1. **Problem statement:** Find when the ball is at a height of 25 m using the quadratic formula. 2. **General form:** The height $h$ of a ball thrown upwards can be modeled by a quadratic equation of the form $$h = at^2 + bt + c$$ where $t$ is time. 3. **Given:** We want to solve for $t$ when $h=25$ m. 4. **Quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. **Apply to the problem:** Rearrange the height equation to standard form $at^2 + bt + (c - 25) = 0$. 6. **Calculate discriminant:** $$\Delta = b^2 - 4a(c - 25)$$ 7. **Calculate roots:** $$t = \frac{-b \pm \sqrt{\Delta}}{2a}$$ 8. **Interpret roots:** The two values of $t$ correspond to the times when the ball is at 25 m on the way up and on the way down. --- 9. **Problem statement:** Two boats leave the dock at the same time, speeds 120 km/h and 30 km/h, after 0.75 hours they are 100 km apart. Find the angle between their paths. 10. **Use Law of Cosines:** If $d$ is distance between boats, $v_1 t$ and $v_2 t$ are distances traveled, $$d^2 = (v_1 t)^2 + (v_2 t)^2 - 2(v_1 t)(v_2 t) \cos \theta$$ 11. **Plug in values:** $$100^2 = (120 \times 0.75)^2 + (30 \times 0.75)^2 - 2 \times 120 \times 0.75 \times 30 \times 0.75 \cos \theta$$ 12. **Simplify:** $$10000 = 90^2 + 22.5^2 - 2 \times 90 \times 22.5 \cos \theta$$ 13. **Calculate:** $$10000 = 8100 + 506.25 - 4050 \cos \theta$$ 14. **Rearrange:** $$4050 \cos \theta = 8100 + 506.25 - 10000 = -1393.75$$ 15. **Solve for angle:** $$\cos \theta = \frac{-1393.75}{4050} \approx -0.3441$$ 16. **Find angle:** $$\theta = \arccos(-0.3441) \approx 110^\circ$$ (rounded to nearest degree)