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Re: beech proxy returns
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Re: beech proxy returns - 10-26-2007, 04:07 PM
Quote:
Originally Posted by sillysally
That's what I was hoping you would say.
But not hoping what I can do? But if there is a will there is a way.

Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(t)·r(t)=c²
  
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