Algebra 1

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

Chapter 1 - Foundations for Algebra - Chapter Review - 1-5 and 1-6 Operations with Real Numbers - Page 71: 73

Answer

225

Work Step by Step

$(pq)^2\longrightarrow$ substitute the given values for p and q =$(5\times-3)^2\longrightarrow$ multiply in parentheses =$(-15)^2\longrightarrow$ simplify the exponent =$-15\times-15\longrightarrow$ multiply =$225\longrightarrow$ since there are an even number of negative factors, the product is positive.
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