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(a) \(y = ax^{2} - bx - 12\) When a = 1, b = 2 and y = 3. \(3 = x^{2} - 2x - 12\) \(x^{2} - 2x - 12 - 3 = 0 \implies x^{2} - 2x - 15 = 0\) \(x^{2} - 5x + 3x - 15 = 0\) \((x - 5)(x + 3) = 0\) \(\text{x = 5 or -3}\) (b) \(\sqrt{x^{2} + 1} = \frac{5}{4}\) Squaring both sides, \(x^{2} + 1 = \frac{25}{16}\) \(x^{2} = \frac{25}{16} - 1 = \frac{9}{16}\) \(x = \sqrt{\frac{9}{16}} = \pm \frac{3}{4}\) The positive value of x = \(\frac{3}{4}\).

(a) x -3 -2 -1 0 1 2 3 4 5 \(x^{2}\) 9 4 1 0 1 4 9 16 25 \(-2x\) 6 4 2 0 -2 -4 -6 -8 -10 -5 -5 -5 -5 -5 -5 -5 -5 -5 -5 y 10 3 -2 -5 -6 -5 -2 3 10 (b) see graph above (c) x -3 -2 -1 0 1 2 3 4 5 \(2x\) -6 -4 -2 0 2 4 6 8 10 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 y -9 -7 -5 -3 -1 1 3 5 7 (d) \(x^{2} - 2x - 5 = 2x - 3\) \(x^{2} - 2x - 2x - 5 + 3 = 0\) \(x^{2} - 4x - 2 = 0\) (e) x = -0.8 or 4.3

more than means greater than WHILE not more than means less than or equal to therefore the answer is 18 < d \le 23 meaning d is more than 8 but can be equal to or less than 23

\(2x - 2 \leq\) y = \( y \geq\) 2x - 2 when x = 0 , y = -2 and when y = 0, x = 1 \(2y - 2 \leq\) x when x = 0, y = 1 and when y = 0, x = 1 Use the diagram above for the OPTIONS

\(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}\) \(\frac{1}{x-3}-\frac{3(x-1)}{(x-3)(x+3)}\) \(\frac{x+3-3x+3}{(x-3)(x+3)};\frac{-2x+6}{(x-3)(x+3)}\) \(\frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3}\)

\(\frac{1}{x} + \frac{4}{3x} - \frac{5}{6x} + 1 = 0\) \(\frac{6 + 8 - 5 + 6x}{6x} = 0\) \(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\) \(6x = -9 \implies x = \frac{-3}{2}\)

(a) \(T_{n} = a + (n - 1)d\) (terms of an AP) Given a = -8; \(T_{7} = a + 6d = -8 + 6d\) \(T_{9} = a + 8d = -8 + 8d\) \(\frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\) \(5(-8 + 8d) = 8(-8 + 6d)\) \(-40 + 40d = -64 + 48d\) \(-40 + 64 = 48d - 40d \times 24 = 8d\) \( d = 3\) (b) Given r = 2 cm, m = 11.2g (i) \(V = \frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 2^{3}\) = \(\frac{704}{21}\) = \(33.52 cm^{3} = 33.52 \times 10^{-6} m^{3}\) = \(3.352 \times 10^{-5} m^{3}\) (ii) \(Density = \frac{mass}{volume}\) = \(\frac{11.2}{33.52}\) = \(0.334 g/cm^{3}\) (iii) \(V = \frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 3^{3}\) = \(\frac{2376}{21}\) = \(113.14 cm^{3}\) \(mass = density \times volume\) \(0.334 \times 113.14 = 37.789g\)

The statement that is not true about a rectangle is: II and III only The diagonals of a rectangle do not intersect at right angles. and does not have four lines of symmetry but two.

