[MATH DUMP] page=1
\oint_C \vec{F}\cdot d\vec{r} = 0
11n² + 2n + 5 = 0
4t² + 3t + 1 = 0
\binom{n}{k} = \frac{n!}{k!(n-k)!}
Fe₂O₃ + 3CO → 2Fe + 3CO₂
iħ ∂ψ/∂t = Ĥψ
Fe₂O₃ + 3CO → 2Fe + 3CO₂
det| 3 7 ; 0 4 | = 12
det| 6 3 ; 7 7 | = 21
det| 8 4 ; 7 9 | = 44
det| 2 4 ; 6 3 | = -18
∑_{k=1}^{n} k = n(n+1)/2
d/dy [y^12] = 12y^11
CaCO₃ → CaO + CO₂
F = G m₁m₂ / r²
10θ² + 6θ + 1 = 0
8n² + 4n + 7 = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
AgNO₃ + NaCl → AgCl↓ + NaNO₃
λ = h/p
d/dγ [γ^5] = 5γ^4
e^{i\pi} + 1 = 0
det| 3 5 ; 1 4 | = 7
\oint_C \vec{F}\cdot d\vec{r} = 0
\lim_{α\to 0} \frac{\sin α}{α} = 1
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Fe₂O₃ + 3CO → 2Fe + 3CO₂
det| 2 6 ; 1 3 | = 0
det| 4 3 ; 4 5 | = 8
det| 8 2 ; 5 9 | = 62
\oint_C \vec{F}\cdot d\vec{r} = 0
d/dn [n^11] = 11n^10
∇ × E = −∂B/∂t
∂α/∂r = 8α
det| 2 9 ; 7 3 | = -57
∑_{k=1}^{n} k = n(n+1)/2
F = G m₁m₂ / r²
CH₄ + 2O₂ → CO₂ + 2H₂O
e^{i\pi} + 1 = 0
\binom{n}{k} = \frac{n!}{k!(n-k)!}
PV = nRT
AgNO₃ + NaCl → AgCl↓ + NaNO₃
det| 6 3 ; 5 7 | = 27
4r² + 3r + 5 = 0
\oint_C \vec{F}\cdot d\vec{r} = 0
12ω² + 4ω + 1 = 0
iħ ∂ψ/∂t = Ĥψ
N₂ + 3H₂ ⇌ 2NH₃
∫₀^∞ e^(-α²) dα = √π / 2
ΔS ≥ 0
\lim_{ψ\to 0} \frac{\sin ψ}{ψ} = 1
8r² + 9r + 7 = 0
∇ × E = −∂B/∂t
CaCO₃ → CaO + CO₂
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
∑_{k=1}^{n} k = n(n+1)/2
det| 6 7 ; 6 7 | = 0
2r² + 4r + 5 = 0
11ψ² + 7ψ + 2 = 0
12z² + 7z + 0 = 0
F = G m₁m₂ / r²
∑_{k=1}^{n} k = n(n+1)/2
\binom{n}{k} = \frac{n!}{k!(n-k)!}
det| 11 1 ; 3 12 | = 129
6y² + 7y + 2 = 0
e^{i\pi} + 1 = 0
∇²α = 0
ω = 2πf
∇²γ = 0
∂θ/∂n = 1θ
11φ² + 6φ + 3 = 0
\lim_{x\to 0} \frac{\sin x}{x} = 1
∑_{k=1}^{n} k = n(n+1)/2
CaCO₃ → CaO + CO₂
N₂ + 3H₂ ⇌ 2NH₃
det| 12 9 ; 3 13 | = 129
det| 2 5 ; 2 3 | = -4
det| 3 6 ; 6 4 | = -24
11r² + 4r + 2 = 0
iħ ∂ψ/∂t = Ĥψ
λ = h/p
det| 10 4 ; 6 11 | = 86
1z² + 5z + 2 = 0
det| 8 3 ; 4 9 | = 60
d/dy [y^2] = 2y^1
∫₀^∞ e^(-r²) dr = √π / 2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
∇²r = 0