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\(y \propto \frac{1}{x^2}\) \(y = \frac{k}{x^2}\) \(x^2 = \frac{k}{y}\) \(x = \frac{\sqrt{k}}{\sqrt{y}}\) Since k is a constant, then \(\sqrt{k}\) is also a constant. \(\therefore x \propto \frac{1}{\sqrt{y}}\)

Numbers: 3, 6, x, 7, 5 Mean = \(\frac{(21 + x)}{5}\) Standard deviation = 2, so the sum of squared deviations = 20. Solving: 2x² - 21x + 27 = 0 x = 9 or x = 1.5 ≈ 2 The **least possible value of x is 1.5

If the side = 10 cm, correct to the nearest whole number, then The side ranges from 9.5 cm to 10.5 cm. (9.5 \(\leq\) s < 10.5) Perimeter of a rhombus = 4 x side \(\therefore\) 4 x 9.5 \(\leq\) perimeter < 4 x 10.5 = 38 cm \(\leq\) P < 42 cm

Total angle in a circle = 360^\circ \(\therefore\) 105 + 75 + 2k + k + 3k = 360^\circ 6k = 360 - 180 = 180 k = 180/6 = 30^\circ

Let the exterior angle = d Note: Exterior angle + Interior angle = 180^\circ \(\implies\) d + 2d = 180^\circ 3d = 180^\circ \(\implies\) d = 60^\circ Recall, exterior angle = \(\frac{360}{\text{no of sides}}\) \(\therefore \text{No of sides} = \frac{360}{60}\) = 6 sides

Isosceles triangle and Kite shapes have 1 line of symmetry each while the rhombus has 2 lines of symmetry.

< QOR = 50^\circ x 2 = 100^\circ Reflex < QOR = 360^\circ - 100^\circ = 260^\circ \(\therefore\) 30^\circ + 50^\circ + 260^\circ + m = (4 - 2) x 180^\circ 340^\circ + m = 360^\circ m = 360^\circ - 340^\circ = 20^\circ

\(\cos x = \frac{\sqrt{3}}{2}\) \(\sin x = \frac{1}{2}\) \(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)

From the diagram above: \(\tan \theta = \frac{20}{75} = 0.267\) \(\theta = \tan^{-1} 0.267 = 14.93^\circ\) \(\approxeq\) 14.9^\circ

In the diagram given in the question, < ABC = 90º ( < in a semi-circle) In triangle ABC, Sin 40º = \(\frac{x}{|AC|}\) |AC| = \(\frac{x}{sin40º}\) Since |AC| is a diameter, radius is half of |AC| = \(\frac{x}{2sin40º}\)

< ABC = 40^\circ (alternate angles) \(\therefore\) < ACB = \(\frac{180^\circ - 40^\circ}{2}\) = 70^\circ \(\therefore\) Bearing of A from C = 360^\circ - 70^\circ = 290^\circ

PR = 4 + 4 = 8 cm Area of trapezium = \(\frac{1}{2} (a + b) h\) = \(\frac{1}{2} (4 + 8) \times 5\) = 30 cm\(^2\)

\(3.72 \times 0.025 = 0.093 = 9.3\times 10^{-2}\)

In the diagram:

Source: Myschool Past Questions Archive (Record #26420)