Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises: 20

Answer

(a) $-2^{3}\times(-2)^{0} =-8$ (b) $-2^{-3}\times(-2)^{0}=-\frac{1}{8}$ (c) $(\frac{-3}{5})^{-3}=-\frac{125}{27}$

Work Step by Step

(a) $-2^{3}\times(-2)^{0} =-(2^{3})\times(-2)^{0}$ Note: anything to the power of zero is 1 hence $(-2)^{0}=1$ $-(2)^{3}=-8$ hence $-2^{3}\times(-2)^{0} =-8\times1=-8$ (b) $-2^{-3}\times(-2)^{0} =-(2^{-3})\times(-2)^{0}$ Note: anything to the power of zero is 1 hence $(-2)^{0}=1$ $-(2)^{-3}=-\frac{1}{8}$ hence $-2^{-3}\times(-2)^{0} =-\frac{1}{8}\times1=-\frac{1}{8}$ (c) $(\frac{-3}{5})^{-3}=\frac{1}{(\frac{-3}{5})^{3}}=\frac{1}{(-\frac{27}{125})}=-\frac{125}{27}$
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.