# Convolution including $\delta(t-5)$

I know the two properties of convolution that are related to my question

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

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

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