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Length of an arc = \(\frac{\theta}{360}\) x 2\pir \(\frac{90}{360}\) x 2 x \pi x 7 = \(\frac{7\pi}{2}\)cm\(^2\)

If the line 3y = 4x – 1 is parallel t[ line qy = x + 3 Implies gradient of 3y = 4x – 1 Y = 4/3x - 1/3 ∴Gradient = 4/3 Gradient of line qy = x + 3 Y = 1/qx + 3/q ∴ Gradient = 1/q 4/3 = 1/q 4q = 3 Q = 3/4

x = 30o (alternate ∠s b/c PQ//Sr) Y = 50o (vertical opposite ∠s) r = 30o (alternate ∠s b/c PQ//SR) But = y + q + t = 180o (∠s on a straight line) 50 + q + 30 = 180 q + 80 = 180 q = 180 -80 q = 100o

Volume of bowl \(\frac{2}{3}\pi r^3\\ 718\frac{2}{3}=\frac{2}{3}\pi r^3\\ \frac{2156}{3}=\frac{2}{3} \times \frac{22}{7} \times r^3 ∴r^3 = \frac{2156 \times 3 \times 7}{3 \times 2 \times 22}\\ r^3 = 343\\ r = \sqrt[3]{343}\\ r= 7.0cm\)

When a particle P moves such that the angle ∠SPT is constant while ST remains fixed, it traces a circular path. Specifically, the locus of P is a semicircle because points on the circle subtended by the line segment ST maintain that angle.

If the point of intersection is (1, -2), it implies that x = 1 and y = -2 when the two equation are solved simultaneously. ∴ substitute x = 1 and y = -2 in any of the equations 2y - k x + 2 = 0 2(-2) - k(1) + 2 = 0 -4 - k + 2 = 0 -4 + 2 = k -2 = k

Tan 30 = \(\frac{h}{40}\) \(\frac{1}{\sqrt{3}}=\frac{h}{40}\) h\(\sqrt{3}\) = 40 h = \(\frac{40}{\sqrt{3}}\) h = \(\frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\) h = \(\frac{40\sqrt{3}}{3}\)

Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0

(1-k+2) / 2 = - k and -4 + k + 1 = k 3-k = -2k and -3 + k = 2k K = -3 and k = -3

Each Exterior ∠ = 360/n = 360/8 = 45o

\(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}= \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\\ =\left(\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}\right)\div \left(\frac{\sqrt{3}\sqrt{3}+1}{\sqrt{3}}\right)\\ =\frac{(3-1)}{(3+1)}\\ =\frac{2}{4}=\frac{1}{2}\)

Area of a square = x * x 144 = x2 \sqrt144 = x 12 = x But d2 = 122 + 122 = 144 + 144 = 288 d = \sqrt288 = \sqrt144 x 2 = 12\sqrt2

(a) TSA of a closed cylinder = \(2\pi r^{2} + 2\pi rh\) = \(2\pi r (r + h)\) \(2 \times \frac{22}{7} \times 3.5 (3.5 + h) = 209\) \(22(3.5 + h) = 209 \implies 77 + 22h = 209\) \(22h = 209 - 77 = 132\) \(h = \frac{132}{22} = 6 cm\) (b) \(< EDB = 90^\circ\) (angle subtended at the circumference by a semi-circle) \(\therefore < EDF = 90^\circ - 66^\circ = 24^\circ\) \(< EDF = < EBF = 24^\circ\) (angles in the same segment) (ii) \(< DEB = < DBC = 57^\circ\) (Alternate segment theorem) \(\therefore < BGF = 180 - (57 + 24) = 180 - 81\) \(< BGF = 99^\circ\)

0.0409 x 0.0021 = 409 x 10\(^{-4}\) x 21 x 10\(^{-4}\) = 409 x 21 x 10\(^{-4 * -4}\) = 8589 x 10\(^{-8}\) = 8.589 x 10\(^{3}\) x 10\(^{-8}\) = 8.6 x 10\(^{3-8}\) = 8.6 x 10\(^{-5}\)

Vol. of cylinder = \(\pi r^2h\) = 1200cm2 Area of base = \(\pi(r^2)\) = 150cm2 h = \(\frac{\text{volume of cylinder}}{\text{base area}} = \frac{1200}{150}\) = 8.00cm