Original volume: $$V_1 = \frac{4}{3}\pi(6^3) = \frac{4}{3}\pi(216) = 288\pi$$. Remaining volume: $$V_2 = V_1 - \frac{7}{8}V_1 = \frac{1}{8}V_1 = \frac{288\pi}{8} = 36\pi$$. For the new radius: $$\frac{4}{3}\pi r^3 = 36\pi$$, so $$\frac{4}{3}r^3 = 36$$, giving $$r^3 = 36 \times \frac{3}{4} = 27$$, so $$r = 3$$ inches. Choice A is correct. Choice D might result from thinking the radius decreases by $$\frac{7}{8}$$: $$6 \times \frac{1}{8} = 0.75$$. Choice B could come from $$6 \times \frac{2}{3} = 4$$. Choice C might result from $$6 \times \frac{1}{3} = 2$$.

First, find the vertical height using the Pythagorean theorem. Radius = 7 feet, slant height = 25 feet. $$h^2 + 7^2 = 25^2$$, so $$h^2 + 49 = 625$$, giving $$h^2 = 576$$ and $$h = 24$$ feet. Volume: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(7^2)(24) = \frac{1}{3}\pi(49)(24) = \frac{1,176\pi}{3} = 392\pi \approx 1,232$$ cubic feet. Choice A might result from using slant height instead of vertical height. Choice C could come from calculation errors. Choice D might result from rounding errors or using approximate values for π.

Let C = capacity of each tank. Tank A contains 0.75C gallons, Tank B contains 0.60C gallons. The difference is: $$0.75C - 0.60C = 0.15C = 450$$ gallons. Solving: $$C = \frac{450}{0.15} = \frac{450}{\frac{15}{100}} = \frac{450 \times 100}{15} = \frac{45,000}{15} = 3,000$$ gallons. Choice B gives a difference of $$0.15 \times 2,500 = 375$$. Choice C gives $$0.15 \times 3,600 = 540$$. Choice D gives $$0.15 \times 4,200 = 630$$.

This involves similar cones. The water forms a smaller cone with height $$\frac{3}{4} \times 12 = 9$$ cm. By similar triangles, if the original radius is 4 cm, the water cone's radius is $$\frac{9}{12} \times 4 = 3$$ cm. Volume scales as the cube of the linear scale factor. The scale factor is $$\frac{3}{4}$$, so the volume fraction is $$(\frac{3}{4})^3 = \frac{27}{64}$$. Choice C incorrectly assumes volume scales linearly with height. Choice B uses $$(\frac{3}{4})^2$$. Choice D is $$\frac{1}{2}$$, which doesn't relate to the given fraction.

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=3 cm and h=10 cm, V=π(3²)(10)=π(9)(10)=90π≈282.7 cm³. Common errors include using the cone formula by mistake (adding 1/3), treating radius as diameter, incorrect squaring (like 3²=6), or forgetting to cube the units (cm instead of cm³). Steps: (1) identify shape as cylinder (circular base and constant height), (2) gather dimensions (r=3 cm, h=10 cm), (3) select formula πr²h, (4) substitute values, (5) calculate 910π=90π, (6) add cubic units cm³. Always double-check arithmetic and formula selection to avoid mistakes like confusing cylinder with cone.

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.