| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.25 |
| Score | 0% | 65% |
If a rectangle is twice as long as it is wide and has a perimeter of 48 meters, what is the area of the rectangle?
| 98 m2 | |
| 8 m2 | |
| 72 m2 | |
| 128 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 48 meters so the equation becomes: 2w + 2h = 48.
Putting these two equations together and solving for width (w):
2w + 2h = 48
w + h = \( \frac{48}{2} \)
w + h = 24
w = 24 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 24 - 2w
3w = 24
w = \( \frac{24}{3} \)
w = 8
Since h = 2w that makes h = (2 x 8) = 16 and the area = h x w = 8 x 16 = 128 m2
What is \( \frac{8}{8} \) - \( \frac{5}{10} \)?
| \(\frac{1}{2}\) | |
| \( \frac{7}{10} \) | |
| 1 \( \frac{9}{40} \) | |
| 1 \( \frac{3}{40} \) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [40, 80] making 40 the smallest multiple 8 and 10 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{8 x 5}{8 x 5} \) - \( \frac{5 x 4}{10 x 4} \)
\( \frac{40}{40} \) - \( \frac{20}{40} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{40 - 20}{40} \) = \( \frac{20}{40} \) = \(\frac{1}{2}\)
Which of the following is an improper fraction?
\(1 {2 \over 5} \) |
|
\({7 \over 5} \) |
|
\({a \over 5} \) |
|
\({2 \over 5} \) |
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.
4! = ?
4 x 3 |
|
5 x 4 x 3 x 2 x 1 |
|
3 x 2 x 1 |
|
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.
If \( \left|a - 1\right| \) - 5 = 0, which of these is a possible value for a?
| -4 | |
| -13 | |
| 19 | |
| -5 |
First, solve for \( \left|a - 1\right| \):
\( \left|a - 1\right| \) - 5 = 0
\( \left|a - 1\right| \) = 0 + 5
\( \left|a - 1\right| \) = 5
The value inside the absolute value brackets can be either positive or negative so (a - 1) must equal + 5 or -5 for \( \left|a - 1\right| \) to equal 5:
| a - 1 = 5 a = 5 + 1 a = 6 | a - 1 = -5 a = -5 + 1 a = -4 |
So, a = -4 or a = 6.