| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.65 |
| Score | 0% | 53% |
Convert a-4 to remove the negative exponent.
| \( \frac{-4}{-a} \) | |
| \( \frac{-1}{-4a} \) | |
| \( \frac{1}{a^4} \) | |
| \( \frac{4}{a} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.
What is 5\( \sqrt{3} \) x 3\( \sqrt{3} \)?
| 8\( \sqrt{3} \) | |
| 15\( \sqrt{3} \) | |
| 45 | |
| 15\( \sqrt{6} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
5\( \sqrt{3} \) x 3\( \sqrt{3} \)
(5 x 3)\( \sqrt{3 \times 3} \)
15\( \sqrt{9} \)
Now we need to simplify the radical:
15\( \sqrt{9} \)
15\( \sqrt{3^2} \)
(15)(3)
45
What is \( 4 \)\( \sqrt{32} \) - \( 9 \)\( \sqrt{2} \)
| 36\( \sqrt{64} \) | |
| -5\( \sqrt{32} \) | |
| 7\( \sqrt{2} \) | |
| 36\( \sqrt{32} \) |
To subtract these radicals together their radicands must be the same:
4\( \sqrt{32} \) - 9\( \sqrt{2} \)
4\( \sqrt{16 \times 2} \) - 9\( \sqrt{2} \)
4\( \sqrt{4^2 \times 2} \) - 9\( \sqrt{2} \)
(4)(4)\( \sqrt{2} \) - 9\( \sqrt{2} \)
16\( \sqrt{2} \) - 9\( \sqrt{2} \)
Now that the radicands are identical, you can subtract them:
16\( \sqrt{2} \) - 9\( \sqrt{2} \)What is x4 + 3x4?
| 4x-8 | |
| -2x-4 | |
| 4x4 | |
| 4x16 |
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:
1x4 + 3x4
(1 + 3)x4
4x4
| 9.0 | |
| 1 | |
| 2.4 | |
| 0.8 |
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