$$\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=$$ A. \(\frac{12}{9}\)
B. \(\frac{49}{144}\)
C. \(\frac{12}{7}\)
D. \(7\)
E. \(9\)
[spoiler]OA=E[/spoiler]
Source: Veritas Prep
\(\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=\)
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Simplifying the numerator we have:Gmat_mission wrote:$$\frac{\left(3^{-1}+4^{-1}\right)^{-2}}{\frac{4^2}{7^2}}=$$ A. \(\frac{12}{9}\)
B. \(\frac{49}{144}\)
C. \(\frac{12}{7}\)
D. \(7\)
E. \(9\)
[spoiler]OA=E[/spoiler]
Source: Veritas Prep
(1/3 + 1/4)^-2
(4/12 + 3/12)^-2
(7/12)^-2 = (12/7)^2 = 12^2/7^2
Now we have:
(12^2/7^2)/(4^2/7^2)
12^2/4^2 = (12/4)^2 = 9
Answer: E
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