Find the volume of a right circular cone that has a height of 2m and a base with a diameter of 11.1m. Round your answer to the nearest tenth of a cubic meter. Answer: m^(3)

Find Radius: To find the volume of a cone, we use the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height of the cone. First, we need to find the radius of the base. The diameter of the base is given as 11.1 meters, so the radius r is half of the diameter. r = diameter / 2 r = 11.1m / 2 r = 5.55mSubstitute values: Now that we have the radius, we can substitute the values into the volume formula. V = (1/3)πr^2h V = (1/3)π(5.55m)^2(2m)Calculate area: Next, we calculate the square of the radius and multiply by the height. (5.55m)^2 = 30.8025m^2 V = (1/3)π(30.8025m^2)(2m)Multiply by height: We then multiply the area by the height and divide by 3 to get the volume. V = π(30.8025m^2)(2m) / 3 V = π(61.605m^2) / 3Find volume: Now we multiply by π and divide by 3 to find the volume. V ≈ 3.14159 * 61.605m^2 / 3 V ≈ 193.672m^3 / 3 V ≈ 64.557m^3Round answer: Finally, we round the answer to the nearest tenth of a cubic meter. V ≈ 64.6m^3

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Find the volume of a right circular cone that has a height of
2m and a base with a diameter of
11.1m. Round your answer to the nearest tenth of a cubic meter.
Answer:
m^(3)

Find the volume of a right circular cone that has a height of $2 \mathrm{~m}$ and a base with a diameter of $11.1 \mathrm{~m}$ . Round your answer to the nearest tenth of a cubic meter. $\newline$ Answer: $\mathrm{m}^{3}$

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Grade 7

Convert between customary and metric systems

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Q. Find the volume of a right circular cone that has a height of

2 \mathrm{~m}

and a base with a diameter of

11.1 \mathrm{~m}

. Round your answer to the nearest tenth of a cubic meter.

\newline

Answer:

\mathrm{m}^{3}

Find Radius: To find the volume of a cone, we use the formula $V = \frac{1}{3}\pi r^2 h$ , where $r$ is the radius of the base and $h$ is the height of the cone. First, we need to find the radius of the base. $\newline$ The diameter of the base is given as $11.1$ meters, so the radius $r$ is half of the diameter. $\newline$ $r = \frac{\text{diameter}}{2}$ $\newline$ $r = \frac{11.1\,\text{m}}{2}$ $\newline$ $r = 5.55\,\text{m}$
Substitute values: Now that we have the radius, we can substitute the values into the volume formula. $\newline$ $V = \frac{1}{3}\pi r^2h$ $\newline$ $V = \frac{1}{3}\pi(5.55\,\text{m})^2(2\,\text{m})$
Calculate area: Next, we calculate the square of the radius and multiply by the height. $\newline$ $(5.55\,\text{m})^2 = 30.8025\,\text{m}^2$ $\newline$ $V = \frac{1}{3}\pi(30.8025\,\text{m}^2)(2\,\text{m})$
Multiply by height: We then multiply the area by the height and divide by $3$ to get the volume. $\newline$ $V = \frac{\pi(30.8025\,\text{m}^2)(2\,\text{m})}{3}$ $\newline$ $V = \frac{\pi(61.605\,\text{m}^2)}{3}$
Find volume: Now we multiply by $\pi$ and divide by $3$ to find the volume. $\newline$ $V \approx 3.14159 \times 61.605\,\text{m}^2 / 3$ $\newline$ $V \approx 193.672\,\text{m}^3 / 3$ $\newline$ $V \approx 64.557\,\text{m}^3$
Round answer: Finally, we round the answer to the nearest tenth of a cubic meter. $V \approx 64.6\,\text{m}^3$