Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 16:36Z
∇ × E = −∂B/∂t
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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\frac{8}{9} \sum_{i=1}^{n} i^{2}
\lim_{t\to 0} \frac{\sin t}{t} = 1
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F = ma
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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10n² + 4n + 5 = 0
\oint_C \vec{F}\cdot d\vec{r} = 0
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det| 11 9 ; 5 12 | = 87
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∫₀^∞ e^(-φ²) dφ = √π / 2
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