What is 4^{t-6}\H(t-6)e^{-4t}cosh(9t) ?

The solution to 4^{t-6}\H(t-6)e^{-4t}cosh(9t) is (4^{t-6}\H(t-6)cosh(9t))/(e^{4t)}

Popular >

simplify 4^{t-6}\H(t-6)e^{-4t}cosh(9t)

simplify

4^{t - 6} H (t - 6) e^{- 4 t} cosh (9 t)

\frac{4 ^{t - 6} H ( t - 6 ) cosh ( 9 t )}{e ^{4 t}}

Solution steps

4^{t - 6} H (t - 6) e^{- 4 t} cosh (9 t)

Apply exponent rule:

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

= 4^{t - 6} \cdot \frac{1}{e ^{4 t}} H (t - 6) cosh (9 t)

Multiply fractions:

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

= \frac{1 \cdot 4 ^{t - 6} H ( t - 6 ) cosh ( 9 t )}{e ^{4 t}}

Multiply:

1 \cdot 4^{t - 6} = 4^{t - 6}

= \frac{4 ^{t - 6} H ( t - 6 ) cosh ( 9 t )}{e ^{4 t}}

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

5.25 \cdot 1 0^{- 5}

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

2.25 \cdot 1 0^{- 10}

What is 4^{t-6}\H(t-6)e^{-4t}cosh(9t) ?
The solution to 4^{t-6}\H(t-6)e^{-4t}cosh(9t) is (4^{t-6}\H(t-6)cosh(9t))/(e^{4t)}