Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 1 - Review: 95



Work Step by Step

$\frac{5(6^2-3)}{3^2+2}\ \longrightarrow$exponents are resolved first =$\frac{5(36-3)}{9+2}\ \longrightarrow$resolve expressions in parentheses =$\frac{5(33)}{9+2}\ \longrightarrow$multiply =$\frac{165}{9+2}\ \longrightarrow$add =$\frac{165}{11}\ \longrightarrow$reduce =$\frac{165\div11}{11\div11}\ \longrightarrow$divide =$\frac{15}{1}=15$
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