Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 2 - Test - Page 134: 35

Answer

$$11.7 \times 10^{24}$$

Work Step by Step

We have $$\frac{(3.03\times 10^{12})^2}{\sqrt{615 \times 10^{-3}}} =\frac{(3.03\times 10^{12})^2}{\sqrt{61.5 \times 10^{-2}}} \\=\frac{9.181\times 10^{24}}{7.842\times 10^{-1}} =1.1707\times 10^{24+1}\\=11.7 \times 10^{24}$$ which is in the engineering notation.
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