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2 years ago

Please help

Mathematics

2 answers:

MissTica2 years ago

6 0

Answer:

Step-by-step explanation:

11 square root of 3 end root minus 7 square root of 6

Got it right on my test

oksian1 [2.3K]2 years ago

5 0

Answer:

1: 11√3 - 7√6

2: 11√3 - 7√6

3: -9

4:12

Step-by-step explanation:

To add radicals they need to have the same radical part for the first one we have

7√3- 4√6 + √48 - √54

We can simplify the last two into 4√3 and 3√6

So we have 7√3 - 4√6 + 4√3 - 3√6

adding similar radicals we get

11√3 - 7√6

For the second one we have 11√3 - 7√6

There's nothing we can do from here so keep that as your answer

This one is quite easy -3√9

square root of 9 is 3

so we have -3*3 which is -9

next is

4√9

same deal as the one before

3*4=12

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Mrs Smith has 12 times as many markers as color pencils the total number of markers and colored pencils is 78 how many markers d

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x = markers
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12x = y

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Okay, so what I did was divide 78 by 12 (to find the number of colored pencils) which gave me 6.5. You can't have half a colored pencil, so I rounded the number down to get 6. To find the number of markers, I multiplied 12 and 6 (12x = 6) which gave me 72. To check my answer I added the two amounts, 72 + 6, which gave me 78.

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$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

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Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

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$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

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$\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S$

that is, the integral of $\vec F$ over the disk, oriented downward. Since $z=0$ in $D$ , we have

$\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k$

Parameterize $D$ by

$\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

where $0\le u\le 3$ and $0\le v\le2\pi$ . Take the normal vector to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k$

Then taking the dot product of $\vec F$ with the normal vector gives

$\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v$

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