Pyramid scheme
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Hyperbolic
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Isn't the function 2^n for n starting at 0
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Mathematical
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Originally Posted by SquirrelDip
[((x-h)^2)/(a^2)]-[((y-k)^2)/(b^2)] = 1
or
[((y-k)^2)/(a^2)]-[((x-h)^2)/(b^2)] = 1
actually the formula 1+X(a)^2 doesn't yield the results listed...therefore the edit of the reply to "Retarded"
Calculus
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makntraksIn the theater of the mind...
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Originally Posted by pcexpress-guy
makntraksIn the theater of the mind...
It's always good to know where the exits are... -
boot
His name was MackemX
What kind of a man are you? The guy is unconscious in a coma and you don't have the guts to kiss his girlfriend? -
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Bum-Floss
@pcexpress-guy: No, your first statement says no same values for a, b, c, d. Second equation states that c=d which is in contradiction to the first statement. -
marshmellow
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(Edited because I got the URL wrong... Never mind, eh?)
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