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From the graph; when x = -2, y = 4 (-2, 4) x = 3.0, y = 9; (3,9) (-2, 4)(3,9)

r = 3.5 cm; l = 8.2 cm. Curved surface area = \(\pi rl\) = \(\frac{22}{7} \times \frac{7}{2} \times \frac{41}{5}\) = 90.2 cm\(^2\)

\(\log_{a} x = p\) \(x = a^p\)

Circumference of circle = 2πr = 2 x 3.14 x 4 = 25.1cm 360o = 25.1cm 70o = 25.1/360 x 70 = 4.89

Reflex POR = 2PQR Reflex PQR = 360o - 114o = 246o PQR = 246/2 = 123o

Tan 69.7 = 40/h h = 40/tan 69.7 = 40/2.703 h = 14.79 = 14.8

x\(^2\) + 2x + 1 = 25 x\(^2\) + 6x - 4x -24= 0 x(x + 6) - 4(x + 6) = 0 (x - 4)(x + 6) = 0 x = 4 or -6

log p = 2log x + 3log q log p = logx 2 + logq 3 logp = log x 2q 3 p = x 2q 3

In the 2nd Quadrant, SINE is the only Positive. Where \theta is 30 for the first quadrant and 150 in the 2nd quadrant, which makes COSINE negative Sin \theta = \(\frac{1}{2}\) \theta = Sin\(^{-1}(\frac{1}{2}\)) = 30º which is positive in the first quadrant Sin 30 = Sin 150, i.e, (180 - 30), which is also positive in the second quadrant. BUT Cos \theta = -\(\frac{\sqrt{3}}{2}\) \theta = Cos\(^{-1}\)(-\(\frac{\sqrt{3}}{2}\)) = 150º which is in the second quadrant but negative Hence, \theta = 150º

Probability that 3 customers are attended to in the sequence they arrive = prob of attending to the 1st arrival x prob of attending to the second last x prob of attending to the last = \(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\) = \(\frac{1}{6}\)

80o = 2 x 5x(angle at the center) =80o = 10x; x = 80o/10 = 8o

1st term a = 2; 6th term = a + (6 - 1)d = 2 + 5d 2 + (2 + 5d) = 16\(\frac{1}{2}\); 4 + 5d = 16\(\frac{1}{2}\) 5d = 12.5 ∴ = 2.5 the 4th term is 2 + (4 - 1)2.5= 9\(\frac{1}{2}\)

\(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) = \(\frac{18, 16, 15, 17}{20}\) \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) = \(\frac{16, 18, 15, 17}{20}\) \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) = \(\frac{18, 17, 16, 15}{20}\) ∴ \(\frac{9}{10}; \frac{17}{20}; \frac{4}{5}; \frac{3}{4}\) is in descending order.