1. **State the problem:** We need to find the mass of fuel required for the model rocket to reach a height of 20 meters. 2. **Analyze the graph:** The graph shows a linear relationship between the mass of fuel ($x$) and the height reached ($y$). The line passes through the origin $(0,0)$ and approximately $(30,55)$. 3. **Write the equation of the line:** Since the line passes through the origin, the equation is of the form $$y = mx$$ where $m$ is the slope. 4. **Calculate the slope $m$:** $$m = \frac{\text{change in } y}{\text{change in } x} = \frac{55 - 0}{30 - 0} = \frac{55}{30} = \frac{11}{6}$$ 5. **Write the equation with the slope:** $$y = \frac{11}{6}x$$ 6. **Find the mass of fuel for $y=20$ meters:** $$20 = \frac{11}{6}x$$ 7. **Solve for $x$:** $$x = \frac{20}{\frac{11}{6}} = 20 \times \frac{6}{11} = \frac{120}{11}$$ 8. **Simplify the fraction:** $$x = \frac{120}{11} \approx 10.91$$ **Answer:** Approximately 10.91 units of fuel must be put into the rocket to reach a height of 20 meters.