If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.
2
3
4
5
Correct answer is C
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10
\(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10
\(\frac{6x - 4}{2}\) = 10
6x - 4 = 2 x 10
= 20
6x = 20 + 4
6x = 20
x = \(\frac{24}{6}\)
x = 4
40.0%
42.2%
50.0%
52.5%
Correct answer is D
Population of school = 250 + 150 = 400
60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150
40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60
Total number of students who plays football;
150 + 60 = 210
Percentage of school that play football;
\(\frac{210}{400}\) x 100% = 52.5%
Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)
\(\frac{5}{9}\)
1\(\frac{1}{5}\)
1\(\frac{1}{4}\)
1\(\frac{4}{5}\)
Correct answer is D
2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)
= \(\frac{9}{4} \times \frac{7}{2} \div \frac{35}{8}\)
= \(\frac{9}{4} \times \frac{7}{2} \div \frac{8}{35}\)
= \(\frac{9}{5}\)
= 1 \(\frac{4}{5}\)
Find the value of x for which \(32_{four} = 22_x\)
three
five
six
seven
Correct answer is C
\(32_4 = 22_x\)
\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)
12 + 2 x 1 = 2x + 2 x 1
14 = 2x + 2
14 - 2 = 2x
12 = 2x
x = \(\frac{12}{2}\)
x = 6
Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)
121
144
169
196
Correct answer is C
\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)
\([8 + 5]^2\) = \([13]^2\)
= 169