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 399: 77

Answer

see graph

Work Step by Step

With $ k=\dfrac{1}{2} ,$ then the given variation model, $y=\dfrac{k}{x},$ becomes \begin{array}{l}\require{cancel} y=\dfrac{1/2}{x} \\\\ y=\dfrac{1}{2}\div x \\\\ y=\dfrac{1}{2}\cdot\dfrac{1}{x} \\\\ y=\dfrac{1}{2x} .\end{array} If $x=\dfrac{1}{4},$ then \begin{array}{l}\require{cancel} y=\dfrac{1}{2(1/4)} \\\\ y=\dfrac{1}{1/2} \\\\ y=2 .\end{array} If $x=\dfrac{1}{2},$ then \begin{array}{l}\require{cancel} y=\dfrac{1}{2(1/2)} \\\\ y=\dfrac{1}{1} \\\\ y=1 .\end{array} If $x=1,$ then \begin{array}{l}\require{cancel} y=\dfrac{1}{2(1)} \\\\ y=\dfrac{1}{2} .\end{array} If $x=2,$ then \begin{array}{l}\require{cancel} y=\dfrac{1}{2(2)} \\\\ y=\dfrac{1}{4} .\end{array} If $x=4,$ then \begin{array}{l}\require{cancel} y=\dfrac{1}{2(4)} \\\\ y=\dfrac{1}{8} .\end{array} The completed table and the corresponding graph are shown above.
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