Melody

The web2 calc of course.     

It depends what help you want.  It won't just give you answers without you understanding anything.

Ok I have simplified it and made it a 2x by 2y rectangle (which is a special case parallelogram)

This is a rough outline:

Triangle DPC and triangle EPB are similar.  DC=2x   and  EB=x    so the little one is half the size of the big one.

So  JP=2*IP

so 3*IP= 2y

IP = 2y/3

so  PG= y-  2y/3  = y/3

Now look at similar triangles  AF and i forgot to give it a name .. . ok look at the big green triangle and the similar little grey one.

PG is y/3  and  A? is  y  so the big one is 3 times as big as the little one.

So AP:AF = 2:3

I don't know but i have an interactive graph here:

I have no idea, sorry.

I have opened the question though, so other people can answer here. 

If it starts with an even

There are 4 first digit possibilities (not 5 becasue it can's start with 0)  , 4 third digits 3 fifth digits etc  so that is    4*4!  *   (now for the odd digits)  4!

If it starts with an odd then there will be  5!*4! different ones

Now add them together.

4*4!*4! + 5!*4!

Oh, I only just saw the 'same guest here' bit.

It is really good that you are discussing your own problem.  We really like that.

Why don't you become a member and get known to us.  

Members who make a good impression get preferential treatment. 

Hi Guest, your logic sounds good to me.   

It can definitely be a square or a rhombus.  

I think the translation condition means that opposite sides must be parallel and equal length.

The second condition means that the diagonals must intersect at right angles.

there are no other quadrilaterals that meet both these conditions.

Let p(x) = x^(2), q(x) = 2x^(3)+ 1, and r(x) = 7x^(4) - 5x^(3) - 6x - 2. Find p(q(x)) + r(x).

\( p(x) = x^2, \quad q(x) = 2x^3+ 1,\quad and \quad r(x) = 7x^4 - 5x^3- 6x - 2. \\Find \qquad p(q(x)) + r(x).\\ \\~\\ p(q(x)) + r(x)=(2x^3+ 1)^2\quad +7x^4 - 5x^3- 6x - 2\)

You can finish it.

Thanks guys