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For all numbers x and y, the operation & is defined by x&y
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27 Aug 2018, 09:32
27 Aug 2018, 09:32
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Question Stats:
90% (00:43) correct
10% (00:35) wrong based on 30 sessions
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For all numbers x and y, the operation & is defined by \(x&y = \frac{x-y}{x-1}\) . If \((-3)&q=2\), then q =
A. -11
B. -8
C. -5
D. 5
E. 11
A. -11
B. -8
C. -5
D. 5
E. 11
Re: For all numbers x and y, the operation & is defined by x&y
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28 Aug 2018, 06:54
28 Aug 2018, 06:54
(-3-q)/(-3-1) = 2
solve this and you get q = 5
solve this and you get q = 5
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Re: For all numbers x and y, the operation & is defined by x&y
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01 Jul 2022, 04:31
01 Jul 2022, 04:31
1
Given that the operation & is defined by \(x\&y = \frac{x-y}{x-1}\) and \((-3)\&q=2\) and we need to find the value of q
To find \((-3)\&q\) we need to compare what is before and after the & in \((-3)\&q\) and \(x\&y\)
=> We need to substitute x with -3 and y with q in \(x\&y = \frac{x-y}{x-1}\) to get the value of \((-3)\&q\)
=> \(-3\&q = \frac{-3-q}{-3-1}\) = \(\frac{-(3+q)}{-4}\) = \(\frac{3+q}{4}\) = 2 (Given)
=> 3+q = 4*2 = 8
=> q = 8-3 = 5
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
To find \((-3)\&q\) we need to compare what is before and after the & in \((-3)\&q\) and \(x\&y\)
=> We need to substitute x with -3 and y with q in \(x\&y = \frac{x-y}{x-1}\) to get the value of \((-3)\&q\)
=> \(-3\&q = \frac{-3-q}{-3-1}\) = \(\frac{-(3+q)}{-4}\) = \(\frac{3+q}{4}\) = 2 (Given)
=> 3+q = 4*2 = 8
=> q = 8-3 = 5
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
