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The volume of a cone is V = 1/3(pir^2(h)). If the height is twice the radius, express the volume V as a function of r.

Also, is there an easier way to say
"express the volume V as a function of r" in English?

2007-12-12 23:02:59 · 6 answers · asked by journey 1 in Science & Mathematics Mathematics

6 answers

Expressing a variable as a function of another variable just means something like, for the above:

V = f(r)

Where f(r) is a function which does something to 'r'.

Because you know that the height is twice the radius, just substitute '2r' instead of 'h', so you get:

V = 1/3*pi*r^2*2r
V = 2/3*pi*r^2*r
V = 2/3*pi*r^3

2007-12-12 23:12:54 · answer #1 · answered by Anonymous · 1 0

V = 1/3(πr^2)h

height is twice the radius is h = 2r

Hence,

V = 1/3(πr^2)(2r)
V = 2/3(πr^3) ANS

Well, the expression "express the volume V as a function of r"
might sound too 'technical' for most people but it tells you exactly by what variable you want volume to be defined. It's already a shortcut way of saying it without losing the precise meaning.

teddy boy

2007-12-12 23:21:42 · answer #2 · answered by teddy boy 6 · 0 0

Since the height is twice the radius we just put in 2r for the h. It will look like this:

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

That's it -- the volume is expressed as a function of r and I don't know a better way to say it! :)

2007-12-12 23:10:36 · answer #3 · answered by Marley K 7 · 0 0

Volume of Cone:

V=1/3 x Ah

V=Volume,A=the base area,h=the height

2007-12-13 00:52:41 · answer #4 · answered by An ESL Learner 7 · 0 0

the quantity of a cone is a million/3(component of Base)(height) = a million/3 ? r^2 h the quantity of a cylinder equals the (component of the backside)*height = ? r^2 h consequently the quantity of a cone(a million/3 ? r^2 h) is a million/3 the quantity of a corresponding cylinder(? r^2 h).consequently the quantity of the cone is 30cm^3 divided by utilising 3=10cm^3

2016-10-11 04:52:58 · answer #5 · answered by ? 3 · 0 0

1/3pir2 ( 2 is squared )

2007-12-12 23:17:17 · answer #6 · answered by Anonymous · 0 0

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