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(x^2 - 9)/3
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Updated on: 27 Dec 2018, 05:43
Updated on: 27 Dec 2018, 05:43
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Question Stats:
72% (01:08) correct
27% (01:10) wrong based on 244 sessions
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Given \(x >4\)
Quantity A |
Quantity B |
\(\left[ \begin{array}{cc|r} \frac{\frac{(x^2 - 9)}{3}}{\frac{(x+3)}{8} \end{array} \right] ^{-1}\) |
\(\frac{3}{8}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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(x^2 - 9)/3
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25 Mar 2021, 22:48
25 Mar 2021, 22:48
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Carcass wrote:
Given \(x >4\)
Quantity A |
Quantity B |
\(\left[ \begin{array}{cc|r} \frac{\frac{(x^2 - 9)}{3}}{\frac{(x+3)}{8} \end{array} \right] ^{-1}\) |
\(\frac{3}{8}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Col. A: \(\left[ \begin{array}{cc|r} \frac{\frac{(x^2 - 9)}{3}}{\frac{(x+3)}{8} \end{array} \right] ^{-1}\)
Col. B: \(\frac{3}{8}\)
Col. A: \(\left[ \begin{array}{cc|r}{\frac{(x+3)(x-3)}{3}\frac{8}{(x+3)} \end{array} \right] ^{-1}\)
Col. B: \(\frac{3}{8}\)
Col. A: \(\frac{3}{8(x-3)}\)
Col. B: \(\frac{3}{8}\)
Multiplying both sides by \(\frac{8}{3}\);
Col. A: \(\frac{1}{(x-3)}\)
Col. B: \(1\)
Since, \(x > 4, (x-3) > 0\)
Multiplying both sides by \((x-3)\);
Col. A: \(1\)
Col. B: \((x-3)\)
Adding \(3\) to both sides;
Col. A: \(4\)
Col. B: \(x\)
Hence, option B
