heureka

√((0.3071-〖0.0929〗^2)/700)  = 0.0206490951029683

What is the y intercept of the line AB ?

$$\small{\text{ $ \begin{array}{rcl} \dfrac{y_{\rm{intercept}} - y_a } { x_a} &=& \dfrac{ y_a - y_b } { x_b-x_a} \\ \\ y_{ \rm{intercept} } &=& y_a - x_a \cdot { \dfrac{ y_b - y_a } { x_b-x_a} } \end{array} $ }}\\\\\\ \text{Example}: \small{\text{ $ \begin{array}{rcl} x_a=1 & y_a = 2 &\\ x_b=2 & y_b = 1 &\\ && y_{ \rm{intercept} } = 2 - 1 \cdot { \dfrac{ 1-2 } { 2-1} } = 2 + 1 \cdot { \dfrac{ 1 } { 1} } =2 + 1 = 3 \end{array} $}}$$

sin(3pi/2/2) ?

$$\sin_{(rad)} \left(\dfrac{ \dfrac{3\cdot \pi}{2} }{2} \right) =\sin_{(rad)} \left( \dfrac{3\cdot \pi}{4} \right)=\frac{1}{2}\sqrt{2}\\\\ \dfrac{3\cdot \pi}{4}\; \rm{rad} \equiv 135 \ensurement{^{\circ}}\\\\ \sin_{(360 \ensurement{^{\circ}} )} \left(135 \ensurement{^{\circ}} \right) = \frac{1}{2}\sqrt{2}\\\\$$

(°)	(rad)

how many inches is 1.3 feet = 15.6[in]

Rotate the axis to eliminate the xy term for 2x^2-72xy+23y^2-80x-60y-125=0

$$\small{\text{$ \begin{array}{ccccccccccc} \textcolor[rgb]{1,0,0}{2}\cdot x^2 &+& \textcolor[rgb]{1,0,0}{23} \cdot y^2 &-& \textcolor[rgb]{1,0,0}{72} \cdot x \cdot y &-& \textcolor[rgb]{1,0,0}{80} \cdot x &-& \textcolor[rgb]{1,0,0}{60} \cdot y &=& 125\\ \textcolor[rgb]{1,0,0}{a} \cdot x^2 &+& \textcolor[rgb]{1,0,0}{b} \cdot y^2 &+& \textcolor[rgb]{1,0,0}{2\cdot c} \cdot x \cdot y &+& \textcolor[rgb]{1,0,0}{2\cdot d} \cdot x &+& \textcolor[rgb]{1,0,0}{2\cdot e} \cdot y &=& 125\\ \end{array} $}}$$

$$\\ \small{\text{$ \begin{array}{rcr} a &=& 2 \\ b &=& 23 \\ c &=& -36 \\ d &=& -40 \\ e &=& -30 \end{array} $}}$$

$$\boxed{ \small{\text{$ \tan{(\varphi)}=\dfrac{ \sqrt{ (a-b)^2+4\cdot c^2 }-(a-b) } {2\cdot c} $}} }\\\\ \small{\text{$ \tan{(\varphi)}=\dfrac{ \sqrt{ (2-23)^2+4\cdot (-36)^2 }-(2-23) } {2\cdot (-36) } = \dfrac{ \sqrt{ (-21)^2+4\cdot (-36)^2 }+21 } { -72 } $}}\\\\ \small{\text{$ \tan{(\varphi)} = \dfrac{ \sqrt{ 441+4\cdot 1296 }+21 } { -72 } $}}\\\\ \small{\text{$ \tan{(\varphi)} = \dfrac{ 96 } { -72 } $}}\\\\ \small{\text{$ \tan{(\varphi)} = - \dfrac{ 4 } { 3 } $}}\\$$

$$\boxed{ \small{\text{ $ \sin{ (\varphi) } =\dfrac { \tan{(\varphi)} } { \sqrt{1+\tan{(\varphi)}^2 } } \qquad \cos{ (\varphi) } =\dfrac { 1 } { \sqrt{1+\tan{(\varphi)}^2 } } $}} }\\\\ \small{\text{ $ \sin{ (\varphi) } =\dfrac { -\frac{4}{3} } { \sqrt{1+ (-\frac{4}{3} )^2 } } \qquad \cos{ (\varphi) } =\dfrac { 1 } { \sqrt{1+ (-\frac{4}{3} )^2 } } $}}\\\ \small{\text{ $ \sin{ (\varphi) } =-\dfrac { 4 } { 5 } \qquad \cos{ (\varphi) }=\dfrac { 3 } { 5 } $}}$$

