| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.84 |
| Score | 0% | 57% |
What is 8\( \sqrt{2} \) x 6\( \sqrt{9} \)?
| 14\( \sqrt{18} \) | |
| 48\( \sqrt{9} \) | |
| 144\( \sqrt{2} \) | |
| 48\( \sqrt{2} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
8\( \sqrt{2} \) x 6\( \sqrt{9} \)
(8 x 6)\( \sqrt{2 \times 9} \)
48\( \sqrt{18} \)
Now we need to simplify the radical:
48\( \sqrt{18} \)
48\( \sqrt{2 \times 9} \)
48\( \sqrt{2 \times 3^2} \)
(48)(3)\( \sqrt{2} \)
144\( \sqrt{2} \)
What is \( \frac{3}{3} \) + \( \frac{6}{7} \)?
| 2 \( \frac{6}{10} \) | |
| 2 \( \frac{2}{21} \) | |
| \( \frac{7}{21} \) | |
| 1\(\frac{6}{7}\) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 7 are [7, 14, 21, 28, 35, 42, 49, 56, 63, 70]. The first few multiples they share are [21, 42, 63, 84] making 21 the smallest multiple 3 and 7 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{3 x 7}{3 x 7} \) + \( \frac{6 x 3}{7 x 3} \)
\( \frac{21}{21} \) + \( \frac{18}{21} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{21 + 18}{21} \) = \( \frac{39}{21} \) = 1\(\frac{6}{7}\)
A bread recipe calls for 2\(\frac{3}{8}\) cups of flour. If you only have 1\(\frac{3}{4}\) cups, how much more flour is needed?
| 1\(\frac{5}{8}\) cups | |
| 1\(\frac{1}{4}\) cups | |
| \(\frac{5}{8}\) cups | |
| 2\(\frac{5}{8}\) cups |
The amount of flour you need is (2\(\frac{3}{8}\) - 1\(\frac{3}{4}\)) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{19}{8} \) - \( \frac{14}{8} \)) cups
\( \frac{5}{8} \) cups
\(\frac{5}{8}\) cups
If all of a roofing company's 15 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 10 complete crews out on jobs?
| 8 | |
| 15 | |
| 7 | |
| 4 |
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 15 workers at the company now and that's enough to staff 5 crews so there are \( \frac{15}{5} \) = 3 workers on a crew. 10 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 10 x 3 = 30 total workers to staff the crews during the busy season. The company already employs 15 workers so they need to add 30 - 15 = 15 new staff for the busy season.
Solve for \( \frac{5!}{6!} \)
| 7 | |
| \( \frac{1}{120} \) | |
| \( \frac{1}{6} \) | |
| \( \frac{1}{20} \) |
A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:
\( \frac{5!}{6!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6} \)
\( \frac{1}{6} \)