pet Packet and Place Value Activities. More information about those below.
The Addition Bundle includes 7 pdf files, 250+ pages
1. the Pirate Pack which includes 65 pages of math activities
3. Pet Pack addition worksheets and games
5. Dragon Math Packet (Addition, Subtraction worksheets & games)
The price for the Addition Bundle is still the same: \$5.50.
Note: After payment, you will receive an email from SendOwl, the delivery service I use. It will go to your PayPal email address. If you have any trouble with your download remember you can always contact me! ~Liesl
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1. Finding the Equation of a quadratic function
A jet flew from tokyo to bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flight was 2 hours, what was the jets speed from bangkok to tokyo?
2. Hello, Chr2010!
A jet flew from Tokyo to Bangkok, a distance of 4800km.
On the return trip, the speed was decreased by 200 km/hr.
If the difference in the times of the flight was 2 hours,
what was the jet's speed from Bangkok to Tokyo?
The jet flew 4800 km from Bangkok to Tokyo at $\,x$ km/hr.
. . This took $\dfrac{4800}{x}$ hours.
The jet flew 4800 km from Tokyo to Bangkok at $x + 200$ km/hr.
. . This took $\dfrac{4800}{x+200}$ hours.
The difference in time was 2 hours: . $\displaystyle \frac{4800}{x} - \frac{4800}{x+200} \;=\;2$
Multiply by $x(x+200)\!:\;\;4800(x+200) - 4800x \;=\;2x(x+200)$
This equation simplifies to: . $x^2 + 200x - 480,\!000 \:=\:0$
. . whoch factors: . $(x - 600)(x + 800) \:=\:0$
. . and has roots: . $x \:=\:600,\:\text{-}800$
The jet's speed from Bangkok to Tokyo was 600 km/hr.
3. thanks so much!
4. Hi,
Thanks so much for your response, i understand the concept but could you walk me through on how u knew where to start and how u got the simplified equation of x^2+200x-480000. thanks
5. Soroban started by choosing a letter to represent the quantity he wanted to find- in this case he let "x" represent the speed flying from Bangcock to Tokyo. You are told that the distance is 4800 km and should know that "speed= distance/time" so if we let $t_2$ be the time the flight took, $x= \frac{4800}{t_2}$ so, solving that for $t_2$, $t_2= \frac{4800}{x}$.
We don't know $t_2$ so we can use that equation alone to find x. But we also know that "On the return trip, the speed was decreased by 200 km/h". Since is the speed on the return trip, the speed on the first leg must be 200 more: x+ 200. The time to fly the same 4800 is $t_1= \frac{4800}{x+ 200}$. That flight was at a faster speed so of course, it takes less time- 2 hours less: $t_2- t_1$, the difference in times, is 2 hours:
$t_2- t_1= \frac{4800}{x}- \frac{4800}{x+ 200}= 2$
Get rid of the fractions by multiplying both sides of that equation by x(x+ 200) and you get the quadratic equation Soroban gave.
6. thank-you!
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# Introductory Analysis
Watch
Announcements
#1
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.
a) {m/n: where m and n are natural with m < n}
I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?
b) {m/m+n: where m and n are natural}
I don't think this has a supremum and infimum. But its a guess. Nothing concrete.
c) { ((-1)^m)/n: where m and n are natural}
Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.
a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.
These questions are from Abott Understanding Analysis. I am to start uni in October so I am self-teaching analysis, and hence I am a completely rookie/beginner. Just wanted to put it out there if my answers look ridiculous and stupid
0
1 year ago
#2
(Original post by Rohan77642)
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.
a) {m/n: where m and n are natural with m < n}
I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?
Agreed, 0 and 1.
We can see that m/n is always less than 1. So, 1 is an upper bound. Is it the least upper bound? Well yes, because we can think of the elments (n-1)/n for n>1, which belong to this set, and taking n as large as we like, we can get as close to 1 as we like, from below.
As to 0.
Well we know m/n is greater than 0. So, 0 is a lower bound. Is it the greatest lower bound? Again yes, since we can consider the elements 1/n which belong to the set, and in a similar fashion we can get as close to 0 as we like, from above.
b) {m/m+n: where m and n are natural}
I don't think this has a supremum and infimum. But its a guess. Nothing concrete.
It has both.
m,n are both positive integers. so, m/(n+n) is always positive. Also m < m+n, so m/(m+n) < 1. So we know the set is bounded below and above. What might the sup and inf be?
c) { ((-1)^m)/n: where m and n are natural}
Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.
Sup is 1, since the largest positive element is 1.
However inf is not 0. If m is odd, then the element is negative. What's the lowest negative number it can be?
a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.
It might help to divide top and bottom by n, think about what you have and consider the first few values of n.
The sup is indeed 1/3, since all terms are less than 1/3, and making n as large as we like we can get as close to 1/3 as we like, from below.
Have a further think on the inf.
Last edited by ghostwalker; 1 year ago
0
1 year ago
#3
(Original post by Rohan77642)
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.
a) {m/n: where m and n are natural with m < n}
I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?
Sounds good. The infimum is indeed 0. You can make the denominator as large as you want and the fraction will tend to zero.
b) {m/m+n: where m and n are natural}
I don't think this has a supremum and infimum. But its a guess. Nothing concrete.
It does. What happens when you fix and let get very large? Or what happens when and you let get very large?
c) { ((-1)^m)/n: where m and n are natural}
Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.
Supremum looks good, but infimum not so much. I don't see why you set being even here; we lose generality.
What happens when and is odd??
a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.
Supremum is good.
But notice that this is clearly an increasing sequence. Therefore you can find the smallest element of this set.
2
#4
(Original post by ghostwalker)
Agreed, 0 and 1.
We can see that m/n is always less than 1. So, 1 is an upper bound. Is it the least upper bound? Well yes, because we can think of the elments (n-1)/n for n>1, which belong to this set, and taking n as large as wel like, be can get as close to 1 as we like.
As to 0.
Well we know m/n is greater than 0. So, it's a lower bound. Is it the greatest lower bound? Again yes, since we can consider the elements 1/n which belong to the set, and in a similar fashion we can get as close to 0 as we like.
It has both.
m,n are both positive integers. so, m/(n+n) is always positive. Also m < m+n, so m/(m+n) < 1. So we know the set is bounded below and above. What might the sup and inf be?
Sup is 1, since the largest positive element is 1.
However inf is not 0. If m is odd, then the element is negative. What's the lowest negative number it can be?
It might help to divide top and bottom by n, think about what you have and consider the first few values of n.
The sup is indeed 1/3, since all terms are less than 1/3, and making n as large as we like we can get as close to 1/3 as we like.
Have a further think on the inf.
Thank you for helping me .
So I have gathered some thoughts based on your advice and updated my solutions below.
b) we have m < m+n, therefore m/m+n < 1. Therefore sup = 1. As for infimum, I think its 0. Because if we say m is 1, then we have 1/1+n where now if we say n tends to infinity then we get the limit tends to 0. So I am guessing its 0. It obviously cant be negative cause m, n are naturals.
c) Yes. Inf is not 0. My mistake. I think its -1.
d) for inf by using your advice if I divide up and down by n then I have 1/(3+1/n). Now, 1/n is maximized when n is 1. So inf is 1/4?
