The line joining the points (0,3) and (5,-2) | vedclass

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(C) The slope of the line joining the points $(0,3)$ and $(5,-2)$ is $m = \frac{-2-3}{5-0} = \frac{-5}{5} = -1$. The equation of this line is $(y-3) =

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$4$

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(C) The slope of the line joining the points $(0,3)$ and $(5,-2)$ is $m = \frac{-2-3}{5-0} = \frac{-5}{5} = -1$. The equation of this line is $(y-3) = -1(x-0)$,which simplifies to $y = -x+3$. For the curve $y = \frac{c}{x+1}$,the slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{-c}{(x+1)^2}$. Since the line is tangent to the curve,the slope of the tangent must equal the slope of the line: $\frac{-c}{(x+1)^2} = -1$,which implies $c = (x+1)^2$. At the point of tangency,the curve and the line intersect,so $\frac{c}{x+1} = -x+3$. Substituting $c = (x+1)^2$ into this equation: $\frac{(x+1)^2}{x+1} = -x+3$,which simplifies to $x+1 = -x+3$. Solving for $x$: $2x = 2$,so $x = 1$. Substituting $x=1$ back into $c = (x+1)^2$,we get $c = (1+1)^2 = 2^2 = 4$.

The line joining the points $(0,3)$ and $(5,-2)$ is a tangent to the curve $y=\frac{c}{x+1}$. Then,$c=$

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