Question

(x-2)^2+(y+1)^2>1(x-2)^2+(y+1)^2<9​

Answer 1

Объяснение:

1.

${(x - 2)}^{2} + {(y + 1)}^{2} > 1$

${(x - 2)}^{2} + {(y + 1)}^{2} = {1}^{2}$

- уравнение окружности с центром в точке А(2;-1) и радиусом R=1

решение неравенства - часть плоскости "за окружностью" - "солнышко"

2.

${(x - 2)}^{2} + {(y + 1)}^{2} < 9$

${(x - 2)}^{2} + {(y + 1)}^{2} = {3}^{2}$

- уравнение окружности с центром в точке А(2;-1) и радиусом R=3

решение неравенства - "внутренность" окружности

3. решение системы неравенств - " кольцо" - пересечение двух штриховок.

обе окружности пунктиром, т.к. неравенства строгие

Answer 2

The given inequality is:

(x - 2)^2 + (y + 1)^2 > 1 (x - 2)^2 + (y + 1)^2 < 9

Let's break it down step by step.

1. (x - 2)^2 + (y + 1)^2 > 1: This represents a circle centered at (2, -1) with a radius greater than 1. In other words, any point (x, y) outside this circle satisfies the inequality.

2. (x - 2)^2 + (y + 1)^2 < 9: This represents a circle centered at (2, -1) with a radius less than 3. In other words, any point (x, y) within this circle satisfies the inequality.

To find the solution, we need to determine the region that satisfies both conditions simultaneously. This corresponds to the area between the two circles, excluding the boundaries.

Therefore, the solution to the inequality is the annular region (or the ring-shaped region) between the two circles.