1. **Problem 14: Find the length and width of the Parthenon's base.** The length $L$ is 8 meters more than twice the width $W$, so: $$L = 2W + 8$$ The area $A$ of the rectangular base is given by: $$A = L \times W = 2170$$ Substitute $L$: $$W(2W + 8) = 2170$$ 2. **Set up the quadratic equation:** $$2W^2 + 8W = 2170$$ Bring all terms to one side: $$2W^2 + 8W - 2170 = 0$$ 3. **Simplify by dividing all terms by 2:** $$\cancel{2}W^2 + \cancel{8}W - \cancel{2170} = 0 \Rightarrow W^2 + 4W - 1085 = 0$$ 4. **Use the quadratic formula:** $$W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=4$, $c=-1085$. Calculate the discriminant: $$\Delta = 4^2 - 4(1)(-1085) = 16 + 4340 = 4356$$ Calculate the square root: $$\sqrt{4356} = 66$$ 5. **Find the two possible values for $W$:** $$W = \frac{-4 \pm 66}{2}$$ Calculate each: - $$W = \frac{-4 + 66}{2} = \frac{62}{2} = 31$$ - $$W = \frac{-4 - 66}{2} = \frac{-70}{2} = -35$$ Since width cannot be negative, $W = 31$ meters. 6. **Find the length $L$:** $$L = 2(31) + 8 = 62 + 8 = 70$$ meters. --- 7. **Problem 15: Interpret the constant term and find how long the diver is in the air.** Given height function: $$h = -16t^2 + 8t + 80$$ - The constant term 80 represents the initial height of the diver above the water at time $t=0$ seconds. 8. **Find when the diver hits the water:** Set $h=0$: $$-16t^2 + 8t + 80 = 0$$ Divide all terms by -8 to simplify: $$\cancel{-8} \times 2t^2 - \cancel{8} t - \cancel{8} \times 10 = 0 \Rightarrow 2t^2 - t - 10 = 0$$ 9. **Use quadratic formula with $a=2$, $b=-1$, $c=-10$:** $$t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-10)}}{2(2)} = \frac{1 \pm \sqrt{1 + 80}}{4} = \frac{1 \pm \sqrt{81}}{4}$$ Calculate square root: $$\sqrt{81} = 9$$ 10. **Calculate the two possible times:** $$t = \frac{1 + 9}{4} = \frac{10}{4} = 2.5$$ $$t = \frac{1 - 9}{4} = \frac{-8}{4} = -2$$ Negative time is not physically meaningful, so the diver is in the air for $2.5$ seconds. **Final answers:** - Problem 14: Width = 31 meters, Length = 70 meters. - Problem 15: The constant term 80 is the initial height in feet; the diver is in the air for 2.5 seconds.