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Answer to Question #223815 in Chemical Engineering for Lokika
Question #223815
L {t^{2}*H'(t-3)}
Expert's answer
"L = ( {t^{2}(t-3)})\\\\\nt-3 =L;\\quad t\u00b2=1\\\\\nt-3 =L;\\quad t=\\sqrt{1\u00b2 },=1\\\\\nt-3 =L;\\quad t=1\\\\\n1-3 =L;\\quad t=1\\\\\nL =-2; \\quad \\ \\ \\ t=1"
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