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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)'  ?

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

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

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

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

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