Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 4 - Section 4.2 - Factors and Simplest Form - Exercise Set - Page 236: 49

Answer

$14a^2$

Work Step by Step

$\frac{224a^3b^4c^2}{16ab^4c^2}=\frac{224a^3b^4c^2\div16ab^4c^2}{16ab^4c^2\div16ab^4c^2}=\frac{14a^2}{1}=14a^2$
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