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x = 3, y = 2, z = 4 3x2 - 2y + z = 3(3)2 - 2(2) + 4 3(9) - 4 + 4 = 27

x + y = 2........... (i) 3x - 2y = 1......(ii) From (i): x = 2 - y ..........(iii) subt. (2 - y) for x in (ii) 3(2 - y) 2y = 1 6 - 3y - 2y = 1 -5y = 1 - 6 = -5 y = \(\frac{-5}{-5}\) = 1 subst. (i) for y in (iii) x = 2 - 1 = 1 x = 1, y = 1

8x-1 = \(\frac{1}{4}\) 23(x - 1 = 2-2 3x - 3 = -2 3x = -2 + 3 3x = 1 x = \(\frac{1}{3}\)

Distance(d) = d\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9 = 16 d = \(\sqrt{16}\) = 4cm

Area = 110.252 f2 = 110.25(square) l = \(\sqrt{110.25}\) = 10.5m Since N75.00 per metre N75.00 = 1 metre; x = 10.5m x = 787.5 for one side, then four sides = N787.5 x 4 = N3,150.00

Length of arc = circumference of the base of the cone \(\frac{\theta}{360} \times 2\pi R = 2 \pi r\) \(\frac{\theta R}{360}\) = r r = \(\frac{60 \times 14}{360}\) = \(\frac{7}{3} = 2\frac{1}{3}\)cm

volume of a cylinder = \(\pi r^2h\) r = 10cm h = 14cm = \(\frac{22}{7} \times 10 \times 10 \times 14\) = 4400cm3

\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides) 162n = 180n - 360 162n - 180n = -360 -18n = - 360 n = \(\frac{360}{18}\) n = 20

Statement ii: △PQR is an isosceles triangle. This statement is correct. Since two angles are equal (45º ), the triangle must have at least two equal sides. Statement iii: △PQR is a right-angled triangle. This statement is correct. The sum of the angles in a triangle is 180º Therefore, the third angle ∠QRP can be calculated as: ∠QRP = 180º −(45º + 45º ) = 90º

No expanded working has been attached for this question yet. The verified answer is Option B.

If p is the probability of an event happening(occurring) and the probability of an event not occurring is p1 then p + p1 = 1 p1 = 1 - p x1 = 1 - x

Mean(age) = \(\frac{(1 \times 2) + (2 \times 3) + (3 \times 5) + (4 \times 6) + (5 \times 4)}{2 + 3 + 5 + 6 + 4}\) = \(\frac{2 + 6 + 15 + 24 + 20}{20}\) = \(\frac{67}{20}\) = 3.35yrs

Median = (\(\frac{n + 1}{2})^{th}\) = \(\frac{20 + 1}{2} = \frac{21}{2}\) = 10.5 = \(\frac{{\text{10th term and 11th term}}}{2}\) = \(\frac{3 + 4}{2} = \frac{7}{2}\) = 3.5yrs

Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 mean = \(\frac{39}{9}\) = 4.33 then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\) = 9

△PQR is a right-angle triangle /QR/\(^2\) = 32\(^2\) + 25\(^2\) = 1024 + 625 = 1649 /QR/ = \(\sqrt{1649}\) = 40.6Km Let < PQR = \theta \(\frac{25}{sin \theta}\) = \(\frac{40.6}{sin 90}\) Sin \theta = \(\frac{25 Sin 90}{40.6}\) = 38º Bearing of R from Q = 270 - (48 + 38) = 184º