Question

\begin{aligned} &f(t)=\dfrac{2t+7}{t-3} \\\\ &h(n)=-2n^2-3n+1 \end{aligned} ​ f(t)= t−3 2t+7 ​ h(n)=−2n 2 −3n+1 ​ f(h(-1))=f(h(−1))=

Answer 1

Answer:

$f(h(-1))=-11$

Step-by-step explanation:

Given:

$\begin{aligned} &f(t)=\dfrac{2t+7}{t-3} \\\\ &h(n)=-2n^2-3n+1 \end{aligned}$

We want to determine the value of $f(h(-1))$

First, we evaluate h(-1)

$h(-1)=-2(-1)^2-3(-1)+1 =-2+3+1\\h(-1)=2$

Therefore: $f(h(-1))=f(2)$

$f(2)=\dfrac{2(2)+7}{2-3}\\=\dfrac{4+7}{-1}\\=-11$

Therefore: f(h(-1))=-11