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: 20

Answer

$\frac{1}{9}$

Work Step by Step

(i) $a^{-m} =\dfrac{1}{a^m}, a \ne 0, m \gt 0$ Use rule (i) to find: $3^{-2}=\dfrac{1}{3^2}=\dfrac{1}{9}$
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