College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2: 57

Answer

$\frac{3y^{14}}{4x^{4}}$

Work Step by Step

$$\frac{24x^{3}y^{5}}{32x^{7}y^{-9}}$$ Group factors with like bases: $$=(\frac{24}{32}))(\frac{x^{3}}{x^{7}})(\frac{y^{5}}{y^{-9}})$$ $$=\frac{3}{4}x^{(3-7)}y^{(5-(-9))}$$ $$=\frac{3}{4}x^{-4}y^{(5+9)}$$ $$=\frac{3}{4}x^{-4}y^{14}$$ Write as a fraction, move the base with the negative exponent to the other side of the fraction, and make the exponent positive: $$=\frac{3y^{14}}{4x^{4}}$$
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