Answer
$h(6)-h(-3)=27$
Work Step by Step
We are given:
$h(t)=t^{2}+5$
We evaluate:
$h(-3)=(-3)^{2}+5=9+5=14$
$h(6)=6^{2}+5=36+5=41$
$h(6)-h(-3)=41-14=27$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 193: 42
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