Apply the translation theorem to find the inverse Laplace transform of the following function. 6s + 10 F(S) = $2 - 10s + 89 Reference Click the icon to view the table of Laplace transforms. f (t ) = 2-{F(s)} eff(t)} = F(s) f (t ) = 2 - 1( F ( s ) ) eff(t) } = F(s) S 2- 1{F (s)} = (s > 0) cos kt - ol - s- + 1 2 (s > 0) (Type an expression using t as the variable.) K (s > 0) sin kt s + K- (s > 0 ) n! th S n+ 1 (s > 0) cosh kt S (s > [kl) T(a + 1) ta K (s > 0) sinh kt s |kl) 1 n! e at (s> a) e atn s - a (s-a) +1 (s>a)