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?