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(C) Calculation of $A$:Let $S$ be the area of a square of side $a$. Then $S = a^2$.The relative error is $\frac{\Delta S}{S} = \frac{2a \Delta a}{a^2}

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$C, D, A, B$

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(C) Calculation of $A$:Let $S$ be the area of a square of side $a$. Then $S = a^2$.The relative error is $\frac{\Delta S}{S} = \frac{2a \Delta a}{a^2} = 2 \frac{\Delta a}{a}$.Given $\frac{\Delta a}{a} = 0.4$,so $A = 2 	imes 0.4 = 0.8$.Calculation of $B$:Let $V$ be the volume of a sphere of radius $r$. Then $V = \frac{4}{3} \pi r^3$.The relative error is $\frac{\Delta V}{V} = \frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} = 3 \frac{\Delta r}{r}$.Given $\frac{\Delta r}{r} = 0.3$,so $B = 3 	imes 0.3 = 0.9$.Calculation of $C$:Let $S'$ be the surface area of a closed cylinder with radius $r$ and height $h$. Since $r = h$,$S' = 2 \pi r h + 2 \pi r^2 = 2 \pi h^2 + 2 \pi h^2 = 4 \pi h^2$.The relative error is $\frac{\Delta S'}{S'} = \frac{8 \pi h \Delta h}{4 \pi h^2} = 2 \frac{\Delta h}{h}$.Given $\frac{\Delta h}{h} = 0.2$,so $C = 2 	imes 0.2 = 0.4$.Calculation of $D$:Given $y = x^2 + x - 3$,then $\frac{dy}{dx} = 2x + 1$.The approximate error $\Delta y \approx \frac{dy}{dx} \Delta x = (2x + 1) \Delta x$.For $x = 2$ and $\Delta x = 0.1$,$\Delta y = (2(2) + 1) 	imes 0.1 = 5 	imes 0.1 = 0.5$.So $D = 0.5$.Comparing the values: $C = 0.4, D = 0.5, A = 0.8, B = 0.9$.The ascending order is $C < D < A < B$.

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