Answer to Question #223815 in Chemical Engineering for Lokika

Question

L {t^{2}*H'(t-3)}

Answer 1

$L = ( {t^{2}(t-3)})\\ t-3 =L;\quad t²=1\\ t-3 =L;\quad t=\sqrt{1² },=1\\ t-3 =L;\quad t=1\\ 1-3 =L;\quad t=1\\ L =-2; \quad \ \ \ t=1$