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Given: \(x + y = 90^\circ ... (1)\) \((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\) = \(\sin^{2} x + \sin^{2} y ... (2)\) Recall: \(\sin x = \cos (90 - x) ... (a)\) From (1), \(y = 90 - x ... (b)\) Putting (a) and (b) in (2), we have \(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\) = 1

His annual salary = N2000 His allowances = N600 So his taxable income = Annual salary − allowance = N2000 − N600 = N1400 He pay at 5% Then, his allowance income tax 5/100 \times 1400 = N70

Using the quadratic formula, we have \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\)

nth term of a linear sequence (AP) = a+(n − 1)d first term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 \(S_n\)= n/2(2a + (n − 1)d or Sn = \div2 (a + l) \(S_n\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\) = \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\)

\( A = P \left(1 + \frac{r}{100}\right)^n \) Where P = 126, r = 4,n = 2 A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \) =126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \) =126 \( \left(1.04^2 \right) \) = 126 * 1.04 * 1.04 =136.28 A = 136.30 (approx.) The Amount A, = N136.30k

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

log520 = x 5x = 20(Take log10 of both sides) log5x = log20 xlog5 = log20 x= [log20 ÷ log5] [1.30103 ÷ 0.69897] x = 1.861

I = A − P = N588 − N525 ∴I = N63 I = PRT ÷ 100 R = [100I ÷PT] R = [(100 × 63 ) ÷ (525 × 3)] = (6300 ÷ 1575) = 4 ∴ The rate = 4 %

{A ∪ B ∪ C} {A ∪ B} = {1, 3, 5, 7, 9, 12, 15} ∪ C={2,4,6,8} {A ∪ B} ∪ C = {1, 3, 5, 7, 9, 12, 15} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15} {A ∪ B} ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

Let the extension be E and the tension be T. Then \(E \propto T\) \(E = kT\) when T = 8N, E = 2cm \(2 = k \times 8\) \(k = \frac{2}{8} = 0.25\) \(\therefore E = 0.25T\) when T = 12N, \(E = 0.25 \times 12 = 3cm\)

x2 − 2x − 15 (x2 − 5x) + (3x − 15) x(x − 5)+ 3(x − 5) (x − 5)(x + 3)

The coordinate of the midpoint is \((\frac{-3 + 4}{2}, \frac{5 + (-1)}{2})\) = \((\frac{1}{2}, 2)\)

I =[ PRT ÷ 100] [(325 × 5 × 3) ÷ (100)] N(195 ÷ 4) N48.75K

(0.09)2 = 0.09 × 0.09 = 0.0081 = 0.008100 to 4 significant figures Please, start counting first from the non-zero digits i.e. 8