Thanks again for helping. Really appreciate it.
0
#5
(Original post by RDKGames)
Sounds good. The infimum is indeed 0. You can make the denominator as large as you want and the fraction will tend to zero.
It does. What happens when you fix and let get very large? Or what happens when and you let get very large?
Supremum looks good, but infimum not so much. I don't see why you set being even here; we lose generality.
What happens when and is odd??
Supremum is good.
But notice that this is clearly an increasing sequence. Therefore you can find the smallest element of this set.
Thanks. I have updated some of my answers in a recent post.
Thanks for the help. Appreciate it
0
1 year ago
#6
(Original post by Rohan77642)
b) we have m < m+n, therefore m/m+n < 1. Therefore sup = 1.
The second "therefore" needs a little more explanation. 1 is clearly an upper bound. If we consider the subset formed by m/(m+1) we can see that this gets as close as we like to 1 as m is increased. And therefore the sup is 1.
As for infimum, I think its 0. Because if we say m is 1, then we have 1/1+n where now if we say n tends to infinity then we get the limit tends to 0. So I am guessing its 0. It obviously cant be negative cause m, n are naturals.
Yes.
c) Yes. Inf is not 0. My mistake. I think its -1.
Yes.
d) for inf by using your advice if I divide up and down by n then I have 1/(3+1/n). Now, 1/n is maximized when n is 1. So inf is 1/4?
Yes.
Last edited by ghostwalker; 1 year ago
0
#7
(Original post by ghostwalker)
The "therefore" needs a little more explanation. 1 is clearly an upper bound. If we consider the subset formed by m/(m+1) we can see that this gets as close as we like to 1 as m is increased. And therefore the sup is 1.
Yes.
Yes.
Yes.
Thanks a lot.
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# Relationship between hyperbolic arctan and logarithm
## Homework Statement
The relationship between arctanh and log is:
arctanh(x)=$\frac{1}{2}log(\frac{1+x}{1-x})$
but if i take x=1.5,
I have:
arctanh(1.5)=0.8047 + 1.5708i
and
$\frac{1}{2}log(\frac{1+1.5}{1-1.5})$=0.8047 + 1.5708i
as expected, but using the laws of logarithm, why does this not equal to:
$(\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2)$=0.8047
?
## The Attempt at a Solution
Related Calculus and Beyond Homework Help News on Phys.org
I like Serena
Homework Helper
Hi sara_87!
You have applied the log to a negative number yielding a complex number.
So implicitly you have used the complex logarithm instead of the regular logarithm.
The complex logarithm happens to be a multivalued function:
$$\log re^{i\phi} = \log r + i (\phi + 2k\pi)$$
This means you have to consider the other solutions.
By taking the square you introduce a new solution.
In your example there is another solution, which is the right one:
$$(\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2) = (\frac{1}{2})\frac{1}{2}(\log 25 + i 2\pi)=\frac{1}{2}\log 5 + i \frac \pi 2$$
thank you :)
so, in general, can i say that if x is less than 1,
$arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)$
if x is more than 1, then
$arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)+i*pi/2$
I like Serena
Homework Helper
Yes.
Thanks again :)
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# City Lots fractions task keeps ALL students thinking
Here’s a 7th grade fractions task called City Lots. My colleagues designed it with two great extensions. I differentiated it by creating resource cards plus I modified the directions to challenge students who need it.
The basic task requires students to determine which of four companies owns the most land. They also have to calculate how much land each company owns based on four quadrants.
The activity was a hit with my 6th grade advanced math students. Most groups began with a strategy of cutting out the shapes to find which company owned the most land. While that was an effective strategy, it’s a difficult way to calculate exactly how much land was owned by each company. That’s where the resource cards came in.
Then the mathematical conversations shifted from cutting shapes to discussing how the property was divided. I asked probing questions such as, “What’s the minimum number of squares do you see?” Many didn’t see one unit divided into four quadrants.
We’re trying to do much more reading and writing in math class, and this task is another opportunity for students to articulate their mathematical thinking in writing.
In terms of time, I presented the task midway into the block. In those 40 minutes, no groups got to the extensions, four groups successfully completed questions 1 and 2 and the others are not far behind. We’ll get to the extensions the next time I see them. I did not hand out the challenge task but at least I was prepared.
I don’t think I’ll have an opportunity to try the challenge task, but if you do I would love to hear how it goes.
## 5 thoughts on “City Lots fractions task keeps ALL students thinking”
1. Thanks for sharing this resource! I’m going to try it on Thursday with my advanced 6th graders.
I’ve done similar tasks with my students — ones where a larger square is divided into smaller squares, rectangles, and triangles — but I like that this task requires addition of fractions as well as just the identification of the various fractional parts. And the extensions are great! I hope to get to the them and will let you know how it goes if I do!
1. Yes, please do let me know! There was an incredible amount of mathematical thinking with my students. We continued the task the next class period and two groups presented their strategies to the class. One group was misguided and was corrected by a classmate. I also appreciate the students’ taking risks by sharing their thinking to the rest of the class. They sooo want to be right, but we learn so much more from our mistakes!
2. Sarah R says:
I have been holding onto this activity since you first posted it. I am hoping to try after our break. Thanks for sharing!
3. Katherine says:
Ah dang, I found this page too late into my unit to set aside a whole day for it, but the picture was great to talk about! It was a nice context to start thinking about fraction multiplication since the kids can see what “1/8 of 1/4” looks like etc. and then make sense of why that’s the same as 1/8 x 1/4. Thanks!!
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# How can I find the roots of a quartic equation, knowing one of its roots?
I need to decompose (in $\Bbb{C}[x]$) the function
$$f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15$$
in its simplest form, knowing that $1 - 2i$ is one of its roots. Any ideas?
-
Since the coefficients are real, $1+2i$ is another of its roots. Dividing $f$ by $(x-(1+2i))(x-(1-2i))$ will reduce it to a quadratic, which you can solve easily. – Nick Thompson Jun 15 '14 at 17:51
If $\alpha +i\beta$ is a root, so is $\alpha -i\beta$. Thus $\underbrace{(x-(\alpha +i\beta))(x-(\alpha -i\beta))}_{\in \mathbb Q[x]}$ is a factor of the given polynomial. Proceed with polynomial division to get the other factor of degree $2$. – Git Gud Jun 15 '14 at 17:51
You know a second root. $1+2i$ multiply together the two linear factors with these roots and divide into your polynomial. – Rene Schipperus Jun 15 '14 at 17:52
After diving diving f by (x-(1+2i))(x-(1-2i)) = x^2 -2x + 5 I got a remainder of -10. But i'm pretty sure the long division is correct. Did I do something wrong? I got the polynomial x^2+6x+3 – user157246 Jun 15 '14 at 18:03
The polynomial $x^2+6x+3$ is right. – André Nicolas Jun 15 '14 at 18:04
Hint: Another root is $1+2i$. The sum and product of the two roots we know are $2$ and $5$.
