Benutzer:Broli

$s(t)=0.25t^{3}-1,5t^{2}+2,25t\,$

$d(t;h)={\frac {s(t+h)-s(t)}{h}}$ $={\frac {0,25(t+h)^{3}-1,5(t+h)^{2}+2,25(t+h)-(0,25t^{3}-1,5t^{2}+2,25t)}{h}}$ $={\frac {0,25(t^{3}+3t^{2}h+3th^{2}+h^{3})-1,5(t^{2}+2th+h^{2})+2,25(t+h)}{h}}$ $={\frac {0,75t^{2}h+0,75th^{2}+0,75h^{3}-3th-1,5h^{2}+2,25h}{h}}$ $=0,75t^{2}+0,75th+0,25h^{2}-3t-1,5h+2,25\,$ $=0,75t^{2}-3t+2,25+0,75th-1,5h+0,25h^{2}\,$

$d(2;3)=0,75\cdot 2^{2}-3\cdot 2+2,25+0,75\cdot 2\cdot 3-1,5\cdot 3+0,25\cdot 3^{2}$ $=1,5\,$

$f(x)=x^{2}\,$ $f'(x)=1\,$ $f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ $=\lim _{h\to 0}{\frac {(x+h)^{2}-x^{2}}{h}}$ $=\lim _{h\to 0}{\frac {x^{2}+2xh+h^{2}-x^{2}}{h}}$ $=\lim _{h\to 0}2x+h$ $=2x+0\,$ $=2x\,$

$f'(x)=1\,$ $2x=1\qquad \vert \quad :2\,$ $x={\frac {1}{2}}$ $f({\frac {1}{2}})={\frac {1}{4}}$

$f'(x)=2x\,$ $2x=1\,$

${\frac {1}{f}}={\frac {1}{g}}+{\frac {1}{b}}$

$\gamma ={\frac {B}{G}}={\frac {b}{g}}$

$\gamma _{M}=\gamma _{\text{Ob}}\cdot \gamma _{\text{Ok}}=[{\frac {t}{f_{\text{Ob}}}}\cdot {\frac {s_{0}}{f_{\text{Ok}}}}]$

$d={\frac {\lambda }{n\cdot \sin \varphi }}$

$R={\frac {1}{d}}={\frac {n\cdot \sin \varphi }{\lambda }}$

$\gamma ={\frac {b}{g}}=b\cdot {\frac {1}{g}}=b\cdot ({\frac {1}{f}}-{\frac {1}{b}})$

$\gamma =8cm\cdot ({\frac {1}{2cm}}-{\frac {1}{8cm}})={\underline {\underline {3}}}$

${\begin{aligned}{\frac {G_{\text{Testobj}}}{S_{\text{Testobj}}}}&={\frac {G_{\text{Zelle}}}{S_{\text{Zelle}}}}\qquad \vert \cdot S_{\text{Zelle}}\\G_{\text{Zelle}}&={\frac {G_{\text{Testobj}}\cdot S_{\text{Zelle}}}{S_{\text{Testobj}}}}\\G_{\text{Zelle}}&={\frac {30\mu m\cdot 7}{23}}\approx {\underline {\underline {9{,}13\mu m}}}\end{aligned}}$

${\begin{aligned}d&={\frac {\lambda }{n\cdot \sin \varphi }}\\&={\frac {550\cdot 10^{-9}m}{0{,}25}}\\&=2{,}2\cdot 10^{-6}m\\d&={\underline {\underline {2{,}2\mu m}}}\end{aligned}}$

${\begin{aligned}d&={\frac {\lambda }{n\cdot \sin \varphi }}\\&={\frac {550\cdot 10^{-9}m}{0{,}65}}\\&=0{,}846\cdot 10^{-6}m\\d&\approx {\underline {\underline {0{,}8\mu m}}}\end{aligned}}$

Fehler beim Parsen (Unbekannte Funktion „\begin{align}“): {\displaystyle \begin{align} \text{Erythrozytenanzahl\ pro \mu l\ Volumen} &= \frac{\mathrm{ausgez\ddot a hlte\ Erythrozyten}}{\mathrm{ausgez\ddot a hlte\ Fl\ddot a che\cdot Kammertiefe \cdot Verd\ddot u nnung}}\\ &= \frac{570}{5 \cdot 0{,}04mm^2 \cdot 0{,}1mm \cdot \frac{1}{200}}\\ &= 5{.}700{.}000 \frac{Eryhrozyten}{mm^3}\\ &= \underline{\underline{5{,}7\cdot10^6 \frac{Erythrozyten}{\mu m}}} \end{align} }

