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03 Oct 2024, 02:21
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
58% (02:01) correct
41% (01:16) wrong based on 12 sessions
History
Date
Time
Result
Not Attempted Yet
If \(-1 < x < y < 0\), which of the following shows the expressions \(xy, x^2y,\) and \(xy^2\) listed in order from least to greatest?
\(A. xy, x^2y, xy^2\)
\(B. xy, xy^2, x^2y\)
\(C. xy^2, xy, x^2y\)
\(D. xy^2, x^2y, xy\)
\(E. x^2y, xy^2, xy\)
\(A. xy, x^2y, xy^2\)
\(B. xy, xy^2, x^2y\)
\(C. xy^2, xy, x^2y\)
\(D. xy^2, x^2y, xy\)
\(E. x^2y, xy^2, xy\)
23 Feb 2025, 00:40
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OE
You are given that \(-1 < x < y < 0.\) Since x and y are both negative numbers, it follows that xy is positive and both \(x^2y\) and \(xy^2\) are negative. So xy is greater than both \(x^2y\) and \(xy^2\). Now you need to determine which is greater, \(x^2y\) or \(xy^2\). You can do this by multiplying the inequality \(x < y\) by the positive number xy to get \(x^2y < xy^2\). Thus \(x^2y < xy^2 < xy,\) and the correct answer is Choice E.
You are given that \(-1 < x < y < 0.\) Since x and y are both negative numbers, it follows that xy is positive and both \(x^2y\) and \(xy^2\) are negative. So xy is greater than both \(x^2y\) and \(xy^2\). Now you need to determine which is greater, \(x^2y\) or \(xy^2\). You can do this by multiplying the inequality \(x < y\) by the positive number xy to get \(x^2y < xy^2\). Thus \(x^2y < xy^2 < xy,\) and the correct answer is Choice E.
