The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
<h3>Equation of a circle</h3>
The standard equation of a circle is expressed as:
x^2 + y^2 + 2gx + 2fy + c = 0
where:
(-g, -f) is the centre of the circle
Given the equations
x^2 +y^2 – 12x – 2y +12 = 0
Compare
2gx = -12x
g = -6
Simiarly
-2y = 2fy
f = -1
Centre = (6, 1)
Hence the center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
Learn more on equation of a circle here: brainly.com/question/1506955
Answer:
The solution is (2,6)
Step-by-step explanation:
Let
x-----> the number of hours to change a fuel injection unit
y-----> the number of hours to change a transmission
we know that
5x+10y=70 -----> equation A
8x+8y=64 -----> equation B
Solve the system of equations by graphing
Remember that the solution of the system of equations is the intersection point both graphs
The solution is the point (2,6)
see the attached figure
Step-by-step explanation:
the first one is 2 that appears most while the second is 20 and 3
Answer:
hi I was wondering if I could come in
Answer:
y = (x+3)^2 -1
Step-by-step explanation:
The vertex form of the equation is
y = a(x-h)^2 +k where (h,k) is the vertex
The vertex is (-3,-1)
y = a(x- -3)^2 -1
y = a(x+3)^2 -1
Pick another point (-2,0) and substitute it into the equation
0 = a(-2+3)^2 -1 to find a
0 =a(1)^2 -1
0 = a-1
1 = a
y = (x+3)^2 -1
