heureka

What is the value of x? A^3a^x = a^12

$$\small{\text{$ \begin{array}{rcl} A^3a^x &=& a^{12} \\ a^x &=& \dfrac{a^{12}}{A^3} \qquad | \qquad b=\dfrac{a^{12}}{A^3} \\\\ a^x &=& b \qquad | \qquad \ln{()} \\ x\ln{(a)} &=& \ln{(b)}\\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ \ln{(b)} }{\ln{(a)} } }\quad $ and $ \mathbf{ \quad b=\dfrac{a^{12}}{A^3} }\\\\ \end{array} $}}}$$

see the solution in german forum:

Solve the following system of equations.

8x + 5y = -84

3x + y = -21

A. x = -12, y = -3

B. x = 5, y = -24.8

C. x = 12, y = -57

D. x = -3, y = -12

The answer is D. x = -3 and y = -12

$$\small{ \begin{array}{lrcl} (1) & 8x + 5y &=& -84 \\ \hline (2) & 3x + y &=& -21 \qquad | \qquad \cdot 5\\ (2) & 15x + 5y &=& -105 \\ \hline (2)-(1)& 15x-8x +5y-5y &=& -105 -(-84) \\ & 15x-8x &=& -105 + 84 \\ & 7x &=& -21 \qquad | \qquad :7\\ & \mathbf{x} & \mathbf{=} & \mathbf{-3} \\ \hline &3x+y &=& -21\\ &3\cdot(-3) + y &=& -21\\ &-9 + y &=& -21\\ & y &=& -21 + 9 \\ & \mathbf{y} &\mathbf{=} & \mathbf{-12} \end{array} }$$

How to calculate ( 1/a - 2/b ) / ( 2/a - 1/b )

$$\dfrac{ ( \frac{1}{a} - \frac{2}{b} ) } { ( \frac{2}{a} - \frac{1}{b} ) }\cdot \left( \dfrac{ab}{ab} \right) \\\\\\ = \dfrac{ ( \frac{1}{a} - \frac{2}{b} ) \cdot ab} { ( \frac{2}{a} - \frac{1}{b} ) \cdot ab }\\\\\\ = \dfrac{ ( \frac{1}{a}\cdot ab - \frac{2}{b}\cdot ab ) } { ( \frac{2}{a}\cdot ab - \frac{1}{b}\cdot ab ) }\\\\\\ = \dfrac{ ( b- 2a) } { (2b - a) }\\\\\\$$

Thank you Melody, i don't need to divide by 16, my mistake!

What is the slope of a line that passes through (-2, 2) and (-4, 5) ?

$$\\x_1=-2,~y_1=2 \\ x_2 = -4,~y_2=5\\\\ m=\dfrac{y_2-y_1}{x_2-x_1}\\\\ m=\dfrac{5-2}{(-4)-(-2)}\\\\ m=\dfrac{5-2}{(-4)+2}\\\\ m=\dfrac{5-2}{2-4}\\\\ m=\dfrac{3}{-2}\\\\ m=-\dfrac{3}{2}\\\\ \mathbf{m=-1.5}$$

In general:

The choice of the parametres "a" and "b" determine the angel - orientation:

$$\\\small{\mathrm{atan2}{(\Delta y, \Delta x )} \quad \text{ angle counterclockwise direction start from 'x'-axis }}\\ \small{\mathrm{atan2}{(-\Delta y, \Delta x )} \quad \text{ angle clockwise direction start from 'x'-axis }}\\ \small{\mathrm{atan2}{(\Delta x, \Delta y )} \quad \text{ angle clockwise direction start from 'y'-axis }}\\ \small{\mathrm{atan2}{(-\Delta x, \Delta y )} \quad \text{ angle counterclockwise direction start from 'y'-axis }}\\$$

$$\\\small{\mathrm{atan2}{(\Delta y, -\Delta x )} \quad \text{ angle clockwise direction start from '-x'-axis }}\\ \small{\mathrm{atan2}{(-\Delta y, -\Delta x )} \quad \text{ angle counterclockwise direction start from '-x'-axis }}\\ \small{\mathrm{atan2}{(-\Delta x, -\Delta y )} \quad \text{ angle clockwise direction start from '-y'-axis }}\\ \small{\mathrm{atan2}{(\Delta x, -\Delta y )} \quad \text{ angle counterclockwise direction start from '-y'-axis }}\\$$

