The answer is x^6(2X^6-3)(4X^12+6X^6+9)
Answer:
Step-by-step explanation:
By definition of Laplace transform we have
L{f(t)} = ![L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\](https://tex.z-dn.net/?f=L%7B%7Bf%28t%29%7D%7D%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7Df%28t%29dt%5C%5C%5C%5CGiven%5C%5Cf%28t%29%3D7t%5E%7B3%7D%5C%5C%5C%5C%5Ctherefore%20L%5B7t%5E%7B3%7D%5D%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7D7t%5E%7B3%7Ddt%5C%5C%5C%5C)
Now to solve the integral on the right hand side we shall use Integration by parts Taking 
as first function thus we have
![\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7D7t%5E%7B3%7Ddt%3D7%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7Dt%5E%7B3%7Ddt%5C%5C%5C%5C%3D%20%5Bt%5E3%5Cint%20e%5E%7B-st%7D%20%5D_%7B0%7D%5E%7B%5Cinfty%7D-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5B%283t%5E2%29%5Cint%20e%5E%7B-st%7Ddt%5Ddt%5C%5C%5C%5C%3D0-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B3t%5E%7B2%7D%7D%7B-s%7De%5E%7B-st%7Ddt%5C%5C%5C%5C%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B3t%5E%7B2%7D%7D%7Bs%7De%5E%7B-st%7Ddt%5C%5C%5C%5C)
Again repeating the same procedure we get
![=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\](https://tex.z-dn.net/?f=%3D0-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B3t%5E%7B2%7D%7D%7B-s%7De%5E%7B-st%7Ddt%5C%5C%5C%5C%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B3t%5E%7B2%7D%7D%7Bs%7De%5E%7B-st%7Ddt%5C%5C%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B3t%5E%7B2%7D%7D%7Bs%7De%5E%7B-st%7Ddt%3D%20%5Cfrac%7B3%7D%7Bs%7D%5Bt%5E2%5Cint%20e%5E%7B-st%7D%20%5D_%7B0%7D%5E%7B%5Cinfty%7D-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5B%28t%5E2%29%5Cint%20e%5E%7B-st%7Ddt%5Ddt%5C%5C%5C%5C%3D%5Cfrac%7B3%7D%7Bs%7D%5B0-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B2t%5E%7B1%7D%7D%7B-s%7De%5E%7B-st%7Ddt%5D%5C%5C%5C%5C%3D%5Cfrac%7B3%5Ctimes%202%7D%7Bs%5E%7B2%7D%7D%5B%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7Dte%5E%7B-st%7Ddt%5D%5C%5C%5C%5C)
Again repeating the same procedure we get
![\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Ctimes%202%7D%7Bs%5E2%7D%5B%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7Dte%5E%7B-st%7Ddt%5D%3D%20%5Cfrac%7B3%5Ctimes%202%7D%7Bs%5E%7B2%7D%7D%5Bt%5Cint%20e%5E%7B-st%7D%20%5D_%7B0%7D%5E%7B%5Cinfty%7D-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5B%28t%29%5Cint%20e%5E%7B-st%7Ddt%5Ddt%5C%5C%5C%5C%3D%5Cfrac%7B3%5Ctimes%202%7D%7Bs%5E2%7D%5B0-%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%5Cfrac%7B1%7D%7B-s%7De%5E%7B-st%7Ddt%5D%5C%5C%5C%5C%3D%5Cfrac%7B3%5Ctimes%202%7D%7Bs%5E%7B3%7D%7D%5B%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7Ddt%5D%5C%5C%5C%5C)
Now solving this integral we have
![\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7Ddt%3D%5Cfrac%7B1%7D%7B-s%7D%5B%5Cfrac%7B1%7D%7Be%5E%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B1%7D%5D%5C%5C%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-st%7Ddt%3D%5Cfrac%7B1%7D%7Bs%7D)
Thus we have
![L[7t^{3}]=\frac{7\times 3\times 2}{s^4}](https://tex.z-dn.net/?f=L%5B7t%5E%7B3%7D%5D%3D%5Cfrac%7B7%5Ctimes%203%5Ctimes%202%7D%7Bs%5E4%7D)
where s is any complex parameter
Answer:
C. It is not a good fit because there are no points on the line.
Step-by-step explanation:
In order for a line to be a good fit for a data set represented as a scatterplot, the line must follow the general trend of the data in the scatterplot. This line does not follow the general trend of the data on the scatterplot, thus option (C) is the best statement to describe the situation.
C. It is not a good fit because there are no points on the line.
