What is (18e^{-3t})(2e^{3t}+6e^{3t}t) ?

The solution to (18e^{-3t})(2e^{3t}+6e^{3t}t) is 36(3t+1)

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simplify (18e^{-3t})(2e^{3t}+6e^{3t}t)

simplify

(18 e^{- 3 t}) (2 e^{3 t} + 6 e^{3 t} t)

36 (3 t + 1)

Solution steps

(18 e^{- 3 t}) (2 e^{3 t} + 6 e^{3 t} t)

Apply exponent rule:

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

= 18 \cdot \frac{1}{e ^{3 t}} (6 e^{3 t} t + 2 e^{3 t})

Multiply fractions:

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

= \frac{1 \cdot 18 ( 2 e ^{3 t} + 6 e ^{3 t} t )}{e ^{3 t}}

Multiply the numbers:

1 \cdot 18 = 18

= \frac{18 ( 6 e ^{3 t} t + 2 e ^{3 t} )}{e ^{3 t}}

Factor

18 (2 e^{3 t} + 6 e^{3 t} t) : 36 e^{3 t} (1 + 3 t)

= \frac{36 e ^{3 t} ( 1 + 3 t )}{e ^{3 t}}

Cancel the common factor:

e^{3 t}

= 36 (3 t + 1)

- 8.921 x 1 0^{(- 3)} (s + (- 2.23077))

\frac{x ^{3}}{∣ x ∣ ^{3}}

\frac{0.9 6 ^{2}}{1.5 1 ^{2}}

0.142 x 1 0^{- 3}

1.369 e^{- 2}

What is (18e^{-3t})(2e^{3t}+6e^{3t}t) ?
The solution to (18e^{-3t})(2e^{3t}+6e^{3t}t) is 36(3t+1)