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vitfil [10]
3 years ago
12

Used DVDs are on sale for a price of 5 for $20. At that price, how much will is cost to buy 12 used DVDs?

Mathematics
2 answers:
Levart [38]3 years ago
6 0
If you reduce the 5 for 20, you will get 1 for $4. Therefore, 12×4 will be the answer ($48)
siniylev [52]3 years ago
5 0

Hey there!

We know that 5 are $20, if we have the same rate then:

$20/5= 4

Each movie costs $4.

$4 * 12= $48

12 movies will cost $48 dollars.

I hope this helps!

~kaikers

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A university found that 10% of its students withdraw without completing the introductory statistics course. assume that 20 stude
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Which algebraic expression is equivalent to the expression below 6(2x+5)+4x
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12x+30+4
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A farm has chickens and cows. All the cows have 4 legs and all the chickens have 2 legs.
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Answer:

21 chickens

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3 years ago
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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