| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.16 |
| Score | 0% | 63% |
Simplify \( \frac{20}{68} \).
| \( \frac{6}{17} \) | |
| \( \frac{4}{7} \) | |
| \( \frac{5}{17} \) | |
| \( \frac{8}{19} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 20 are [1, 2, 4, 5, 10, 20] and the factors of 68 are [1, 2, 4, 17, 34, 68]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{20}{68} \) = \( \frac{\frac{20}{4}}{\frac{68}{4}} \) = \( \frac{5}{17} \)
The __________ is the smallest positive integer that is a multiple of two or more integers.
least common factor |
|
least common multiple |
|
greatest common factor |
|
absolute value |
The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.
Christine scored 75% on her final exam. If each question was worth 4 points and there were 240 possible points on the exam, how many questions did Christine answer correctly?
| 45 | |
| 35 | |
| 47 | |
| 44 |
Christine scored 75% on the test meaning she earned 75% of the possible points on the test. There were 240 possible points on the test so she earned 240 x 0.75 = 180 points. Each question is worth 4 points so she got \( \frac{180}{4} \) = 45 questions right.
What is \( \sqrt{\frac{9}{4}} \)?
| 1\(\frac{1}{2}\) | |
| 1\(\frac{1}{8}\) | |
| 1 | |
| 2 |
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:
\( \sqrt{\frac{9}{4}} \)
\( \frac{\sqrt{9}}{\sqrt{4}} \)
\( \frac{\sqrt{3^2}}{\sqrt{2^2}} \)
\( \frac{3}{2} \)
1\(\frac{1}{2}\)
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
commutative property for multiplication |
|
distributive property for division |
|
distributive property for multiplication |
|
commutative 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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).