Answer:
Cable attached height (Perpendicular) = 10√3 m
Step-by-step explanation:
Given:
Length of cable (Hypotenuse) = 20 m
Angle θ = 60°
Find:
Cable attached height (Perpendicular)
Computation:
Perpendicular/Hypotenuse = Sinθ
Perpendicular/20 = Sin 60
Perpendicular/20 = √3 / 2
Perpendicular = 10√3
Cable attached height (Perpendicular) = 10√3 m
Answer: Yes, ![\bold{f^{-1}(x)=\dfrac{x+6}{3}}](https://tex.z-dn.net/?f=%5Cbold%7Bf%5E%7B-1%7D%28x%29%3D%5Cdfrac%7Bx%2B6%7D%7B3%7D%7D)
<u>Step-by-step explanation:</u>
f(x) = 3x - 6 is a line so it is a function because
it passes the vertical line test
and the horizontal line test
To find the inverse, swap the x's and y's and solve for y
![y=3x-6\\\\\\\text{Swap the x's and y's:}\qquad x=3y-6\\\\\text{Add 6 to both sides:}\qquad x+6=3y\\\\\text{Divide both sides by 3:}\qquad \dfrac{x+6}{3}=y](https://tex.z-dn.net/?f=y%3D3x-6%5C%5C%5C%5C%5C%5C%5Ctext%7BSwap%20the%20x%27s%20and%20y%27s%3A%7D%5Cqquad%20x%3D3y-6%5C%5C%5C%5C%5Ctext%7BAdd%206%20to%20both%20sides%3A%7D%5Cqquad%20x%2B6%3D3y%5C%5C%5C%5C%5Ctext%7BDivide%20both%20sides%20by%203%3A%7D%5Cqquad%20%5Cdfrac%7Bx%2B6%7D%7B3%7D%3Dy)
is a closed surface with interior
, so you can use the divergence theorem.
![\vec F(x,y,z)=x\,\vec\imath+y\,\vec\jmath+9\,\vec k\implies\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(x)}{\partial x}+\dfrac{\partial(y)}{\partial y}+\dfrac{\partial(9)}{\partial z}=2](https://tex.z-dn.net/?f=%5Cvec%20F%28x%2Cy%2Cz%29%3Dx%5C%2C%5Cvec%5Cimath%2By%5C%2C%5Cvec%5Cjmath%2B9%5C%2C%5Cvec%20k%5Cimplies%5Cnabla%5Ccdot%5Cvec%20F%28x%2Cy%2Cz%29%3D%5Cdfrac%7B%5Cpartial%28x%29%7D%7B%5Cpartial%20x%7D%2B%5Cdfrac%7B%5Cpartial%28y%29%7D%7B%5Cpartial%20y%7D%2B%5Cdfrac%7B%5Cpartial%289%29%7D%7B%5Cpartial%20z%7D%3D2)
By the divergence theorem, the flux of
across
is given by the integral of
over
:
![\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D%5Ciiint_R%28%5Cnabla%5Ccdot%5Cvec%20F%29%5C%2C%5Cmathrm%20dV)
Convert to cylindrical coordinates, setting
![x=u\cos v](https://tex.z-dn.net/?f=x%3Du%5Ccos%20v)
![y=y](https://tex.z-dn.net/?f=y%3Dy)
![z=u\sin v](https://tex.z-dn.net/?f=z%3Du%5Csin%20v)
The integral is then
![\displaystyle2\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}\int_{y=0}^{y=8-u\cos v}u\,\mathrm dy\,\mathrm du\,\mathrm dv=\boxed{16\pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Cint_%7Bv%3D0%7D%5E%7Bv%3D2%5Cpi%7D%5Cint_%7Bu%3D0%7D%5E%7Bu%3D1%7D%5Cint_%7By%3D0%7D%5E%7By%3D8-u%5Ccos%20v%7Du%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B16%5Cpi%7D)