annoying math in a riddle game....need help

W

Wwzdrgn

Guest
The outside radius of the pipe is 5, The inside radius is 4, If the pipe is 400 long, What is the volume of the metal used for this length of pipe? round your answer to the nearest whole number.

radius used to be diameter i saw a forum saying it was really radius...........................but anyway i REALLY need some help
 

risen_jihad

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just find the area of whole pip-area of inside pipe

AREA=PI*R^2*length

so
areainside=(PI*5^2*400)-(pi*4^2*400)

so (pi*10000)-(pi*6400)

don't have a calculator handy, :/ as much as i can do
 
W

Wwzdrgn

Guest
-.-

i have no idea what you mean...im in 9th grade i dont know this stuff :banghead:
 
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These things was in the 8th grade in our school.. I just finished it (the grade).

But I have really no idea how to calculate that..
 

Demonwrath

Happy[ExtremelyOverCommercializ ed]HolidaysEveryon
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just find the area of whole pip-area of inside pipe

AREA=PI*R^2*length

so
areainside=(PI*5^2*400)-(pi*4^2*400)

so (pi*10000)-(pi*6400)

don't have a calculator handy, :/ as much as i can do

let me write this differently:

To figure this out we need to find three things:
1. The total volume of the metal used for the pipe (z)
2. The total volume of the inside of the pipe (y)
3. The total area of the pipe (x)

we can find all of this with the following equations:
1. z = x - y
2. x = (pie)x(radius)^2x(length)
3. y = (pie)x(radius)^2x(length)

Now lets find our knowns, and unknowns:
Total Length of the Pipe (l) = 400 units
Inside Radius of the Pipe (r) = 4 units
Outside Radius of the Pipe (s) = 5 units

So know our equations become:
1. z = x - y
2. x = 3.14*s^2*l
3. y = 3.14*r^2*l

Now lets solve for x

x = 3.14*s^2*l
x = 3.14*5^2*400
x = 3.14*25*400
x = 78.5*400
x = 31400 units

Now lets solve for y

y = 3.14*r^2*l
y = 3.14*4^2*400
y = 3.14*16*400
y = 50.24*400
y = 20096 units

Finally, we solve for z

z = x - y
z = 31400 - 20096
z = 11304 units

Therefore the total volume of the metal used for the pipe is 11304 units


And that my friends...is how it's done :D
 
W

Wwzdrgn

Guest
...

let me write this differently:

To figure this out we need to find three things:
1. The total volume of the metal used for the pipe (z)
2. The total volume of the inside of the pipe (y)
3. The total area of the pipe (x)

we can find all of this with the following equations:
1. z = x - y
2. x = (pie)x(radius)^2x(length)
3. y = (pie)x(radius)^2x(length)

Now lets find our knowns, and unknowns:
Total Length of the Pipe (l) = 400 units
Inside Radius of the Pipe (r) = 4 units
Outside Radius of the Pipe (s) = 5 units

So know our equations become:
1. z = x - y
2. x = 3.14*s^2*l
3. y = 3.14*r^2*l

Now lets solve for x

x = 3.14*s^2*l
x = 3.14*5^2*400
x = 3.14*25*400
x = 78.5*400
x = 31400 units

Now lets solve for y

y = 3.14*r^2*l
y = 3.14*4^2*400
y = 3.14*16*400
y = 50.24*400
y = 20096 units

Finally, we solve for z

z = x - y
z = 31400 - 20096
z = 11304 units

Therefore the total volume of the metal used for the pipe is 11304 units


And that my friends...is how it's done :D


yea um......apprently 11304 isnt it...did you round it etc etc?
 

AceHart

Your Friendly Neighborhood Admin
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> did you round it?

Yes.
He used PI to two digits only, once for x and once for y.

Just plugging the entire thing in a calculator says 11309.7336.
So, try with 11309 (rounded down), or 11310 (rounded to nearest).
 
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