heureka

$$\small{\text{
$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
\mathrm{child} & #1 & #2 & #3 & #4 & #5 & #6 & #7 \\
\hline
#1 & x & \textcolor[rgb]{1,0,0}{(1,2)} & \textcolor[rgb]{1,0,0}{(1,3)} & \textcolor[rgb]{1,0,0}{(1,4)} & \textcolor[rgb]{1,0,0}{(1,5)} & \textcolor[rgb]{1,0,0}{(1,6)} & \textcolor[rgb]{1,0,0}{(1,7)} \\
\hline
#2 & & x & \textcolor[rgb]{1,0,0}{(2,3)} & \textcolor[rgb]{1,0,0}{(2,4)} & \textcolor[rgb]{1,0,0}{(2,5)} & \textcolor[rgb]{1,0,0}{(2,6)} & \textcolor[rgb]{1,0,0}{(2,7)} \\
\hline
#3 & & & x & \textcolor[rgb]{1,0,0}{(3,4)} & \textcolor[rgb]{1,0,0}{(3,5)} & \textcolor[rgb]{1,0,0}{(3,6)} & \textcolor[rgb]{1,0,0}{(3,7)} \\
\hline
#4 & & & & x & \textcolor[rgb]{1,0,0}{(4,5)} & \textcolor[rgb]{1,0,0}{(4,6)} & \textcolor[rgb]{1,0,0}{(4,7)} \\
\hline
#5 & & & & & x & \textcolor[rgb]{1,0,0}{(5,6)} & \textcolor[rgb]{1,0,0}{(5,7)} \\
\hline
#6 & & & & & & x & \textcolor[rgb]{1,0,0}{(6,7)} \\
\hline
#7 & & & & & & & x \\
\hline
\end{array}=\textcolor[rgb]{1,0,0}{(21)} \ \mathrm{handshakes}
$}}$$

1 1/2 divided by 1/2

$$\dfrac{
1\frac{1}{2} }
{\frac{1}{2} }\\\\
=\dfrac{
1+\frac{1}{2} }
{\frac{1}{2} }\\\\
=\dfrac{ 1 } {\frac{1}{2} }
+\dfrac{ \frac{1}{2} } {\frac{1}{2} }\\\\
=\dfrac{ 1 } {\frac{1}{2} }
+1\\\\
=1\cdot \frac{2}{1} }
+1\\
= 2+1\\
=3$$

or

$$\dfrac{
1\frac{1}{2} }
{\frac{1}{2} }\\\\
=
\dfrac{
\frac{3}{2} }
{\frac{1}{2}
}\\\\
=
\frac{3}{2} \cdot
\frac{2}{1} \\
=3$$

Wie gestaltet sich der Rechenweg bei (lg(2))^2+lg(5)*lg(20) ?

$$\small{\text{
$
\boxed{\
[\log(2)]^2+\log(5)\cdot \log(20)\
}
$
}}\\\\
\small{\text{
$
=[\log(2)]^2+\log(5)\cdot \log(2\cdot 2 \cdot 5)
$
}}\\
\small{\text{
$
=[\log(2)]^2+\log(5)\cdot [ \log(2) + \log(2) + \log(5) ]
$
}}\\
\small{\text{
$
=[\log(2)]^2+2\log(2)\cdot\log(5)+[\log(5)]^2
$
}}\\
\small{\text{
$
=[\ \log(2) + \log(5) \ ]^2
$
}}\\
\small{\text{ $ =[\ \log(2\cdot 5)\ ]^2 $ }}\\
\small{\text{ $ =[\ \log(10)\ ]^2
\quad | \quad \log(10) = 1 $ }}\\
\small{\text{ $ =1^2 $ }}\\
\small{\text{ $ =1 $ }}\\$$

|2 -1 2|

|-4 0 1|

|-1 -1 -1|

(the vertical line at the front and back of each one are all suppose to be connected vertically)

$$\small{\text{
$
\left|
\begin{array}{rrr}
2 &-1 &2 \\
-4 &0 &1 \\
-1 &-1 &-1 \\
\end{array}
\right|
$
}}\\\\\\
\begin{array}{ll}
= &
2\cdot 0 \cdot (-1) + (-1)\cdot(-1)\cdot 1 + (-4)\cdot(-1)\cdot 2\\
&- (-1)\cdot 0 \cdot 2 - 2\cdot(-1)\cdot 1 - (-4)\cdot(-1)\cdot(-1)\\
= &
0 + 1 + 8\\
&-0 + 2 +4\\
= &
9\\
&+6\\
= 15
\end{array}
$
}}$$

y=3x+1 ( line )

y=x^2+1 ( parabola )

$$\small{\text{
$
\begin{array}{rcl}
y_{parabola} &=& y_{line} \\
x^2+1 &=& 3x+1\\
x^2+\not{1} &=& 3x+\not{1}\\
x^2-3x &=& 3x \\
x^2-3x &=& 0 \\
\underbrace{x}_{x_1=0}\cdot\underbrace{(x-3)}_{=0} &=& 0 \\
x_2-3 &=& 0 \\
x_2 &=& 3
\end{array}
$
}}$$

$$\small{\text{
$
\begin{array}{rcl}
\\y_1 &=& 3\cdot x_1 + 1 \\
y_1 &=& 3 \cdot 0 + 1 \\
y_1 &=& 1\\\\
y_2 &=& 3\cdot x_2 + 1 \\
y_2 &=& 3 \cdot 3 + 1 \\
y_2 &=& 10
\end{array}
$
}}$$

