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Law Incorporation [45]
3 years ago
13

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
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