heureka

8 schüler der theathergruppe bestuhlen die Aula. jeder muss dabei 30 stühle tragen.

wie viele stühle muss jeder tragen, wenn es 10 schüler sind?

Anzahl der Stühle in der Aula = 8*30 = 240 Stühle.

In der Aula sind 240 Stühle.

240 Stühle durch 10 Schüler sind 24 Stühle.

Jeder muss 24 Stühle tragen, wenn es 10 Schüler sind.

Wie rechnet der TR z.B. 2.5!, kann man das irgendwie selber nachrechnen?

$\Gamma = \quad Gamma -Funktion$

$\begin{array}{rcll} 2.5! &=& \Gamma(3.5) \\ &=& \Gamma(2.5+1) \\ &=& 2.5\cdot\Gamma(2.5) \\ &=& 2.5\cdot\Gamma(1.5+1) \\ &=& 2.5\cdot 1.5\cdot\Gamma(1.5) \\ &=& 2.5\cdot 1.5\cdot\Gamma(0.5+1) \\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\Gamma(0.5) \qquad \Gamma(0.5) = \sqrt{\pi}\\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\sqrt{\pi}\\\\ &=& 1.875\cdot\sqrt{\pi}\\ &=& 1.875\cdot 1.77245385091\\ &=& 3.32335097045\\ \mathbf{2.5!} & \mathbf{=} & \mathbf{3.323350970447842551184064031264647217745405230229475865400\ldots} \end{array}$

An engineer estimates the angle of elevation to the top of the building to be 50°.
After moving 1.5 meters further away, the angle of elevation was 40°.
How high is the top of the building?

$\begin{array}{|lrcll|} \hline (1) & \tan(50^{\circ}) &=& \frac{h}{x} \\ & x &=& \frac{h}{\tan(50^{\circ})} \\\\ (2) & \tan(40^{\circ}) &=& \frac{h}{1.5+x} \\ & \tan(40^{\circ}) &=& \frac{h}{1.5+\frac{h}{\tan(50^{\circ})}} \\ & \tan(40^{\circ})\cdot \left( 1.5+\frac{h}{\tan(50^{\circ})}\right) &=& h \\ & 1.5\cdot \tan(40^{\circ}) + h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& h \\ & h-h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left(1- \frac{ \tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left( \frac{ \tan(50^{\circ})-\tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h &=& 1.5\cdot \left( \frac{ \tan(50^{\circ})\cdot\tan(40^{\circ}) }{ \tan(50^{\circ})-\tan(40^{\circ} ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1.19175359259\cdot 0.83909963118}{ 1.19175359259-0.83909963118 ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1 }{ 0.35265396142 } \right) \\ & h &=& 1.5\cdot 2.83564090981 \\ & h &=& 4.25346136471\ m \\ \hline \end{array}$

The top of the building is 4.25 m high.

How I should resolve (1/(2))^x=16 without the calculator?

$\begin{array}{|rcll|} \hline \left(\frac12 \right)^x &=& 16 \\ \left(\frac12 \right)^x &=& 2^4 \\ \left(\frac12 \right)^x &=& \frac{1}{2^{-4}} \\ \left(\frac12 \right)^x &=& \left(\frac12 \right)^{-4} \\ x &=& -4 \\ \hline \end{array}$

Need help with this continuous fraction integral.

integral \limits_{x=0}^1  \frac{x\pm\sqrt{x^2+4x} }{2}  \ dx

Continuous Fraction.

$\begin{array}{|rcll|} \hline \begin{equation*} sum=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \end{equation*}\\\\ sum &=& x + \frac{x}{sum} \\ sum - \frac{x}{sum} &=& x \\ \frac{sum^2-x}{sum} &=& x \\ sum^2-x &=& x\cdot sum \\ sum^2-x\cdot sum -x &=& 0 \\ sum &=& \frac{x\pm\sqrt{x^2-4\cdot(-x)} }{2} \\ \mathbf{sum} & \mathbf{=} & \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \\ \hline \end{array}$

