72
100
90
200
125
Correct answer is D
P \(\alpha\) \(\frac{q^2}{r}\)
P = \(\frac{kq^2}{r}\)
k = \(\frac{pr}{q^2}\)
= \(\frac{36 x 4}{(3)^2}\)
p = \(\frac{16q^2}{r}\)
= \(\frac{16 \times 25}{2}\)
= 200
Simplify \(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\)
1
6
\(\frac{1}{27}\)
\(\frac{4}{3}\)
Correct answer is C
\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\)
= \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\)
= \(\frac{1}{27}\)
Rationalize \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\)
-2\(\sqrt{35}\)
4\(\sqrt{7}\) - 6\(\sqrt{5}\)
-\(\sqrt{35}\)
4\(\sqrt{7}\) - 8\(\sqrt{5}\)
\(\sqrt{35}\)
Correct answer is C
\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\)
= \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\)
= \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\)
= \(\frac{-2\sqrt{35}}{2}\)
= - \(\sqrt{35}\)
1035
10305
1025
20025
30325
Correct answer is B
Total cost of 1035 oranges at N145 each
= 1035 x 145
= 20025
Total selling price at N245 each
= (103)5 x 245
= 30325
Hence his gain = 30325 - 20025
= 10305
-3 = \(\frac{y + 4}{x + 1}\)
4y = -3 + x
\(\frac{y}{x}\) = \(\frac{-3}{4}\)
4x = y + 3
Correct answer is D
P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4
\(\frac{x}{1}\) = \(\frac{y}{4}\)
y = 4x
Hence the sign of the graph is y + 3 = 4x