| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.08 |
| Score | 0% | 62% |
What is 2\( \sqrt{9} \) x 4\( \sqrt{6} \)?
| 24\( \sqrt{6} \) | |
| 8\( \sqrt{6} \) | |
| 6\( \sqrt{6} \) | |
| 6\( \sqrt{54} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
2\( \sqrt{9} \) x 4\( \sqrt{6} \)
(2 x 4)\( \sqrt{9 \times 6} \)
8\( \sqrt{54} \)
Now we need to simplify the radical:
8\( \sqrt{54} \)
8\( \sqrt{6 \times 9} \)
8\( \sqrt{6 \times 3^2} \)
(8)(3)\( \sqrt{6} \)
24\( \sqrt{6} \)
What is -4a5 + a5?
| -5a5 | |
| -5a-5 | |
| -3a5 | |
| -3a25 |
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:
-4a5 + 1a5
(-4 + 1)a5
-3a5
A triathlon course includes a 300m swim, a 40.1km bike ride, and a 11.0km run. What is the total length of the race course?
| 33.3km | |
| 49.6km | |
| 64.4km | |
| 51.4km |
To add these distances, they must share the same unit so first you need to first convert the swim distance from meters (m) to kilometers (km) before adding it to the bike and run distances which are already in km. To convert 300 meters to kilometers, divide the distance by 1000 to get 0.3km then add the remaining distances:
total distance = swim + bike + run
total distance = 0.3km + 40.1km + 11.0km
total distance = 51.4km
Roger loaned Jennifer $100 at an annual interest rate of 1%. If no payments are made, what is the total amount owed at the end of the first year?
| $104 | |
| $101 | |
| $107 | |
| $105 |
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 $100
i = 0.01 x $100
No payments were made so the total amount due is the original amount + the accumulated interest:
total = $100 + $1If \( \left|y - 3\right| \) + 4 = 8, which of these is a possible value for y?
| -20 | |
| 16 | |
| 7 | |
| -9 |
First, solve for \( \left|y - 3\right| \):
\( \left|y - 3\right| \) + 4 = 8
\( \left|y - 3\right| \) = 8 - 4
\( \left|y - 3\right| \) = 4
The value inside the absolute value brackets can be either positive or negative so (y - 3) must equal + 4 or -4 for \( \left|y - 3\right| \) to equal 4:
| y - 3 = 4 y = 4 + 3 y = 7 | y - 3 = -4 y = -4 + 3 y = -1 |
So, y = -1 or y = 7.