1. **State the problem:** We need to find the positive number $a$ such that the area enclosed by the asymptotes of the graphs of the functions $$y=\frac{x}{x+a}$$ and $$y=\frac{ax+1}{x-2}$$ is 20 square units. 2. **Identify the asymptotes:** - For $$y=\frac{x}{x+a}$$, the vertical asymptote is where the denominator is zero: $$x+a=0 \Rightarrow x=-a$$. - The horizontal asymptote is the limit as $$x \to \pm \infty$$: $$y=\lim_{x \to \infty} \frac{x}{x+a} = 1$$. - For $$y=\frac{ax+1}{x-2}$$, the vertical asymptote is where the denominator is zero: $$x-2=0 \Rightarrow x=2$$. - The horizontal asymptote is $$y=\lim_{x \to \pm \infty} \frac{ax+1}{x-2} = a$$. 3. **Asymptotes are lines:** - From the first function: vertical asymptote $$x=-a$$ and horizontal asymptote $$y=1$$. - From the second function: vertical asymptote $$x=2$$ and horizontal asymptote $$y=a$$. 4. **Area enclosed by these asymptotes:** The figure is a rectangle bounded by vertical lines $$x=-a$$ and $$x=2$$, and horizontal lines $$y=1$$ and $$y=a$$. 5. **Calculate the area of the rectangle:** $$\text{Area} = \text{width} \times \text{height} = (2 - (-a)) \times (a - 1) = (2 + a)(a - 1)$$. 6. **Given area is 20:** $$ (2 + a)(a - 1) = 20 $$ 7. **Expand and simplify:** $$ 2a - 2 + a^2 - a = 20 $$ $$ a^2 + a - 2 = 20 $$ $$ a^2 + a - 22 = 0 $$ 8. **Solve quadratic equation:** Using the quadratic formula $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1$$, $$b=1$$, $$c=-22$$: $$ a = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-22)}}{2} = \frac{-1 \pm \sqrt{1 + 88}}{2} = \frac{-1 \pm \sqrt{89}}{2} $$ 9. **Find positive root:** $$ a = \frac{-1 + \sqrt{89}}{2} \approx \frac{-1 + 9.433}{2} = \frac{8.433}{2} = 4.2165 $$ 10. **Final answer:** The positive number $$a$$ is approximately $$4.2165$$.