What is the volume of a cylinder, in cubic cm, with a height of 8cm and a base diameter of 14cm? Round to the nearest tenths place
The radius of the base is half of the diameter, so r = 14/2 = 7 cm.
The formula for the volume of a cylinder is V = πr^2h.
Substituting the given values, we get V = π(7)^2(8) = 1372.39 cubic cm.
Rounding to the nearest tenths place, the volume is 1372.4 cubic cm.
What is the volume of a cylinder, in cubic inches, with a height of 2 inches and a base diameter of 18 inches? Round to the nearest tenths place.
The radius of the base is half of the diameter, so r = 18/2 = 9 inches.
The formula for the volume of a cylinder is V = πr^2h.
Substituting the given values, we get V = π(9)^2(2) ≈ 508.94 cubic inches.
Rounding to the nearest tenths place, the volume is 508.9 cubic inches.
Find the volume of a pyramid with a square base, where the area of the base is 19.6 ft^2 and the height is 11.6 ft. Round your answer to the nearest tenth of a cubic foot.
The formula for the volume of a pyramid with a square base is V = (1/3)Bh, where B is the area of the base and h is the height.
Substitute the given values into this formula to get:
V = (1/3)(19.6 ft^2)(11.6 ft)
V ≈ 81.5 ft^3
Rounding to the nearest tenth of a cubic foot, the volume is 81.5 ft^3.
Your answer was not correct. Please try again.
I apologize for the mistake in my previous answer. Here's the correct solution:
The formula for the volume of a pyramid with a square base is V = (1/3)Bh, where B is the area of the base and h is the height.
Substitute the given values into this formula to get:
V = (1/3)(19.6 ft^2)(11.6 ft)
V ≈ 83.3 ft^3
Rounding to the nearest tenth of a cubic foot, the volume is 83.3 ft^3.
The answer was 75.8 ft^3. I got it wrong. Thanks for nothing.
I apologize for the confusion and incorrect answer. It seems that I made a mistake in my calculations.
The correct answer would be:
V = (1/3)Bh
V = (1/3)(19.6 ft^2)(11.6 ft)
V = 75.8 ft^3 (rounded to the nearest tenth)
Once again, I'm sorry for the previous error, and I hope this answer is helpful.