anonymous
  • anonymous
Lost in step 3: 2. Apply the formula of a right circular cylinder (V = r2h) to find the volume of the object. 3. Rewrite the formula using the variable x for the radius. Substitute the value of the volume found in step 2 for V and express the height of the object in terms of x plus or minus a constant. For example, if the height measurement is 4 inches longer than the radius, then the expression for the height will be (x + 4). Diameter: 3in Height: 4.5in Volume: ≈31.8in
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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KingGeorge
  • KingGeorge
I think what it's saying is to let height be the radius (x) plus some value. Since your height is 4.5, and your diameter is 3, the radius is 1.5, so height = x+3. Also, V~31.8, so you would rewrite your equation as V=x2(x+3)
anonymous
  • anonymous
ooops thinking SA
anonymous
  • anonymous
v=pi r^2 h

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anonymous
  • anonymous
did you forget pi ?
anonymous
  • anonymous
Yes, pi should be there. :)
KingGeorge
  • KingGeorge
I'm a little confused by your original formula though. I though that \[V=\pi r^2 h\]Like mathg8 said.
anonymous
  • anonymous
Sorry, copy and pasted. Didn't realize that the pi symbol didn't transfer.
KingGeorge
  • KingGeorge
So then, using what I said earlier, \[V=\pi x^2 (x+3)\]
anonymous
  • anonymous
Now I: Simplify the equation and write it in standard form. If the equation contains decimals, multiply each term by a constant that will make all coefficients integers. Example: 0 = x3 + 2.75x2 – 6.25 If this equation is multiplied by 4, the decimals will become integers. 4(0 = x3 + 2.75x2 – 6.25) 0 = 4x3 + 11x2 – 25
KingGeorge
  • KingGeorge
Since \(V \sim 31.8\) you have \(31.8=\pi x^3+\pi 3x^2\). Here, \(31.8\) is the only decimal, so multiply by \(5\) to get rid of the decimal.
anonymous
  • anonymous
159=5pix^2+5pi3x^2 ?
KingGeorge
  • KingGeorge
So you get \(0 \sim 5\pi x^3 + 15\pi x^2 -159\). Now, the only problem I'm seeing, is that \(\pi\) isn't an integer, but there's no way to get rid of it. However, if you kept the original value of \(V\) in terms of \(\pi\), we could get rid of pi.
KingGeorge
  • KingGeorge
So, assuming that, we can use the original diameter and height to find that \[V=10.125 \pi = \pi x^3 +3 \pi x^2\]This is much better since we can now divide by \(\pi\) and multiply by \(8\) since \(0.125 = {1/8}\). Finally, that means that \[0=8x^3+24x^2-81\]
anonymous
  • anonymous
Thank you! I'm taking this class online, and have e-mailed my teacher asking for help multiple times. They don't seem as though they are willing to help, as they have never responded. Lol. :(
KingGeorge
  • KingGeorge
You're very welcome.
KingGeorge
  • KingGeorge
Which class is this?
anonymous
  • anonymous
Algebra II
KingGeorge
  • KingGeorge
I've always found algebra interesting.
anonymous
  • anonymous
Math is not a skill of mine, so I would have to disagree! Thank you again. :)
KingGeorge
  • KingGeorge
You're welcome. Feel free to ask any more questions.

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