Pratikum 1 hardiansyah
1
x 1,2 = −b± √ b 2 −4 ac 2 a A = [ 1 2 3 4 ] ∫ 1 3 ( x 2 +5 x +6 ) dx f ( x )= x+ log x 2 ❑ 2 x 2 + 1
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Transcript of Pratikum 1 hardiansyah
x1,2=−b±√b2−4ac
2a
A=[1 23 4 ]
∫1
3
(x2+5 x+6 )dx
f ( x )= x+ log x2❑2
x2+1
