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ln(e*sqrt5(e)  ?

$$\small{\text{ $ \ln\left( e*\sqrt[5]{e} \right) \qquad \boxed{\ln(a*b) = \ln(a) + \ln(b) } $ }} $\\$ \small{\text{ $=\ln(e) + \ln(\sqrt[5]{e}) \qquad \boxed{\ln(e) = 1} $ }} $\\$ \small{\text{ $=1 + \ln(\sqrt[5]{e}) \qquad \boxed{\sqrt[5]{e} = e^{\frac{1}{5}}} $ }} $\\$ \small{\text{ $=1 + \ln( e^{\frac{1}{5}} ) \qquad \boxed{ \ln( e^{\frac{1}{5}} ) = \frac{1}{5} * \ln(e) } $ }} $\\$ \small{\text{ $=1 + \frac{1}{5} * \ln(e) \qquad \boxed{\ln(e) = 1} $ }} $\\$ \small{\text{ $=1 + \frac{1}{5} $ }} $\\$ \small{\text{ $=\frac{6}{5} $ }}$$

the sum of two consecutive numbers is 57. what are the numbers

a + ( a+1 ) = 57

2a + 1 = 57

2a = 57 - 1

2a = 56

a = 56 / 2

a = 28

28 + 29 = 57

Find x for the equation: x+1/x = -1 (x plus one divided by x equals negative one)

$$\small{\text{ set $\frac{1}{x} = -1-x$ and see there is no cut between function $\frac{1}{x}$ and line $-1-x$ }}$$

Suppose R id (3, 5) and S is (8, -3).  Find each point on the line through R and S that is three times as far from R as it is from S

$$\vec{r}=\left(\begin{array}{r} 3 & 5 \end{array}\right) \quad \vec{s}=\left(\begin{array}{r} 8 & -3 \end{array}\right) \quad \lambda=\frac{3}{4}\\ \boxed{\vec{x}=\vec{r}+\lambda\left( \vec{s}-\vec{r}\right) \small{ \text{ line equation in vector-form}}}\\ \vec{x} =\left(\begin{array}{r} 3 & 5 \end{array}\right) +\frac{3}{4}\left( \left(\begin{array}{r} 8 & -3 \end{array}\right)- \left(\begin{array}{r} 3 & 5 \end{array}\right) \right)\\ \vec{x} =\left(\begin{array}{r} 3 & 5 \end{array}\right) +\frac{3}{4} \left(\begin{array}{r} 5 & -8 \end{array}\right) \\ \vec{x} =\left(\begin{array}{r} 3 +\frac{3}{4}* 5 & 5 +\frac{3}{4}*( -8 )\end{array}\right) \\ \vec{x} =\left(\begin{array}{r} 6.75 & -1\end{array}\right) \\$$

Offsetwinkel = arccos Breite/Höhe

$$\small{ \text{ \textcolor[rgb]{1,0,0}{Offsetwinkel$\ensurement{^{\circ}} = \arccos\left(\dfrac{Breite}{ H\ddot{o}he } \right) $}} $\\$ \small{\text{ auch Offsetwinkel $\ensurement{^{\circ}} = acos\left(\dfrac{Breite}{ H\ddot{o}he } \right) $ }} $\\$ \small{\text{ auch Offsetwinkel $\ensurement{^{\circ}} = \cos^{-1}\left(\dfrac{Breite}{ H\ddot{o}he } \right) $ }}$$

$$\small{\text{ Der $arccos(x)$ auch als $acos(x)$ bezeichnet und auch als $\cos^{-1}(x)$ bezeichnet } } $\\$ \small{\text{ ist eine Winkelfunktion und zwar ist er die Umkehrung des $\cos(\alpha)$ . Der $\cos(\alpha)$ }}$\\$ \small{\text{ berechnet aus dem Winkel $\alpha$ die Winkelfunktion. }}$\\$ \small{\text{ Zum Beispiel aus $20,36\ensurement{^{\circ}}$ den Wert $\cos(\alpha)=\cos(20,36\ensurement{^{\circ}})=0.9375$ }} }}$\\$ \small{\text{ Nach der obigen Formel muss man die Breite durch die $ H\ddot{o}he $ teilen. }} }}$\\$ \small{\text{ $ \dfrac{Breite}{ H\ddot{o}he } = \frac{75\ cm}{80\ cm} = 0.9375 $ . Nun muss man nur noch}}$\\$ \small{\text{ zum Wert $ 0.9375 $ den entsprechenden Winkel $ \alpha $ suchen. }}$\\$ \small{\text{ Dazu verwendet man die Umkehrfunktion zum $\cos{(\alpha)}$ eben den $ \arccos{(x)} $. }} }}$\\$ \small{\text{ \arccos{(0.9375 )} = 20.36\ensurement{^{\circ}} $ }}$$

