1. The problem asks which table of values represents a linear function. 2. A linear function has a constant rate of change (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 calculate the slope between consecutive points in each table to check if it is constant. 5. Table A: Between $(-1,7)$ and $(1,3)$: $$m = \frac{3 - 7}{1 - (-1)} = \frac{-4}{2} = -2$$ Between $(1,3)$ and $(3,-1)$: $$m = \frac{-1 - 3}{3 - 1} = \frac{-4}{2} = -2$$ Between $(3,-1)$ and $(5,-5)$: $$m = \frac{-5 - (-1)}{5 - 3} = \frac{-4}{2} = -2$$ All slopes are $-2$, so Table A is linear. 6. Table B: Between $(-6,-6)$ and $(-2,-4)$: $$m = \frac{-4 - (-6)}{-2 - (-6)} = \frac{2}{4} = 0.5$$ Between $(-2,-4)$ and $(2,-2)$: $$m = \frac{-2 - (-4)}{2 - (-2)} = \frac{2}{4} = 0.5$$ Between $(2,-2)$ and $(6,1)$: $$m = \frac{1 - (-2)}{6 - 2} = \frac{3}{4} = 0.75$$ Slopes are not constant, so Table B is not linear. 7. Table C: Between $(-4,4)$ and $(-1,2)$: $$m = \frac{2 - 4}{-1 - (-4)} = \frac{-2}{3} = -\frac{2}{3}$$ Between $(-1,2)$ and $(3,0)$: $$m = \frac{0 - 2}{3 - (-1)} = \frac{-2}{4} = -\frac{1}{2}$$ Slopes differ, so Table C is not linear. 8. Table D: Between $(1,3)$ and $(3,5)$: $$m = \frac{5 - 3}{3 - 1} = \frac{2}{2} = 1$$ Between $(3,5)$ and $(5,6)$: $$m = \frac{6 - 5}{5 - 3} = \frac{1}{2} = 0.5$$ Slopes differ, so Table D is not linear. 9. Conclusion: Only Table A has a constant slope and represents a linear function. **Final answer:** Table A represents a linear function.