heureka

Solve for each variable:

-4x-6y+5z=21

3x+4y-2z=-15

-7x-5y+3z=15

$$\small{\text{$
\begin{array}{rrrrcrl}
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 3x&+4y&-2z&=&-15 \quad & | \quad \cdot \frac{4}{3}\\
(3) &-7x&-5y&+3z&=&15 \quad & | \quad \cdot \frac{4}{-7}\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 3\cdot\frac{4}{3}x &+4\cdot \frac{4}{3} y&-2\cdot \frac{4}{3} z&=&-15\cdot \frac{4}{3} \quad &\\
(3) &-7\cdot \frac{4}{-7}x&-5\cdot \frac{4}{-7}y&+3\cdot \frac{4}{-7}z&=&15\cdot \frac{4}{-7} \quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 4x &+\frac{16}{3} y&-\frac{8}{3} z&=&-20 \quad &\\
(3) & 4x &+\frac{20}{7}y&-\frac{12}{7}z&=&-\frac{60}{7}\quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2)+(1) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3)+(1) & 0 &-\frac{22}{7}y&+\frac{23}{7}z&=&\frac{87}{7}\quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 &-\frac{22}{7}y&+\frac{23}{7}z&=&\frac{87}{7}\quad & | \quad \cdot \frac{-7}{22}\cdot \frac{2}{3}\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 &-\frac{22}{7}\cdot \frac{-7}{22}\cdot\frac{2}{3}y&+\frac{23}{7} \cdot \frac{-7}{22}\cdot \frac{2}{3}z&=&\frac{87}{7} \cdot \frac{-7}{22}\cdot\frac{2}{3}\quad & \\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 & +\frac{2}{3}y& -\frac{23}{22} \cdot \frac{2}{3}z&=&-\frac{87}{22}\cdot \frac{2}{3}\quad & \\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3)+(2) & 0 & 0 & \frac{108}{66}z&=&-\frac{108}{66}\quad & \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\frac{108}{66}z &=& -\frac{108}{66}\\\\
z &=& \dfrac{ -\frac{108}{66} } {\frac{108}{66}} \\\\
z &=& - \frac{108}{66 }\cdot \frac{66}{108} \\\\
\boxed{\, z = -1 \, }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
-\frac{2}{3} y+\frac{7}{3} z&=&1 \\\\
-\frac{2}{3} y+\frac{7}{3} \cdot(-1)&=&1 \\\\
-\frac{2}{3} y-\frac{7}{3} &=&1 \\\\
-\frac{2}{3} y &=&1+\frac{7}{3} \\\\
-\frac{2}{3} y &=&\frac{10}{3} \\\\
y &=& \frac{10}{3}\cdot ( -\frac{3}{2} ) \\\\
\boxed{\, y = -5 \, }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
-4x-6y+5z&=&21 \\\\
-4x-6\cdot(-5)+5\cdot(-1)&=&21 \\\\
-4x+30-5&=&21 \\\\
-4x+25&=&21 \\\\
-4x &=&-4 \\\\
4x &=&4 \\\\
\boxed{\, x= 1 \, }
\end{array}
$}}$$

ln(e^0.047t) ?

$$\ln{(e^{0.047\cdot t})}\\
= 0.047\cdot t \cdot \ln{(e)} \quad | \quad \boxed{\;\ln{(e)} =1\;}\\
= 0.047\cdot t$$

What is the sum of the geometric sequence 1, 3, 9, ... if there are 14 terms?

$$$$

$$\small{\text{
geometric sequence: $a_1 = 1 \quad r = 3$
}}\\
\small{\text{
$
\begin{array}{lcll}
\hline
\\
s_{14} &=& \textcolor[rgb]{150,0,0}{1} + & 3^1 + 3^2 + 3^3
+ 3^4 + 3^5 + 3^6
+ 3^7 + 3^8 + 3^9
+ 3^{10} + 3^{11} + 3^{12} + 3^{13} \\
3\cdot s_{14} &=& & 3^1 + 3^2 + 3^3
+ 3^4 + 3^5 + 3^6
+ 3^7 + 3^8 + 3^9
+ 3^{10} + 3^{11} + 3^{12} + 3^{13} + \textcolor[rgb]{150,0,0}{3^{14}}\\
\hline
\\
s_{14}-3\cdot s_{14} &=& \textcolor[rgb]{150,0,0}{1-}&\textcolor[rgb]{150,0,0}{3^{14}}\\
s_{14}\cdot(1-3) &=& 1- &3^{14}\\
-2\cdot s_{14} &=& 1- &3^{14}\\
\end{array}
$}}$$

$$\small{\text{
$
\begin{array}{rcl}
s_{14} & = & \dfrac { 1- 3^{14} } {-2}\\\\
s_{14} & = & \dfrac { 3^{14}-1 } {2} = 2\,391\,484 \\
\end{array}
$
}}$$

