what i did:

402 = 2πr^2+2πr(3r)

how do i separate r?

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what i did:

402 = 2πr^2+2πr(3r)

how do i separate r?

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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