Let B equiv (0,3) and C equiv (4,0). The poin | vedclass

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(A) Let $A = (h, 2h)$.The distance $OA = \sqrt{h^2 + (2h)^2} = \sqrt{5}h$.Given that the point $A$ moves at a rate of $2 	ext{ units/sec}$,we have $\

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$\frac{11}{\sqrt{5}} 	ext{ (units)}^2/	ext{sec}$

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(A) Let $A = (h, 2h)$.The distance $OA = \sqrt{h^2 + (2h)^2} = \sqrt{5}h$.Given that the point $A$ moves at a rate of $2 	ext{ units/sec}$,we have $\frac{d(OA)}{dt} = 2$.Thus,$\sqrt{5} \frac{dh}{dt} = 2$,which implies $\frac{dh}{dt} = \frac{2}{\sqrt{5}}$.The area $\alpha$ of $	riangle ABC$ with vertices $A(h, 2h)$,$B(0, 3)$,and $C(4, 0)$ is given by the determinant formula:$\alpha = \frac{1}{2} |h(3-0) + 0(0-2h) + 4(2h-3)| = \frac{1}{2} |3h + 8h - 12| = \frac{1}{2} |11h - 12|$.Assuming the area is increasing,we consider $\alpha = \frac{11h - 12}{2}$.Differentiating with respect to $t$: $\frac{d\alpha}{dt} = \frac{11}{2} \frac{dh}{dt}$.Substituting $\frac{dh}{dt} = \frac{2}{\sqrt{5}}$,we get $\frac{d\alpha}{dt} = \frac{11}{2} \cdot \frac{2}{\sqrt{5}} = \frac{11}{\sqrt{5}} 	ext{ (units)}^2/	ext{sec}$.

Let $B \equiv (0,3)$ and $C \equiv (4,0)$. The point $A$ is moving on the line $y=2x$ at the rate of $2 \text{ units/second}$. The area of $\triangle ABC$ is increasing at the rate of

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