The sum of all the roots is $-4$, and the product of all the roots is $15$. So the sum of the missing roots is $-6$, and the product is $3$.
-
So we have the function $$f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15$$ One root has been found at $x = 1-2i$.
Since the coefficients are real, there must be another root at $x = 1+2i$. We can now reduce this equation to a quadratic one, by dividing those out.
\begin{align} & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-(1-2i))(x-(1+2i))} \\ = \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-1+2i)(x-1-2i)} \\ = \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{x^2-2x+5} \\ = \ & x^2+6x+3 \end{align}
The roots for that quadratic equation can be found at $x = -3-\sqrt{6}$ and $x = -3 + \sqrt{6}$.
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# Goodness of fit test - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Goodness of fit test
Kruskal-Wallis test
Cochran's Q test
One way ANOVA
One sample $z$ test for the mean
Two sample $t$ test - equal variances not assumed
Independent variableIndependent/grouping variableIndependent/grouping variableIndependent/grouping variableIndependent variableIndependent/grouping variable
NoneOne categorical with $I$ independent groups ($I \geqslant 2$)One within subject factor ($\geq 2$ related groups)One categorical with $I$ independent groups ($I \geqslant 2$)NoneOne categorical with 2 independent groups
Dependent variableDependent variableDependent variableDependent variableDependent variableDependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesis
• H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
• H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: $\pi_1 = \pi_2 = \ldots = \pi_I$
Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$
ANOVA $F$ test:
• H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
• H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
• H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
• H1: the population proportions are not all as specified under the null hypothesis
or equivalently
• H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
H1: not all population proportions are equalANOVA $F$ test:
• H1: not all population means are equal
$t$ Test for contrast:
• H1 two sided: $\Psi \neq 0$
• H1 right sided: $\Psi > 0$
• H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
• H1 - usually two sided: $\mu_g \neq \mu_h$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• Scores are normally distributed in the population
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statisticTest statisticTest statisticTest statisticTest statisticTest statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$
Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.
Note: if ties are present in the data, the formula for $H$ is more complicated.
If a failure is scored as 0 and a success is scored as 1:
$Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$
Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores.
Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.
ANOVA $F$ test:
• \begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned}
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
• $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
• $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
n.a.n.a.n.a.Pooled standard deviationn.a.n.a.
---\begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}
Here $s^2_i$ is the variance in group $i.$
--
Sampling distribution of $X^2$ if H0 were trueSampling distribution of $H$ if H0 were trueSampling distribution of $Q$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $t$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedom
For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.
For small samples, the exact distribution of $H$ should be used.
If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedomSampling distribution of $F$:
• $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - I$ degrees of freedom
Standard normal distributionApproximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1
First definition of $k$ is used by computer programs, second definition is often used for hand calculations.
Significant?Significant?Significant?Significant?Significant?Significant?
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)
$t$ Test for contrast two sided:
$t$ Test for contrast right sided:
$t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
• Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
• Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
• Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
n.a.n.a.n.a.$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for \mu_i$$C\% confidence interval for \muApproximate C\% confidence interval for \mu_1 - \mu_2 ---Confidence interval for \Psi (contrast): • c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}} where the critical value t^* is the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). Note that n_i is the sample size of group i, and N is the total sample size, based on all the I groups. Confidence interval for \mu_g - \mu_h (multiple comparisons): • (\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}} where t^{**} depends upon C, degrees of freedom (N - I), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, t^{**} = t^* = the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^*. Note that n_g is the sample size of group g, n_h is the sample size of group h, and N is the total sample size, based on all the I groups. Confidence interval for single population mean \mu_i: • \bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}} where \bar{y}_i is the sample mean in group i, n_i is the sample size of group i, and the critical value t^* is the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). Note that n_i is the sample size of group i, and N is the total sample size, based on all the I groups. \bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval). The confidence interval for \mu can also be used as significance test. (\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}} where the critical value t^* is the value under the t_{k} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu_1 - \mu_2 can also be used as significance test. n.a.n.a.n.a.Effect sizeEffect sizen.a. --- • Proportion variance explained \eta^2 and R^2: Proportion variance of the dependent variable y explained by the independent variable:$$ \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} $$Only in one way ANOVA \eta^2 = R^2. \eta^2 (and R^2) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population. • Proportion variance explained \omega^2: Corrects for the positive bias in \eta^2 and is equal to:$$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$\omega^2 is a better estimate of the explained variance in the population than \eta^2. • Cohen's d: Standardized difference between the mean in group g and in group h:$$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$Cohen's d indicates how many standard deviations s_p two sample means are removed from each other. Cohen's d: Standardized difference between the sample mean and \mu_0:$$d = \frac{\bar{y} - \mu_0}{\sigma}$Cohen's$d$indicates how many standard deviations$\sigma$the sample mean$\bar{y}$is removed from$\mu_0.$- n.a.n.a.n.a.n.a.Visual representationVisual representation ---- n.a.n.a.n.a.ANOVA tablen.a.n.a. --- Click the link for a step by step explanation of how to compute the sum of squares. -- n.a.n.a.Equivalent toEquivalent ton.a.n.a. --Friedman test, with a categorical dependent variable consisting of two independent groups.OLS regression with one categorical independent variable transformed into$I - 1$code variables: •$F$test ANOVA is equivalent to$F$test regression model •$t$test for contrast$i$is equivalent to$t$test for regression coefficient$\beta_i$(specific contrast tested depends on how the code variables are defined) -- Example contextExample contextExample contextExample contextExample contextExample context Is the proportion of people with a low, moderate, and high social economic status in the population different from$\pi_{low} = 0.2,\pi_{moderate} = 0.6,$and$\pi_{high} = 0.2$?Do people from different religions tend to score differently on social economic status? Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?Is the average mental health score different between people from a low, moderate, and high economic class?Is the average mental health score of office workers different from$\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is$\sigma = 3.$Is the average mental health score different between men and women? SPSSSPSSSPSSSPSSn.a.SPSS Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square... • Put your categorical variable in the box below Test Variable List • Fill in the population proportions / probabilities according to$H_0$in the box below Expected Values. If$H_0$states that they are all equal, just pick 'All categories equal' (default) Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples... • Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable • Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow • Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum • Continue and click OK Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples... • Put the$k$variables containing the scores for the$k$related groups in the white box below Test Variables • Under Test Type, select Cochran's Q test Analyze > Compare Means > One-Way ANOVA... • Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor or Analyze > General Linear Model > Univariate... • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s) -Analyze > Compare Means > Independent-Samples T Test... • Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2 • Continue and click OK JamoviJamoviJamoviJamovin.a.Jamovi Frequencies > N Outcomes -$\chi^2$Goodness of fit • Put your categorical variable in the box below Variable • Click on Expected Proportions and fill in the population proportions / probabilities according to$H_0$in the boxes below Ratio. If$H_0$states that they are all equal, you can leave the ratios equal to the default values (1) ANOVA > One Way ANOVA - Kruskal-Wallis • Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The$p$value resulting from this Friedman test is equivalent to the$p$value that would have resulted from the Cochran's Q test. Go to: ANOVA > Repeated Measures ANOVA - Friedman • Put the$k$variables containing the scores for the$k\$ related groups in the box below Measures
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
-T-Tests > Independent Samples T-Test
• Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
• Under Tests, select Welch's
• Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questions
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A303212 Number of minimum total dominating sets in the n X n rook complement graph. 2
%I
%S 0,1,6,96,600,2400,7350,18816,42336,86400,163350,290400,490776,794976,
%T 1242150,1881600,2774400,3995136,5633766,7797600,10613400,14229600,
%U 18818646,24579456,31740000,40560000,51333750,64393056,80110296,98901600
%N Number of minimum total dominating sets in the n X n rook complement graph.
