Which equation represents a circle with a center at (-3,-5) and a radius of 6 units?

Mathematics

2 answers:

brilliants [131]3 years ago

7 0

Answer:

Step-by-step explanation:

The equation of a circle is:

(x - h)² + (y - k)² = r²

where (h, k) is the center and r is the radius.

Given that (h, k) is (-3, -5) and r = 6:

(x + 3)² + (y + 5)² = 6²

(x + 3)² + (y + 5)² = 36

Tema [17]3 years ago

3 0

Answer:

$(x+3)^2+(y+5)^2=36$

Step-by-step explanation:

edgenuity 2020

hope this helps!

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Use cylindrical coordinates. find the volume of the solid that lies within both the cylinder x2 y2 = 9 and the sphere x2 y2 z2 =

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In Cartesian coordinates, the region is given by $-3\le x\le3$ , $-\sqrt{9-x^2}\le y\le\sqrt{9-x^2}$ , and $-\sqrt{16-x^2-y^2}\le z\le\sqrt{16-x^2-y^2}$ . Converting to cylindrical coordinates, using

$\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}$

we get a Jacobian determinant of $r$ , and the region is given in cylindrical coordinates by $0\le\theta\le2\pi$ , $0\le r\le3$ , and $-\sqrt{16-r^2}\le z\le\sqrt{16-r^2}$ .

The volume is then

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=-\sqrt{16-r^2}}^{z=\sqrt{16-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\dfrac{4(64-7\sqrt7)\pi}3$