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xy > 0
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23 Jun 2018, 12:08
23 Jun 2018, 12:08
Expert Reply
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Question Stats:
81% (00:53) correct
18% (01:12) wrong based on 139 sessions
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\(xy > 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\((2x - y)(x + 4y)\) |
\(2x^2\) + \(8xy\) – \(4y^2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: xy > 0
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29 Jun 2018, 05:22
29 Jun 2018, 05:22
sandy wrote:
\(xy > 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\((2x - y)(x + 4y)\) |
\(2x^2\) + \(8xy\) – \(4y^2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Here;
Statement 1: \((2x - y)(x + 4y)\) = \(2x^2\)+ \(8xy\) - \(xy\)– \(4y^2\) = \(2x^2\) + \(7xy\) –\(4y^2\)
Statement 2:\(2x^2\)+ \(8xy\) – \(4y^2\)
Since xy>0
Therefore Statement 1 < Statement 2
Re: xy > 0
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16 Jul 2018, 18:54
16 Jul 2018, 18:54
Expert Reply
1
Bookmarks
Explanation
the terms in Quantity A:
\((2x - y)(x + 4y) = 2x^2 + 8xy - xy - 4y^2 = 2x^2 + 7xy - 4y2\)
Since \(2x^2\) and \(-4y^2\) appear in both quantities, eliminate them.
Quantity A is now equal to 7xy and Quantity B is now equal to 8xy.
Because xy > 0, Quantity B is greater. (Don’t assume! If xy were zero, the two quantities would have been equal. If xy were negative, Quantity A would have been greater.)
the terms in Quantity A:
\((2x - y)(x + 4y) = 2x^2 + 8xy - xy - 4y^2 = 2x^2 + 7xy - 4y2\)
Since \(2x^2\) and \(-4y^2\) appear in both quantities, eliminate them.
Quantity A is now equal to 7xy and Quantity B is now equal to 8xy.
Because xy > 0, Quantity B is greater. (Don’t assume! If xy were zero, the two quantities would have been equal. If xy were negative, Quantity A would have been greater.)
