heureka

Wen Ein Fernfahrer fuhr in 15 Arbeitstagen 9810km.

Wie viele km fuhr er dann durchschnittlich an einem Tag

\(\begin{array}{|rcll|} \hline \frac{9810}{ 15 }\ km \\ = 654\ km \\ \hline \end{array}\)

Er fuhr an einem Tag durchschnittlich 654 km.

Thank you Melody.

1/4lg16+lg25-lg(1/2)+5*lg100 ohne TR lösen.

\(\begin{array}{|rcll|} \hline && \frac{1}{4}\cdot \lg(16) + \lg(25)- \lg( \frac{1}{2} )+5\cdot \lg(100) \\ &=& \frac{1}{4}\cdot \lg(2^4) + \lg(5^2)- \lg(\frac{1}{2})+ 5\cdot \lg(10^2) \\ &=& \frac{4}{4}\cdot \lg(2) + 2\cdot \lg(5)- \lg(\frac{1}{2})+ 5\cdot 2\cdot \lg(10) \\ &=& \lg(2) + 2\cdot \lg(5)- \lg(\frac{1}{2})+ 10\cdot \lg(10) \quad & | \quad \lg(10) = 1\\ &=& \lg(2) + 2\cdot \lg(5)- \lg(\frac{1}{2})+ 10\cdot 1 \\ &=& \lg(2) + 2\cdot \lg(5)- \lg(\frac{1}{2})+ 10 \quad & | \quad 5 = \frac{10}{2}\\ &=& \lg(2) + 2\cdot \lg(\frac{10}{2})- \lg(\frac{1}{2})+ 10 \quad & | \quad \lg(\frac{10}{2}) = \lg(10) - \lg(2) \\ &=& \lg(2) + 2\cdot [\lg(10) - \lg(2) ] - \lg(\frac{1}{2})+ 10 \quad & | \quad \lg(\frac{1}{2}) = \lg(1) - \lg(2) \\ &=& \lg(2) + 2\cdot [\lg(10) - \lg(2) ] - [ \lg(1) - \lg(2) ]+ 10 \quad & | \quad \lg(10) = 1 \qquad \lg(1) = 0\\ &=& \lg(2) + 2\cdot [1 - \lg(2) ] - [ 0 - \lg(2) ]+ 10 \\ &=& \lg(2) + 2 - 2\cdot \lg(2) + \lg(2)+ 10 \\ &=& 2\cdot \lg(2)- 2\cdot \lg(2) + 2 + 10 \\ &=& 2 + 10 \\ &\mathbf{=}& \mathbf{ 12 } \\ \hline \end{array}\)

How using L'Hospital's rule?

\(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\)

Formula: \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} f(x)\right) &=& \displaystyle\lim \limits_{x\to 0} \ln f(x) \\ \hline \end{array} \)

1. We  take the logarithm:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& \displaystyle\lim \limits_{x\to 0} \ln \left\{ \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } \right\}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \ln \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -\ln(e) \right] \quad &|\quad \ln(e) = 1\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -\frac{x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{\ln(1+x)-x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{\ln(1+x)-x}{x^2} \quad &|\quad \text{L'Hospital's rule}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{(\ln(1+x)-x)'}{(x^2)'} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1}{1+x}-1 }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1-1-x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{ -x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -x}{2x(1+x)} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -1}{2(1+x)} \quad &|\quad x = 0\\ &=& \frac{ -1}{2(1+0)} \\ &=& - \frac{1}{2} \\ \hline \end{array} \)

2. We revert the logarithm:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& - \frac{1}{2} \\\\ e^{\ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) } &=& e^{- \frac{1}{2} } \\\\ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } &=& e^{- \frac{1}{2} } \\ &=& \frac{1}{ e^{ \frac{1}{2} } } \\ \mathbf{ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } } &\mathbf{=}& \mathbf{\frac{1}{ \sqrt{e} } } \\ \hline \end{array}\)

Fundamental theorem of calculus

\(\begin{array}{|rcll|} \hline g(s) &=& \displaystyle\int\limits_{8}^{s} f(t)\ dt =\displaystyle \int\limits_{8}^{s} (t-t^4)^5\ dt \\ &=& F(s) - F(8) \\\\ g'(s) &=& \dfrac{d}{ds} \displaystyle\int\limits_{8}^{s} f(t)\ dt \\\\ &=& \dfrac{d\ F(s)}{ds} - \dfrac{d\ F(8)}{ds} \quad &| \quad \dfrac{d\ F(8)}{ds} = 0, \text{ because } F(8) \text{ is constant.} \\\\ &=& \dfrac{d\ F(s)}{ds} \\\\ &=& f(s) \quad & |\quad f(t)=(t-t^4)^5 \\\\ \mathbf{g'(s)}& \mathbf{=} & \mathbf{(s-s^4)^5} \\ \hline \end{array}\)

