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∠PSQ = 90^\circ (∠ in semi circle) ∠PSQ + ∠QSR = 145^\circ 90 + ∠QSR = 145 ∠QSR = 145 - 90 ∠QSR = 55^\circ BUt x^\circ = ∠QSR (∠s in the same segment) ∴x^\circ = 55^\circ

The set of points equidistant from two parallel lines can be determined by the perpendicular distance from a point to each line. Given the lines are of the form ax + by + k\(_1\) = 0 and ax + by + k\(_2\) = 0, the locus of points equidistant from these lines is another line parallel to them. To find this line, which represents the equation of the locus, we need to find the average k\(_1\) and k\(_2\), which we shall be denoted by q This line will have the same coefficients for x and y as the original lines. Thus, the equation of the locus is: ax + by + q = 0

Each ∠of a regular polygon \(=\frac{(n-2)180}{n}\\ =\frac{(6-2)180}{6}\\ =\frac{(4)180}{6}\) = 120o ΔQSR is isosceles ∴ Q = S = 30o Also ΔTSR is isosceles ∴ T = R = 30o ∠TSV + ∠VSR = 120 ∠TSV + 30 = 120 ∠TSV = 120 - 30 ∠TSV = 90o ∴∴VTS + ∴VST + ∴TVS = 180 (sum of ∠s of a Δ) 30 + 90 + ∠TVS = 180 120 + ∠TVS = 180 ∠TVS = 180 - 120 ∠TVS = 60o

Locus of point equidistant from points P and Q is a perpendicular bisector of the straight joining P and Q ∴PN = NQ

In ΔPST; cos 60 = ST/8 ST = 8cos 60 ST = 8 x 1/2 = 4 TR = 10 cm (opp. sides of a rectangle PQRT) SR = ST + TR SR = 4 + 10 SR = 14 cm

y = x implies y \le x y + x = 4 implies y = 4 – x ∴y = 4 – x ∴y \le x, y + x \le 4 and y \ge 0

x ⊕ y = xy/6 = xe/6 = x xe = 6x e = 6 x ⊕ y = xx1/6 where x1 = 20 x x 20/6 = 6 x = 36/20 x = 9/5

P ∝ 1/q3 P = K/q3 q3 = K/P q = K/p1/3 But q ∝ r2 q = Kr2 K/p1/3 = Kr2 r2 = K/p1/3 x 1/K r2 = 1/p1/3 r = (1/p1/3)2 r = 1/p1/6 r ∝ 1/p1/6 = p = \(\frac{1}{r^6}\) ∴ p varies inversely as \(r^6\)

\(x \ast y = \frac{x^2 - y^2}{2xy}\) =\(\frac{(x+y)(x-y)}{2xy}\) -5 \(\ast 3 =\frac{(-5+3)(-5-3)}{2(-5\times3)}\) =\(\frac{-2 \times -8}{2(-5\times3)}\) =\(\frac{-8}{15}\)

F(x) = Q x D + R Kx3 + x2 - 5x – 2 = Q(2x+1)+R If 2x+1 = 0 implies x = -1/2 ∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2 = Q(2(-1/2) + 1) +2 K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2 -k/8 + 1/4 + 5/2 - 2 = 0+2 (-k+2+20-16) / 8 = 2 (-k+6) / 8 = 2 -k+6 = 2*8 -k+6 = 16 -k = 16-6 -k = 10 ∴k = -10

\(\frac{x+4}{3}-\frac{x-3}{2}\) < 4 \(\frac{(2(x+4))-(3(x-3))}{6}\) < 4 2(x+4) – 3(x-3) < 4 x 6 2x + 8 – 3x+9 < 24 -x + 17 < 24 -x < 24 – 17 -x < 7 x > -7

Cost of removing a 6m\(^2\) room = N540 Cost of renovating a 1m\(^2\) room = 540 / 6 Cost of removing a 9m\(^2\) room = (540 / 6) * 9 = N810

Let the positive integers be 1, 2, ,3, 4, .....n ∴ a = 1, d = 1 and n = n Sn = n/2(2a + (n-1)d) = n/2 (2 + n – 1) = 1/2n(n + 1)

\(T = 2\pi \sqrt{\frac{l}{g}}\) \(\frac{T}{2\pi} = \sqrt{\frac{l}{g}}\) \((\frac{T}{2\pi})^{2} = \frac{l}{g}\) \(\therefore g = \frac{4 \pi^{2} l}{T^{2}}\)

y = x\(^2\) - x - 12 = (x - 4)(x + 3) ∴ x = 4 or x = -3 Checking the cases for y \( \geq \) 0 We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0. We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.