Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-4 - Cumulative Review: 8

Answer

$50a^4b^7$

Work Step by Step

RECALL: (1) $(ab)^m=a^mb^m$ (2) $(a^m)^n=a^{mn}$ Use rule (1) above to obtain: $(2a^2b)(5ab^3)^2 \\=(2a^2b)[5^2a^2(b^3)^2] \\=(2a^2b)[25a^2(b^3)^2]$ Use rule (2) above to obtain: $(2a^2b)[25a^2(b^3)^2] \\=(2a^2b)(25a^2b^{3(2)}) \\=(2a^2b)(25a^2b^{6})$ Multiply the coefficients together. Multiply the variables using the product rule for exponents $(a^m \cdot a^n=a^{m+n})$ to obtain: $=(2\cdot 25)a^{2+2}b^{1+6} \\=50a^4b^7$
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.