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Sum of interior ∠s = 1800o ∴(n - 2) 180o = 1800o 180n -360o = 1800o 180n = 1800o + 360o 180n = 2160o n = 2160o/180o n = 12 sides Each exterior ∠ = 360o/n = 360o/12 = 30o

∠OUT = ∠OTU = a (Base ∠s of 180 \Delta) ∴ a + a + 70 = 180o (sum of ∠s of a \Delta) 2a = 180o - 70o 2a = 110o a = 55o But ∠RUT + ∠RBT = 180o (opposite ∠s of a Cyclic quad) ∴ x + a = 100 = 180 x + 55 + 100 = 180 x = 180 - 155 x = 25 so ∠RUO = x = 25o

Area of a sector = \(\frac{\theta}{360}\times \pi r^2\) 55 = \(\frac{\theta}{360}\times \frac{22}{7}\times \frac{10\times 10}{1}\) \(\theta = \frac{360 \times 55 \times 7}{22 \times 10 \times 10}\) \(\theta = 63^{\circ}\)

L2 = 122 + 52 = 144 + 25 = 169 L = √169 = 13 Curved surface Area = πrL = 5/1 * 13/1 * π = 65πcm2

The locus of points equidistant from lines PQ and ST lies on the angle bisector of the angles formed by the intersection of those lines.

Gradient of the line y + 2x = 5 = -2 Gradient of the line perpendicular to y + 2x + 5 = 1/2 equation of a line perpendicular to y + 2x = 5 at the point (4, 3) y - y1 = m(x - x1) y - 3 = 1/2(x - 4) 2y - 6 = x -4 2y - x = 2

Chord Length (L) = \(2 \times Radius \times (sin \frac{angle}{2})\) = \(2 \times 14 \times (sin \frac{60}{2})\) = \( 28 \times (sin 30)\) = \(28 \times \frac{1}{2}\) = \(14cm\)

sin θ = -1/2 = -0.5 θ = sin-1 (0.5) θ = 30o Since θ is negative θ = 180 + 30 = 210o θ = 360 - 30 = 330o ∴θ = 210o and 330o

y = (1 - 2x)\(^3\) let u = 1-2x du/dx = -2 ∴y = u\(^3\) dy/du = 3u\(^2\) But dy/dx = du/dx * dy/du = -2 * 3u\(^2\) = -6u\(^2\) = -6(1-2x)\(^2\) At x = -1: dy/dx = -6(1 - 2(-1))\(^2\) dy/dx = -6(1 + 2)\(^2\) = -6 * 3\(^2\) = -6 * 9 = -54

y = sin (2x\(^3\) + 3x - 4) let u = 2x\(^3\) + 3x - 4 ∴du/dx = 6x\(^2\) y = sin u dy/du = cos u dy/dx = du/dx * dy/du ∴dy/dx = (6x\(^2\) + 3) cos u = (6x\(^2\) + 3)cos(2x\(^3\) + 3x - 4)

Increase in radius dr/dt = 0.5 Area of the circular disc = πr2 Increase in area: dA/dr = 2πr Rate of increase in area: dA/dt = dA/dr * dr/dt when r = 6cm = 2πr * 0.5 = 2 * π * 6 * 0.5 2 * π * 6 * 1/2 = 6 π cm2/sec

f(x) = 2 + x - x2 dy/dx = 1-2x As dy/dx = 0 1-2x = 0 2x = 1 x = 1/2 At x = 1/2 f(x) = 2 + x - x2 = 2 + 1/2 -(1/2)2 = 2 + 1/2 - 1/4 = (8+2-1) / 4 = 9/4

Area bounded by y = \(x^2\) - x + 3 and y = 3 The intersection is given as \(x^2\) - x + 3 = 3 \(x^2\) - x = 0 x(x -1) = 0 The graph is sketched between x= 0 and x = 1 \(\int^{1}_{0}\)\(x^2\) - x + 3)dx - \(\int^{1}_{0}\)3dx = \(\int^{1}{0}(x^2\) - x)dx [\(\frac{2x^3}{3} - \frac{x^2}{2}\)] = f(1) - f(0) = [\(\frac{2}{3} - \frac{1}{2}\)] - [0 - 0] = \(\frac{4 - 3}{6}\) = \(\frac{1}{6}\)unit(sq)

\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\) =[-1/2 cos 2 * \pi/2 + C] - [-1/2 cos 2 * 0] = [-1/2 cos \pi] - [-1/2 cos 0] = [-1/2x - 1] - [-1/2 * 1] = 1/2 -(-1/2) = 1/2 + 1/2 = 1

Workers who earn above #700 = 4 + 6 + 2 = 12