Question

5.4. 1) x2 + y2 < 4;​

Answer 1

Ответ:

Х^2 + y^2 = 4 X - y = 2 Решение Х = y + 2 X^2 = y^2 + 4y + 4 y^2 + 4y + 4 + y^2 = 4 2y^2 + 4y = 0 2y( y + 2 ) = 0 y1 = 0 ; y2 = - 2 X1 = 0 + 2 ... 

Answer 2

The inequality $x^2 + y^2 < 4$ represents a circular region in the Cartesian coordinate plane.

This inequality describes all points $(x, y)$ that are inside the circle centered at the origin $(0, 0)$ with a radius of $2$ units. In other words, it includes all points whose distance from the origin is less than $2$ units.

If you have any specific questions or if you'd like to know more about this inequality, feel free to ask!