Answer:
L[f(t)=s/(1+s^2)]
Step-by-step explanation:
The Laplace Transform is given by the integral:
![L[f(t)]=\int_0^\infty e^{-st}\ f(t)dt](https://tex.z-dn.net/?f=L%5Bf%28t%29%5D%3D%5Cint_0%5E%5Cinfty%20e%5E%7B-st%7D%5C%20f%28t%29dt)
by replacing f(t)=cost we get
![\int_0^{\infty} e^{-st}costdt=[e^{-s(\infty)}sin(\infty)-1sin(0)]-s\int_{0}^{\infty}e^{-st}sintdt\\\\=0+s[-e^{-s(\infty)}cos(\infty)+e^{s(0)}cost(0)-s\int_0^{\infty}e^{-st}costdt]\\\\=0+s[0+1-s\int_0^{\infty}e^{-st}costdt]=s-s^2\int_0^{\infty}e^{-st}costdt\\\\(1+s^2)\int_0^{\infty}e^{-st}costdt=s\\\\\int_0^{\infty}e^{-st}costdt=\frac{s}{1+s^2}](https://tex.z-dn.net/?f=%5Cint_0%5E%7B%5Cinfty%7D%20e%5E%7B-st%7Dcostdt%3D%5Be%5E%7B-s%28%5Cinfty%29%7Dsin%28%5Cinfty%29-1sin%280%29%5D-s%5Cint_%7B0%7D%5E%7B%5Cinfty%7De%5E%7B-st%7Dsintdt%5C%5C%5C%5C%3D0%2Bs%5B-e%5E%7B-s%28%5Cinfty%29%7Dcos%28%5Cinfty%29%2Be%5E%7Bs%280%29%7Dcost%280%29-s%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-st%7Dcostdt%5D%5C%5C%5C%5C%3D0%2Bs%5B0%2B1-s%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-st%7Dcostdt%5D%3Ds-s%5E2%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-st%7Dcostdt%5C%5C%5C%5C%281%2Bs%5E2%29%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-st%7Dcostdt%3Ds%5C%5C%5C%5C%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-st%7Dcostdt%3D%5Cfrac%7Bs%7D%7B1%2Bs%5E2%7D)
hope this helps!!
X + (x + 2) + (x + 4) + (x + 6) = 156
x + x + x + x + 2 + 4 + 6 = 156
4x + 12 = 156
- 12 - 12
4x = 144
4 4
x = 36
x = 36
x + 2 = 38
x + 4 = 40
x + 6 = 42
S=-4r-2 tell me if I'm wrong cuz I'm not very great a math but I want to try the question out
Answer:
8/3
Step-by-step explanation: