lp = πr(r + 2t)
105π = πr(r + 2(8)
105 = r(r + 16)
105 = (r(r) + (r(16)
105 = r² + 16r
r² + 16r = 105
r² + 16r - 105 = 0
(r + 21)(r - 5) = 0
r + 21 = 0 V r - 5 = 0
r = 0 - 21 V r = 0 + 5
r = -21 V r = 5
kita gunakan yang positif => r2 = 5 dm
maka volume :
πr²t
= π(5²(8)
= π(5(5(8)
= π(5(40)
= π(200)
= 200π dm³
[tex]\mathcal{ \color{cyan} \:INFINITE \: WORLD}[/tex]
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lp = πr(r + 2t)
105π = πr(r + 2(8)
105 = r(r + 16)
105 = (r(r) + (r(16)
105 = r² + 16r
r² + 16r = 105
r² + 16r - 105 = 0
(r + 21)(r - 5) = 0
r + 21 = 0 V r - 5 = 0
r = 0 - 21 V r = 0 + 5
r = -21 V r = 5
kita gunakan yang positif => r2 = 5 dm
maka volume :
πr²t
= π(5²(8)
= π(5(5(8)
= π(5(40)
= π(200)
= 200π dm³
[tex]\mathcal{ \color{cyan} \:INFINITE \: WORLD}[/tex]