Answer
$\frac{8x^2y^{-3}}{30x^{-1}y^2}=\frac{4x^3}{15y^5}$
Work Step by Step
$\frac{8x^2y^{-3}}{30x^{-1}y^2}=\frac{(2)4x^2\frac{1}{y^3}}{(2)15\frac{1}{x}y^2}=\frac{4x^2}{15y^2}\frac{1}{y^3}\frac{x}{1}=\frac{4x^3}{15y^5}$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 2 - Cumulative Test for Chapters P-2 - Page 236: 1
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