Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set: 98

Answer

$1.331928\times10^{13} \text{ tons}$

Work Step by Step

Using $D=\dfrac{M}{V}$ or the density of an object, then \begin{array}{l}\require{cancel} D=\dfrac{M}{V} \\ 3.12\times10^{-2}=\dfrac{M}{4.269\times10^{14}} .\end{array} By cross-multiplication, then \begin{array}{l}\require{cancel} M=(3.12\times10^{-2})(4.269\times10^{14}) \\ M=(3.12)(4.269)\times(10^{-2})(10^{14}) \\ M=13.31928\times(10^{-2})(10^{14}) .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} M=13.31928\times10^{-2+14} \\ M=13.31928\times10^{12} \\ M=1.331928\times10^{13} .\end{array} Hence, the mass is $ 1.331928\times10^{13} \text{ tons} .$
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