| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.92 |
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This property states taht the order of addition or multiplication does not mater. For example, 2 + 5 and 5 + 2 are equivalent.
commutative |
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associative |
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PEDMAS |
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distributive |
The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 20% off." If Monty buys two shirts, each with a regular price of $41, how much will he pay for both shirts?
| $32.80 | |
| $59.45 | |
| $73.80 | |
| $8.20 |
By buying two shirts, Monty will save $41 x \( \frac{20}{100} \) = \( \frac{$41 x 20}{100} \) = \( \frac{$820}{100} \) = $8.20 on the second shirt.
So, his total cost will be
$41.00 + ($41.00 - $8.20)
$41.00 + $32.80
$73.80
Convert b-4 to remove the negative exponent.
| \( \frac{-4}{-b} \) | |
| \( \frac{-1}{b^{-4}} \) | |
| \( \frac{4}{b} \) | |
| \( \frac{1}{b^4} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.
Solve 5 + (3 + 5) ÷ 3 x 3 - 42
| \(\frac{8}{9}\) | |
| -3 | |
| 1 | |
| 1\(\frac{1}{3}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
5 + (3 + 5) ÷ 3 x 3 - 42
P: 5 + (8) ÷ 3 x 3 - 42
E: 5 + 8 ÷ 3 x 3 - 16
MD: 5 + \( \frac{8}{3} \) x 3 - 16
MD: 5 + \( \frac{24}{3} \) - 16
AS: \( \frac{15}{3} \) + \( \frac{24}{3} \) - 16
AS: \( \frac{39}{3} \) - 16
AS: \( \frac{39 - 48}{3} \)
\( \frac{-9}{3} \)
-3
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for division |
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commutative property for division |
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distributive property for multiplication |
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commutative property for multiplication |
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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).