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2x\(^2\) - 21x + 45 = 2x\(^2\) - 15x - 6x + 45 = x(2x - 15) - 3(2x - 15) = (x - 3)(2x - 15)

y = 3x ... (i); 4y - 5x =14 ... (ii) Put 3x for y in (ii). 4(3x) - 5x = 14 12x - 5x = 14 7x = 14 \(\implies\) x = 2. y = 3x = 3(2) = 6 (x, y) = (2, 6)

Area of a sector = \theta/360 x \pir\(^2\) = 120/360 x 22/7 x 81/1 = 84.86 cm\(^2\) \(\approxeq\) 85cm\(^2\) (to the nearest cm)

\pirl + \pir2 = \pir(l + r) \(\frac{22}{7}\) x 5(13 + 5) 282\(\frac{6}{7}\)cm\(^2\)

Volume of a cone = \(\frac{1}{3} \pi r^2 h\) r = 4\(\frac{1}{2}\) cm; h = 14 cm Volume of cone = \(\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 14\) = 297 cm\(^3\) When half- filled, the volume of the water = \(\frac{297}{2} = 148.5 cm^3\)

Area of rhombus = \(\frac{pq}{2}\) where p and q are the two diagonals of the rhombus. \(\therefore 60 = \frac{12 \times q}{2}\) 6q = 60 \(\implies\) q = 10 cm

Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm

< STQ = < SQR = 68^\circ (alternate segment) \(\therefore\) < STQ = 68^\circ < TQS = \(\frac{180^\circ - 68^\circ}{2}\) = \(\frac{112}{2} = 56^\circ\) \(\therefore\) < PQT = 180^\circ - (68^\circ + 56^\circ) = 180^\circ - 124^\circ = 56^\circ

Sum of interior angles in a polygon = \((2n - 4) \times 90^\circ\) \(\therefore (2n - 4) \times 90^\circ = 30 \times 90^\circ\) \(\implies 2n - 4 = 30 \) \(2n = 34 \implies n = 17\) The polygon has 17 sides.

\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles) \(\therefore \frac{PS + 10}{12} = \frac{12}{10}\) \(10 (PS + 10) = 12 \times 12\) \(10 PS + 100 = 144 \implies 10 PS = 44\) \(PS = 4.4 cm\)

Join PR QRP = QPR = 180 - 80 = \(\frac{100}{20}\) = 50º SRP = SPR = 180 - 30 = \(\frac{150}{2}\) = 75º ∴ SPQ = SPR - QPR = 75 - 50 = 25º

QPS - QRK = 31o QRK + RKQ + KQR = 180 31 + 42 + KQR = 180o KQR = 180 - 73 = 107o

< PQR = 180^\circ - (84^\circ + 56^\circ) = 40^\circ \(\therefore\) x = 40^\circ (corresponding angles QP// RT)

\(\frac{2.25}{0.015}\) = \(\frac{2250}{15}\) = \(150\) = \(1.5 \times 10^{2}\)

\(-2, -3, -4\frac{1}{2}, ...\) This is a G.P with r = 1\(\frac{1}{2}\). \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (-2)(\frac{3}{2})^{5 - 1}\) = \(-2 \times \frac{81}{16}\) = \(-\frac{81}{8}\) \(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\) = \(-2 \times \frac{729}{64}\) = \(-\frac{729}{32}\)