A coast guard cutter detects an unidentified ship at 20.0km in the direction 15.0 degrees east of north. The ship is traveling at 26.0km/h on a course at 40.0 degrees east of north. The Coast Guard wishes to send a speed boat to intercept and inveatigate the vessel. If the speedboat travels at 50.0km/h , in what direction should it head?

We set t = time

$$\small{ \begin{array}{lcl} \text{Ship position: } \vec{p}= 20\cdot\binom { \sin{(15\ensurement{^{\circ}})} } { \cos{(15\ensurement{^{\circ}})} } \\\\ \text{Ship traveling: } \vec{s}= 26\cdot t \cdot\binom { \sin{(40\ensurement{^{\circ}})} } { \cos{(40\ensurement{^{\circ}})} } \\\\ \text{Guard cutter traveling: } \vec{c}= 50\cdot t\cdot\binom { \sin{(x)} } { \cos{(x)} } \\\\ \vec{p} + \vec{s} & = &\vec{c} \\\\ 20\cdot\binom { \sin{(15\ensurement{^{\circ}})} } { \cos{(15\ensurement{^{\circ}})} } + 26\cdot t \cdot\binom { \sin{(40\ensurement{^{\circ}})} } { \cos{(40\ensurement{^{\circ}})} } &=& 50\cdot t\cdot\binom { \sin{(x) } } { \cos{(x)} } \\\\ \hline 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t &=&50\cdot t\cdot \sin{(x)} \\ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t &=&50\cdot t\cdot \cos{(x})} \\ \hline \text{Square both sides }\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=&50^2\cdot t^2 \sin^2{(x)} \\ \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=&50^2\cdot t^2 \cos^2{(x})} \\ \hline \text{Add }\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 \\ + \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=& 50^2\cdot t^2 [ \sin^2{(x)} + \cos^2{(x})} ] \\ \hline \sin^2{(x)} + \cos^2{(x})} = 1\\ \left[ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t \right]^2 \\ + \left[ 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t \right]^2 &=& 50^2\cdot t^2\\ \hline 20^2\cdot \sin^2{(15\ensurement{^{\circ}})} + 2\cdot 20\cdot 26 \cdot \sin{(15\ensurement{^{\circ}})} \cdot \sin{(40\ensurement{^{\circ}})} \cdot t + 26^2\cdot t^2 \sin^2{(40\ensurement{^{\circ}})} \\ +20^2\cdot \cos^2{(15\ensurement{^{\circ}})} + 2\cdot 20\cdot 26 \cdot \cos{(15\ensurement{^{\circ}})} \cdot \cos{(40\ensurement{^{\circ}})} \cdot t +26^2 \cdot t^2 \cos^2{(40\ensurement{^{\circ}})} &=& 50^2\cdot t^2\\ \hline \sin^2{ (15\ensurement{^{\circ}}) } + \cos^2{ (15\ensurement{^{\circ}}) } = 1\\ \sin^2{ (40\ensurement{^{\circ}}) } + \cos^2{ (40\ensurement{^{\circ}}) } = 1\\ 20^2 + 2\cdot 20\cdot 26 \cdot [ \underbrace{ \cos{ (15\ensurement{^{\circ}}) } \cdot \cos{ (40\ensurement{^{\circ}}) } + \sin{ (15\ensurement{^{\circ}}) } \cdot \sin{ (40\ensurement{^{\circ}}) } }_{=\cos{(40\ensurement{^{\circ}}-15\ensurement{^{\circ}})}=\cos{25\ensurement{^{\circ}}}} ] \cdot t +26^2 t^2 &=& 50^2\cdot t^2\\ \hline \end{array} }$$

