Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:17Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
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F = ma
det| 9 4 ; 1 10 | = 86
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∑_{k=1}^{n} k = n(n+1)/2
390K
∑_{k=1}^{n} k = n(n+1)/2
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∫₀^∞ e^(-φ²) dφ = √π / 2
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