WILL GIVE BRAINLIEST IF RIGHT IG....PLEASE get it RIGHT....
find the volume of a cone with a diameter of 12m and a height of 8m.
ANSWER CHOICES
oA. 96m^3
oB. 384 pi m^3
oC. 96 pi m^3
oD. 288 pi m^3

Answers

Answer 1

Answer:

V = 96 pi m^3

Step-by-step explanation:

V = (1/3) pi(r^2)h

Radius = 1/2(Diameter)

R = 1/2(12)

R = 6

Input.  

V = (1/3) pi(6^2)8

Solve.

V = (1/3) pi(36)8

V = (1/3) 904.32

V = 301.44

V = 301.44/pi

V = 96 pi m^3

Answer 2

Answer:

The correct answer is option (C) 96π m³.

Step-by-step explanation:

As per given question we have provided that :

Diameter = 12 mRadius = 12/2 = 6 mHeight = 8 m

Here's the required formula to find the volume of cone :

[tex]{\longrightarrow{\pmb{\sf{V_{(Cone)} = \dfrac{1}{3}\pi{r}^{2}h}}}}[/tex]

➝ V = Volume➝ π = 22/7➝ r = radius➝ h = height

Substituting all the given values in the formula to find the volume of cone :

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{3}\pi{r}^{2}h}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{3}\pi{(6)}^{2}8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{3}\pi{(6 \times 6)}8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{3}\pi \times {(36)} \times 8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{3} \times \pi \times 36 \times 8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \dfrac{1}{\cancel{3}} \times \pi \times \cancel{36} \times 8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \pi \times 12 \times 8}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} = \pi \times 96}}}[/tex]

[tex]{\implies{\sf{V_{(Cone)} =96\pi}}}[/tex]

[tex]\star{\underline{\boxed{\sf{\red{V_{(Cone)} =96\pi \: {m}^{3}}}}}}[/tex]

Hence, the volume of cone is 96π m³.

[tex]\rule{300}{2.5}[/tex]


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stay safe healthy and happy...

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Answers

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Answer:

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Given:

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For Coordinate (6, -3)

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Answers

Answer:

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Step-by-step explanation:

Given

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

It’s the answer :>

Answer:

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Answers

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1, 2, 3

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Sorry I answered so late, hope this helps even though you probably don't need it anymore.

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Answers


Yes ...........................

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Answers

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Answer:

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Step-by-step explanation:

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