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29 Jul 2024, 11:14
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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0% (00:00) correct
100% (00:07) wrong based on 1 sessions
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If a<b<0, which of the following numbers must be positive?
Indicate all such numbers.
A. \(a-b\)
B. \(a^2-b^2\)
C. \(ab\)
D. \(a^2b\)
E. \(a^2b+ab^2\)
Indicate all such numbers.
A. \(a-b\)
B. \(a^2-b^2\)
C. \(ab\)
D. \(a^2b\)
E. \(a^2b+ab^2\)
13 Nov 2024, 10:38
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OE
In this question, you are given that and \(a<b<c\) are asked to determine which of the answer choices must be positive. Note that the condition \(a<b<c\) means that a and b are negative and that \(a<b\).
Choice A: a - b. In the question, it is given that \(a<b\) . Subtracting b from both sides of the inequality gives the inequality \(a-b<0\). Therefore, a - b must be negative.
Choice B: \(a^2-b^2\). Since a and b are negative, you can square both sides of the inequality a<b to get the inequality \(a^2-b^2\). Then you can subtract \(b^2\) from both sides of the inequality \(a^2-b^2\) to conclude that \(a^2-b^2>0\). So \(a^2-b^2\) must be positive. Alternatively, note that \(a^2-b^2\) can be factored as (a - b)(a + b). The factor a - b is Choice A, which must be negative, and the factor a + b is the sum of two negative numbers, which also must be negative. Thus, \(a^2-b^2\) is the product of two negative numbers, so it must be positive.
Choice C: \(ab\). Because a and b are negative, you can conclude that their product \(ab\) must be positive.
Choice D: \(a^2b\). Because \(a^2b\) can be written as (a)(a)(b), which is the product of three negative numbers, you can conclude that \(a^2b\) must be negative.
Choice E: \(a^2b+ab^2\). By the reasoning in the explanation of Choice D, Choice E is the sum of two negative numbers. Therefore, you can conclude that \(a^2b+ab^2\) must be negative. Choices B and C must be positive, and Choices A, D, and E must be negative.
The correct answer consists of Choices B and C.
In this question, you are given that and \(a<b<c\) are asked to determine which of the answer choices must be positive. Note that the condition \(a<b<c\) means that a and b are negative and that \(a<b\).
Choice A: a - b. In the question, it is given that \(a<b\) . Subtracting b from both sides of the inequality gives the inequality \(a-b<0\). Therefore, a - b must be negative.
Choice B: \(a^2-b^2\). Since a and b are negative, you can square both sides of the inequality a<b to get the inequality \(a^2-b^2\). Then you can subtract \(b^2\) from both sides of the inequality \(a^2-b^2\) to conclude that \(a^2-b^2>0\). So \(a^2-b^2\) must be positive. Alternatively, note that \(a^2-b^2\) can be factored as (a - b)(a + b). The factor a - b is Choice A, which must be negative, and the factor a + b is the sum of two negative numbers, which also must be negative. Thus, \(a^2-b^2\) is the product of two negative numbers, so it must be positive.
Choice C: \(ab\). Because a and b are negative, you can conclude that their product \(ab\) must be positive.
Choice D: \(a^2b\). Because \(a^2b\) can be written as (a)(a)(b), which is the product of three negative numbers, you can conclude that \(a^2b\) must be negative.
Choice E: \(a^2b+ab^2\). By the reasoning in the explanation of Choice D, Choice E is the sum of two negative numbers. Therefore, you can conclude that \(a^2b+ab^2\) must be negative. Choices B and C must be positive, and Choices A, D, and E must be negative.
The correct answer consists of Choices B and C.