%C For n > 2, the minimum total dominating sets are any three vertices such that no two are in the same row or column. - _Andrew Howroyd_, Apr 20 2018
%H Colin Barker, <a href="/A303212/b303212.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RookComplementGraph.html">Rook Complement Graph</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TotalDominatingSet.html">Total Dominating Set</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F a(n) = A179058(n) for n > 2. - _Andrew Howroyd_, Apr 20 2018
%F From _Colin Barker_, Apr 20 2018: (Start)
%F G.f.: x^2*(1 - x + 75*x^2 + 19*x^3 + 41*x^4 - 21*x^5 + 7*x^6 - x^7) / (1 - x)^7.
%F a(n) = n^2*(2 - 3*n + n^2)^2 / 6 for n>2.
%F a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>9.
%F (End)
%t Table[If[n == 2, 1, 6 Binomial[n, 3]^2], {n, 20}]
%t Join[{0, 1}, LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 0, 6, 96, 600, 2400, 7350}, {3, 20}]]
%t CoefficientList[Series[x (-1 + x - 75 x^2 - 19 x^3 - 41 x^4 + 21 x^5 - 7 x^6 + x^7)/(-1 + x)^7, {x, 0, 20}], x]
%o (PARI) a(n) = if(n<3, n==2, 6*binomial(n,3)^2) \\ _Andrew Howroyd_, Apr 20 2018
%o (PARI) concat(0, Vec(x^2*(1 - x + 75*x^2 + 19*x^3 + 41*x^4 - 21*x^5 + 7*x^6 - x^7) / (1 - x)^7 + O(x^60))) \\ _Colin Barker_, Apr 20 2018
%Y Cf. A179058.
%K nonn,easy
%O 1,3
%A _Eric W. Weisstein_, Apr 19 2018
%E a(6)-a(30) from _Andrew Howroyd_, Apr 20 2018
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Last modified October 14 00:32 EDT 2019. Contains 327991 sequences. (Running on oeis4.)
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# Calculate $\oint_C y^2\, dx + x\, dy$ using Green's Theorem?
Let $C$ be the curve parametrized by the equation $r(t) = 2\cos^3(t) i + 2\sin^3(t) j$ for $t \in [0,2\pi]$. I want to find the line integral
$$\oint_C y^2 \,dx + x \,dy .$$
I evaluated it directly and the answer appears to be $\frac{3\pi}{2}$. But I'm supposed to use Green's Theorem and I don't know how to set up the integral. If the region enclosed by $C$ is $D$, then the integral is
$$\iint\limits_D 1 + 2y \, dy\, dx$$
But I have no idea how to find the region $D$. I tried to set $x = 2 \cos^3(t)$, which implies $y \in [-k,k]$ for
$$k = 2(1- (\frac{x}{2})^{2/3})^{3/2}$$
So the integral would become
$$\int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} 1 - 2y\, dy\, dx .$$
But this is too messy to evaluate by hand.
So, how do I evaluate the line integral using Green's Theorem?
Note that by symmetry $$I:=\int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} (1 - 2y)\ dy dx=4\int_{0}^2 \int_{0}^{2(1- (x/2)^{2/3})^{3/2}} 1dy dx\\ =8\int_{0}^2 (1- (x/2)^{2/3})^{3/2} dx=48\int_{0}^{\pi/2} \cos^4(t) \sin^2(t) dt\\ =48\int_{0}^{\pi/2} \cos^4(t)dt-48\int_{0}^{\pi/2} \cos^6(t)dt$$ where we let $x=(2\sin(t))^3$. Now, for $n\geq 1$, by integration by parts, $$\int_{0}^{\pi/2} \cos^{2n}(t)dt=\frac{2n-1}{2n}\int_{0}^{\pi/2} \cos^{2n-2}(t)dt$$ which implies that $I=3\pi/2$.
• Yes, but this just reduces to integrating $(1-x^{2/3})^{3/2}$, which is also tricky...unless I'm missing something. – aras Jun 30 '17 at 11:07
• OK, thank you for explicitly showing me the substitution $x = (2\sin(t))^3$. I see why this solution works. However, isn't this essentially just reducing the computation back to the line integral? If I compute $\int_{0}^{2\pi} \langle y^2, x\rangle \dot \langle x'(t), y'(t) \rangle dt$ I get $$\int_{0}^{2\pi} -24 \sin^7(t) \cos^2(t) + 12 \sin^2(t) \cos^4(t)$$ The $\sin^7 \cos^2$ term integrates to $0$, and the $12 \sin^2 \cos^4$ term simplifies to $$12 \int_{0}^{2\pi} \cos^4(t) - \cos^6(t) dt$$ which is what you evaluated in your answer. – aras Jun 30 '17 at 11:22
• @aras Yes, so are able to find the final result? – Robert Z Jun 30 '17 at 11:24
• Yes, but I was wondering if there was a cleaner way to apply Green's Theorem that avoided the computational mess of explicitly computing the line integral. – aras Jun 30 '17 at 11:25
• Finally you have to evaluate the area inside the curve. I don't know if here you can avoid the computation of $\int_{0}^{2\pi} \cos^4(t) - \cos^6(t) dt$. – Robert Z Jun 30 '17 at 11:35
Note that at any point $x$, we have $$y = 2\left(1-(x/2)^{2/3}\right)^{3/2}$$ so it sounds like you have the equation $$\left(\frac{y}{2}\right)^{2/3}+ \left(\frac{x}{2}\right)^{2/3} = 1$$
• How does this help me find the limits of integration of $D$? – aras Jun 30 '17 at 11:09
• @aras since you have the description of the boundary, you can solve for whatever you like -- I even solved for $y$ for you... – gt6989b Jun 30 '17 at 14:46
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Algebra and Trigonometry 10th Edition
The property was proved for $n=1$ The property is correct if n is changed by $n+1$
Let's prove the property for $n=1$: $(ab)^n=(ab)^1=ab=a^1b^1$ It is correct! Suppose that the propety is correct, that is: $(ab)^n=a^nb^n$ for all integers $n\geq1$ Now, let's prove the property for $n+1$: $(ab)^{n+1}=(ab)^n(ab)^1=a^nb^nab=(a^na)(b^nb)=a^{n+1}b^{n+1}$ That is the given property if $n$ is changed by $n+1$
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# A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and the second bond pays 7% per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds if the trust fund must obtain an annual total interest of (i) Rs 1800 and (ii) Rs 2000.
