Ask question

Ask question

All categories

3 years ago

6

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = xyi + yzj + zxk

, C is the boundary of the part of the paraboloid z = 1 − x2 − y2 in the first octant.

1 answer:

garri49 [273]3 years ago

6 0

Answer:

hi I don't know the answer but really sorry I did because I needed some points so that I can ask my questions from another people and thank you for free points

You might be interested in

A, B and C are partners sharing profits in the ratio 3:2:1. During the year A withdraws Rs. 3000 at the start of each month, B w

andre [41]

Answer:

what is the question

Step-by-step explanation:

i cant see any question marks

5 0

3 years ago

the ratio of black bags to blue bags is 5 to 3. If there are a total of 185 black bags, then how many blue bags are there?

nlexa [21]

Answer:

There are 111 blue bags

Step-by-step explanation:

Let

x ----> the number of black bags

y ---> the number of blue bags

we know that

$\frac{x}{y}=\frac{5}{3}$

isolate the variable y

$y=\frac{3}{5}x$ -----> equation A

$x=185$ ----> equation B

substitute equation B in equation A

$y=\frac{3}{5}(185)$

$y=111$

therefore

There are 111 blue bags

8 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Geometry help please how do I do these?

trapecia [35]

Answer:

Question 7:

∠L = 124°

∠M = 124°

∠J = 118°

Question 8:

∠Q = 98°

∠T = 98°

∠R = 82°

Question 15:

m∠G = 110°

Question 16:

∠G = 60°

Question 17:

∠G = 80°

Question 18:

∠G = 70°

Step-by-step explanation:

The angles can be solving using Symmetry.

Question 7.

The sum of interior angles in an isosceles trapezoid is 360°, and because it is an isosceles trapezoid

∠K = ∠J = 118°

∠L = ∠M

∠K+∠J+∠L +∠M = 360°

236° + 2 ∠L = 360°

Therefore,

∠L = 124°

∠M = 124°

∠J = 118°

Question 8.

In a similar fashion,

∠Q+∠T+∠S +∠R = 360°

and

∠R = ∠S = 82°

∠Q = ∠T

∠Q+∠T + 164° = 360°

2∠Q + 164° = 360°

2∠Q = 196°

∠Q = ∠T =98°.

Therefore,

∠Q = 98°

∠T = 98°

∠R = 82°

Question 15.

The sum of interior angles of a kite is 360°.

∠E + ∠G + ∠H + ∠F = 360°

Because the kite is symmetrical

∠E = ∠G.

And since all the angles sum to 360°, we have

∠E +∠G + 100° +40° = 360°

2∠E = 140° = 360°

∠E = 110° = ∠G.

Therefore,

m∠G = 110°

Question 16.

The angles

∠E = ∠G,

and since all the interior angles sum to 360°,

∠E + ∠G + ∠F +∠H = 360°

∠E + ∠G + 150 + 90 = 360°

∠E + ∠G = 120 °

∠E = 60° = ∠G

therefore,

∠G = 60°

Question 17.

The shape is a kite; therefore,

∠H = ∠F = 110°

and

∠H + ∠F + ∠E +∠G = 360°

220° + 60° + ∠G = 360°,

therefore,

∠G = 80°

Question 18.

The shape is a kite; therefore,

∠F = ∠H = 90°

and

∠F +∠H + ∠E + ∠G = 360°

180° + 110° + ∠G = 360°

therefore,

∠G = 70°.

3 0

3 years ago

3y-3.5/3y+6=17/25.5. find the value of y. give the scale factor of the polygon

Mekhanik [1.2K]

Hi the answer to your question is

((3y - 3.5) / (3y + 6)) * 25.5 = (17 / 25.5) * 25.5

((3y - 3.5) / (3y + 6)) * 25.5 = 17 (76.5y - 89.25) / (3y + 6) = 17

((76.5y - 89.25) / (3y + 6)) * (3y + 6) = 17 * (3y + 6) 76.5y - 89.25 = 51y + 102 76.5y - 51y = 102 + 89.25

25.5y = 191.25

y = 191.25 / 25.5

<span>y = 7.5</span>

3 0

3 years ago