$\small{ \begin{array}{llcl} \int \limits_{x=0}^{1} { x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \ dx} \\\\ = \int \limits_{x=0}^{1} { sum \ dx} \\ = \int \limits_{x=0}^{1} { \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \ dx} \\ = \frac12 \int \limits_{x=0}^{1} {x \ dx } \pm \frac12 \int \limits_{x=0}^{1} { \sqrt{x^2+4x} \ dx } \\ = \frac12 \left[\frac{x^2}{2}\right]_{x=0}^{1} \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ = \frac14 \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ & \text{substitute:}\\ & \boxed{~ u=x+2\\ du = dx ~}\\ & \text{new limits:}\\ & \boxed{~ x=0: \qquad u=0+2 \Rightarrow u = 2 \\ x=1: \qquad u=1+2 \Rightarrow u = 3 ~}\\ = \frac14 \pm \frac12 \int \limits_{u=2}^{3} { \sqrt{u^2-4} \ du } \\ & \text{substitute:}\\ & \boxed{~ u=2 \cosh(z) \qquad z = \text{arcosh} \left(\frac{u}{2}\right)\\ du = 2 \sinh(z)\ dz ~}\\ & \text{new limits:}\\ & \boxed{~ u=2: \qquad z=\text{arcosh} \left(\frac{2}{2}\right) \\ \Rightarrow z = \text{arcosh}(1) = 0 \\ u=3: \qquad z=\text{arcosh} \left(\frac{3}{2}\right) \\ \Rightarrow z = \text{arcosh}(1.5) ~}\\ = \frac14 \pm \frac12 \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot 2 \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot \sinh(z) \ dz } \\ & \boxed{~ \cosh^2(z) - \sinh^2(z) = 1 \\ \cosh^2(z) - 1 = \sinh^2(z) \quad | \quad \cdot 4 \\ 4\cosh^2(z) - 4 = 4\sinh^2(z) \\ ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\sinh^2(z)} \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh(z) \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh^2(z) \ dz } \\ \end{array} }$

$\small{ \begin{array}{llcl} & \boxed{~ \cosh(2z) = \cosh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+\sinh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+2\sinh^2(z) \\ 2\sinh^2(z) = \cosh(2z)-1 ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \Big( \cosh(2z)-1 \Big) \ dz } \\ = \frac14 \pm \Big( \int \limits_{z=0}^{\text{arcosh}(1.5)} {\cosh(2z) \ dz } -\int \limits_{z=0}^{\text{arcosh}(1.5)} {\ dz } \Big) \\ = \frac14 \pm \Big( \left[\frac12\cdot \sinh(2z) \right]_{z=0}^{\text{arcosh}(1.5)} -\left[ z \right]_{z=0}^{\text{arcosh}(1.5)} \Big) \\ & \boxed{~ \sinh(2z) = 2\sinh(z)\cosh(z) ~}\\ = \frac14 \pm \Big( \left[\frac12\cdot 2\sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big( \left[ \sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big[ \sinh\Big(\text{arcosh}(1.5)\Big)\cdot 1.5 - \text{arcosh}(1.5) \Big] \\ & \boxed{~ \cosh^2(x) - \sinh^2(x) = 1 \\ \sinh^2(x) = \cosh^2(x) - 1 \\ \sinh(x) = \sqrt{\cosh^2(x) - 1} \\ ~}\\ \boxed{~ \sinh\Big(\text{arcosh}(x)\Big)= \sqrt{\cosh^2(\text{arcosh}(x)) - 1} = \sqrt{x^2 - 1} \\ \sinh\Big(\text{arcosh}(1.5)\Big) = \sqrt{1.5^2 - 1} = \sqrt{1.25} = \frac{ \sqrt{5} }{2} ~}\\ = \frac14 \pm \Big[ \frac{\sqrt{5}}{2} \cdot \frac{3}{2} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm \Big[ \frac{3\sqrt{5}}{4} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm ( 1.6770509831248424 - 0.9624236501192069 ) \\ = \frac14 \pm 0.7146273330056354 \\ \end{array} }$

$\begin{array}{|rcll|} \hline = \frac14 + 0.7146273330056354 \qquad &\text{or}&\qquad = \frac14 - 0.7146273330056354 \\ = 0.9646273330056354 \qquad &\text{or}& \qquad =-0.4646273330056354 \\ \hline \end{array}$

Definite Integral Problem - Having trouble

One more :)

Continuous Fraction.

$\begin{array}{|rcll|} \hline \begin{equation*} sum=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cdots}}} \end{equation*}\\\\ sum &=& x + \frac{1}{sum} \\ sum - \frac{1}{sum} &=& x \\ \frac{sum^2-1}{sum} &=& x \\ sum^2-1 &=& x\cdot sum \\ sum^2-x\cdot sum -1 &=& 0 \\ sum &=& \frac{x\pm\sqrt{x^2-4\cdot(-1)} }{2} \\ \mathbf{sum} & \mathbf{=} & \mathbf{ \frac{x\pm\sqrt{x^2+4} }{2} } \\ \hline \end{array}$

