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: $$f(t)\star \delta(t-t_0)=f(t-t_0)=\ldots$$


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