$$\small{ \begin{array}{lcl} 50^2\cdot t^2 - 26^2 \cdot t^2 -2\cdot 20\cdot 26 \cdot \cos{ (25\ensurement{^{\circ}}) }\cdot t -20^2 &=& 0\\\\ 1824\cdot t^2 - 1040 \cdot \cos{ (25\ensurement{^{\circ}}) }\cdot t - 400 &=& 0 \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{ 1040^2\cdot \cos^2{ (25\ensurement{^{\circ}}) } - 4\cdot 1824 \cdot (-400) } } { 2\cdot 1824 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{ 1040^2\cdot \cos^2{ (25\ensurement{^{\circ}}) } + 2918400 } } { 2\cdot 1824 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm \sqrt{3806819.53932 } } { 3648 } \\\\ t_{1,2} &=& \frac { 1040\cdot \cos{ (25\ensurement{^{\circ}}) } \pm 1951.10725982 } { 3648 } \\\\ t_{1,2} &=& \frac{ 942.560098518 \pm 1951.10725982 } { 3648 } \\\\ t &=& 0.79322021884~ \text{hours}\\ t &=& 47.5932131305~ \text{minutes}\\ \hline \end{array} }$$

$$\small{ \begin{array}{llcl} \text{Azimuth angle: } & \tan{(\alpha_{\text{azimuth}})} &=& \frac{ \sin{ (x) } }{ \cos{ (x) } } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& \dfrac{ 20\cdot \sin{(15\ensurement{^{\circ}})} + 26\cdot \sin{(40\ensurement{^{\circ}})}\cdot t } { 20\cdot \cos{(15\ensurement{^{\circ}})} + 26\cdot \cos{(40\ensurement{^{\circ}})}\cdot t } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& \dfrac{ 18.4330562411 } { 35.1172069869 } \\\\ & \tan{(\alpha_{\text{azimuth}})} &=& 0.52490097655 \\\\ &\alpha_{\text{azimuth}} &=& \arctan{ ( 0.52490097655 ) } \\ &\mathbf{\alpha_{\text{azimuth}} } & \mathbf{=} & \mathbf{27.6950249044\ensurement{^{\circ}} }\\ \end{array} }$$

The speed boat needs to set a course for

    $$N\;27^0~41'~42"~E$$

 Thank you Melody, i don't need to divide by 16, my mistake!

was kostet eine zigarete wen in pakung 30 zigareten und pakung kostet 7 €

$$\left( \dfrac{7~\rm{Euro} }{30~\rm{Zigaretten}} \right) \cdot 1 ~\rm{Zigarette} = 0.23\overline{3}~\rm{Euro}$$

Eine Zigarette kostet rund 23 Cent

$$\small{\text{$ 12345678^4-(12345679^2+1)(12345677^2+1) $}}$$

$$\small{\text{ We set $x = 12345678 $, so we have $x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] $}}$$

$$\small{\text{$ \begin{array}{lcl} x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] \\ &= & x^4 -[ x^2+2x+1+1 ]\cdot [ x^2-2x+1+1 ] \\ &= & x^4 -[ x^2+2x+2 ]\cdot [ x^2-2x+2 ] \\ &= & x^4 -[ (x^2+2)+2x ]\cdot [ (x^2+2)-2x ] \\ &= & x^4 -[ (x^2+2)^2 -(2x)^2 ] \\ &= & x^4 -[ x^4 +4x^2 + 4 -4x^2 ] \\ &= & x^4 -[ x^4 + 4 ] \\ &= & x^4 -x^4 - 4 \\ \mathbf{ x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ]}& \mathbf{= } & \mathbf{ - 4 } \\ \end{array} $}}$$