heureka

could you explain WHY this works ?

$$\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\
\small{\text{direction vectors: }}\\
\begin{array}{crcl}
& \vec{d_1} &=& \vec{P_2} - \vec{P_1}\\
& \vec{d_2} &=& \vec{P_3} - \vec{P_2} \\
& \vec{d_3} &=& \vec{P_1} - \vec{P_3}\\
\end{array}\\\\\\
\small{\text{$\begin{array}{lclclclcl}
\vec{P_4} = \vec{P_1}+\vec{d_1}-\vec{d_3}
&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3} =\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_2}+\vec{d_2}-\vec{d_1}
&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_3}+\vec{d_3}-\vec{d_2}
&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\\end{array}$}}$$

(a+2b)²-(a-2b)*(a+2b)-(3a-b)² = ???

$$\small{\text{
$
(a+2b)^2-(a-2b)*(a+2b)-(3a-b)^2 = 7b^2+10ab-9a^2
$}}$$

Whats the area of a regular hexagon with a height of 2 times the square root of 3 ?

$$\boxed{
\mathbf{
A = 2\cdot h^2\cdot\sqrt{3}
}
}\\\\
\small{\text{
$
\begin{array}{lrcl}
A = 2\cdot \textcolor[rgb]{150,0,0}{h^2}\cdot\sqrt{3} \qquad & h &=& 2\cdot\sqrt{3}\\
& h^2 &=& 4\cdot 3 \\
& h^2 &=& 12 \\
A = 2\cdot \textcolor[rgb]{150,0,0}{12}\cdot\sqrt{3}\\
A = 24 \cdot \sqrt{3}\\
A = 41.5692193817 \\
A \approx 41.569
\end{array}
$}}$$

Question:

 u(x,0) = e^{(-x^2)}  ,  u_t(x,0) = d/dx(e^{-x^2})

Solve  using d'Alembert's solution

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

$$u(x,0) = g(x) = e^{-x^2} \qquad
\boxed{
g(x-t) = e^{-(x-t)^2} \qquad
g(x+t) = e^{-(x+t)^2} }$$

$$u_t(x,0) = h(x) = \dfrac{d\left( e^{-x^2} \right)}{dx} \qquad
\boxed{
\int_{x-t}^{x+t} h(\xi) \, d\xi. = \int_{x-t}^{x+t} \dfrac{d\left( e^{-\xi^2} \right)}{d\xi} = \left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}\\
}$$

$$u(x,t) = \frac{1}{2}\left[g(x-t) + g(x+t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(\xi) \, d\xi.\\\\
u(x,t) = \frac{1}{2}\left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2}\left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}$$

$$u(x,t) = \frac{1}{2} \left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} - e^{-(x-t)^2} \right]\\\\
u(x,t) = \frac{1}{2} \left[ e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} } \right]\\\\
\boxed{ u(x,t) = e^{-(x+t)^2} }\\\\
u(x,0) = e^{-(x)^2}\\
u(x,1) = e^{-(x+1)^2}\\
u(x,10) = e^{-(x+10)^2}$$

Will wisssen wie ich in Taschenrechner diesen Formel eingebe

A=1/2*g*h

Suche A=

Geben ist: g=4cm h=7cm wie gebe ich in Taschenrechner ein

$$\small{\text{
$
\begin{array}{cccccccccccccccccc}
&&&&&\boxed{
32=}\\
&&&&&\boxed{
8~$Herz$~+~8~$Karo$~+~8~$Pik$~+~8~$Treff$}\\
&&&&& | &&&& \\
&& + &-& + & - & + &-& + \\
&& | && | && | && | \\
&& \boxed{
1.~$Herz$ =\frac{8}{32}} && \boxed{
1.~$Karo$=\frac{8}{32}} && \boxed{
1.~$Pik$ =\frac{8}{32}} && \boxed{
1.~$Treff$ = \frac{8}{32}}\\
&& | && | && | && | \\
&&\boxed{
2.~$Herz$ =\frac{7}{31}} && \boxed{
2.~$Karo$ =\frac{7}{31}} && \boxed{
2.~$Pik$=\frac{7}{31} } && \boxed{
2.~$Treff$ =\frac{7}{31} }
\end{array}
$}}$$

