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2 years ago

What is the volume of this cone in cubic inches

Mathematics

2 answers:

Ainat [17]2 years ago

6 0

Answer:

84.78 inches cubed

Step-by-step explanation:

You need to comprehend the volume formula for cones

Volume Formula for Cones: V=1/3hπr²

Your measurements:

9 inches for height

3 inches for radius

**Radius is half of a circle and diameter is full of the circle/base.***

Equation:V=1/3 x 9in x π x 3²

You should do 3 squared first so that you can input it into your calculator easily.

V=1/3 x 9in x π x 9

Now on a calculator do 1 divided by 3 which is the "1/3" in the formula.

Then multiply 9, 3.14 for pi, and 9 for radius.

Note: You should do 3.14 for pi not the whole pi symbol on the calculator.

Your answer: 84.78 in cubed

Solnce55 [7]2 years ago

4 0

Answer:

84.78 inches cubed - hope this helped :)

Step-by-step explanation:

formula - πr^2 $\frac{h}{3}$

3 x 3 x 3.14 = 28.26

28.26 x 9/3

28.26 x 3

answer = 84.78

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Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Fiona and Pip win some money and share it in the ratio 6:1. Fiona gets £35 more than Pip. How much did they get altogether?

Softa [21]

$\boxed{ \mathfrak { question↷}}$

Fiona and Pip win some money and share it in the ratio 6:1. Fiona gets £35 more than Pip. How much did they get altogether?

$\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}$

$\boxed{ \mathfrak { solution↷}}$

Let Pip get be x and fiona get be x+35

Ratio = 6:1

$\tt \frac{x + 35}{x } = 6$

$\tt \: x + 35 = 6x$

$\tt \: 5x = 35$

$\tt \: x = 7$

$\boxed{ \sf \: Total \: money = x + x + 35 = 14+ 35 = £49}$

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garri49 [273]

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27/60= .45 miles per minute

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