Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\) 3 + \(\frac{2}{3} \log_{10} 27\)
3 \(\log_{10}^2\)
\(\log_{10}^2\)
\(\log_{10}^3\)
2 \(\log_{10}^3\)
Correct answer is B
log\(_{10}\) 6 - log\(_{10}\)3\(^3\) + log\(_{10}\) (\(\sqrt[3]{27}\))\(^2\)
= log \(_{10}\) 6 - log \(_{10}\) 27 + log\(_{10}\) 9
= log\(_{10}\) \(\frac{6 \times 9}{27}\)
= log\(_{10}\)2
Solve 4x^{2}\) - 16x + 15 = 0.
x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\)
x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)
x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\)
x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\)
Correct answer is B
4x\(^2\) - 16x + 15 = 0
(2x - 3)(2x - 5) = 0
x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)
H = \(\frac{p}{4y^2}\)
H = \(\frac{2p}{y^2}\)
H = \(\frac{p}{2y^2}\)
H = \(\frac{p}{y^2}\)
Correct answer is C
H \(\propto\) \(\frac{p}{y^2}\)
H = \(\frac{pk}{y^2}\)
1 = \(\frac{8k}{2^2}\)
k = \(\frac{4}{8}\)
= \(\frac{1}{2}\)
H = \(\frac{p}{2y^2}\)
If m : n = 2 : 1, evaluate \(\frac{3m^2 - 2n^2}{m^2 + mn}\)
\(\frac{4}{3}\)
\(\frac{5}{3}\)
\(\frac{3}{4}\)
\(\frac{3}{5}\)
Correct answer is B
m = 2, n = 1
\(\frac{3m^2 - 2n^2}{m^2 _ mn}\)
= \(\frac{3(2)^2 - 2(1)^2}{2^2 + 2(1)}\)
= \(\frac{12 - 2}{4 + 2} = \frac{10}{6}\)
= \(\frac{5}{3}\)
Evaluate: 2\(\sqrt{28} - 3\sqrt{50} + \sqrt{72}\)
4\(\sqrt{7} - 21 \sqrt{2}\)
4\(\sqrt{7} - 11 \sqrt{2}\)
4\(\sqrt{7} - 9 \sqrt{2}\)
4\(\sqrt{7} + \sqrt{2}\)
Correct answer is C
2\(\sqrt{28} - 3\sqrt{50} + \sqrt{22}\)
4\(\sqrt{7} - 15\sqrt{2} + 6\sqrt{2}\)
6\(\sqrt{7} - 9\sqrt{2}\)