| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.71 |
| Score | 0% | 54% |
What is \( 5 \)\( \sqrt{125} \) - \( 8 \)\( \sqrt{5} \)
| -3\( \sqrt{125} \) | |
| 17\( \sqrt{5} \) | |
| 40\( \sqrt{25} \) | |
| 40\( \sqrt{125} \) |
To subtract these radicals together their radicands must be the same:
5\( \sqrt{125} \) - 8\( \sqrt{5} \)
5\( \sqrt{25 \times 5} \) - 8\( \sqrt{5} \)
5\( \sqrt{5^2 \times 5} \) - 8\( \sqrt{5} \)
(5)(5)\( \sqrt{5} \) - 8\( \sqrt{5} \)
25\( \sqrt{5} \) - 8\( \sqrt{5} \)
Now that the radicands are identical, you can subtract them:
25\( \sqrt{5} \) - 8\( \sqrt{5} \)In a class of 26 students, 9 are taking German and 11 are taking Spanish. Of the students studying German or Spanish, 5 are taking both courses. How many students are not enrolled in either course?
| 16 | |
| 25 | |
| 26 | |
| 11 |
The number of students taking German or Spanish is 9 + 11 = 20. Of that group of 20, 5 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 20 - 5 = 15 who are taking at least one language. 26 - 15 = 11 students who are not taking either language.
Solve 3 + (5 + 3) ÷ 4 x 4 - 32
| 1\(\frac{1}{5}\) | |
| 2 | |
| \(\frac{3}{4}\) | |
| 1 |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
3 + (5 + 3) ÷ 4 x 4 - 32
P: 3 + (8) ÷ 4 x 4 - 32
E: 3 + 8 ÷ 4 x 4 - 9
MD: 3 + \( \frac{8}{4} \) x 4 - 9
MD: 3 + \( \frac{32}{4} \) - 9
AS: \( \frac{12}{4} \) + \( \frac{32}{4} \) - 9
AS: \( \frac{44}{4} \) - 9
AS: \( \frac{44 - 36}{4} \)
\( \frac{8}{4} \)
2
What is \( \frac{4}{5} \) - \( \frac{4}{9} \)?
| \( \frac{7}{45} \) | |
| 1 \( \frac{7}{16} \) | |
| \( \frac{1}{4} \) | |
| \(\frac{16}{45}\) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{4 x 9}{5 x 9} \) - \( \frac{4 x 5}{9 x 5} \)
\( \frac{36}{45} \) - \( \frac{20}{45} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{36 - 20}{45} \) = \( \frac{16}{45} \) = \(\frac{16}{45}\)
If all of a roofing company's 6 workers are required to staff 3 roofing crews, how many workers need to be added during the busy season in order to send 5 complete crews out on jobs?
| 12 | |
| 4 | |
| 6 | |
| 11 |
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 3 crews so there are \( \frac{6}{3} \) = 2 workers on a crew. 5 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 5 x 2 = 10 total workers to staff the crews during the busy season. The company already employs 6 workers so they need to add 10 - 6 = 4 new staff for the busy season.