3. Find the inverse Laplace transform of F(s) = (-4s-9) / (s^2 + 25-8) f(t) =

[tex]f(s)\Longrightarrow L^{-1}=\{\frac{-4s-9}{s^2+25-8}\}[/tex]First dismantle,[tex]L^{-1}=\{-\frac{4s}{s^2+25-8}-\frac{9}{s^2+25-8}\}[/tex]Now use the linearity property of Inverse Laplace Transform which states, For functions [tex]f(s),g(s)[/tex] and constants [tex]a, b[/tex] rule applies,[tex]L^{-1}=\{a\cdot f(s)+b\cdot g(s)\}=aL^{-1}\{f(s)\}+bL^{-1}\{f(s)\}[/tex]Hence,[tex]-4L^{-1}\{\frac{s}{s^2+25-8}\}-9L^{-1}\{\frac{1}{s^2+25-8}\}[/tex]The first part simplifies to,[tex]L^{-1}\{\frac{s}{s^2+25-8}\} \\\frac{d}{dt}(\frac{1}{\sqrt{17}}\sin(t\sqrt{17})) \\\cos(t\sqrt{17})[/tex]The second part simplifies to,[tex]L^{-1}\{\frac{1}{s^2+25-8}\} \\\frac{1}{\sqrt{17}}\sin(t\sqrt{17})[/tex]And we result with,[tex]\boxed{-4\cos(t\sqrt{17})-\frac{9}{\sqrt{17}}\sin(t\sqrt{17})}[/tex]Hope this helps.If you have any additional questions please ask. I made process of solving as quick as possible therefore you might be left over with some uncertainty.Hope this helps.r3t40