| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.24 |
| Score | 0% | 65% |
A bread recipe calls for 3\(\frac{1}{2}\) cups of flour. If you only have 1 cup, how much more flour is needed?
| 3\(\frac{1}{8}\) cups | |
| 1\(\frac{1}{4}\) cups | |
| \(\frac{1}{8}\) cups | |
| 2\(\frac{1}{2}\) cups |
The amount of flour you need is (3\(\frac{1}{2}\) - 1) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{28}{8} \) - \( \frac{8}{8} \)) cups
\( \frac{20}{8} \) cups
2\(\frac{1}{2}\) cups
What is 2b2 - 4b2?
| -2b2 | |
| 6b-4 | |
| 6b4 | |
| 6b2 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:
2b2 - 4b2
(2 - 4)b2
-2b2
If all of a roofing company's 10 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 8 complete crews out on jobs?
| 6 | |
| 5 | |
| 2 | |
| 18 |
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 10 workers at the company now and that's enough to staff 5 crews so there are \( \frac{10}{5} \) = 2 workers on a crew. 8 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 8 x 2 = 16 total workers to staff the crews during the busy season. The company already employs 10 workers so they need to add 16 - 10 = 6 new staff for the busy season.
What is -a6 x 7a5?
| 6a5 | |
| -7a11 | |
| 6a6 | |
| -7a-1 |
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:
-a6 x 7a5
(-1 x 7)a(6 + 5)
-7a11
If \( \left|c - 6\right| \) - 3 = -6, which of these is a possible value for c?
| 9 | |
| 1 | |
| -8 | |
| -3 |
First, solve for \( \left|c - 6\right| \):
\( \left|c - 6\right| \) - 3 = -6
\( \left|c - 6\right| \) = -6 + 3
\( \left|c - 6\right| \) = -3
The value inside the absolute value brackets can be either positive or negative so (c - 6) must equal - 3 or --3 for \( \left|c - 6\right| \) to equal -3:
| c - 6 = -3 c = -3 + 6 c = 3 | c - 6 = 3 c = 3 + 6 c = 9 |
So, c = 9 or c = 3.