Convolution including $\delta(t-5)$ I know the two properties of convolution that are related to my question


$\quad x(t)*\delta(t)=x(t) $
$\quad x(t)*\delta(t-t_0)=x(t-t_0)$


But my question is, how can I use those two to calculate $$ y(t)=x(5t)*\delta(t-5) $$
HINT:

Set $f(t)=x(5t)$ and use your rule number 2: $$f(t)\star \delta(t-t_0)=f(t-t_0)=\ldots$$