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The greatest possible value of 3/x-y, where
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27 Nov 2019, 11:07
27 Nov 2019, 11:07
1
Expert Reply
00:00
Question Stats:
74% (00:44) correct
25% (00:54) wrong based on 87 sessions
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Quantity A |
Quantity B |
The greatest possible value of \(\frac{3}{x-y}\), where \(6 \leq x \leq 8\) and \(0 \leq y \leq 5\) |
\(3\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: The greatest possible value of 3/x-y, where
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27 Nov 2019, 12:26
27 Nov 2019, 12:26
2
The answer is C.
To get greatest value of A, x-y has to be the least. To minimize x-y, we have to select lowest x-value and greatest y-value.
A=3/(6-5) = 3
So A and B are equal.
To get greatest value of A, x-y has to be the least. To minimize x-y, we have to select lowest x-value and greatest y-value.
A=3/(6-5) = 3
So A and B are equal.
Re: The greatest possible value of 3/x-y, where
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21 Sep 2020, 04:14
21 Sep 2020, 04:14
Expert Reply
Bump for further disscussiona