The Formula is:

\(\begin{array}{|rcll|} \hline \dfrac{d}{ds} \displaystyle\int\limits_{a}^{s} f(t)\ dt = f(s) \\ \hline \end{array}\)

First State Bridge-Painting Costs

\(\begin{array}{|rcll|} \hline \text{Painting Costs } = 1000\cdot \left[ -24 + \frac{351}{6} \cdot \left(\frac{\text{Length}}{100}\right) - 19 \cdot \left(\frac{\text{Length}}{100}\right)^2 + \frac{15}{6} \cdot \left(\frac{\text{Length}}{100}\right)^3 \right] \\ \hline \end{array}\)

Hello Melody,

in firefox we can do the login without checkmark the Policy "Secure Login" "check box".

   ^------------------------  no checkmark here!

Hallo asinus,

übergangsweise könntest du mit http://web2.0rechner.de anstelle von https://web2.0rechner.de die Seite aufrufen.

Gleicherweise kannst du den login verwenden, indem du ganz unten den Haken unter Policy Secure Login entfernst,

dann ist die Verbindung wie üblich wieder da. Siehe Abbildung:

    ^------------------------  Haken entfernen

How much is an acre?

1 Acre (International):

\(\small{ \begin{array}{|rcll|} \hline 1\ \text{Acre (International)} &=& 4\ \text{roods} \cdot \dfrac{ 40\ \text{perches}} { 1\ \text{rood} } \cdot \dfrac{ 30\frac14\ \text{sq. yds} } { 1\ \text{perch} } \cdot \dfrac{ 9\ \text{sq. ft} } { 1\ \text{sq. yds} } \cdot \dfrac { 12^2\ \text{sq. in} } { 1\ \text{sq. ft} } \cdot \dfrac { 0.0254^2\ m^2 } { 1\ \text{sq. in} }\\\\ &=& 4 \cdot 40 \cdot 30\frac14 \cdot 9 \cdot 12^2 \cdot 0.0254^2\ m^2 \\\\ &=& 4 \cdot 40 \cdot 30.25 \cdot 9 \cdot 12^2 \cdot 0.0254^2\ m^2 \\\\ \mathbf{1\ \text{Acre (International)}} &\mathbf{=}& \mathbf{4046.85642240\ m^2 } \\ \hline \end{array} } \)

1 Acre (U.S. Survey):

\(\small{ \begin{array}{|rcll|} \hline 1\ \text{Acre (U.S. Survey)} &=& 4\ \text{roods} \cdot \dfrac{ 40\ \text{perches}} { 1\ \text{rood} } \cdot \dfrac{ 30\frac14\ \text{sq. yds} } { 1\ \text{perch} } \cdot \dfrac{ 9\ \text{sq. ft} } { 1\ \text{sq. yds} } \cdot \dfrac { \left( \frac{1200}{3937} \right)^2\ \ m^2 } { 1\ \text{sq. Survey ft} } \cdot \\\\ &=& 4 \cdot 40 \cdot 30\frac14 \cdot 9 \cdot \left( \frac{1200}{3937} \right)^2\ m^2 \\\\ &=& 4 \cdot 40 \cdot 30.25 \cdot 9 \cdot \left( \frac{1200}{3937} \right)^2\ m^2 \\\\ \mathbf{1\ \text{Acre (U.S. Survey)}} &\mathbf{=}& \mathbf{4046.87260987\ m^2 } \\ \hline \end{array} }\)

two dice are rolled.

what is the probability that the total number achieved

is greater than 4 but less than or equal to ten?

\(\color{green}{\text{In green dice one}}\\ \color{blue}{\text{In blue dice two}}\\ \color{red}{\text{In red the total number } >4 \text{ and } \le 10 }\)

\(\begin{array}{|r|r|r|r|r|r|r|} \hline + & \color{blue}1 & \color{blue}2 & \color{blue}3 & \color{blue}4 & \color{blue}5 & \color{blue}6 \\ \hline \color{green}1 & 2 & 3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 \\ \hline \color{green} 2 & 3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 \\ \hline \color{green}3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 \\ \hline \color{green}4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10 \\ \hline \color{green}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10& 11 \\ \hline \color{green}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10& 11& 12 \\ \hline \end{array} \)

\(\text{The probability of } > 4 \text{ but } \le 10 \text{ is } \dfrac{27}{36} \text{ or } 75\ \% \)