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

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

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

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