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 - Page 33: 58

Answer

$\frac{y^{12}}{3x^{8}}$

Work Step by Step

$$\frac{10x^{4}y^{9}}{30x^{12}y^{-3}}$$ Group factors with like bases: $$=(\frac{10}{30}))(\frac{x^{4}}{x^{12}})(\frac{y^{9}}{y^{-3}})$$ $$=\frac{1}{3}x^{(4-12)}y^{(9-(-3))}$$ $$=\frac{1}{3}x^{-8}y^{(9+3)}$$ $$=\frac{1}{3}x^{-8}y^{12}$$ Write as a fraction, move the base with the negative exponent to the other side of the fraction, and make the exponent positive: $$=\frac{y^{12}}{3x^{8}}$$
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