Given that Rs 30000 must be invested into two types of bonds with 5% and 7% interest rates.
Let Rs x be invested in bonds of the first type.
Thus, Rs (30000 – x) will be invested in the other type.
Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as -
(i) Annual interest obtained is Rs 1800.
We know the formula to calculate the interest on a principal of Rs P at a rate R% per annum for t years is given by,
Here, the time is one year and thus T = 1.
Hence, the interest obtained after one year can be expressed in matrix representation as -
5x + 7(30000 – x) = 1800 × 100
5x + 210000 – 7x = 180000
–2x = 180000 – 210000
–2x = –30000
x = 15000
Amount invested in the first bond = x = Rs 15000
Amount invested in the second bond = 30000 – x
Amount invested in the second bond = 30000 – 15000
Amount invested in the second bond = Rs 15000
Thus, the trust has to invest Rs 15000 each in both the bonds in order to obtain an annual interest of Rs 1800.
(ii) Annual interest obtained is Rs 2000.
As in the previous case, the interest obtained after one year can be expressed in matrix representation as -
5x + 7(30000 – x) = 2000 × 100
5x + 210000 – 7x = 200000
–2x = 200000 – 210000
–2x = –10000
x = 5000
Amount invested in the first bond = x = Rs 5000
Amount invested in the second bond = 30000 – x
Amount invested in the second bond = 30000 – 5000
Amount invested in the second bond = Rs 25000
Thus, the trust has to invest Rs 5000 in the first bond and Rs 25000 in the second bond in order to obtain an annual interest of Rs 2000.
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# Quick Answer: How Much Interest Will 5000 Earn In A Year?
How much will an investment of \$5,000 be worth in the future?
At the end of 20 years, your savings will have grown to \$16,036.
You will have earned in \$11,036 in interest.
Interest Calculator for \$5,000.
RateAfter 10 YearsAfter 30 Years
1.25%5,6617,258
1.50%5,8037,815
1.75%5,9478,414
2.00%6,0959,057
54 more rows
## How much interest does 1000 earn in a year?
In the simplest of words, \$1,000 at 1% interest per year would yield \$1,010 at the end of the year. But that is simple interest, paid only on the principal. Money in savings accounts will earn compound interest, where the interest is calculated based on the principal and all accumulated interest.
## How much interest will I earn if I have a million in the bank?
That’s quite a spread, which can make a significant difference in how much interest your million dollars can earn. For example, one million dollars earning 0.01% in a savings account would generate \$100 of interest after a year, while a CD paying 2.5% would generate \$25,000 of interest.
## How much interest will 2 million dollars earn in a year?
At the end of 20 years, your savings will have grown to \$6,414,271. You will have earned in \$4,414,271 in interest.
Interest Calculator for \$2,000,000.
RateAfter 10 YearsAfter 30 Years
0.00%2,000,0002,000,000
0.25%2,050,5662,155,567
0.50%2,102,2802,322,800
0.75%2,155,1652,502,544
54 more rows
## How do you figure out how much interest you will earn?
Simple interest formula:
• p x r x t = i.
• \$100 deposit x 5% interest x 1 year term = \$5.
• \$100 x 0.05 x 1 = \$5.
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# RD Sharma Class 12 Maths Solutions Chapter 11 - Differentiation
Last updated date: 14th Sep 2024
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## RD Sharma Solutions for Class 12 Maths Chapter 11 - Differentiation - Free PDF Download
Differentiation in Class 12 Maths is not only a scoring but an easily understandable chapter. It lays the foundation for the complete concept of calculus for the ongoing class and even the advanced concepts of calculus in higher classes. Students need to focus on this chapter to perform well in maths, also a few topics in Physics and chemistry as well.
RD Sharma is a good book, which is almost the most preferred and reliable book after completing the Class 12 NCERT textbook. RD Sharma is the toughest book to cover at school Maths level. However if you wish to aim higher and also are preparing for entrance exams this book is a must and will help you sail through the subject with excellent results.
Vedantu provides a free PDF version of RD Sharma. After clearing basic concepts and completing the syllabus from the NCERT textbook, Class 12 students should definitely download and practice RD Sharma.
In addition to integration, differentiation is one of the two essential principles in calculus.
Differentiation is the tactic to work out the rate of change in function on the basis of per unit change of its variables. A common example is finding the rate of change of velocity with respect to time is called acceleration. A derivative of a function is found by differentiation. Functions can be of linear and non-linear categories. The opposite of finding derivatives is referred to as anti-differentiation. The general expression of differentiation is dy/dx which determines the rate of change of x with respect to y.
Competitive Exams after 12th Science
## Basic Differentiation Rules
The differentiation rules are clubbed up as follows with a brief explanation:
• Sum and Difference Rule which determines the sum or difference of two functions and the derivative is basically the sum or difference of the individual functions.
• Product Rule which determines the product of the first function by the derivative of the second plus second function is multiplied by the derivative of the first.
• Quotient Rule determines the derivative of a function that is the ratio of two differentiable functions.
• Chain Rule determines the formula for computing the derivative of the composition of two or more differentiable functions.
• Derivative Rules
Common Functions Function Derivative Difference Rule f - g f' − g' Product Rule fg f g' + f' g Quotient Rule f/g f' g − g' fg2 Reciprocal Rule 1/f −f'/f2
### Conclusion:
All solutions are explained clearly in a step by step manner to provide a unique learning experience to students. The RD Sharma Solutions for Class 12 Math Chapter 11 also have a lot of exercise and practice problems so that students will have enough practice of the subject before their exams. Students can download a free PDF of RD Sharma Class 12 Differentiation Solutions free of cost.
## FAQs on RD Sharma Class 12 Maths Solutions Chapter 11 - Differentiation
1. What is the importance of differentiation in real life?
Differentiation helps in solving many daily problems in life. For instance to determine the maximum and minimum values of certain specific functions like cost, profit, loss etc we use derivatives. To determine the speed or acceleration of a moving vehicle we need differentiation. Differentiation is the platform of calculation for most engineering and science problems. Differentiation even helps us understand how stars, planets and satellites move with respect to each other. There are many applications of derivatives in Physics and even in Chemistry.
2. How to score well in differentiation in Class 12?
Differentiation is an important chapter in class 12 Math and also for competitive exams like JEE Main and Advanced. Students need to have a clear concept about this topic to score high marks in Math of Class 12 and JEE. A few important tips that can help them to score high on this topic are as follows:
• Understand and clear the concept from the definition itself
• The standard formulas in differentiation are very important. Revise them regularly
• The nature of functions need to be well understood
• Application of derivatives are important
• Revising and practicing this topic on regular basis at least for a few minutes is important
3. What is the weightage of differentiation in different exams of class 12?
Differentiation is an important chapter in both Class 12 Boards and JEE Main. In JEE Main Differentiation comes within the first four top chapters with weightage of expected marks of around 28-30 from this particular chapter. Almost 2 questions will surely appear from this chapter. Above that this topic is easy to score and takes less time to be solved. So, students must focus on this chapter from the very beginning both in their Class 12 Maths preparation and JEE preparation. Differentiation is also important and easily scoring in the board exam. The unit of calculus including differentiation and integration also has a great weightage in the board paper bearing 35 marks.
