To determine which points are solutions to the inequality \(y > \frac{1}{9}x - \frac{2}{9}\), we can evaluate the right side of the inequality for each point where \(x = 0\).
-
For the point \((0, -\frac{4}{9})\): \[ y = -\frac{4}{9}, \quad \text{and } y > \frac{1}{9}(0) - \frac{2}{9} = -\frac{2}{9} \] \(-\frac{4}{9} > -\frac{2}{9}\) is false.
-
For the point \((0, -\frac{1}{3})\): \[ y = -\frac{1}{3}, \quad \text{and } y > \frac{1}{9}(0) - \frac{2}{9} = -\frac{2}{9} \] \(-\frac{1}{3} > -\frac{2}{9}\) is false (since \(-\frac{1}{3} \equiv -\frac{3}{9}\), \(-\frac{3}{9} > -\frac{2}{9}\) is not true).
-
For the point \((0, -\frac{1}{9})\): \[ y = -\frac{1}{9}, \quad \text{and } y > \frac{1}{9}(0) - \frac{2}{9} = -\frac{2}{9} \] \(-\frac{1}{9} > -\frac{2}{9}\) is true.
-
For the point \((0, -\frac{2}{9})\): \[ y = -\frac{2}{9}, \quad \text{and } y > \frac{1}{9}(0) - \frac{2}{9} = -\frac{2}{9} \] \(-\frac{2}{9} > -\frac{2}{9}\) is false.
The only point that satisfies the inequality is:
\((0, -\frac{1}{9})\).