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Determine if the following is a function: 2x2 + 3y2 = 16
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Let X = {1,2,3,4}, where R is a relation on X. Determine whether the relation is a function: R = {(3, 1), (1, 3), (4, 4), (3, 2)}
\[2x^2+3y^2=16\\ 3y^2=16-2x^2 \\y^2=\frac{ 1 }{ 3 }(16-2x^2)\\y=\pm \sqrt{\frac{ 1 }{ 3 }(16-2x^2)}\\so\\if\\you\\put\\x\\into\\it \\you\\will\\have\\2\\y\\so\\it\\is\\not\\a\\function\\\]
what about the other ?
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