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Vinil7 [7]
2 years ago
8

What is the solution set of 4x=yand 2x - y=0?

Mathematics
1 answer:
lidiya [134]2 years ago
4 0

Answer:

the \: answer \: is \: b \\ 4x = y \\ 2 {x}^{2}  - y = 0 \\ 2 {x}^{2}  - 4x = 0 \\ 2x(x - 2)  = 0 \\ x = 0 \: and \: x = 2 \\ y = 0 \: and \: y = 8

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Find the equation of a circle with a center at (7,2) and a point on the circle at (2,5)?
monitta

Answer:

(x-7)^2+(y-2)^2=34

Step-by-step explanation:

We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).

First, recall that the equation of a circle is given by:

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

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:

(x-7)^2+(y-2)^2=r^2

Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:

(2-7)^2+(5-2)^2=r^2

Solve for <em>r: </em>

<em />(-5)^2+(3)^2=r^2<em />

Simplify. Thus:

25+9=r^2

Finally, add:

r^2=34

We don't need to take the square root of both sides, as we will have the square it again anyways.

Therefore, our equation is:

(x-7)^2+(y-2)^2=34

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