$\small{ \begin{array}{llcl} \int \limits_{x=0}^{1} { x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cdots}}} \ dx} \\\\ = \int \limits_{x=0}^{1} { sum \ dx} \\ = \int \limits_{x=0}^{1} { \mathbf{ \frac{x\pm\sqrt{x^2+4} }{2} } \ dx} \\ = \frac12 \int \limits_{x=0}^{1} {x \ dx } \pm \int \limits_{x=0}^{1} { \frac{ \sqrt{x^2+4} }{2} \ dx } \\ = \frac12 [\frac{x^2}{2}]_{x=0}^{1} \pm \int \limits_{x=0}^{1} { 2\cdot \frac{ \sqrt{(\frac{x}{2})^2+1} }{2} \ dx } \\ = \frac14 \pm \int \limits_{x=0}^{1} { \sqrt{(\frac{x}{2})^2+1} \ dx } \\ & \text{substitute:}\\ & \boxed{~ \frac{x}{2}=\sinh(z) \qquad z = \text{arsinh}\left(\frac{x}{2}\right)\\ dx = 2\cdot \cosh(z)\ dz ~}\\ & \text{new limits:}\\ & \boxed{~ x=0: \qquad z=\text{arsinh}\left(\frac{0}{2}\right) \Rightarrow z = 0 \\ x=1: \qquad z=\text{arsinh}\left(\frac{1}{2}\right) ~}\\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \sqrt{\sinh^2(z)+1} \cdot 2\cdot \cosh(z)\ dz } \\ & \boxed{~ \cosh^2(z) = 1+\sinh^2(z) ~}\\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \sqrt{\cosh^2(z)}\cdot 2\cdot \cosh(z)\ dz } \\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh(z)\cdot 2\cdot \cosh(z) \ dz } \\ = \frac14 \pm 2 \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh^2(z) \ dz } \\ & \boxed{~ \cosh^2(z) = \frac12+\frac12\cosh(2z) ~}\\ = \frac14 \pm 2 \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \frac12+\frac12\cosh(2z) \ dz } \\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { 1+ \cosh(2z) \ dz } \\ = \frac14 \pm \Big( \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \ dz } +\int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh(2z) \ dz } \Big) \\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[\frac{\sinh(2z)}{2} \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ & \boxed{~ \sinh(2z) = 2\sinh(z) \cosh(z) ~}\\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[\frac{2\sinh(z) \cosh(z)}{2} \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[ \sinh(z) \cosh(z) \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ = \frac14 \pm \Big( \text{arsinh}\left(\frac{1}{2}\right) + \frac12 \cdot \cosh\left(\text{arsinh}\left(\frac{1}{2}\right)\right) \Big) \\ = \frac14 \pm \Big( 0.4812118250596 + \frac12 \cdot \cosh\left( 0.4812118250596\right) \Big) \\ = \frac14 \pm \Big( 1.0402288194345508 \Big) \\ \end{array} }$

$\begin{array}{|rcll|} \hline = \frac14 + 1.0402288194345508 \qquad &\text{or}&\qquad = \frac14 - 1.0402288194345508 \\ = 1.2902288194345508 \qquad &\text{or}& \qquad =-0.7902288194345508 \\ \hline \end{array}$

The number of positive integers less than 5000 that are divisible by neither 3 nor 7.

Number  of integers divisible by 3  =   1666

Number of integers divisble by 7  =  714

Number of integers divisible by 3 and 7  =  5000/21    =  238  (Double Counting... )

So.....The number of positive integers less than 5000 that are divisible by neither 3 nor 7 =

4999 - 1666 - 714  + 238   =  2857

Thanks Melody

Something Mathematical to do.

