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**Titus****Member**- Registered: 2005-03-07
- Posts: 10

Really need some help with my maths homework, any help would be greatly appreciated

by the way "-²" is to the power of negative 2, same deal with negative 3.

^ is just to the power of.

These ones have to go to positive indicies, thats all:

(c-²)/4

(3x^-1 y-²)/z²

(x-³)/(3^-1)

(P^½)-½

(2x-³)³

thanks

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,555

(c-²)/4 = 1/4c²

(3x^-1 y-²)/z² = 3/xy²z²

(x-³)/(3^-1) = 3/x³

(P^½)-½ = p^(-1/4)=1/p^(1/4)

(2x-³)³ = 2x^(-9)=2/(x^9)

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