Intermediate Algebra for College Students (7th Edition)

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

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 81: 133

Answer

The distance between Mercury and the sun is $ 0.55 $ astronomical unit.

Work Step by Step

The given formula is $ d=\frac{3(2^{n-2})+4}{10} $ Where, $ d $ is measured in astronomical units and $ n $ is the number of planet from the sun. Substitute $ n=1 $ into the formula. $ d=\frac{3(2^{1-2})+4}{10} $ $ d=\frac{3(2^{-1})+4}{10} $ $ d=\frac{3 \cdot \frac{1}{2}+4}{10} $ $ d=\frac{ \frac{3}{2}+4}{10} $ $ d=\frac{ \frac{3+8}{2}}{10} $ $ d=\frac{ \frac{11}{2}}{10} $ $ d=\frac{11}{20} $ $ d=0.55 $
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