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x = 10 cot35o - 10 cot55o = 10(cot35o - cot55o)

4 sin2x - 3 = 0 sin2x = \(\frac{3}{4}\) Sin x = \(\pm\) \(\sqrt{\frac{3}{4}}\) = \(\pm\) \(\sqrt{\frac{3}{4}}\) = 60o

Total floor area = 720 x 420 Area of one tile = 30 x 30 No. of tile required = \(\frac{720 \times 420}{30 \times 30}\) = 24 x 14 = 336 tiles

To find the volume of metal used for the cylindrical pipe, we calculate the volume of the outer cylinder and subtract the volume of the inner cylinder. Convert dimensions to consistent units - Outer diameter = \(7.2 \, \text{cm}\), so outer radius \( R = \frac{7.2}{2} = 3.6 \, \text{cm} \) - Inner diameter = \(2.8 \, \text{cm}\), so inner radius \( r = \frac{2.8}{2} = 1.4 \, \text{cm} \) - Height \( h = 1 \, \text{m} = 100 \, \text{cm} \) Calculate the volume of the outer cylinder The formula for the volume of a cylinder is: \(V = \pi R^2 h\) For the outer cylinder: \(V_{\text{outer}} = \pi (3.6^2)(100) = \pi (12.96)(100) = 1296\pi \, \text{cm}^3\) Calculate the volume of the inner cylinder For the inner cylinder: \(V_{\text{inner}} = \pi (1.4^2)(100) = \pi (1.96)(100) = 196\pi \, \text{cm}^3\) Calculate the volume of metal used The volume of metal is given by: \(V_{\text{metal}} = V_{\text{outer}} - V_{\text{inner}} = 1296\pi - 196\pi = 1100\pi \, \text{cm}^3\) Thus, the volume of metal used for the cylinder is: 1100\(\pi\)cm\(^3\)

xz\(^2\) = 6\(^2\) + 6\(^2\) 36 + 36 = 72 xz = \(\sqrt{72}\) 6\(\sqrt{2}\) = XT \(\frac{6\sqrt{2}}{2}\) = \(\frac{3}{\sqrt{2}}\) OT\(^2\) = 5\(^2\) + (3\(\sqrt{2}\))2 = 25 + 18 OT = 4\(\sqrt{3}\)

Rice = 75g, Margarine = 40g, Meat = 105g, Bread = 20g, Total = 240 Angle of sector represented by meat = \(\frac{105}{240}\) x \(\frac{360^\circ}{1}\) = 157.5^\circ

Total age of the whole family = 21 x 14 = 294 Total age without the grandfather = 20 x 12 = 240 Age of grandfather = 294 - 240 = 54 years

Arrange the numbers in order, 1.6, 2.1, 2.1| 2.4, 2.6| 2.6, 2.6, 3.7 n = median = \(\frac{2.4 + 2.6}{2}\) = 2.5 m = mode = 2.6 ∴ (n, m) = (2.5, 2.6)

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

QMN = 60º MRQ = 60º(angle in the alternate segment are equal) MQN = 80º(angle sum of a \(\bigtriangleup\) = 180º) 60 + x = 80º(exterior angle = sum of opposite interior angles) x = 80º - 60º = 20º RMQ = 20º

In \(\bigtriangleup\)PHK and \(\bigtriangleup\)PAR = \(\frac{PH}{PQ} = \frac{PK}{PR}\) \(\frac{4}{7} \times \frac{PR-KR}{PR}\) 4PR = 7(PR - KR) = 7PR - 7KR \(\frac{KR}{PR} = \frac{3}{7}\) KR:PR = 3:7

Since QS = QR then, angle SQR = angle SRQ 2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62o QTP = 62o QTP = 62o, corresponding angle 3y + 56 + 62 = 180 = 3y = 180 - 118 3y = 62 = 180 3y = 180 - 118 3y = 62 y = \(\frac{62}{3}\) = 20\(\frac{3}{2}\)

∠PSQ = ∠SPQ = 24 ( base ∠s of iso △) ∠SQR =

Source: Myschool Past Questions Archive (Record #32904)