heureka

$$\vec{v} = \begin{pmatrix}-5\\7 \end{pmatrix} \| v\| = \sqrt{(-5)^2+7^2}= \sqrt{25+49}= \sqrt{74} \approx 8.60$$

$$\\85000000\;mm^3=85000000\;mm*mm*mm\\ \boxed{1dm = 100mm}\\ 85000000\;mm*mm*mm*\dfrac{1\;dm}{100\;mm}*\dfrac{1\;dm}{100\;mm}*\dfrac{1\;dm}{100\;mm}\\$$

$$\\=\dfrac{85000000\;*dm*dm*dm}{100*100*100}\\\\ =\dfrac{85*100*100*100\;*dm*dm*dm}{100*100*100}\\\\ =85\;*dm*dm*dm\\\\ =85\;dm^3$$

(1)

      4 * cars + 2 * bicycle = 110 | : 2

      2 * cars +       bicycle = 55

(2)  

            cars +       bicycle = 40

(1) - (2)

       2 * cars - cars + bicycle - bicycle = 55 - 40

       cars + 0 = 15

       cars = 15

       bicycle = 40 - cars = 40 - 15 = 25

cars = 15 and bicycle = 25

$$\\\boxed{\vec{v}*\vec{u} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{v_1}\\ \textcolor[rgb]{0,0,1}{v_2} \end{pmatrix} * \begin{pmatrix} \textcolor[rgb]{1,0,0}{u_1}\\ \textcolor[rgb]{0,0,1}{u_2} \end{pmatrix} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{v_1}*\textcolor[rgb]{1,0,0}{u_1} + \textcolor[rgb]{0,0,1}{v_2}*\textcolor[rgb]{0,0,1}{u_2} \end{pmatrix}}\\ \boxed{\vec{v}*\vec{u} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{3}\\ \textcolor[rgb]{0,0,1}{8} \end{pmatrix} * \begin{pmatrix} \textcolor[rgb]{1,0,0}{2}\\ \textcolor[rgb]{0,0,1}{-6} \end{pmatrix} = \begin{pmatrix} \textcolor[rgb]{1,0,0}{3}*\textcolor[rgb]{1,0,0}{2} + \textcolor[rgb]{0,0,1}{8}*\textcolor[rgb]{0,0,1}{(-6)} \end{pmatrix}=\textcolor[rgb]{1,0,0}{6}\textcolor[rgb]{0,0,1}{-48}=-42}$$

This is wrong: $$\sqrt{(3*2)^2+(8*(-6))^2} = 48.374 \;?!$$

( 11.000000, 0.000000, 0.000000, 0.000000, 0.000000,

0.000000, 11.000000, 1.000000, 1.000000, 1.000000,

0.000000, 1.000000, 11.000000, 1.000000, 1.000000,

0.000000, 1.000000, 1.000000, 11.000000, 1.000000,

0.000000, 1.000000, 1.000000, 1.000000, 11.000000 )

Matrix D: eigenvalue
( 10.000000, 0.000000, 0.000000, 0.000000, 0.000000,

0.000000, 10.000000, 0.000000, 0.000000, 0.000000,

0.000000, 0.000000, 10.000000, 0.000000, 0.000000,

0.000000, 0.000000, 0.000000, 11.000000, 0.000000,

0.000000, 0.000000, 0.000000, 0.000000, 14.000000 )

Matrix V: eigenvector
( 0.000000, 0.000000, 0.000000, 1.000000, 0.000000,

0.007685, -0.707107, 0.499941, 0.000000, 0.500000,

0.007685, 0.707107, 0.499941, 0.000000, 0.500000,

0.699338, 0.000000, -0.510809, 0.000000, 0.500000,

-0.714708, 0.000000, -0.489073, 0.000000, 0.500000 )

Vector D: eigenvalue - real
( 10.000000,

10.000000,

10.000000,

11.000000,

14.000000 )

Vector e: eigenvalue - imag
( 0.000000,

0.000000,

0.000000,

0.000000,

0.000000 )

Matix: eigenvalue - real,imag
( 10.000000, 0.000000,

10.000000, 0.000000,

10.000000, 0.000000,

11.000000, 0.000000,

14.000000, 0.000000 )

$$\\x=\dfrac{2m}{3+2n}\\\\\\ x=\dfrac{2m}{3*\dfrac{2}{2}+2n}\\\\\\ x=\dfrac{\not{2}m}{\not{2}*\left( \dfrac{3}{2}+n\right)}\\\\\\ \boxed{x=\dfrac{m}{ 1.5+n} }$$

Hi Melody,

your answer is not wrong!

But r ist also a position in the Pascal triangle too.

$$\\\mbox{r = 0: } (\frac{3}{x^2})^{10}=\dfrac{59049}{x^{20}}\\ \mbox{r = 1: } \dfrac{ -787320*x^{5} }{x^{20}}\\ \mbox{r = 2: } \dfrac{ 4723920*x^{10} }{x^{20}}\\ \mbox{r = 3: } \dfrac{ -16796160*x^{15} }{x^{20}}\\ \textcolor[rgb]{1,0,0}{\mbox{r = 4: } \dfrac{ 39191040*x^{20} }{x^{20}} = 39191040} \\ \mbox{r = 5: } \dfrac{ -62705664*x^{25} }{x^{20}}\\ \mbox{r = 6: } \dfrac{ 69672960*x^{30} }{x^{20}}\\ \mbox{r = 7: } \dfrac{ -53084160*x^{35} }{x^{20}}\\ \mbox{r = 8: } \dfrac{ 26542080*x^{40} }{x^{20}}\\ \mbox{r = 9: } \dfrac{ -7864320*x^{45} }{x^{20}}\\ \mbox{r = 10: } (-4x^3)^{10}=\dfrac{ 1048576*x^{50} }{x^{20}}$$

Many Greetings

Heureka

$$\mbox{simplify } (12c+9b)-(6c-7b)$$

(12c+9b)-(6c)-(-7b)

(12c+9b)-(6c)+(7b)

(12c)+(9b)-(6c)+(7b)

(12c)-(6c)+(9b)+(7b)

(12c-6c)+(9b+7b)  

(12-6)c+(9+7)b

(6c)+(16b)

$$\boxed{6c+16b}$$

The derivate of   $$y=\dfrac{t}{\sqrt{(ct^2-1)}} = t*\left( ct^2-1\right)^{-\frac{1}{2}}$$

Product Rule: $$y^\prime=t * \left( \frac{d(ct^2-1)^{-\frac{1}{2}}}{dt} \right)+\frac{d(t)}{dt}*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=t *(-\frac{1}{\not{2}})(ct^2-1)^{-\frac{1}{2}-1}}*\not{2}ct + 1*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=-ct^2(ct^2-1)^{-\frac{3}{2}}} + (ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}} -\dfrac{ct^2}{ \left( \sqrt{(ct^2-1)} \right)^3 }$$

$$\boxed{y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}} \left( 1-\dfrac{ct^2}{ct^2-1} \right)}$$

$$\\y=\dfrac{1}{4}*\cos^{-1}(2x)\\ \mbox{Now change y and x }\quad \textcolor[rgb]{1,0,0}{x}=\dfrac{1}{4}*\cos^{-1}(2\textcolor[rgb]{1,0,0}{y})\\ 4x=\cos^{-1}(2y)\quad | \quad \cos\\$$

$$\\\cos{(4x)}= 2y\\ 2y = \cos{(4x)} \quad | \quad :2\\ \boxed{y=\dfrac{1}{2}*\cos{(4x)}} \quad inverse$$