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

The solution to 2e^{-3t}[cos(4t)+sin(4t)] is (2(cos(4t)+sin(4t)))/(e^{3t)}

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simplify 2e^{-3t}[cos(4t)+sin(4t)]

simplify

2 e^{- 3 t} [cos (4 t) + sin (4 t)]

\frac{2 ( cos ( 4 t ) + sin ( 4 t ) )}{e ^{3 t}}

Solution steps

2 e^{- 3 t} [cos (4 t) + sin (4 t)]

Apply exponent rule:

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

= 2 \cdot \frac{1}{e ^{3 t}} [cos (4 t) + sin (4 t)]

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

5.25 \cdot 1 0^{- 5}

e^{- 2 a r c t a n (x)}

2.25 \cdot 1 0^{- 10}

A e^{- b x}

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