| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.03 |
| Score | 0% | 61% |
Which of the following statements about exponents is false?
b1 = 1 |
|
b0 = 1 |
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b1 = b |
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all of these are false |
A number with an exponent (be) consists of a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).
What is \( \frac{35\sqrt{15}}{5\sqrt{3}} \)?
| 7 \( \sqrt{5} \) | |
| \(\frac{1}{7}\) \( \sqrt{\frac{1}{5}} \) | |
| 7 \( \sqrt{\frac{1}{5}} \) | |
| \(\frac{1}{5}\) \( \sqrt{\frac{1}{7}} \) |
To divide terms with radicals, divide the coefficients and radicands separately:
\( \frac{35\sqrt{15}}{5\sqrt{3}} \)
\( \frac{35}{5} \) \( \sqrt{\frac{15}{3}} \)
7 \( \sqrt{5} \)
9 members of a bridal party need transported to a wedding reception but there are only 2 3-passenger taxis available to take them. How many will need to find other transportation?
| 3 | |
| 4 | |
| 2 | |
| 1 |
There are 2 3-passenger taxis available so that's 2 x 3 = 6 total seats. There are 9 people needing transportation leaving 9 - 6 = 3 who will have to find other transportation.
What is \( \frac{7}{5} \) - \( \frac{4}{7} \)?
| 1 \( \frac{6}{35} \) | |
| 1 \( \frac{7}{11} \) | |
| 2 \( \frac{8}{35} \) | |
| \(\frac{29}{35}\) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 7 are [7, 14, 21, 28, 35, 42, 49, 56, 63, 70]. The first few multiples they share are [35, 70] making 35 the smallest multiple 5 and 7 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{7 x 7}{5 x 7} \) - \( \frac{4 x 5}{7 x 5} \)
\( \frac{49}{35} \) - \( \frac{20}{35} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{49 - 20}{35} \) = \( \frac{29}{35} \) = \(\frac{29}{35}\)
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?
| 2 | |
| 3 | |
| 5 | |
| 2 |
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