Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter R - Section R.3 - Exponents, Roots, and Order of Operations - R.3 Exercises - Page 30: 85



Work Step by Step

We are given that $a=-3$, $b=64$, and $c=6$. Therefore, $\frac{2c+a^{3}}{4b+6a}=\frac{2(6)+(-3)^{3}}{4(64)+6(-3)}=\frac{12+(-27)}{256+(-18)}=\frac{12-27}{256-18}=\frac{-15}{238}=-\frac{15}{238}$.
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