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