#### Answer

$h(6)-h(-3)=27$

#### Work Step by Step

We are given:
$h(t)=t^{2}+5$
We evaluate:
$h(-3)=(-3)^{2}+5=9+5=14$
$h(6)=6^{2}+5=36+5=41$
$h(6)-h(-3)=41-14=27$

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$h(6)-h(-3)=27$

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