College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.5 - Rational Expressions - R.5 Exercises - Page 47: 25

Answer

$\dfrac{2}{9}$

Work Step by Step

The given expression, $ \dfrac{2k+8}{6}\div\dfrac{3k+12}{2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2k+8}{6}\cdot\dfrac{2}{3k+12} \\\\= \dfrac{2(k+4)}{6}\cdot\dfrac{2}{3(k+4)} \\\\= \dfrac{\cancel{2}(\cancel{k+4})}{\cancel{2}(3)}\cdot\dfrac{2}{3(\cancel{k+4})} \\\\= \dfrac{2}{9} .\end{array}
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