# use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=2(t-5)u_5(t)

use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$f\left(t\right)=2\left(t-5\right){u}_{5}\left(t\right)$
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$\text{Step 1}$

$f\left(t\right)=2\left(t-5\right){u}_{5}\left(t\right)$
$L\left[f\left(t\right)\right]-?$
$\text{Step 2}$

$\text{Step 3}$

$L\left[f\left(t\right)\right]=L\left[2\left(t-5\right){u}_{5}\left(t\right)\right]$
$L\left[f\left(t\right)\right]=2L\left[\left(t-5\right){u}_{5}\left(t\right)\right]$
$L\left[f\left(t\right)\right]=2{e}^{-5s}L\left(t\right)⇒\left[L\left[f\left(t-c\right){u}_{c}\left(t\right)\right]={e}^{-cs}L\left[f\left(t\right)\right]\right]$
$L\left[f\left(t\right)\right]=2{e}^{-5s}\left(\frac{1!}{{s}^{1+1}}\right)⇒\left[L\left({t}^{n}\right)=\frac{n!}{{s}^{n+1}}\right]$
$L\left[f\left(t\right)\right]=2{e}^{-5s}\left(\frac{1}{{s}^{2}}\right)$
$\begin{array}{|c|}\hline L\left[f\left(t\right)\right]=2\frac{{e}^{-5s}}{{s}^{2}}\\ \hline\end{array}$
$\text{Step 4}$