heureka

2sinpi/3cospi/3 ?

$$2\cdot\sin{ (\frac{\pi}{3}) }\cdot \cos{ (\frac{\pi}{3}) } = \sin{( ~2\cdot(\frac{\pi }{3})~ ) } =\frac{1}{2}\sqrt{3} = 0.86602540378$$

Hallo Melody,

$$\small{\begin{array}{rcl}\mathbf{ \varphi(12)} &\mathbf{=}& \mathbf{4}$\\
\end{array}}$$,

because

$$\small{
\begin{array}{r|rcl|c}
\hline
\text{number} &\text{greatest common devisor} &&& \text{relative prime, if 1} \\
\hline
1 & \gcd{(12,1)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
2 & \gcd{(12,2)} &=& 2\\
3 & \gcd{(12,3)} &=& 3\\
4 & \gcd{(12,4)} &=& 4\\
5 & \gcd{(12,5)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
6 & \gcd{(12,6)} &=& 6\\
7 & \gcd{(12,7)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
8 & \gcd{(12,8)} &=& 4\\
9 & \gcd{(12,9)} &=& 3\\
10 & \gcd{(12,10)} &=& 2\\
11 & \gcd{(12,11)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
12 & \gcd{(12,12)} &=& 12\\
\hline
&&& \text{sum} &\textcolor[rgb]{1,0,0}{4}\\
\hline
\end{array}
}$$

6e^x/(3e^x+1)=5/(e^(-x)+2) x=-ln3, aber wie komme ich da hin ?

$$\small{
\begin{array}{rcl}
\dfrac{ 6e^x } { 3e^x+1 } &=& \dfrac{ 5 } { e^{(-x)}+2 } \\\\
6e^x (e^{(-x)}+2) &=& 5 ( 3e^x+1 ) \\\\
6e^x e^{(-x)} + 2\cdot 6e^x &=& 5 \cdot 3e^x + 5\cdot 1 \\\\
6e^x e^{(-x)} + 12\cdot e^x &=& 15 \cdot e^x + 5 \\\\
6e^{(x-x)} + 12\cdot e^x &=& 15 \cdot e^x + 5 \\\\
6e^{(0)} + 12\cdot e^x &=& 15 \cdot e^x + 5 \qquad |\qquad e^0 = 1\\\\
6\cdot 1 + 12\cdot e^x &=& 15 \cdot e^x + 5 \\\\
6 + 12\cdot e^x &=& 15 \cdot e^x + 5 \\\\
6 + 12\cdot e^x &=& 15 \cdot e^x + 5 \qquad |\qquad -12\cdot e^x\\\\
6 &=& 15 \cdot e^x -12\cdot e^x + 5 \\\\
6 &=& 3 \cdot e^x + 5 \qquad |\qquad -5\\\\
6-5 &=& 3 \cdot e^x \\\\
1 &=& 3 \cdot e^x \\\\
3 \cdot e^x &=& 1 \qquad |\qquad :3\\\\
e^x &=& \dfrac{1}{3} \\\\
e^x &=& \dfrac{1}{3} \qquad |\qquad \ln{()}\\\\
\ln{ (e^x) } &=& \ln{ \left( \dfrac{1}{3} \right) } \\\\
x\cdot \ln{(e)} &=& \ln{ \left( \dfrac{1}{3} \right) } \qquad | \qquad \ln{(e)} = 1 \\\\
x &=& \ln{ \left( \dfrac{1}{3} \right) }\\\\
x &=& \ln{(1)} - \ln{(3)} \qquad | \qquad \ln{(1)} = 0 \\\\
\mathbf{x} &\mathbf{=}& \mathbf{ - \ln{(3)} } \\\\
\end{array}
}$$

What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98 ?

$$\small{
\begin{array}{rcl}
a_n &=& a_1\cdot r^{n-1} \\
a_6 &=& a_1\cdot r^5 = 4 \\
a_{100} &=& a_1\cdot r^{99} = 98\\
\\
\hline
\\
\dfrac{a_{100}}{a_6} &=&
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} = \dfrac{98}{4} \\\\
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} &=& \dfrac{98}{4} \\\\
\dfrac{r^{99}}{r^{5}} &=& 24.5 \\\\
r^{99-5} &=& 24.5 \\\\
r^{94} &=& 24.5 \\\\
r &=& \sqrt[94]{24.5} \\\\
r &=& 1.03461402806 \\
\\
\hline
\\
a_6 &=& a_1\cdot r^5 = 4 \\\\
a_1 \cdot r^{5} &=& 4 \\\\
a_1 &=& \dfrac{4} { r^5 } \\\\
a_1 &=& \dfrac{4} { 1.03461402806^5 } \\\\
a_1 &=& 3.37417946747\\
\\
\hline
\\
a_{55} &=& a_1\cdot r^{54} \\\\
a_{55} &=& 3.37417946747 \cdot 1.03461402806^{54}\\\\
a_{55} &=& 21.1933572378
\end{array}
}$$

