Mathematics
Mathematics, 26.05.2020 21:05, dancemomsrule1

\begin{aligned} &f(t)=\dfrac{2t+7}{t-3} &h(n)=-2n^2-3n+1 \end{aligned} f(t)= tβˆ’3 2t+7 h(n)=βˆ’2n 2 βˆ’3n+1 f(h(-1))=f(h(βˆ’1))=

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\begin{aligned} &f(t)=\dfrac{2t+7}{t-3} &h(n)=-2n^2-3n+1 \end{aligned} f(t)= tβˆ’3 2t+7 h(n...

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