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Consider the curve given by the equation xy^2+5xy=50xy 2 +5xy=50x, y, squared, plus, 5, x, y, equals, 50. it can be shown that \dfrac{dy}{dx}=\dfrac{-y(y+5)}{x(2y +5)} dx dy = x(2y+5) −y(y+5) start fraction, d, y, divided by, d, x, end fraction, equals, start fraction, minus, y, left parenthesis, y, plus, 5, right parenthesis, divided by, x, left parenthesis, 2, y, plus, 5, right parenthesis, end fraction. write the equation of the vertical line that is tangent to the curve

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Consider the curve given by the equation xy^2+5xy=50xy 2 +5xy=50x, y, squared, plus, 5, x, y, equals...
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