heureka

Hallo

$$\boxed{\binom{n}{k}=\frac{n(n-1)(n-2)\dots(n-(k-1))}{1*2*3 \dots k}}$$

$$\\\binom{\frac{2}{3}}{3}=
\frac{
\left(\frac{2}{3}\right)
\left(\frac{2}{3}-1\right)
\left(\frac{2}{3}-2\right)
}
{1*2*3}=
\frac{
\left(\frac{2}{3}\right)
\left(-\frac{1}{3}\right)
\left(-\frac{4}{3}\right)
}
{1*2*3}=
\frac{ \frac{8}{27} }
{6}=
\frac{8}{27}*\frac{1}{6}=
\frac{4}{27*3}=
\frac{4}{81}=
\left( \frac{2}{9}\right)^2$$

Viele Grüße

Division von Potenzen mit gleicher Basis:  $$\boxed{\frac{a^n}{a^m}=a^{n-m}}$$

$$\frac{a^{\frac{1}{2}}} {a^{\frac{1}{6}}}=a^{\frac{1}{2}-\frac{1}{6}}=a^{\frac{3}{6}-\frac{1}{6}}= a^{\frac{2}{6}}=a^{\frac{1}{3}}$$

Lösung: $$a^{\frac{1}{3}}=\sqrt[3]{a}$$

...continued.

if x is in degrees:

$$\\cos(x) = cos\left[180\textcolor[rgb]{1,0,0}{\ensuremath{^\circ}}-cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]
\quad | \quad \pm cos^{-1}\\
\pm x=\pm \left[ 180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]\\
x=180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\\
cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]=180\ensuremath{^\circ}}-x
\quad | \quad \cos{}\\\\
\frac{0.25}{6,76}*\cos{(x)}=\cos{(180\ensuremath{^\circ}}-x)}\\
\frac{0.25}{6,76}*\cos{(x)}=\underbrace{\cos{180\ensuremath{^\circ}}}_{-1}\cos{(x)}
+\underbrace{\sin{180\ensuremath{^\circ}}}_{0}\sin{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}=-\cos{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}+\cos{(x)}=0\\
\cos{(x)}\left(\frac{0.25}{6,76}+1\right)=0\\
\cos{(x)}=0\quad | \quad \pm \cos^{-1}\\
\pm x=\frac{\pi}{2}\\\\
\boxed{x=\frac{\pi}{2} \pm 2\pi*k\qquad or \qquad x=-\frac{\pi}{2} \pm 2\pi*k}$$

\\cos(x) = cos\left[180\textcolor[rgb]{1,0,0}{\ensuremath{^\circ}}-cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]
\quad | \quad \pm cos^{-1}\\
\pm x=\pm \left[ 180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]\\
x=180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\\
cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]=180\ensuremath{^\circ}}-x
\quad | \quad \cos{}\\\\
\frac{0.25}{6,76}*\cos{(x)}=\cos{(180\ensuremath{^\circ}}-x)}\\
\frac{0.25}{6,76}*\cos{(x)}=\underbrace{\cos{180\ensuremath{^\circ}}}_{-1}\cos{(x)}
+\underbrace{\sin{180\ensuremath{^\circ}}}_{0}\sin{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}=-\cos{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}+\cos{(x)}=0\\
\cos{(x)}\left(\frac{0.25}{6,76}+1\right)=0\\
\cos{(x)}=0\quad | \quad  \pm \cos^{-1}\\
\pm x=\frac{\pi}{2}\\\\
\boxed{x=\frac{\pi}{2} \pm 2\pi*k\qquad or \qquad x=-\frac{\pi}{2} \pm 2\pi*k}

cosx = cos[180-cosinverse((0.25*cosx)/6.76)]        x in rad!!!

The Solution for  $$\cos{(x)} = \cos{(180\textcolor[rgb]{1,0,0}{rad}-cos^{-1}{(\;(0.25*\cos{(x)})/6.76})\;)}$$      x in rad is:

\cos{(x)} = \cos{(180\textcolor[rgb]{1,0,0}{rad}-cos^{-1}{(\;(0.25*\cos{(x)})/6.76})\;)}  

$$x=\pm cos^{-1}
\left(
\;\frac
{
\sin{ ( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
}
{
\sqrt{
\left( \frac{0.25}{6.76} - \cos{
( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
\right) ^2
+ ( \sin{ (180\;\textcolor[rgb]{1,0,0}{rad}) } )^2
}
}
\;\right)$$

x=\pm cos^{-1}
\left(
    \;\frac
{
  \sin{ ( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
}
{
\sqrt{
     \left( \frac{0.25}{6.76} - \cos{
( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
     \right) ^2
+ ( \sin{ (180\;\textcolor[rgb]{1,0,0}{rad}) }  )^2
}
}
\;\right) 

$$\\x=\pm (2.47103630997\pm 2\pi*k)\\\\
Example:\\
x=-3.81215\\
x=-2.47104\\
x=2.47104\\
x=3.81245\\
\dots$$

\\x=\pm (2.47103630997\pm 2\pi*k)\\\\
Example:\\
x=-3.81215\\
x=-2.47104\\
x=2.47104\\
x=3.81245\\
\dots

$$\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11$$

latex code:

\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11

$$\lfloor \dots \rfloor = floor function$$

latex code:

\lfloor \dots \rfloor = floor function

$$\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11$$

latex code:

\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11

$$\lfloor \dots \rfloor = floor function$$

latex code:

\lfloor \dots \rfloor = floor function

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}$$

\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

x^3+4x^2+x-6=(x-1)(x^2+5x+6)

We solve:

$$\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}$$

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

We solve:

$$\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

1. Lösungsweg:

( 1 - 20% ) * 159,99

= ( 1 - 0,2 ) * 159,99

= 0,8 * 159,99

= 127,992

2.Lösungsweg:

159,99 - 20% * 159,99

= 159,99 - 0.2 * 159,99

= 159,99 - 31,998

= 127,992

$$\\12037367386=2*11*29*461*40927\\2*11*\not{\textcolor[rgb]{1,0,0}{2}}\not{\textcolor[rgb]{1,0,0}{9}}*461*40927=415081634\\\boxed{\frac{12037367386}{29}=415081634}$$