Mitglied: heureka

$-1.7 \mathrm{~rad~} \cdot \frac{360\ensurement{^{\circ}}}{2\pi \mathrm{~rad~}}\\\\ =-1.7 \cdot \frac{180\ensurement{^{\circ}}}{\pi} \\\\ = -97.4028251722\ensurement{^{\circ}}\\ \mathrm{ }\\ \hline \\$

heureka22.06.2015

+26397

Die Fläche A eines Dreiecks mit Determinante berechnen:

Gegeben sind drei Punkte :

$\small{\text{$ S_1(x_1|y_1) = (4|-3) \qquad S_2 (x_2|y_2)=(-2|0) \qquad S_3(x_3|y_3)=(-1|2) $}}$

$\small{\text{$ \begin{array}{|l|l|l|} \hline & x & y \\ \hline S_1 & x_1=4 & y_1=-3 \\ S_2 & x_2=-2 & y_2= 0 \\ S_3 & x_3=-1 & y_3=2 \\ \hline \end{array} $}}\\\\$

$\small{\text{$ A = \dfrac{1}{2}\cdot \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} $ }}\\\\$

Auflösung nach der 1. Spalte:

$\small{\text{$ \begin{array}{lll} A &=& \dfrac{1}{2}\cdot \begin{vmatrix} \textcolor[rgb]{1,0,0}{x_1} & y_1 & 1 \\ \textcolor[rgb]{1,0,0}{x_2} & y_2 & 1 \\ \textcolor[rgb]{1,0,0}{x_3} & y_3 & 1 \\ \end{vmatrix} =\frac12\cdot \left( \textcolor[rgb]{1,0,0}{x_1}\cdot \begin{vmatrix} y_2 & 1 \\ y_3 & 1 \\ \end{vmatrix} -\textcolor[rgb]{1,0,0}{x_2}\cdot \begin{vmatrix} y_1 & 1 \\ y_3 & 1 \\ \end{vmatrix} +\textcolor[rgb]{1,0,0}{x_3}\cdot \begin{vmatrix} y_1 & 1 \\ y_2 & 1 \\ \end{vmatrix} \right)\\\\ &=& \frac12\cdot \left(~~ [~x_1\cdot (y_2-y_3)~]-[~x_2\cdot (y_1-y_3)~]+[~x_3\cdot (y_1-y_2)~] ~~\right)\\\\ &=& \mathbf{ \frac12\cdot \left(~~ [~x_1\cdot (y_2-y_3)~]+[~x_2\cdot (y_3-y_1)~]+[~x_3\cdot (y_1-y_2)~] ~~\right) } \\ \\ &=& \frac12\cdot \left(~~ [~4\cdot (0-2)~]-[~2\cdot (2-(-3))~]-[~1\cdot (-3-0)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ [~4\cdot (-2)~]-[~2\cdot (2+3))~]-[~1\cdot (-3)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ -8-10+3 ~~\right)\\\\ &=& -7,5\\\\ A&=& 7,5 \end{array} $ }}\\\\$

Auflösung nach der 2. Spalte:

$\small{\text{$ \begin{array}{lll} A &=& \dfrac{1}{2}\cdot \begin{vmatrix} x_1 & \textcolor[rgb]{1,0,0}{y_1} & 1 \\ x_2 & \textcolor[rgb]{1,0,0}{y_2} & 1 \\ x_3 & \textcolor[rgb]{1,0,0}{y_3} & 1 \\ \end{vmatrix} =\frac12\cdot \left( -\textcolor[rgb]{1,0,0}{y_1}\cdot \begin{vmatrix} x_2 & 1 \\ x_3 & 1 \\ \end{vmatrix} +\textcolor[rgb]{1,0,0}{y_2}\cdot \begin{vmatrix} x_1 & 1 \\ x_3 & 1 \\ \end{vmatrix} -\textcolor[rgb]{1,0,0}{y_3}\cdot \begin{vmatrix} x_1 & 1 \\ x_2 & 1 \\ \end{vmatrix} \right)\\\\ &=& \frac12\cdot \left(~~ -[~y_1\cdot (x_2-x_3)~] +[~y_2\cdot (x_1-x_3)~] -[~y_3\cdot (x_1-x_2)~] ~~\right)\\\\ &=& \mathbf{ \frac12\cdot \left(~~ [~y_1\cdot (x_3-x_2)~] +[~y_2\cdot (x_1-x_3)~] +[~y_3\cdot (x_2-x_1)~] ~~\right) }\\\\ &=& \frac12\cdot \left(~~ [~-3\cdot (-1-(-2))~] +[~0\cdot (4-(-1))~] +[~2\cdot (-2-4)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ [~-3\cdot 1~] +[~2\cdot (-6)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ -3 - 12 ~~\right)\\\\ &=& \frac12\cdot \left(~~ -15~~\right)\\\\ &=& -7,5\\\\ A&=& 7,5 \end{array} $ }}\\\\$

heureka22.06.2015

+26397

Vielen Dank radix

Heureka

heureka19.06.2015

+26397

heureka19.06.2015

+26397

Beispiel: Wie löse ich das nach x auf?

