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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

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

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

The overall order of a reaction is the sum of the exponents in the rate law. The rate equation is \(\frac{-\Deltax}{\Deltat}\) = K[X]\(^2\)[Y]. The exponent for [X] = 2 The exponent for [Y] = 1 The overall order is the sum of the exponents = 2 + 1 = 3

Chlorine and hydrogen do not react vigorously without heating unless exposed to sunlight or a flame. The reaction is spontaneous and explosive when exposed to these conditions. However, chlorine, bromine, and iodine all dissolve in alkali to form alkali metal halides. Alkali metals are electropositive elements that bond ionically with halogens, which are electronegative elements.

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.