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Each interior Angle of the polygon = (n-2)180_n = (6-2)180_6 = 4 x 30 = 120o i.e ∠PQR = 120o ∴ RQD = 120/2 = 60o also ∴ Δ QOR is equilateral Δ ∴ QO = 7cm ∴ PO = QO = RO ...... = 7cm radius of the circumference of the circle = 2πr = 2 x 22/7 x 7/1 = 44 cm

\(202^2_{three}\)when converted to base ten \(=(202_3)^2\) 202_3 = 2 \(\times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\) =20\(_{ten}; (202_3)^2 = (20)^2_{ten} = 400\) 112\(^2_{three}\)when converted to base ten \(= (112_3)^2\) 112_3 = 1 \(\times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\) (112_3)\(^2 = (14)^2_{ten} = 196_{ten}\) Evaluate \(\Longrightarrow 400-196 = 204\) Reconvert to base three 204 \div 3 = 68 remainder 0 68 \div 3 = 22 remainder 2 22 \div 3 = 7 remainder 1 7 \div 3 = 2 remainder 1 2 \div 3 = 0 remainder 2 = 21120\(_3\)

\(23_{five} = X_{ten}; X_{ten}\) = 2\(\times 5^1 + 3\times 5^0 = 10 + 3 = 13\) 101\(_{three}\) = P\(_{ten}\); P\(_{ten}\) = 1\(\times 3^2 + 0\times 3^1 + 1\times 3^0 = 9 + 0 + 1 = 10_{ten}\) Y = 13+10=23\(_{ten}\); Converting to base two \(\begin{matrix} 2 & 23\\ 2 & 11 &R1\\ 2 & 5 & R1\\ 2 & 2 & R1\\ 2 & 1 & R0\\ & 0& R1 \uparrow\\ \end{matrix}\) = y = 10111\(_2\)

Sin (5x – 28)o = cos (3x - 50)o Since by the trigonometry relation Sin(5x – 28)o = cos[90 – (5x – 28)]o Hence cos(3x – 50)o = cos[90 – (5x – 28)]o 3x – 50 = 90 - (5x-28) 3x – 50 = 90 – 5x + 28 3x + 5x = 90 + 28 + 50 8x = 168 \(x = \frac{168}{8}=21^{\circ}\)

\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\ \frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\ 9t + 84 – 4t = 132; 5t = 132 – 84\\ 5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)

Let the girls pocket money be rep. by x. The amount spent on books = \(\frac{1}{4}of\hspace{1mm}x = \frac{x}{4}\) The amount spent on dress \(=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}\) ∴The fraction that remains = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\ \frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (SSS Congruence) \(\frac{QR}{4} = \frac{8}{5}\) \(QR = \frac{4 \times 8}{5}\) = 6.4 cm

- Number of favourable outcomes = number of girls = 12 - Total number of possible outcomes = total students = m boys + 12 girls = m + 12 Therefore, Probability of selecting a girl = \(\dfrac{\text{Number of girls}}{\text{Total number of students}} = \dfrac{12}{m + 12}\).

Where H is the height of the tower H = ? \(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\\ H = 36 \times 0.5774 = 20.79\)

\(x + y = 7 ------ I\\ 3x – y = 5 ------ II\) Add I and II; \(4x = 12 => x = \frac{12}{4} = 3\\ Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\\ \frac{4}{2} – 3 => 2-3 = -1\)

∠PSQ = 90^\circ (angle in a semi-circle) (when SR is joined to SP) ∠SPQ = 180 – (90+35) = 180 – 125 = 55^\circ ∠QRS + ∠SPQ = 180^\circ (opposite angles in a cyclic quad is supplementary) ∠QRS = 180^\circ - 55^\circ = 125^\circ

\(2^{-2(2-y)}-x=2^{0}; -4 + 2y = 0\\ 2y = 4; y = 2\)

Total surfaced area of a cupboard = 2(lb + lh + bh) l = length h = height b = base l = 12cm h = 10cm b = 8cm T.S.A = 2(\(12 \times 8 \times + 12 \times 10 + 8 \times 10\)) = 2\((96 + 120 + 80)cm^2 = 592cm^2\)

If \(\frac{bx+y(a+1)}{b(a+1)}=1\\ bx + ya + y = ba + b; y(a+1) = ba+b-bx\\ y(a+1)=b(a+1-x); y = \frac{b(1+a-x}{a+1}\)

If \(log_q p\) = r convert from logarithm to indices P = q\(^r\)