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Mathematics 19 Online
Lazorwolf64:

List the order of the greatest volume to the least volume. When calculating use the pi key. 1.Cylinder with a height of 3 units and a radius of 4 units 2.Cone with a height of 2.74 units and a radius of 7 units. 3.Sphere with a radius of 3.2 units

ILOVESPAGHETTI:

Do you know how to do this?

Lazorwolf64:

No

ShoaibChaudhry:

OK lazor shall Ur friend teach u

Lazorwolf64:

Yus i really need help

ShoaibChaudhry:

Ok so for the first one

ShoaibChaudhry:

for the cilinder its hight times the radius

Lazorwolf64:

so 3 x 4?

ShoaibChaudhry:

Wait

ShoaibChaudhry:

its pi times radius to the power of 2 times the hight

ShoaibChaudhry:

V=πr2h

Lazorwolf64:

thats the equation?

ShoaibChaudhry:

yes

Lazorwolf64:

okie i will solve it

ShoaibChaudhry:

Plug what u know in that equation

Lazorwolf64:

would it be 6.3?

ShoaibChaudhry:

no it would not

ShoaibChaudhry:

Post what u did

ShoaibChaudhry:

π·42·3≈150.79645

ShoaibChaudhry:

I dont think u Did 4 to the second power

ShoaibChaudhry:

So its 3.14 times 16 and then that times 3

ILOVESPAGHETTI:

Alright here's how you find out the volume of a cylinder first. Our formula for this problem is \[V=\pi r ^{2}h\] All you have to do is plug in the numbers (''r'' represents radius, ''h'' for height) \[\pi(4)^{2}(3)\] You can use a calculator like desmos to figure this out, here's the link: https://www.desmos.com/scientific The answer would be \[V=150.7964474\]

ILOVESPAGHETTI:

Our formula for the next problem is, \[V=\frac{ 1 }{ 3 }\pi r ^{2}h\] Just as before, plug in the choices \[V=\frac{ 1 }{ 3 }\pi(7)^{2}(2.74)\] You can use a calculator like desmos to figure this out, here's the link: https://www.desmos.com/scientific The answer would be \[V=140.5967432\]

Lazorwolf64:

thank you!!!

ILOVESPAGHETTI:

@lazorwolf64 wrote:
thank you!!!
Did you solve the question?

Lazorwolf64:

yes

ILOVESPAGHETTI:

Alright, have a good one and if you need any help just ping me!

Lazorwolf64:

@ilovespaghetti wrote:
Alright, have a good one and if you need any help just ping me!
Thank you for the help!!!

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