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\(x^4 * y < 0\) -> \(y<0\), as a variable (that's not 0) powered to an even is greater than 0. Hence, x must be positive and y negative
\(y^4 * x > 0\) -> \(x>0\), same reasoning. Hence, x must be positive.

Consequently, x > y

Option A
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The two inequalities simply mean that y<0 and x>0. Therefore it must be true that x>y.

Final answer is (A)

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From the first information, x^4 ∗y<0
x^4 is always positive
y is negative.

From the second information, x∗y^4>0
y^4 is always positive
x is also positive.


From the above discussions, we can conclude as x>y (A)
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If \(x^4∗y<0\) and \(x∗y^4>0\) which of the following must be true?

A. x>y
B. y>x
C. x=y
D. x<0
E. y>0

\(x^4∗y<0\) suggests y < 0
AND
\(x∗y^4>0\) suggests that x > 0

So, x > y

Answer A.
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If x^4 * y < 0, then we conclude that y<0, since irrespective of the value of x, x^4 will be positive, hence for x^4 * y to be negative, then y<0.
y^4 * x > 0, then we can conclude that x>0, since irrespective of the value of y, y^4 > 0, hence x>0 for y^4 * x > 0.

Scanning through the answer choices, only A x>y must be true because a positive number is greater than a negative number.

The answer is A.
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\(x^{4}∗y<0 => y<0, x \neq{0}\)
\(x∗y^{4}>0 => x>0\)
=> x>y
=> Choice A
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