how can i solve matrix equation

how to check that 2814749767109 and 59482661568307 are relatively prime

GCD(59482661568307, 2814749767109 ) with Euclidean algorithm:

$$\frac{59482661568307 }{2814749767109} = \frac{59482661568307 -21*2814749767109}{2814749767109} = \frac{372916459018 }{2814749767109}\\\\ = \frac{372916459018 }{2814749767109} = \frac{372916459018 }{2814749767109-7*372916459018} = \frac{372916459018 }{204334553983} \\\\ = \frac{372916459018 }{204334553983} = \frac{372916459018-204334553983 }{204334553983} = \frac{168581905035 }{204334553983}\\\\ = \frac{168581905035 }{204334553983} = \frac{168581905035 }{204334553983-168581905035} = \frac{168581905035 }{35752648948} \\\\ \dots$$

a                             b                                q           r

59482661568307  2814749767109     21          372916459018

2814749767109      372916459018       7           204334553983

372916459018        204334553983       1           168581905035

204334553983        168581905035       1             35752648948

168581905035          35752648948       4              25571309243

35752648948             25571309243       1             10181339705

25571309243             10181339705       2               5208629833

10181339705               5208629833       1               4972709872

5208629833                 4972709872       1                 235919961

4972709872                   235919961     21                   18390691

235919961                       18390691     12                   15231669

18390691                         15231669       1                     3159022

15231669                           3159022       4                     2595581

3159022                             2595581       1                       563441

2595581                               563441       4                       341817

563441                                 341817       1                       221624

341817                                 221624       1                       120193

221624                                 120193       1                       101431

120193                                 101431       1                         18762

101431                                   18762       5                           7621

18762                                       7621       2                           3520

7621                                         3520       2                              581

3520                                           581       6                                34

581                                               34      17                                 3

34                                                    3     11                                   $$\textcolor[rgb]{1,0,0}{1}$$

3                                                      1       3                                  0

GCD = 1, so 2814749767109 and 59482661568307 are relatively prime

f(x) = tan^-1(2x+1)      f(x)'  ?

$$\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ ] = 2x+1 \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\ (1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ ) ) \times \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \left[ 1+(2x+1)^2} \right] \times \left[\textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]' = \frac{2} {1+(2x+1)^2} } }$$

$$$\int \left(\frac{x}{\sqrt{3-4x^2}}\right)\ dx \quad \text { ?}$$ $

$$\small{ \text{ $ \begin{array}{rcl} &=&\int \left(\frac{ \big{x} }{\sqrt{3\left(1-\frac{4}{3}x^2\right)}} \right) \ dx \\ \\ &=&\frac{1 }{\sqrt{3}}\int \left(\frac{ \big{x} }{\sqrt{ 1- \left( \frac{ \big{x} }{ \sqrt \frac{3}{4} } \right)^2 } } \right) \ dx \end{array} $ }} $\\\\$ \small\text{ we substitue: $ \frac{x} {\sqrt{ \frac{3}{4} } } = \sin(u) \quad \Rightarrow \quad \frac{ \ dx} {\sqrt{ \frac{3}{4} } } = \cos(u) \ du$ }} $\\\\$ \small\text{ and set also: $ x = ( \sqrt{ \frac{3}{4} } ) * \sin(u) \quad $ and $\quad \ dx = ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{1 }{\sqrt{3}}\int \left(\frac{ ( \big{ \sqrt{ \frac{3}{4} } ) * \sin(u) } }{\sqrt{ 1- \big{ \left( \sin(u) \right)^2 } } } \right) ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }}$$

$$$\\\\$ \small\text{ $ =\frac{1 }{\sqrt{3}}\int \left( \frac{ ( \big{ \sqrt{ \frac{3}{4} } ) * \sin(u) } } { \big{\cos(u) } } \right) ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }} $\\\\$ \small\text{ $ =\dfrac{\frac{3}{4} }{ \sqrt{3} }\int \left( \sin(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}\int \left( \sin(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}\int \left( \sin(u) \ du \quad | \quad \int\sin(u)\ du = -\cos(u)$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}(-\cos(u))$ }} $\\\\$ \small\text{ $ =-\frac{ \sqrt{3}}{4}\cos(u) \quad | \quad cos(u) = \sqrt{1-\sin(u)^2 }= \sqrt{1-\frac{4}{3}x^2 } $ }} $\\\\$ \small\text{ $ =-\frac{ \sqrt{3}}{4}\sqrt{1-\frac{4}{3}x^2 } }$ }} $\\\\$ \small\text{ $ =-\frac{1}{4}\sqrt{3-4x^2 } }$ }}$$

$$\boxed{\int \left(\frac{x}{\sqrt{3-4x^2}}\right)\ dx =-\frac{1}{4}\sqrt{3-4x^2 } \quad + c }$$