| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
| Score | 0% | 65% |
What is \( \frac{3}{6} \) x \( \frac{4}{5} \)?
| \(\frac{9}{25}\) | |
| \(\frac{9}{35}\) | |
| \(\frac{2}{5}\) | |
| \(\frac{4}{63}\) |
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{3}{6} \) x \( \frac{4}{5} \) = \( \frac{3 x 4}{6 x 5} \) = \( \frac{12}{30} \) = \(\frac{2}{5}\)
The __________ is the greatest factor that divides two integers.
greatest common multiple |
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greatest common factor |
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absolute value |
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least common multiple |
The greatest common factor (GCF) is the greatest factor that divides two integers.
Solve for \( \frac{4!}{5!} \)
| \( \frac{1}{5} \) | |
| 42 | |
| 3024 | |
| 15120 |
A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:
\( \frac{4!}{5!} \)
\( \frac{4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{5} \)
\( \frac{1}{5} \)
In a class of 19 students, 8 are taking German and 13 are taking Spanish. Of the students studying German or Spanish, 7 are taking both courses. How many students are not enrolled in either course?
| 18 | |
| 17 | |
| 5 | |
| 15 |
The number of students taking German or Spanish is 8 + 13 = 21. Of that group of 21, 7 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 21 - 7 = 14 who are taking at least one language. 19 - 14 = 5 students who are not taking either language.
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
commutative property for multiplication |
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distributive property for division |
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commutative property for division |
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distributive property for multiplication |
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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).