Khan-Aufgaben für Mathematik II
Gegeben sei das Rechteck D⊂R2\color{orange}D \subset \mathbb R^2D⊂R2.
D⊂R2\color{orange}D \subset \mathbb R^2D⊂R2
D=D1∪D2∪D3{\orange{D}} = {\red{D_1}} \cup {\red{D_2}} \cup {\red{D_3}} D=D1∪D2∪D3
D1={(x,y) ∣ 1≤x≤3, −12x−112≤y≤−12x−12} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D1={(x,y)∣1≤x≤3,−21x−211≤y≤−21x−21} und
D1={(x,y) ∣ 1≤x≤3, −12x−112≤y≤−12x−12} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D1={(x,y)∣1≤x≤3,−21x−211≤y≤−21x−21}
D2={(x,y) ∣ 3≤x≤9, −12x+112≤y≤2x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 9, \; -\frac{1}{2} x + \frac{11}{2} \leq y \leq 2 x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤9,−21x+211≤y≤2x−21} und
D2={(x,y) ∣ 3≤x≤9, −12x+112≤y≤2x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 9, \; -\frac{1}{2} x + \frac{11}{2} \leq y \leq 2 x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤9,−21x+211≤y≤2x−21}
D3={(x,y) ∣ 9≤x≤11, −12x−112≤y≤−12x−12}. {\red{D_3}} = \left\{ (x,y) \, | \, 9 \leq x \leq 11, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}.D3={(x,y)∣9≤x≤11,−21x−211≤y≤−21x−21}.
D1={(x,y) ∣ 1≤x≤3, −12x+112≤y≤2x−8} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x + \frac{11}{2} \leq y \leq 2 x - 8\right\}D1={(x,y)∣1≤x≤3,−21x+211≤y≤2x−8} und
D1={(x,y) ∣ 1≤x≤3, −12x+112≤y≤2x−8} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x + \frac{11}{2} \leq y \leq 2 x - 8\right\}D1={(x,y)∣1≤x≤3,−21x+211≤y≤2x−8}
D2={(x,y) ∣ 3≤x≤9, −12x−112≤y≤−12x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 9, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤9,−21x−211≤y≤−21x−21} und
D2={(x,y) ∣ 3≤x≤9, −12x−112≤y≤−12x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 9, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤9,−21x−211≤y≤−21x−21}
D3={(x,y) ∣ 9≤x≤11, 2x−28≤y≤−12x−12}. {\red{D_3}} = \left\{ (x,y) \, | \, 9 \leq x \leq 11, \; 2 x - 28 \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}.D3={(x,y)∣9≤x≤11,2x−28≤y≤−21x−21}.
D1={(x,y) ∣ 1≤x≤9, −12x−112≤y≤2x−12} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 9, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq 2 x - \frac{1}{2}\right\}D1={(x,y)∣1≤x≤9,−21x−211≤y≤2x−21} und
D1={(x,y) ∣ 1≤x≤9, −12x−112≤y≤2x−12} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 9, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq 2 x - \frac{1}{2}\right\}D1={(x,y)∣1≤x≤9,−21x−211≤y≤2x−21}
D2={(x,y) ∣ 3≤x≤11, −12x−112≤y≤−12x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 11, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤11,−21x−211≤y≤−21x−21} und
D2={(x,y) ∣ 3≤x≤11, −12x−112≤y≤−12x−12} {\red{D_2}} = \left\{ (x,y) \, | \, 3 \leq x \leq 11, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}D2={(x,y)∣3≤x≤11,−21x−211≤y≤−21x−21}
D3={(x,y) ∣ 1≤x≤11, 2x−28≤y≤−12x−12}. {\red{D_3}} = \left\{ (x,y) \, | \, 1 \leq x \leq 11, \; 2 x - 28 \leq y \leq -\frac{1}{2} x - \frac{1}{2}\right\}.D3={(x,y)∣1≤x≤11,2x−28≤y≤−21x−21}.
D1={(x,y) ∣ 1≤x≤3, −12x−112≤y≤2x−8} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq 2 x - 8\right\}D1={(x,y)∣1≤x≤3,−21x−211≤y≤2x−8} und
D1={(x,y) ∣ 1≤x≤3, −12x−112≤y≤2x−8} {\red{D_1}} = \left\{ (x,y) \, | \, 1 \leq x \leq 3, \; -\frac{1}{2} x - \frac{11}{2} \leq y \leq 2 x - 8\right\}D1={(x,y)∣1≤x≤3,−21x−211≤y≤2x−8}