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 - Page 320: 39

Answer

$-\dfrac{6x^2}{y^3}$

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: $=-6x^{3+(-1)}y^{-4+1} \\=-6x^2y^{-3} \\=-6x^2(\frac{1}{y^3}) \\=-\dfrac{6x^2}{y^3}$
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