| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.43 |
| Score | 0% | 69% |
What is (z4)3?
| z7 | |
| z | |
| 4z3 | |
| z12 |
To raise a term with an exponent to another exponent, retain the base and multiply the exponents:
(z4)3Convert b-3 to remove the negative exponent.
| \( \frac{-3}{-b} \) | |
| \( \frac{-3}{b} \) | |
| \( \frac{1}{b^3} \) | |
| \( \frac{-1}{-3b^{3}} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.
Which of the following is a mixed number?
\({5 \over 7} \) |
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\({7 \over 5} \) |
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\(1 {2 \over 5} \) |
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\({a \over 5} \) |
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?
commutative property for multiplication |
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commutative property for division |
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distributive property for multiplication |
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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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).
If all of a roofing company's 20 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 10 complete crews out on jobs?
| 4 | |
| 18 | |
| 9 | |
| 20 |
In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 20 workers at the company now and that's enough to staff 5 crews so there are \( \frac{20}{5} \) = 4 workers on a crew. 10 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 10 x 4 = 40 total workers to staff the crews during the busy season. The company already employs 20 workers so they need to add 40 - 20 = 20 new staff for the busy season.