What is 5e^{-2t}cos(3t+θ) ?

The solution to 5e^{-2t}cos(3t+θ) is (5cos(3t+θ))/(e^{2t)}

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simplify 5e^{-2t}cos(3t+θ)

simplify

5 e^{- 2 t} cos (3 t + θ)

\frac{5 cos ( 3 t + θ )}{e ^{2 t}}

Solution steps

5 e^{- 2 t} cos (3 t + θ)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= 5 \cdot \frac{1}{e ^{2 t}} cos (3 t + θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 5 cos ( 3 t + θ )}{e ^{2 t}}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5 cos ( 3 t + θ )}{e ^{2 t}}

3 X^{2} \cdot 4 X^{3}

(6.673 \cdot 1 0^{- 11}) \cdot (5.9736 \cdot 1 0^{24})

3.6 \times 1 0^{- 2}

e^{- \frac{400}{360}}

(14 2 y)^{10}

What is 5e^{-2t}cos(3t+θ) ?
The solution to 5e^{-2t}cos(3t+θ) is (5cos(3t+θ))/(e^{2t)}