Melody

$${\frac{{\mathtt{0.003}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{9\,000\,000}} = {\mathtt{270}}$$

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$$\\tan 17^0 = \frac{h}{13}\\\\
13\times tan17^0=h$$

$${\mathtt{13}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{17}}^\circ\right)} = {\mathtt{3.974\: \!498\: \!858\: \!967}}$$

So the height is 3.97 feet correct to 2 dec places.

you can find the slant height using the pythagorean theorem.   

Your question can be interpreted in many different ways.  Each one will give you a slightly different answer.

I have looked at just one of these.  I decided to consider a regular octagon.  The border is a concentric octagon where the width of 2 feet is at the corners.  the width between the sides would actually be a little less than this.  If you wanted the perpendicular width of the border to be the same all around then the outside shape would not be an octagon.

now, triangles OAB and triangle OQP are similar figures so their sides must be in the same ratio.

I have determined the length of QP to be  $$L_2=\frac{L_1(r+2)}{r}$$

Now  $$L_1$$   is be worked out as a function of r (original radius)  I think CPhill may have done this for you.

BUT   $$L_2$$    is still going to be dependant on the original radius of the octagon.  So I do not think that there is any set answer to you questions.  

It is an interesting question though.  

Yes there are 39C5 different combinations that could be chosen

on the web2 calc you can enter it like this

nCr(39,5)

draw a right angled triangle.

mark one of the acute angles $$\theta$$

$$\theta$$ = asin (x/1)

now

$$sin\theta = \frac{opp}{hyp}=\frac{1}{x}$$

put 1 on the opp side and x on the hypotenuse

find the adjacent side using the pythagorean theorem.

so you can read cos $$\theta$$ off the triangle.

BUT

You should consider the possible quadrants

if x is positive $$\theta$$ will be in the first quad so cos$$\theta$$ is also positive.

if x is negative $$\theta$$ will be in the 4ht quad so cos$$\theta$$ will be positive also.

You can divide both sides by -2 to simplify it.

BUT it is the equation of a line so you cannot 'solve' it.

do you mean that y = that and you want to find dy/dx

or

 are you looking for something else?

Where should the brackets be  ???

Please ask you question properly and post again.

x2/a2+y2/b2=1

the meaning is not clear.

x^2 =  $${{\mathtt{x}}}^{{\mathtt{2}}}$$

or

x_2    =  $$x_2$$