heureka

Jim bought a round picnic table. The diameter of this table is 12.

What is the area and circumference of Jim’s picnic table?

\(\begin{array}{rcll} \text{diameter }~ d &=& 12 \\\\ \text{radius } ~r &=& \frac{d}{2} \\ r &=& \frac{12}{2} \\ r &=& 6\\\\ \text{area } ~ a &=& \pi\cdot r^2 \\ a &=& \pi\cdot 6^2 \\ a &=& \pi\cdot 36 \\\\ \text{circumference } ~ c &=& 2\cdot \pi\cdot r \\ c &=& 2\cdot \pi\cdot 6 \\ c &=& \pi\cdot 12 \\\\ a &=& 36\cdot \pi \\ a &=& 36\cdot 3.14159265359\\ \mathbf{a} &\mathbf{=}& \mathbf{113.097335529} \\\\ c &=& 12\cdot \pi \\ c &=& 12\cdot 3.14159265359\\ \mathbf{c} &\mathbf{=}& \mathbf{37.6991118431} \end{array}\)

normalcdf(1.09,10000) = 0.13785657203203

law of cosine

a=6 b=9 c=10 find the angles

\(\begin{array}{rcll} \hline a^2 &=& b^2+c^2-2bc\cdot \cos{(A)} \\ 2bc\cdot \cos{(A)} &=& b^2+c^2-a^2 \\ \cos{(A)} &=& \frac{ b^2+c^2-a^2 } {2bc} \\ \cos{(A)} &=& \frac{ 9^2+10^2-6^2 } {2\cdot 9 \cdot 10} \\ \cos{(A)} &=& \frac{ 81+100-36 } {180} \\ \cos{(A)} &=& \frac{ 145 } {180} \\ \cos{(A)} &=& 0.80555555556 \\ A &=& \arccos{(0.80555555556)} \\ A &=& 36.3360575146^{\circ} \\\\ \hline b^2 &=& c^2+a^2-2ca\cdot \cos{(B)} \\ 2ca\cdot \cos{(B)} &=& c^2+a^2-b^2 \\ \cos{(B)} &=& \frac{ c^2+a^2-b^2 } {2ca} \\ \cos{(B)} &=& \frac{ 10^2+6^2-9^2 } {2\cdot 10 \cdot 6} \\ \cos{(B)} &=& \frac{ 100+36-81 } {120} \\ \cos{(B)} &=& \frac{ 55 } {120} \\ \cos{(B)} &=& 0.45833333333 \\ B &=& \arccos{(0.45833333333)} \\ B &=& 62.7203872640^{\circ}\\\\ \hline c^2 &=& a^2+b^2-2ab\cdot \cos{(C)} \\ 2ab\cdot \cos{(C)} &=& a^2+b^2-c^2 \\ \cos{(C)} &=& \frac{ a^2+b^2-c^2 } {2ab} \\ \cos{(C)} &=& \frac{ 6^2+9^2-10^2 } {2\cdot 6\cdot 9} \\ \cos{(C)} &=& \frac{ 36+81-100 } {108} \\ \cos{(C)} &=& \frac{ 17 } {108} \\ \cos{(C)} &=& 0.15740740741 \\ C &=& \arccos{(0.15740740741)} \\ C &=& 80.9435552214^{\circ} \\ \hline \end{array}\)

The distance s, in meters, of an object from the origin at time t=0
seconds is given by\( s=s(t)=A\cdot cos(\omega t+\phi)\), where A, \(\omega\), and \(\phi\) are constant.

\(\small{ \begin{array}{lrcll} & s(t) &=& A\cdot cos(\omega t+\phi) \quad & ( \text{distance} ) \\ (a) \text{ Find the velocity v of the object at time t. } \\ & v(t) &=& \frac{ds}{dt} = -A\cdot \omega \cdot sin(\omega t+\phi) \quad & ( \text{velocity} ) \\ (c) \text{ Find the acceleration a of the object at time t.}\\ & a(t) &=& \frac{d^2s}{dt^2} = -A\cdot \omega^2 \cdot cos(\omega t+\phi)\quad & ( \text{acceleration} ) \end{array} }\)

\((b) \text{ When is the velocity of the object 0?}\\ \begin{array}{rcll} v(t) = -A\cdot \omega \cdot sin(\omega t+\phi) &=& 0 \\ sin(\omega t+\phi) &=& 0 \\ \omega t+\phi &=& \arcsin{(0)} \pm k\cdot \pi \\ \omega t+\phi &=& 0 \pm k\cdot \pi \\ \omega t &=& -\phi \pm k\cdot \pi \\ t &=& \frac{ -\phi \pm k\cdot \pi }{ \omega } \qquad k = 0,1,2,\dots\\ \end{array}\)

