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\(0.5x + 2.6 = 5x + 0.35\\ 0.5x - 5x = 0.35-2.6\\ -4.5x = -2.25\\ x = \frac{-2.25}{-4.5}\\ 0.5\)

Sum of exterior angle of any polygon is 360o (2x+5)o + 2xo + (x-20)o + xo + (3x+10)o + (x + 15)o = 360o; 10x = 350 x = 35

\(M5_{ten} = 1001011_{two}\\ =1 \times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 \\ =64+8+2+1=75_{ten}\\ ∴ m = 7 \)

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

Draw a straight horizontal line across the y-axis at y = 3\(\frac{1}{3}\). let the line cut through the curve. the points where the horizontal line meets the curve, trace it down to the x-axis to obtain the values of x when y = 3\(\frac{1}{3}\) take only the positive value of x.

The longest side of a right angle triangle is the hypotenuse. \((x + 2)^{2} = x^{2} + (x + 1)^{2}\) \(x^2 + 4x + 4 = x^2 + x^2 + 2x + 1\) \(2x^{2} - x^{2} + 2x - 4x + 1 - 4 = 0\) \(x^{2} - 2x - 3 = 0\) \(x^{2} - 3x + x - 3 = 0 \implies x(x - 3) + 1(x - 3) = 0\) \((x - 3)(x + 1) = 0 \implies \text{x = 3 or -1}\) \(x > 0 \implies x = 3\) The longest side = 3 + 2 = 5.

\(log2 = 0.3010\hspace{1mm}given\\ log2^y = 1.8062\\ ∴ ylog2 = 1.8062\\ y=\frac{1.8062}{logy}=\frac{1.8062}{0.3010}=6\)

ZY\(^2\) = 12\(^2\) + 6\(^2\) ZY\(^2\) = 144 + 36 = 180 ZY = \(\sqrt{180}\) = \(13.416\) = 13.4 km

20 members in the ratio 3:1 Number of women = \(\frac{1}{4} \times 20\) = 5 Let the number of women to be added = x Total number of members in the committee = 20 + x \(\frac{5 + x}{20 + x} = \frac{2}{5}\) \(5(5 + x) = 2(20 + x)\) \(25 + 5x = 40 + 2x \implies 5x - 2x = 40 - 25\) \(3x = 15 \implies x = 5\)

\((x+3) \propto y\\ ∴x + 3 = ky\hspace{1mm}when\hspace{1mm}x = 3, y = 12\\ 3+3 = 12k\\ ∴ k = \frac{1}{2}\\ \Longrightarrow x + 3 = \frac{1}{2}y\) to find x when y = 8 \(x + 3 = \frac{1}{2}\times 8\\ x=4-3\\ x = 1\)

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

\(T_{1} = 4; T_2 = 10; T_3 = 16\) \(T_{2} - T_1 = T_3 - T_1 = 6\) \(T_n = a + (n - 1) d\) = \(4 + (n - 1) \times 6\) = \(4 + 6n - 6\) = \(6n - 2\) = 2(3n - 1)

If (-3, -4) is a point on the line then -4 = -3m + 2 -4 - 2 = -3m -6 = -3m m = 2

From the graph, the zeros of the equation exist at x = -2 and x = 3 \(\therefore\) (x + 2) = 0 and (x - 3) = 0 \(\implies (x + 2)(x - 3) = 0\) \(x^2 - 3x + 2x - 6 = 0\) \(x^2 - x - 6 = 0\) is the equation represented on the graph.

< XYQ = 180^\circ - (80^\circ + 56^\circ) = 44^\circ < PYQ = 56^\circ (alternate angles, XP||YQ) < QPY = 90^\circ < PQY = 180^\circ - (90^\circ + 56^\circ) = 34^\circ