Answer
a) $\frac{7}{x}$
b) $\frac{4(x+y)^2}{3}$
Work Step by Step
a) $\frac{7x^2}{x^3}=\frac{7x^2}{x^2\cdot x}=\frac{7}{x}$
b) $\frac{12(x+y)^3}{9(x+y)}=\frac{3(x+y)4(x+y)^2}{3\cdot3(x+y)}=\frac{4(x+y)^2}{3}$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Prerequisites - P.2 - Exponents and Radicals - Exercises - Page 24: 24
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