Rotation: $$\boxed{ \small{\text{ $ \begin{array}{rcl} x &=& x'\cdot \cos{(\varphi)}-y'\cdot \sin{(\varphi)} \\ y &=& x'\cdot \sin{(\varphi)}+y'\cdot \cos{(\varphi)} \end{array} $}}}$$

$$\small{\text{ $ \begin{array}{rcl} x &=& x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} \\ y &=& -x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5} \end{array} $}}$$

We substitute:

$$\\\small{\text{ $ x^2 = \left( x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} \right)^2 = \frac{9}{25}x'^2 + 2 \frac{12}{25}x'y'+\frac{16}{25}y'^2 $}}\\ \small{\text{ $ y^2 = \left( -x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5} \right)^2 =\frac{16}{25}x'^2 - 2 \frac{12}{25}x'y'+\frac{9}{25}y'^2 $}}\\ \small{\text{ $ xy =\left( x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} \right)\cdot \left( -x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5} \right) = -\frac{12}{25}x'^2 + \frac{12}{25}y'^2 - \frac{7} {25} \cdot x'y' $}}\\$$

$$\small{\text{$ \begin{array}{rcl} 2\cdot (\frac{9}{25}x'^2 + 2 \frac{12}{25}x'y'+\frac{16}{25}y'^2) \\\\ + 23 \cdot (\frac{16}{25}x'^2 - 2 \frac{12}{25}x'y'+\frac{9}{25}y'^2) \\\\ -72 \cdot (-\frac{12}{25}x'^2 +\frac{12}{25}y'^2 - \frac{7}{25} \cdot x'y') \\\\ - 80 \cdot ( x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} ) \\\\ -60 \cdot (-x'\cdot \dfrac{4}{5}-y'\cdot \dfrac{3}{5}) &=& 125 \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} \frac{18}{25}x'^2 +\frac{32}{25}y'^2 + \frac{368}{25}x'^2 +\frac{207}{25}y'^2+\frac{864}{25}x'^2 -\frac{864}{25}y'^2 -\frac{240}{5} x' - \frac{320}{5} y' +\frac{240}{5} x'+\frac{180}{5} y' &=& 125\\\\ 50\cdot x'^2 -25\cdot y'^2 +0 \cdot x' -28\cdot y' &=& 125\\\\ 50\cdot x'^2 -25\cdot y'^2 -28\cdot y' &=& 125 \end{array} $}}$$

5∙8^(x+1)=〖16〗^(x-1)

$$\begin{array}{rcl} 5\cdot 8^{(x+1)} &=& 16^{(x-1)} \\ 5\cdot 8^x\cdot 8^1 &=& 16^x \cdot 16^{-1} \\ 40\cdot 8^x &=& \dfrac{16^x}{16}\\\\ 40\cdot 16 &=& \dfrac{16^x}{8^x}\\\\ \dfrac{16^x}{8^x} &=& 40\cdot 16 \\\\ \dfrac{16^x}{8^x} &=& 640\\\\ \left( \dfrac{16}{8} \right)^x &=& 640\\\\ 2^x &=& 640 \quad | \quad \log_{10}\\\\ \log_{10}(2^x) &=& \log_{10}(640) \quad | \quad \log_{10}\\\\ x\cdot \log_{10}(2) &=& \log_{10}(640) \\\\ x &=& \dfrac{ \log_{10}(640) } { \log_{10}(2) }\\\\ x &=& \dfrac{ 2.80617997398} {0.30102999566}\\\\ x &=& 9.32192809489 \end{array}$$

Wie löse ich die folgenden Potenzgleichungen? 

3∙5^2x=7^(x+4)

