1. **State the problem:** We are given the function $$r(x) = -\frac{2}{3}x + 4$$ and need to find the values of $$r(-6)$$ and $$r\left(\frac{1}{2}\right)$$. 2. **Recall the formula:** The function is a linear function of the form $$r(x) = mx + b$$ where $$m = -\frac{2}{3}$$ and $$b = 4$$. 3. **Calculate $$r(-6)$$:** $$r(-6) = -\frac{2}{3} \times (-6) + 4$$ Multiply: $$= -\frac{2}{3} \times (-6) + 4 = \cancel{-\frac{2}{3}} \times \cancel{(-6)} + 4 = 4 + 4$$ $$= 8$$ 4. **Calculate $$r\left(\frac{1}{2}\right)$$:** $$r\left(\frac{1}{2}\right) = -\frac{2}{3} \times \frac{1}{2} + 4$$ Multiply: $$= -\frac{2}{3} \times \frac{1}{2} + 4 = -\frac{2 \times 1}{3 \times 2} + 4 = -\frac{2}{6} + 4 = -\frac{1}{3} + 4$$ Simplify: $$= \frac{-1}{3} + \frac{12}{3} = \frac{11}{3}$$ 5. **Final answers:** $$r(-6) = 8$$ $$r\left(\frac{1}{2}\right) = \frac{11}{3}$$ These values are found by substituting the given x-values into the linear function and simplifying step-by-step.