Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set: 104

Answer

$\dfrac{-12y^{15}}{x^3}$

Work Step by Step

RECALL: (i) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne0$ (ii) $a^{-m} = \dfrac{1}{a^m}, a \ne0$ Divide the coefficients by each other and the variables by each other to obtain: $=\dfrac{24}{-2} \cdot x^{2-5}y^{13-(-2)} \\=-12 \cdot x^{-3}y^{13+2} \\=-12x^{-3}y^{15}$ Use rule (ii) above to obtain: $=-12 \cdot \dfrac{1}{x^3} \cdot y^{15} \\=\dfrac{-12y^{15}}{x^3}$
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