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2 years ago

9

What is the approximate volume of a cone with a height of 9 ft and radius of 3 ft?

2 answers:

Nesterboy [21]2 years ago

7 0

Answer: The answer is 84 ft³.

Step-by-step explanation: Given that a cone has a height of 9 ft and a radius of 3 feet. We are to find the approximate volume of the cone.

Also, it is given to assume that

$\pi=3.14.$

We know that the volume of a cone with height 'h' and radius 'r' is given by

$V=\dfrac{1}{3}\pi r^2h.$

Here, r = 3 ft and h=9 ft.

Therefore, the volume of the cone will be

$V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times 3.14\times 3^2\times 9=\dfrac{254.34}{3}=84.78\sim 84~\textup{ft}^3.$

Thus, the required volume will be 84 ft³.

mr Goodwill [35]2 years ago

4 0

<span>Cone Volume = (<span>π<span> • r² •<span> h) ÷ 3

</span></span></span></span><span>Cone Volume = (3.14 * 3^2 * 9) / 3
</span>
<span>Cone Volume = (3.14 * 27)

</span><span><span>Cone Volume = </span>84.82 cubic feet

</span>

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The coordinates of B' and C' are $B'(x,y) = (4, -7)$ and $C'(x,y) = (-1, -8)$ , respectively.

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From the Linear Algebra, we define the translation of a given point as:

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$O(x,y)$ - Original point, dimensionless.

$T(x,y)$ - Translation vector, dimensionless.

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