What is e^{-2(t-t^0)}u(t-2t^0) ?

The solution to e^{-2(t-t^0)}u(t-2t^0) is (u(t-2))/(e^{2(t-1))}

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simplify e^{-2(t-t^0)}u(t-2t^0)

simplify

e^{- 2 (t - t^{0})} u (t - 2 t^{0})

\frac{u ( t - 2 )}{e ^{2 (t - 1)}}

Solution steps

e^{- 2 (t - t^{0})} u (t - 2 t^{0})

Apply exponent rule:

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

= \frac{1}{e ^{2 (t - t^{0})}} u (t - 2 t^{0})

Apply rule

a^{0} = 1, a \neq = 0

t^{0} = 1

= \frac{1}{e ^{2 (t - 1)}} u (t - 2 \cdot 1)

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1}{e ^{2 (t - 1)}} u (t - 2)

Multiply fractions:

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

= \frac{1 \cdot u ( t - 2 )}{e ^{2 (t - 1)}}

Multiply:

1 \cdot u = u

= \frac{u ( t - 2 )}{e ^{2 (t - 1)}}

(\frac{0.04}{0.46})^{- 1.78} \cdot 2.93

⎩ ⎨ ⎧ [(- \frac{2}{3} + x)^{5}]^{0} ⎭ ⎬ ⎫^{- 2}

1.79 \cdot 1 0^{- 9}

What is e^{-2(t-t^0)}u(t-2t^0) ?
The solution to e^{-2(t-t^0)}u(t-2t^0) is (u(t-2))/(e^{2(t-1))}