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(D) The slope of the tangent to the curve $y = f(x)$ is given by $\frac{dy}{dx}$.The given line is $2x - 3y = 5$,which can be rewritten as $3y = 2x -

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$-3/2$

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(D) The slope of the tangent to the curve $y = f(x)$ is given by $\frac{dy}{dx}$.The given line is $2x - 3y = 5$,which can be rewritten as $3y = 2x - 5$ or $y = \frac{2}{3}x - \frac{5}{3}$.The slope of this line is $m_1 = \frac{2}{3}$.Since the tangent is perpendicular to this line,the product of their slopes must be $-1$. Let $m_2$ be the slope of the tangent.$m_1 	imes m_2 = -1$$\frac{2}{3} 	imes m_2 = -1$$m_2 = -\frac{3}{2}$.Therefore,$\frac{dy}{dx} = -\frac{3}{2}$.

If the tangent to the curve $y = f(x)$ at a point is perpendicular to the line $2x - 3y = 5$,then the value of $\frac{dy}{dx}$ at that point is:

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