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# Triangle wavefunction - position uncertainty and probability amplitude
1. Sep 30, 2013
### Dixanadu
1. The problem statement, all variables and given/known data
Hey guys. Basically I have a wavefunction that looks like this
http://imageshack.com/a/img843/3691/22r3.jpg [Broken]
I have to find:
(a) The normalization constant N, of course by normalizing it
(b) Find <x> and <x^2> and use this to find Δx.
2. Relevant equations
I'm just gona insert images, as i'm writing it all with my tablet:
http://imageshack.com/a/img33/5705/czbs.jpg [Broken]
3. The attempt at a solution
Okay, so I've found the normalization constant, N, but I've got 2 different answers depending on the way in which I solve the problem, and I want to know which is correct (and why, if possible). Please view the image below:
http://imageshack.com/a/img11/7232/0hmb.jpg [Broken]
So...which is correct? :S
I need to know the explicit expression for the probability density to proceed further.
Thanks a lot guys!
Last edited by a moderator: May 6, 2017
2. Sep 30, 2013
### Päällikkö
The square of a function that is defined in parts, is still a function defined in parts, eg.
From
$f(x) = \left\{\begin{array}(g(x); x<0 \\ h(x); x\geq0\end{array}\right.$
It follows that
$(f(x))^2 = \left\{\begin{array}((g(x))^2 &; x<0 \\ (h(x))^2&; x\geq0\end{array}\right.$,
that is to say
$(f(x))^2 \neq (g(x))^2 + (h(x))^2$
This is a notational problem: your $|\Psi|^2$ is incorrect, although you do manage to integrate it correctly in parts. Your first approach is correct as it is the integral of the square that you want to have as 1 (probability of the particle being anywhere is 1).
3. Sep 30, 2013
### Dixanadu
Great, thanks a lot dude! But my next question is this - how do i go about finding <x> ? i mean I need a single expression for psi mod squared, right?
4. Sep 30, 2013
### Päällikkö
I'm not exactly sure what you mean by a single expression. Could you elaborate?
There should be no problem integrating over functions that are piece-wise defined.
For example if
f(x) = 1 for x < 0 and f(x) = 2 for x > 0, the integral $\int_{-1}^1f(x)dx = \int_{-1}^0 1 dx + \int_0^1 2 dx = 1 + 2 = 3$.
Start by writing out the function $\Psi^* x \Psi$. It will be a piece-wise defined function. Then proceed to integrate it, and you will get <x> by definition.
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# Root of 154017505
#### [Root of one hundred fifty-four million seventeen thousand five hundred five]
square root
12410.3789
cube root
536.0311
fourth root
111.4019
fifth root
43.4024
In mathematics extracting a root is declared as the determination of the unknown "x" in the equation $y=x^n$ The result of the extraction of the root is seen as a root. In the case of "n is 2", one talks about a square root or sometimes a second root also, another possibility could be that n = 3 then one would call it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 154017505 is represented as this: $$\sqrt[]{154017505}=12410.378922499$$
On top of this it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 154017505 is 12410.378922499. The cube root of 154017505 is 536.03114956901. The fourth root of 154017505 is 111.40188024669 and the fifth root is 43.402427145314.
Look Up
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Home » Math Theory » Algebra » Addition & Subtraction Of Square Roots
# Addition & Subtraction Of Square Roots
## What are square roots?
A square root of a number is a part of the number that can be multiplied by itself to get back to its original value. For example, if we solve 2 x 2, we will get 4. Thus, we can say that 4 is the square of 2 and that 2 is the square root of 4.
Some square roots can be easily identified, like 2 and 4, while others can be more complex. For instance, if we try to find the square root of 2, we will get an ”irrational” number. This means that there is no simple way to write it except for its radical form, which is $\sqrt2$.
## How to simplify square roots?
Just like simplifying other mathematical expressions, we can simplify square roots in radical form by combining “like terms” or “like radicals” through addition or subtraction
One essential key when combining radicals by addition or subtraction is to look at the radicand. If the given square roots have the same radicand, they are like radicals, and combining them is feasible. If not, then you cannot combine the two square roots. This is like combining like terms in an algebraic expression. You can treat the square roots in radical form as variables and combine like terms by adding or subtracting their numerical coefficients and attaching their common variable. For example, to combine $\sqrt{x}$ and 2$\sqrt{x}$, we can have $\sqrt{x}$+ 2$\sqrt{xy}$ . The radical $\sqrt{x}$ has a numerical coefficient of 1 so we can visualize it as 1$\sqrt{xy}$ while the radical 2$\sqrt{x}$ has a numerical coefficient of 2. To combine, simply add their numerical coefficients. So, $\sqrt{x}+ 2\sqrt{x}=(1+2)\sqrt{x}=3\sqrt{x}$.
Simplifying square roots is the process of writing them in the most efficient and compact structure possible while maintaining the value of the original expression. It is a useful mathematical skill because it transforms complicated or difficult-to-read radical expressions into simpler ones. Below are some rules and steps in simplifying square roots expressions.
### Simplifying square roots without variables
1. Factor the radicand whenever possible. Find any factors of the radicand that is a perfect square. For example, in the expression $\sqrt{50}$ we can factor 50 such that one factor is a perfect square, that is, $\sqrt{50} = \sqrt{25 x 2}$. The perfect square here is 25 because 25 = 5 5.
2. Bring out the square root of the perfect square factor. Get the square root of the perfect square and bring it in front of the radical sign. Leave the remaining factor inside the radical. So, in the example above, $\sqrt{50} = \sqrt{25 x 2}=5\sqrt{2}$.
### Simplifying square roots with variables
Let us assume that all variables represent nonnegative real numbers to make the simplification rules simpler.
Simplifying square roots with variables is like simplifying square roots without variables. We can treat the variable as a factor. If the variable appears twice, such as x2, we can bring out the variable x to the front of the radical sign. If the variable appears three times, such as x3 , we can factor it as x2 × x , and then bring out the variable x from x2 to the front of the radical sign leaving the single x inside the radical sign.
In general,
1. If the exponent of a variable is even, divide the exponent by two and write the result in front of the radical sign leaving no variable in the radicand. For example, in the radical expression $\sqrt{b^8}$, the exponent of the variable is even, which is 8. Hence, to find the square root of b8, simply bring out b in front of the radical sign, and its exponent will be half of the exponent of b in the radicand. Thus, $\sqrt{b^8}$= b4.