$\begin{array}{llcl} \int \limits_{x=0}^{\infty} { \frac{1}{ (x+\sqrt{1+x^2})^2 } \ dx}\\\\ & \text{substitute:}\\ & \boxed{~ x=\tan(z) \\ dx = ( 1+\tan^2(z))\ dz \\ ~}\\ & \text{new limits:}\\ & \boxed{~ z=\arctan(0) \Rightarrow z = 0 \\ z=\arctan(\infty) \Rightarrow z = \frac{\pi}{2} \\\\ ~}\\ =\int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1+\tan^2(z)}{ \Big(\tan(z)+\sqrt{1+\tan^2(z)}\Big)^2 } \ dz} \\ & \boxed{~ 1+\tan^2(z) = \frac{1}{\cos^2(z)} \\ ~}\\ =\int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{\cos^2(z)}\cdot \left( \frac{1}{\frac{1}{\cos^2(z)}+ \tan^2(z)+2\cdot \frac{\sin(z)}{\cos^2(z)} } \right) \ dz} \\ = \int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{1+2\cdot \sin(z) + \sin^2(z) } \ dz} \\ = \int \limits_{z=0}^{\frac{\pi}{2}} { \frac{1}{ \Big(1+\sin(z)\Big)^2 } \ dz} \\ & \text{substitute:}\\ & \boxed{~ t=\tan(\frac{z}{2}) \\ dt = \frac12 \cdot \left( 1+\tan^2(\frac{z}{2}) \right)\ dz\\ =\frac12 \cdot (1+t^2)\ dz \\ ~}\\ & \text{new limits:}\\ & \boxed{~ t=\tan(\frac{0}{2}) \Rightarrow t = 0 \\ t=\tan(\frac{ \frac{\pi}{2}}{2}) \Rightarrow t = 1 \\\\ ~}\\ = \int \limits_{t=0}^{1} \frac{2\ dt}{1+t^2} \cdot \left( \frac{1}{ \left(1+ \sin(z)\right)^2 } \right) \\ & \boxed{~ \sin(z) = \frac{2t}{1+t^2} \\ ~}\\ = \int \limits_{t=0}^{1} \frac{2\ dt}{1+t^2} \cdot \left( \frac{1}{ \left(1+ \frac{2t}{1+t^2}\right)^2 } \right) \\ = 2\cdot \int \limits_{t=0}^{1} \frac{1+t^2}{1+4t+6t^2+4t^3+t^4} \ dt\\ = 2\cdot \int \limits_{t=0}^{1} \frac{1+t^2}{(1+t)^4} \ dt\\ \end{array}$

$\begin{array}{llcl} & \text{Partial fraction decomposition:}\\ & \boxed{~ \begin{array}{lcll} & \frac{1+t^2}{(1+t)^4} &=& \frac{A}{1+t} + \frac{B}{(1+t)^2} + \frac{C}{(1+t)^3} + \frac{D}{(1+t)^4} \\\\ & 1+t^2 &=& A\cdot (1+t)^3 +B\cdot (1+t)^2 +C\cdot (1+t) +D \\ (1)\quad t=-1 : & \Rightarrow 2 &=& D \\ (2)\quad t= 0 : & \Rightarrow -1 &=& A+B+C \\ (3)\quad t= 1 : & \Rightarrow 0 &=& 8A+4B+2C \quad | \quad : 2 \\ & 0 &=& 4A+2B+C \\ (4)\quad t= -2 : & \Rightarrow 3 &=& -A+B-C \\\\ (2)+(4) : & -1+3 &=& 2 B \qquad \Rightarrow B =1 \\ (3)+(4) : & 0+3 &=& 3A + 3B \quad | \quad : 3 \\ & 1 &=& A + B \quad | \quad B=1 \\ & 1 &=& A + 1 \qquad \Rightarrow A =0 \\ (2): & C &=& -1-A-B \\ & &=& -1-0-1 \qquad \Rightarrow C =-2 \\ \end{array} ~}\\ \end{array}$

$\begin{array}{llcl} = 2 \int \limits_{t=0}^{1} \left( \frac{1}{(1+t)^2} -2\cdot \frac{1}{(1+t)^3} +2\cdot \frac{1}{(1+t)^4 } \right) \ dt\\ = 2 \int \limits_{t=0}^{1} \frac{1}{(1+t)^2}\ dt -4 \int \limits_{t=0}^{1} \frac{1}{(1+t)^3}\ dt +4 \int \limits_{t=0}^{1} \frac{1}{(1+t)^4}\ dt \\ = 2 \left[ -\frac{1}{1+t} \right]_{t=0}^{1} -\frac{4}{2} \left[ -\frac{1}{(1+t)^2} \right]_{t=0}^{1} +\frac{4}{3} \left[ -\frac{1}{(1+t)^3} \right]_{t=0}^{1} \\ = 2 \left[ -\frac{1}{2} -(-1) \right] -\frac{4}{2} \left[ -\frac{1}{4}-(-1) \right] +\frac{4}{3} \left[-\frac{1}{8}-(-1) \right] \\ = 2 \left[ -\frac{1}{2} +1 \right] -2 \left[ -\frac{1}{4}+1 \right] +\frac{4}{3} \left[-\frac{1}{8}+1 \right] \\ = 2 \cdot \frac12 -2\cdot \frac34 +\frac43 \cdot \frac78 \\ = 1 - \frac32+ \frac76 \\ = \frac66 - \frac96 +\frac76 \\ = \frac{13}{6} - \frac96 \\ = \frac46 \\ = \frac23 \\ \end{array}$

$\begin{array}{llcl} \int \limits_{x=0}^{\infty} { \frac{1}{ (x+\sqrt{1+x^2})^2 } \ dx} = \frac23 \end{array}$