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
It is expressed as x=a when a equals the x-intercept.
e.g. a straight vertical line through (0,4) has the equation x=4.
Answer:
square root of 125 rounded to 11.2 milies to the nearest tenth
Answer:
x = -41
y = 9
Step-by-step explanation:
Begin by multiplying the top equation by 2
2x = -8y - 10
The second equation requires a little more care.
Start by moving 10y to the right. Normally the y value is on the left, but we are going to be substituting for 2x.
2x = - 10y + 8
What I've done is not the usual practice, but there is nothing wrong with it.
Now since both right sides are equal to 2x, you can substitute for them
-8y - 10 = -10y + 8 Add 10y to both sides
-8y + 10y - 10 = 8
2y - 10 = 8 Add 10 to both sides
2y = 8+10
2y = 18 Divide by 2
y = 18/2
y = 9
Use the second original equation to solve for x
2x + 10y = 8
2x + 10*9 = 8
2x + 90 = 8 Subtract 20 from both sides
2x = 8 - 90
2x = - 82 Divide by 2
x = -822/2
x = - 41
Answer:
<em>Answer is</em><em> </em><em>given below</em><em> </em><em>with explanations</em><em>. </em>
Step-by-step explanation:
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
