Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises: 25

Answer

$\color{blue}{\dfrac{8}{9}}$

Work Step by Step

Factor the numerator and the denominator then cancel the common factors to obtain: $\require{cancel} =\dfrac{8(k+2)}{9(k+2)} \\=\dfrac{8\cancel{(k+2)}}{9\cancel{(k+2)}} \\=\color{blue}{\dfrac{8}{9}}$
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