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The equation of the circle whose center is at (4,4) and whose radius is 5 is
Question
Asked 4/14/2014 3:29:14 PM
Updated 4/14/2014 8:49:35 PM
2 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by yeswey [4/14/2014 8:43:30 PM]
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Original conversation
User: The equation of the circle whose center is at (4,4) and whose radius is 5 is

Weegy: (x+4)^2+(y+6)^2=49 or x^2+y^2+8x+12y=3
sterces011|Points 166|

User: Solve the following system. x 2 - y = 3 x - y = -3

Weegy: -(1/3)(-6y)(3)(-3)(1) = -18y




Question
Asked 4/14/2014 3:29:14 PM
Updated 4/14/2014 8:49:35 PM
2 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by yeswey [4/14/2014 8:43:30 PM]
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The equation of the circle whose center is at (4, 4) and whose radius is 5 is (x - 4)^2 + (y - 4)^2 = 25
Added 4/14/2014 8:44:37 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [4/14/2014 8:47:17 PM]
3
x^2 - y = 3 equation 1;
x - y = -3 equation 2
Equation 1 - 2 we get:
x^2 - x = 6
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 or x = -2
y = x^2 - 3 = 3^2 - 3 = 6 or y = (-2)^2 - 3 = 1
The solution for the system of equations x^2 - y = 3, x - y = -3 is (3, 6), (-2, 1)
Added 4/14/2014 8:49:35 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [4/14/2014 8:58:03 PM]
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