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the graph of y = sin x and the line y = 1/6 over the interval [0,2 pi]. Where do the two graphs intersect? Give exact answers (no decimals) in radians

$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& \frac16 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =\frac16 \\&&&\\ x_1 &=& \arcsin{( \frac16 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&\frac16 \\&&&\\ x_1 &=& 9.59406822686 \ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin{( \frac16 )} \\&&&\\x_1 &=& 0.05330037904\cdot \pi = 0.16744807922 \qquad &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( \frac16 )} \\
&&&\\
&&\qquad & x_2 &=& 180\ensurement{^{\circ}}-9.59406822686 \ensurement{^{\circ}}\\
&&&\\
&&\qquad & x_2 &=& 170.405931773\ensurement{^{\circ}}\\
&&&\\
&&\qquad & x_2 &=& 0.94669962096\cdot \pi = 2.97414457437
\end{array}$}}$$

the graph of y = sin x and the line y = (-1)/2 over the interval [0 degree, 360 degree]. Where do the two graphs intersect? Give exact answers in degrees

$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& -\frac12 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =-\frac12 \\&&&\\ x_1 &=& \arcsin{( -\frac12 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&-\frac12 \\&&&\\x_1 &=& -30\ensurement{^{\circ}} + 360\ensurement{^{\circ}} =330\ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin(-\frac12) \\&&&\\&& &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( -\frac12 )} \\&&&\\&&\qquad & x_2 &=& 180\ensurement{^{\circ}} +30\ensurement{^{\circ}} \\&&&\\&&\qquad & x_2 &=& 210\ensurement{^{\circ}}\\\end{array}$}}$$

Above is the graph of y = sin x and the line y = (-1)/2 over the interval [-180^@, 180^@]. Where do the two graphs intersect? Give exact answers in degrees,

$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& -\frac12 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =-\frac12 \\&&&\\ x_1 &=& \arcsin{( -\frac12 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&-\frac12 \\&&&\\
x_1 &=& -30\ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin(-\frac12) \\&&&\\
&& &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( -\frac12 )} \\&&&\\&&\qquad & x_2 &=& 180\ensurement{^{\circ}} +30\ensurement{^{\circ}} - 360 \ensurement{^{\circ}} \\
&&&\\
&&\qquad & x_2 &=& -150\ensurement{^{\circ}}\\
\end{array}$}}$$

the graph of y = cos x and the line y = 1/2 over the interval [0 degree,360 degree]. Where do the two graphs intersect? Give exact answers in degrees

$$\small{\text{$ \begin{array}{rcll}\cos{(x)} &=& \frac12 \\\\\cos{(x)} &=& \frac12 & \qquad | \qquad \pm\arccos{} \\\\x_{1,2} &=& \pm \arccos{(\frac12 )}\\\\x_{1,2} &=& \pm 60 \ensurement{^{\circ}} \\\\x_{1} &=& 60 \ensurement{^{\circ}} \\\\\\
x_2 &=& -60\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\
x_2 &=& 300 \ensurement{^{\circ}}
\end{array}$}}$$

Two airplanes are spotted on a radar map located at the coordinates (2√2 , - √5 ) and (3√2 , 5√5 ). Determine the distance the airplanes are apart using a calculator to state your answer in decimal form to the nearest hundreds. Assume the coordinates are given so that the distance is in units of miles

$$\binom{x_1}{y_1}=\binom{2\sqrt{2}}{-\sqrt{5}} \qquad
\binom{x_2}{y_2}=\binom{3\sqrt{2}}{5\sqrt{5}}$$

$$\boxed{\rm{distance}~&=& \sqrt { (x_2-x_1)^2 + (y_2-y_1)^2 }}\\\\
\small{\text{$
\begin{array}{rcl}
\rm{distance}~&=& \sqrt { ( 3\sqrt{2} - 2\sqrt{2})^2
+ [ 5\sqrt{5} - (-\sqrt{5}) ]^2 } \\\\
\rm{distance}~&=& \sqrt { ( 3\sqrt{2} - 2\sqrt{2})^2
+ [ 5\sqrt{5} + \sqrt{5} ]^2 } \\\\
\rm{distance}~&=& \sqrt { (\sqrt{2})^2 + ( 6\sqrt{5} )^2 } \\\\
\rm{distance}~&=& \sqrt { 2 + 36\cdot 5 } \\\\
\rm{distance}~&=& \sqrt { 182} \\\\
\rm{distance}~&=& 13.4907375632 \\\\
\rm{distance}~&\approx& 13.49 ~\rm{miles}
\end{array}
$}}$$

