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3 years ago

HELP ASAP! Right/Best Answer gets Brainliest!

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

4 0

X= 14 + 9y over 2
y= -x - 7 over 6

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g Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 7yi + xzj + (

Sati [7]

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

where $S$ is any oriented surface with boundary $C$ . We have

$\vec F(x,y,z)=7y\,\vec\imath+xz\,\vec\jmath+(x+y)\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-7)\,\vec k$

Take $S$ to be the ellipse that lies in the plane $z=y+9$ with boundary on the cylinder $x^2+y^2=1$ . Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+9)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

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

Then we have

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\big((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+2)\,\vec k\big)\cdot\big(-u\,\vec\jmath+u\,\vec k\big)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(3u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{3\pi}$

5 0

3 years ago

Which of the following are reasons used in the proof that the angle-bisector construction can be used to bisect any angle? Check

Bad White [126]

The answer for apex users are:

All of the radii of a circle are congruent

CPCTC

SSS triangle congruence postulate

3 0

3 years ago

Read 2 more answers

Someone please help me

GalinKa [24]

Answer:

A. 9 sqrt 2

Explanation:

Plug in hypotenuse and given side into
a^2 + b^2 = c^2 and solve.

8 0

3 years ago

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in

Kryger [21]

Answer:

boat: 16 mph
current: 4 mph

Step-by-step explanation:

speed = distance / time

The rate downriver is (60 mi)/(3 h) = 20 mi/h.

The rate upriver is (60 mi)/(5 h) = 12 mi/h.

The rate of the boat in still water is the average of these:

(20 +12)/2 mi/h = 16 mi/h

The rate of the current is the difference between the boat speed and actual speed:

16 mi/h - 12 mi/h = 4 mi/h

3 0

3 years ago

I keep coming up with different answers please help me?

allsm [11]

The answer is 1,600.

3 0

2 years ago

Read 2 more answers