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ValentinkaMS [17]
4 years ago
6

A store offers 20% off all items. If x is the original purchase price, which expression represents the final price with the disc

ount? Select all that apply. A. 1.2x B. 0.8x C. x D. x – 0.2x E. x + 0.2x
Mathematics
2 answers:
denis23 [38]4 years ago
6 0

Answer:

B & D

Step-by-step explanation:

We use percents in decimal form to multiply it with the price. We convert percents into decimals by dividing the percent number by 100. For example, 78% divided by 100 becomes 0.78.

There are two ways to look at it:

  • For finding the price we pay during a sale, we focus on the percent we pay. If 22% off is the sale, then we spend 78% or 100-22=78. If 20% off is the sale, then we pay 80% or 0.80. Multiply that by x an unknown price and we have 0.8x.
  • We can find the percent off by multiplying the price by the percent conversion. So 20% is 0.20. Then subtract it from the original price to find the leftover that we pay. This is x-0.2x.


likoan [24]4 years ago
3 0

Answer:

B + D

Step-by-step explanation:

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Answer:

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Jeff is buying several packs of batteries. He purchased 16 packs of batteries and his total is $28. What is the cost of each pac
trapecia [35]
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3 years ago
Which expression is a sum of cubes?
erik [133]

we know that

A polynomial in the form a^{3} +b^{3} is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> -64x^{6} y^{12} +125x^{16} y^{3}

we know that

-64=-4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case A) is not a sum of cubes

<u>case B)</u> -32x^{6} y^{12} +125x^{16} y^{3}

we know that

-32=-2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case B) is not a sum of cubes

<u>case C)</u> 32x^{6} y^{12} +125x^{9} y^{3}

we know that

32=2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

therefore

the case C) is not a sum of cubes

<u>case A)</u> 64x^{6} y^{12} +125x^{9} y^{3}

we know that

64=4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

Substitute

4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}

(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}

therefore

<u>the answer is</u>

64x^{6} y^{12} +125x^{9} y^{3} is a sum of cubes

6 0
3 years ago
Read 2 more answers
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