Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 7 - Exponents and Exponential Functions - Chapter Review - 7-1 Zero and Negative Exponents: 13

Answer

$45$

Work Step by Step

We rewrite the given expression as a division problem: $5x^0\div y^{-2}$ The zero as an exponent rule states that for every nonzero number $a$, $a^0=1$. Since $x^0=1$, and anything multiplied by $1$ is itself, we can remove that term. $5\div y^{-2}$ The negative exponent rule states that for every nonzero number $a$ and integer $n$, $a^{-n}=\frac{1}{a^n}$. We use this rule to rewrite the expression: $\frac{5}{1}\div\frac{1}{y^2}$ To divide by a fraction, we multiply by the reciprocal: $5\times y^2$ We plug in the value for $y$: $5(-3)^2$ We simplify powers: $5(9)$ We multiply: $45$
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