\((d) \text{ When is the acceleration of the object 0?}\\ \begin{array}{rcll} a(t) = -A\cdot \omega^2 \cdot cos(\omega t+\phi) &=& 0\\ cos(\omega t+\phi) &=& 0 \\ \omega t+\phi &=& \arccos{(0)} \pm k\cdot \pi \\ \omega t+\phi &=& \frac{\pi}{2} \pm k\cdot \pi \\ \omega t &=& \frac{\pi}{2} -\phi \pm k\cdot \pi \\ t &=& \frac{ \frac{\pi}{2} -\phi \pm k\cdot \pi }{ \omega } \qquad k = 0,1,2,\dots\\ \end{array}\)

calculate the area bounded by the graphs of f(x)=x2 , g(x)=1/x2, x>0 and the line x=3    :)

1. cut\( f(x) \text{ and } g(x) , x \gt 0\)

\(\begin{array}{rcll} f(x) &=& g(x)\\ x^2 &=& \frac{1}{x^2} \\ x^4 &=& 1 \\ x &=& 1 \end{array}\)

2. calculate the area

\(\begin{array}{rcll} \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } &=& \int \limits_{1}^{3} { f(x) \ dx } -\int \limits_{1}^{3} { g(x) \ dx } \\ &=& \int \limits_{1}^{3} { x^2 \ dx } -\int \limits_{1}^{3} { \frac{1}{x^2} \ dx } \\ &=& [\frac{x^3}{3}]_{1}^{3} - [-\frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot [x^3]_{1}^{3} + [ \frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot (3^3-1^3) + ( \frac{1}{3}-\frac{1}{1} )\\ &=& \frac13\cdot (27-1) + ( -\frac{2}{3} )\\ &=& \frac13\cdot (26) -\frac{2}{3} \\ &=& \frac{26-2}{3} \\ &=& \frac{24}{3} \\ &=& \frac{3\cdot 8}{3} \\ \mathbf{ \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } } & \mathbf{=} & \mathbf{8}\\ \end{array}\)

in the number sequence 1,1,2,3,5,8,13.... What is the remainder when the 1995th number is divided by 8? 

\(f_{1995} =\)

\(f_{1995} \pmod 8 = 2\)

Please give the next 5 terms of this sequence and explain your conclusions:

1, 4, 6, 9, 12, 16, 21, 28, 38, 53....... Thanks to all and have fun.

\(\small{ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline &1&4&6&9&12&16&21&28&38&53&{\color{grey}76}&{\color{grey}112}&{\color{grey}169}&{\color{grey}260}&{\color{grey}406} \\ \hline \text{Fibonacci numbers}&0&1&1&2&3&5&8&13&21&34&55&89&144&233&377 \\ \hline \text{odd numbering}&1&3&5&7&9&11&13&15&17&19&21&23&25&27&29\\ \hline \text{add }& \\ \hline \end{array} }\)

4sinx+5cosx in the form ksin(x+a)

\(\boxed{~ \begin{array}{rcll} A\cdot \sin{(x)}+ B \cdot \cos{(x)} &=& k\cdot \sin{(x+a)} \\ && k = \sqrt{A^2+B^2} \qquad a = \tan^{-1}{ \left(\frac{B}{A} \right)} \end{array} ~}\)

\(\begin{array}{rcll} A\cdot \sin{(x)}+ B \cdot \cos{(x)} &=& k\cdot \sin{(x+a)} \qquad A = 4 \qquad B = 5 \\ \qquad && k = \sqrt{4^2+5^2} \qquad a = \tan^{-1}{ \left(\frac{5}{4} \right)} \\ \qquad && k = \sqrt{16+25} \qquad a = \tan^{-1}{ (1.25 )} \\ \qquad && k = \sqrt{41} \qquad \qquad a = 51.3401917459^\circ\\\\ 4\cdot \sin{(x)}+ 5 \cdot \cos{(x)} &=& \sqrt{41}\cdot \sin{(x+51.3401917459^\circ )} \\ 4\cdot \sin{(x)}+ 5 \cdot \cos{(x)} &=& 6.40312423743\cdot \sin{(x+51.3401917459^\circ )} \\ \end{array}\)

If there are 41 blades of grass per square centimeter how many blades of grass would you have on a bit of land that is 4 square meters?

\(\begin{array}{rcll} 4\ m^2 &=& 4\ m^2 \cdot \frac{100\ cm}{1\ m}\cdot \frac{100\ cm}{1\ m}\\ &=& 40000\ cm^2\\\\ \frac{41\ \text{Blades of grass}}{ 1\ cm^2} \cdot 40000\ cm^2 &=& 41\cdot 40000 \quad \text{Blades of grass}\\ &=& 1640000 \quad \text{Blades of grass}\\ \end{array}\)

wieso ergibt -x/-x^2 nicht 1/-x ?

\(\begin{array}{rcll} \dfrac{ -x } {-x^2} &=& \dfrac{ (-1)\cdot x } {(-1) \cdot x^2} \\\\ &=& \dfrac{ (-1)} {(-1)} \cdot \dfrac{ x } { x^2} \qquad & | \qquad \dfrac{ (-1)} {(-1)}= 1 \\\\ &=& 1 \cdot \dfrac{ x } { x^2} \\\\ &=& \dfrac{ 1 } { x} \\ \end{array}\)