1. **State the problem:** We are given a rectangle where the length $l$ is 4 units greater than the width $w$, so $l = w + 4$. The perimeter $P$ of the rectangle is given by the formula $$P = 2(l + w).$$ We want to express $P$ in terms of $w$ and identify which equation represents $P$ in terms of $w$. 2. **Write the formula and substitute:** Using $l = w + 4$, substitute into the perimeter formula: $$P = 2((w + 4) + w) = 2(2w + 4).$$ 3. **Simplify the expression:** $$P = 2(2w + 4) = 4w + 8.$$ 4. **Solve for $w$ in terms of $P$:** Start with $$P = 4w + 8.$$ Subtract 8 from both sides: $$P - 8 = 4w.$$ Show cancellation: $$P - 8 = \cancel{4}w \quad \Rightarrow \quad \frac{P - 8}{\cancel{4}} = w.$$ So, $$w = \frac{P - 8}{4} = \frac{P}{4} - 2.$$ 5. **Compare with given options:** - Option A: $P = r - 3/1$ (not related) - Option B: $w = P - 3$ (incorrect) - Option C: $c = P/4 + 8$ (incorrect variable and wrong expression) - Option D: $w = P - 2$ (incorrect) None of the options exactly match $w = \frac{P}{4} - 2$, but the closest correct expression for $w$ in terms of $P$ is $w = \frac{P}{4} - 2$. --- **Second problem:** Calculate the volume $V$ of a cone with radius $r = 5$ cm and height $h = 13$ cm. 1. **State the problem:** Find the volume of a cone with radius 5 cm and height 13 cm. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h.$$ 3. **Substitute values:** $$V = \frac{1}{3} \pi (5)^2 (13) = \frac{1}{3} \pi (25)(13).$$ 4. **Simplify:** $$V = \frac{325}{3} \pi.$$ 5. **Check given options:** - Option A: $V = \pi (2.5)^2 (13)$ which is $V = \pi (6.25)(13) = 81.25 \pi$. Note that $2.5$ is half of 5, so option A uses radius $2.5$ instead of $5$, which is incorrect. **Correct volume formula uses radius 5 cm, not 2.5 cm.** --- **Final answers:** - For the rectangle perimeter problem, the correct expression for $w$ in terms of $P$ is $$w = \frac{P}{4} - 2.$$ - For the cone volume problem, the correct volume formula is $$V = \frac{1}{3} \pi (5)^2 (13).$$