| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.44 |
| Score | 0% | 69% |
What is (x3)3?
| x0 | |
| x9 | |
| 3x3 | |
| x6 |
To raise a term with an exponent to another exponent, retain the base and multiply the exponents:
(x3)3Find the average of the following numbers: 14, 6, 12, 8.
| 10 | |
| 14 | |
| 8 | |
| 11 |
To find the average of these 4 numbers add them together then divide by 4:
\( \frac{14 + 6 + 12 + 8}{4} \) = \( \frac{40}{4} \) = 10
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 17 small cakes per hour. The kitchen is available for 2 hours and 38 large cakes and 110 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?
| 13 | |
| 11 | |
| 7 | |
| 10 |
If a single cook can bake 3 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 3 x 2 = 6 large cakes during that time. 38 large cakes are needed for the party so \( \frac{38}{6} \) = 6\(\frac{1}{3}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 17 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 17 x 2 = 34 small cakes during that time. 110 small cakes are needed for the party so \( \frac{110}{34} \) = 3\(\frac{4}{17}\) 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 7 + 4 = 11 cooks.
Simplify \( \frac{32}{60} \).
| \( \frac{5}{9} \) | |
| \( \frac{8}{15} \) | |
| \( \frac{9}{11} \) | |
| \( \frac{3}{4} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 32 are [1, 2, 4, 8, 16, 32] and the factors of 60 are [1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{32}{60} \) = \( \frac{\frac{32}{4}}{\frac{60}{4}} \) = \( \frac{8}{15} \)
What is \( \frac{49\sqrt{27}}{7\sqrt{9}} \)?
| \(\frac{1}{7}\) \( \sqrt{3} \) | |
| \(\frac{1}{3}\) \( \sqrt{7} \) | |
| 7 \( \sqrt{3} \) | |
| 3 \( \sqrt{7} \) |
To divide terms with radicals, divide the coefficients and radicands separately:
\( \frac{49\sqrt{27}}{7\sqrt{9}} \)
\( \frac{49}{7} \) \( \sqrt{\frac{27}{9}} \)
7 \( \sqrt{3} \)