> From: Xaq <xaq@INDIRECT.COM>
>
> I while back I posted a question about a simple math problem, that everyone
> was very helpful in solving.  I hope you guys can help me with this one, it
> is a little harder.
>
> I need to solve this equation:
>
>         sqrt(x^2+y^2)/(x*y)
>
>      Both x, y, and the answer are all byte(0-255) values.
>
> [cut]

That's easy:

if x or y are zero, return 255 (nearest thing to infinity).
if x == 1 or y == 1, return 1 (nearest thing to sqrt(x^2+1)/x).
else return 0 (nearest thing to any other case).

But seriously! I presume you want to scale the result to a meaningful
precision?

Regards,
SJH