Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:02Z
∑_{k=1}^{n} k = n(n+1)/2
\binom{n}{k} = \frac{n!}{k!(n-k)!}
75K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
479K
3β² + 6β + 2 = 0
F = ma
∑_{k=1}^{n} k = n(n+1)/2
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
153K
1x² + 5x + 7 = 0
det| 9 4 ; 1 10 | = 86
473K
∇²ψ = 0
3φ² + 8φ + 6 = 0
\binom{n}{k} = \frac{n!}{k!(n-k)!}