1. The problem asks which set of ordered pairs $(x, y)$ could represent a linear function. 2. A linear function has a constant rate of change or slope between any two points. 3. The slope formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 4. We will calculate the slope between consecutive points in each set and check if the slope is constant. 5. For set A: $\{(-6, 9), (-3, 7), (0, 5), (6, 1)\}$ - Between $(-6, 9)$ and $(-3, 7)$: $$m = \frac{7 - 9}{-3 - (-6)} = \frac{-2}{3} = -\frac{2}{3}$$ - Between $(-3, 7)$ and $(0, 5)$: $$m = \frac{5 - 7}{0 - (-3)} = \frac{-2}{3} = -\frac{2}{3}$$ - Between $(0, 5)$ and $(6, 1)$: $$m = \frac{1 - 5}{6 - 0} = \frac{-4}{6} = -\frac{2}{3}$$ All slopes are $-\frac{2}{3}$, so set A is linear. 6. For set B: $\{(-5, 3), (-2, 5), (2, 7), (5, 9)\}$ - Between $(-5, 3)$ and $(-2, 5)$: $$m = \frac{5 - 3}{-2 - (-5)} = \frac{2}{3}$$ - Between $(-2, 5)$ and $(2, 7)$: $$m = \frac{7 - 5}{2 - (-2)} = \frac{2}{4} = \frac{1}{2}$$ Slopes differ, so set B is not linear. 7. For set C: $\{(-2, 3), (1, 0), (3, -1), (5, -3)\}$ - Between $(-2, 3)$ and $(1, 0)$: $$m = \frac{0 - 3}{1 - (-2)} = \frac{-3}{3} = -1$$ - Between $(1, 0)$ and $(3, -1)$: $$m = \frac{-1 - 0}{3 - 1} = \frac{-1}{2} = -\frac{1}{2}$$ Slopes differ, so set C is not linear. 8. For set D: $\{(-4, -1), (-1, 2), (2, 5), (5, 7)\}$ - Between $(-4, -1)$ and $(-1, 2)$: $$m = \frac{2 - (-1)}{-1 - (-4)} = \frac{3}{3} = 1$$ - Between $(-1, 2)$ and $(2, 5)$: $$m = \frac{5 - 2}{2 - (-1)} = \frac{3}{3} = 1$$ - Between $(2, 5)$ and $(5, 7)$: $$m = \frac{7 - 5}{5 - 2} = \frac{2}{3}$$ Slopes differ, so set D is not linear. 9. Conclusion: Only set A has a constant slope and represents a linear function. **Final answer:** Set A