$$\begin{array}{rcl} 3\cdot5^{2x} &=& 7^{(x+4)}\\ 3\cdot 5^{2x} &=& 7^x\cdot 7^4\\ \dfrac{5^{2x} }{7^x} &=& \dfrac{7^4}{3} \quad | \quad \log_{10}\\\\ \log_{10}\left( \dfrac{5^{2x} }{7^x} \right) &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ \log_{10}\left( 5^{2x} \right) - \log_{10}\left( 7^x \right) &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ 2x\cdot \log_{10}\left( 5\right) - x \cdot \log_{10}\left( 7\right) &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ x \cdot \left[ 2\cdot \log_{10}( 5 )- \log_{10}( 7 ) \right] &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ x \cdot \left[ \log_{10}( 5^2 )- \log_{10}( 7 ) \right] &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ x \cdot \left[ \log_{10} \left( \dfrac{5^2}{ 7 }\right) \right] &=& \log_{10}\left( \dfrac{7^4}{3} \right) \\ \\ x &=& \dfrac{ \log_{10} \left( \dfrac{7^4}{ 3 }\right) } { \log_{10}\left( \dfrac{5^2}{7} \right) } \\ \\ x &=& \dfrac{ \log_{10} \left( 800.333333333 \right) } { \log_{10}\left( 3.57142857143 \right) } \\ \\ x &=& \dfrac{ 2.90327090534 } { 0.55284196866 } \\ \\ x &=& 5.25153854073 \end{array}$$

36x^2+36y^2-24x-12y=31

$\begin{array}{rcll} 36x^2+36y^2-24x-12y &=& 31 \\ (36x^2-24x)+(36y^2-12y) &=& 31 \quad & | \quad : 36 \\\\ (x^2-\frac{2}{3}x)+(y^2-\frac{1}{3}y) &=& \frac{31}{36} \\ \\ (x-\frac{2}{6})^2-\frac{4}{36}+(y-\frac{1}{6})^2-\frac{1}{36} &=& \frac{31}{36} \\ \\ (x-\frac{1}{3})^2 +(y-\frac{1}{6})^2&=& \frac{31}{36} +\frac{4}{36}+\frac{1}{36} \\ \\ (x-\frac{1}{3})^2 +(y-\frac{1}{6})^2&=& 1 \\ \\ \end{array}$

36X^2+36y^2-24x-12y=31

This is a circle with center

$\left(\dfrac{1}{3} ,\; \dfrac{1}{6}\right)$

and the radius = 1

How to find x in the question Sinh(x-3)=1 even if i expand it into e terms i cant seem to get it

I.

$$\small{\text{ $ \begin{array}{rcl} \sinh{(x-3)} &=& 1 \quad | \quad \sinh^{-1} \\ x-3 &=& \sinh^{-1}(1) \\ x &=& 3 + \sinh^{-1}(1) \\ x &=& 3 + 0.88137358702\\ x &=& 3.88137358702\\ \end{array} $ }}$$

II.

$$\boxed{ \small{\text{ $ \sinh(x)=\frac{1}{2}\cdot \left( e^x - e^{-x}\right) \qquad \sinh(x-3)=\frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right) $ }}}$$

$$\small{\text{$ \begin{array}{rcl} \sinh(x-3) &=& 1\\ \frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right) &=& 1\\ e^{x-3} - e^{-(x-3)} &=& 2 \\ e^{x-3} - \frac{1}{e^{x-3}} &=& 2 \quad | \quad u = e^{x-3}\\ u-\frac{1}{u} &=& 2 \quad | \quad \cdot u\\ u^2-1 &=& 2\cdot u \\ u^2 -2\cdot u - 1 &=& 0 \\ u_{1,2} &=& \frac{ 2\pm \sqrt{4-4\cdot(-1) } }{2}\\ u_{1,2} &=& \frac{ 2\pm \sqrt{2\cdot4 } }{2}\\ u_{1,2} &=& \frac{ 2\pm 2\cdot\sqrt{ 2 } }{2}\\ u_{1,2} &=& 1\pm \sqrt{ 2 }\\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcll} u &=& e^{x-3} \quad & | \quad \ln\\ \ln(u) &=& (x-3) \cdot \ln(e) \quad & | \quad \ln(e) = 1 \\ \ln(u) &=& x-3 \\ \boxed{x = 3 + \ln(u)} \end{array} $ }}$$

$$\small{\text{$ \begin{array}{rcl|rcl} u_1 &=& 1+\sqrt{2} \quad & \quad u_2 &=& 1-\sqrt{2}\\ x_1 &=& 3 + \ln(u_1) \quad & \quad x_2 &=&3 + \ln(u_2) \\ x_1 &=& 3 + \ln(1+\sqrt(2)) \quad & \quad x_2 &=&3 + \ln(\underbrace{1-\sqrt(2)}_{<0\text{ no solution!}}) \\ x &=& 3 + \ln(1+\sqrt(2)) \\ x &=& 3 + \ln( 2.41421356237 ) \\ x &=& 3 + 0.88137358702\\ x &=& 3.88137358702\\ \end{array} $ }}$$

How many miles are in a kilometer?