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Crazy boy [7]
4 years ago
9

Find the volume of a right circular cone that has a height of 17.7 m and a base with a diameter of 18.2 m. Round your answer to

the nearest tenth of a cubic meter.
Mathematics
1 answer:
Korvikt [17]4 years ago
4 0

Steps:

1: Use the formula V=πr2h 3 to calculate the volume for Right circular cone

2: H is the height,  R is the Radius

3: Use 17.7 to calculate the Height.

4: Use  18.2  to calculate the Radius

Description:

Radius: 18.2

Height: 17.7

Shape: Right circular cone

Solved for volume

Formula: V=πr2h 3

Answer: V≈6139.66

Please mark brainliest

<em><u>Hope this helps.</u></em>

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\sqrt{(12xy^2)^2+\left( \cfrac{7x+3}{2}\right)^2}\implies \sqrt{(12xy^2)^2+ \cfrac{(7x+3)^2}{2^2}} \\\\\\ \sqrt{12^2x^2y^4+ \cfrac{49x^2+42x+9}{4}}\implies \sqrt{ \cfrac{4\cdot 12^2x^2y^4+49x^2+42x+9}{4}} \\\\\\ \cfrac{\sqrt{576x^2y^4+49x^2+42x+9}}{\sqrt{4}}\implies \cfrac{\sqrt{576x^2y^4+49x^2+42x+9}}{2}

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\cfrac{\sqrt{576x^2y^4+49x^2+42x+9}}{2}+\cfrac{\sqrt{576x^2y^4+49x^2+42x+9}}{2}+(7x+3) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\large perimeter}}{7x+3+\sqrt{576x^2y^4+49x^2+42x+9}}~\hfill

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