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Principal = N255.00, Interest = 27.00 year = x Rate: 4% ∴ 1 = \(\frac{PRT}{100}\) 27 = \(\frac{225 \times 4 \times T}{100}\) 2700 = 900T T = 3 years

\(\sqrt{x^2 + 9}\) = x + 1 x\(^2\) + 9 = (x + 1)\(^2\) 0 = x\(^2\) + 2x + 1 - x\(^2\) - 9 = 2x - 8 = 0 2(x - 4) = 0 x = 4

\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication, q(1 + ax) = p(1 - ax) q + qax = p - pax qax + pax = p - q ∴ x = \(\frac{p - q}{a(p + q)}\)

\(\left(x + \frac{1}{x} + 1\right)^2 - \left(x - \frac{1}{x} - 1\right)^2,\) We can use the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b),\) where \(a = x + \frac{1}{x} + 1 \quad \text{and} \quad b = x - \frac{1}{x} - 1.\) Calculating \( a - b \): \(a - b = \left(x + \frac{1}{x} + 1\right) - \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 - x + \frac{1}{x} + 1 = 2 \cdot \frac{1}{x} + 2 = 2\left(\frac{1}{x} + 1\right).\) Calculating \( a + b \): \(a + b = \left(x + \frac{1}{x} + 1\right) + \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 + x - \frac{1}{x} - 1 = 2x.\) Substituting back into the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b) = \left(2\left(\frac{1}{x} + 1\right)\right)(2x) = 4x\left(\frac{1}{x} + 1\right).\) Distributing \(4x\): \(= 4x \cdot \frac{1}{x} + 4x \cdot 1 = 4 + 4x.\) Thus, the evaluated expression is: 4 + 4x = 4(1 + x)

f(x + 2) = f(-1) f(-1) is gotten when x = -3. \(2x^{2} + 7x - 5\) at x = -3. f(-1) = \(2(-3)^{2} + 7(-3) - 5\) = \(18 - 21 - 5 = -8\)

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

\(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) \div \(\frac{p^2 - q^2}{pq}\) \(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\) \(\frac{-(p - q)}{(p + q)(p - q)}\) = \(\frac{-1}{p + q}\)

\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3) x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 0, x -3, x 0 = x -4 = x < -0, -4 < x < -3(solution) combining the solutions -4 < x < -3

x \(\ast\) y = xy x \(\ast\) 2 = x2 x \(\ast\) 2 = x ∴ x2 - x = 0 x(x - 1) = 0 x = 0 or 1

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\) \((k + 1)(3k + 1) = (2k - 1)(2k - 1)\) \(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\) \(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\) \(k^{2} - 8k = 0\) \(k(k - 8) = 0\) \(\therefore \text{k = 0 or 8}\) The terms of the sequence given k = 0: (1, -1, 1) \(\implies \text{The common ratio r = -1}\) The terms of the sequence given k = 8: (9, 15, 25) \(\implies \text{The common ratio r = } \frac{5}{3}\) The possible values of the common ratio are -1 and \(\frac{5}{3}\).

Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 years and 6 months amount(increment) = N36 x 5 = N180 The man's new salary = N540 + N180 = N720.00

Surface area = 154cm2 (area of sphere) 4\(\pi\)r2 = 154 r\(\sqrt{\frac{154}{4\pi}}\) = 3.50cm

Number of students scoring at least 50 marks = Number of students scoring 50 and above From the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors itself and 1 The prime numbers among the group are 23, 29. Probability of choosing a prime number = \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\) = \(\frac{2}{11}\)

Chance of x = \(\frac{1}{2}\) Change of Y = \(\frac{2}{3}\) Chance of Z = \(\frac{1}{4}\) Chance of Y and Z only occurring = Pr (Y ∩ Z ∩ Xc) where Xc = 1 - Pr(X) 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\) = Pr(Y) x Pr(Z) x Pr(Xc) = \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\) = \(\frac{1}{12}\)