$\mathbf{ 50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) } }\\\\\\ \begin{array}{rcl} 50 &=& -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) } \qquad | \qquad : -10 \\\\ -5 &=& \log_{10}{ \left[10^{(-5,7)} + 10^{(\frac{x}{10})} \right] } \\\\ -5 &=& \log_{10}{ \left[10^{(-5,7)} + 10^{(\frac{x}{10})} \right] } \qquad | \qquad : 10^x \\\\ 10^{-5} &=& 10^{(-5,7)} + 10^{(\frac{x}{10})} \\\\ 10^{(\frac{x}{10})} &=&10^{(-5)} - 10^{(-5,7)} \qquad | \qquad \log_{10} \\\\ \log_{10}{ (10^{(\frac{x}{10})}) } &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) }\\\\ \frac{x}{10}\cdot \log_{10}{ (10) } &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \qquad | \qquad \log_{10} =1 \\\\ \frac{x}{10} &=&\log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \\\\ \end{array}$

$\begin{array}{rcl} x &=& 10 \cdot \log_{10}{ (10^{(-5)} - 10^{(-5,7)}) } \\\\ x &=& 10 \cdot \log_{10}{ ( 0,00001 - 0,00000199526 ) } \\\\ x &=& 10 \cdot \log_{10}{ ( 0,00000800474) } \\\\ x &=& 10 \cdot -5,09665289533 \\\\ \mathbf{x} &\mathbf{=}&\mathbf{-50,9665289533} \end{array}$

Probe:

$50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{x}{10})} \right) } \\\\ 50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{(\frac{-50,9665289533}{10})} \right) } \\\\ 50= -10 \cdot \log_{10}{ \left( 10^{(-5,7)} + 10^{( -5,09665289533 )} \right) } \\\\ 50= -10 \cdot \log_{10}{ \left( 0,00000199526 + 0,00000800474 \right) } \\\\ 50= -10 \cdot \log_{10}{ \left( 0,00001\right) } \\\\ 50= -10 \cdot (-5) \\\\ 50= 50 \qquad \mathrm{okay} \\\\$

heureka19.06.2015

+26397

+10

Integrate (cos^-1x)^2 /(1-x^2)^1/2

$\small{\text{$ \begin{array}{lrcl} & \int{ \dfrac{[ \cos^{-1}{(x)} ]^2 } { (1-x^2)^{\frac{1}{2}} } \ dx } = \mathbf{ \int{ [\arccos{(x)}]^2 \cdot \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \ dx } }\\\\ \end{array} $}}\\\\$

$\small{\text{$ \boxed{ \begin{array}{lrcl} \mathrm{Formula:~} & y &=& \dfrac{ f(x)^{n+1} } {n+1} \\\\ & y' &=& \left(\dfrac{n+1}{n+1}\right) \cdot f(x)^{n+1-1}\cdot f'(x) \\\\ & y' &=& f(x)^{n}\cdot f'(x) \\\\ &\mathbf{ \int{ f(x)^{n}\cdot f'(x) \ dx } } & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} } \end{array}} $}}$

$\small{\text{$ \begin{array}{lrcl} &\mathbf{ \int{ f(x)^{n}\cdot f'(x) \ dx } } & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} } \qquad f(x) = \arccos{(x)} \qquad f'(x)= -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }\\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( -\dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & \dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & -\int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & \dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & \int{ \left[ \arccos{(x)} \right] ^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{1}{3} \cdot \left[ \arccos{(x)} \right]^{3} \\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{1}{3} \cdot \left[\cos^{-1}{(x)}\right]^3 \\\\ \end{array} $}}$

heureka19.06.2015

+26397

Equation X^6-X^3=2

$\small{\text{$ \begin{array}{rcl} x^6-x^3&=&2 \qquad \mathrm{we ~substitute~and ~set~} z = x^3 \\ \mathrm{then~we~have~} \quad z^2-z &=& 2 \\ z^2-z -2 &=& 0\\ z_{1,2} &=& \dfrac{1\pm \sqrt{1-4\cdot(-2)}}{2} \\ z_{1,2} &=& \dfrac{1\pm \sqrt{9}}{2} \\\\ z_{1,2} &=& \dfrac{1\pm 3 }{2} \\\\ z_1 &=& \frac{1 + 3 }{2} = 2 \\\\ z_2 &=& \frac{1 - 3 }{2} = - 1\\\\\\ x_1 & =& \sqrt[3]{z_1}= \sqrt[3]{2}= 1.25992104989\\\\ \mathbf{x_1} & \mathbf{=}& \mathbf{1.25992104989}\\\\\\ x_2 &=& \sqrt[3]{z_2}= \sqrt[3]{-1}= -1\\\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1}\\\\ \end{array} $}}$

heureka19.06.2015

+26397

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b oder y - mx = b

$\small{\text{$ \begin{array}{llll} (1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\ (2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\ (3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\ \end{array} $}}$