(Use c=f*lamda) calculate the frequency of green light with 545 nm (1nm = 10 to the -9th power)

$$\boxed{\; c = f\cdot \lambda \qquad \text{ or }\qquad \lambda=\dfrac{c}{f} \qquad \text{ or }\qquad f=\dfrac{c}{\lambda}
\qquad
\begin{array}{rcl}
c &=& \small{\text{ speed of light in vacuum }} \\
\lambda &=& \small{\text{ wavelength }} \\
f &=& \small{\text{ wave's frequency }}
\end{Array}
\; }$$

$$\small{\text{
$
\begin{array}{rclcc}
f &=& \dfrac{c}{\lambda} \quad & \quad c = 299\,792\,458 ~ \mathrm{\dfrac{m}{s}} \quad & \quad \lambda = 545 ~ \mathrm{nm} = 545\cdot 10^{-9} ~ \mathrm{m} \\\\
f&=& \dfrac{ 299\,792\,458 ~ \mathrm{\dfrac{m}{s}} }{545\cdot 10^{-9} ~ \mathrm{m} } \\\\
f&=& \dfrac{ 2.99\,792\,458\cdot 10^{8} ~ \mathrm{\dfrac{m}{s}} }{545\cdot 10^{-9} ~ \mathrm{m} } \\\\
f&=& \dfrac{ 2.99\,792\,458 }{545}\cdot 10^{8}\cdot 10^{9} ~ \mathrm{\dfrac{1}{s}} \\\\
f &=& 0.00550077905\cdot 10^{17} ~ \mathrm{\dfrac{1}{s}} \\\\
f &=& 0.550077905\cdot 10^{-2}\cdot 10^{17} ~ \mathrm{\dfrac{1}{s}} \\\\
f &=& 0.550077905\cdot 10^{15} ~ \mathrm{\dfrac{1}{s}} \\\\
f &=& 0.550077905\cdot 10^{15} ~ \mathrm{ Hz } \\\\
f &=& 0.550077905 ~ \mathrm{ PHz } \quad & \quad \mathrm{ PHz } = \mathrm{ Petahertz }\\\\
f &=& 550.077905 ~ \mathrm{ THz } \quad & \quad \mathrm{ THz } = \mathrm{ Terahertz }
\end{array}
$}}$$

Wieso ist 1/√3 = √3/3 ? Wie kommt man drauf ?

 $$\small{\text{
$
\begin{array}{rcl}
\dfrac{1}{\sqrt{3}} \\
&=& \sqrt { \dfrac13 } \\\\
&=& \sqrt { \dfrac13 \cdot \dfrac33 }\\\\
&=& \sqrt { \dfrac{3}{3\cdot 3} }\\\\
&=& \sqrt { \dfrac{3}{3^2} }\\\\
&=& \dfrac{ \sqrt { 3 } }
{ \sqrt { 3^2 } } \\\\
&=& \dfrac{ \sqrt { 3 } } { 3 }
\end{array}
$}}$$

Wie lautet der Lösungsweg für die Gleichung 3^(2*x)+3^x=90 ?

$$\small{\text{
$
\begin{array}{rcl}
3^{(2\cdot x)}+3^x &=& 90 \\
3^{(x\cdot 2)}+3^x &=& 90 \\
(3^x)^2 +3^x &=& 90 \\
\end{array}
$}}$$

Wir setzen $$u = 3^x$$ und erhalten $$u^2 + u = 90$$

Jetzt lösen wir nach u auf:

$$\small{\text{
$
\begin{array}{rcl}
u^2 + u &=& 90 \\
u^2 + u - 90 &=& 0 \\
u_{1,2} &=& -\frac{1}{2}\pm\sqrt{\frac{1}{4}+90} \\
u_{1,2} &=& -\frac{1}{2}\pm\sqrt{\frac{1+4\cdot 90}{4}} \\
u_{1,2} &=& -\frac{1}{2}\pm\sqrt{\frac{361}{4}} \\
u_{1,2} &=& -\frac{1}{2}\pm \frac{19}{2} \\
\end{array}
$}}$$

