Melody

I am sorry that you are having this problem.

I am not involved with the technical side of the forum at all.

I  will try to get input from admin for you but maybe just try again every few days and see if it works.

I am sorry that I cannot be more helpful.

Actually,  You can lodge a complaint, which admin will probably see, and then he will have your email address because I think you will have to include it

There is a link at the bottom of the page saying 'report a problem'  try doing that. 

I hope this is helpful.  Get back to me with an update.  I hope you can get your membership to activate.    

They work

the second one is the proper link.

I like your song.

What do you want help with?

18/23

Dear guest,

I would suggest you not use terms such as 'asap' in your questions.

It implies that you think your question is more important than other people's.

Or worse still that you have left your homework till the last minute, or you are doing a test and hence late help is equal no help at all.

Terms such as this, make me less inclined to respond quickly.

= \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }

\(z= \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\\ z= \frac{ 3 - 2 *2 }{\sqrt{3} - 2\sqrt{2} }\\ z= \frac{ -1 }{\sqrt{3} - 2\sqrt{2} }\\ z= \frac{ -1 }{\sqrt{3} - 2\sqrt{2} }*\frac{\sqrt{3} + 2\sqrt{2} }{\sqrt{3} + 2\sqrt{2} }\\ z= \frac{-1(\sqrt{3} + 2\sqrt{2}) }{3 - 8}\\ z= \frac{(\sqrt{3} + 2\sqrt{2}) }{5}\\ z= \frac{(almost 2 + almost3)}{5}\\ z= \frac{almost 5 }{5}\\0

This would not display before so here is a pic of my answer.

Become a member and then maybe someone might respond appropriately.

With a private message I mean.

It will be a girl with fat box feet and arms up in the air.

What is      A/xyz      I do not understand what I am being asked to find ???

The ratio of the areas is 9:16  but I really do not know if that is what you want......

I count 7

Diveneology:

Thank you very much for your answer.