2. If the exponent of the variable is odd, subtract 1 from the exponent, divide it by two, and write the result in front of the radical sign, leaving the variable inside the radical sign without an exponent. For example, in the radical expression $\sqrt{c^7}$, the exponent of the variable is odd, which is 7. Hence, to find the square root of c7, we must factor it as c6 ×c and bring out the square root of c6 in front of the radical sign leaving just c inside the radical sign Thus, $\sqrt{c^7}= c^3\sqrt{c}$.
Example
Simplify $\sqrt{45 x^5 y^6}$.
Solution
$\sqrt{45 x^5 y^6}=\sqrt{(9 × 5)(x^4 × x){y^6}}\sqrt{45 x^5 y^6}=(3)(x^2)(y^3)\sqrt{(5)(x)}\sqrt{45 x^5 y^6}=3x^{2}y^{3}\sqrt{(5)(x)}$
## How to add square roots?
In adding square roots, we combine the like radicals, and the unlike radicals are written as they are. So, if we add 4$\sqrt{2}$ and 2$\sqrt{2}$, it is possible because they are like radicals. We can think of this as follows: “If we have four square roots of 2, and we add that to two square roots of 2, then how many square roots of 2 in total do we have?” Well, four of them plus two of them are a total of six of them. So, we get 6$\sqrt{2}$. In other words, we simply added their coefficient and attached it to their common radical.
If the radicands of the square roots are not the same, then they are not like radicals, and we cannot combine them. So, if we have $\sqrt{2}$ and $\sqrt{3}$, we cannot add or subtract one of them from the other. The radicands are different (one is 2 and the other is 3), so they are not like radicals and cannot be combined.
It is not always true that square roots with different radicands cannot be added or subtracted. Sometimes, we need to simplify radicals to end up with like radicals. For example, it appears that $\sqrt{3}+\sqrt{27}$ cannot be simplified since the radicands of the terms are different. But $\sqrt{27}$ can be simplified as 3$\sqrt{3}$. Now, $\sqrt{3}$ and 3$\sqrt{3}$ are like radicals and can be therefore combined as 4$\sqrt{3}$.
Even the radicands are different; sometimes, one or more radicals in an expression can be rewritten such that the radicands will be the same. So, it is important to look out for opportunities to rewrite the radicals before concluding that they are not like radicals and cannot be combined.
Example #1
What is the sum of 2$\sqrt{5}$ and 5$\sqrt{5}$?
Solution
Example #2
Find the sum of 2$\sqrt{x}$ and 2$\sqrt{y}$?
Solution
Example #3
Add $x\sqrt{3}\:and\:2\sqrt{12}$?
Solution
Example #4
What is the result of adding $\sqrt{9a^{2}bc^{5}}- 2\sqrt{ab^2}+ 7\sqrt{8a^{2}b}\:and\:3b\sqrt{a^3}- \sqrt{b^{3}c}$?
Solution
## How to subtract square roots?
The process for subtracting square roots is the same as the process for adding them. The only difference is that when subtracting one square root term from another, you must change the signs of each term in the expression being subtracted and then combine like radicals.
But why do we have to change the signs of a mathematical expression when being subject to subtraction? Let us take an example of two numbers, 2 and −1. Suppose if we must subtract −1 from 2, we write it as 2−(−1). We know that the product of two positive signs or two negative signs is positive, and the product of two unlike signs is negative. In the example: 2−(−1) = 2 + 1 = 3. If we do not change the sign, we will have 2 – 1 = 1, which is a completely different result. So, changing the signs of the subtrahend when subtracting expressions is necessary. In other words, the subtraction symbol (−) must be distributed to each term of the subtrahends before combining mathematical terms.
Example #1
Subtract -$\sqrt{3}$ from $3\sqrt{3}$.
Solution
Example #2
What is the result of subtracting $\sqrt{2y}$ from $\sqrt{2x}$?
Solution
Example #3
Find the difference between $2\sqrt{8x^2}\:and\:3x\sqrt{32}$?
Solution
Example #4
Solve $(\sqrt{9abc^{7}}- 2\sqrt{ab^{3}c}+ 7a\sqrt{a^{2}b}) – (3\sqrt{a^{4}b} – \sqrt{b^{2}c^{8}})$.
Solution
## How to solve problems involving addition and subtraction of square roots?
Problem #1
Mark and John both have a square garden. The area of Marks’s garden is 18x + 9 square feet, while John’s garden has an area of 8x + 4 square feet. What is the difference between the length of the sides of their garden?
Solution
Problem #2
Find the perimeter of an isosceles triangle whose legs measures $2\sqrt{2ab}$ inches and a base that is $\sqrt{8a^{3}b}$ inches.
Solution
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# Negative feedback stability
paulmdrdo
TL;DR Summary
Trying to make sense of the stability of a negative feedback.
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LvW
Most important: Show the following three functions in one diagram: f(x), g(x) and h(x)=f(x)-g(x),
berkeman
paulmdrdo
Most important: Show the following three functions in one diagram: f(x), g(x) and h(x)=f(x)-g(x),
willem2
Starting with the calculation for h1 you actually apply a phase lag of 180 degrees. The period of the signal is π so h(x-π/2) will lag h(x) by half a period.
paulmdrdo
Starting with the calculation for h1 you actually apply a phase lag of 180 degrees. The period of the signal is π so h(x-π/2) will lag h(x) by half a period.
What is the proper way of doing it then? Isn't the difference f(x)-g(x) supposed to be delayed by 90 deg when it passes the loop?
The open loop gain term A does not appear in your equations.
The open loop gain of the amplifier should be very high.
The loss and the phase shift in the feedback loop should be low for low frequencies.
The amplifier is an op-amp, the output changes, hopefully to make the inputs equal.
The feedback gain above the frequency where 180° phase shift occurs must be less than 1/A. You should only test with frequencies below that frequency.
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willem2
What is the proper way of doing it then? Isn't the difference f(x)-g(x) supposed to be delayed by 90 deg when it passes the loop?
If you want that, you should use a delay of a quarter of the full period, so you must use h1(x) = h(x-π/4). Your calculation of g(x) is still correct because sin (2x - π/2) = sin (2 (x-π/4)) so you have the correct delay there.
Gold Member
Hi Paul;
This approach uses a circuit simulator such as the free LTSpice. The advantage of such a simulator is that it uses circuit components and automatically handles the appropriate equations based on the components used. Hopefully this will give you enough of a handle on the situation that you can (eventually) write the equations directly and do a straight mathematical simulation.
I suggest the following:
1. Start with a frequency low enough that any phase shifts do not interfere with operation
2. Get the circuit working without feedback, signal to (+) input, (-) input thru a 10k resistor to Ground. Define the amplifier gain as G=2
3. The amplifier output should be +2f(x), that is twice the input amplitude
4. Now implement the feedback circuit β with a gain of 0.5. Connect a 10k resistor between the amplifier output and the (-) input
5. The amplifier output should be 1.33f(x)
• Note: the 1.33f(x) output rather than the expected 2.0f(x) output is because of the low amplifier gain
6. Increase the amplifier gain, G, to 1,000,000
7. The amplifier output should be 2.0f(x)
Once the circuit is working as above, you can start playing with the gain, or frequency response and phase shifts to see what happens.
