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2 years ago

I'll give brainliest to the correct answer! (Picture below)

Mathematics

1 answer:

Lina20 [59]2 years ago

6 0

9514 1404 393

Answer:

D. 7

Step-by-step explanation:

The equation is ...

x = a(y +1)² -3

For the point (x, y) = (4, 0), we have ...

4 = a(0 -1)² -3

7 = a . . . . . . . . . . . . add 3, simplify

The leading coefficient is 7.

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

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

so we first compute the curl:

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

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

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

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

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

2 years ago

Triangle KLM was dilated according to the rule

Anna35 [415]

Answer:

The statement that is true is;

The vertices of the image are closer to the origin than those of the pre-image

Step-by-step explanation:

The dilation rule is 0.75(x,y)= (0.75x, 0.75y)

K (-4,4)----------( 0.75*-4,0.75*4)-------K' (-3,3)

L (2,4)-----------(0.75*2,0.75*4)---------L' (1.5,3)

M (-2,2)----------(0.75*-2,0.75*2)--------M' (-1.5,1.5)

4 0

2 years ago

Read 2 more answers

Colton has a stone paperweight composed of a triangular pyramid on top of a triangular prism with the dimensions shown below. Us

Eva8 [605]

The area of the base would be found using the area of a triangle formula which is 1/2 x base x height.

The base and height are the two sides perpendicular to each other, which are both 5 inches.

The area of the base = 1/2 x 5 x 5 = 12.5 square inches.

The volume of the triangular prism is the area of the base times the height, which is 4 inches.

Volume of the triangular prism is 12.5 x 4 = 50 cubic inches.

Volume of the triangular prism is 1/3 x area of base x height, which is 7:

Volume of the triangular prism = 1/3 x 12.5 x 7 = 29.17 cubic inches.

Total volume = 29.17 + 50 = 79.17 cubic inches.

3 0

2 years ago

Write an expression that can be used to multiply 6x198 mentally

RoseWind [281]

Do a mental sum for 6 x 200 (1200) then subtract 6<span>x 2 (12) to give you 1188.</span>

8 0

2 years ago

Read 2 more answers

Given the formula K = LMN, what is the formula for M?

Andre45 [30]

To start off devide both sides by M i.e
K/M=LMN/M
-> K/M=LN
Then devide by K to remove it on the left , this will end you up with M=LN/K
So the answer is (c).
Hope this helps :).

5 0

3 years ago