Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.7 - Variation and Problem Solving - Exercise Set - Page 398: 36

Answer

$k=1.1$, $y=\frac{1.1}{x^{2}}$

Work Step by Step

Since $y$ varies inversely as the square of $x$, then, $ y=\dfrac{k}{x^2} .$ Substituting the given values of the variables, then the value of $k$ is, \begin{array}{l} 0.011=\dfrac{k}{10^2} \\\\ 0.011=\dfrac{k}{100} \\\\ 0.011(100)=k \\\\ 1.1=k \\\\ k=1.1 .\end{array}
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