heureka

Um 600 Seiten auszudrucken benötigt ein Drucker 12 Minuten.

Wie lange braucht der Drucker für 900 Seiten?

$\begin{array}{|rcll|} \hline && \dfrac{12\ \text{Minuten}} {600\ \text{Seiten}} \cdot 900\ \text{Seiten} \\\\ &=& \dfrac{12\cdot 900} {600}\ \text{Minuten} \\\\ &=& \dfrac{12\cdot 9} {6}\ \text{Minuten} \\\\ &=& 2\cdot 9\ \text{Minuten} \\\\ &=& 18\ \text{Minuten} \\ \hline \end{array}$

Der Drucker braucht für 900 Seiten 18 Minuten.

how many digits does 2^1224 have

Number of precomma places:

$\begin{array}{|rcll|} \hline && \lfloor 1 + \log_{10}{(2^{1224})} \rfloor \\ &=& \lfloor 1 + 1224\cdot \log_{10}{(2)} \rfloor \quad & | \quad \log_{10}{(2)} = 0.30102999566 \\ &=& \lfloor 1 + 1224\cdot 0.30102999566 \rfloor \\ &=& \lfloor 1 + 368.460714693 \rfloor \\ &=& \lfloor 369.460714693 \rfloor \\ &=& 369 \\ \hline \end{array}$

2¹²²⁴ has 369 digits

point H is located at (-7,5).

Where is H after a 270 degrees clockwise rotation?

$\begin{array}{|rcll|} \hline \begin{pmatrix} -7 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} \cos(270^{\circ}) & -\sin(270^{\circ}) \\ \sin(270^{\circ}) & \cos(270^{\circ}) \\ \end{pmatrix} &=& \begin{pmatrix} -7 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0\\ \end{pmatrix} \\ &=& \begin{pmatrix} -7\cdot 0 + 5\cdot(-1) \\ -7\cdot 1 + 5\cdot 0 \end{pmatrix} \\ &=& \begin{pmatrix} -5 \\ -7 \end{pmatrix} \\ \hline \end{array}$

H after a 270 degrees clockwise rotation is at  (-5,-7)

0.0000015390771693 As a fraction

$\begin{array}{|rcll|} \hline \dfrac{15390771693} { 10000000000000000 }&=& 1.5390771693 \times 10^{-6} \\\\ \dfrac{1}{ 649740} &\approx & 1.539077169329\ldots \times 10^{-6} \\ \hline \end{array}$

Guten Tag, ich habe frage zum berechnen vom Auftreffwinkel beim blutspuren.

Die Formel ist breite/länge=Sin Alfa Alfa = arcsin (b/l) ZB tropfenbreite 14 und tropfenlänge 19 ergibt 47 grad auftreffwinkel (excel-Tabelle).

Wie tippe ich hier das ein, damit das ergebnis von 47 rauskommt?

the surface area of a sphere is 75 square centimeters,

what is the volume of the sphere in cubic cm?

Formula:
The volume inside a sphere is: $V = \frac43\pi r^3 \qquad V= ?$.

The surface area of a sphere is: $A = 4\pi r^2 \qquad A=75\ cm^2$.

$\begin{array}{|rcll|} \hline \dfrac{V}{A} &=& \dfrac{\frac43\pi r^3} {4\pi r^2} \\\\ \dfrac{V}{A} &=& \dfrac13\cdot \dfrac{4\pi}{4\pi}\cdot \dfrac{r^2}{r^2}\cdot r \\\\ \dfrac{V}{A} &=& \dfrac13\cdot r \\\\ V &=& \dfrac13 \cdot r \cdot A \\ \hline \end{array}$

Radius r:
$\begin{array}{|rcll|} \hline A &=& 4\pi r^2 \\ 4\pi r^2 &=& A \quad &| \quad : 4\pi \\ r^2 &=& \dfrac{A}{4\pi} \quad &| \quad \sqrt{} \\ r &=& \sqrt{ \dfrac{A}{4\pi} } \\ r &=& \dfrac12 \cdot \sqrt{ \dfrac{A}{\pi} } \\ \hline \end{array}$

Volume V:

$\begin{array}{|rcll|} \hline V &=& \dfrac13 \cdot r \cdot A \quad &| \quad r = \dfrac12 \cdot \sqrt{ \dfrac{A}{\pi} } \\\\ V &=& \dfrac13 \cdot \dfrac12 \cdot \sqrt{ \dfrac{A}{\pi} } \cdot A \\\\ \mathbf{V} &\mathbf{=}& \mathbf{ \dfrac16 \cdot \sqrt{ \dfrac{A^3}{\pi} } } \quad &| \quad A=75\ cm^2 \\\\ V &=& \dfrac16 \cdot \sqrt{ \dfrac{ ( 75\ cm^2)^3}{\pi} } \\\\ V &=& \dfrac16 \cdot \sqrt{ \dfrac{ 75^3}{\pi} }\ cm^3 \\\\ V &=& \dfrac16 \cdot \sqrt{ 134286.983234 }\ cm^3 \\\\ V &=& \dfrac16 \cdot 366.451883927 \ cm^3 \\\\ \mathbf{V} &\mathbf{=}& \mathbf{ 61.0753139879 \ cm^3 } \\ \hline \end{array}$

The volume of the sphere is $61.1\ cm^3$

was ist i hoch 19032003

$\begin{array}{|rcll|} \hline i^0 &=& 1 \\ i^1 &=& i \\ i^2 &=& -1 \\ i^3 &=& -i \\ \hline \end{array}$

Allgemein:

$\begin{array}{|rcll|} \hline i^{0+n\cdot 4} &=& 1 \\ i^{1+n\cdot 4} &=& i \\ i^{2+n\cdot 4} &=& -1 \\ i^{3+n\cdot 4} &=& -i \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && i^{19032003} \\ &=& i^{3+4758000\cdot 4} \\ &=& -i \\ \hline \end{array}$

Thanks Heureka, but I wanted both sides up against the equal sign.

I wanted right aligned on the left side

and left aligned on the right side.    ://

Do you know how to do that?

$\begin{array}{rcl} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}$

\begin{array} {rcl}... &=&...\end{array}
left side:         r = right aligned
Second field:   c = center
right side:          l = left aligned

Thanks Heureka, but I wanted both sides up against the equal sign.

I wanted right aligned on the left side

and left aligned on the right side.    ://

Do you know how to do that?

$\begin{array}{rcl} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}$

$$\begin{array}{rcl}  log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\   log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\    log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\    2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\     \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\      log_2(x)&=&2(10-2log_2(x))\\       log_2(x)&=&20-4log_2(x)\\        5log_2(x)&=&20\\         log_2(x)&=&4\\          2^{log_2(x)}&=&2^4\\           x&=&16\\ \end{array}$$

\begin{array} {rcl}... &=&...\end{array}
left side:         r = right aligned
Second field:   c = center
right side:          l = left aligned

Heureka, could you please tell me how I can get my Latex to right align in front of the equal sign. ??

right alignment :

$\begin{array}{lcr} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}$

The code:

\begin{array}{lcr}
 log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\
  log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\
   log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\
   2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\
    \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\
     log_2(x)&=&2(10-2log_2(x))\\
      log_2(x)&=&20-4log_2(x)\\
       5log_2(x)&=&20\\
        log_2(x)&=&4\\
         2^{log_2(x)}&=&2^4\\
          x&=&16\\
\end{array}

\begin{array} {lcr}... &=&...\end{array}

First field:       l = left alignment

Second field:   c = center

Third field:         r = right alignment