| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.21 |
| Score | 0% | 64% |
What is \( \frac{1}{5} \) x \( \frac{2}{5} \)?
| \(\frac{8}{35}\) | |
| \(\frac{2}{27}\) | |
| \(\frac{8}{21}\) | |
| \(\frac{2}{25}\) |
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{1}{5} \) x \( \frac{2}{5} \) = \( \frac{1 x 2}{5 x 5} \) = \( \frac{2}{25} \) = \(\frac{2}{25}\)
Alex loaned Monica $1,300 at an annual interest rate of 8%. If no payments are made, what is the total amount owed at the end of the first year?
| $1,339 | |
| $1,365 | |
| $1,391 | |
| $1,404 |
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 $1,300
i = 0.08 x $1,300
No payments were made so the total amount due is the original amount + the accumulated interest:
total = $1,300 + $104What is \( \sqrt{\frac{81}{4}} \)?
| \(\frac{5}{8}\) | |
| 1\(\frac{2}{3}\) | |
| 1 | |
| 4\(\frac{1}{2}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{81}{4}} \)
\( \frac{\sqrt{81}}{\sqrt{4}} \)
\( \frac{\sqrt{9^2}}{\sqrt{2^2}} \)
\( \frac{9}{2} \)
4\(\frac{1}{2}\)
How many 1\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 9 gallon tank to fill it exactly halfway?
| 2 | |
| 6 | |
| 3 | |
| 5 |
To fill a 9 gallon tank exactly halfway you'll need 4\(\frac{1}{2}\) gallons of fuel. Each fuel can holds 1\(\frac{1}{2}\) gallons so:
cans = \( \frac{4\frac{1}{2} \text{ gallons}}{1\frac{1}{2} \text{ gallons}} \) = 3
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
commutative property for multiplication |
|
commutative property for division |
|
distributive property for multiplication |
|
distributive property for division |
The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).