What is expand 2sin(3t)[u(t-2)] ?

The solution to expand 2sin(3t)[u(t-2)] is 2tusin(3t)-4usin(3t)

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expand 2sin(3t)[u(t-2)]

expand

2 sin (3 t) [u (t - 2)]

2 t u sin (3 t) - 4 u sin (3 t)

Solution steps

2 sin (3 t) (u (t - 2))

Remove parentheses:

(a) = a

= 2 sin (3 t) u (t - 2)

= 2 u sin (3 t) (t - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = 2 sin (3 t) u, b = t, c = 2

= 2 sin (3 t) u t - 2 sin (3 t) u \cdot 2

= 2 t u sin (3 t) - 2 \cdot 2 u sin (3 t)

Multiply the numbers:

2 \cdot 2 = 4

= 2 t u sin (3 t) - 4 u sin (3 t)

\frac{30 + s}{s ( s + 10 )}

e^{x} (1 - e^{- x})

\frac{\partial}{\partial x} ((z) (yz - ln (x + z)))

\frac{6 x ^{5} + 3 x ^{4} - 7 x ^{3} + 5 x ^{2} - 8}{x ^{2} \cdot ( 3 x ^{2} - 2 )}

\frac{( 3 x ) !}{( 3 ( x + 1 ) ) !}

What is expand 2sin(3t)[u(t-2)] ?
The solution to expand 2sin(3t)[u(t-2)] is 2tusin(3t)-4usin(3t)