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Circumstances of latitude 0\(^o\)s where R is the radius of the earth 2\(\pi\)r cos \(\theta\)

ABCD is the floor. By pathagoras AC\(^2\) = 144 + 81 = \(\sqrt{225}\) AC = 15cm Height of room 8m, diagonal of floor is 15m Therefore, the cosine of the angle which a diagonal of the room makes with the floor is EC\(^2\) = 15\(^2\) + 8\(^2\) cosine \(\frac{adj}{Hyp} = \frac{15}{17}\) EC\(^2\) = \(\sqrt{225 + 64}\) EC = \(\sqrt{289}\) = 17

Let x represent number of goals \(\begin{array}{c|c|} x & F & Frequency(fx)\\\hline 0 & 4 & 0 \\\hline 1 & 3 & 3\\\hline 2 & 15 & 30 \\\hline 3 & 16 & 48\\ \hline 4 & 1 & 4 \\\hline 5 & 0 & 0 \\\hline 6 & 1 & 6 \\\hline & \sum Fx = 91& \end{array}\) \(\sum fx\) = 91

No of students offering physics are \(\frac{12}{360}\) x 150 = 5

log\(_{10}\)2 = 0.3010 and log\(_{10}\)3 = 04771 log\(_{10} 4.5 = log_{10}\) (\(\frac{3 \times 3}{2}\)) log\(_{10}\) 3 + log\(_{10}\) 3 - log\(_{10}\)2 = 0.4471 + 0.771 - 0.3010 = 0.6532

234 - 4x\(^2\) = 18\(^2\) - (2x)\(^2\) = (18 - 2x)(18 + 2x) 2x + 18 = 2x + 18 = (2x + 18) 18 - 2x = 2(9 - x) or -2(x - 9)

Rice = 75g, Margarine = 40g, Meat = 105g Bread = 20g Total = 240 Angle of sector represented by meat = \(\frac{105}{240} \times \frac{360^o}{1}\) = 157.5

The lenght of Arc AB = \(\frac{Q}{360}\) 2\(\pi\)r = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5 = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm

\(\frac{27}{5} \times 3^{-3} \times \frac{(1)^{-1}}{5}\) = \(\frac{27}{5} \times \frac{1}{3^3} \times \frac{1}{\frac{1}{5}}\) \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{5}{1}\) = 1

Coca-Cola = 10 bottles, Fanta = 8 bottles, Spirite = 6 bottles Total = 24 P(Coca-Cola) = \(\frac{10}{24}\); P(not Coca-Cola) 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24} = \frac{14}{24} = \frac{7}{12}\)

Since PQRS is quadrilateral 2y + 2x + QPS = 360\(^o\) i.e. (y + x) + QPS = 360\(^o\) QPS = 360\(^o\) - 2 (y + x) But x + y + 68\(^o\) = 180\(^o\) There; x + y = 180\(^o\) - 68\(^o\) = 112\(^o\) QPS = 360 - 2(112\(^o\)) = 360\(^o\) - 224 = 136\(^o\)

Equation through (5,7) parallel to the line 7x + 5y = 12 5y = 7x + 12 y = \(\frac{-7x}{5} + \frac{12}{5}\) Gradient = \(\frac{-7}{5}\) Required equation = \(\frac{y - 7}{x - 5} = \frac{-7}{5}\) i.e. 5y - 35 = -7x + 35 5y + 7x = 70

Principal = N225.00, interest = N27.00 Year = x, Rate = 4% 1 = \(\frac{PRT}{100}\) 27 = \(\frac{225 \times 4 \times x}{100}\) = 2700 = 900T T = \(\frac{2700}{900}\) = 3 years

x x - x (x - x)\(^2\) 7 8 9 10 11 12 13 -3 -2 -1 0 1 2 3 9 4 1 1 0 4 9 -------------- 28 S.D = \(\frac{\sqrt\sum (x - x)^2}{N} = \frac{\sum d^2}{N} = \frac{\sqrt{28}}{7}\) = \(\sqrt{4}\) = 2

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers have only two factors, itself and 1. The prime numbers among the group are 23, 29. The probability of choosing a prime number = \(\frac{\text{Number of Prime}}{\text{No. of Total Possible Outcome}}\) = \(\frac{2}{11}\)