ChemCalc Dump /sym/1a8a5d
id=790497 · host=br.gridbatch.net · 2026-09-03 04:02Z
∇ × E = −∂B/∂t
\oint_C \vec{F}\cdot d\vec{r} = 0
252K
\oint_C \vec{F}\cdot d\vec{r} = 0
464K
2r² + 8r + 2 = 0
ω = 2πf
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
202K
6φ² + 9φ + 1 = 0
10r² + 2r + 7 = 0
10z² + 8z + 7 = 0
\lim_{r\to 0} \frac{\sin r}{r} = 1
∂ψ/∂y = 4ψ
∑_{k=1}^{n} k = n(n+1)/2
det| 8 5 ; 4 9 | = 52
\lim_{ψ\to 0} \frac{\sin ψ}{ψ} = 1