u = \(\frac{8}{v^3}\)
v = \(\frac{8}{u^2v^3}\)
u = 8v3
v = 8u2
Correct answer is A
W \(\alpha\) \(\frac{1}{v}\)u \(\alpha\) w3
w = \(\frac{k1}{v}\)
u = k2w3
u = k2(\(\frac{k1}{v}\))3
= \(\frac{k_2k_1^2}{v^3}\)
k = k2k1k2
u = \(\frac{k}{v^3}\)
k = uv3
= (1)(2)3
= 8
u = \(\frac{8}{v^3}\)
Simplify log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7
1
\(\frac{7}{6}\)log10 a
zero
10
Correct answer is C
log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\)
= log10 a\(\frac{7}{12}\) - log10 a\(\frac{7}{12}\)
= log10 1 = 0
Find x if (x\(_4\))\(^2\) = 100100\(_2\)
6
12
100
210
110
Correct answer is B
x\(_4\) = x \(\times\) 4\(^0\) = x in base 10.
100100\(_2\) = 1 x 2\(^5\) + 1 x 2\(^2\)
= 32 + 4
= 36 in base 10
\(\implies\) x\(^2\) = 36
x = 6 in base 10.
Convert 6 to a number in base 4.
| 4 | 6 |
| 4 | 1 r 2 |
| 0 r 1 |
= 12\(_4\)
47
50
48\(\frac{1}{2}\)
48
49
Correct answer is C
By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median = \(\frac{47 + 50}{2}\)
\(\frac{97}{2}\)
= 48\(\frac{1}{2}\)
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x
2
6
18
12
5
Correct answer is D
\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)
\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\)
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
The sides cannot be 0 hence x = 12.