${\begin{aligned}\mathrm {Erythrozytenanzahl\ pro\ \mu l\ Volumen} &={\frac {\mathrm {ausgez{\ddot {a}}hlte\ Erythrozyten} }{\mathrm {ausgez{\ddot {a}}hlte\ Fl{\ddot {a}}che\cdot Kammertiefe\cdot Verd{\ddot {u}}nnung} }}\\&={\frac {570}{5\cdot 0{,}04mm^{2}\cdot 0{,}1mm\cdot {\frac {1}{200}}}}\\&=5{.}700{.}000{\frac {Erythrozyten}{mm^{3}}}\\&={\underline {\underline {5{,}7\cdot 10^{6}{\frac {Erythrozyten}{\mu m}}}}}\end{aligned}}$

$D={\frac {I}{I_{0}}}$

$I(x)=I_{0}\cdot e^{-{\tilde {\varepsilon }}_{\lambda }\cdot c\cdot x}$

$\ln {({\frac {I(x)}{I_{0}}})}=-{\tilde {\varepsilon }}_{\lambda }\cdot c\cdot x$

$E=\log {({\frac {I_{0}}{I(x)}})}$

$E=\varepsilon _{\lambda }\cdot c\cdot x$

${\begin{aligned}\ln {(10)}^{-1}\approx 0{,}4343\\\varepsilon _{\lambda }={\tilde {\varepsilon }}_{\lambda }\cdot 0{,}4343\end{aligned}}$

${\begin{aligned}E&=\log {({\frac {I_{0}}{I(x)}})}\\0{,}8&=\log {({\frac {1}{I(x)}})}\qquad \vert 10^{x}\\10^{0{,}8}&=({\frac {1}{I(x)}})\qquad \vert \cdot I(x)\div 10^{0{,}8}\\I(x)&=10^{-0{,}8}\\&\approx 0{,}158\\&={\underline {\underline {15,8\%}}}\end{aligned}}$

${\begin{aligned}E&=\log {({\frac {I_{0}}{I(x)}})}\\E&=\log {({\frac {1}{0{,}8}})}\\E&\approx {\underline {\underline {0{,}01}}}\end{aligned}}$

${\begin{aligned}E&=\varepsilon _{\lambda }\cdot c\cdot x\\1&=\varepsilon _{\lambda }\cdot 10^{-2}\cdot 1\qquad \vert \div 10^{-2}\\\varepsilon _{380}&={\frac {1}{10^{-2}}}\\&={\underline {\underline {100{\frac {l}{g\cdot cm}}}}}\end{aligned}}$

${\begin{aligned}n&={\frac {m}{M_{g}}}\\n&={\frac {10^{-2}}{500}}\\&=2\cdot 10^{-5}mol\\\hookrightarrow c&={\underline {\underline {2\cdot 10^{-5}mol/l}}}\end{aligned}}$

${\begin{aligned}{\frac {E_{0}}{x_{0}}}&={\frac {E_{1}}{x_{1}}}\\{\frac {1{,}2}{1}}&={\frac {1}{x_{1}}}\qquad \vert Kehrwert\\x_{1}&={\frac {1}{1{,}2}}\\x_{1}&={\underline {\underline {0{,}83cm}}}\end{aligned}}$

${\begin{aligned}\varepsilon _{550}&=13{,}23{\frac {l}{g\cdot cm}}\cdot 378{,}90{\frac {g}{mol}}\\\varepsilon _{550}&=5012{,}847{\frac {l}{mol\cdot cm}}\end{aligned}}$

$\varepsilon _{550}={\underline {\underline {5012{,}847{\frac {1}{M\cdot cm}}}}}$

${\begin{aligned}x\pm 2s_{x}&=13{,}23\pm 2\cdot 0{,}6957\\x+2s_{x}&=14{,}6214\\x-2s_{x}&=11{,}8386\end{aligned}}$

$\varepsilon _{\lambda }={\frac {E}{c\cdot x}}$