Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Review - Page 330: 37



Work Step by Step

Using laws of exponents and concepts of scientific notation, then, \begin{array}{l} \dfrac{(0.00012)(144,000)}{0.0003} \\\\= \dfrac{(1.2\times10^{-4})(1.44\times10^{5})}{3.0\times10^{-4}} \\\\= (1.2\times1.44\div3.0)(10^{-4+5-(-4)}) \\\\= (0.576)(10^{5}) \\\\= 5.76\times10^{4} .\end{array}
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