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: $$(x-2)^2+(y-5)^2=25 \qquad | \quad(x-x_{center})^2+(y-y_{center})^2=r^2$$