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dmitriy555 [2]
1 year ago
11

4va olleckpoiniDiscrete Mathematics OL v1 A - Koscianski, Megan / Chapter 2: Set TheoryRD 2. Use the accompanying Venn diagram t

hat shows the number of elements in regions I through IV to answer the question.10MHow many elements belong to neither set A nor set B?BU Font FamilyAAIFEEOOOE•"'NE

Mathematics
1 answer:
Blababa [14]1 year ago
8 0

The elements that are neither in set A nor set B are: {10}

Answer: 10

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The total amount that Makayla needs to pay is equal to $530 + $760 which is equal to $1290. This value is lesser compared to the amount that Makayla is able to take home. Therefore, Makayla will not be in danger of credit overload and she will be able to provide for her other needs. 
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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

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