What is expand e^{2(t-3)}cos(w)(t-3)\H(t-3) ?

The solution to expand e^{2(t-3)}cos(w)(t-3)\H(t-3) is e^{2t-6}t\H(t-3)cos(w)-3e^{2t-6}\H(t-3)cos(w)

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expand e^{2(t-3)}cos(w)(t-3)\H(t-3)

expand

e^{2 (t - 3)} cos (w) (t - 3) H (t - 3)

e^{2 t - 6} t H (t - 3) cos (w) - 3 e^{2 t - 6} H (t - 3) cos (w)

Solution steps

e^{2 (t - 3)} cos (w) (t - 3) H (t - 3)

= e^{2 (t - 3)} H (t - 3) cos (w) (t - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = e^{2 (t - 3)} cos (w) H (t - 3), b = t, c = 3

= e^{2 (t - 3)} cos (w) H (t - 3) t - e^{2 (t - 3)} cos (w) H (t - 3) \cdot 3

= e^{2 (t - 3)} t H (t - 3) cos (w) - 3 e^{2 (t - 3)} H (t - 3) cos (w)

Simplify

e^{2 (t - 3)} t H (t - 3) cos (w) - 3 e^{2 (t - 3)} H (t - 3) cos (w) : e^{2 t - 6} t H (t - 3) cos (w) - 3 e^{2 t - 6} H (t - 3) cos (w)

= e^{2 t - 6} t H (t - 3) cos (w) - 3 e^{2 t - 6} H (t - 3) cos (w)

\frac{x ( x - 1 )}{( x + 7 ) ( x - 6 )}

\frac{x ^{2} + 2 x + k}{2 ( x + 2 )}

(2, - 3) y (- 5, 4)

\frac{2 x + 3}{( x - k ) ( x + 4 )}

50 (1 - 0.4 e^{4 x})

What is expand e^{2(t-3)}cos(w)(t-3)\H(t-3) ?
The solution to expand e^{2(t-3)}cos(w)(t-3)\H(t-3) is e^{2t-6}t\H(t-3)cos(w)-3e^{2t-6}\H(t-3)cos(w)