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\(y = 2x \\ V = \int\pi y^{2}dy \\ but\hspace{1mm}y = 2x \\ V = \int\pi4x^{2}dx\\ V = \frac{4(3)^{3}\pi}{3} - \frac{4(0)^{3}\pi}{3}\) V = \(\frac{4*27\pi}{3}\) = 36\(\pi \hspace{1mm}cubic\hspace{1mm}units\) (Note: the limit of integration x = 0 → x = 3)

t2 sin (3t - 5) = 2t sin ( 3t - 5) + t2 x 3 cos (3t - 5) = 2t sin (3t - 5) + 3t2 cos (3t - 5). Detailed answer below was provided by Ifechuks, a female prospective student of Okopoly.

To evaluate the integral \(\int_{-2}^{1} (x - 1)^{2} \, dx,\) We first expand the integrand: \((x - 1)^{2} = x^{2} - 2x + 1\) Thus, the integral becomes: \(\int_{-2}^{1} (x^{2} - 2x + 1) \, dx = \int_{-2}^{1} x^{2} \, dx - 2\int_{-2}^{1} x \, dx + \int_{-2}^{1} 1 \, dx\) Now, we evaluate each integral: For \(\int_{-2}^{1} x^{2} \, dx\): \(\left[ \frac{x^{3}}{3} \right]_{-2}^{1} = \frac{1}{3} - \left(-\frac{8}{3}\right) = 3\) For \(-2\int_{-2}^{1} x \, dx\): \(-2\left[ \frac{x^{2}}{2} \right]_{-2}^{1} = -2\left(\frac{1}{2} - 2\right) = 3\) For \(\int_{-2}^{1} 1 \, dx\): \([x]_{-2}^{1} = 1 + 2 = 3\) Combining the results: 3 + 3 + 3 = 9.

We are given the function: \(y = x^3 - x\) To find the critical points, we set the derivative equal to zero: \(3x^2 - 1 = 0\) Solving this gives: \(x = \pm \frac{1}{\sqrt{3}}\) Using the second derivative test, we find that: \(\frac{d^2y}{dx^2} = 6x\) - For \( x = \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \) (minimum). - For \( x = -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} < 0 \) (maximum). Thus, the function \( y = x^3 - x \) has a minimum value at: \(x = \frac{1}{\sqrt{3}} \quad \text{or approximately} \quad x \approx 0.577\)

To find \( h \) for the function \(y = 1 + hx - 3x^2\) with a minimum value of 13: 1. The vertex \( x \) is given by \(x = \frac{h}{6}\) 2. Substitute into the function: \(y = 1 + \frac{h^2}{6} - 3\left(\frac{h}{6}\right)^2 = 1 + \frac{h^2}{6} - \frac{h^2}{12}\) 3. Combine terms: \(y = 1 + \frac{2h^2}{12} - \frac{h^2}{12} = 1 + \frac{h^2}{12}\) 4. Set equal to 13: \(1 + \frac{h^2}{12} = 13 \implies \frac{h^2}{12} = 12 \implies h^2 = 144 \implies h = 12 \text{ or } -12\) \( h = 12 \) or \( h = -12 \).

Tope bought x oranges at ₦5.00 each and 2x mangoes at ₦4.00 each. Total amount spent = \5x + 4(2x) = 5x + 8x = 13x Given: ₦65.00 \le 13x \le ₦130.00 Divide through by 13: 5 \(\leq\) x \(\leq\) 10. (or x = 5, 6, 7, 8, 9, 10) Verification: - When x = 5: 13 \times 5 = ₦65.00 (minimum) - When x = 10: 13 \times 10 = ₦130.00 (maximum)

We are to form two-digit numbers using the digits 0, 1, 2, 3, with repetition allowed, but the number cannot begin with 0. - Tens place (first digit): Cannot be 0 → Possible digits = 1, 2, 3 → 3 choices - Units place (second digit): Can be any of the 4 digits → 0, 1, 2, 3 → 4 choices Total numbers = 3 \times 4 = 12

\(y \propto \frac{1}{x^2}\) \(y = \frac{k}{x^2}\) \(x^2 = \frac{k}{y}\) \(x = \frac{\sqrt{k}}{\sqrt{y}}\) Since k is a constant, then \(\sqrt{k}\) is also a constant. \(\therefore x \propto \frac{1}{\sqrt{y}}\)

