[(x^2 y^-2)^-1]^-1= (1/x^2 y^-2)^-1= (y^2/x^2)-1= x^2/y^2 this solution is right according to the back of the book, are the steps also correct?
Last edited by dymand68; Apr 1st 2010 at 06:05 PM.
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Originally Posted by dymand68 [(x^2y^-2)^-1]^-1= (1/x^2y^-2)^-1= (y^2/x^2)-1= x^2/y^2 this solution is right according to the back of the book, are the steps also correct? $\displaystyle [(x^2y^{-2})^{-1}]^{-1}$ $\displaystyle = (x^2y^{-2})^1$ $\displaystyle = x^2y^{-2}$ $\displaystyle = \frac{x^2}{y^2}$ $\displaystyle = \left(\frac{x}{y}\right)^2$.
ty.........so the correct order of operations starts with the negative exponent OUTSIDE the brackets?
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