I. Berechnung der Schnittpunkte:

1. Schnitt

$\small{\text{$S_1$}}$ Gerade (1) mit Gerade (2 ):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot \frac{1}{2} - (-1)\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { \frac{1}{2} + 1 } { \frac{1}{2} - 1 } =\dfrac { \frac{3}{2} } { -\frac{1}{2} } = -3\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot (-1) - 1\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { -1-1 } { \frac{1}{2} - 1 } =\dfrac { -2 } { -\frac{1}{2} } = 4 \end{array} $}}$

2. Schnitt

$\small{\text{$S_2$}}$ Gerade (2) mit Gerade (3):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { -1\cdot 2 - 4 \cdot(- \frac{1}{2}) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { -2+2} { -2 - \frac{1}{2} } =\dfrac { 0 } { -\frac{5}{2} } = 0\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { 1\cdot 4 - 1\cdot (- 1) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { 4+1 } { -2 - \frac{1}{2} } =\dfrac { 5 } { -\frac{5}{2} } = -2 \end{array} $}}$

3. Schnitt

$\small{\text{$S_3$}}$ Gerade (3) mit Gerade (1):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 4\cdot 1 - 1 \cdot (-2) } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 4+2} { 1+2 } =\dfrac { 6 } { 3 } = 2\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 1\cdot 1 - 1\cdot 4 } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 1-4 } { 1+2 } =\dfrac { -3 } { 3 } = -1 \end{array} $}}$

Die 3 Schnittpunkte zusammengefasst:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}$

Die Strecke von

$\small{\text{$S_1 $ nach $ S_2$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$

Die Strecke von

$\small{\text{$S_2 $ nach $ S_3$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$

Die Strecke von

$\small{\text{$S_3 $ nach $ S_1$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt:

$\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$

Die Fläche des Dreiecks nach Heron:

$\small{\text{$ \boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad s = (a + b + c)/2 ~~} $}}\\\\ \small{\text{$ \begin{array}{lcl} a= \overline{S_1S_2} = 6,70820393250\\ b= \overline{S_2S_3} = 2,23606797750\\ c= \overline{S_3S_1} = 7,07106781187 \\\\ s = \dfrac{16,0153397219}{2}= 8,00766986093\\ \end{array} $}}\\\\\\ \small{\text{$ \begin{array}{lcl} A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\ A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\ A =\sqrt{ 56.2499999994 }\\ A=7,5 \end{array} $}}\\\\$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

Probe der Fläche:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}\\\\ \small{\text{$ \boxed{ ~~ A = \dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ x_{s_1} & x_{s_2} & x_{s_3} \\ y_{s_1} & y_{s_2} & y_{s_3} \\ \end{vmatrix} ~~ } $}}\\\\ \small{\text{$ \begin{array}{rcl} A &=&\dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ 4 & -2 & -1 \\ -3 & 0 & 2 \\ \end{vmatrix} \\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ -4 + 3 - 6 \cdot 1 - 8]\\\\ &=&\frac{1}{2} \cdot [ -15 ]\\\\ &=& - 7,5 \\\\ |A| &=& 7,5 \qquad \mathrm{okay} \end{array} $}}$

heureka19.06.2015

+26397

+15

8(9s+15)=6(4s-4)

$\small{\text{$ \begin{array}{rcl} 8(9s+15) &=& 6(4s-4) \\ 8\cdot 9s+8\cdot 15 &=& 6\cdot 4s - 6\cdot 4 \\ 72s + 120 &=& 24s - 24 \quad | \quad - 24s\\ 72s-24s + 120 &=& -24 \\ 48s +120 &=& - 24 \\ 48s +120 &=& - 24 \quad | \quad - 120\\ 48s &=& - 24- 120\\ 48s &=& -144\\ 48s &=& -144 \quad | \quad :48\\\\ s & =& -\dfrac{144}{48} \\\\ \mathbf{s} & \mathbf{=} &\mathbf{ - 3 } \end{array} $}}\\\\\\\\ \small{\text{$\mathbf{check:} \begin{array}{rcl} 8\cdot [9(-3)+15] &=& 6\cdot [4(-3)-4]\\ 8\cdot (-27+15) &=& 6\cdot (-12-4)\\ 8\cdot (-12) &=& 6\cdot (-16) \\ -96 &=& -96 \qquad \mathrm{okay} \end{array} $}}$

heureka19.06.2015

+26397

What is the slope of the line through the points (-2, -5) and (1, -7)

The formula of a line is y=mx+b

and m is the slop.

We have point 1 with x = -2 and y = -5 ... so we have -5 = m*(-2) + b or -5 = -2m+b

We have point 2 with x = 1 and y = -7... so we have -7 = m*(1) + b or -7= m+b

Now we subtract the equation 2 of the equation 1: -7 -(-5) = m+b - (-2m+b)

so we have -7+5 = m+b+2m-b

or -2 = 3m

and the slope m is

$\small{\text{$-\dfrac{2}{3}$}}$

heureka19.06.2015

Benutzername	heureka
Punkte	26397
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