In triangle RXY, < RXY = < RYX (base angles of an isosceles triangle) \(\implies\) 180^\circ - 62^\circ = 2 < RYX 118^\circ = 2 < RYX \(\implies\) < RYX = 59^\circ In triangle RYZ, < RZY = 62^\circ (base angles of an isosceles triangle) \(\therefore\) < RYZ = 180^\circ - (62^\circ + 62^\circ) = 180^\circ - 124^\circ = 56^\circ \(\therefore\) < XYZ = 56^\circ + 59^\circ = 115^\circ

Let \( u = 3x^2 \) and \( v = (x^3 + 1)^{1/2} \) Then \( y = u \cdot v \) - \( \frac{du}{dx} = 6x \) - For \( v \): \(\frac{dv}{dx} = \frac{1}{2}(x^3 + 1)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 1}}\) Now apply the product rule: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \) \(\frac{dy}{dx} = 3x^2 \cdot \frac{3x^2}{2\sqrt{x^3 + 1}} + \sqrt{x^3 + 1} \cdot 6x\) \(= \frac{9x^4}{2\sqrt{x^3 + 1}} + 6x \sqrt{x^3 + 1}\) Simplify (common denominator \( 2\sqrt{x^3 + 1} \)): \(\frac{dy}{dx} = \frac{9x^4 + 12x (x^3 + 1)}{2\sqrt{x^3 + 1}}\) \(= \frac{9x^4 + 12x^4 + 12x}{2\sqrt{x^3 + 1}} = \frac{21x^4 + 12x}{2\sqrt{x^3 + 1}}\) Factor out 3x: \(\frac{dy}{dx} = \frac{3x(7x^3 + 4)}{2\sqrt{x^3 + 1}}\)

The student calculated the mean as 45.3 for 5 numbers. Therefore, the **wrong total (sum) he used was: 45.3 × 5 = 226.5 He discovered that his total was short by 20.5. This means the actual (correct) sum is: 226.5 + 20.5 = 247 Correct mean = Correct sum ÷ Number of numbers 247 ÷ 5 = 49.4

The probabilities of the three boys winning are independent and are given as: - Boy A: \( \frac{1}{2} \), - Boy B: \( \frac{1}{3} \), - Boy C: \( \frac{1}{4} \) "One and only one wins" means exactly one boy wins (the other two lose). The three possible cases are: A wins, B loses, C loses \( P = \frac{1}{2} \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \) B wins, A loses, C loses \( P = \left(1 - \frac{1}{2}\right) \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} \) C wins, A loses, B loses \( P = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \frac{1}{4} = \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{2}{24} \) Add them up: \( \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24} \)

Q + 2P = Q + 2(Q + P - Q) = 2(Q + P) - Q Q + P = \(\begin{pmatrix} 7 & -2 \\ 6 & 8 \end{pmatrix}\) 2(Q + P) = \(\begin{pmatrix} 14 & -4 \\ 12 & 16 \end{pmatrix}\) Q + 2P = 2(Q + P) - Q = \(\begin{pmatrix} 14-6 & -4-0 \\ 12-4 & 16-5 \end{pmatrix} = \begin{pmatrix} 8 & -4 \\ 8 & 11 \end{pmatrix}\) Determinant: |Q + 2P| = 8 \(\times\) 11 - (-4)\(\times\) 8 = 88 + 32 = 120.

Numbers: 3, 6, x, 7, 5 Mean = \(\frac{(21 + x)}{5}\) Standard deviation = 2, so the sum of squared deviations = 20. Solving: 2x² - 21x + 27 = 0 x = 9 or x = 1.5 ≈ 2 The **least possible value of x is 1.5

\DeltaABE and \DeltaACD are similar therefore \(\frac{ x}{ x + 4} = \frac{3}{6}\) 6x = 3(x+4) 6x = 3x + 12 3x = 12 x = 4 In \DeltaACD; \(L^2\) = \((4+4)^2 + 6^2\) = \(8^2 + 6^2\) = 64 + 36 = 100 L = \(\sqrt{100}\) = 10cm In ABE, \((AE)^2 = 4^2 + 3^2\) = 16 + 9 = 25 AE = 5cm ∴L = 10 - 5 = 5cm