| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.98 |
| Score | 0% | 60% |
What is the distance in miles of a trip that takes 3 hours at an average speed of 45 miles per hour?
| 150 miles | |
| 270 miles | |
| 210 miles | |
| 135 miles |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)Solving for distance:
distance = \( \text{speed} \times \text{time} \)
distance = \( 45mph \times 3h \)
135 miles
What is \( \sqrt{\frac{25}{64}} \)?
| \(\frac{2}{5}\) | |
| \(\frac{3}{4}\) | |
| \(\frac{5}{8}\) | |
| 2\(\frac{1}{2}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{25}{64}} \)
\( \frac{\sqrt{25}}{\sqrt{64}} \)
\( \frac{\sqrt{5^2}}{\sqrt{8^2}} \)
\(\frac{5}{8}\)
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 20 small cakes per hour. The kitchen is available for 2 hours and 36 large cakes and 380 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?
| 6 | |
| 8 | |
| 10 | |
| 16 |
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. 36 large cakes are needed for the party so \( \frac{36}{6} \) = 6 cooks are needed to bake the required number of large cakes.
If a single cook can bake 20 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 20 x 2 = 40 small cakes during that time. 380 small cakes are needed for the party so \( \frac{380}{40} \) = 9\(\frac{1}{2}\) 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 + 10 = 16 cooks.
What is 2x3 + x3?
| 3x9 | |
| x-3 | |
| -x-3 | |
| 3x3 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:
2x3 + 1x3
(2 + 1)x3
3x3
What is \( 7 \)\( \sqrt{175} \) + \( 6 \)\( \sqrt{7} \)
| 42\( \sqrt{25} \) | |
| 41\( \sqrt{7} \) | |
| 42\( \sqrt{1225} \) | |
| 42\( \sqrt{175} \) |
To add these radicals together their radicands must be the same:
7\( \sqrt{175} \) + 6\( \sqrt{7} \)
7\( \sqrt{25 \times 7} \) + 6\( \sqrt{7} \)
7\( \sqrt{5^2 \times 7} \) + 6\( \sqrt{7} \)
(7)(5)\( \sqrt{7} \) + 6\( \sqrt{7} \)
35\( \sqrt{7} \) + 6\( \sqrt{7} \)
Now that the radicands are identical, you can add them together:
35\( \sqrt{7} \) + 6\( \sqrt{7} \)