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**Correct Option:** A **Reasoning:** * First derivative: \( \frac{dy}{dx} = -3x^2 + 6x + 8 \) * Set second derivative to zero: \( \frac{d^2y}{dx^2} = -6x + 6 = 0 \implies x = 1 \) * Substitute \( x=1 \) into original equation: \( y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -10 \). The point is (1, -10).

The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$. The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Substitute the expression for $r$: $V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3$ $V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right)$ $V = \frac{125\pi}{48}(3x+1)^3$ To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3\right)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$ $\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$ $\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$ $\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$ Now, we evaluate this rate of change when $x=1$: $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(4)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16} \cdot 16$ $\left.\frac{dV}{dx}\right|_{x=1} = 375\pi$ The final answer is $\boxed{\text{375\pi}}$.

Let the side of the square be $s$. The perimeter of the square is $P = 4s$. We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$. We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$. Differentiate the perimeter equation with respect to time $t$: $\frac{dP}{dt} = \frac{d}{dt}(4s)$ $\frac{dP}{dt} = 4 \frac{ds}{dt}$ Substitute the given value of $\frac{ds}{dt}$: $\frac{dP}{dt} = 4 (-1.5~cm/s)$ $\frac{dP}{dt} = -6~cm/s$ The rate of decrease of its perimeter is $6~cm/s$. The final answer is $\boxed{6~cm/s}$.

The position vector of the midpoint \(D\) is the average of the position vectors of \(B\) and \(C\) relative to \(A\):\[\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\] \[\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\] The length of the median is the magnitude of the vector \(\vec{AD}\), denoted as \(|\vec{AD}|\):\[|\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(0\right)^2 + \left(\frac{5}{2}\right)^2}\] \[|\vec{AD}| = \frac{\sqrt{34}}{2}\]