9. In $\triangle A B C$, $\sin A=\frac{3}{5}, \cos B=\frac{5}{13}$, then the value of $\cos C$ is $\qquad$
$\sin B=\frac{12}{13}>\sin A=\frac{3}{5} \Rightarrow B>A$, thus $A$ is an acute angle $\Rightarrow \cos A=\frac{4}{5}$.
So $\cos C=-\cos (A+B)=\sin A \sin B-\cos A \cos B=\frac{16}{65}$.