| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.03 |
| Score | 0% | 61% |
How many 7-passenger vans will it take to drive all 87 members of the football team to an away game?
| 13 vans | |
| 4 vans | |
| 6 vans | |
| 8 vans |
Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:
vans = \( \frac{87}{7} \) = 12\(\frac{3}{7}\)
So, it will take 12 full vans and one partially full van to transport the entire team making a total of 13 vans.
What is \( 5 \)\( \sqrt{18} \) + \( 3 \)\( \sqrt{2} \)
| 15\( \sqrt{18} \) | |
| 18\( \sqrt{2} \) | |
| 15\( \sqrt{2} \) | |
| 8\( \sqrt{2} \) |
To add these radicals together their radicands must be the same:
5\( \sqrt{18} \) + 3\( \sqrt{2} \)
5\( \sqrt{9 \times 2} \) + 3\( \sqrt{2} \)
5\( \sqrt{3^2 \times 2} \) + 3\( \sqrt{2} \)
(5)(3)\( \sqrt{2} \) + 3\( \sqrt{2} \)
15\( \sqrt{2} \) + 3\( \sqrt{2} \)
Now that the radicands are identical, you can add them together:
15\( \sqrt{2} \) + 3\( \sqrt{2} \)Diane scored 87% on her final exam. If each question was worth 4 points and there were 240 possible points on the exam, how many questions did Diane answer correctly?
| 38 | |
| 52 | |
| 39 | |
| 66 |
Diane scored 87% on the test meaning she earned 87% of the possible points on the test. There were 240 possible points on the test so she earned 240 x 0.87 = 208 points. Each question is worth 4 points so she got \( \frac{208}{4} \) = 52 questions right.
What is \( \frac{2}{7} \) ÷ \( \frac{2}{8} \)?
| \(\frac{4}{35}\) | |
| \(\frac{8}{63}\) | |
| 1\(\frac{1}{7}\) | |
| 2\(\frac{2}{7}\) |
To divide fractions, invert the second fraction and then multiply:
\( \frac{2}{7} \) ÷ \( \frac{2}{8} \) = \( \frac{2}{7} \) x \( \frac{8}{2} \)
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{2}{7} \) x \( \frac{8}{2} \) = \( \frac{2 x 8}{7 x 2} \) = \( \frac{16}{14} \) = 1\(\frac{1}{7}\)
What is \( \frac{-8y^6}{9y^2} \)?
| -1\(\frac{1}{8}\)y4 | |
| -1\(\frac{1}{8}\)y8 | |
| -\(\frac{8}{9}\)y4 | |
| -\(\frac{8}{9}\)y3 |
To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:
\( \frac{-8y^6}{9y^2} \)
\( \frac{-8}{9} \) y(6 - 2)
-\(\frac{8}{9}\)y4