1.) a) D = R ∖ { − 1 } {\displaystyle \mathbb {D} =\mathbb {R} \backslash \{-1\}} D = R ∖ [ − ∞ | − 1 [ {\displaystyle \mathbb {D} =\mathbb {R} \backslash [-\infty |-1[} D = R ∖ [ − ∞ | + 1 ] {\displaystyle \mathbb {D} =\mathbb {R} \backslash [-\infty |+1]} D = R {\displaystyle \mathbb {D} =\mathbb {R} }
b) D = { y ∈ R ∖ { 0 } } {\displaystyle \mathbb {D} =\{y\in \mathbb {R} \backslash \{0\}\}} D = { x ∈ R 0 + } , { y ∈ R ∖ [ − 1 | 0 ] } {\displaystyle \mathbb {D} =\{x\in \mathbb {R} _{0}^{+}\},\{y\in \mathbb {R} \backslash [-1|0]\}} D = { x ∈ R ∖ { 1 } } , { y ∈ R ∖ { − 1 } } {\displaystyle \mathbb {D} =\{x\in \mathbb {R} \backslash \{1\}\},\{y\in \mathbb {R} \backslash \{-1\}\}}
2.) a) U = 2 r π {\displaystyle U=2r\pi \,} r ( U ) = U 2 π {\displaystyle r(U)={\frac {U}{2\pi }}}
b) V = 4 3 π r 3 {\displaystyle V={\frac {4}{3}}\pi r^{3}} V ( d ) = 4 3 π ⋅ ( d 2 ) 3 {\displaystyle V(d)={\frac {4}{3}}\pi \cdot \left({\frac {d}{2}}\right)^{3}} V ( d ) = 4 3 π ⋅ 1 8 d 3 {\displaystyle V(d)={\frac {4}{3}}\pi \cdot {\frac {1}{8}}d^{3}} V ( d ) = π 6 ⋅ d 3 {\displaystyle V(d)={\frac {\pi }{6}}\cdot d^{3}}
c) h = 2 r {\displaystyle h=2r\,} V = r 2 π h = r 2 π ⋅ 2 r = 2 r 3 π {\displaystyle V=r^{2}\pi h\quad =r^{2}\pi \cdot 2r\quad =2r^{3}\pi } O = 2 π r h + 2 π r 2 = 2 π r ( h + r ) = 2 π r ( 3 r ) = 6 r 3 π {\displaystyle O=2\pi rh+2\pi r^{2}\quad =2\pi r(h+r)\quad =2\pi r(3r)\quad =6r^{3}\pi } f ( r ) = O V {\displaystyle f(r)={\frac {O}{V}}} f ( r ) = 6 r 2 π 2 r 3 π {\displaystyle f(r)={\frac {6r^{2}\pi }{2r^{3}\pi }}} f ( r ) = 3 r {\displaystyle f(r)={\frac {3}{r}}}
3.) Betrachtet man die Höhe und die Breite des Schirms als Katheten eines rechtwinkeligen Dreiecks, so ist die Bilddiagonale die Hypothenuse. ⇒ {\displaystyle \Rightarrow } c = a 2 + b 2 ⇒ c = 5 d m {\displaystyle c={\sqrt {a^{2}+b^{2}}}\qquad \Rightarrow \qquad c=5dm}