Gleichung nach f auflösen

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Gleichung nach f auflösen

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Gleichung nach f auflösen

Guest 01.02.2015

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Hallo Anonymous,

könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$

Gruß radix !

radix 02.02.2015

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radix · Answer 1 · 2015-02-02T09:21:37

Hallo Anonymous,

könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$

Gruß radix !

Omi67 · Answer 2 · 2015-02-01T08:49:23

Wie heißet die Gleichung????

Gleichung nach f auflösen

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Hallo Anonymous,

könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$

Gruß radix !

2⁺⁰ Answers

Hallo Anonymous,

könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$

Gruß radix !

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Gleichung nach f auflösen

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Hallo Anonymous,

könnte es die Linsengleichung sein ? 1f=1g+1b{\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}

Dann so : 1f=b(g×b)+g(g×b){\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}

1f=(b+g)(g×b){\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}

dann "kippen" => f=g×b(g+b){\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}

Gruß radix !

2+0 Answers

Hallo Anonymous,

könnte es die Linsengleichung sein ? 1f=1g+1b{\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}

Dann so : 1f=b(g×b)+g(g×b){\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}

1f=(b+g)(g×b){\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}

dann "kippen" => f=g×b(g+b){\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}

Gruß radix !

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könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$

2⁺⁰ Answers

könnte es die Linsengleichung sein ? ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{1}}}{{\mathtt{g}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{b}}}}$

Dann so : ${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{g}}}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

${\frac{{\mathtt{1}}}{{\mathtt{f}}}} = {\frac{\left({\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{g}}\right)}{\left({\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}}$

dann "kippen" => ${\mathtt{f}} = {\frac{{\mathtt{g}}{\mathtt{\,\times\,}}{\mathtt{b}}}{\left({\mathtt{g}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}}$