heureka

$$\boxed{\begin{array}{lrcrcrr} (1) &3y & + & 4x &=& 11 &\\ (2)& 3y& + & 2x &=& 13 & \\&--&-&--&-&--& \\(1) - (2)& 0 y&+&2x&=&-2 & \\&&&2x&=&-2 & \quad | \quad :2 \\ &&&\textcolor[rgb]{1,0,0}{x}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{-1} & \end{array}}$$

Hi Melody,

0 + 2 = 2    n o t  0

$$\\A=\left( \begin{array}{cc}1 & 2 \\3 & 4 \end{array} \right)\qquad B=\left( \begin{array}{cc}2 & 6 \\-1 & 0 \end{array} \right)$$

$$\boxed{AB+B=(A+\textcolor[rgb]{0,0,1}{I})B}$$

$$\mbox{Identity Matrix }\textcolor[rgb]{0,0,1}{I=\left( \begin{array}{cc}1 & 0 \\0 & 1 \end{array} \right)}\\$$

$$\mbox{A+\textcolor[rgb]{0,0,1}{I}} =\left( \begin{array}{cc}1 & 2 \\3 & 4 \end{array} \right) + \textcolor[rgb]{0,0,1}{\left( \begin{array}{cc}1 & 0 \\0 & 1 \end{array} \right)} } = \left( \begin{array}{cc}1+\textcolor[rgb]{0,0,1}{1} & 2+\textcolor[rgb]{0,0,1}{0} \\3+\textcolor[rgb]{0,0,1}{0} & 4+\textcolor[rgb]{0,0,1}{1} \end{array} \right) = \left( \begin{array}{cc}2 & 2 \\3 & 5 \end{array} \right)$$

$$\mbox{(A+\textcolor[rgb]{0,0,1}{I})B}=\left( \begin{array}{cc}2 & 2 \\3 & 5 \end{array} \right) } \left( \begin{array}{cc}2 & 6 \\-1 & 0 \end{array} \right) = \left( \begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18 \end{array} \right)$$

$$\boxed{AB+B= \left( \begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18 \end{array} \right) }$$

Hallo radix,

ich sehe diese Frage eher als "Fun" oder Quiz - Frage.

Jemand setzt sich wirklich mit der Mathematik auseinander.

Lernen heißt auch spielen mit den Möglichkeiten.

Don't worry be happy! 

Heureka

$$(1+i)^{102} \quad?$$  

$$(1+i)^{102} = \left((1+i)^2\right)^{51}$$

$$\\(1+i)^2=1^2+2*1*i+i^2=1+2i+i^2 \quad | \quad \textcolor[rgb]{0,0,1}{i^2 = \left(\sqrt{-1}\right)^2 = -1} \\ (1+i)^2=1+2i-1\\ \textcolor[rgb]{1,0,0}{(1+i)^2=2i }$$

$$(1+i)^{102}=(2i)^{51} =2^{51}\times i^{51}$$

$$\\i^{51} = \left(i^4\right)^{12}\times (i^3)$$

$$\\i^4=i^2*i^2=(-1)*(-1)=1\\ i^3=i^2*i=(-1)*i=-i$$

$$\\i^{51} = \left(1\right)^{12}\times (-i) =-i$$

$$(1+i)^{102} =2^{51}\times (-i) =-2^{51}i$$

$$\boxed{(1+i)^{102}=-2^{51}i=-2251799813685248i}$$

$$\\3\dfrac{7}{10}\\\\ =3+\dfrac{7}{10}\\\\ =3*\dfrac{10}{10}+\dfrac{7}{10}\\\\ =\dfrac{3*10}{10}+\dfrac{7}{10}\\\\ =\dfrac{30}{10}+\dfrac{7}{10}\\\\ =\dfrac{30+7}{10}\\\\ =\dfrac{37}{10}\\\\ =3.7$$

1 135 263 408 000 000 000 metres per hour.

Use the calculator: 120[ly/h] in [m/h]

$$\\P(-1, 1)\quad and \quad Q(5, 9)\\ \boxed{Center = \frac{1}{2}\left( P+Q\right)} =\frac{1}{2}* \left[ \left(\begin {array} {c} -1 \\1 \end {array} \right) + \left(\begin {array} {c} 5 \\ 9 \end {array} \right) \right]\\ =\frac{1}{2}* \left(\begin {array} {c} -1+5 \\1+9 \end {array} \right)\\ =\frac{1}{2}* \left(\begin {array} {c} 4 \\10 \end {array} \right)\\ =\left(\begin {array} {c} \frac{1}{2}* 4 \\\frac{1}{2}* 10 \end {array} \right)\\ \boxed{Center=\left(\begin {array} {c} 2 \\5 \end {array} \right) =(2, 5) }$$

$$\\\boxed{\vec{radius}= \frac{1}{2}\left( P-Q\right)} =\frac{1}{2}* \left[ \left(\begin {array} {c} -1 \\1 \end {array} \right) - \left(\begin {array} {c} 5 \\ 9 \end {array} \right) \right]\\ =\frac{1}{2}* \left(\begin {array} {c} -1-5 \\1-9 \end {array} \right)\\ =\frac{1}{2}* \left(\begin {array} {c} -6 \\-8 \end {array} \right)\\ =\left(\begin {array} {c} \frac{1}{2}* -6 \\\frac{1}{2}* -8 \end {array} \right)\\ \vec{radius}= \left(\begin {array} {c} -3 \\-4 \end {array} \right)\\ radius=\|\vec{radius}\|=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \boxed{radius=5}$$

The equation for the circle is: $$(x-2)^2+(y-5)^2=25 \qquad | \quad(x-x_{center})^2+(y-y_{center})^2=r^2$$