Prealgebra (7th Edition)

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

Chapter 4 - Test - Page 325: 34

Answer

$\frac{4}{31}$

Work Step by Step

x $\div$ y; x = $\frac{1}{2}$, y = 3$\frac{7}{8}$ Plug $\frac{1}{2}$ in for x and 3$\frac{7}{8}$ in for y: x $\div$ y = $\frac{1}{2}$ $\div$ 3$\frac{7}{8}$ To solve, convert 3$\frac{7}{8}$ into an improper fraction and then multiply by the reciprocal of this improper fraction: $\frac{1}{2}$ $\div$ $\frac{31}{8}$ $\frac{1}{2}$ $\times$ $\frac{8}{31}$ = $\frac{8}{62}$ Reduce by dividing numerators and denominators by 2: $\frac{8\div2}{62\div2}$ = $\frac{4}{31}$
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