Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 16:32Z
∇ × E = −∂B/∂t
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
75K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
479K
\frac{8}{9} \sum_{i=1}^{n} i^{2}
F = ma
det| 3 1 ; 7 4 | = 5
390K
10n² + 4n + 5 = 0
∑_{k=1}^{n} k = n(n+1)/2
1K
det| 1 4 ; 3 2 | = -10
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\binom{n}{k} = \frac{n!}{k!(n-k)!}
∫₀^∞ e^(-γ²) dγ = √π / 2
det| 9 4 ; 1 10 | = 86
188K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
416K