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3 years ago

HELP ASAP PLEASE! 100 POINTS!

Mathematics

2 answers:

Feliz [49]3 years ago

8 0

Answer:

The third one is correct

Step-by-step explanation:

I took the test and got it correct.

irina1246 [14]3 years ago

4 0

It’s the first one.

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Use the Law of Cosines to determine the indicated side x. (Assume a = b = 20. Round your answer to one decimal place.)

zhuklara [117]

Answer:

37.6

Step-by-step explanation:

x² = a² + b² - 2ab cos 140°

x² = 20² + 20² - 2(20)(20)(-0.766)

x² = 400 + 400 - 800(-0,766)

x² = 800 - (-612.835)

x² = 800 + 612.835

x² = 1412.835

x =√1412.835 = 37.6

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3 years ago

Complete for y=x-4. How do I do this pls help

Hitman42 [59]

Answer:

-5 = -9

-2 = -6

0 = -4

4 = 0

Step-by-step explanation:

You substitute each value for x and then solve the equation. (Please give me brainliest)

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3 years ago

ratio of the number of boy campers to the number of girl campers was 8:7. if there were a total of 195 campers how many boy camp

elixir [45]

About 90 I think maybe

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3 years ago

The following lock needs a correct 4 digit passcode to open.

vova2212 [387]

The answer for this question is 12 56

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2 years ago

F(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 3)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 18 that l

Cerrena [4.2K]

$\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k$
$\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1$

By the divergence theorem, the flux of $\mathbf F$ across the *closed* surface $\mathcal S$ combined with the plane $z=2$ is given by a volume integral over the closed region:

$\displaystyle\iint_{\mathcal S}\mathbf F\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}\nabla\cdot\mathbf F\,\mathrm dV$

So in fact, to find the flux over $\mathcal S$ alone, we'll need to subtract the flux of $\mathbf F$ over the planar portion, oriented outward. First, compute the volume integral by converting to cylindrical coordinates:

$x^2+y^2+z=18$
$z=2\implies x^2+y^2=16\implies r^2=16\implies r=4$

$\displaystyle\iiint_{\mathcal R}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=2}^{z=18-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=128\pi$

If the surface does actually contain $z=2$ , then you can stop here; otherwise, continue.

Now, parameterize the part of the *closed* surface in $z=2$ by

$\mathbf s(r,\theta)=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j+2\,\mathbf k$

where $0\le r\le4$ and $0\le\theta\le2\pi$ . We get a surface element

$\mathrm d\mathbf S=(\mathbf s_r\times\mathbf s_\theta)\,\mathrm dr\,\mathrm d\theta=(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$

We don't need to worry about the first two components of

and so the surface integral over this region is

$\displaystyle\iint_{z=2\,\land\,x^2+y^2\le16}\mathbf F\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}2r\,\mathrm dr\,\mathrm d\theta=32\pi$

Then the total flux over $\mathcal S$ alone is $(128-32)\pi=96\pi$ .

4 0

3 years ago