An ice ball melts at a rate proportional to t | vedclass

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(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -kx$ (where $k > 0$).Int

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$\frac{1}{4}$

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(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -kx$ (where $k > 0$).Integrating this,we get $x(t) = x_0 e^{-kt}$.Given that half the ice melts in $20 	ext{ minutes}$,at $t = 20$,$x(20) = \frac{x_0}{2}$.So,$\frac{x_0}{2} = x_0 e^{-20k}$,which implies $e^{-20k} = \frac{1}{2}$.We need to find the amount of ice left after $40 	ext{ minutes}$,which is $x(40) = x_0 e^{-40k}$.$x(40) = x_0 (e^{-20k})^2 = x_0 \left(\frac{1}{2}ight)^2 = \frac{x_0}{4}$.Since the amount left is $Kx_0$,we have $Kx_0 = \frac{x_0}{4}$,so $K = \frac{1}{4}$.

An ice ball melts at a rate proportional to the amount of ice present at that instant. Half the quantity of ice melts in $20 \text{ minutes}$. Let $x_0$ be the initial quantity of ice. If after $40 \text{ minutes}$ the amount of ice left is $Kx_0$,then $K=$

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