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

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

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

simplify

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

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

Solution steps

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(t - 3) (t - 3) = (t - 3)^{1 + 1}

= e^{4 (t - 3)} cos (5) u (t - 3)^{1 + 1}

Add the numbers:

1 + 1 = 2

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

φ^{- 1} (0.9)

e^{- s i n (x)} e^{s i n (x)}

2000 (1 + \frac{0.24}{4})^{- 2}

44.8 \cdot 1 0^{- 6}

(6 x^{3} y^{2}) (3 x^{2} y^{5})

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