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x, y, r, and t are integers such that x^y is negative and r t is posit
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13 Jan 2023, 06:21
13 Jan 2023, 06:21
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x, y, r, and t are integers such that \(x^y\) is negative and \(r^t\) is positive. If t is not a multiple of 2, then which of the following statements must be true?
(A) xr > 0
(B) x − r < 0
(C) y is a multiple of 2
(D) x^t > 0
(E) None of the above
(A) xr > 0
(B) x − r < 0
(C) y is a multiple of 2
(D) x^t > 0
(E) None of the above
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Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
Re: x, y, r, and t are integers such that x^y is negative and r t is posit
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21 Jan 2023, 10:50
21 Jan 2023, 10:50
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Re: x, y, r, and t are integers such that x^y is negative and r t is posit
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22 Jan 2023, 11:01
22 Jan 2023, 11:01
2
x^y = -ve; this means that x is a -ve number and y is an odd number. If y is an even number, x^y will be a +ve number.
r^t = +ve; this means that if r = -ve, t = even number (2 or more) and if r = +ve, t = can be even or odd numbers. And since we know that t is not a multiple of 2, then t cannot be an even number. So, r = +ve and t = odd
Now, to go over the options:
(A) xr > 0 --> x = -ve; r = +ve. xr < 0 so this is incorrect.
(B) x − r < 0 --> -ve number minus +ve number will remain negative. This is correct.
(C) y is a multiple of 2 --> according to the explaination above, y is an odd number. Therefore, this is incorrect.
(D) x^t > 0 --> -ve number to the power of an odd number will remain -ve. So, this is also incorrect.
(E) None of the above --> Since B is possible, this option is incorrect.
r^t = +ve; this means that if r = -ve, t = even number (2 or more) and if r = +ve, t = can be even or odd numbers. And since we know that t is not a multiple of 2, then t cannot be an even number. So, r = +ve and t = odd
Now, to go over the options:
(A) xr > 0 --> x = -ve; r = +ve. xr < 0 so this is incorrect.
(B) x − r < 0 --> -ve number minus +ve number will remain negative. This is correct.
(C) y is a multiple of 2 --> according to the explaination above, y is an odd number. Therefore, this is incorrect.
(D) x^t > 0 --> -ve number to the power of an odd number will remain -ve. So, this is also incorrect.
(E) None of the above --> Since B is possible, this option is incorrect.