So we have: $$\small{\text{$(\ x_1 = 0 \quad y_1 = 1 \ )$}}$$ and $$\small{\text{$(\ x_2 = 3 \quad y_2 = 10 \ )$}}$$

Cramer's Rule

$$\begin{array}{rcrcrcr}
1\cdot x &+& 4 \cdot y &-&3 \cdot z &=& -8 \\
3\cdot x &-& 1 \cdot y &+& 3 \cdot z &=& 12 \\
1\cdot x &+& 1 \cdot y &+& 6 \cdot z &=& 1 \\
\end{array}$$

$$\small{\text{
$
x=
\frac
{\left|\begin{array}{rrr}
-8 & 4 & -3 \\
12 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}} \quad
\small{\text{
$
y=
\frac
{\left|\begin{array}{rrr}
1 & -8 & -3 \\
3 & 12 & 3 \\
1 & 1 & 6
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}} \quad
\small{\text{
$
z=
\frac
{\left|\begin{array}{rrr}
1 & 4 & -8 \\
3 & -1 & 12 \\
1 & 1 & 1
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}}$$

$$\small{\text{
$
x=
\frac
{
(-8)\cdot(-1)\cdot6+1\cdot4\cdot3+12\cdot 1\cdot(-3)
-1\cdot(-1)\cdot(-3)-(-8)\cdot 1\cdot 3 - 12\cdot 4\cdot 6
}
{
1\cdot(-1)\cdot 6 + 1\cdot 4 \cdot 3 + 3\cdot 1 \cdot (-3)
-1\cdot(-1)\cdot(-3) - 1\cdot 1 \cdot 3- 3\cdot 4 \cdot 6
}
=\frac
{48+12-36-3+24-288}
{-6+12-9-3-3-72}
=\frac
{-243}
{-81}
=3
$
}}$$

$$\small{\text{
$
y=
\frac
{
1\cdot 12 \cdot 6 + 1\cdot(-8)\cdot3 + 3\cdot 1 \cdot (-3)
-1\cdot 12 \cdot (-3) - 1\cdot 1 \cdot 3 - 3 \cdot (-8) \cdot 6
}
{
-81
}
=\frac
{72-24-9+36-3+144}
{-81}
=\frac
{216}
{-81}
=-2\frac{2}{3}
$
}} \quad$$

$$\small{\text{
$
z=
\frac
{
1\cdot(-1)\cdot 1 +1\cdot 4 \cdot 12 + 3 \cdot 1 \cdot(-8)
-1\cdot(-1)\cdot (-8) -1\cdot 1 \cdot 12 - 3\cdot 4 \cdot 1
}
{
-81
}
=\frac
{-1+48-24-8-12 -12}
{-81}
=\frac
{-9}
{-81}
=\frac{1}{9}
$
}} \quad$$

Gegeben: fk(x)=x2-7x+k

b) y=-2x+t und k=12,25 Für welches t berührt die Gerade die Parabel ?

$$\small{\text{
$
\boxed{\ y'_{Parabel} = y'_{Gerade} \ }
$}} \qquad
\small{\text{
$
y'_{Parabel} = 2x-7
\qquad y'_{Gerade} = -2
$
}} \\\\
\small{\text{
Der Ber\"uhrpunkt
$
x_b:
\begin{array}{rcl}
\\
2\cdot x_b-7 &=& -2 \\
2\cdot x_b &=& 7 -2 \\
2\cdot x_b &=& 5 \\
x_b &=& \frac{5}{2} = 2,5
\end{array}
$
}} \\\\
\small{\text{
Der Ber\"uhrpunkt
$
y_b:
\begin{array}{rcl}
\\
y_b &=& x_b^2-7\cdot x_b+12,25 \\
y_b &=& 2,5^2-7\cdot 2,5+12,25 \\
y_b &=& 6,25-17,5+12,25 \\
y_b &=& 1
\end{array}
$
}} \\\\
\small{\text{
Berechnung von
$
t:
\begin{array}{rcl}
\\
y_b &=& -2\cdot x_b + t\\
1 &=& -2\cdot2,5 + t\\
1 &=& -5 + t\\
t &=& 6
\end{array}
$
}} \\\\$$

If I completely factored -4x^3+24x^2-36x would the answer be -4x(x-3)^2   ?

$$-4x^3+24x^2-36x \\
= -4x \cdot (x^2-6x+9) \\
= -4x \cdot (x-3)\cdot (x-3) \\
= -4x \cdot (x-3)^2 \\
= (-4)\cdot x \cdot (x-3)\cdot (x-3)$$

wie viele Diagonalen hat ein 7-Eck und wie berechnet man Diagonalen in n-Ecken? bitte so schreiben das ein 7. Klässler es versteht danke

1+2+3+4+5+6 - 7 = 6*(1+6)/2 - 7 = 7 * 3 - 7 =  2*7 = 14

a^5*a^X=a^2

$$\small{\text\
$
\begin{array}{rcl}
a^5\cdot a^x &=& a^2 \\
a^{5+x} &=& a^2 \\
5+x &=& 2 \\
x &=& 2-5 \\
x &=&-3
\end{array}
$
}}$$