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Just to tie up loose ends here, I have the generalization.
[math]\frac{b\sqrt{a^2+b^2}}{a}[/math]
This is only true where:
[math]a>b[/math]
I think that's about as much math that we can suck from that problem :lol:
[math]\frac{b\sqrt{a^2+b^2}}{a}[/math]
This is only true where:
[math]a>b[/math]
I think that's about as much math that we can suck from that problem :lol:
