Question Library

tan\(\theta\) = 9/20 = 0.45 \(\theta\) = tan\(^{-1}\) 0.45 = 24.2^\circ the bearing of y from x = 180 0 - 0 = 180^\circ - 24.2^\circ the bearing = 155.8^\circ \(\approxeq\) 156^\circ

In \(\Delta DOB\), let < DOB = \(\alpha\) In \(\Delta DOB\), \(5^2 = 3^2 + s^2\) \(s^2 = 25 - 9 = 16\) \(s = 4cm\) \(\sin \alpha = \frac{4}{5}\) \(\alpha = \frac{< AOB}{2}\) Length of chord = \(2r \sin (\frac{\theta}{2})\) |OB| = r = 5cm L = \(2(5)(\frac{4}{5})\) = 8 cm

since it is an open cylinder; r= 3.5cm = 7/2cm h= 8cm; S.A = 2πrh + πr 2 = πr(2h + r) (16 + 3.5); 11 x 19.5 = 214.5cm2

6x\(^2\) - 7x - 5 = 0 6x\(^2\) - 10x + 3x - 5 = 0 2x(3x - 5) + 1(3x - 5) = 0 (2x + 1)(3x - 5) = 0 x = \(\frac{5}{3}\) or x = \(-\frac{1}{2}\)

Total population = 1400 + 550 + 200 + 150 + 400 = 2700 grown up girls = 200 = \(\frac {200}{2700}\) x \(\frac {360}{1}\) =80/3 = 26.7^\circ

Married = 550 + 400 = 950 Married% = \(\frac {950}{2700}\) x \(\frac {100}{1}\) =35.2%

A = {a, b, c}; B = {a, b, c, d, e}; C = {a, b, c, d, e, f} (A \(\cup\) B) = {a, b, c, d, e} (A \(\cup\) C) = {a, b, c, d, e, f} (A \(\cup\) B) \(\cap\) (A \(\cup\) C) = {a, b, c, d, e}

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

If the data is arranged either in ascending or descending order, the 15th and 16th scores are in the middle, which is 5 and 5, respectively median = \(\frac{5 + 5}{2}\) = 5 since n is even i.e, 30

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

< PBQ = 180^\circ - 75^\circ (opp angles of a cyclic quadrilateral) = 105^\circ

\(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) = \(\frac{3^{-1}}{3^2}\) = \(\frac{1}{27}\) In full: To simplify the expression : \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) We start by rewriting the bases in terms of powers of 3. We know that: \(9 = 3^2 \quad \text{and} \quad 27 = 3^3\) Thus, we can express \(9^{-\frac{1}{2}}\) as follows: \(9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1}\) Next, we rewrite \(27^{\frac{2}{3}}\): \(27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{2}\) Substitute back into the expression: Now we substitute these values back into the original expression: \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{3^{-1}}{3^{2}}\) This simplifies to: \(\frac{1/3}{9} = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}\) Thus, the simplified form of \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) is \(\frac{1}{27}\).

< ABC = 80^\circ (alternate angles) < ACB = 180^\circ - (80^\circ + 40^\circ) = 60^\circ

from the graph, the solution sets represented by OPQ is the shaded region where y \le x, y \ge 0, and y + x \le 7

The volume of a cone = \(\frac{1}{3}\) x \(\pi\) x r\(^2\) x h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 12 = 154cm\(^3\)