Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 54

Answer

$\frac{5}{8}$

Work Step by Step

According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$. Therefore, $2^{-1}+8^{-1}=\frac{1}{2}+\frac{1}{8}=\frac{1\times4}{2\times4}+\frac{1}{8}=\frac{4+1}{8}=\frac{5}{8}$.
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