| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.37 |
| Score | 0% | 67% |
What is \( 6 \)\( \sqrt{80} \) - \( 4 \)\( \sqrt{5} \)
| 2\( \sqrt{16} \) | |
| 2\( \sqrt{5} \) | |
| 20\( \sqrt{5} \) | |
| 24\( \sqrt{5} \) |
To subtract these radicals together their radicands must be the same:
6\( \sqrt{80} \) - 4\( \sqrt{5} \)
6\( \sqrt{16 \times 5} \) - 4\( \sqrt{5} \)
6\( \sqrt{4^2 \times 5} \) - 4\( \sqrt{5} \)
(6)(4)\( \sqrt{5} \) - 4\( \sqrt{5} \)
24\( \sqrt{5} \) - 4\( \sqrt{5} \)
Now that the radicands are identical, you can subtract them:
24\( \sqrt{5} \) - 4\( \sqrt{5} \)Which of the following is an improper fraction?
\(1 {2 \over 5} \) |
|
\({a \over 5} \) |
|
\({2 \over 5} \) |
|
\({7 \over 5} \) |
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.
What is the next number in this sequence: 1, 9, 17, 25, 33, __________ ?
| 35 | |
| 41 | |
| 34 | |
| 45 |
The equation for this sequence is:
an = an-1 + 8
where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:
a6 = a5 + 8
a6 = 33 + 8
a6 = 41
Bob loaned Monty $700 at an annual interest rate of 9%. If no payments are made, what is the interest owed on this loan at the end of the first year?
| $9 | |
| $63 | |
| $24 | |
| $6 |
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 $700
i = 0.09 x $700
i = $63
If \( \left|x + 5\right| \) - 1 = 5, which of these is a possible value for x?
| -13 | |
| -17 | |
| -4 | |
| -11 |
First, solve for \( \left|x + 5\right| \):
\( \left|x + 5\right| \) - 1 = 5
\( \left|x + 5\right| \) = 5 + 1
\( \left|x + 5\right| \) = 6
The value inside the absolute value brackets can be either positive or negative so (x + 5) must equal + 6 or -6 for \( \left|x + 5\right| \) to equal 6:
| x + 5 = 6 x = 6 - 5 x = 1 | x + 5 = -6 x = -6 - 5 x = -11 |
So, x = -11 or x = 1.