$$\\
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 Herzen: $ \\
\mathbf{
P(2 ~Herzen)= \frac{8}{32}\cdot \frac{7}{31} = \frac{56}{992}
}
\end{array}
$}}\\\\
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 Karos: $ \\
\mathbf{
P(2 ~Karos)= \frac{8}{32}\cdot \frac{7}{31} = \frac{56}{992}
}
\end{array}
$}}\\\\
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 Piks: $ \\
\mathbf{
P(2 ~Piks)= \frac{8}{32}\cdot \frac{7}{31} = \frac{56}{992}
}
\end{array}
$}}\\\\
\boxed{
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 Herzen, 2 Karos und 2 Piks: $ \\
\mathbf{
P(2 ~Herzen)+P(2 ~Karos)+P(2 ~Piks)= \frac{56}{992} +\frac{56}{992}+\frac{56}{992}=3\cdot \frac{56}{992}=0,16935483871
}
\end{array}
$}}
}\\\\
\boxed{
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 Treffs: $ \\
\mathbf{
P(2 ~Treffs)= \frac{8}{32}\cdot \frac{7}{31} = \frac{56}{992} =0,05645161290
}
\end{array}
$}}
}\\\\\\
\boxed{
\small{\text{$\begin{array}{lcl}$Die Wahrscheinlichkeit f\"ur 2 verschiedene Farben: $ \\
\mathbf{
P(2~verschiedene~ Farben)= 12\cdot\frac{8}{32}\cdot \frac{8}{31} = \frac{768}{992} = 0,77419354839
}
\end{array}
$}}
}\\\\$$

Gewinn und Verlust:

Gewinn Alex:

$$\small{\text{$
\begin{array}{lcl}
&0,16935483871 \cdot 0,50~ \mathrm{Euro} \\
+& 0,05645161290 \cdot 1,00 ~\rm{Euro} \\\\
=& 0.08467741935\cdot ~\rm{Euro} \\
+& 0,05645161290\cdot ~\rm{Euro}\\\\
=& \mathbf{0,14112903226~Euro }
\end{array}
$}}$$

Verlust Alex:

$$\small{\text{$
\begin{array}{lcl}
&0,77419354839 \cdot 0,20~ \mathrm{Euro} \\
=& \mathbf{-0,15483870968~Euro }
\end{array}
$}}$$

Alex macht insgesamt einen Verlust von

0,14112903226 € - 0,15483870968 € = - 0,01370967742 €  $$\small{\text{$\mathbf{
\approx -1,37 ~Cent}$}}$$

If three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate ?

$$\small{\text{$
\begin{array}{rclclclcl}
\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad
\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad
\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\\\
\vec{P_4}&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}
=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\
\end{array}
$}}$$

In the figure below, AB = AC = 50, AD = 52, and bc = 28. Determine cd.

I. Herons Formula: Area A (ABC):

$$\small{\text{$
\begin{array}{l|l}
A=\sqrt{s\cdot (s-a)\cdot(s-b)\cdot (s-c)} & \qquad \overline{AB} = a=50, ~ \overline{AC} = b=50,~ \overline{AD} =c=28\\\\
& \qquad s = \frac{a+b+c}{2}=\frac{50+50+28}{2}=\frac{128}{2}=64 \\
A=\sqrt{s\cdot (s-50)\cdot(s-50)\cdot (s-28)} \\
A= \sqrt{64\cdot (64-50)\cdot(64-50)\cdot (64-28)} \\
A= \sqrt{64\cdot 14 \cdot 14 \cdot 36} \\
A= \sqrt{8^2\cdot 14^2 \cdot 6^2} \\
A= 8\cdot 14 \cdot 6 \\
A=672
\end{array}
$}}$$$$$$

II: Height h of (ABC):

$$\small{\text{$
\begin{array}{rcl}
c\cdot h &=& 2\cdot A\\\\
h&=& \dfrac{2\cdot A}{c} \\\\
h&=& \dfrac{2\cdot 672}{28}\\\\
h &= &48
\end{array}
$}}$$

III. Pythagoras:

$$\small{\text{$
x = \overline{CD}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
h^2 + \left( \dfrac{c}{2}+x \right) ^2 &=& 52^2 \\\\
48^2 + \left( 14 +x \right) ^2 &=& 52^2 \\\\
\left( 14 +x \right) ^2 &=& 52^2 - 48^2\\\\
\left( 14 +x \right) ^2 &=& 20^2\\\\
14 +x &=& 20\\\\
x &=& 20-14\\\\
x &=& 6
\end{array}
$}}$$

CD = 6

Inverse Matrix Method

Example:

Step 1: Rewrite the system using matrix multiplication:

and writing the coefficient matrix as A, we have

.

Step 2: FInd the inverse of the coefficient matrix A. In this case the inverse is

Step 3: Multiply both sides of the equation (that you wrote in step #1) by the matrix A-1.

On the left you'll get

.

On the right, you get

and so the solution is

3x-2y=4

-6x+4y=7 solve with inversion method

$$\small{\text{
$
A = \begin{pmatrix} 3 & -2 \\ -6 &4 \end{pmatrix} \qquad
det ~ A = \begin{vmatrix} 3& -2 \\ -6& 4 \end{vmatrix}=3 \cdot 4 -(-6)\cdot (-2) = 12 - 12 = 0
$}}$$

det A = 0, no solution