Problem: Find the volume of each figure. Round the answer to two decimal places. Use $\pi=3.14$ for problems 1–5 and use the exact symbol $\pi$ (with an exact expression) for problem 6, and show all work. 1. Figure 1: A rectangular prism with length 16 ft, width 4 ft, height 6 ft and a square pyramid on top with base side 12 ft and pyramid height 9 ft. Prism volume: $V_{prism}=16\times4\times6=384$ ft^3. Pyramid base area: $A_{base}=12\times12=144$ ft^2. Pyramid volume: $V_{pyr}=\frac{1}{3}A_{base}\times9=\frac{1}{3}\times144\times9=432$ ft^3. Total volume: $V=384+432=816$ ft^3. Rounded: $V\approx816.00$ ft^3. 2. Figure 2: A cylinder of height 11 m and radius 6 m with hemispheres attached top and bottom (together equal one full sphere). Cylinder volume: $V_{cyl}=\pi r^2 h=\pi\times6^2\times11=396\pi$ m^3. Two hemispheres = one sphere: $V_{sph}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\times6^3=288\pi$ m^3. Total symbolic: $V=396\pi+288\pi=684\pi$ m^3. Numeric using $\pi=3.14$: $V=684\times3.14=2147.76$ m^3. Rounded: $V\approx2147.76$ m^3. 3. Figure 3: A rectangular box length 12 cm, width 8 cm, height 8 cm with a hemisphere of radius 4 cm on top. Box volume: $V_{box}=12\times8\times8=768$ cm^3. Hemisphere volume: $V_{hemi}=\frac{1}{2}\times\frac{4}{3}\pi r^3=\frac{1}{2}\times\frac{4}{3}\pi\times4^3=\frac{128}{3}\pi$ cm^3. Numeric hemisphere using $\pi=3.14$: $V_{hemi}=\frac{128}{3}\times3.14\approx133.973333...$ cm^3. Total: $V=768+\frac{128}{3}\pi$ cm^3 $\approx768+133.973333...=901.973333...$ cm^3. Rounded: $V\approx901.97$ cm^3. 4. Figure 4: A rectangular prism base length 14 in, width 9 in, height 8 in with a four-sided pyramid roof of peak height 5 in. Prism volume: $V_{prism}=14\times9\times8=1008$ in^3. Pyramid base area: $A_{base}=14\times9=126$ in^2. Pyramid volume: $V_{pyr}=\frac{1}{3}A_{base}\times5=\frac{1}{3}\times126\times5=210$ in^3. Total volume: $V=1008+210=1218$ in^3. Rounded: $V\approx1218.00$ in^3. 5. Figure 5: Two stacked rectangular prisms: lower prism 13 mm by 7 mm by 4 mm, upper prism centered of 5 mm by 5 mm by 5 mm. Lower prism volume: $V_{lower}=13\times7\times4=364$ mm^3. Upper prism volume: $V_{upper}=5\times5\times5=125$ mm^3. Total volume: $V=364+125=489$ mm^3. Rounded: $V\approx489.00$ mm^3. 6. Figure 6: A cone of radius 10 cm and height 15 cm on top of a hemisphere of radius 10 cm. I will explain and give the exact expression using $\pi$ and then a numeric approximation. Cone volume formula: $V_{cone}=\frac{1}{3}\pi r^2 h$. Substitute $r=10$ and $h=15$: $V_{cone}=\frac{1}{3}\pi\times10^2\times15=500\pi$ cm^3. Hemisphere volume formula: $V_{hemi}=\frac{1}{2}\times\frac{4}{3}\pi r^3$. Substitute $r=10$: $V_{hemi}=\frac{1}{2}\times\frac{4}{3}\pi\times10^3=\frac{2000}{3}\pi$ cm^3. Add them for the exact total: $V_{total}=500\pi+\frac{2000}{3}\pi=\frac{3500}{3}\pi$ cm^3. Explanation: compute each solid separately with its formula, substitute the radius and height, simplify algebraically, and then add because the cone and hemisphere occupy disjoint volumes stacked vertically. Numeric approximation using the exact $\pi$ value (for rounding only): $V_{total}=\frac{3500}{3}\pi\approx3665.19$ cm^3 when using $\pi\approx3.141592653589793$ and rounding to two decimals. Final answers (rounded to two decimals): 1) $816.00$ ft^3. 2) $2147.76$ m^3. 3) $901.97$ cm^3. 4) $1218.00$ in^3. 5) $489.00$ mm^3. 6) Exact: $\frac{3500}{3}\pi$ cm^3; Approximate: $3665.19$ cm^3.