Answer
$t^{29}$
Work Step by Step
Given: $y^8(y^{-7})^{-3}$
Firstly multiply the power exponents and then add the exponents of the common base term.
$y^8(y^{-7})^{-3}=y^8(y^{-7 \times {-3}})$
$=y^{8+21}$
$=y^{29}$
Algebra 1
by Hall, Prentice
Published by
Prentice Hall
ISBN 10:
0133500403
ISBN 13:
978-0-13350-040-0
Chapter 9 - Quadratic Functions and Equations - 9-5 Completing the Square - Mixed Review - Page 566: 63
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