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

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

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

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