Algebra 1

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

Chapter 7 - Exponents and Exponential Functions - 7-3 Multipying Powers with the Same Base - Practice and Problem-Solving Exercises - Page 430: 49


$4y^{5}$ + $8y^{2}$

Work Step by Step

The area of a parallelogram, Area = $base$ X $height$ Hence , Area = $4y^{2}$ X $(y^{3}+2)$ By distributive property, Area = $4y^{2}$ X $y^{3}+ 4y^{2}$ X $2$ This becomes : Area = $4y^{2+3}$ + $8y^{2}$ (since the exponents are added while multipliying with the same base) Final result : Area = $4y^{5}$ + $8y^{2}$
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