What is the center of the circle with equation x^2 + y^2 - 10x +6y-2= 0

C(5, -3)

Step-by-step explanation:

Given the equation of circle;

x² + y² - 10x + 6y - 2 = 0

The general equation for a circle is given by the formula;

x² + y² + 2gx + 2fy + c = 0

Where the center is C(-g, -f)

To find the center, we would compare the two equations;

For the value of g.

2gx = -10x

2g = -10

g = -10/2

g = -5

To find the value of f.

2fy = 6y

2f = 6

f = 6/2

f = 3

Therefore, the center C(-g, -f) = (-(-5), -(3))

C(-g, -f) = C(5, -3).

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

idk

step-by-step explanation: