| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.50 |
| Score | 0% | 50% |
What is \( 9 \)\( \sqrt{28} \) - \( 3 \)\( \sqrt{7} \)
| 27\( \sqrt{7} \) | |
| 6\( \sqrt{196} \) | |
| 6\( \sqrt{7} \) | |
| 15\( \sqrt{7} \) |
To subtract these radicals together their radicands must be the same:
9\( \sqrt{28} \) - 3\( \sqrt{7} \)
9\( \sqrt{4 \times 7} \) - 3\( \sqrt{7} \)
9\( \sqrt{2^2 \times 7} \) - 3\( \sqrt{7} \)
(9)(2)\( \sqrt{7} \) - 3\( \sqrt{7} \)
18\( \sqrt{7} \) - 3\( \sqrt{7} \)
Now that the radicands are identical, you can subtract them:
18\( \sqrt{7} \) - 3\( \sqrt{7} \)Solve for \( \frac{5!}{6!} \)
| \( \frac{1}{6} \) | |
| 336 | |
| \( \frac{1}{210} \) | |
| \( \frac{1}{4} \) |
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:
\( \frac{5!}{6!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6} \)
\( \frac{1}{6} \)
A triathlon course includes a 300m swim, a 40.3km bike ride, and a 7.6km run. What is the total length of the race course?
| 65.1km | |
| 48.2km | |
| 32.1km | |
| 56.2km |
To add these distances, they must share the same unit so first you need to first convert the swim distance from meters (m) to kilometers (km) before adding it to the bike and run distances which are already in km. To convert 300 meters to kilometers, divide the distance by 1000 to get 0.3km then add the remaining distances:
total distance = swim + bike + run
total distance = 0.3km + 40.3km + 7.6km
total distance = 48.2km
Cooks are needed to prepare for a large party. Each cook can bake either 2 large cakes or 12 small cakes per hour. The kitchen is available for 2 hours and 21 large cakes and 500 small cakes need to be baked.
How many cooks are required to bake the required number of cakes during the time the kitchen is available?
| 12 | |
| 9 | |
| 27 | |
| 6 |
If a single cook can bake 2 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 2 x 2 = 4 large cakes during that time. 21 large cakes are needed for the party so \( \frac{21}{4} \) = 5\(\frac{1}{4}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 12 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 12 x 2 = 24 small cakes during that time. 500 small cakes are needed for the party so \( \frac{500}{24} \) = 20\(\frac{5}{6}\) cooks are needed to bake the required number of small cakes.
Because you can't employ a fractional cook, round the number of cooks needed for each type of cake up to the next whole number resulting in 6 + 21 = 27 cooks.
What is \( 3 \)\( \sqrt{27} \) + \( 5 \)\( \sqrt{3} \)
| 8\( \sqrt{3} \) | |
| 14\( \sqrt{3} \) | |
| 8\( \sqrt{9} \) | |
| 15\( \sqrt{27} \) |
To add these radicals together their radicands must be the same:
3\( \sqrt{27} \) + 5\( \sqrt{3} \)
3\( \sqrt{9 \times 3} \) + 5\( \sqrt{3} \)
3\( \sqrt{3^2 \times 3} \) + 5\( \sqrt{3} \)
(3)(3)\( \sqrt{3} \) + 5\( \sqrt{3} \)
9\( \sqrt{3} \) + 5\( \sqrt{3} \)
Now that the radicands are identical, you can add them together:
9\( \sqrt{3} \) + 5\( \sqrt{3} \)