Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Review: 39

Answer

$\dfrac{3x^{7}}{2y^{2}}$

Work Step by Step

Using laws of exponents, then, \begin{array}{l} \dfrac{27x^{-5}y^{5}}{18x^{-6}y^{2}}\cdot\dfrac{x^4y^{-2}}{x^{-2}y^3} \\\\= \dfrac{3x^{-5-(-6)}y^{5-2}}{2}\cdot x^{4-(-2)}y^{-2-3} \\\\= \dfrac{3x^{}y^{3}}{2}\cdot x^{6}y^{-5} \\\\= \dfrac{3x^{1+6}y^{3+(-5)}}{2} \\\\= \dfrac{3x^{7}y^{-2}}{2} \\\\= \dfrac{3x^{7}}{2y^{2}} .\end{array}
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