$$\small{\text{
$
\begin{array}{rcl|rcl}
u_1 &=& -\frac{1}{2} +\frac{19}{2} \quad &
\quad u_2 &=& -\frac{1}{2} - \frac{19}{2}\\
&&&\\
u_1 &=& \frac{18}{2} \quad & \quad u_2 &=& -\frac{20}{2}\\
&&&\\
u_1 &=& 9 \quad & \quad u_2 &=& -10\\
&&&\\
u_1 &=& 3^{x_1} \quad & \quad u_2 &=& 3^{x_2}\\
&&&\\
x_1\cdot \ln{(3)} &=& \ln{(u_1)} \quad & \quad x_2\cdot \ln{(3)} &=& \ln{(u_2)} \\
&&&\\
x_1&=& \dfrac{\ln{(u_1)} }{ \ln{(3)} } \quad & \quad x_2&=& \dfrac{\ln{(u_2)} }{ \ln{(3)} } \\
&&&\\
x_1&=& \dfrac{\ln{(9)} }{ \ln{(3)} } \quad & \quad x_2&=& \dfrac{\ln{(-10)} }{ \ln{(3)} } $ keine L\"osung$ \\
&&&\\
x_1&=& \dfrac{\ln{(3^2)} }{ \ln{(3)} } \quad & \\
&&&\\
x_1&=& \dfrac{2\ln{(3)} }{ \ln{(3)} } \quad & \\
&&&\\
x_1&=& 2 &
\end{array}
$}}$$

n über k

$$\small{\text{
$
\begin{array}{rcl}
\left(\begin{array}{c}n\\
k
\end{array} \right)
&=& \dfrac{n!}{k!\cdot(n-k)!}\\\\
&=&\dfrac{n\cdot(n-1)\cdot(n-2) \cdots (n-(k-1)) }{k!} \\\\
&=& \left( \dfrac {n}{1} \right) \cdot
\left( \dfrac {n-1}{2}\right) \cdot
\left( \dfrac {n-2}{3}\right) \cdots
\left( \dfrac {(n-(k-1)) }{k} \right)
\end{array}
$}}$$

Eine Ware um 8% reduziert danach nochmal um 5% erhöht jetzt beträgt der Preis 483,00€ wie hoch war der Preis ursprünglich ?

$$\small{
\begin{array}{rcl}
\text{Preis}_{\text{(urspr\"unglich)}}
\cdot
\underbrace{(100\%-8\%)}_{\text{reduziert}}
\cdot
\underbrace{(100\%+5\%)}_{\text{erh\"oht}}
&=& 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}
\cdot
\underbrace{(92\%)}_{\text{reduziert}}
\cdot
\underbrace{(105\%)}_{\text{erh\"oht}}
&=& 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}
\cdot
(92\%)\cdot (105\%) &=& 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}
\cdot
\left (\dfrac{92}{100} \right)
\cdot
\left (\dfrac{105}{100} \right) &=& 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}&=&
\left (\dfrac{100}{92} \right)
\cdot
\left (\dfrac{100}{105} \right) \cdot 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}&=&
1.08695652174\cdot
0.95238095238 \cdot 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}&=&
1.03519668737\cdot 483,00 \\\\
\text{Preis}_{\text{(urspr\"unglich)}}&=&
500,00
\end{array}
}$$ 

Der ursprüngliche Preis war 500,00 €

13a)

Wenn der Kosinus-Satz bekannt ist, dann kann b folgendermaßen berechnet werden:

Wir verschieben die Seite b parallel von C nach D und erhalten ein Dreieck (D=C).

$$\boxed{b^2 = d^2 + (a-c)^2 - 2\cdot d \cdot(a-c) \cdot \cos{(70 \ensurement{^{\circ}})}}}\\\\
\small{\text{
$
\begin{array}{rcl}
b^2 &=& (5,5)^2 + (10-4)^2 - 2\cdot 5,5 \cdot (10-4)
\cdot \cos{(70 \ensurement{^{\circ}}) }\\
b^2 &=& 30,25 + (6)^2 - 2\cdot 5,5\cdot (6) \cdot \cos{(70 \ensurement{^{\circ}}) } \\
b^2 &=& 30,25 + 36- 11\cdot (6) \cdot \cos{(70 \ensurement{^{\circ}}) } \\
b^2 &=& 66,25 - 66\cdot \cos{(70 \ensurement{^{\circ}}) } \\
b^2 &=& 66,25 - 66\cdot 0.34202014333 \\
b^2 &=& 66,25 -22.5733294595\\
b^2 &=& 43.6766705405\\
b &=& 6.60883276687\\
b &\approx& 6.6 \; \rm{cm}
\end{array}
$
}}$$

