A certain quantity of water cools from 70,^{c | vedclass

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)$,where $T_0

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$45$

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)$,where $T_0$ is the surrounding temperature.For the first interval: $\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_0 \right) \Rightarrow 2 = K(65 - T_0)$  --- $(1)$For the second interval: $\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_0 \right) \Rightarrow 1.2 = K(57 - T_0)$ --- $(2)$Dividing $(1)$ by $(2)$: $\frac{2}{1.2} = \frac{65 - T_0}{57 - T_0} \Rightarrow \frac{5}{3} = \frac{65 - T_0}{57 - T_0}$.$5(57 - T_0) = 3(65 - T_0) \Rightarrow 285 - 5T_0 = 195 - 3T_0$.$285 - 195 = 5T_0 - 3T_0 \Rightarrow 90 = 2T_0 \Rightarrow T_0 = 45\,^{\circ}C$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

$A$ certain quantity of water cools from $70\,^{\circ}C$ to $60\,^{\circ}C$ in the first $5\, \text{minutes}$ and to $54\,^{\circ}C$ in the next $5\, \text{minutes}$. The temperature of the surroundings is ......... $^{\circ}C$.

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