Evaluate, using logarithm tables \(\frac{5.34 \times 67.4}{2.7}\)
1.332
13.32
133.2
1332
13.320
Correct answer is C
| No | Log |
| 5.34 | 0.7275 + |
| 67.4 | 1.8287 |
| 2.5562 - | |
| 2.7 | 0.4314 |
| Antilog = 133.2 | 2.1248 |
\(\therefore \frac{5.34 \times 67.4}{2.7} = 133.2\)
In the diagram O is the center of the circle, if ∠QRS = 62o, find the value of ∠SQR.
14o
28o
31o
45o
90o
Correct answer is B
< QSR = 90° (angle in a semi-circle)
\(\therefore\) < SQR = 180° - (90° + 62°)
= 28°
39cm2
28cm2
24cm2
21cm2
19cm2
Correct answer is B
Area of \(\Delta\)QST = 52cm\(^2\); height = 8cm
b=base length; but A = 1/2 b x h
=52=1/2 x b x 8cm = 104 =8b
b=104/8 = 13cm; ST= 13cm; PS = 6cm
TP = 13 - 6 = 7cm;
area of \(\Delta\)QST = 1/2 b x h
= 1/2 x 7 x 8/1 = 28cm\(^2\)
28
40
48
60
68
Correct answer is D
ar = 8 .....(1)
ar3 = 32 .....(2)
divide (2) by (1)
r2 = 4 r = 2 ∴ a = 4
If 3loga + 5loga - 6loga = log64, what is a?
4
6
8
16
32
Correct answer is C
loga3 + loga5 - loga6
a2 = 64 a = 8
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