| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.46 |
| Score | 0% | 69% |
What is \( \frac{1}{9} \) x \( \frac{2}{8} \)?
| \(\frac{1}{4}\) | |
| \(\frac{1}{36}\) | |
| \(\frac{4}{25}\) | |
| \(\frac{1}{7}\) |
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{1}{9} \) x \( \frac{2}{8} \) = \( \frac{1 x 2}{9 x 8} \) = \( \frac{2}{72} \) = \(\frac{1}{36}\)
Find the average of the following numbers: 14, 6, 12, 8.
| 6 | |
| 10 | |
| 15 | |
| 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
What is \( \sqrt{\frac{64}{16}} \)?
| \(\frac{2}{5}\) | |
| 2 | |
| \(\frac{5}{8}\) | |
| \(\frac{2}{3}\) |
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:
\( \sqrt{\frac{64}{16}} \)
\( \frac{\sqrt{64}}{\sqrt{16}} \)
\( \frac{\sqrt{8^2}}{\sqrt{4^2}} \)
\( \frac{8}{4} \)
2
What is \( \frac{-8y^8}{1y^3} \)?
| -8y-5 | |
| -8y11 | |
| -\(\frac{1}{8}\)y11 | |
| -8y5 |
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:
\( \frac{-8y^8}{y^3} \)
\( \frac{-8}{1} \) y(8 - 3)
-8y5
Which of the following is an improper fraction?
\({a \over 5} \) |
|
\(1 {2 \over 5} \) |
|
\({2 \over 5} \) |
|
\({7 \over 5} \) |
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.