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

Radius of the circle r = 6cm, Length of the arc = 8cm Area of sector = \(\frac{\theta}{360}\) x \(\pi\)r2........(i) Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r........(ii) from eqn. (ii) \(\theta\) = \(\frac{240}{\pi}\), subt. for \(\theta\) in eqn (i) Area x \(\frac{240}{1}\) x \(\frac{1}{360}\) x \(\frac{\pi 6}{1}\) = 24cm\(^2\)

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

Total surface area(s) = 2(4 x 3) + 2(4 x 4) = 2(12) + 2(16) = 24 + 32 = 56cm2 1m2 costs N2.00 ∴ 56m∴ will cost 56 x N2.00 = N112.00

Anambra = \(\frac{185}{900}\) x \(\frac{360}{1}\) = 74o

PT x QT = TR x TS 24 x 8 = TR x 12 TR = \(\frac{24 \times 8}{12}\) = = 16cm

Volume of solid = cross section x H Since the cross section is a trapezium = \(\frac{1}{2} (6 + 11) \times 12 \times 8\) = 6 x 17 x 8 = 816m3

From the circle centre 0, if PQ & PR are tangents from P and QRP = 34o Then the angle marked x i.e. QOP 34o x 2 = 68o

Given \(\bigtriangleup\) isosceles PQ = QT, SRQ = 35o TPQ = 20o PQR = is a straight line Since PQ = QT, angle P = angle T = 20o Angle PQR = 180o - (20 + 20) = 140o TQR = 180o - 140o = 40o < on a straight line QSR = 180o - (40 + 35)o = 105o TSR = 180o - 105o = 75o

Draw a line WX through R and parallel to \(\overline{PQ}\) and \(\overline{ST}\) as shown in the diagram above X\(\widehat{R}\)Q = P\(\widehat{Q}\)R = 35º X\(\widehat{R}\)Q + X\(\widehat{R}\)S = 65º 35 + x = 65 x = 65 - 35 = 30º R\(\widehat{S}\)T = 30º (Alternative angle) S\(\widehat{T}\)V = 30º (Alternative angle)

cos z = \(\frac{y^2 + x^2 -z^2}{2yx}\) = \(\frac{9 + 36 - 16}{2(3)(6)}\) = \(\frac{29}{36}\)

C = a + k \(\frac{1}{N}\) = c = \(\frac{aN + k}{N}\) CN = aN + K 30(100) = a(100) + k 3000 = 100a + k.......(i) 60(40) = a(40) + k 2400 = 40a + k.......(ii) eqn (i) - eqn (ii) 600 = 60a a = 10 subt. for a in eqn (i) 3000 = 100(10) + K 3000 - 1000 = k k = 2000 CN = 10N + 2000. when N = 50, 50C = 10(50) + 2000 50C = 500 + 2000 C = \(\frac{2500}{50}\) = 50k

cos\(\theta\) = \(\frac{13^2 + 9^2 - 11^2}{2(13)(9)}\) = \(\frac{169 + 81 - 21}{26 \times 9}\) cos\(\theta\) = \(\frac{129}{26 \times 9}\) = \(\frac{43}{78}\)

When x = -3, y = 4 - 3(3) - (-3)3 = 4 + 9 + 27 = 13 + 27 = 40

45o = \(\frac{x}{2}\), Since 45o = \(\frac{1}{\sqrt{2}}\) x = 2 x \(\frac{1}{\sqrt{2}}\) = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)