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8 points are located on a circle. how many line segments can be drawn with these points as endpoints?

$\begin{array}{|c|c|c|} \hline \text{Point } Number & \text{from Point} & \text{to Point} & \text{line number}\\ \hline 1 & 1 & 2 & 1\\ & 1 & 3 & 2\\ & 1 & 4 & 3\\ & 1 & 5 & 4\\ & 1 & 6 & 5\\ & 1 & 7 & 6\\ & 1 & 8 & 7\\ \hline 2\text{ new lines} & 2 & 3 & 8\\ & 2 & 4 & 9\\ & 2 & 5 & 10\\ & 2 & 6 & 11\\ & 2 & 7 & 12\\ & 2 & 8 & 13\\ \hline 3\text{ new lines} & 3 & 4 & 14\\ & 3 & 5 & 15\\ & 3 & 6 & 16\\ & 3 & 7 & 17\\ & 3 & 8 & 18\\ \hline 4\text{ new lines} & 4 & 5 & 19\\ & 4 & 6 & 20\\ & 4 & 7 & 21\\ & 4 & 8 & 22\\ \hline 5\text{ new lines} & 5 & 6 & 23\\ & 5 & 7 & 24\\ & 5 & 8 & 25\\ \hline 6\text{ new lines} & 6 & 7 & 26\\ & 6 & 8 & 27\\ \hline 7\text{ new line} & 7 & 8 & 28\\ \hline \end{array}$

If we have 8 Points there are $7 + 6 + 5 + 4 + 3 + 2 + 1 = \mathbf{28}$ line segments.

if we have n Points there are$1 + 2 + 3+ 4 + 5 + \dots + (n-1) = \frac{ [ 1+(n-1) ] \cdot (n-1) } {2} = \mathbf{ \frac{ n\cdot (n-1) } {2} }$line segments.

ich hätte eine Frage zur Trigonometrie, die wie folgt lautet.

Wenn ich die Gleichung 0 = 2*sin(t) * Wurzel(1-sin^2(t)) - sin(t) habe, würde ich als nächstes quadrieren, um die Wurzel aufzulösen. Meine Frage ist, wie quadriert man 2*sin(t) .

$\begin{array}{rcll} ( 2\cdot \sin{(t)} )^2 = 2^2\cdot \sin^2{(t)} = 4\cdot \sin^2{(t)} \end{array}$

Ich vermute mal, Sie wollen nach t auflösen.

In diesem Falle würde man die Wurzel ersetzen:  $\sqrt{ 1-sin^2{(t)} } = \cos{(t)}$ da

$\sin^2{(t)}+\cos^2{(t)} = 1$

Wir hätten dann:

$\begin{array}{rcll} 2\cdot \sin{(t)} \cdot \sqrt{ 1-sin^2{(t)} } - \sin{(t)} &=& 0 \\ 2\cdot \sin{(t)} \cdot \cos{(t)} - \sin{(t)} &=& 0 \\ \sin{(t)} \cdot ( 2\cdot \cos{(t)} - 1 ) &=& 0 \\ \end{array}$

1. Lösung für t: 

$\begin{array}{rcll} \sin{(t)} &=& 0 \qquad & | \qquad \arcsin() \\ t &=& \arcsin{(0)} \\ t &=& 0 \pm k\cdot 180^\circ \qquad & k \in N \quad k= 0,1,2,3,\dots \end{array}$

2. Lösung für t:  

$\begin{array}{rcll} 2\cdot \cos{(t)} - 1 &=& 0 \\ 2\cdot \cos{(t)} &=& 1 \\ \cos{(t)} &=& \frac12 \qquad & | \qquad \pm \arccos()\\ t &=& \pm \arccos(\frac12) \\ t &=& \pm 60^\circ \pm k \cdot 360^\circ\\ t_1 &=& 60^\circ \pm k \cdot 360^\circ \qquad & k \in N \quad k= 0,1,2,3,\dots\\ t_2 &=& -60^\circ \pm k \cdot 360^\circ \qquad & k \in N \quad k= 0,1,2,3,\dots\\ \end{array}$

what is the kinetic energy of a van that has a mass of 4200kg and is moving at 40m/s ?

 $\begin{array}{cll} E_k = \frac12 \cdot m \cdot v^2 \\ \text{where }~ m = \text{ mass of object} \\ v = \text{ speed of object} \\\\ 1\ \text{ Joule } = 1\ kg \cdot \frac{ m^2 } { s^2 } \end{array}$

$\begin{array}{rcl} E_k &=& \frac12 \cdot m \cdot v^2 \qquad & m = 4200\ kg \qquad v = 40\ \frac{m}{s} \\ E_k &=& \frac12 \cdot 4200\ kg \cdot ( 40\ \frac{m}{s} )^2 \\ E_k &=& \frac12 \cdot 4200 \cdot 40^2 \frac{kg\cdot m^2}{s^2} \\ E_k &=& 3360000\ \text{ Joule } \\ E_k &=& 3360\ \text{ kJ } \\ E_k &=& 3.360\ \text{ MJ } \end{array}$

An art student uses dilations in all her art. She first plans the art piece on a coordinate grid. Determine the vertices of the image of the triangle with vertices A(1,1) B(2,4) and C(3,9) after a dilation with scale factor 1.5.

