Make h the subject of the formula A=πr²tπr√h²+r²
To make h the subject of the formula A=πr²tπr√h²+r², we need to isolate h on one side of the equation. Here are the steps to do that:
1. Start with the given formula: A = πr²tπr√h² + r²
2. Rearrange the equation by subtracting r² from both sides: A - r² = πr²tπr√h²
3. Divide both sides of the equation by πr²t: (A - r²) / (πr²t) = √h²
4. Square both sides of the equation to eliminate the square root: [(A - r²) / (πr²t)]^2 = h²
5. Take the square root of both sides to solve for h: h = √[(A - r²) / (πr²t)]^2
Therefore, the equation for h as the subject is h = √[(A - r²) / (πr²t)]^2.
Did you mean A=πr²tπr√(h²+r²) , really different from what you wrote!!
A = π^2 r^3 t √(h²+r²)
√(h²+r²) = A/(π^2 r^3 t)
square both sides
h^2 + r^2 = A^2/(π^4 r^6 t^2)
h^2 = A^2/(π^4 r^6 t^2) - r^2
h = √( A^2/(π^4 r^6 t^2) - r^2 )
somehow I don't think this is what you meant
It looks like the area of a cone
A = πr² + πr√h²+r²
A - πr² = πr√(h²+r²)
(A - πr²)/(πr²) = √(h²+r²)
h²+r² = ((A - πr²)/(πr²))²
h² = ((A - πr²)/(πr²))² - r²
h = √(((A - πr²)/(πr²))² - r²)