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[m](3^{−1} − 2^{−1})^{−1}[/m]
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22 Jan 2021, 09:37
22 Jan 2021, 09:37
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Question Stats:
90% (00:42) correct
9% (00:43) wrong based on 11 sessions
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\((3^{−1} − 2^{−1})^{−1}\)
A. -6
B. -5
C. -1/6
D. 16
E. 6
A. -6
B. -5
C. -1/6
D. 16
E. 6
[m](3^{−1} − 2^{−1})^{−1}[/m]
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22 Jan 2021, 09:41
22 Jan 2021, 09:41
1
(\(\frac{1}{3}\) - \(\frac{1}{2}\))^-1
(\(\frac{2-3}{6}\))^-1
(\(\frac{-1}{6}\))^-1
-6
Answer A
(\(\frac{2-3}{6}\))^-1
(\(\frac{-1}{6}\))^-1
-6
Answer A
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Re: [m](3^{−1} − 2^{−1})^{−1}[/m]
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22 Jan 2021, 09:50
22 Jan 2021, 09:50
Carcass wrote:
\((3^{−1} − 2^{−1})^{−1}\)
A. -6
B. -5
C. -1/6
D. 16
E. 6
A. -6
B. -5
C. -1/6
D. 16
E. 6
\(3^{−1}\) = 1/3
\(2^{−1}\) = 1/2
And 1/3 - 1/2 = 2/6 - 3/6 = -1/6
So, \((3^{−1} − 2^{−1})^{−1}\) = (1/3 - 1/2)^(-1)
= (-1/6)^(-1)
Finally, there's a nice rule that says (a/b)^(-n) = (b/a)^n
So, (-1/6)^(-1) = (-6/1)^(1)
= -6/1
= -6
Answer: A
Cheers,
Brent
