1. Problem 11(a): Find an equation for the quadratic curve modeling the mouse hole entrance. - The entrance is symmetrical and 6 cm wide at the top, so the curve passes through points (-6,0), (0,6), and (6,0). - Let the quadratic be $y = ax^2 + bx + c$. - Since the curve is symmetrical about the y-axis, $b=0$. - Using point (0,6): $6 = a\cdot0^2 + c \Rightarrow c = 6$. - Using point (6,0): $0 = a\cdot6^2 + 6 \Rightarrow 36a + 6 = 0 \Rightarrow a = -\frac{1}{6}$. - Equation: $$y = -\frac{1}{6}x^2 + 6$$ 2. Problem 11(b): Can a cuboid cheese 4cm by 4cm by 9cm slide into the hole with one face flat on the ground? - The hole width at height $y$ is given by solving for $x$ in the equation: $y = -\frac{1}{6}x^2 + 6$. - Rearranged: $x^2 = 6(6 - y)$, so half-width at height $y$ is $x = \sqrt{6(6 - y)}$. - The cheese must fit within the hole width and height. - The cheese height is 4 cm (if placed flat), so check hole width at $y=4$: $$x = \sqrt{6(6 - 4)} = \sqrt{6 \times 2} = \sqrt{12} \approx 3.46 \text{ cm}$$ - Total hole width at $y=4$ is $2x = 6.92$ cm. - Cheese width is 4 cm, which is less than 6.92 cm, so it fits width-wise. - Cheese length is 9 cm, but hole height is 6 cm max, so cheese height must be 4 cm (flat face on ground), which fits. - Therefore, the cheese can slide in flat. 3. Problem 12: Find equation $y = ax^2 + bx + c$ for the suspension bridge cable. - The cable is symmetrical about midpoint between towers, so place origin at midpoint: $x=0$ at center. - Towers are 200 m apart, so towers at $x = -100$ and $x = 100$. - Heights at towers: $y(-100) = 80$, $y(100) = 80$. - Minimum height at center: $y(0) = 5$. - Equation: $y = ax^2 + bx + c$. - Symmetry implies $b=0$. - Using $y(0) = c = 5$. - Using $y(100) = a(100)^2 + 5 = 80 \Rightarrow 10000a + 5 = 80 \Rightarrow 10000a = 75 \Rightarrow a = \frac{75}{10000} = 0.0075$. - Final equation: $$y = 0.0075x^2 + 5$$ Note: Since the cable hangs downwards, $a$ should be positive for a parabola opening upwards, but the problem states minimum height at center, so parabola opens upwards, consistent with $a>0$. Final answers: (a) $y = -\frac{1}{6}x^2 + 6$ (b) Yes, the cheese can slide in flat. (12) $y = 0.0075x^2 + 5$