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At resonance, inductive reactant XL is equal to the capacitive reactance XC i.e. XL = XC and XL = VL/IL and XC = VC/IC Again since the components are in series, the same current flows in the series arrangment i.e. IL = IC at resonance, VL/IL = VC/IC ∴ VL = VC = 40V

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

2311Na + Proton → pqX + alpha particle Represent; 2311Na + 11P → pqX + 42He Thus from the above equations, p = 20 and q = 10

A rectifier of an alternating current is a device which allows the flow of current only in one direction

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

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

With P1 and P2 as shown (make a V.R of 2), the weight W = 40N * 2 = 80N

Voltage across 5\Omega resistor = IR = 2 \times 5 = 10V Since voltage in a parallel connection is the same, voltage across 10\Omega is also 10V Current flowing across the 10\Omega resistor I = V/R = 10/10 Current I= 1A Total current flowing = 2 + 1 = 3A Voltage across 20\Omega resistor = IR = 3 \times 20 = 60V

V = -6cm M = 0.6cm Recall m = \(\frac{V}{U}\) U = \(\frac{V}{M}\) U = \(\frac{6}{0.6}\) U = 10 \(\frac{1}{U}\) + \(\frac{1}{V}\) = \(\frac{1}{F}\) \(\frac{1}{10}\) + \(\frac{1}{-6}\) = \(\frac{1}{F}\) \(\frac{-4}{60}\) = \(\frac{1}{F}\) F = \(\frac{60}{-4}\) = -15cm

The frequency f = V/\lambda, Where \lambda = wavelength From the above, the distance OA = one wavelength; AB = one wave length and BC = one wavelength. => OC = 3 wavelength = 3\lambda ∴3\lambda = 0.3M \lambda = 0.3/3 = 0.1m ∴ f = V/\lambda = (3.0 x 108) / 0.1 = 3.0 x 109Hz