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the graph of y = cos x and the line y = (-1)/2 over the interval [0,2 pi]. Where do the two graphs intersect? Give exact answers in radians,

$$\small{\text{$
\begin{array}{rcll}
\cos{(x)} &=& -\frac12 \\\\
\cos{(x)} &=& -\frac12 & \qquad | \qquad \pm\arccos{} \\\\
x_{1,2} &=& \pm \arccos{(-\frac12 )}\\\\
x_{1,2} &=& \pm 120 \ensurement{^{\circ}} \\\\
x_{1} &=& 120 \ensurement{^{\circ}} \\\\
x_1 &=& \frac{2}{3}\cdot \pi \\\\\\
x_2 &=& -120\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\
x_2 &=& 240 \ensurement{^{\circ}} \\\\
x_2 &=& \frac{4}{3}\cdot \pi
\end{array}
$}}$$

Wie rechne ich 36 geht in 100 wie oft? die antwort ist wohl 2.78, komme aber nicht auf die formel.

$$\dfrac{100}{36}\\\\
= \frac{ 2\cdot 50 } { 2\cdot 18 } \\\\
= \frac{ \not{2}\cdot 50 } { \not{2}\cdot 18 }\\\\
= \frac{ 50 } { 18 }\\\\
= \frac{ 2\cdot 25 } { 2\cdot 9 } \\\\
= \frac{ \not{2}\cdot 25 } { \not{2}\cdot 9 }\\\\
= \frac{ 25 } { 9 }\\\\
= \frac{ 2\cdot 9 + 7} {9}\\\\
= \frac{2\cdot 9}{9} \frac{ 7} {9}\\\\
= \frac{2\cdot \not{9}}{ \not{9} } + \frac{ 7} {9}\\\\
= 2 + \frac{ 7} {9}$$

$$\\= 2 + 0,77777777778 \\\\
= 2,77777777778 \\\\
=2,78$$

can you explain me step by step how to solve this equations of 2 variables" x+y=170,x-y=20"

$$\small{\text{$
\begin{array}{lrclrclr}
(1) &\qquad x &+& y &=& 170\\
(2) &\qquad x &-& y &=& 20 \\
\hline
(1)+(2) & \qquad x+x &+&y-y &=& 170+20\\
& \qquad x+x &+&0 &=& 170+20\\
& \qquad 2x &+&0 &=& 170+20\\
& \qquad 2x &+&0 &=& 190\\
& \qquad 2x && &=& 190 &\qquad | \qquad :2\\
& \qquad \textcolor[rgb]{1,0,0}{x} &&&\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{95}\\
\hline
(1) &\qquad x &+& y &=& 170\\
&\qquad 95 &+& y &=& 170\\
&\qquad 95 &+& y &=& 170&\qquad | \qquad -95\\
&\qquad & & y &=& 170-95 \\
& \qquad && \textcolor[rgb]{1,0,0}{y} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{75}\\
\end{array}
$}}$$

bove is the graph of y = sin x and the line y = 1/2 over the interval [0,2 pi]. Where do the two graphs intersect? Give exact answers in rad

$$\small{\text{$
\begin{array}{rcl|rcl}
\sin{(x_1)} &=& \frac12 \qquad &
\qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =\frac12 \\
&&&\\
x_1 &=& \arcsin{( \frac12 )} \qquad &
\qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&\frac12 \\
&&&\\
x_1 &=& 30\ensurement{^{\circ}} \qquad &
\qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin{( \frac12 )} \\
&&&\\
x_1 &=& \frac{\pi}{6} \qquad &
\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( \frac12 )} \\
&&&\\
&&\qquad & x_2 &=& 180\ensurement{^{\circ}}-30\ensurement{^{\circ}}\\
&&&\\
&&\qquad & x_2 &=& 150\ensurement{^{\circ}}\\
&&&\\
&&\qquad & x_2 &=& \frac56 \cdot \pi\\
\end{array}
$}}$$