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: 33

Answer

$-13$

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(2)^4}{12}+\dfrac{5(-3)^3}{15} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{-3(2)(2)(2)(2)}{12}+\dfrac{5(-3)(-3)(-3)}{15} \\\\= \dfrac{-3(2)(2)(2)(2)}{2(2)(3)}+\dfrac{5(-3)(-3)(-3)}{3(5)} \\\\= \dfrac{-\cancel{3}(\cancel{2})(\cancel{2})(2)(2)}{\cancel{2}(\cancel{2})(\cancel{3})}+\dfrac{\cancel{5}(-3)(-3)(-\cancel{3})}{\cancel{3}(\cancel{5})} \\\\= -4+(-9) \\\\= -4-9 \\\\= -13 .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.