To change the gain, change the 10k resistor from (-) input to ground to a voltage divider. The amplifier (-) input goes to the tap of the divider. Try to keep the total divider resistance to ground around 10k, this makes it easier to play with the phase shift.
Probably the easiest way to change the phase shift is to change the 10k feedback resistor to three resistors in series, each of them 3.3k. Then connect a capacitor across each of the three resistors. The reason for three RC networks is to get 180° phase shift with three 60° shifts. If I got the calculation correct, at 1kHz about 30ηF should yield 180° phase shift. Capacitor values will scale inversely with frequency, i.e. use 3ηF for 10kHz. (Hmm... further thought says to double the capacitor values, oh well, try it and see what works )
Of course the drawback to all this is you have to learn a new program! LTSpice or similiar.
Hope this helps, it got more involved than I expected!
Cheers,
Tom
This approach uses a circuit simulator such as the free LTSpice. The advantage of such a simulator is that it uses circuit components and automatically handles the appropriate equations based on the components used.
One problem with Spice simulations is that the frequency response may look good, but the circuit will be unstable in the time domain. It is easy to get it wrong if it becomes non-linear.
An op-amp with an integrator (= 90° delay) in the feedback path, makes a high gain differentiator in the forward direction, not a stable amplifier. I would expect an output switching between high and low saturated limits.
Increase the amplifier gain, G, to 1,000,000
When the input signal is a unit phasor.
The feedback signal is the output phasor, rotated back, or delayed by 90°.
The question then becomes, what output phasor satisfies (+)input = (–)input.
The circuit can only be stable if the amplified differential input moves the output in a safe direction at each point in the cycle. You must solve for a stable output phasor.
I do not believe there is a stable solution.
LvW
Quote Baluncore: One problem with Spice simulations is that the frequency response may look good, but the circuit will be unstable in the time domain.
I want to restore confidence in Spice-based simulations. The problem is generally not the simulator, but in most cases the user, if he does not know exactly what to expect from the simulator.
In an AC analysis, an operating point is calculated for the given supply voltage, which is assumed to be ideal, constant and free of disturbances. Even a calculation by hand would - in case of an unstable arrangement - result in an apparently stable circuit.
This even applies to a positive DC feedback (by hand as well as simulation).
Only a TRAN analysis in the time domain with a supply voltage switched on at t=0 can reveal a case of instability.
I have been dealing with SPICE analyses for more than 20 years. In case of errors I was always the cause (errors in interpretation, evaluation, polarities, grounding, ...).
But there is an indication of instability in AC analysis: The phase response shows an abnormal (increasing) behaviour in the region of instability.
eq1
Summary:: Trying to make sense of the stability of a negative feedback
I don’t think the math in the image is correct which might be why you’re confused.
The output of the open gain amp should be Ah(x) but it is drawn as g(x) which means the f(x) term was dropped.
Also, one can definitely analyze this structure with math but to do this analysis you need to define what definition of “stability“ you’re using. There are options.
The usual solution is:
Ah(x)=g(x)
A(f(x)-Bg(x))=g(x)
Af(x)=(1+AB)g(x)
(1/(1/A)+B))f(x)=g(x), giving the output in terms of the input
In your example A and B are not equal to 1. They are an all pass filter with constant phase delay.
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# Two Numbers
Find two numbers whose sum is 9 and product is 20.
Find two numbers whose sum is 18 and product is 77.
Find two numbers whose sum is 17.5 and product is 76.
## A Mathematics Lesson Starter Of The Day
• Mr Jankowski, Haydock Sports College
•
• Very nice as it can be used to lead into factorising quadratics.
• Matt Robey, Brighton Hill School
•
• This was really easy. I got the last one on the first guess!
• Chris And Phillip, Swansea
•
• The first question is really easy and can be spotted quite quickly.
The third question always has to be an even number multiplied by a number ending in 0.5
chow.
• Mr Spanion, Ejs
•
• I thought it was brilliant.
• Mr M Shepherd, St Lawrence Academy, Scunthorpe
•
• Easiest way to solve these we found was to double the "sum" number. This means that both the two original numbers have doubled so we need to multiply the product number by 4. This gave us all whole numbers to deal with. Once solved, half both the numbers.
(Prime factor breakdown is the best way to look at combinations systematically if needed to, although most of year 8 relied on trial and error to find the abswer once they'd found the whole numbers).
• Transum,
•
• This one of those starters that is generated with random numbers. Each time you load this page a new set of numbers appear in the statements above. You can use it many times over the course of the year so that your pupils are reminded of the techniques and have regular practice applying them.
How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.
Previous Day | This starter is for 3 December | Next Day
Note to teacher: Doing this activity once with a class helps students develop strategies. It is only when they do this activity a second time that they will have the opportunity to practise those strategies. That is when the learning is consolidated. Click the button above to regenerate another version of this starter from random numbers.
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## How Not To Be Wrong
The maths we learn in school can seem like an abstract set of rules, laid down by the ancients and not to be questioned. In fact, Jordan Ellenberg shows us, maths touches on everything we do, and a little mathematical knowledge reveals the hidden structures that lie beneath the world's messy and chaotic surface. In How Not to be Wrong, Ellenberg explores the mathematician's method of analyzing life, from the everyday to the cosmic, showing us which numbers to defend, which ones to ignore, and when to change the equation entirely. Along the way, he explains calculus in a single page, describes Gödel's theorem using only one-syllable words, and reveals how early you actually need to get to the airport.
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Educational Technology on Amazon
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Here is the URL which will take them to a related student activity.
Transum.org/go/?to=FactorPairs
For Students:
For All:
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# Thread: question for "find the coordinates of the centroid "
1. ## question for "find the coordinates of the centroid "
::: find the coordinates of the centroid of each triangle with given vertices.
A(-1,11) B(3,1) C(7,6)
how can i find it?? what formula can i use for this ?
thanks.
2. Wikipedia has a nice article on centroids which contains the solution to your type of problem:
Centroid - Wikipedia, the free encyclopedia
3. Originally Posted by sarahhh
::: find the coordinates of the centroid of each triangle with given vertices.
A(-1,11) B(3,1) C(7,6)
how can i find it?? what formula can i use for this ?
thanks.
First, use mid-point theorem to find the mid-points (X, Y, Z) of each side of triangle. For example, mid-point of AB:
$x = \frac{3 + (-1)}{2}$ = 1
$y = \frac{1 + 11}{2} = 6$
hence mid-point X = (1, 6)
Then use points X(1,6) and C(7,6) to work out the equation for line XC:
$\frac{y - 6}{x - 1} = \frac{6 - 6}{7 - 1}$
and simplify it into the form of
Ax + By + C = 0 or
y = mx + c
where A, B, C, m and c are positive/negative constants or coefficients.
do the same for AZ and BY, and finally put all the three equations for XC, AZ and BY together and solve it for the values of x and y, which are the coordinates of the centriod O.
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