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

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