U = a + b + c + d = 10,0 cm + 6,6 cm + 4,0 cm + 5,5 cm = 26,1 cm

$$\boxed{A = \left(
\dfrac{a+c}{2}\right)
\cdot h
\qquad h = d \cdot \sin{(70 \ensurement{^{\circ}})}} }\\\\
\small{\text{
$
\begin{array}{rcl}
A &=& \left(
\dfrac{a+c}{2}\right)
\cdot d \cdot \sin{(70 \ensurement{^{\circ}})} \\
A &=& \left(
\dfrac{10+4}{2}\right)
\cdot 5,5 \cdot \sin{(70 \ensurement{^{\circ}})} \\
A &=& 7\cdot 5,5 \cdot \sin{(70 \ensurement{^{\circ}})} \\
A &=& 38,5 \cdot \sin{(70 \ensurement{^{\circ}})} \\
A &=& 38,5 \cdot 0.93969262079\\
A &=& 36.1781659003\\
A &\approx& 36,2\,\rm{cm}^2
\end{array}
$
}}$$

Find the value of C for the pic below

$$0.828= \dfrac{
\left(\dfrac{2-c}{2}\right)^{\dfrac{1}{2}}
\cdot
\left( \dfrac{4-c}{2}\right)^{\dfrac{1}{2}}
}
{c}
\cdot
\underbrace{
\left( \dfrac{2}{3.1}\right)^{
\underbrace{ \left(\dfrac{1}{2}+\dfrac{1}{2}-1 \right)
}_{=0} }
}_{=1}
\qquad \boxed{a^0 = 1}$$

$$\small{\text{
$
\begin{array}{rcl}
0.828 &=& \dfrac{
\left(\dfrac{2-c}{2}\right)^{\dfrac{1}{2}}
\cdot \left( \dfrac{4-c}{2}\right)^{\dfrac{1}{2}}
}
{c}\cdot 1\\ \\
0.828 &=& \dfrac{
\left(\dfrac{2-c}{2}\right)^{\dfrac{1}{2}}
\cdot
\left( \dfrac{4-c}{2}\right)^{\dfrac{1}{2}}
}
{c}\\\\
0.828\cdot c &=& \left(\dfrac{2-c}{2}\right)^{\dfrac{1}{2}}
\cdot
\left( \dfrac{4-c}{2}\right)^{\dfrac{1}{2}}\\\\
0.828\cdot c &=& \sqrt{ \dfrac{2-c}{2} }
\cdot \sqrt{ \dfrac{4-c}{2}\right) } \quad | \quad ()^2\\\\
0.828^2\cdot c^2 &=&
\left( \dfrac{2-c}{2} \right)
\cdot
\left( \dfrac{4-c}{2}\right) \\\\
4\cdot 0.828^2\cdot c^2 &=&
\left( 2-c \right)
\cdot
\left( 4-c \right) \\\\
2.742336 \cdot c^2 &=& 8-2c-4c+c^2 \\\\
2.742336 \cdot c^2 &=& 8-6c+c^2 \\\\
2.742336 \cdot c^2 - c^2 + 6c - 8 &=& 0 \\\\
c^2 \cdot (2.742336 -1 ) + 6c - 8 &=& 0 \\\\
1.742336 \cdot c^2 + 6c - 8 &=& 0 \\\\
c_{1,2} &=& \dfrac{-6 \pm \sqrt{6^2-4\cdot 1.742336 \cdot (-8) } }{ 2\cdot 1.742336 } \\\\
c_{1,2} &=& \dfrac{-6 \pm \sqrt{36+55.7547520000} }{ 2\cdot 1.742336 } \\\\
c_{1,2} &=& \dfrac{-6 \pm \sqrt{91.7547520000} }{ 2\cdot 1.742336 } \\\\
c_{1,2} &=& \dfrac{-6 \pm 9.57887007950}{ 2\cdot 1.742336 } \\\\
c_{1,2} &=& \dfrac{-6 \pm 9.57887007950}{ 3.484672 } \\\\
\end{array}
$}}$$

$$\small{\text{
\begin{array}{rcl|rcl}
$
c_1 &=& \dfrac{-6 + 9.57887007950}{ 3.484672 } \quad & \quad
c_2 &=& \dfrac{-6 - 9.57887007950}{ 3.484672 } \\\\
c_1 &=& 1.02703212225 \quad & \quad
c_2 &=& -4.47068478167$ no solution $
$
\end{array}
}}\\\\
c = 1.02703212225$$