How do you evaluate #h(2)# given #h(t)=t^3+5t^2#?

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Answer 1 · 2016-08-14T09:42:06

#h(2)=28#

#h(t)=t^3+5t^2#

#h(2)=2^3+5(2^2)#

#=8+5times4#

#=8+20#

#=28#

Answer 2 · 2016-08-14T09:47:10

h(2) = 28

To evaluate h(2), substitute t = 2 into h(t).

#h(color(red)(2))=(color(red)(2))^3+5(color(red)(2))^2#

#rArrh(2)=(2xx2xx2)+(5xx2xx2)=8+20=28#