Vectors A and B are at angles α = 44.9° and β = 26.2° up from the x-axis respectively. If the vector sum A B C = 0, what are the magnitudes of A and B?

a = magnitude of A,   b = magnitude of B

$$\small{
\vec{A} = a\cdot \begin{pmatrix}\cos{\alpha}\\\sin{\alpha}\end{pmatrix}\quad
\vec{B} = b\cdot \begin{pmatrix}\cos{\beta}\\\sin{\beta}\end{pmatrix}
\qquad
\vec{A}_{\bot }= \begin{pmatrix}-\sin{\alpha}\\\cos{\alpha}\end{pmatrix}\quad
\vec{B}_{\bot } =\begin{pmatrix}-\sin{\beta}\\\cos{\beta}\end{pmatrix}
}\\\\
\small{
\begin{array}{rcl}
a\vec{A}+b\vec{B}+\vec{c}&=&\vec{0}\quad | \quad \cdot \vec{B}_{\bot }\\
a\vec{A}\vec{B}_{\bot }+b\vec{B}\vec{B}_{\bot }+\vec{c}\vec{B}_{\bot }&=&\vec{0}\quad | \quad \vec{B}\cdot {\vec{B}_{\bot }=0 \\
a\vec{A}\vec{B}_{\bot }+\vec{c}\vec{B}_{\bot }&=&\vec{0} \\
a &=&\dfrac { -\vec{c}\vec{B}_{\bot } } {\vec{A}\vec{B}_{\bot }}\\
a &=& \vec{c}\cdot \begin{pmatrix}\frac{ \sin{\beta} }{ \sin{ (\alpha-\beta) }} \\\\ \frac{-\cos{\beta}}{\sin{ (\alpha-\beta) }} \end{pmatrix}
= \vec{c}\cdot \begin{pmatrix}1.37706788482\\ -2.79857147409\end{pmatrix}\\
\end{array}
}\\\\\\
\small{
\begin{array}{rcl}
a\vec{A}+b\vec{B}+\vec{c}&=&\vec{0}\quad | \quad \cdot \vec{A}_{\bot }\\
a\vec{A}\vec{A}_{\bot }+b\vec{B}\vec{A}_{\bot }+\vec{c}\vec{A}_{\bot }&=&\vec{0}\quad | \quad \vec{A}\cdot {\vec{A}_{\bot }=0 \\
b\vec{B}\vec{A}_{\bot }+\vec{c}\vec{A}_{\bot }&=&\vec{0} \\
b &=&\dfrac { -\vec{c}\vec{A}_{\bot } } {\vec{B}\vec{A}_{\bot }}\\
b &=& \vec{c}\cdot \begin{pmatrix}\frac{ -\sin{\alpha} }{ \sin{(\alpha-\beta)}} \\\\ \frac{\cos{\alpha}}{\sin{(\alpha-\beta)}} \end{pmatrix}
= \vec{c}\cdot \begin{pmatrix}-2.20163122333\\ 2.20932981047\end{pmatrix}\\
\end{array}
}$$

Hallo fiora,

yes, this question challange me.

Thank you for this question!

find the sum of the first 100 term of the geometric series with the first term 5 and the common ratio is -2

$$\\\small{
\begin{array}{rccl}
s_{100} &=& \textcolor[rgb]{1,0,0}{5}+&5(-2)^1+5(-2)^2+5(-2)^3+5(-2)^4+\dots+5(-2)^{98}+5(-2)^{99}
}\\\\
(-2)s_{100}&=& & 5(-2)^1+5(-2)^2+5(-2)^3+5(-2)^4+\dots+5(-2)^{98}+5(-2)^{99}+\textcolor[rgb]{1,0,0}{5(-2)^{100}}
}\\\\
\hline
s_{100}-(-2)s_{100}&=& 5-&5(-2)^{100}\\\\
s_{100}+2s_{100}&=& 5-&5(-2)^{100}\\\\
3s_{100}&=& 5-&5(-2)^{100}\\\\
\end{array}
}\\\\
\small
\begin{array}{rcl}
s_{100} &=& \dfrac{ 5-5(-2)^{100} } {3}\\\\
s_{100} &=& -2112751000380382335827838675625
\end{array}
}$$