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN OE = OG = r(radii) (r - 8)\(^2\) + 12\(^2\) = r\(^2\) r\(^2\) - 16r + 64 + 144 = r\(^2\) 16r = 64 + 144 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm

(a) SEE FIGURE A above., Longitude difference = 123^\circ - 67^\circ = 56^\circ (i) Distance between X and Y along the equator = \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 0\) = \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(6257.78km \approxeq 6258km\) (ii) Latitude difference = 90^\circ Distance from the North pole = \(\frac{90}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(10057.14km \approxeq 10057km\) (b) SEE FIGURE B above,., \(OP^{2} = PN^{2} + ON^{2}\) \(4^{2} + 3^{2} = 25\) \(OP = \sqrt{25} = 5cm\) \(OP = OR = 5cm\) \(\Delta NOR = scalene\) \(\frac{OR}{\sin 20} = \frac{3}{\sin < ORN}\) \(\frac{5}{\sin 20} = \frac{3}{\sin < ORN}\) \(\sin < ORN = \frac{3 \times \sin 20}{5}\) \(\sin < ORN = 0.2052\) \(< ORN = \sin^{-1} (0.2052)\) \(< ORN = 11.84^\circ \approxeq 12^\circ\)

Area of trapezium = \(\frac{1}{2} (a + b)h\) \(120 = \frac{1}{2} (5 + 7) \times h\) \(120 = 6h\) \(h = 20 cm\)

The given figure is a kite. Here are the properties of a kite: Two Pairs of Adjacent Sides: A kite has two pairs of adjacent sides that are equal in length. Diagonals: The diagonals of a kite are perpendicular to each other. One diagonal bisects the other. Symmetry: A kite has one line of symmetry, which runs along the longer diagonal. Angles: The angles between the unequal sides are equal. Specifically, the angles opposite the equal sides are also equal. Therefore, ∠PQR = ∠PSR is correct.

(a) See Figure A above. (i) r = 1m, h = 8m. \(V = \pi r^{2} h = \frac{22}{7} \times 1^{2} \times 8\) = \(\frac{176}{7}\) = \(25.14 m^{3}\) (ii) \(V = L \times B \times H\) \(\frac{176}{7} = 4 \times 12 \times h\) \(h = \frac{176}{7} \div 48\) \(h = \frac{11}{21}m\) = \(\frac{1100}{21} cm = 52.38 cm\) \(\approxeq 52 cm\) (nearest whole number) (b)(i) see figure B. \(3.3^{2} + 5.6^{2} = BD^{2}\) \(BD = \sqrt{3.3^{2} + 5.6^{2}}\) = \(\sqrt{10.89 + 31.36}\) = \(\sqrt{42.25} = 6.5 cm\) (ii) See Figure C If < BCD is right-angled, then Pythagoras rule holds true. \(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\) = \(42.25\) \(BD = \sqrt{42.25} = 6.5 cm\) Hence, \(\Delta BCD\) is right-angled.

\(\pi r^2\) = 38.5 \(\frac{22}{7} r^2\) = 38.5 r\(^2\) = \(\frac{38.5 \times 7}{22}\) r = \(\sqrt{12.25}\) = 3.5 diameter = 2r = 2 x 3.5 = 7cm

(a) Mean \(\bar{x} = \frac{\sum x}{n} = 8\) \(\frac{1 + 2 + x + 11 + y + 14}{6} = 8\) \(28 + x + y = 48 \implies x + y = 20 .... (1) \) The middle numbers are x and 11 therefore, \(\frac{x + 11}{2} = 9\) \(x + 11 = 18 \implies x = 18 - 11 = 7\) Putting x = 7 in (1), \(7 + y = 20 \implies y = 20 - 7 = 13\) (x, y) = (7, 13). (b) \(\frac{LM}{LP} = \frac{3}{4}\) \(\frac{\text{Area of LMN}}{\text{Area of LPQ}} = \frac{3^{2}}{4^{2}}\) Area of \(\Delta\) LPQ = \(\frac{16}{9} \times 18 = 32 cm^{2}\) \(\therefore \text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\)