0.20 mm/s
0.08 mm/s
0.25 mm/s
0.05 mm/s
Correct answer is B
Given: \(\frac{\mathrm d A}{\mathrm d t} = 4 mm^{2}/s\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
\(A = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(\implies 4 = 2\pi r \times \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d r}{\mathrm d t} = \frac{4}{2\pi r} = \frac{4 \times 7}{2 \times 22 \times 8}\)
= \(0.07954 mm/s \approxeq 0.08 mm/s\)
GH¢ 5.65
GH¢ 5.66
GH¢ 6.5
GH¢ 6.56
Correct answer is B
| \(x\) | \(x - \bar{x}\) | \((x - \bar{x})^{2}\) |
|
26.00 |
-6 | 36 |
| 39.00 | 7 | 49 |
| 33.00 | 1 | 1 |
| 25.00 | -7 | 49 |
| 37.00 | 5 | 25 |
| \(\sum (x - \bar{x})^{2}\)=160 |
\(S.D = \sqrt{\frac{(x - \bar{x})^{2}}{n}}\)
\(S.D = \sqrt{\frac{160}{5}} = \sqrt{32}\)
= \(GH¢ 5.656 \approxeq GH¢ 5.66\)
13N
43N
57N
95N
Correct answer is A
Given a force F, the horizontal component = \(F\cos \theta\)
R = \(-50\cos 30 + 80\cos 45\)
= \(-50(\frac{\sqrt{3}}{2}) + 80(\frac{\sqrt{2}}{2})\)
= \( -25\sqrt{3} + 40\sqrt{2} = -43.30 + 56.67 \)
= \(13.37N \approxeq 13N\)
35
30
18
12
Correct answer is C
Since Ozo is a boy and must be in the sub-committee, we have 2 spaces for the boys and 2 for the girls.
= \(^{4}C_{2} \times ^{3}C_{2} \)
= \(\frac{4!}{(4 - 2)! 2!} \times \frac{3!}{(3 - 2)! 2!} = 6 \times 3 = 18\)
28
26
-26
-28
Correct answer is A
\(f(4) = 3(4) - 2 = 12 - 2 = 10\)
\(f(-3) = 5(-3) - 3 = -15 - 3 = -18\)
\(f(4) - f(-3) = 10 - (-18) = 28\)
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