Question 1021146: if xy<0 and y>0 which of the following must be positive?
A.x-y
B.2x+3y
C.x+10/y+2
D.(-y-2)/x
E.2y^2+x
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i would say selection D.
that is:
(-y-2) / x
you are given that y > 0.
you are also given that xy < 0.
if y > 0, and xy < 0, then x < 0.
has to be because a negative times a positive is a negative which is less than 0.
so, if y is positive, x must be negative in order for x times y to be negative.
of all your options, the only one that has to be positive is eelction D.
this is because y is positive, therefore -y - 2 has to be negative.
since x also has to be negative, you get a negative divided by a negative which has to be positive.
none of the others have to be positive.
we'll go through each one to show you why.
A.x-y
if y is positive and x is negative, then x - y will have to be negative because when you subtract a positive you are adding a negative and a negative (x) plus a negative (-y) is a negative.
B.2x+3y
whether thi is positive or negative depends on the values of x and y.
if x is -20 and y is + 1, then the sum will be negative.
C.x+10/y+2
this also depends on the values of x and y.
if x is more negative than -10, then the numerator will be negative and the denominator will be positive, resulting in a negative.
D.(-y-2)/x
y is positive and so -y is negative.
the numerator has to be negative.
the denominator has to be negative.
the result is positive.
E.2y^2+x
if x is large enough, then the result will be negative.
for example, y is 1 and x is - 100.
y^2 is 1.
add 1 and -100 and you get a negative number.
the only one that has to be positive is selection D.
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