Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 2 - Real Numbers - 2.4 - Exponents - Problem Set 2.4: 35

Answer

$7$

Work Step by Step

Recall, in order to solve problems involving order of operations, we use the PEMDAS rule. First Priority: P - parentheses and other grouping symbols (including fraction bars) Second Priority: E - exponents Third Priority: M/D - Multiplication or division, whichever comes first from the left to the right Fourth Priority: A/S - Addition or subtraction, whichever comes first from the left to the right We follow order of operations to obtain that the expression, $ \dfrac{3(4-2)^2}{4}+\dfrac{2(-1+3)^3}{4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3(4-2)(4-2)}{4}+\dfrac{2(-1+3)(-1+3)(-1+3)}{4} \\\\= \dfrac{3(2)(2)}{4}+\dfrac{2(2)(2)(2)}{4} \\\\= \dfrac{3(\cancel{2})(\cancel{2})}{\cancel{2}(\cancel{2})}+\dfrac{\cancel{2}(\cancel{2})(2)(2)}{\cancel{2}(\cancel{2})} \\\\= 3+4 \\\\= 7 .\end{array}
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