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(a) \(\log_{10} 54 \) \(54 = 2 \times 3^{3}\) \(\log_{10} 54 = \log_{10} (2 \times 3^{3})\) = \(\log_{10} 2 + \log_{10} 3^{3}\) = \(\log_{10} 2 + 3\log_{10} 3\) = \(0.3010 + (3 \times 0.4771)\) = \(0.3010 + 1.4313\) = \(1.7323\) (b) \(\log_{10} 0.24\) \(0.24 = \frac{24}{100} = \frac{2^{3} \times 3}{100}\) \(\log_{10} 0.24 = \log_{10} 24 - \log_{10} 100\) \(\log_{10} (2^{3} \times 3) - \log_{10} (10^{2})\) \(\log_{10} 2^{3} + \log_{10} 3 - 2\log_{10} 10\) = \(3\log_{10} 2 + \log_{10} 3 - 2\) = \((3 \times 0.3010) + 0.4771 - 2\) = \(0.9030 + 0.4771 - 2\) = \(-0.6199\)

(a) \((x + 8)^{2} = x^{2} + (x + 7)^{2}\) \(x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49\) \(x^{2} + 16x + 64 - 2x^{2} - 14x - 49 = 0\) \(15 + 2x - x^{2} = 0\) \(x^{2} - 2x - 15 = 0\) \((x + 3)(x - 5) = 0\) x = 5m since measurements cannot be negative. (b) \(Area = length \times breadth\) Length = (x + 7) = (5 + 7) = 12 m Breadth = x m = 5 m \(Area = 12 \times 5 = 60 m^{2}\)

(a) see figure above a \(< QTR = < TRP = 70^\circ\) (alternate angle) \(< MNR = < TRP = 70^\circ\) (alternate angle; MN || ST) \(< MNR + < PNM = 180^\circ\) \(70^\circ + < PNM = 180^\circ \implies < PNM = 110^\circ\) \(\therefore x = 110^\circ\) (b)see figure b above \(< BDC = < BCD = 50^\circ\) (BD = BC) \(\therefore \Delta BDC = isosceles\) \(< ABD = 2(50^\circ) = 100^\circ\) (exterior angle) \(55^\circ + < BDA + 100^\circ = 180^\circ\) \(< BDA = 180^\circ - 155^\circ = 25^\circ\)

(a) Frequency Distribution Table Birth 45 - 54 55 - 64 65 - 74 75 - 84 85 - 94 95 - 104 105 - 114 Freq(f) 1 2 15 21 7 3 1 (b) see histogram above. (c) Modal class is 75 - 84 Mode = \(\frac{75 + 84}{2} = 79.5\) (d) Birth freq (f) class midpoint (x) \(fx\) 45 - 54 1 49.5 49.5 55 - 64 2 59.5 119 65 - 74 15 69.5 1042.5 75 - 84 21 79.5 1669.5 85 - 94 7 89.5 626.5 95 - 104 3 99.5 298.5 105 - 114 1 109.5 109.5 Total 50 3915 Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3915}{50}\) = \(78.3\).

< PAB = < ABE = 53^\circ (alternate angles) < CBE = 180^\circ - < DBC = 180^\circ - 161^\circ = 19^\circ < ABC = < ABE + < CBE = 53^\circ + 19^\circ = 72^\circ (a) In \(\Delta ABC\), \(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\) \(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\) = \(225 + 324 - 540 \cos 72\) = \(549 - 166.869\) \(AC^{2} = 382.131\) \(AC = \sqrt{382.131} = 19.548m\) \(\approxeq 19.5m\) (b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\) \(\sin A = \frac{18 \times \sin 72}{19.548}\) \(\sin A = 0.8757\) \(A = \sin^{-1} (0.8757) = 61.13^\circ\) The bearing of C from A = 61.13^\circ + 53^\circ = 114.13^\circ \(\approxeq\) 114^\circ. (c) In \(\Delta ATB\), \(\frac{15}{BT} = \tan 58\) \(BT = \frac{15}{\tan 58}\) \(9.373 m \approxeq 9.37m\)

(a) \(d = 15 cm \implies r = \frac{15}{2} cm\) \(V = \frac{4\pi r^{3}}{3}\) \(V = \frac{4}{3} \times \frac{22}{7} \times (\frac{15}{2})^{3}\) \(V = \frac{1125 \times 22}{7 \times 2}\) \(V = 1,767.857 cm^{3}\) (b)(i) Longitude difference : 105^\circ + 35^\circ = 140^\circ Distance along the parallel of latitude = \(\frac{140}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 28\) = \(\frac{140 \times 44 \times 6400 \times 0.8829}{360 \times 7}\) = \(13812.48 km\) \(\approxeq 13800 km\) (ii) Distance from the equator : \(\frac{28}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3128.89 km\) \(\approxeq 3130 km\)

Length of arc = \(\frac{Ø}{360} \times 2\pi\) 5.5 = \(\frac{360 - 210}{360}\) x 2 x \(\frac{22}{7}\) \(\frac{ 5.5 \times 360\times7}{150\times2\times22}\) = 2.1 length of major arc = \(\frac{210}{360}\) x 2 x \(\frac{22}{7}\) x 2.1 7.7m ≈ 8m

See Figure above. Since ST is parallel to UV => angle< U = 50^\circ(alternate to Angle UST = 180^\circ - 50^\circ = 130^\circ (angle on a straight line)

Since the inscribed triangle is equilateral, therefore the angles at all the points = 60^\circ Using the formula for inscribed circle, 2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) where R = radius of the circle; a, b and c are the sides of the triangle. ⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\) 2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\) 2R = 2 R = 1cm

2x + x = 180^\circ, => 3x = 180^\circ, and thus x = 60^\circ Each exterior angle = 60^\circ but size of ext. angle = 360^\circ/n Therefore 60^\circ = 360^\circ/n n = 360^\circ/60^\circ = 6 sides

x2 + y2 = (\sqrt2)2 x2 + y2 = 2 but y = x Thus; x2 + x2 = 2 2x2 = 2 x2 = + or - 1 But x2 + y2 = 2 12 + y2 = 2 1 + y2 = 2 y2 = 2 - 1 y2 = 1 y = + or - 1 Thus point (x,y) = (1,1) only.

\(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta\) = \(\int^{\pi}_{0}\frac{-sin^{2}\theta}{sin^{2}\theta}\) = \(\int^{\pi}_{0}d\theta = -\pi\)

y = 2x cos2x - sin2x \(\frac{dy}{dx}\) = 2 cos2x +(-2x. 2sin2x) - 2cos2x = 2 cos2x - 2x sin2x - cos2x = -2x 2sin2x = -4 x (\(\frac{\pi}{4}\)) sin2 x (\(\frac{\pi}{4}\)) = -(\pi) x 1 = - \pi

\(V = \frac{2}{3} \pi r^{3}\) Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\)

Angle corresponding to 4 is 40/240 x 360/1 = 60^\circ Note that total angle in a circle is 360^\circ. Note also that sum of the frequencies given is 240.