| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.57 |
| Score | 0% | 71% |
What is y7 x 7y5?
| 8y5 | |
| 7y7 | |
| 8y35 | |
| 7y12 |
To multiply terms with exponents, the base of both exponents must be the same. In this case they are so multiply the coefficients and add the exponents:
y7 x 7y5
(1 x 7)y(7 + 5)
7y12
What is \( \frac{2}{7} \) x \( \frac{1}{8} \)?
| \(\frac{1}{28}\) | |
| \(\frac{1}{4}\) | |
| \(\frac{1}{9}\) | |
| \(\frac{1}{14}\) |
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{2}{7} \) x \( \frac{1}{8} \) = \( \frac{2 x 1}{7 x 8} \) = \( \frac{2}{56} \) = \(\frac{1}{28}\)
What is \( \sqrt{\frac{81}{16}} \)?
| 1\(\frac{1}{3}\) | |
| 2\(\frac{1}{4}\) | |
| 3\(\frac{1}{2}\) | |
| \(\frac{3}{7}\) |
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:
\( \sqrt{\frac{81}{16}} \)
\( \frac{\sqrt{81}}{\sqrt{16}} \)
\( \frac{\sqrt{9^2}}{\sqrt{4^2}} \)
\( \frac{9}{4} \)
2\(\frac{1}{4}\)
Which of the following is a mixed number?
\({7 \over 5} \) |
|
\({5 \over 7} \) |
|
\(1 {2 \over 5} \) |
|
\({a \over 5} \) |
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.
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for multiplication |
|
commutative property for division |
|
commutative property for multiplication |
|
distributive property for division |
The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).