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Paul [167]
3 years ago
10

Someone help me I forgot about this I need the answer

Mathematics
1 answer:
bekas [8.4K]3 years ago
8 0
The answer is B because -3.62 is less than -3.5 and greater than -3.8.
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Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straig
Brilliant_brown [7]

Stokes' theorem equates the line integral of \vec F along the curve to the surface integral of the curl of \vec F over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it T) by

\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)

\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to T to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of T is traversed in the counterclockwise direction when viewed from above.

Compute the curl of \vec F:

\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)

Then by Stokes' theorem,

\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S

where

\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS

\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

The integral thus reduces to

\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}

3 0
3 years ago
Please answer i’ll give brainliest
dangina [55]

Answer:

B

Step-by-step explanation:

The domain is the x axis and if you look at the x axis it shows up to 11 months. So that means that the domain shows flower production up to 11 months

8 0
4 years ago
Read 2 more answers
ANSWER ALL PLEASE!!!! Thanks in advanced
Evgen [1.6K]

Answer:

you add 5 for the compound investments each time

you add 6 for the simple investments each time

7 0
3 years ago
Read 2 more answers
I will AWARD BRANLIST TO YOU! Its A Math Question Help! <br><br> Thank you alot.
Molodets [167]

The best description for point F is: b. Circumcenter of triangle ABC.

<h3>What is the Circumcenter of a Triangle?</h3>

The perpendicular bisector is the line is drawn from the vertex of a triangle and bisects the opposite side at right angle. A triangle has three perpendicular bisectors.

The point where the three perpendicular bisectors of a triangle intersects is called the circumcenter of a triangle.

The triangle in the image above has three perpendicular bisector that meets at point F.

Therefore, the best description for point F is: b. Circumcenter of triangle ABC.

Learn more about the circumcenter on:

brainly.com/question/21299234

#SPJ1

6 0
2 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
dem82 [27]

Answer:

Converges at -2

Step-by-step explanation:

We solve this by using integral by parts.

\int\limits^a_b {(2x-1)/x^2 - x} \, dx

let:

u=2x-1             dv=(1/x^2)dx

du=2xdx         v=-(1/x)

We use the following expression for integral by parts:

\int\limits {u} \, dv=uv-\int\limits{v} \, du

\int\limits^a_b {(2x-1)/x^2} \, dx = -2-1/x+\int\limits {2} \, dx

We can take the limit with x tending to -1

\lim_{x \to \(-1} -2-1/x+x = -2+1-1=-2

The integral converges at -2

3 0
4 years ago
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