Answer:
Option A
Step-by-step explanation:
We need to find two expressions that, when simplified, give the same results.
1) First, simplify the expression stated in the question. Multiply each of the terms in the parentheses by the number that is next to them. This would mean you have to multiply both 9x and -6 by
. You also have to multiply
and
by 4. Then, simplify.
![\frac{2}{3}(9x-6) + 4 (\frac{1}{2} x - \frac{1}{2})\\\frac{18}{3}x- \frac{12}{3} + \frac{4}{2}x -\frac{4}{2} \\6x - 4 + 2x - 2](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%289x-6%29%20%2B%204%20%28%5Cfrac%7B1%7D%7B2%7D%20x%20-%20%5Cfrac%7B1%7D%7B2%7D%29%5C%5C%5Cfrac%7B18%7D%7B3%7Dx-%20%5Cfrac%7B12%7D%7B3%7D%20%2B%20%5Cfrac%7B4%7D%7B2%7Dx%20-%5Cfrac%7B4%7D%7B2%7D%20%5C%5C6x%20-%204%20%2B%202x%20-%202)
2) Now, combine the like terms.
![6x - 4 + 2x - 2](https://tex.z-dn.net/?f=6x%20-%204%20%2B%202x%20-%202)
![8x - 6](https://tex.z-dn.net/?f=8x%20-%206)
So, we need to find which of the expressions listed equal 8x - 6.
3) Let's try option A. Do the same as before. Multiply each of the terms in the parentheses by the number that is next to them. So, multiply 4x and -12 by
. Also, multiply 30x and 18 by
. Then, combine like terms and simplify.
![\frac{3}{4} (4x-12) + \frac{1}{6}(30x + 18)\\\frac{12}{4}x-\frac{36}{4} + \frac{30}{6} x + \frac{18}{6} \\3x - 9 + 5x + 3 \\8x - 6](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20%284x-12%29%20%2B%20%5Cfrac%7B1%7D%7B6%7D%2830x%20%2B%2018%29%5C%5C%5Cfrac%7B12%7D%7B4%7Dx-%5Cfrac%7B36%7D%7B4%7D%20%2B%20%5Cfrac%7B30%7D%7B6%7D%20x%20%2B%20%5Cfrac%7B18%7D%7B6%7D%20%20%20%5C%5C3x%20-%209%20%2B%205x%20%2B%203%20%5C%5C8x%20-%206)
This also equals 8x - 6. Therefore, option A is the answer.
Answer:
20
Step-by-step explanation:
5(-2)+30 = (-10)+30 = 20
Answer:
let W and R represent white toothpaste and red toothpaste represently
n(W)=143
n(R)=135
n(WnR)=70
n( WuR)=?
Now,
n(WuR)=n(W)+n(R)_n(WnR)
=143+135_70
=208
that's all don't forget to write therefore.
Out of the six, only “u” wouldn’t work so the answer is d. 5
Hope this helps and hope you have a great day! Please make brainiest
Apparently my answer was unclear the first time?
The flux of <em>F</em> across <em>S</em> is given by the surface integral,
![\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cmathbf%20F%5Ccdot%5Cmathrm%20d%5Cmathbf%20S)
Parameterize <em>S</em> by the vector-valued function <em>r</em>(<em>u</em>, <em>v</em>) defined by
![\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20r%28u%2Cv%29%3D7%5Ccos%20u%5Csin%20v%5C%2C%5Cmathbf%20i%2B7%5Csin%20u%5Csin%20v%5C%2C%5Cmathbf%20j%2B7%5Ccos%20v%5C%2C%5Cmathbf%20k)
with 0 ≤ <em>u</em> ≤ π/2 and 0 ≤ <em>v</em> ≤ π/2. Then the surface element is
d<em>S</em> = <em>n</em> • d<em>S</em>
where <em>n</em> is the normal vector to the surface. Take it to be
![\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}](https://tex.z-dn.net/?f=%5Cmathbf%20n%3D%5Cdfrac%7B%5Cfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20u%7D%7D%7B%5Cleft%5C%7C%5Cfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20u%7D%5Cright%5C%7C%7D)
The surface element reduces to
![\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20S%3D%5Cmathbf%20n%5C%2C%5Cmathrm%20dS%3D%5Cmathbf%20n%5Cleft%5C%7C%5Cdfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20u%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20r%7D%7B%5Cpartial%20v%7D%5Cright%5C%7C%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbf%20n%5C%2C%5Cmathrm%20dS%3D-49%28%5Ccos%20u%5Csin%5E2v%5C%2C%5Cmathbf%20i%2B%5Csin%20u%5Csin%5E2v%5C%2C%5Cmathbf%20j%2B%5Ccos%20v%5Csin%20v%5C%2C%5Cmathbf%20k%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
so that it points toward the origin at any point on <em>S</em>.
Then the integral with respect to <em>u</em> and <em>v</em> is
![\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cmathbf%20F%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cmathbf%20F%28x%28u%2Cv%29%2Cy%28u%2Cv%29%2Cz%28u%2Cv%29%29%5Ccdot%5Cmathbf%20n%5C%2C%5Cmathrm%20dS)
![=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle-49%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B%5Cpi%2F2%7D%287%5Ccos%20u%5Csin%20v%5C%2C%5Cmathbf%20i-7%5Ccos%20v%5C%2C%5Cmathbf%20j%2B7%5Csin%20u%5Csin%20v%5C%2C%5Cmathbf%20%29%5Ccdot%5Cmathbf%20n%5C%2C%5Cmathrm%20dS)
![=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}](https://tex.z-dn.net/?f=%3D-343%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Ccos%5E2u%5Csin%5E3v%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B-%5Cfrac%7B343%5Cpi%7D6%7D)