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1st convert to base 10 2232\(_4\) = 2 x 4\(^3\) + 2 x 4\(^2\) + 3 x 4\(^1\) + 2 x 4\(^0\) = 2 x 64 + 2 x 16 + 3 x 4 + 2 x 1 = 128 + 32 + 12 + 2 = 174 convert to base 6 6/174 6/29 R 0 6/4 R 5 6/0 R 4 = 450\(_6\)

Area of rectangle PQRS = \(10 \times 7 = 70cm^2\) Area of semi-circle: \(\frac{\pi r^{2}}{2}\) r = \(\frac{7}{2} cm\) Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77}{4} = 19.25 cm^{2}\) Area of figure = \((70 + 19.25) cm^{2}\) = 89.25 cm\(^{2} \approxeq\) 89 cm\(^{2}\)

Tan θ = 5/4 x2 = 52 + 42 = 25 + 16 = 41 x = √41 sin2θ - cos2θ = (5/√41)2 - (4/√41)2 = 25/41 - 16/41 = 9/41

Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their gradients are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2

∠yxz = 180 - (45 + 15) = 180 - 60 = 120o Using sine rule \(\frac{x}{sinx}=\frac{7}{sinz}\) \(\frac{x}{sin 120}=\frac{7}{sin 45}\) x = \(\frac{7 sin 120}{sin45}\) x = \(\frac{7sin(180-120)}{sin 45}\) x = \(\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = \(\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = 7\(\left(\frac{\sqrt{6}}{2}\right)\)

y = 55 (alternate angles) x = x (same reason) 260 + y + x = 360 (∠s at a point) 260 + 55 + x = 360 315 + x = 360 x = 360 - 315 x = 45o

∠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 \(\ast 3 =\frac{(-5+3)(-5-3)}{2(-5\times3)}\) =\(\frac{-2 \times -8}{2(-5\times3)}\) =\(\frac{-8}{15}\)