Answer:
![y = \frac{ - 4}{3} x + 3](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%20-%204%7D%7B3%7D%20x%20%2B%203)
Step-by-step explanation:
Answer:
A. Measure of angle A + measure of angle B
Step-by-step explanation:
Because angle 4 is an exterior angle of given triangle which will be equal to the Measure of angle 2 + measure of angle 3
Answer:
Non-repeating decimal
Step-by-step explanation:
As a example, 32.93729, would be a never ending number. Which means it's not a rational number, it is a irrational number.
Consider the closed region
![V](https://tex.z-dn.net/?f=V)
bounded simultaneously by the paraboloid and plane, jointly denoted
![S](https://tex.z-dn.net/?f=S)
. By the divergence theorem,
![\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20dS%3D%5Ciiint_V%5Cnabla%5Ccdot%5Cmathbf%20f%28x%2Cy%2Cz%29%5C%2C%5Cmathrm%20dV)
And since we have
![\nabla\cdot\mathbf f(x,y,z)=1](https://tex.z-dn.net/?f=%5Cnabla%5Ccdot%5Cmathbf%20f%28x%2Cy%2Cz%29%3D1)
the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have
![\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_V%5Cnabla%5Ccdot%5Cmathbf%20f%28x%2Cy%2Cz%29%5C%2C%5Cmathrm%20dV%3D%5Ciiint_V%5Cmathrm%20dV)
![=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Cint_%7Bz%3D2%7D%5E%7Bz%3D3-r%5E2%7Dr%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle2%5Cpi%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%283-r%5E2-2%29%5C%2C%5Cmathrm%20dr)
![=\dfrac\pi2](https://tex.z-dn.net/?f=%3D%5Cdfrac%5Cpi2)
Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by
![D](https://tex.z-dn.net/?f=D)
, we have
![\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS-D%7D%5Cmathbf%20f%5Ccdot%5Cmathrm%20dS%3D%5Cfrac%5Cpi2-%5Ciint_D%5Cmathbf%20f%5Ccdot%5Cmathrm%20dS)
Parameterize
![D](https://tex.z-dn.net/?f=D)
by
![\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20s%28u%2Cv%29%3Du%5Ccos%20v%5C%2C%5Cmathbf%20i%2Bu%5Csin%20v%5C%2C%5Cmathbf%20j%2B2%5C%2C%5Cmathbf%20k)
![\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbf%20s_u%5Ctimes%5Cmathbf%20s_v%3Du%5C%2C%5Cmathbf%20k)
which would give a unit normal vector of
![\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20k)
. However, the divergence theorem requires that the closed surface
![S](https://tex.z-dn.net/?f=S)
be oriented with outward-pointing normal vectors, which means we should instead use
![\mathbf s_v\times\mathbf s_u=-u\,\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%3D-u%5C%2C%5Cmathbf%20k)
.
Now,
![\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_D%5Cmathbf%20f%5Ccdot%5Cmathrm%20dS%3D%5Cint_%7Bu%3D0%7D%5E%7Bu%3D1%7D%5Cint_%7Bv%3D0%7D%5E%7Bv%3D2%5Cpi%7D%5Cmathbf%20f%28x%28u%2Cv%29%2Cy%28u%2Cv%29%2Cz%28u%2Cv%29%29%5Ccdot%28-u%5C%2C%5Cmathbf%20k%29%5C%2C%5Cmathrm%20dv%5C%2C%5Cmathrm%20du)
![=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle-4%5Cpi%5Cint_%7Bu%3D0%7D%5E%7Bu%3D1%7Du%5C%2C%5Cmathrm%20du)
![=-2\pi](https://tex.z-dn.net/?f=%3D-2%5Cpi)
So, the flux over the paraboloid alone is
A "rational expression" is a fraction in which the numerator and the denominator are both polynomials. The sum of two such expressions will always produce another such expression, viz.,
<span>p(x)/q(x) + r(x)/s(x) = [p(x)s(x) + q(x)r(x)] / [q(x)*s(x)]. </span>
<span>The numerator and denominator of the sum will both be polynomials when fully expanded</span>