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infiniti007
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|xy| > x²y² only if x and y are NONZERO, with the result that both sides of the inequality are POSITIVE.infiniti007 wrote:Is |xy| > x²y²?
1.) 0 < x² < 1/4
2.) 0 < y² < 1/9
Since |xy| > x²y² only if both sides are positive, we can safely square the inequality:
(|xy|)² > (x²y²)²
x²y² > x�y�.
Since x and y are nonzero, x²y²>0.
Thus, we can safely divide each side by x²y²:
x²y²/x²y² > x�y�/x²y²
1 > x²y².
Question stem, rephrased:
Is x²y² < 1?
Statement 1: 0 < x² < 1/4
If x² = 1/10 and y² = 1/10, then x²y² < 1.
If x² = 1/10 and y² = 100, then x²y² > 1.
INSUFFICIENT.
Statement 2: 0 < y² < 1/9
If y² = 1/10 and x² = 1/10, then x²y² < 1.
If y² = 1/10 and x² = 100, then x²y² > 1.
INSUFFICIENT.
Statements combined:
Since x² and y² are both POSITIVE FRACTIONS, x²y² < 1.
SUFFICIENT.
The correct answer is C.
