Answer:
No
Step-by-step explanation:
The square root of 34 does not multiply by itself
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-
<em>The inverse of
is </em>

<u>Solution-</u>
The given function is,

We can get the inverse by interchanging he variable x and y among themselves and then separating each variables.
So in the inverse would be,

Taking log of both sides,

As,

Applying the same,


As,

Applying the same,

Therefore, the inverse of
is
.
Answer:
QN = 43
Step-by-step explanation:
3x-5 = 2x+8+3
3x-5 = 2x+11
x-5 = 11
x = 16
MP = 3x-5
MP = 3(16)-5
MP = 48-5
MP = 43
MP and QN are ≅ so QN = 43