#### Answer

$\dfrac{81}{4} \text{ and } \dfrac{9}{2}$

#### Work Step by Step

The third term of a perfect square trinomial is equal to the square of one-half the coefficient of the middle term. Hence, the third term of the given expression, $
t^2-9t
+\underline{ }
,$ is
\begin{array}{l}\require{cancel}
t^2-9t+\left( \dfrac{-9}{2} \right)^2
\\\\=
t^2-9t+\dfrac{81}{4}
.\end{array}
Since the factored form of a perfect square trinomial is a binomial where the terms are the square roots of the first and third terms, respectively, and the operator between the first and second terms is the same as the sign of the middle term of the trinomial, the factored form of $
t^2-9t+\dfrac{81}{4}
$ is
\begin{array}{l}\require{cancel}
\left( t-\dfrac{9}{2} \right)^2
.\end{array}
Hence, the expressions appropriate in the blanks are $
\dfrac{81}{4} \text{ and } \dfrac{9}{2}
.$