| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.95 |
| Score | 0% | 59% |
What is \( \frac{3}{7} \) ÷ \( \frac{4}{5} \)?
| \(\frac{1}{15}\) | |
| \(\frac{1}{36}\) | |
| \(\frac{15}{28}\) | |
| 2\(\frac{1}{7}\) |
To divide fractions, invert the second fraction and then multiply:
\( \frac{3}{7} \) ÷ \( \frac{4}{5} \) = \( \frac{3}{7} \) x \( \frac{5}{4} \)
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{3}{7} \) x \( \frac{5}{4} \) = \( \frac{3 x 5}{7 x 4} \) = \( \frac{15}{28} \) = \(\frac{15}{28}\)
If a mayor is elected with 80% of the votes cast and 79% of a town's 34,000 voters cast a vote, how many votes did the mayor receive?
| 19,339 | |
| 15,847 | |
| 21,488 | |
| 16,653 |
If 79% of the town's 34,000 voters cast ballots the number of votes cast is:
(\( \frac{79}{100} \)) x 34,000 = \( \frac{2,686,000}{100} \) = 26,860
The mayor got 80% of the votes cast which is:
(\( \frac{80}{100} \)) x 26,860 = \( \frac{2,148,800}{100} \) = 21,488 votes.
Which of these numbers is a factor of 48?
| 12 | |
| 43 | |
| 32 | |
| 10 |
The factors of a number are all positive integers that divide evenly into the number. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
What is \( 7 \)\( \sqrt{32} \) - \( 6 \)\( \sqrt{2} \)
| \( \sqrt{64} \) | |
| 42\( \sqrt{64} \) | |
| 42\( \sqrt{16} \) | |
| 22\( \sqrt{2} \) |
To subtract these radicals together their radicands must be the same:
7\( \sqrt{32} \) - 6\( \sqrt{2} \)
7\( \sqrt{16 \times 2} \) - 6\( \sqrt{2} \)
7\( \sqrt{4^2 \times 2} \) - 6\( \sqrt{2} \)
(7)(4)\( \sqrt{2} \) - 6\( \sqrt{2} \)
28\( \sqrt{2} \) - 6\( \sqrt{2} \)
Now that the radicands are identical, you can subtract them:
28\( \sqrt{2} \) - 6\( \sqrt{2} \)Convert y-2 to remove the negative exponent.
| \( \frac{1}{y^2} \) | |
| \( \frac{-1}{-2y^{2}} \) | |
| \( \frac{-2}{y} \) | |
| \( \frac{2}{y} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.