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If x < y and 0 < x + y, which of the following must be negat
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01 Feb 2020, 09:33
01 Feb 2020, 09:33
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56% (01:10) correct
44% (01:23) wrong based on 25 sessions
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If \(x < y\) and \(0 < x + y\), which of the following must be negative?
Indicate all possible values.
\(\square\) –x
\(\square\) –y
\(\square\) x – y
\(\square\) \((x – y)^2\)
\(\square\) 2x – y
Indicate all possible values.
\(\square\) –x
\(\square\) –y
\(\square\) x – y
\(\square\) \((x – y)^2\)
\(\square\) 2x – y
Re: If x < y and 0 < x + y, which of the following must be negat
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03 Feb 2020, 10:03
03 Feb 2020, 10:03
1
We know: x < y (or x-y < 0) and 0 < x+y or if we subtract x --> -x < y --> |x| < y --> therefore x can be negative or positive, but y has to be positive
A) see above, x can be positive or negative --> therefore -x can be positive or negative --> not true
B) -y has to be negative, since |x| < y and that is only true, if y is positive --> true
C) see beginning (x-y <0 ) --> true
D) a squared number is always positive
E) can be negative, e.g. when x<0 or if 2x < y, but we don't know that
--> B and C are correct
A) see above, x can be positive or negative --> therefore -x can be positive or negative --> not true
B) -y has to be negative, since |x| < y and that is only true, if y is positive --> true
C) see beginning (x-y <0 ) --> true
D) a squared number is always positive
E) can be negative, e.g. when x<0 or if 2x < y, but we don't know that
--> B and C are correct
Re: If x < y and 0 < x + y, which of the following must be negat
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18 Jun 2023, 11:05
18 Jun 2023, 11:05
1
A) If you consider the case where x>0, y>0 then -x is negative as x is some positive number.
If you consider the case where x<0, then -x>0. Not always true.
B) If you consider the case where x<0, y must be some positive integer greater x (ex: x=-2, y must be at least 3 so that the second inequality x+y>0 holds). In this case, as y is some positive number, -y will always be negative.
The other case is where both are positive integers, and again y will be negative. Always True.
C) Simply move the first inequality (x-y)<0. Always true.
D) (x-y)^2 will always be >=0, so it can never be negative
E) 2x-y
Consider the case where x=4, y=5
and the case where x=-5, y=6. Not always true.
B and C are always true, and are therefore the correct options.
If you consider the case where x<0, then -x>0. Not always true.
B) If you consider the case where x<0, y must be some positive integer greater x (ex: x=-2, y must be at least 3 so that the second inequality x+y>0 holds). In this case, as y is some positive number, -y will always be negative.
The other case is where both are positive integers, and again y will be negative. Always True.
C) Simply move the first inequality (x-y)<0. Always true.
D) (x-y)^2 will always be >=0, so it can never be negative
E) 2x-y
Consider the case where x=4, y=5
and the case where x=-5, y=6. Not always true.
B and C are always true, and are therefore the correct options.
