heureka

What are the coordinate Points on the graph and the Intersection

x-Intersection = 0

y-Intersection = 0

What are the Points on the graph and the Intersection?

x- Intersection ( 0.5 ,  0 )

y- Intersection ( 0 ,  -1 )

Draw the graph of the relation defined by y=2x-1

 Draw the graph of the relation defined by:  3y^2 = x

an=(n+1)/(2n+3) find the " n " in this que
an-an-1=1/99

$$a_n=\dfrac{n+1}{2n+3} \qquad \text{ and } \qquad a_n-a_{n-1}=\dfrac{1}{99}$$

$$\begin{array}{rcl} \frac{n+1}{2n+3}-\frac{(n-1)+1}{2(n-1)+3} &=& \frac{1}{99} \\\\ \frac{n+1}{2n+3}-\frac{n}{2n+1} &=& \frac{1}{99} \\\\ \frac{(2n+1)(n+1)-n(2n+3)}{(2n+3)(2n+1)} &=& \frac{1}{99} \\\\ (2n+3)(2n+1) &=& 99[(2n+1)(n+1)-n(2n+3)]\\\\ (2n+3)(2n+1) &=& 99(2n+1)(n+1)-99n(2n+3)]\\\\ 4n^2+2n+6n+3 &=& 99(2n^2+2n+n+1)-99(2n^2+3n)\\\\ 4n^2+8n+3 &=& 99(2n^2+3n) +99 -99(2n^2+3n)\\\\ 4n^2+8n+3 &=& 99 \\\\ 4n^2+8n-96 &=& 0 \quad | \quad :4 \\\\ n^2 +2n-24 &=& 0 \\\\ n_{1,2}=\frac{-2\pm\sqrt{4-4(-24)} } {2} \\\\ n_{1,2}=\frac{-2\pm\sqrt{100}} {2} \\\\ n &=& \frac{-2+10}{2} \\\\ n &=& \frac{8}{2} \\\\ n &=& 4 \end{array}$$

Draw the graph of relation defined by: 3y^2 = x 

(a-b)/(c-d)=e wie stelle ich nach d um? bitte mit lösungsweg

$$\small{\text{ $ \begin{array}{rcl} \dfrac{a-b}{c-d} &=& \dfrac{e}{1} \quad | \quad \updownarrow\\\\ \dfrac{c-d}{a-b} &=& \dfrac{1}{e} \quad | \quad *(a-b)\\\\ c-d &=&\dfrac{a-b}{e} \quad | \quad *(-1)\\\\ -c+d &=& -\dfrac{a-b}{e} \quad | \quad +c\\\\ d &=& c-\dfrac{a-b}{e} \end{array} $ }}$$

Wie geht es nun weiter?

$$\frac{3}{6}\times x + \frac{2} {6} \times x =5$$

$$\\\small{\text{ Jetzt gleiche Teile ausklammern: $\frac{3}{\textcolor[rgb]{1,0,0}{6}}\times \textcolor[rgb]{1,0,0}{x} + \frac{2} {\textcolor[rgb]{1,0,0}{6}} \times \textcolor[rgb]{1,0,0}{x} =5 $ }} \\\\ \small{\text{ $ \begin{array}{rclc} \dfrac{x}{6}\left( 3+2\right)&=&5\\\\ \dfrac{x}{6}\left(5\right)&=&5\\\\ \dfrac{x}{6}5&=&5 &\quad | \quad :5\\\\ \dfrac{x}{6}&=&1 &\quad | \quad *6\\\\ x&=&6 \end{array} $ }}$$

Hi Melody,

use \limits.

Example:

\sum \limits_{n=1}^{\infty} : $$\sum \limits_{n=1}^{\infty}$$

What is 19 1/6 - 4 1/2 - 4 1/4 ?

$$\begin{array}{rcl} 19 \frac{1} {6} - 4 \frac{1} {2} - 4 \frac{1} {4} \\ &=& 19+ \frac{1} {6} - 4 -\frac{1} {2} - 4 -\frac{1} {4} \\\\ &=& 19-4-4 +\frac{1} {6} -\frac{1} {2} -\frac{1} {4} \\\\ &=& 11 +\frac{1} {6} -\frac{1} {2} *\frac{3}{3} -\frac{1} {4} \\\\ &=& 11 +\frac{1-3} {6} -\frac{1} {4} \\\\ &=& 11 -\frac{2} {6} -\frac{1} {4} \\\\ &=& 11 -\frac{1} {3} -\frac{1} {4} \\ \\ &=& 11 -(\frac{1} {3} + \frac{1} {4}) \\ \\ &=& 11 -(\frac{1} {3}*\frac{4}{4} + \frac{1} {4}*\frac{3}{3} ) \\ \\ &=& 11 -\frac{7} {12}\\ \\ &=& 10+1 -\frac{7} {12}\\ \\ &=& 10+\frac{12}{12} -\frac{7} {12}\\ \\ &=& 10 +\frac{12-7} {12}\\ \\ &=& 10 +\frac{5} {12}\\ \\ &=& 10 \frac{5} {12}\\ \\ \end{array}$$