(5) For sets $M$ and $N$, define $M-N=\{x \mid x \in M$, and $x \notin N\}$,
$M \oplus N=(M-N) \cup(N-M)$, let $A=\left\{y \mid y=x^{2}-3 x, x \in \mathbf{R}\right\}, B=$ $\left\{y \mid y=-2^{x}, x \in \mathbf{R}\right\}$, then $A \oplus B=(\quad)$.
(A) $\left(-\frac{9}{4}, 0\right]$
(B) $\left[-\frac{9}{4}, 0\right)$
(C) $\left(-\infty,-\frac{9}{4}\right) \cup[0,+\infty)$
(D) $\left(-\infty,-\frac{9}{4}\right) \cup(0,+\infty)$
(5) From the problem, we have $A=\left[-\frac{9}{4},+\infty\right), B=(-\infty, 0)$, so
$$
A \oplus B=\left(-\infty,-\frac{9}{4}\right) \cup[0,+\infty) \text {. }
$$

Therefore, the correct choice is C.