| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.07 |
| Score | 0% | 61% |
What is -8x5 + 7x5?
| -x-10 | |
| -x5 | |
| -x10 | |
| -15x5 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:
-8x5 + 7x5
(-8 + 7)x5
-x5
If a rectangle is twice as long as it is wide and has a perimeter of 24 meters, what is the area of the rectangle?
| 8 m2 | |
| 2 m2 | |
| 32 m2 | |
| 50 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 24 meters so the equation becomes: 2w + 2h = 24.
Putting these two equations together and solving for width (w):
2w + 2h = 24
w + h = \( \frac{24}{2} \)
w + h = 12
w = 12 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 12 - 2w
3w = 12
w = \( \frac{12}{3} \)
w = 4
Since h = 2w that makes h = (2 x 4) = 8 and the area = h x w = 4 x 8 = 32 m2
4! = ?
3 x 2 x 1 |
|
4 x 3 |
|
4 x 3 x 2 x 1 |
|
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.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 50% off." If Roger buys two shirts, each with a regular price of $46, how much money will he save?
| $6.90 | |
| $16.10 | |
| $18.40 | |
| $23.00 |
By buying two shirts, Roger will save $46 x \( \frac{50}{100} \) = \( \frac{$46 x 50}{100} \) = \( \frac{$2300}{100} \) = $23.00 on the second shirt.
What is \( 9 \)\( \sqrt{32} \) - \( 3 \)\( \sqrt{2} \)
| 6\( \sqrt{16} \) | |
| 27\( \sqrt{2} \) | |
| 33\( \sqrt{2} \) | |
| 6\( \sqrt{-12} \) |
To subtract these radicals together their radicands must be the same:
9\( \sqrt{32} \) - 3\( \sqrt{2} \)
9\( \sqrt{16 \times 2} \) - 3\( \sqrt{2} \)
9\( \sqrt{4^2 \times 2} \) - 3\( \sqrt{2} \)
(9)(4)\( \sqrt{2} \) - 3\( \sqrt{2} \)
36\( \sqrt{2} \) - 3\( \sqrt{2} \)
Now that the radicands are identical, you can subtract them:
36\( \sqrt{2} \) - 3\( \sqrt{2} \)