1. Problem 1: Calculate the radius of a circle sector with central angle 133° and arc length 246 cm. Formula: Arc length $s = r \theta$ where $\theta$ is in radians. Convert angle to radians: $\theta = 133^\circ \times \frac{\pi}{180} = \frac{133\pi}{180}$. Calculate radius: $$r = \frac{s}{\theta} = \frac{246}{\frac{133\pi}{180}} = \frac{246 \times 180}{133\pi} \approx 106.01\text{ cm}.$$ Answer: Radius $\approx 106.01$ cm. 2. Problem 2: Two sectors have the same arc length $x$. Sector 1 has radius 9 cm and angle 66°. Sector 2 has radius 14 cm and unknown angle $\theta$. Formula: $s = r \theta$ (angle in radians). Calculate arc length $x$ for sector 1: $$x = 9 \times \frac{66\pi}{180} = 9 \times \frac{11\pi}{30} = \frac{99\pi}{30}.$$ Calculate angle $\theta$ for sector 2: $$\theta = \frac{x}{14} = \frac{99\pi/30}{14} = \frac{99\pi}{420} = \frac{33\pi}{140}.$$ Convert to degrees: $$\theta = \frac{33\pi}{140} \times \frac{180}{\pi} = \frac{33 \times 180}{140} = 42.4^\circ.$$ Answer: Central angle $\approx 42.4^\circ$. 3. Problem 3: Rectangular garden with length $(2x+3)$ m, width $(x-1)$ m, area 30 m². Formula: Area $= \text{length} \times \text{width}$ Set up equation: $$(2x+3)(x-1) = 30$$ Expand: $$2x^2 - 2x + 3x - 3 = 30$$ $$2x^2 + x - 3 = 30$$ Bring all terms to one side: $$2x^2 + x - 33 = 0$$ Answer: Quadratic equation is $2x^2 + x - 33 = 0$. 4. Problem 4: Stephanie thinks of a positive fraction $x$. She squares it and multiplies by 6, then adds original number to get 40: $$6x^2 + x = 40$$ Rewrite: $$6x^2 + x - 40 = 0$$ Use quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 6 \times (-40)}}{2 \times 6} = \frac{-1 \pm \sqrt{1 + 960}}{12} = \frac{-1 \pm \sqrt{961}}{12} = \frac{-1 \pm 31}{12}.$$ Positive solution: $$x = \frac{-1 + 31}{12} = \frac{30}{12} = 2.5.$$ Answer: Stephanie's fraction is $\frac{5}{2}$ or 2.5. 5. Problem 5: Trapezium with top base $(x+6)$ cm, bottom base $(x+4)$ cm, height $(x-1)$ cm, area 27 cm². Formula: Area $= \frac{(\text{base}_1 + \text{base}_2)}{2} \times \text{height}$ Set up equation: $$27 = \frac{(x+6) + (x+4)}{2} \times (x-1) = \frac{2x + 10}{2} (x-1) = (x+5)(x-1).$$ Expand: $$x^2 - x + 5x - 5 = x^2 + 4x - 5 = 27$$ Bring all terms to one side: $$x^2 + 4x - 32 = 0$$ Answer: Quadratic equation is $x^2 + 4x - 32 = 0$. 6. Problem 6: Inequalities defining shaded region bounded by lines $y = x$ and $x = 2$. Answer: $$y \leq x$$ $$x \leq 2$$ 7. Problem 7: Inequalities defining shaded rectangle with horizontal boundaries $y=1$ and $y=5$, vertical boundaries $x=-3$ and $x=2$. Answer: $$-3 \leq x \leq 2$$ $$1 \leq y \leq 5$$ 8. Problem 8: Shaded region above line $y = -x + 6$ and to the left of $x=1$. Answer: $$y \geq -x + 6$$ $$x \leq 1$$ 9. Problem 9: Unshaded region satisfies inequalities $x + y < a$ and $x + y > b$ between two dashed lines. From graph description, lines are: Line 1: $x + y = a$ passing through (-6,6) and (6,-6) so slope -1, intercept 0, so $a=0$. Line 2: $x + y = b$ passing through (-6,5) and (5,-6), slope -1, intercept -1, so $b = -1$. Answer: $$a = 0$$ $$b = -1$$