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3 years ago

Write the standard equation for the circle.

Mathematics

1 answer:

Sidana [21]3 years ago

5 0

Answer:

the first one

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (10, - 6) and r = 6, hence

(x - 10)² + (y - (- 6))² = 6², that is

(x - 10)² + (y + 6)² = 36

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Aleksandr-060686 [28]

Answer:

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Step-by-step explanation:

6 0

3 years ago

VERY IMPORTANT... write the equation of the line that has a y intercept of (0.4) and a slope of -3/2

shutvik [7]

Answer:

y = -3/2 x + 4

Step-by-step explanation:

y-4 = -3/2 ( x-0)

y = -3/2 x + 4

7 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

Distributed property (2×40) + (2 x3)

SpyIntel [72]

Answer:

Step-by-step explanation:

4 0

3 years ago

GEOMETRY: Find JL of the line.

Yuliya22 [10]

Answer:

JL = 31

Step-by-step explanation:

JL is the total length of the line segment. We know that JK is 15 units long and that KL is 16 units long. So, you would add these numbers (15 & 16) together to get the total length of the line segment. Hope this helps! :)

6 0

3 years ago