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Otrada [13]
3 years ago
5

IM IN DESPERATE NEEED!!! What is the solution to this system of equations?

Mathematics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

x + y - x = 6 - y  - 4 \\ y = 2 - y \\ 2y = y \\ y = 1 \\ x + y = 6 \\ x + 1 = 6 \\ x = 5

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Answer:

center = (-2, 6)

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Step-by-step explanation:

<u>Equation of a circle</u>

(x-a)^2+(y-b)^2=r^2

(where (a, b) is the center and r is the radius)

Therefore, we need to rewrite the given equation into the standard form of an equation of a circle.

Given equation:

x^2+y^2+4x-12y+4=0

Collect like terms and subtract 4 from both sides:

\implies x^2+4x+y^2-12y=-4

Complete the square for both variables by adding the square of half of the coefficient of x and y to both sides:

\implies x^2+4x+\left(\dfrac{4}{2}\right)^2+y^2-12y+\left(\dfrac{-12}{2}\right)^2=-4+\left(\dfrac{4}{2}\right)^2+\left(\dfrac{-12}{2}\right)^2

\implies x^2+4x+4+y^2-12y+36=-4+4+36

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\implies (x+2)^2+(y-6)^2=36

Therefore:

  • center = (-2, 6)
  • radius = √36 = 6
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2 years ago
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