Melody

I'm am just writing the same as guest here. Thanks guest (below)

\(\begin{pmatrix} n\\ k-1 \end{pmatrix}\\=\frac{n!}{(k-1)!(n-[k-1])!}\\ =\frac{n!}{(k-1)!(n-k+1)!}\\\)

I 

Thanks Heureka   

I do not think the earlier answer is correct.

Here is a graphical solution;

It shows and answer of approximately 2.04 units

the inverse of 

\(y=-\frac{1}{2x}\qquad where \qquad x<0\\is\\ x=-\frac{1}{2y} \qquad where \qquad x>0\\ rearrange\\ y=-\frac{1}{2x} \qquad where \qquad x>0\\ \)

so

\(k(x)=-\frac{1}{2x}\)

\(\begin{pmatrix} n\\ k \end{pmatrix}\;means \;\; \frac{n!}{k!(n-k)!}\)

You need to know this!

Then you need to do some study on your own.

This is very basic to this topic.

\((x-y)^4 \\=\begin{pmatrix}4\\0\end{pmatrix}(x)^4(-y)^0+ \begin{pmatrix} 4\\ 1 \end{pmatrix}(x)^3(-y)^1 +\begin{pmatrix} 4\\ 2 \end{pmatrix} (x)^2(-y)^2 +\begin{pmatrix}4\\3\end{pmatrix}(x)^1(-y)^3 +\begin{pmatrix} 4\\ 4 \end{pmatrix}(x)^0(-y)^4 \)

Work out the pattern and then finish it yourself.

You can ask questions if you want.

I do not know that formula so here is my method.

The radius is r units.  The encircle centre is the point (r,r)

To find r, the perpendicular distance of the hypotenuse to (r,r) must equal r.

\(r=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+b^2}}\\ r=\frac{|8r+15r-120|}{\sqrt{8^2+15^2}}\\ r=\frac{|23r-120|}{17}\\ r=\frac{120-23r}{17}\\ 17r+23r=120\\ 40r=120\\ r=3 \)

Radius is 3

Diameter is 6

No just at that one point. At the point of inflextion the gradient is neither increasing nor decreasing. 

If ithe rate of change is not + or - then it must be 0. Which means that  the gradient must be a constant at that point. 

It is not something I would usually add.  (I can see the confusion)

I would normally have just left the statement out altogether.

It is true though.