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(6*x^3*y^-7 divided by 9xy)^-1 ????

Guest Aug 23, 2016

1⁺⁰ Answers

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$\left(\dfrac{6x^3y^{-7}}{9xy}\right)^{-1}\\ = \dfrac{9xy}{6x^3y^{-7}}\\ = \dfrac{9xy^8}{6x^3}\\ =\dfrac{3y^8}{2x^2}$

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MaxWong Aug 23, 2016

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