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option A CANNOT BE BECAUSE THE LAST NUMBER BEFORE 1 CAN ONLY BE ROUNDED DOWN TO ZERO.

\(\frac{3^{(1-n)}}{9^{-2n}}\) =\(\frac{1}{9}\) 3\(^{1-n}\times 3^{-2(-2n)} = 3^{-2}\) 1 - n - 2(-2n) = -2 1 - n + 4n = -2 n = -1

Using the formula, \((n - 2) \times 180^\circ\) to get the sum of the interior angles. Then we can have \((n - 2) \times 180^\circ = 108n\) ... (1) \((n - 2) \times 180^\circ = 116n\) ... (2) \((n - 2) \times 180^\circ = 120n\) ... (3) Solving the above given equations, where n is not a positive integer then that angle is not the interior for a regular polygon. (1): \(180n - 360 = 108n \implies 72n = 360\) \(n = 5\) (regular pentagon) (2): \(180n - 360 = 116n \implies 64n = 360\) \(n = 5.625\) (3): \(180n - 360 = 120n \implies 60n = 360\) \(n = 6\) (regular hexagon) Hence, 116^\circ is not an angle of a regular polygon.

\(sin \theta = \frac{opp}{hyp}\) sin 60^o = \(\frac{|YZ|}{|XZ|}=\frac{6}{P}\) P sin 60^o = 6 P = \(\frac{6}{sin60^o}\) =\(\frac{6}{\sqrt{\frac{3}{2}}}=4\sqrt{3}\)

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

P \(\cap\) Q = {f, e} = 2

Let the number be x. \(\sqrt{x} = 2k \implies x = (2k)^2\) = \(4k^2\) \(\frac{1}{2} \times 4k^2 = 2k^2\)

\(p \propto q^2\) \(q\propto\frac{1}{\sqrt{r}\) \(p = kq^2\) \(q = \frac{c}{\sqrt{r}}\) where c and k are constants. \(q^2 = \frac{c^2}{r}\) \(p = \frac{kc^2}{r}\) If k and c are constants, then kc\(^2\) is also a constant, say z. \(p = \frac{z}{r}\) p varies inversely as r.

Prob (passing English) = x Prob (passing Maths) = Y Prob (failing English) = 1 - x Prob (failing Maths) = 1 - y Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y) =1 - y - x + xy =1 - (y + x) + xy

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

\(x = -\frac{2}{3} \implies x + \frac{2}{3} = 0\) \(x = 3 \implies x - 3 = 0\) \(\implies (x - 3)(x + \frac{2}{3}) = 0\) \(x^2 - 3x + \frac{2}{3}x - 2 = 0\) \(x^2 - \frac{7}{3}x - 2 = 0\) \(3x^2 - 7x - 6 = 0\)

\(cos45^o \times cos30^o - sin45^o \times sin30^o\) \(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\) \(\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

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

Considering the triangle in the diagram \(tan\theta = \frac{opp}{hyp}\) tan 63\(^{\circ} = \frac{h}{4}\) h = 4 tan 63\(^{\circ}\) Area of trapezium = \(\frac{1}{2}h(a+b)\) \(\frac{1}{2} \times 4 tan 63^{\circ}[5+9]\right)\) = 28\(\times 1.963 = 54.96 = 55cm^2\)

Volume of a cone = \(\frac{\pi r^2 h}{3}\) r = 2h. V = \(\frac{\pi \times (2h)^2 \times h}{3}\) = \(\frac{4}{3} \pi h^3\)