MO MAP Algebra II Geometry. Practice it free.

Set up the proportion $\dfrac{8\text{ pounds}}{1\text{ cubic foot}} = \dfrac{1{,}920\text{ pounds}}{x\text{ cubic feet}}\textsf{.}$ Cross-multiplying gives $8x = 1{,}920\textsf{,}$ so $x = 240\text{ ft}^{3}\textsf{.}$

The endpoints share the same $y$-coordinate, so the diameter is horizontal with length $|18 - 2| = 16\textsf{.}$ The radius is half the diameter, so $r = \dfrac{16}{2} = 8\textsf{.}$

Circle A has center $(4, 2)$ and radius $6$. Shifting left $5$ moves the center to $(-1, 2)$, so Circle B is $(x + 1)^2 + (y - 2)^2 = 36$.

The center is $(-8, 6)$ and the radius is $13\textsf{.}$ A point on the circle must satisfy $6 - 13 \le b \le 6 + 13\textsf{,}$ so $-7 \le b \le 19\textsf{.}$ The value $11$ is in this interval.

Set $y = 0$: then $x^{2} = 16$, so the vertices are $(\pm 4, 0)$ and the positive x-coordinate is $4$.

The radius measures $6$ units, so the diameter measures $12$ units.

Since $\dfrac{8 + Q_y}{2} = 2$, we get $Q_y = 4 - 8 = -4$.

Perpendicular lines have slopes that are negative reciprocals, so $m = -2$. Substitute $(2, -1)$ into $y = -2x + b$: $-1 = -2(2) + b = -4 + b$, so $b = 3$. The equation is $y = -2x + 3\textsf{.}$