1. **Problem:** A water balloon is catapulted from a building with height function $$h(t) = -5t^2 + 30t + 10$$. We want to determine: - How long the balloon is in the air. - The maximum height reached. 2. **Time in the air:** When the balloon hits the ground, height $$h(t) = 0$$. Solve $$-5t^2 + 30t + 10 = 0$$. 3. Divide through by -5 to simplify: $$t^2 - 6t - 2 = 0$$ 4. Use quadratic formula $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1$$, $$b=-6$$, $$c=-2$$: $$t = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-2)}}{2} = \frac{6 \pm \sqrt{36 +8}}{2} = \frac{6 \pm \sqrt{44}}{2}$$ 5. Simplify $$\sqrt{44} = 2\sqrt{11}$$: $$t = \frac{6 \pm 2\sqrt{11}}{2} = 3 \pm \sqrt{11}$$ Since time cannot be negative, take $$t = 3 + \sqrt{11} \approx 6.32$$ seconds. 6. **Maximum height:** Occurs at vertex $$t = -\frac{b}{2a} = -\frac{30}{2(-5)} = \frac{30}{10} = 3$$ seconds. Calculate $$h(3) = -5(3)^2 + 30(3) + 10 = -45 + 90 + 10 = 55$$ metres. 1. **Problem:** Hotel has 250 rooms, $150 price, 100 rented on average. For every $20 price drop, 25 more rooms rented. Find price to maximize revenue. 2. Define variables: Let $$x$$ be number of $20 drops. Then price $$p = 150 - 20x$$. Rooms rented $$r = 100 + 25x$$ but max 250 rooms. 3. Revenue $$R = p \times r = (150 - 20x)(100 + 25x)$$. 4. Expand: $$R = 150 \cdot 100 + 150 \cdot 25x - 20x \cdot 100 - 20x \cdot 25x = 15000 + 3750x - 2000x - 500x^2 = 15000 + 1750x - 500x^2$$ 5. Rewrite: $$R = -500x^2 + 1750x + 15000$$ 6. Max revenue at vertex $$x = -\frac{b}{2a} = -\frac{1750}{2(-500)} = \frac{1750}{1000} = 1.75$$ 7. Price: $$p=150 - 20(1.75) = 150 - 35 = 115$$ dollars. 1. **Problem:** Right triangle perimeter 60 cm. Hypotenuse $$c$$ is 6 cm more than twice one side $$a$$. Find sides $$a$$, $$b$$, and $$c$$. 2. Equations: $$c = 2a + 6$$ Perimeter: $$a + b + c = 60$$ 3. Substitute $$c$$: $$a + b + 2a + 6 = 60 \Rightarrow b = 54 - 3a$$ 4. Use Pythagorean theorem: $$a^2 + b^2 = c^2$$ Substituting $$b$$ and $$c$$: $$a^2 + (54 - 3a)^2 = (2a + 6)^2$$ 5. Expand: $$a^2 + (2916 - 324a + 9a^2) = 4a^2 + 24a + 36$$ $$a^2 + 2916 - 324a + 9a^2 = 4a^2 + 24a + 36$$ 6. Combine like terms: $$10a^2 - 324a + 2916 = 4a^2 + 24a + 36$$ 7. Bring all terms to left: $$10a^2 - 324a + 2916 - 4a^2 - 24a - 36 = 0$$ $$6a^2 - 348a + 2880 = 0$$ 8. Divide by 6: $$a^2 - 58a + 480 = 0$$ 9. Solve using quadratic formula: $$a = \frac{58 \pm \sqrt{58^2 - 4 \cdot 480}}{2} = \frac{58 \pm \sqrt{3364 - 1920}}{2} = \frac{58 \pm \sqrt{1444}}{2} = \frac{58 \pm 38}{2}$$ 10. Two solutions: $$a=\frac{58 + 38}{2} = 48$$ (discard since $$b=54-3a$$ becomes negative) $$a = \frac{58 - 38}{2} = 10$$ cm 11. Find $$b$$: $$b=54 - 3(10) = 24$$ cm 12. Find $$c$$: $$c = 2(10) + 6 = 26$$ cm 1. **Problem:** Rachel and Taylor plan to sell scarves. Cost per scarf: $6 Initial price: $10, sold 40 scarves For every $0.50 increase in price, sell 4 fewer scarves. Find price to maximize profit and maximum profit. 2. Let $$x$$ be number of $0.50 increases. Price $$p = 10 + 0.5x$$ Scarves sold $$q = 40 - 4x$$ 3. Profit per scarf $$= p - 6 = 4 + 0.5x$$ 4. Total profit $$P = (4 + 0.5x)(40 - 4x)$$ 5. Expand: $$P = 160 - 16x + 20x - 2x^2 = 160 + 4x - 2x^2$$ 6. Rewrite: $$P = -2x^2 + 4x + 160$$ 7. Max profit at vertex $$x = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1$$ 8. Calculate price: $$p = 10 + 0.5(1) = 10.5$$ dollars 9. Calculate profit: $$P = -2(1)^2 + 4(1) + 160 = -2 + 4 + 160 = 162$$ dollars 1. **Problem:** Parabola tunnel opening is 32 m wide. Vertex is at the top. 2. The width at the ground (base of parabola) means zeros at $$x = -16$$ and $$x = 16$$ if vertex at origin. 3. General form: $$y = a x^2 + k$$ with vertex at top. 4. Because it opens down and zeros at $$x = \pm 16$$, select vertex at origin $$y=0$$ at $$x=\pm16$$. 5. Use form $$y = a(x-0)^2 + k$$, and since vertex at top, plug zero height at $$x=16$$: $$0 = a(16)^2 + k = 256a + k$$ 6. If we take vertex at $$y = k$$, width 32 means zeros at $$x=\pm 16$$. This is setup for $$k=0$$ as vertex at top, we write: $$y = a x^2 + k$$ with $$k$$ maximum height. This problem description incomplete for further details, so the best we can say is the parabola has zeros at $$x = \pm 16$$ and vertex at the top, implying a quadratic with roots at $$\pm16$$ and negative leading coefficient. --- **Final answers:** 1. Balloon in air $$\approx 6.32$$ seconds, max height $$55$$ metres. 2. Max revenue price: $115. 3. Triangle sides: $$a=10$$ cm, $$b=24$$ cm, $$c=26$$ cm. 4. Max profit price: $10.5, profit $162. 5. Parabola zeros $$x = \pm 16$$ with vertex at top.