This question tests applying volume formulas: cylinder $V=\pi r^2 h$, cone $V=\frac{1}{3} \pi r^2 h$, sphere $V=\frac{4}{3} \pi r^3$, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $\pi r^2$ times height h giving $V=\pi r^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3} \pi r^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3} \pi r^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$), calculate (exponents first, multiply, approximate $\pi \approx 3.14$ or leave exact). For cone with d=8 cm so r=4 cm, h=6 cm, $V=\frac{1}{3} \pi(4^2)(6)=\frac{1}{3} \pi(16)(6)=32 \pi \text{ cm}^3$, matching C. Common errors: using diameter as r for 128$\pi$, cylinder formula for 96$\pi$, sphere for $\frac{256}{3} \pi$, or missing 1/3. Steps: (1) identify cone, (2) gather r=4 cm from d=8, h=6 cm, (3) select $\frac{1}{3} \pi r^2 h$, (4) substitute, (5) calculate 32$\pi$, (6) add cm$^3$. Halving diameter is crucial for accurate radius.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). The correct cone formula is V=(1/3)πr²h, matching choice C, as it accounts for the tapering volume. Common errors include confusing with cylinder (no 1/3), sphere (r³), or made-up like πr³h. Steps: (1) identify cone by its point and base, (2) recall need for 1/3 factor, (3) select (1/3)πr²h, (4) distinguish from others. Memorize formulas to avoid mixing cone and cylinder.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cone, use corresponding formula with given dimensions (radius given, no need to convert), calculate (exponents first, multiply, leave exact in terms of π). Example: cone r=4 in, h=9 in using V=(1/3)π(4²)(9)=(1/3)π(16)(9)=48π in³. Correct formula selection and calculation yield 48π in³ for this cone. Error like wrong formula (cylinder π169=144π, missing 1/3), using diameter as radius (if misread), wrong exponent (r instead of r²), arithmetic error (169=128), or units not cubed. Steps: (1) identify shape (cone with circular base and point), (2) gather dimensions (r=4, h=9), (3) select formula ((1/3)πr²h), (4) substitute, (5) calculate (169=144, 144/3=48, times π), (6) units (cubic: in³).

This question tests applying volume formulas: cylinder $V=πr^2 h$, cone $V=\frac{1}{3}πr^2 h$, sphere $V=\frac{4}{3}πr^3$, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $πr^2$ times height $h$ giving $V=πr^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3}πr^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3}πr^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$), calculate (exponents first, multiply, approximate $π≈3.14$ or leave exact). For this sphere with d=10 ft so r=5 ft, $V=\frac{4}{3}π(5^3)=\frac{4}{3}π(125)=\frac{500}{3}π≈523.6 \text{ ft}^3$. Common errors include using diameter as radius ($r=10$), forgetting to halve diameter, wrong exponent ($r^2$), calculation mistakes ($5^3=125$ not 25), or incorrect units. Steps: (1) identify shape as sphere, (2) gather dimensions (r=5 ft from d=10), (3) select formula $\frac{4}{3}πr^3$, (4) substitute values, (5) calculate ($\frac{4}{3}*125π=\frac{500}{3}π$), (6) add cubic units ft³. Always convert diameter to radius to avoid doubling errors.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=2 m and h=5 m, V=π(2²)(5)=π(4)(5)=20π≈62.8 m³. Common errors include adding 1/3 like a cone, using diameter if misread, exponent mistakes (2²=4 not 8), arithmetic errors (45=25), or forgetting cubic units. Steps: (1) identify shape as cylinder, (2) gather dimensions (r=2 m, h=5 m), (3) select formula πr²h, (4) substitute values, (5) calculate 45*π=20π, (6) add cubic units m³. This calculates maximum water capacity assuming full fill.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shapes, calculate both volumes (cylinder: π94=36π cm³; sphere: (4/3)π27=36π cm³), compare. Example: both yield 36π, so equal. Correct calculation shows they have equal volume. Error like wrong formula (using cone for cylinder) or arithmetic (94=32). Steps: (1) identify shapes, (2) gather dimensions, (3) select formulas, (4) calculate each, (5) compare numerically (36π vs 36π).

This question tests applying volume formulas: cylinder $V=\pi r^2 h$, cone $V=\frac{1}{3} \pi r^2 h$, sphere $V=\frac{4}{3} \pi r^3$, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $\pi r^2$ times height h giving $V=\pi r^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3} \pi r^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3} \pi r^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$), calculate (exponents first, multiply, approximate $\pi\approx3.14$ or leave exact). For $V=288\pi$, $(4/3)\pi r^3=288\pi$ so $r^3=288\times(3/4)=216$, $r=6 \text{ m}$ ($6^3=216$), matching B. Common errors: ignoring 4/3, getting r=9 (wrong inverse), or confusing with other formulas. Steps: (1) identify sphere, (2) set $(4/3)\pi r^3=288\pi$, (3) solve $r^3=216$, (4) cube root to r=6, (5) check units m. Reverse calculations test formula understanding.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this sphere with r=6 m, V=(4/3)π(6³)=(4/3)π(216)=288π m³, matching choice B. Common errors include using r² instead of r³ (like 36π times something), forgetting the 4/3 factor for 216π, dividing instead of multiplying, or incorrect cubing like 6³=216. Steps: (1) identify sphere, (2) gather r=6 m, (3) select V=(4/3)πr³, (4) substitute value, (5) calculate 288π, (6) add m³. Spheres use only radius, no height, due to their uniform symmetry.

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use corresponding formula with given dimensions (radius given), calculate (exponents first, multiply, approximate π≈3.14, round). Example: cylinder r=2 m, h=5 m using V=π(4)(5)=20π≈203.14=62.8, rounded to 63 m³. Correct formula selection and calculation yield ≈63 m³ for this cylinder. Error like not squaring radius (π25=10π≈31), using cone formula ((1/3)π45≈20.93), or wrong rounding. Steps: (1) identify shape (cylinder), (2) gather dimensions (r=2, h=5), (3) select formula (πr²h), (4) substitute, (5) calculate (2²=4, 45=20, 20*3.14=62.8, round to 63), (6) units (cubic: m³).

This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). To find r for sphere V=288π, set (4/3)πr³=288π, divide by π, multiply by 3/4, r³=216, r=6 cm. Common errors include using wrong formula (like πr²h), incorrect solving (forgetting cube root), or arithmetic mistakes $(216^{1/3}$=6). Steps: (1) identify sphere, (2) use V=(4/3)πr³, (3) set equal to 288π, (4) solve r³=288*(3/4)=216, (5) r=∛216=6, (6) check units cm. Solving backwards requires isolating r and taking cube root accurately.