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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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(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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