N68.50
N63.00
N60.00
N52.00
Correct answer is B
C = a + kS. If C = 74, S = 20
C = 96, S = 30, C= ? S = 15
74 = a + 20k......(1)
96 = a + 30k......(2)
subtract equation (1) from (2)
96 = a + 30k
-
74 = a + 20k
--------------
22 = 10k
k = 2.2
find a
74 = a + 44
a = 30
C = 30 + 2.2S
when S = 15, C = 30 + 2.2 x 15
= 30 + 33
= N63
Make R the subject of the fomula S = \(\sqrt{\frac{2R + T}{2RT}}\)
R = \(\frac{T}{(TS^2 + 1)}\)
R = \(\frac{T}{2(TS^2 - 2)}\)
R = \(\frac{T}{2(TS^2 + 1)}\)
R = \(\frac{R}{2(TS^2 + 1)}\)
Correct answer is B
S = \(\sqrt{\frac{2R + T}{2RT}}\)
Squaring both sides,
\(S^{2} = \frac{2R + T}{2RT}\)
\(S^{2} (2RT) = 2R + T\)
\(2S^{2} RT - 2R = T\)
\(R = \frac{T}{2TS^{2} - 2}\)
= \(\frac{T}{2(TS^{2} - 1)}
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ?
-1 and 0
-1 and 1
1 and 3
0 and 1
Correct answer is D
\(9^{y} - 4 \times 3^{y} + 3 = 0\)
\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\)
\((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\)
Let \(3^{y}\) be r. Then,
\(r^{2} - 4r + 3 = 0\)
Solving the equation,
\(r^{2} - 3r - r + 3 = 0\)
\(r(r - 3) - 1(r - 3) = 0\)
\((r - 3)(r - 1) = 0\)
\(\therefore \text{r = 3 or 1}\)
Recall, \(3^{y} = r\)
\(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\)
\(\implies \text{y = 1 or 0}\)
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\)
(\(\frac{3}{q}\))3
(\(\frac{q}{3}\))\(\frac{1}{3}\)
(\(\frac{q}{3}\))3
(\(\frac{3}{q}\))\(\frac{1}{3}\)
Correct answer is A
\(\log_{3} p + 3\log_{3} q = 3\)
\(\log_{3} p + \log_{3} q^{3} = 3\)
\(\implies \log_{3} (pq^{3}) = 3\)
\(pq^{3} = 3^{3} = 27\)
\(\therefore p = \frac{27}{q^{3}}\)
= \((\frac{3}{q})^{3}\)
Simplify \(4 - \frac{1}{2 - \sqrt{3}}\)
2\(\sqrt{3}\)
-2 - \(\sqrt{3}\)
-2 + \(\sqrt{3}\)
2 - \(\sqrt{3}\)
Correct answer is D
\(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\)
= \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\)
= \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\)
Rationalizing,
\((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\)
= \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\)
= \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)