Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 16:27Z
∇ × E = −∂B/∂t
∑_{k=1}^{n} k = n(n+1)/2
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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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\frac{8}{9} \sum_{i=1}^{n} i^{2}
det| 4 4 ; 5 5 | = 0
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F = ma
Fe₂O₃ + 3CO → 2Fe + 3CO₂
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∑_{k=1}^{n} k = n(n+1)/2
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10n² + 4n + 5 = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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\binom{n}{k} = \frac{n!}{k!(n-k)!}
\binom{n}{k} = \frac{n!}{k!(n-k)!}
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