What is expand (2e^{2r}sin(3θ))(s-2t) ?

The solution to expand (2e^{2r}sin(3θ))(s-2t) is 2e^{2r}ssin(3θ)-4e^{2r}tsin(3θ)

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expand (2e^{2r}sin(3θ))(s-2t)

expand

(2 e^{2 r} sin (3 θ)) (s - 2 t)

2 e^{2 r} s sin (3 θ) - 4 e^{2 r} t sin (3 θ)

Solution steps

(2 e^{2 r} sin (3 θ)) (s - 2 t)

Apply the distributive law:

a (b - c) = ab - a c

a = (2 e^{2 r} sin (3 θ)), b = s, c = 2 t

= (2 e^{2 r} sin (3 θ)) s - (2 e^{2 r} sin (3 θ)) \cdot 2 t

= s (2 e^{2 r} sin (3 θ)) - 2 t (2 e^{2 r} sin (3 θ))

Simplify

s (2 e^{2 r} sin (3 θ)) - 2 t (2 e^{2 r} sin (3 θ)) : 2 e^{2 r} s sin (3 θ) - 4 e^{2 r} t sin (3 θ)

= 2 e^{2 r} s sin (3 θ) - 4 e^{2 r} t sin (3 θ)

(4 k - x m) (k z^{2} - x \frac{m z ^{2}}{4})

(\frac{7}{9} + \frac{2}{9})^{9}

\frac{1}{x - 2} - \frac{4}{x ^{2} - 4}

\frac{( 2 + j ) ( 0.6 + j ^{0.8} )}{- 4 + j} - j

lo g_{10} (3) (4 x + 6) - lo g_{3} (3 x + 9)

What is expand (2e^{2r}sin(3θ))(s-2t) ?
The solution to expand (2e^{2r}sin(3θ))(s-2t) is 2e^{2r}ssin(3θ)-4e^{2r}tsin(3θ)