In the following fighure ,if triangle ABC is a right triangle,angle A= 90 degrees,point D located on AC ,point E is loacted on BC, $${\mathtt{AB}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{BD}}$$ ,and $${\mathtt{CE}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{EB}}$$. angle BDE=120 degrees,CD=3, then BC=?

We define:  CD = 3,  AB = y,   DB = 2y,   CE = x,   EB =  4x,   BC = 5x,   ED = u,   DA = d

1. Pythagoras:

$$\small{\overline{AB}^2 + \overline{DA}^2 = \overline{DB}^2}\\
\small{y^2+d^2=4y^2\qquad d^2 = 3y^2 \qquad d = \sqrt{3}y }$$

2. Pythagoras:

$$\small{\overline{CA}^2 + \overline{AB}^2 = \overline{BC}^2 \qquad \overline{CA} = 3+\sqrt{3}y}\\\\
\small{ (3+\sqrt{3}y)^2+y^2=(5x)^2\qquad \cdots \qquad
\boxed{x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} ~~(1) }}$$

3. cos-rule:

$$\small{u^2 = 3^2 + x^2 - 2\cdot 3\cdot x \cdot \cos{(C)}\qquad \cos{(C)} = \dfrac{ 3 + \sqrt{3}y }{5x} \quad \cdots \quad \boxed{ u = \sqrt{x^2-1.2\sqrt{3}+5.4} ~~(2)}}$$

4. cos-rule:

$$\small{
\begin{array}{lcl}
(4x)^2 = u^2+(2y)^2-2u2y\cos{ (120\ensurement{^{\circ}}) } \quad \cos{(120\ensurement{^{\circ}}) } = -\frac{1}{2} \quad \cdots \quad 16x^2=u^2+4y^2+2uy \\
\text{we substitute u, formula (2)}\\
\cdots\\
15x^2 = -1.2\sqrt{3}y+5.4+4y^2+2y\sqrt{x^2-1.2\sqrt{3}y+5.4}\\
\text{we substitute first x, formula (1)}\\
\cdots\\
\frac{12}{5}y^2+}\frac{18}{5}\sqrt{3}y=4y^2-1.2\sqrt{3}y+2y\sqrt{x^2-1.2\sqrt{3}y+5.4} \qaud | \quad :y \\
\frac{12}{5}y+}\frac{18}{5}\sqrt{3}=4y-1.2\sqrt{3}+2\sqrt{x^2-1.2\sqrt{3}y+5.4} \\
\cdots\\
x^2-1.2\sqrt{3}y+5.4=(2.4\sqrt{3}-0.8y)^2\\
\text{we substitute again x, formula (1)}\\
\cdots\\
0.16y^2+0.24\sqrt{3}y+0.36-1.2\sqrt{3}y+5.4=17.28-3.84\sqrt{3}y+0.64y^2\\
0.48y^2-2.88\sqrt{3}y+11.52=0 \quad | :0.48 \quad \\
\boxed{y^2-6\sqrt{3}+24 = 0} \\\\
y_{1,2} = \frac{6\sqrt{3}\pm \2\sqrt{3}}{2}\\
y_1 = 4\sqrt{3} \quad \text{no solution}\\
y_2 = 2\sqrt{3} \quad \text{solution}\\\\
x = \dfrac{ \sqrt{ 4y^2+6\sqrt{3}y+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 4 (2\sqrt{3})^2+6\sqrt{3}( 2\sqrt{3})+9 } } {5} \\\\
x = \dfrac{ \sqrt{ 48 +36+9 } } {5} \\\\
x = 1.92873015220\\\\
\mathbf{ \overline{BC} =5x = 9.64365076099 }
\end{array}
}$$

0xaa02<<16

$$AA02_{16} = 43522_{10}\\\\
0\times aa02<<16 =
43522\cdot 2^{16} = 2~852~257~792$$

