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2.7 x 10-4 x 6.3 x 106 = 2.7 x 6.3 x 10-4 x 106 = 17.01 x 10-4 + 6 = 17.01 x 102 = 1.701 x 101 x 102 = 1.701 x 101 + 2 = 1.701 x 103

From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ2 52 = 22 + MQ2 25 = 4 + MQ2 25 - 4 = MQ2 21 - MQ2 MQ2 = 21 MQ2 = \(\sqrt{21}\) Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm

Sin x = \(\frac{5}{13}\) 0o \(\leq\) x \(\leq\) 90o, (cos x - tan x) AC2 = AB2 + BC2 132 = 52 + BC2 169 - 25 + BC2 169 - 25 = BC2 144 = BC2 Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\) BC = \(\sqrt{144}\) BC = 12 tan x = \(\frac{opp}{adj} = \frac{5}{12}\) BC = 12 cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\) \(\frac{144 - 65}{156} = \frac{79}{156}\)

From the diagram above, the bearing of X from Z = 180º + 60º = 240º (Note: alternate angles are equal and measurement here is clockwise)

Let the interior angle = xo interior angle = 5xo (sum of int. angle ann exterior) (angles = angle or straight line) 6x = 180 x = \(\frac{180}{6}\) x = 30o no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\) = \(\frac{360}{30}\) = 12

1km = 100,000cm on the map 1 cm represent every 10 km which is equal to (10 x 100,000cm) = 1,000,000cm the scale is 1:1,000,000

< PQR is a right-angled triangle PR\(^2\) = 24\(^2\) + 10\(^2\) = 676 PR = \(\sqrt{676}\) = 26cm PQRTS is a square and PR = 26cm. meaning the remaining sides of the square are 26cm each Therefore, the perimeter of PQRST = 26 + 26 + 26 + 24 + 10 = 112cm

From the figure above, < PRQ = 55º

\(\bigtriangleup\) ABC, A + 153 = 180 (angles on straight line) o 9m = 180 - 27 9m = 153 m = \(\frac{153}{9}\) = 17

No expanded working has been attached for this question yet. The verified answer is Option C.

Arts = 90o business = 75o, science = 45o Technical 360o - (90o - 75o + 45o) = 360o - 210o Technical = 150o textbooks for technical class = \(\frac{150^o}{360^o}\) x 600 = 250

PQRST is a regular polygon with 5 sides-pentagon n = 5, Each exterior angle = \(\frac{360^o}{n}\) = \(\frac{350}{5}\) = 70o Each interior angle = 180 - 72o (o Since the pentagon is regular, the exterior angles are equal o and o o - (o - (72o + 72o) = 180o - 144o = 36o

Sum of angles on straight line = 180º 80º + 50º + x + 10º = 180º 140º + x = 180º x = 180º - 140º x = 40º total sum of all angles = 360º 2yº + 2xº + 80o + 50º + xº + 10º = 360º 2yº + 2(40)º + 80º + 50º + 40º + 10º = 360º 2yº + 80º + 180º = 360º 2yº + 260º = 360º 2yº = 360º - 260º 2yº = 100º angle 2yº = 100º

R +198 = 360(angles at a point) R = 360 - 198 = 162o C = 72(alternate angle), b + c = 180(angles or straight line) b = 190 - c = 180 - 72 = 108 a + b = R = 162 a = R - b = 162 - 108 = 54, but a = y(alternate angles) y = 54o

(a) \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\) \(\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8}) = \frac{3}{4}(\frac{27}{8} + \frac{13}{8})\) = \(\frac{3}{4}(\frac{40}{8})\) = \(\frac{15}{4}\) \(2\frac{1}{8} - 1\frac{1}{2} = \frac{17}{8} - \frac{3}{2}\) = \(\frac{17}{8} - \frac{12}{8}\) = \(\frac{5}{8}\) \(\therefore \frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}} = \frac{15}{4} \div \frac{5}{8}\) = \(\frac{15}{4} \times \frac{8}{5}\) = \(6\) (b) \(\log_{10} 2 = 0.3010 ; \log_{10} 3 = 0.4771\) \(\log_{10} 1.125 = \log_{10}(\frac{1125}{1000})\) (Dividing through with 125) = \(\log_{10} (\frac{9}{8})\) = \(\log_{10} 9 - \log_{10} 8\) \(\log_{10} 9 = \log_{10} 3^{2} = 2\log_{10} 3 = 2 \times 0.4771 = 0.9542\) \(\log_{10} 8 = \log_{10} 2^{3} = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030\) \(\therefore \log_{10} (\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\) \(\approxeq 0.051\) ( 2 significant figures)