| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.09 |
| Score | 0% | 62% |
What is \( 7 \)\( \sqrt{112} \) - \( 6 \)\( \sqrt{7} \)
| \( \sqrt{16} \) | |
| 42\( \sqrt{16} \) | |
| 22\( \sqrt{7} \) | |
| \( \sqrt{784} \) |
To subtract these radicals together their radicands must be the same:
7\( \sqrt{112} \) - 6\( \sqrt{7} \)
7\( \sqrt{16 \times 7} \) - 6\( \sqrt{7} \)
7\( \sqrt{4^2 \times 7} \) - 6\( \sqrt{7} \)
(7)(4)\( \sqrt{7} \) - 6\( \sqrt{7} \)
28\( \sqrt{7} \) - 6\( \sqrt{7} \)
Now that the radicands are identical, you can subtract them:
28\( \sqrt{7} \) - 6\( \sqrt{7} \)A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 15% off." If Charlie buys two shirts, each with a regular price of $17, how much money will he save?
| $2.55 | |
| $8.50 | |
| $0.85 | |
| 52 |
By buying two shirts, Charlie will save $17 x \( \frac{15}{100} \) = \( \frac{$17 x 15}{100} \) = \( \frac{$255}{100} \) = $2.55 on the second shirt.
If a car travels 120 miles in 6 hours, what is the average speed?
| 20 mph | |
| 40 mph | |
| 65 mph | |
| 50 mph |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)What is \( \frac{-3y^6}{9y^3} \)?
| -3y9 | |
| -\(\frac{1}{3}\)y3 | |
| -\(\frac{1}{3}\)y-3 | |
| -\(\frac{1}{3}\)y\(\frac{1}{2}\) |
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:
\( \frac{-3y^6}{9y^3} \)
\( \frac{-3}{9} \) y(6 - 3)
-\(\frac{1}{3}\)y3
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for multiplication |
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distributive property for division |
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commutative property for multiplication |
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commutative 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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).