Answer:
a
Step-by-step explanation:
Answer:
![y=3x+4](https://tex.z-dn.net/?f=y%3D3x%2B4)
Step-by-step explanation:
Let y be the amount of water in the bottle x seconds after the faucet was turned on.
We have been given that the gauge on the bottle indicates that it contains 19 ounces of water after it has been filling for 5 seconds. After it has been filling for 11 seconds, the gauge indicates the bottle contains 37 ounces of water.
We have two points on the line
and
.
Let us find slope of the line passing through these points.
![m=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
![m=\frac{37-19}{11-5}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B37-19%7D%7B11-5%7D)
![m=\frac{18}{6}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B18%7D%7B6%7D)
![m=3](https://tex.z-dn.net/?f=m%3D3)
We will write our required equation in slope-intercept form
, where, m represents slope and b represents the y-intercept.
Let us find y-intercept by substituting
and coordinates of point
in slope intercept form.
![19=3\cdot 5+b](https://tex.z-dn.net/?f=19%3D3%5Ccdot%205%2Bb)
![19=15+b](https://tex.z-dn.net/?f=19%3D15%2Bb)
![19-15=15-15+b](https://tex.z-dn.net/?f=19-15%3D15-15%2Bb)
![4=b](https://tex.z-dn.net/?f=4%3Db)
Therefore, our required equation would be
.
By Stokes' theorem,
![\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Cpartial%5Cmathcal%20M%7D%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20r%3D%5Ciint_%7B%5Cmathcal%20M%7D%5Cnabla%5Ctimes%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20S)
where
![\mathcal C](https://tex.z-dn.net/?f=%5Cmathcal%20C)
is the circular boundary of the hemisphere
![\mathcal M](https://tex.z-dn.net/?f=%5Cmathcal%20M)
in the
![y](https://tex.z-dn.net/?f=y)
-
![z](https://tex.z-dn.net/?f=z)
plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting
![\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20r%28t%29%3D%5Clangle%200%2C3%5Ccos%20t%2C3%5Csin%20t%5Crangle)
where
![0\le t\le2\pi](https://tex.z-dn.net/?f=0%5Cle%20t%5Cle2%5Cpi)
. Then the line integral is
![\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Cmathcal%20C%7D%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20r%3D%5Cint_%7Bt%3D0%7D%5E%7Bt%3D2%5Cpi%7D%5Cmathbf%20f%28x%28t%29%2Cy%28t%29%2Cz%28t%29%29%5Ccdot%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Clangle%20x%28t%29%2Cy%28t%29%2Cz%28t%29%5Crangle%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Clangle0%2C0%2C3%5Ccos%20t%5Crangle%5Ccdot%5Clangle0%2C-3%5Csin%20t%2C3%5Ccos%20t%5Crangle%5C%2C%5Cmathrm%20dt%3D9%5Cint_0%5E%7B2%5Cpi%7D%5Ccos%5E2t%5C%2C%5Cmathrm%20dt%3D9%5Cpi)
We can check this result by evaluating the equivalent surface integral. We have
![\nabla\times\mathbf f=\langle1,0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%5Ctimes%5Cmathbf%20f%3D%5Clangle1%2C0%2C0%5Crangle)
and we can parameterize
![\mathcal M](https://tex.z-dn.net/?f=%5Cmathcal%20M)
by
![\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20s%28u%2Cv%29%3D%5Clangle3%5Ccos%20v%2C3%5Ccos%20u%5Csin%20v%2C3%5Csin%20u%5Csin%20v%5Crangle)
so that
![\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20S%3D%28%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Clangle9%5Ccos%20v%5Csin%20v%2C9%5Ccos%20u%5Csin%5E2v%2C9%5Csin%20u%5Csin%5E2v%5Crangle%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
where
![0\le v\le\dfrac\pi2](https://tex.z-dn.net/?f=0%5Cle%20v%5Cle%5Cdfrac%5Cpi2)
and
![0\le u\le2\pi](https://tex.z-dn.net/?f=0%5Cle%20u%5Cle2%5Cpi)
. Then,
![\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7B%5Cmathcal%20M%7D%5Cnabla%5Ctimes%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_%7Bv%3D0%7D%5E%7Bv%3D%5Cpi%2F2%7D%5Cint_%7Bu%3D0%7D%5E%7Bu%3D2%5Cpi%7D9%5Ccos%20v%5Csin%20v%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D9%5Cpi)
as expected.
Answer:
![x=12](https://tex.z-dn.net/?f=x%3D12)
Step-by-step explanation:
First, note that the angles shown are vertical angles. In other words, the two equations are equivalent to each other.
Therefore, set the equations equal to each other and solve for x;
![6x+7=8x-17](https://tex.z-dn.net/?f=6x%2B7%3D8x-17)
Add 17 to both sides:
![6x+24=8x](https://tex.z-dn.net/?f=6x%2B24%3D8x)
Subtract 6x from both sides:
![24=2x](https://tex.z-dn.net/?f=24%3D2x)
Divide both sides by 2:
![x=12](https://tex.z-dn.net/?f=x%3D12)
Therefore, the value of x is 12.
And we're done!
Answer:
2.b
3.a
4.c
Step-by-step explanation:
hope it helps you