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3 years ago

Please help! X+(x+2)=72

Mathematics

2 answers:

777dan777 [17]3 years ago

6 0

$x+(x+2)=72\\
2x+2=72\\
2x=70\\
x=35$

Leokris [45]3 years ago

6 0

X + (x + 2) = 72
x + x + 2 = 72
2x + 2 = 72
2x = 72 - 2
2x = 70
x = 70/2
x = 35

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Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

case A) $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> the term is not a perfect cube

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

case B) $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> the term is not a perfect cube

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

case C) $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> the term is not a perfect cube

$a^{22}$ ------> the term is not a perfect cube

$b^{16}$ ------> the term is not a perfect cube

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

case D) $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> the term is not a perfect cube

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

case E) $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

Read 2 more answers

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