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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)

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

\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

% of those above 15 but less than 21 years age = \(\frac{25}{50}\) x \(\frac{100}{1}\) = 50%

log\(_7\) 8 - log\(_7\) 2 + log\(_7\) 4 = log\(_7\) (8/2 x 4) = log\(_7\) 16 = 4 log\(_7\) 2

\(K\sqrt{28}+\sqrt{63}-\sqrt{7}=0\\ 2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0\\ 2K\sqrt{7}=-2\sqrt{7}\\ K=\frac{-2\sqrt{7}}{2\sqrt{7}}\\ K=-1\)

\(A = {3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}}\) n(A) = 5 Let the rational nos = R n(R) = 2 (3, \(\sqrt{9}\)) P(R) = 2/5

Let x represent the total number of hours spent per week ∴ \(\frac{75}{360} \times x = \frac{5}{2}\) ∴ x =\(\frac{360 \times 5}{725 \times 2}\) =12 hours

x = 180^\circ - 123^\circ = 57^\circ z = 180^\circ - 105^\circ = 75^\circ y = 180^\circ - 57^\circ - 75^\circ = 48^\circ

Given that m = -3, n = 2, the value of \(\frac{3n^2 - 2m^3}{m}\) \(\frac{3(2)^2 -2(-3)^3}{-3} = \frac{12+54}{-3}\) = -22

\(\frac{2-18m^2}{1+3m}=\frac{2(1-(3m)^2)}{1+3m}\\ \frac{2(1-(3m)(1+3m)}{1+3m}=2(1-3m)\)

\(y = \sqrt{ax-b}\) \(y^2 = ax-b\) \(y^2 +b = ax\) \(x = \frac{y^2 + b}{a}\)

\(K = 81 \times 2^{2n - 2}\) \(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\) = \(9 \times 2^{n - 1}\)

\(\frac{4}{x+1}-\frac{3}{x-1} \\ =\frac{4x - 4 - 3x - 3}{(x+1)(x-1)}=\frac{x-7}{x^2 - 1}\)