Question

2) x2 + y2 - 6x - 4y > 0;​

Answer 1

Ответ:

2x+yx2x 6x-4y>0

2x+12yx-4y>0

2x+12yx-4y>0

2(x+6yx-2y)>0

x+6yx>2y

x(1+6y)>2y

x>2y/1+6y

Объяснение:

Answer 2

The expression $x^2 + y^2 - 6x - 4y > 0$ represents an inequality in two variables, $x$ and $y$. This inequality defines a region in the $xy$-plane where the sum of the squares of $x$ and $y$ minus the linear terms is greater than zero. Let's break down the steps to understand this inequality:

1. Start with the expression: $x^2 + y^2 - 6x - 4y > 0$.
2. You can rearrange the terms: $x^2 - 6x + y^2 - 4y > 0$.
3. Complete the square for both $x$ and $y$ terms. To complete the square for $x$, you need to add $(6/2)^2 = 9$ inside the square, and for $y$, add $(4/2)^2 = 4$ inside the square: $(x^2 - 6x + 9) + (y^2 - 4y + 4) > 9 + 4$.
4. Simplify: $(x - 3)^2 + (y - 2)^2 > 13$.

The inequality is now in the standard form of a circle equation, with the center at $(3, 2)$ and a radius of $\sqrt{13}$. However, since the inequality asks for the region where this expression is greater than 0, you are looking for the area outside the circle. This region includes all points that are farther away from the center than the radius of the circle.

In summary, the solution to the inequality $x^2 + y^2 - 6x - 4y > 0$ is the region outside the circle with center $(3, 2)$ and radius $\sqrt{13}$ on the $xy$-plane.