1. **State the problem:** We need to find the value of $k$ such that the triangle bounded by the lines $y=0$, $y=2x$, and $y=-0.5x + k$ has an area of 80 square units. 2. **Identify the vertices of the triangle:** The triangle is formed by the intersections of these lines. - Intersection of $y=0$ and $y=2x$ is at $(0,0)$. - Intersection of $y=0$ and $y=-0.5x + k$ is found by setting $0 = -0.5x + k \Rightarrow x = \frac{k}{0.5} = 2k$, so the point is $(2k,0)$. - Intersection of $y=2x$ and $y=-0.5x + k$ is found by setting $2x = -0.5x + k$: $$2x + 0.5x = k \Rightarrow 2.5x = k \Rightarrow x = \frac{k}{2.5} = 0.4k$$ Then $y = 2x = 2 \times 0.4k = 0.8k$, so the point is $(0.4k, 0.8k)$. 3. **Calculate the area of the triangle:** The base lies on the $x$-axis between $(0,0)$ and $(2k,0)$, so base length is $2k$. The height is the vertical distance from the point $(0.4k, 0.8k)$ to the $x$-axis, which is $0.8k$. Area formula for a triangle is: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Substitute base and height: $$80 = \frac{1}{2} \times 2k \times 0.8k$$ 4. **Simplify the equation:** $$80 = \cancel{\frac{1}{2}} \times \cancel{2}k \times 0.8k = 0.8k^2$$ 5. **Solve for $k$:** $$k^2 = \frac{80}{0.8} = 100$$ $$k = \sqrt{100} = 10$$ Since $k$ is positive, $k=10$. **Final answer:** $$\boxed{10}$$