Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:55Z
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
479K
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
∑_{k=1}^{n} k = n(n+1)/2
129K
F = ma
∑_{k=1}^{n} k = n(n+1)/2
191K
∫₀^∞ e^(-ψ²) dψ = √π / 2
153K
det| 7 4 ; 4 8 | = 40
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