| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.13 |
| Score | 0% | 63% |
What is \( \frac{9}{5} \) + \( \frac{7}{9} \)?
| 1 \( \frac{4}{11} \) | |
| \( \frac{6}{45} \) | |
| 2\(\frac{26}{45}\) | |
| \( \frac{3}{45} \) |
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 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:
\( \frac{9 x 9}{5 x 9} \) + \( \frac{7 x 5}{9 x 5} \)
\( \frac{81}{45} \) + \( \frac{35}{45} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{81 + 35}{45} \) = \( \frac{116}{45} \) = 2\(\frac{26}{45}\)
What is \( \frac{5b^6}{3b^3} \)?
| \(\frac{3}{5}\)b3 | |
| 1\(\frac{2}{3}\)b9 | |
| 1\(\frac{2}{3}\)b3 | |
| 1\(\frac{2}{3}\)b\(\frac{1}{2}\) |
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:
\( \frac{5b^6}{3b^3} \)
\( \frac{5}{3} \) b(6 - 3)
1\(\frac{2}{3}\)b3
What is \( \frac{3}{6} \) x \( \frac{3}{6} \)?
| \(\frac{1}{4}\) | |
| \(\frac{1}{7}\) | |
| \(\frac{12}{49}\) | |
| \(\frac{4}{63}\) |
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{3}{6} \) x \( \frac{3}{6} \) = \( \frac{3 x 3}{6 x 6} \) = \( \frac{9}{36} \) = \(\frac{1}{4}\)
| 1 | |
| 1.4 | |
| 5.6 | |
| 1.6 |
1
If \(\left|a\right| = 7\), which of the following best describes a?
a = 7 |
|
a = 7 or a = -7 |
|
a = -7 |
|
none of these is correct |
The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).