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weeeeeb [17]
3 years ago
9

Which expression is not an integer? A) -15 B) 6/14 C) 0 D) 12/6

Mathematics
2 answers:
Minchanka [31]3 years ago
7 0
I believe it is C because I think 0 is actually not an integer
vladimir1956 [14]3 years ago
5 0

Answer : The Expression which is not an integer is  \frac{6}{14}

Reason : Because,\; \frac{6}{14} = 0.42857\;which\;is\;not\;an\;integer

Option - A : -15 is an Integer

Option - C : 0 is an Integer

Option - D : \frac{12}{6} = 2\;which\;is\;an\;integer

So, Option B is the Answer

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Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

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0 x - y+0 z = -2 | (equation 2)

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{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

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{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

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3 years ago
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AleksandrR [38]

Answer:

If she doubles it then it will be 4: 10

If she triples it then it'll be 6: 15

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2 years ago
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Answer:

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Step-by-step explanation:

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10x + 6 = 5x + 11

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