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3 years ago

PLEASE HELP!!! Name the postulate or theorem that you can use to prove.....

Mathematics

1 answer:

bearhunter [10]3 years ago

5 0

Answer:

HL theorem

Step-by-step explanation:

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

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

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

To prove that ΔAED ˜ ΔACB by SAS, Jose shows that StartFraction A E Over A C EndFraction = StartFraction A D Over A B EndFractio

Ira Lisetskai [31]

Answer:

A on Edg.

Step-by-step explanation:

Took the Test

5 0

3 years ago

Read 2 more answers

describe the transformations that produce the graph of g(x)=1/2(x-4)^3+5 from the graph of the parent function f(x)=x^3. Give th

SSSSS [86.1K]

We have been given a parent function $f(x)=x^{3}$ and we need to transform this function into $g(x)=\frac{1}{2}(x-4)^{3}+5$ .

We will be required to use three transformations to obtain the required function from $f(x)=x^{3}$ .

First transformation would be to shift the graph to the right by 4 units. Upon using this transformation, the function will change to $g(x)=(x-4)^{3}$ .

Second transformation would be to compress the graph vertically by half. Upon using the second transformation, the new function becomes $g(x)=\frac{1}{2}(x-4)^{3}$ .

Third transformation would be to shift the graph upwards by 5 units. Upon using this last transformation, we get the new function as $g(x)=\frac{1}{2}(x-4)^{3}+5$ .

6 0

3 years ago

Please help I’m really stuck

motikmotik

Answer:

A is the correct answer

have a great day

7 0

3 years ago

Joy is reading a 352 page novel for her summer reading project on Monday she reads 3/8 of the novel on Tuesday she reads 28 page

MAXImum [283]

Answer:

104 Pages

Step-by-step explanation:

on Monday she reads 3/8 of the novel which means

$\frac{3}{8}$ x 352 = 132 pages

on Tuesday she reads 28 pages, doesn't require any calculations.

on Wednesday she reads 1/4 of the novel,

$\frac{1}{4}$ x 352 = 88

Just add all of that,

132+28+88=248 Pages

Subtract the novel pages by the read pages value

352-248=104pages.

8 0

3 years ago