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x2 x□ log□ √☐ □√☐ ≤ ≥ □□  · ÷ x◦ π
(☐)′ ddx  ∂∂x  ∫ ∫□□ lim ∑ ∞ θ (f ◦ g) f(x)
▭ |▭ ×▭▭ +▭▭ −▭▭ ( ) × ☐☐☐ 
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x2 x□ log□ √☐ □√☐ ≤ ≥ □□  · ÷ x◦ π
(☐)′ ddx  ∂∂x  ∫ ∫□□ lim ∑ ∞ θ (f ◦ g) f(x)
−▭▭ < 7 8 9 ÷ AC
+▭▭ > 4 5 6 × ☐☐☐ 
×▭▭ ( 1 2 3 − x
▭ |▭ ) . 0 = + y
Basic αβγ ABΓ sincos ≥÷→ x ℂ∀ ∑ ∫ ∏ (
☐☐
☐☐
)		 H2O
☐2 x☐ √☐ □√☐ □□  log□ π θ ∞ ∫ ddx 
≥ ≤ · ÷ x◦ (☐) |☐| (f ◦ g) f(x) ln e☐
(☐)′ ∂∂x  ∫□□ lim ∑ sin cos tan cot csc sec
α β γ δ ζ η θ ι κ λ μ
ν ξ π ρ σ τ υ ϕ χ ψ ω
A B Γ Δ E Z H Θ K Λ M
N Ξ Π P Σ T ϒ Φ X Ψ Ω
sin cos tan cot sec csc sinh cosh tanh coth sech
arcsin arccos arctan arccot arcsec arccsc arcsinh arccosh arctanh arccoth arcsech
{
☐
☐
 		 {
☐
☐
☐
 		 = ≠ ÷ · × < > ≤ ≥
(☐) [☐] ▭ |▭ ×▭▭ +▭▭ −▭▭ ☐! x◦ → ⌊☐⌋ ⌈☐⌉
☐ ⃗ ☐ ∈ ∀ ∉ ∃ ℝ ℂ ℕ ℤ ∅
∨ ∧ ¬ ⊕ ∩ ∪ ☐c ⊂ ⊆ ⊃ ⊇
∫ ∫∫ ∫∫∫ ∫☐☐ ∫☐☐∫☐☐ ∫☐☐∫☐☐∫☐☐ ∑ ∏
lim limx→∞ limx→0+ limx→0− ddx  d2dx2  (☐)′ (☐)′′ ∂∂x 
(2×2) (2×3) (3×3) (3×2) (4×2) (4×3) (4×4) (3×4) (2×4) (5×5) matrix button
(1×2) (1×3) (1×4) (1×5) (1×6) (2×1) (3×1) (4×1) (5×1) (6×1) (7×1)
Radians Degrees ☐! ( ) % clear
arcsin sin √☐ 7 8 9 ÷
arccos cos ln 4 5 6 ×
arctan tan log 1 2 3 −
π e x☐ 0 . = +
simplify solve for inverse tangent line
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(
1√6  −1√2  1√3 
1√6  1√2  1√3 
−2√6  0 1√3 
)(
7 0 0
0 1 0
0 0 1
)(
1√6  −1√2  1√3 
1√6  1√2  1√3 
−2√6  0 1√3 
)T
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\begin{pmatrix}\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ -\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\end{pmatrix}\begin{pmatrix}7& 0& 0 \\ 0 & 1& 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ -\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\end{pmatrix}^{T}

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