| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.04 |
| Score | 0% | 61% |
What is \( \frac{2}{5} \) + \( \frac{3}{7} \)?
| 1 \( \frac{6}{13} \) | |
| \( \frac{5}{35} \) | |
| 1 \( \frac{7}{35} \) | |
| \(\frac{29}{35}\) |
To add 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 7 are [7, 14, 21, 28, 35, 42, 49, 56, 63, 70]. The first few multiples they share are [35, 70] making 35 the smallest multiple 5 and 7 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{2 x 7}{5 x 7} \) + \( \frac{3 x 5}{7 x 5} \)
\( \frac{14}{35} \) + \( \frac{15}{35} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{14 + 15}{35} \) = \( \frac{29}{35} \) = \(\frac{29}{35}\)
If the ratio of home fans to visiting fans in a crowd is 3:1 and all 42,000 seats in a stadium are filled, how many home fans are in attendance?
| 24,000 | |
| 31,500 | |
| 28,667 | |
| 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:
42,000 fans x \( \frac{3}{4} \) = \( \frac{126000}{4} \) = 31,500 fans.
A triathlon course includes a 300m swim, a 20.9km bike ride, and a 7.4km run. What is the total length of the race course?
| 39km | |
| 42.9km | |
| 52.9km | |
| 28.6km |
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 + 20.9km + 7.4km
total distance = 28.6km
What is -7c4 + 2c4?
| -5c-8 | |
| -5c4 | |
| -9c4 | |
| 9c4 |
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:
-7c4 + 2c4
(-7 + 2)c4
-5c4
What is \( \frac{-7a^9}{8a^2} \)?
| -\(\frac{7}{8}\)a\(\frac{2}{9}\) | |
| -\(\frac{7}{8}\)a11 | |
| -1\(\frac{1}{7}\)a7 | |
| -\(\frac{7}{8}\)a7 |
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:
\( \frac{-7a^9}{8a^2} \)
\( \frac{-7}{8} \) a(9 - 2)
-\(\frac{7}{8}\)a7