| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.72 |
| Score | 0% | 54% |
How many 1\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 6 gallon tank to fill it exactly halfway?
| 3 | |
| 6 | |
| 2 | |
| 4 |
To fill a 6 gallon tank exactly halfway you'll need 3 gallons of fuel. Each fuel can holds 1\(\frac{1}{2}\) gallons so:
cans = \( \frac{3 \text{ gallons}}{1\frac{1}{2} \text{ gallons}} \) = 2
What is \( \frac{-1c^7}{8c^2} \)?
| -\(\frac{1}{8}\)c3\(\frac{1}{2}\) | |
| -\(\frac{1}{8}\)c5 | |
| -\(\frac{1}{8}\)c-5 | |
| -8c9 |
To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:
\( \frac{-c^7}{8c^2} \)
\( \frac{-1}{8} \) c(7 - 2)
-\(\frac{1}{8}\)c5
What is 3\( \sqrt{9} \) x 7\( \sqrt{5} \)?
| 21\( \sqrt{5} \) | |
| 21\( \sqrt{14} \) | |
| 63\( \sqrt{5} \) | |
| 21\( \sqrt{9} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
3\( \sqrt{9} \) x 7\( \sqrt{5} \)
(3 x 7)\( \sqrt{9 \times 5} \)
21\( \sqrt{45} \)
Now we need to simplify the radical:
21\( \sqrt{45} \)
21\( \sqrt{5 \times 9} \)
21\( \sqrt{5 \times 3^2} \)
(21)(3)\( \sqrt{5} \)
63\( \sqrt{5} \)
Simplify \( \sqrt{8} \)
| 2\( \sqrt{4} \) | |
| 9\( \sqrt{2} \) | |
| 4\( \sqrt{4} \) | |
| 2\( \sqrt{2} \) |
To simplify a radical, factor out the perfect squares:
\( \sqrt{8} \)
\( \sqrt{4 \times 2} \)
\( \sqrt{2^2 \times 2} \)
2\( \sqrt{2} \)
The total water usage for a city is 20,000 gallons each day. Of that total, 17% is for personal use and 40% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?
| 4,600 | |
| 2,500 | |
| 5,100 | |
| 2,400 |
40% of the water consumption is industrial use and 17% is personal use so (40% - 17%) = 23% more water is used for industrial purposes. 20,000 gallons are consumed daily so industry consumes \( \frac{23}{100} \) x 20,000 gallons = 4,600 gallons.