Question

X^2+y^2-6x+4y≥12...​

Answer 1

Ответ:

Объяснение:

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Answer 2

The inequality you've provided is an inequality in two variables, x and y. It appears to be an inequality of a circle centered at (3, -2) with a radius of 2. To rewrite it in standard form:

x^2 + y^2 - 6x + 4y ≥ 12

First, let's complete the square for the x and y terms:

x^2 - 6x + y^2 + 4y ≥ 12

To complete the square for the x terms, we need to add and subtract the square of half the coefficient of x (which is -6/2 = -3)^2 = 9):

x^2 - 6x + 9 + y^2 + 4y ≥ 12 + 9

For the y terms, we add and subtract the square of half the coefficient of y (which is 4/2 = 2)^2 = 4):

x^2 - 6x + 9 + y^2 + 4y + 4 ≥ 12 + 9 + 4

Now we can rewrite the equation:

(x^2 - 6x + 9) + (y^2 + 4y + 4) ≥ 25

Notice that (x^2 - 6x + 9) can be factored as (x - 3)^2, and (y^2 + 4y + 4) can be factored as (y + 2)^2:

(x - 3)^2 + (y + 2)^2 ≥ 25

This is the equation of a circle with center (3, -2) and radius √25 = 5. The inequality represents the region inside or on the circle.