| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.20 |
| Score | 0% | 64% |
What is -6y6 + 7y6?
| y36 | |
| y6 | |
| 13y6 | |
| y-12 |
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:
-6y6 + 7y6
(-6 + 7)y6
y6
If \( \left|a + 4\right| \) + 2 = -1, which of these is a possible value for a?
| 21 | |
| -4 | |
| -5 | |
| -1 |
First, solve for \( \left|a + 4\right| \):
\( \left|a + 4\right| \) + 2 = -1
\( \left|a + 4\right| \) = -1 - 2
\( \left|a + 4\right| \) = -3
The value inside the absolute value brackets can be either positive or negative so (a + 4) must equal - 3 or --3 for \( \left|a + 4\right| \) to equal -3:
| a + 4 = -3 a = -3 - 4 a = -7 | a + 4 = 3 a = 3 - 4 a = -1 |
So, a = -1 or a = -7.
If the ratio of home fans to visiting fans in a crowd is 3:1 and all 31,000 seats in a stadium are filled, how many home fans are in attendance?
| 27,500 | |
| 26,400 | |
| 23,250 | |
| 30,000 |
A ratio of 3:1 means that there are 3 home fans for every one visiting fan. So, of every 4 fans, 3 are home fans and \( \frac{3}{4} \) of every fan in the stadium is a home fan:
31,000 fans x \( \frac{3}{4} \) = \( \frac{93000}{4} \) = 23,250 fans.
How many 14-passenger vans will it take to drive all 37 members of the football team to an away game?
| 9 vans | |
| 8 vans | |
| 3 vans | |
| 11 vans |
Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:
vans = \( \frac{37}{14} \) = 2\(\frac{9}{14}\)
So, it will take 2 full vans and one partially full van to transport the entire team making a total of 3 vans.
What is \( \frac{3}{5} \) - \( \frac{5}{9} \)?
| 1 \( \frac{6}{45} \) | |
| \(\frac{2}{45}\) | |
| 1 \( \frac{3}{12} \) | |
| \( \frac{9}{15} \) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{3 x 9}{5 x 9} \) - \( \frac{5 x 5}{9 x 5} \)
\( \frac{27}{45} \) - \( \frac{25}{45} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{27 - 25}{45} \) = \( \frac{2}{45} \) = \(\frac{2}{45}\)