| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.27 |
| Score | 0% | 65% |
If a rectangle is twice as long as it is wide and has a perimeter of 42 meters, what is the area of the rectangle?
| 162 m2 | |
| 8 m2 | |
| 98 m2 | |
| 18 m2 |
The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 42 meters so the equation becomes: 2w + 2h = 42.
Putting these two equations together and solving for width (w):
2w + 2h = 42
w + h = \( \frac{42}{2} \)
w + h = 21
w = 21 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 21 - 2w
3w = 21
w = \( \frac{21}{3} \)
w = 7
Since h = 2w that makes h = (2 x 7) = 14 and the area = h x w = 7 x 14 = 98 m2
Charlie loaned Alex $900 at an annual interest rate of 5%. If no payments are made, what is the interest owed on this loan at the end of the first year?
| $36 | |
| $88 | |
| $12 | |
| $45 |
The yearly interest charged on this loan is the annual interest rate multiplied by the amount borrowed:
interest = annual interest rate x loan amount
i = (\( \frac{6}{100} \)) x $900
i = 0.05 x $900
i = $45
4! = ?
3 x 2 x 1 |
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4 x 3 |
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4 x 3 x 2 x 1 |
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5 x 4 x 3 x 2 x 1 |
A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
The __________ is the greatest factor that divides two integers.
greatest common factor |
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greatest common multiple |
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least common multiple |
|
absolute value |
The greatest common factor (GCF) is the greatest factor that divides two integers.
If all of a roofing company's 4 workers are required to staff 2 roofing crews, how many workers need to be added during the busy season in order to send 5 complete crews out on jobs?
| 6 | |
| 19 | |
| 12 | |
| 5 |
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 4 workers at the company now and that's enough to staff 2 crews so there are \( \frac{4}{2} \) = 2 workers on a crew. 5 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 5 x 2 = 10 total workers to staff the crews during the busy season. The company already employs 4 workers so they need to add 10 - 4 = 6 new staff for the busy season.