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27 Feb 2025, 01:14
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (00:41) correct
16% (00:36) wrong based on 6 sessions
History
Date
Time
Result
Not Attempted Yet
If \(S^2 > T^2\), which of the following must be true?
(A) \(S > T\)
(B) \(S^2 > T\)
(C) \(ST > 0\)
(D) \(|S| > |T|\)
(E) \(ST < 0\)
(A) \(S > T\)
(B) \(S^2 > T\)
(C) \(ST > 0\)
(D) \(|S| > |T|\)
(E) \(ST < 0\)
01 Mar 2025, 10:00
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For must be true questions, use counterexamples to eliminate wrong choices.
S^2>T^2
(A) S>T
(-3)^2 > (2)^2
But -3 > 2? Nope.
(B) S^2>T
(1/2)^2 > (1/4)^2
But (1/2)^2 > 1/4? Nope.
(C) ST>0
(-3)^2 > (2)^2
But (-3)(2) > 0? Nope.
(E) ST<0
3^2 > 2^2
But 3*2 < 0? Nope.
That leaves us with option D.
S^2>T^2
(A) S>T
(-3)^2 > (2)^2
But -3 > 2? Nope.
(B) S^2>T
(1/2)^2 > (1/4)^2
But (1/2)^2 > 1/4? Nope.
(C) ST>0
(-3)^2 > (2)^2
But (-3)(2) > 0? Nope.
(E) ST<0
3^2 > 2^2
But 3*2 < 0? Nope.
That leaves us with option D.
02 Mar 2025, 11:13
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OE
The square root of a squared variable is equal to the absolute value of that variable: \(\sqrt{x^2}=|x|\), not x. So, taking the square root of both sides of this inequality results in \(|S| > |T|\).
Answer choice (A) does not have to be true because S could be negative while T is positive. For example, if S = –5 and T = 4, then \(S^2\) \(> T^2\).
Testing fractions in answer choice (B) shows that it does not have to be true. If \(S^2 =\frac{1}{9}\) and \(T^2 = \frac{1}{16}\), then \(T = \frac{1}{4}\) or \(\frac{–1}{4}\). \(\frac{1}{4}\) is greater than \(\frac{1}{9}\), which means that T can be greater than \(S^2.\)
Answer choice (C) does not have to be true because S and T could have opposite signs.
Answer choice (E) does not have to be true because S and T could have the same sign.
Answer: D
The square root of a squared variable is equal to the absolute value of that variable: \(\sqrt{x^2}=|x|\), not x. So, taking the square root of both sides of this inequality results in \(|S| > |T|\).
Answer choice (A) does not have to be true because S could be negative while T is positive. For example, if S = –5 and T = 4, then \(S^2\) \(> T^2\).
Testing fractions in answer choice (B) shows that it does not have to be true. If \(S^2 =\frac{1}{9}\) and \(T^2 = \frac{1}{16}\), then \(T = \frac{1}{4}\) or \(\frac{–1}{4}\). \(\frac{1}{4}\) is greater than \(\frac{1}{9}\), which means that T can be greater than \(S^2.\)
Answer choice (C) does not have to be true because S and T could have opposite signs.
Answer choice (E) does not have to be true because S and T could have the same sign.
Answer: D