I asume the center of enlargement is the centroid.

We have: $\begin{array}{rcll} \vec{A}=\binom{1}{1} \qquad \vec{B} = \binom{2}{4} \qquad \vec{C} = \binom{3}{9} \\ \end{array}$

The centroid of the triangle is: $\vec{c} = \frac13\cdot ( \vec{A}+\vec{B}+\vec{C} ) =\dbinom{\frac{x_a+x_b+x_c}{3}}{ \frac{y_a+y_b+y_c}{3}}= \dbinom{\frac{1+2+3}{3}}{ \frac{1+4+9}{3}} = \dbinom{2}{\frac{14}{3}}$

1. Barycentric coordinates:

$\begin{array}{rcll} \vec{A}-\vec{c} &=& \dbinom{1-2}{1-\frac{14}{3}} = \dbinom{-1}{-\frac{11}{3}}\\\\ \vec{B}-\vec{c} &=& \dbinom{2-2}{4-\frac{14}{3}} = \dbinom{0}{-\frac23}\\\\ \vec{C}-\vec{c} &=& \dbinom{3-2}{9-\frac{14}{3}} = \dbinom{1}{\frac{13}{3}} \end{array}$

2. Scale factor 1.5

$\begin{array}{rcll} \dbinom{-1}{-\frac{11}{3}} \cdot 1.5 &=& \dbinom{-1.5}{-\frac{11}{2}}\\\\ \dbinom{0}{-\frac23}\cdot 1.5 &=& \dbinom{0}{-1}\\\\ \dbinom{1}{\frac{13}{3}}\cdot 1.5 &=& \dbinom{1.5}{\frac{13}{2}} \end{array}$

3. The vertices of the image $+\vec{c}$

$\begin{array}{rcll} \dbinom{-1.5}{-\frac{11}{2}} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{0.5}{-\frac{5}{6}}\\\\ \dbinom{0}{-1} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{2}{\frac{11}{3}}\\\\ \dbinom{1.5}{\frac{13}{2}} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{3.5}{\frac{67}{6}} \end{array}$

The vertices of the image are $\begin{array}{rcll} A' (0.5,-\frac{5}{6} ) \qquad B' ( 2, \frac{11}{3}) \qquad C' ( 3.5, \frac{67}{6} ) \end{array}$

use the web 2.0 calc:

input:

log(10,100)/(1/100)* sqrt3(14661^6)*sqrt3(sqrt(16)*(sqrt(64)+16/sqrt(4)))/( sqrt(log(10,100)^4)/( (1/2)^(-1)) *sqrt( 16^(1/(1/2))) *sin(30)*cos(60)  ) + ( sqrt3( ( 0.5/( sqrt((512*0.5)^2) ) )^(-1) ) )^10 - ( 3* (6*sin(30))^3  + sqrt(sqrt((6*0.5)^8) ) /( cos(60) / (21584*cos(60)) )

= 87051515887

-i²= ?

$\boxed{~ \begin{array}{lcll} z &=& a+b\cdot i \\ \bar{z} &=& a-b\cdot i\\ z\cdot \bar{z} &=& a^2+b^2 \\ && \text{where } \bar{z} \text{ is the complex conjugate of } z \end{array} ~}$

$\begin{array}{rcll} -i^2 &=& 0-i^2 \\ &=& (0+i)(0-i) \\ &=& \underbrace{(0+i)}_{=z}\cdot \underbrace{(0-i)}_{=\bar{z}} \qquad a = 0 \qquad b = 1 \\ -i^2 &=& 0^2+1^2 \\ -i^2 &=& 1 \\ \end{array}$

If the first number in a number line is x^2, and the second number is x^2+1, and the third number is X^2+2, then CALCULATE the sum in terms of "x".

$\begin{array}{lrcll} \text{sum of } & ~ x^2 ,x^2 +1,x^2+2 ~\text{ is } ~3x^2 + 3 \\ \text{factored i asume} & ~ 3(x^2 + 1) \end{array}$

Round the next number to 1) Second decimal and 2), Third decimal. The number is $2.456715 * 10^{12}$.

$\begin{array}{lrcll} 1) \quad \text{ Second decimal } & 2.46 * 10^{12}\\ 2) \quad \text{ Third decimal } & 2.457 * 10^{12}\\ \end{array}$

3/4 of a number is 72.what is 1/3 of the number?

$\begin{array}{rcll} x \cdot \dfrac34 &=& 72 \qquad & | \qquad \cdot \frac43 \\ x &=& 72 \cdot \frac43 \\ x &=& 96 \\ x &=& 96 \qquad & | \qquad \cdot \frac13 \\ x\cdot \frac13 &=& 96 \cdot \frac13 \\ \mathbf{ x\cdot \frac13 } &\mathbf{=}& \mathbf{ 32 }\\ \end{array}$

1/3 of the number is 32