Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises: 43

Answer

$a)$ $\Big(\dfrac{3a}{b^{3}}\Big)^{-1}=\dfrac{b^{3}}{3a}$ $b)$ $\Big(\dfrac{q^{-1}r^{-1}s^{-2}}{r^{-5}sq^{-8}}\Big)^{-1}=\dfrac{s^{3}}{q^{7}r^{4}}$

Work Step by Step

$a)$ $\Big(\dfrac{3a}{b^{3}}\Big)^{-1}$ $\Big(\dfrac{3a}{b^{3}}\Big)^{-1}=\dfrac{b^{3}}{3a}$ $b)$ $\Big(\dfrac{q^{-1}r^{-1}s^{-2}}{r^{-5}sq^{-8}}\Big)^{-1}$ $\Big(\dfrac{q^{-1}r^{-1}s^{-2}}{r^{-5}sq^{-8}}\Big)^{-1}=\dfrac{r^{-5}sq^{-8}}{q^{-1}r^{-1}s^{-2}}=q^{-8-(-1)}r^{-5-(-1)}s^{1-(-2)}=...$ $...=q^{-7}r^{-4}s^{3}=\dfrac{s^{3}}{q^{7}r^{4}}$
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