| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.90 |
| Score | 0% | 58% |
What is \( \frac{10\sqrt{20}}{2\sqrt{4}} \)?
| \(\frac{1}{5}\) \( \sqrt{5} \) | |
| \(\frac{1}{5}\) \( \sqrt{\frac{1}{5}} \) | |
| 5 \( \sqrt{\frac{1}{5}} \) | |
| 5 \( \sqrt{5} \) |
To divide terms with radicals, divide the coefficients and radicands separately:
\( \frac{10\sqrt{20}}{2\sqrt{4}} \)
\( \frac{10}{2} \) \( \sqrt{\frac{20}{4}} \)
5 \( \sqrt{5} \)
If the ratio of home fans to visiting fans in a crowd is 5:1 and all 50,000 seats in a stadium are filled, how many home fans are in attendance?
| 35,200 | |
| 30,000 | |
| 41,667 | |
| 24,000 |
A ratio of 5:1 means that there are 5 home fans for every one visiting fan. So, of every 6 fans, 5 are home fans and \( \frac{5}{6} \) of every fan in the stadium is a home fan:
50,000 fans x \( \frac{5}{6} \) = \( \frac{250000}{6} \) = 41,667 fans.
What is the least common multiple of 4 and 10?
| 11 | |
| 12 | |
| 19 | |
| 20 |
The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [20, 40, 60, 80] making 20 the smallest multiple 4 and 10 have in common.
What is 7\( \sqrt{9} \) x 5\( \sqrt{9} \)?
| 105\( \sqrt{9} \) | |
| 12\( \sqrt{9} \) | |
| 35\( \sqrt{9} \) | |
| 35\( \sqrt{18} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
7\( \sqrt{9} \) x 5\( \sqrt{9} \)
(7 x 5)\( \sqrt{9 \times 9} \)
35\( \sqrt{81} \)
Now we need to simplify the radical:
35\( \sqrt{81} \)
35\( \sqrt{9 \times 9} \)
35\( \sqrt{9 \times 3^2} \)
(35)(3)\( \sqrt{9} \)
105\( \sqrt{9} \)
If all of a roofing company's 6 workers are required to staff 2 roofing crews, how many workers need to be added during the busy season in order to send 6 complete crews out on jobs?
| 13 | |
| 10 | |
| 12 | |
| 9 |
In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 6 workers at the company now and that's enough to staff 2 crews so there are \( \frac{6}{2} \) = 3 workers on a crew. 6 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 6 x 3 = 18 total workers to staff the crews during the busy season. The company already employs 6 workers so they need to add 18 - 6 = 12 new staff for the busy season.