what i did:
402 = 2πr^2+2πr(3r)
how do i separate r?
what i did:
402 = 2πr^2+2πr(3r)
how do i separate r?
2π r^{2} + 2π r(3r) = 2π r^{2} + 6π r^{2} = 8π r^{2}. I hope this helped, and if you need more assistance, I or another tutor will be happy to help.
2∏rh + 2∏r^{2 }= 402
h = 4r
2∏r(4r) + 2∏r^{2} = 402
8∏r^{2} + 2∏r^{2} = 402
10∏r^{2} = 402
r^{2 }= 402/10∏
r = √402/√10∏
h = (4√402)/√(10∏)
Nice start, Mathalina!!! Next, you have to simplify equation, you wrote, and solve it for "r"
2(pi)r^{2} + 2(pi)r(3r) = 402
2(pi)r^{2} + 6(pi)r^{2} = 402
8(pi)r^{2} = 402
pi = 22/7
r^{2} = (402/8) · (7/22)
r ≈ √15.99 = 3.0 (we use only positive value of the square root, distance cannot be negative)
h = 3r ≈ 9.0 cm