Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Cumulative Review Exercises - Page 176: 13


$2.1\cdot 10^{-5}$

Work Step by Step

$(7\cdot10^{-8})(3\cdot10^2)=(7\cdot3)\cdot(10^{-8}10^2)=21\cdot10^{-6}=2.1\cdot 10^{-5}$
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