QuestionAugust 27, 2025

h(t)= ) (t)/(3)+6&,&tin (-infty ,-3) t(t+4)&,&tin [-3,3) t^2+t+1&,&tin [3,infty ) h(0)= square

h(t)= ) (t)/(3)+6&,&tin (-infty ,-3) t(t+4)&,&tin [-3,3) t^2+t+1&,&tin [3,infty ) h(0)= square
h(t)= ) (t)/(3)+6&,&tin (-infty ,-3) t(t+4)&,&tin [-3,3) t^2+t+1&,&tin [3,infty ) 
h(0)= square

Solution
4.1(214 votes)

Answer

0 Explanation 1. Identify the correct piece of the function Since 0 \in [-3, 3), use the second piece: h(t) = t(t+4). 2. Substitute t=0 into the function Calculate h(0) = 0(0+4) = 0.

Explanation

1. Identify the correct piece of the function<br /> Since $0 \in [-3, 3)$, use the second piece: $h(t) = t(t+4)$.<br />2. Substitute $t=0$ into the function<br /> Calculate $h(0) = 0(0+4) = 0$.
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