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4 years ago

Need help to solve problem‼️

Mathematics

2 answers:

saveliy_v [14]4 years ago

7 0

Answer:

1. B

2. F

3. D

4. E

5. A

6. C

I hope that this helped

kolbaska11 [484]4 years ago

4 0

Answer:

1. B

2.F

3.D

4.E

5.C

6.A

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The perimeter of a rectangle is 52cm. The length is 12cm longer than the width. Find the dimensions.

alisha [4.7K]

W=7
L=19
Don’t mind the problem above. Hope this helps

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3 years ago

Given the following information, find the length of the missing side. Leave your answer as a simplified radical.

gregori [183]

Answer:

Long = 5√3

Step-by-step explanation:

Hypotenuse = 10

Short = 5

Long = ?

Apply Pythagorean theorem to find Long:

Long² = hyp² - short²

Long² = 10² - 5²

Long² = 100 - 25

Long² = 75

Long = √75

Long = √(25*3)

Long = 5√3

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3 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 $(128-32)\pi=96\pi$ .

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3 years ago

7. Where are temperate grasslands found?

ivolga24 [154]

Answer:

Step-by-step explanation:

We know that grasslands don't thrive in the cold, so we'll cross out the Arctic Circle.

Mid latitudes are 60 degrees, so it would be cold there too.

Tropical latitudes would be pretty hot.

So my guess is near the equator.

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2 years ago

HELP PLSSSSSSSSSSSSSSSSSSSSSSSS

velikii [3]

I would go withhh c but I am not 100% sure

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3 years ago