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 - Test - Page 102: 34


$$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})= \frac{40}{x^{3}y^{8}}$$

Work Step by Step

$$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})$$ Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$ Thus, $$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})$$ $$=(-8)(-5)(x^{-5+2}y^{-3+(-5)}$$ $$=40x^{-3}y^{-8}$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$40x^{-3}y^{-8} = \frac{40}{x^{3}y^{8}}$$
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