26°
13°
80°
102°
Correct answer is D
< BAC = \(\frac{130}{2}\) (angle subtended at the centre)
< BAC = 65°
Also, x = 26° (theorem)
y = 65° - 26° = 39°
< AOC = 180° - (39° + 39°)
= 102°
If \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\), make T the subject of the formula.
\(T = \frac{15R - Q}{P^3}\)
\(T = R - \frac{15P^3}{Q}\)
\(T = \frac{R - 15P^3}{Q}\)
\(T = \frac{R + P^3}{15Q}\)
Correct answer is B
\(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\)
\(P^3 = \frac{Q(R - T)}{15}\)
\(Q(R - T) = 15P^3\)
\(R - T = \frac{15P^3}{Q}\)
\(T = R - \frac{15P^3}{Q}\)
24°
56°
127°
156°
Correct answer is D
\(\tan \theta = \frac{9}{20} = 0.45\)
\(\theta = \tan^{-1} (0.45) \)
= 24.23°
\(\therefore\) The bearing of Y from X = 180° - 24.23°
= 155.77°
= 156° (to the nearest degree)
Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30
x = 3 or 7, y = 12 or 8
x = 6 or 1, y = 11 or 5
x = 2 or 5, y = 15 or 6
x = 1 or 5, y = 10 or 7
Correct answer is C
3x + y = 21 ... (i);
xy = 30 ... (ii)
From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have
3x + \(\frac{30}{x}\) = 21
\(\implies\) 3x\(^2\) + 30 = 21x
3x\(^2\) - 21x + 30 = 0
3x\(^2\) - 15x - 6x + 30 = 0
3x(x - 5) - 6(x - 5) = 0
(3x - 6)(x - 5) = 0
3x - 6 = 0 \(\implies\) x = 2.
x - 5 = 0 \(\implies\) x = 5.
If x = 2, y = \(\frac{30}{2}\) = 15;
If x = 5, y = \(\frac{30}{5}\) = 6.
Simplify \(\frac{0.0839 \times 6.381}{5.44}\) to 2 significant figures.
0.2809
2.51
3.5
0.098
Correct answer is D
No explanation has been provided for this answer.
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