Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - 5.6 Exponents and Scientific Notation - Exercise Set 5.6: 40

Answer

$-\dfrac{20x^3}{y^2}$

Work Step by Step

(i) $a^m \cdot a^n = a^{m+n}$ (ii) $\dfrac{a^m}{a^n}=a^{m-n}$ (iii) $a^{-m} = \dfrac{1}{a^m}, a \ne 0, m \gt0$ (iv) $(a^m)^n=a^{mn}$ Use the applicable rules above to find: $=-20x^{4+(-1)}y^{-3+1} \\=-20x^3y^{-2} \\=-20x^3(\frac{1}{y^2}) \\=-\dfrac{20x^3}{y^2}$
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