Prealgebra (7th Edition)

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

Appendix B - Exercise Set: 7

Answer

$\frac{1}{2}xy^3$

Work Step by Step

$\frac{7x^3y^6}{14x^2y^3}=(\frac{7}{14})(\frac{x^3}{x^2})(\frac{y^6}{y^3})=(\frac{/\!\!7}{2\times/\!\!7})(x^{3-2})(y^{6-3})=\frac{1}{2}xy^3$
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