If x < 0 and y > 0, determine the sign of the real number. (a) x y positive negative (b) xy2 positive negative (c) x y xy positive negative (d) y(y x) positive negative

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If x &lt; 0 and y &gt; 0, determine the sign of the real number. (a) x y positive negative (b) xy2 positive negative (c) x y xy positive negative (d) y(y x) positive negative

kenna17 · Accepted Answer

Answer:From the question it is given that x&lt;0 and y&gt;0 or in other words x is negative and y is positive.(a)x y : x is negative and y is positivenegative x positive = negativeTherefore, xy is negative(b)x[tex]y^{2}[/tex] : x is negative and [tex]y^{2}[/tex] is positive negative x positive = negativeTherefore, x[tex]y^{2}[/tex] is negative(c) (x y)xy : x is negative and [tex]y^{2}[/tex] is positive (x y) is negative and xy is negative {from (a)}negative x negative = positiveTherefore, (x-y)xy is negative(d) y(y x): x is negative and y is positivehere y - x is positivepositive x positive = positiveTherefore, (y-x)y is negative

If x < 0 and y > 0, determine the sign of the real number. (a) x y positive negative (b) xy2 positive negative (c) x − y xy positive negative (d) y(y − x) positive negative

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