1. **State the problem:** We have a zip line descending linearly from height 130 ft at horizontal distance 0 ft to ground level at 100 ft. The waterfall is modeled as a curve starting at height 167 ft at 0 ft and ending at ground level at 75 ft. We want to model both with equations and find their intersection point. 2. **Model the zip line:** It is a straight line from (0,130) to (100,0). Slope $m = \frac{0-130}{100-0} = \frac{-130}{100} = -1.3$. Equation: $$y = -1.3x + 130$$ 3. **Model the waterfall:** It is a curve from (0,167) to (75,0). Assuming a quadratic form $y = ax^2 + bx + c$. Using points: - At $x=0$, $y=167 \Rightarrow c=167$. - At $x=75$, $y=0 \Rightarrow 0 = a(75)^2 + b(75) + 167$. - At $x=0$, derivative $y'(0)$ is unknown, but we can use the vertex or approximate. Alternatively, since the problem states a curved waterfall and the image shows an arc, we can model it as a parabola with vertex at (0,167) and passing through (75,0). Using vertex form: $$y = a(x - h)^2 + k$$ with vertex $(h,k) = (0,167)$, so $$y = a x^2 + 167$$. At $x=75$, $y=0$: $$0 = a (75)^2 + 167 \Rightarrow a = -\frac{167}{75^2} = -\frac{167}{5625} \approx -0.0297$$ So waterfall equation: $$y = -0.0297 x^2 + 167$$ 4. **System of equations:** $$\begin{cases} y = -1.3x + 130 \\ y = -0.0297 x^2 + 167 \end{cases}$$ 5. **Find intersection:** Set equal: $$-1.3x + 130 = -0.0297 x^2 + 167$$ Rearranged: $$-0.0297 x^2 + 167 + 1.3x - 130 = 0$$ $$-0.0297 x^2 + 1.3 x + 37 = 0$$ Multiply both sides by $-1$ to simplify: $$0.0297 x^2 - 1.3 x - 37 = 0$$ 6. **Solve quadratic:** Using quadratic formula: $$x = \frac{1.3 \pm \sqrt{(-1.3)^2 - 4(0.0297)(-37)}}{2(0.0297)}$$ Calculate discriminant: $$1.69 + 4.3956 = 6.0856$$ Square root: $$\sqrt{6.0856} \approx 2.467$$ Calculate roots: $$x_1 = \frac{1.3 + 2.467}{0.0594} = \frac{3.767}{0.0594} \approx 63.4$$ $$x_2 = \frac{1.3 - 2.467}{0.0594} = \frac{-1.167}{0.0594} \approx -19.64$$ Negative distance is invalid, so take $x = 63.4$ ft. 7. **Find height at intersection:** Using zip line equation: $$y = -1.3(63.4) + 130 = -82.42 + 130 = 47.58$$ ft. **Final answer:** The zip line meets the waterfall at approximately horizontal distance $63.40$ ft and height $47.58$ ft.