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(K2)\(_6\) * 3\(_6\) = 3\(_5\)(K4)\(_5\) (K*6\(^1\) + 2*6\(^0\)) 3*6\(^0\) = 3*5\(^0\)(K*5\(^1\) + 4*5\(^0\)) 3(6K + 2) = 3(5K + 4) 6K + 2 = 5K + 4 6K - 5K = 4 - 2 K = 2

Let n(L∩P) = x n(L) only = 350 - x n(P) only = 200 - x ∴350 - x + x + 200 - x = 500 550 - x = 500 x = 550 - 500 x = 50 speak both ∴x% = (50/500) * (100/1) = 10%

N20,000 - N10,000 = N10,000 1st tax on N5000 at N50 per N1000 = 5*50 = N250 2nd tax on the remaining N5000 at N55 per N1000 = 5*55 = N275 Total tax paid = N250 + N275 = N525

\(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\ =\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\ =\frac{12+10\sqrt{3}}{36-25(3)}\\ =\frac{12+10\sqrt{3}}{36-75}\\ =\frac{12+10\sqrt{3}}{-39}\\ =-\left(\frac{12}{39}+\frac{10\sqrt{3}}{39}\right)\)

1100112 + 111112 = 10100102

(25)\(^{\frac{-1}{2}}\) x (27)\(^{\frac{1}{3}}\) + (121)\(^{\frac{-1}{2}}\) x (625)\(^{\frac{-1}{4}}\) 5\(^{2 \times \frac{-1}{2}}\) x 3\(^{3 \times \frac{1}{3}}\) + 11\(^{2 \times \frac{-1}{2}}\) x 5\(^{4 \times \frac{-1}{4}}\) 5\(^{-1}\) x 3\(^1\) x 11\(^{-1}\) x 5\(^{-1}\) \(\frac{1}{5} \times \frac{3}{1} + \frac{1}{11} \times \frac{1}{5}\) \(\frac{3}{5} + \frac{1}{55} = \frac{33+1}{55}\) = \(\frac{34}{55}\)

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