What is the number whose logarithm to base 10 is 2.3482?
223
2228
2.235
22.37
0.2229
Correct answer is A
\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)
N : b Look for the antilog of .3482 which gives 222.9
4
-4
2
-2
None of the above
Correct answer is E
x\(^2\) + 4 = 0
x\(^2\) = -4
You can't apply a square root to a negative number. So the answer is None of the above.
Find the square root of 170 - 20\(\sqrt{30}\)
2 \(\sqrt{10}\) - 5\(\sqrt{3}\)
2 \(\sqrt{5}\) - 5\(\sqrt{6}\)
5 \(\sqrt{10}\) - 2\(\sqrt{3}\)
3 \(\sqrt{5}\) - 8\(\sqrt{6}\)
Correct answer is B
\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\)
Squaring both sides,
\(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\)
Equating the rational and irrational parts, we have
\(a + b = 170 ... (1)\)
\(2 \sqrt{ab} = 20 \sqrt{30}\)
\(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \)
\(ab = 3000 ... (2)\)
From (2), \(b = \frac{3000}{a}\)
\(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\)
\(a^{2} - 170a + 3000 = 0\)
\(a^{2} - 20a - 150a + 3000 = 0\)
\(a(a - 20) - 150(a - 20) = 0\)
\(\text{a = 20 or a = 150}\)
\(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\)
\(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\)
= \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)
N = \(\frac{kp_A}{D^2} + {cp_B}{D^2}\)
N = \(\frac{k P_{A} P_{B} }{D^2}\)
N = \(\frac{kD_AP_D}{B^2}\)
N = \(\frac{kD^2_AP_D}{B}\)
Correct answer is B
\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\)
\(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\)
\(N = \frac{k P_{A} P_{B}}{D^{2}}\)
The sum of the progression is 1 + x + x2 + x3 + ......
\(\frac{1}{1 - x}\)
\(\frac{1}{1 + x}\)
\(\frac{1}{x - 1}\)
\(\frac{1}{x}\)
Correct answer is A
Sum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\)
In the given series, a (the first term) = 1 and r (the common ratio) = x.
\(S_{n} = \frac{1(1 - x^{n})}{1 - x}\)
a = 1, and as n tends to infinity
\(S_{n} = \frac{1}{1 - x}\)