Answer
Yes, see explanations.
Work Step by Step
Yes, it makes sense. As a general rule, if $0\lt a\lt 1$, we have $0\lt a^n\lt 1$. In this exercise, $0\lt 5^{-2}\lt 1$, we have $0\lt (5^{-2})^3\lt 1$.
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter P - Section P.2 - Exponents and Scientific Notation - Exercise Set - Page 31: 130
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