The Scale Factor of 1 2/3?

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B4%20%7D%20%20%2B%205%20%3D%207" id="TexFormula1" title=" \frac{x}{4 } + 5

harina [27]

x=8 is the correct answer. First you had to subtract by 5 from both sides of equation, and it gave us, $\frac{x}{4}+5-5=7-5$ . And then simplify, and it gave us, $\frac{x}{4}=2$ . Then you can multiply by 4 from both sides of equation, and it gave us, $\frac{4x}{4}=2*4$ . And finally simplify, and it gave us the answer is 2*4=8, or x=8 is the correct answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

8 0

2 years ago

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CD is a line parallel to AB and passes through the point (3, 5).

Mrac [35]

Answer:

y=5/3x+b

Step-by-step explanation:

Supposing CD uses the equation y=mb (meaning CD passes through the origin) and passes through (3,5), we then know the slope if 5/3 and the y intercept is 0. We can then write CD's equation as y=5/3x+0, or just y=5/3x.

So if AB is parallel to CB, then their equation is y=5/3x+b, as we don't the y-intercept yet.

Thank you!

4 0

2 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

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