What is e^{-2t}(4cos(3t)+5cos(3t)) ?

The solution to e^{-2t}(4cos(3t)+5cos(3t)) is (9cos(3t))/(e^{2t)}

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simplify e^{-2t}(4cos(3t)+5cos(3t))

simplify

e^{- 2 t} (4 cos (3 t) + 5 cos (3 t))

\frac{9 cos ( 3 t )}{e ^{2 t}}

Solution steps

e^{- 2 t} (4 cos (3 t) + 5 cos (3 t))

Apply exponent rule:

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

= \frac{1}{e ^{2 t}} (4 cos (3 t) + 5 cos (3 t))

Add similar elements:

4 cos (3 t) + 5 cos (3 t) = 9 cos (3 t)

= 9 \cdot \frac{1}{e ^{2 t}} cos (3 t)

Multiply fractions:

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

= \frac{1 \cdot 9 cos ( 3 t )}{e ^{2 t}}

Multiply the numbers:

1 \cdot 9 = 9

= \frac{9 cos ( 3 t )}{e ^{2 t}}

16 I n e^{\frac{1}{4}}

e v a l u t e - 100

60 e^{- 0.00041 \cdot 100}

- e^{- 1} (2)

e^{- α x^{2}}

What is e^{-2t}(4cos(3t)+5cos(3t)) ?
The solution to e^{-2t}(4cos(3t)+5cos(3t)) is (9cos(3t))/(e^{2t)}