Make r the subject of formula t=2πr(r+h)

t=2πr(r+h)

t = 2πr^2 + 2πhr
2πr^2 + 2πhr - t = 0
this is a quadratic in r, where a = 2π , b = 2πh and c = -t

r = (-2πh ± √(4π^2h^2 - 4(2π)(-t) )/(4π)

see if you can simplify that

explain in detail

I don't understand

I don understand

r=A-h

so you will be re-arranging to get r all by itself...

it is a wee bit tricky due to the r inside and outside of the bracket.
Do you usually treat it like a quadratic and factor it?
Because that would be a lovely solution : )

I don't understand anything 🤪 about this

I don't understand

I don’t understand

Not even a single number I can understand

To make "r" the subject of the formula t = 2πr(r + h), we need to isolate "r" on one side of the equation.

Step 1: Expand the equation.
t = 2πr^2 + 2πrh

Step 2: Move the term containing r^2 to the other side of the equation.
2πr^2 = t - 2πrh

Step 3: Divide both sides of the equation by 2π.
r^2 = (t - 2πrh) / (2π)

Step 4: Simplify the right side of the equation.
r^2 = (t - 2πrh) / (2π)

Step 5: Take the square root of both sides of the equation to solve for r.
√(r^2) = √((t - 2πrh) / (2π))

Step 6: Simplify the equation.
r = √((t - 2πrh) / (2π))

Now, "r" is the subject of the formula t = 2πr(r + h).