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(x + y)(x − y) = 0 xy ≠ 0
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Updated on: 31 Oct 2018, 06:21
Updated on: 31 Oct 2018, 06:21
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Question Stats:
48% (01:26) correct
51% (01:26) wrong based on 27 sessions
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\((x + y)(x - y) = 0\)
\(xy \neq 0\)
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.
\(xy \neq 0\)
Quantity A |
Quantity B |
\(6\sqrt{\frac{19}{2x^2}}\) |
\(\sqrt{\frac{342}{y^2}}\) |
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.
Re: (x + y)(x − y) = 0 xy ≠ 0
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31 Oct 2018, 03:56
31 Oct 2018, 03:56
Expert Reply
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)
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.
\(xy \neq 0\)
Quantity A |
Quantity B |
\(\sqrt[6]{\frac{19}{2x^2}}\) |
\(\sqrt{\frac{342}{y^2}}\) |
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.
\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)
I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))
x=y or x=-y or both IMPLIES \(x^2=y^2\)
\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal
C
