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03-01-2008, 03:51 PM
Re: origin of hyperbolic boost

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Originally Posted by mkirkpatrick
we will all find something
I'll be searching for more probable third part time job or lesser probability of winning the lottery.
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Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
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03-01-2008, 03:55 PM
Smile Re: origin of hyperbolic boost

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Originally Posted by AntonioLao View Post
I'll be searching for more probable third part time job or lesser probability of winning the lottery.

My job now is swimming and relaxing and of course posting here,plus moving countries
and starting a new life,thats all.



rgards michael.
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