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3 years ago

Please help me on this I need help and nobody is really helping

Mathematics

2 answers:

Nady [450]3 years ago

3 0

I believe the answer is (0,0)

Anna007 [38]3 years ago

3 0

Answer:

(0,0)

Step-by-step explanation:

That is the middle.

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Which table shows an independent quantity of 3 with a corresponding dependent quantity of 6

Ulleksa [173]

The answer is b because it has the 3 with 6 corresponding.

8 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 F) - (integral over D) = integral over S

==> 486π/5 - (-81π/4) = 2349π/20

5 0

3 years ago

1. A circle with center (-5, 2) is tangent to the y-axis. What is the radius of the circle? What is the equation of the circle?

Bess [88]

Answer:

Step-by-step explanation:

1) The center lies on the vertical line x = -5 and the the circle is tangent to (touches in one place only) the y-axis. Thus, the radius is 5.

2) Starting with (x - h)^2 + (y - k)^2 = r^2 and comparing this to the given

(x - 4)^2 + (y + 3)^2 = 6^2

we see that h = 4, k = -3 and r = 6. The center is at (4, -3) and the radius is 6.

3) Notice that A and B have the same x-coordinate, x = 15. The center of the circle is thus (15, -2), where that -2 is the halfway point between the two given points in the vertical direction. Arbitrarily choose A(15, 4) as one point on the circle. Then the equation of this circle is

(x - 4)^2 + (y + 3)^2 = r^2 = 6^2, where the 6 is one half of the vertical distance between A(15, 4) and B(15, -8) (which is 12).

4 0

2 years ago

HELP I NEED HELP ASAP

Nady [450]

Answer:

Step-by-step explanation:

because the -2 is where the point is and the slope will start there so when you draw a line the line will pass by (0,-2)

5 0

3 years ago

Picture attached, please simplify

Charra [1.4K]

The answer is the second option given

6 0

3 years ago