Explanation

(a) \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\)

= \(\log_{10} (25^{\frac{1}{2}}) - \log_{10} (3^{2}) + \log_{10} 18\)

= \(\log_{10} (\frac{5 \times 18}{9})\)

= \(\log_{10} 10\)

= 1

(b) \(123_{y} = (1 \times y^{2}) + (2 \times y^{1}) + (3 \times y^{0})\)

\(y^{2} + 2y + 3 = 83\)

\(y^{2} + 2y + 3 - 83 = 0 \implies y^{2} + 2y - 80 = 0\)

Equation : \(y^{2} + 2y - 80 = 0\)

\(y^{2} - 8y + 10y - 80 = 0\)

\(y(y - 8) + 10(y - 8) = 0\)

\((y + 10)(y - 8) = 0 \implies y = -10 ; 8\)

y = 8 since y cannot be negative.

(c) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)

\(\implies 9^{2x - 3} = 3^{x + 3}\)

\((3^{2})^{2x - 3} = 3^{x + 3}\)

\(3^{4x - 6} = 3^{x + 3}\)

\(\implies 4x - 6 = x + 3\)

\(4x - x = 3 + 6 \implies 3x = 9\)

\(x = 3\)

Correct Answer: C. 77cm³

Explanation

Volume = \(\frac{1}{3} \pi r^2h = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{6}{1}\)

V = \(\frac{6468}{84}\)

= 77cm³

Explanation

(a) (i) r = 1m, h = 8m.

\(V = \pi r^{2} h = \frac{22}{7} \times 1^{2} \times 8\)

= \(\frac{176}{7}\)

= \(25.14 m^{3}\)

(ii) \(V = L \times B \times H\)

\(\frac{176}{7} = 4 \times 12 \times h\)

\(h = \frac{176}{7} \div 48\)

\(h = \frac{11}{21}m\)

= \(\frac{1100}{21} cm = 52.38 cm\)

\(\approxeq 52 cm\) (nearest whole number)

(b)(i)\(3.3^{2} + 5.6^{2} = BD^{2}\)

\(BD = \sqrt{3.3^{2} + 5.6^{2}}\)

= \(\sqrt{10.89 + 31.36}\)

= \(\sqrt{42.25} = 6.5 cm\)

(ii)

If < BCD is right- angled, then Pythagoras rule holds true.

\(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\)

= \(42.25\)

\(BD = \sqrt{42.25} = 6.5 cm\)

Hence, \(\Delta BCD\) is right-angled.

Correct Answer: C. 22

Correct Answer: A. 234.00 cm³

Correct Answer: A. 1, -2

Explanation

(a)

\(< QPB = 90° - 65° = 25°\)

Also, \(< TQP = 90° + 25° = 115°\)

\(\therefore < QTP = 180° - (25° + 115°) = 40°\)

Using sine rule, \(\frac{PQ}{\sin 40°} = \frac{PT}{\sin 115°}\)

\(PT = \frac{20 \sin 115°}{\sin 40°} = \frac{20 \sin 65°}{\sin 40°}\)

= \(28.199m \approxeq 28.2m\) (to 3 sig. fig)

(b) Height of the tower = ST

\(\sin 65 = \frac{ST}{PT} = \frac{ST}{28.199}\)

\(ST = 28.199 \times \sin 65° = 25.557m\)

\(\approxeq 25.6m\) (to 3 sig. figs)

Correct Answer: D. 115°

Explanation

ACB = ABC [Base angles of isoceles triangle]

ACB + ABC + 50^o = 180^o[sum of angles in a triangle]

ACB + ABC = 180^o - 50^o = 130^o

ACD + ACB = 180^o[sum of ?s on a straight line]

ACD + 65^o = 180^o

ACD = 180^o - 65^o = 115^o

Explanation

(a) From the figure < QRS = 180° - 56° = 124° (opposite angles of a parallelogram)

\(\therefore\) < RQS = < PQS = \(\frac{180° - 124°}{2} = 28°\) (base angles of an isosceles triangle)

\(\therefore\) < SQT = < RSQ = 28° (alternate angles PS // QT)

(ii) < PQT = 90° - 28° = 62° (< PQS is in a semi-circle)

(b)

\(\tan < BDC = \frac{30}{40} = 0.75\)

\(< BDC = \tan^{-1} (0.75) = 36.87°\)

\(\therefore < BDF = 90° - 36.87° = 53.13° \approxeq 53°\)

\(\tan < ADF = \frac{14}{30} = 0.4667\)

\(< ADF = \tan^{-1} (0.4667) = 25.02° \approxeq 25°\)

Correct Answer: D. N5