Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Rainer****Guest**

Hi there,

to caculate the volume of a testtube is not so difficult.

how do I change the formula when only volume and diameter ore hight is given?

V = π*r²*h+(((π*d³)/6)/2)

h = ?

d = ?

Regards,

Rainer

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,659

hi Rainer

Welcome to the forum.

To make h the subject when d and V are known is fairly straight forward:

Getting d as the subject when V and h are known will not be so straight forward, because the 'd' term will occur as a square and a cube.

There are algebraic ways to solve a cubic, but it's not going to be easy to come up with a 'nice' formula. If you have specific values to substitute in, it's probably easiest to use this format and then try to solve for d from there. There are numerical methods that will get d that are easier to use.

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**