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3 years ago

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Garret is in the middle of a 20 -minute bike ride. It takes 4 minutes for him to go a kilometer, so he'll end up going 5 kilomet

ers. Suppose k is the number of kilometers Garret has already biked.

1 answer:

sdas [7]3 years ago

5 0

A x 20 - mintues, 4, a km, 5km, km?

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Will mark brainliest and give 20 points!

Salsk061 [2.6K]

The answer is 3x+2 over (x+2) (x-2)

4 0

3 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

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

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

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

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

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

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$

From this we need to subtract the contribution of

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

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

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

So the contribution of integrating $\vec F$ over $D$ is

$\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4$

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==> 486π/5 - (-81π/4) = 2349π/20

5 0

3 years ago

What is the image point of (0,-9)after a translation right 5 units and down 1 unit?

nekit [7.7K]

Answer:

I think 0 -10 or mabe 0,-11

6 0

2 years ago

convert the linear equation 3x-2y=5 to point slope form. use the point with an x-coordinate equal to 1

Mademuasel [1]

When x = 1, y = (5 - 3) / -2 = -1 , let x1 = 1 and y1 = -1
2y = 3x - 5
so slope = 3/2

y - y1 = m(x - x1)

y -(-1) = (3/2)(x - 1)

y + 1 = 3/2 ( x - 1) <----- answer

6 0

3 years ago

Help me with question 2 plss

stepladder [879]

I cannot do this because it wont let me look at the attachment i am sorry

4 0

3 years ago

Read 2 more answers