Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Section 5.1 - Exponents - Practice: 34

Answer

a. $7$ b. $ 2$ c. $\frac{x^{18}}{4y^{12}} $ d. $ - b^{11}$

Work Step by Step

a. $2^3 - 2^0 = 8 - 1 = 7$ b. $(y^0)^7 + (5^0)^3 = 1^7 + 1^3 = 2$ c. $(\frac{7x^9}{14y^6})^2 = \frac{7^2x^{18}}{14^2y^{12}} = \frac{49x^{18}}{196y^{12}} =\frac{x^{18}}{4y^{12}} $ d. $\frac{(3a^2b^5)^3}{-27a^6b^4} = \frac{27a^6b^{15}}{-27a^6b^4} = \frac{b^{11}}{-1} = - b^{11}$
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