Step-by-step explanation:
Here,let x = 1,2,3 and 4
Then,
Domain =(1,2,3,4)
Again,
f(1)= 3*1 + 5
=3 + 5
=8
f(2)=3*2 + 5
=6 + 5
=11
f(3)=3*3 + 5
=9 + 5
=14
f(4)=3*4 + 5
=12 + 5
=17
Therefore, Range=(8,11,14,17)
Given:
A figure of right triangle with terminal angle
, base u, perpendicular v and hypotenuse r.
To find:
The value of
.
Solution:
In a right angle triangle,
![\sin \theta=\dfrac{Perpendicular}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D%5Cdfrac%7BPerpendicular%7D%7BHypotenuse%7D)
Substituting Perpendicular = v and hypotenuse = r in the above formula, we get
![\sin \theta=\dfrac{v}{r}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D%5Cdfrac%7Bv%7D%7Br%7D)
The value of
is
.
Therefore, the correct option is B.
<span>The standard deviations of 99.1 centimeters and 0.991 meters are equivalent lengths.
The meter measurements will have less variation because of the numbers the variance will be 1m for every 100cm. But the coefficient variation will be the same for both.
</span>
Answer:
![y=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3e%5E%7B-4t%7D)
Step-by-step explanation:
![y''+5y'+4y=0](https://tex.z-dn.net/?f=y%27%27%2B5y%27%2B4y%3D0)
Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)
![s^2L-3s+5sL-3+4L=0](https://tex.z-dn.net/?f=s%5E2L-3s%2B5sL-3%2B4L%3D0)
Solving for ![L](https://tex.z-dn.net/?f=L)
![L(s^2+5s+4)=3s+3](https://tex.z-dn.net/?f=L%28s%5E2%2B5s%2B4%29%3D3s%2B3)
![L=\frac{3s+3}{s^2+5s+4}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B3s%2B3%7D%7Bs%5E2%2B5s%2B4%7D)
![L=\frac{3(s+1)}{(s+1)(s+4)}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B3%28s%2B1%29%7D%7B%28s%2B1%29%28s%2B4%29%7D)
![L=\frac3{s+4}](https://tex.z-dn.net/?f=L%3D%5Cfrac3%7Bs%2B4%7D)
Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)
The graph of the given system of linear inequalities
y < -1/2 + 2
y > -3/2x + 2
is attached below
<h3>Graph of system of linear inequalities </h3>
From the given information, we are to graph the given system of linear inequalities
The given system of linear inequalities is
y < -1/2 + 2
y > -3/2x + 2
The graph of the given system of linear inequalities
y < -1/2 + 2
y > -3/2x + 2
is shown below
Learn more on Graph of system of linear inequalities here: brainly.com/question/26965469
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