Melody

It is for n'th roots.

Like  the the 4th root of 81 is 3 because 3 to the power of4 is 81

Enter like this

81. Y√x.   4 =

Oh this calc says x√y.  I think that is a simple display error.

Thanks Omi,  

Hi 315, Let's see what I would do.

Now that I have already done it all I can see that Omi DID multiply by x^2

Our solutions are almot identical.  

(2/x^2)-(2/x+10)=1/x-1

\(\begin{align}\\ \frac{2}{x^2}-(\frac{2}{x}+10)&=\frac{1}{x}-1\qquad {x\ne0}\\ \frac{2}{x^2}-\frac{2}{x}-10&=\frac{1}{x}-1\\ \frac{2}{x^2}-\frac{3}{x}-9&=0\\ x^2\left(\frac{2}{x^2}-\frac{3}{x}-9\right)&=x^2*0\\~\\ 2-3x-9x^2&=0\\ 9x^2+3x-2&=0\\ x&=\frac{-3\pm\sqrt{9+72}}{18}\\ x&=\frac{-3\pm9}{18}\\ x&=\frac{-1\pm3}{6}\\ \end{align}\\~\\ x=\dfrac{-2}{3}\qquad or \qquad x=\dfrac{1}{3}\)

BINOMIALS

\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2

\(\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2\)

\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \text{ something}.

\(\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \text{ something}.\)

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FRACTIONS

\cfrac   and    \dfrac

\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}

\(\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\)

More here

exponent button ???

x^y

Or if you are typing it in just use ^ 

Hi 315,

I don't know what you mean by paremeter       , and I assume the black dot is the centre,   

Height of the triangle is  \(\frac{D}{2}-d = \frac{D-2d}{2}\)

\(sec \frac{\theta}{2}=r\div \frac{D-2d}{2}\\ ( \frac{D-2d}{2})sec \frac{\theta}{2}=r\\ r=( \frac{D-2d}{2})sec \frac{\theta}{2}\\\)

assuming that theta is in radians:

\(\begin{align}\\Area&=\frac{\theta}{2\pi}\times\pi r^2-\frac{1}{2}r^2sin\theta\\~\\ &=\frac{\theta r^2}{2}-\frac{1}{2}r^2sin\theta\\~\\ &=\frac{ r^2}{2}(\theta-sin\theta)\\~\\ &=\frac{\left [\frac{D-2d}{2}\;sec\frac{\theta}{2}\right]^2}{2}(\theta-sin\theta)\\~\\ &=\frac{(D-2d)^2\;sec^2\left(\frac{\theta}{2}\right)(\theta-sin\theta)}{8}\quad units^2\\~\\ \end{align}\)

*

Hi Itzcubez,

I have no idea what c or d even mean.  Sorry.  Here's the first bit.

(a) Compute the sums of the squares of Rows 1-4 of Pascal's Triangle. 

\(1+1=2\\ 1+4+1=6\\ 1+9+9+1=20\\ 1+16+36+16+1=70\\ \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2} \)

Do these sums appear anywhere else in Pascal's Triangle?   

\( \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2}\)

(b) Guess at an identity based on your observations from part (a). Your identity should be of the form

(You have to figure out what "something" is.) Test your identity for  using your results from part (a).

You can do the testing but yea it works :)

(c) Prove your identity using a committee-forming argument.

(d) Prove your identity using a block-walking argument.

I'm good thanks 315:)

Why are you so confuzzled Tristen   LOL

LaTex discussion on displaying fractions :)

Heureka, I thought maybe you would like to add to this post :/     

I have also been looking at the Latex that is used in this question.

I found this reference to using Latex for fractions but i will admit i have not properly digested it yet.

Hi Alan

I have been toying this question,  I have only just worked out the relevance of the 2 answers :D

Silly me :)

Here is Alan's earlier answer:

what is the fraction of 0.6 out of 20

\(\frac{0.6}{20}=\frac{6}{200}=\frac{3}{100}\)