Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 1 - Section 1.2 - Algebraic Expressions and Sets of Numbers - Exercise Set - Page 15: 6

Answer

$\frac{2}{15}$

Work Step by Step

You are given the expression, $yz$, which means you are multiplying the variable $y$ with the variable $z$, ($y\times z$). The values of each of these variables is given in the question. $y = \frac{2}{3}$ and $z = \frac{1}{5}$ So, substitute these given values into the expression $yz$ and solve! $yz= (\frac{2}{3}) \times(\frac{1}{5})$ Multiply the numbers on top, or numerators with each first, so $2$ x $1$= $2$ Then multiply the numbers at the bottom, or denominators with each other, so $3$ x $5$ = $15$ Put the fraction together again with the products, so your answer will be $\frac{2}{15}$.
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