ChemCalc Dump /sym/e47f8c
id=818200 · host=br.gridbatch.net · 2026-09-03 03:59Z
∇ × E = −∂B/∂t
\oint_C \vec{F}\cdot d\vec{r} = 0
252K
\oint_C \vec{F}\cdot d\vec{r} = 0
464K
8r² + 5r + 5 = 0
∫₀^∞ e^(-α²) dα = √π / 2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
202K
det| 5 9 ; 4 6 | = -6
9ω² + 7ω + 4 = 0
\oint_C \vec{F}\cdot d\vec{r} = 0
\lim_{γ\to 0} \frac{\sin γ}{γ} = 1
E = mc²
∇ × E = −∂B/∂t
5y² + 2y + 6 = 0
\lim_{ω\to 0} \frac{\sin ω}{ω} = 1