Berechnen Sie x, sodass die Dreiecksfläche, die durch diese Vektoren aufgespannt wird, 11,66mm beträgt. Gegeben sind die Vektoren a=(0,3,4) ; b=(x,3,0)

$$\small{
\begin{array}{rcl}
2A_{\text{Dreieck}} &=&\begin{vmatrix} \vec{a} \times \vec{b} \end{vmatrix} \\\\
2A_{\text{Dreieck}} &=& \begin{Vmatrix} 1 & 1 & 1 \\ 0 & 3 & 4 \\ x & 3 & 0 \end{Vmatrix} \\ \\
2A_{\text{Dreieck}} &=&
\left{|} \begin{pmatrix}
\begin{vmatrix} 3 & 4 \\ 3 & 0 \end{vmatrix}\\
-\begin{vmatrix} 0 & 4 \\ x & 0 \end{vmatrix}\\
\begin{vmatrix} 0 & 3 \\ x & 3 \end{vmatrix}
\end{pmatrix} \right{|} \\ \\
2A_{\text{Dreieck}} &=&
| \begin{pmatrix} -12 \\ 4x \\ -3x \end{pmatrix} | \\ \\
\end{array}
}$$

Probe. Der Flächenvektor muss senkrecht stehen zu a und b:

$$\small{
\begin{array}{rcl}
\begin{pmatrix} -12 \\ 4x \\ -3x \end{pmatrix}\cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = 12x-12x = 0 \qquad \text{okay}\\\\
\begin{pmatrix} -12 \\ 4x \\ -3x \end{pmatrix}\cdot \begin{pmatrix} x \\ 3 \\ 0 \end{pmatrix} = -12x+12x = 0 \qquad \text{okay}
\end{array}$$

Die Berechnung von x:

$$\small{
\begin{array}{rcl}
2\cdot A_{\text{Dreieck}} &=& \sqrt{(-12)^2+(4x)^2+(-3)^2 } \\
2\cdot A_{\text{Dreieck}} &=& \sqrt{ 144 + 16x^2 + 9x^2 } \\
2\cdot A_{\text{Dreieck}} &=& \sqrt{ 144 + 25x^2 } \quad | \quad 1^2 \\
25x^2 + 144 &=& 4\cdot A_{\text{Dreieck}}^2\\
25x^2 &=& 4\cdot A_{\text{Dreieck}}^2 + 144 \\
25x^2 &=& 4 \cdot 11,66^2 + 144 \\
25x^2 &=& 543,8224 + 144 \\
25x^2 &=& 399,8224 \\
25x^2 &=& 399,8224 \quad | \quad \sqrt{}\\
5x &=& \pm 19.9955595071 \\
x &=& \pm 3.99911190141 \\\\
\mathbf{x} &\mathbf{\approx}& \mathbf{\pm 4 }
\end{array}
}$$

Probe:  a=(0,3,4) ; b=(4,3,0)

$$\small{
\begin{array}{rcl}
2A_{\text{Dreieck}} &=&\begin{vmatrix} \vec{a} \times \vec{b} \end{vmatrix} \\\\
2A_{\text{Dreieck}} &=& \begin{Vmatrix} 1 & 1 & 1 \\ 0 & 3 & 4 \\ 4 & 3 & 0 \end{Vmatrix} \\ \\
2A_{\text{Dreieck}} &=&
\left{|} \begin{pmatrix}
\begin{vmatrix} 3 & 4 \\ 3 & 0 \end{vmatrix}\\
-\begin{vmatrix} 0 & 4 \\ 4 & 0 \end{vmatrix}\\
\begin{vmatrix} 0 & 3 \\ 4 & 3 \end{vmatrix}
\end{pmatrix} \right{|} \\ \\
2A_{\text{Dreieck}} &=&
| \begin{pmatrix} -12 \\ 16 \\ -12 \end{pmatrix} | \\ \\
\end{array}
}$$

$$\small{
\begin{array}{rcl}
2\cdot A_{\text{Dreieck}} &=& \sqrt{(-12)^2+(16)^2+(-12)^2 } \\
2\cdot A_{\text{Dreieck}} &=& \sqrt{ 144 + 256 + 144 } \\
2\cdot A_{\text{Dreieck}} &=& \sqrt{ 544 } \\
2\cdot A_{\text{Dreieck}} &=& 23,3238075794\\ \\
\mathbf{A_{\text{Dreieck}}} &\mathbf{=}& \mathbf{ 11,66 \quad \text{okay} }
\end{array}
}$$