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Tasya [4]
2 years ago
12

Which equation is an equation of a circle with a radius of 2 and its center is at (4, -2)? (x - 4)² + (y + 2)² = 4 (x - 4)² + (y

+ 2)² = 2 (x + 4)² + (y - 2)² = 4

Mathematics
1 answer:
ozzi2 years ago
3 0

Answer:

(x-4)^2+(y+2)^2=4

Step-by-step explanation:

A circle is the set of all points on a plane at a given distance (radius) from a given point (center). The general equation of a circle with center (h,k) and radius r is given by:

(x-h)^2+(y-k)^2=r^2

Therefore, replacing the data provided into the previous equation:

Center=(4,-2)=(h,k)\\\\h=4\\\\k=-2\\\\Radius=2\\\\r=2

(x-(4))^2+(y-(-2))^2=(2)^2\\\\(x-4)^2+(y+2)^2=4

So, the equation of a circle with a radius of 2 and its center at (4,-2) is:

(x-4)^2+(y+2)^2=4

I attach you the graph.

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A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate
Lyrx [107]

Answer:

Therefore the rate change of height is  \frac{1}{\pi} m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

\frac{dV}{dt}=4 \ m^3/s

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is V=\pi r^2 h

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\therefore V= 4\pi h

Differentiating with respect to t.

\frac{dV}{dt}=4\pi \frac{dh}{dt}

Putting the value \frac{dV}{dt}

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\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}

\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}

The rate change of height does not depend on the height.

Therefore the rate change of height is  \frac{1}{\pi} m/s.

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3 years ago
A rectangular prism has a length of 1/1/4 yards, a width of 2/1/2 yards, and a height of 4 yards. Enter the volume of the prism
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B

Step-by-step explanation:

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