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Re: QOTD # 13 If –1 < x < y < 0, which of the following shows th [#permalink]
13 Sep 2016, 07:06

Expert's post

Explanation

You are given that –1 < x < f < 0. Since x and f are both negative numbers, it follows that xy is positive and both \(x^2 y\) and \(xy^2\) are negative. So xy is greater than both \(x^2 y\) and \(xy^2\) . Now you need to determine which is greater, \(x^2 y\) or \(xy^2\) . You can do this by multiplying the inequality x < y by the positive number xy to get \(x^2 y\) < \(xy^2\) . Thus \(x^2 y\) < \(xy^2\) < \(xy\), and the correct answer is Choice E.
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