what is the binary representation of 2015 ?

$$\begin{array}{rcrr}
& & \rm{q o u t i e n t} & \rm{r e m a i n d e r} \\
2015 & : 2 = & 1007 & 1 \\
1007 & : 2 = & 503 & 1 \\
503 & : 2 = & 251 & 1 \\
251 & : 2 = & 125 & 1 \\
125 & : 2 = & 62 & 1 \\
62 & : 2 = & 31 & 0 \\
31 & : 2 = & 15 & 1 \\
15 & : 2 = & 7 & 1 \\
7 & : 2 = & 3 & 1 \\
3 & : 2 = & 1 & 1 \\
1 & : 2 = & 0 & \textcolor[rgb]{1,0,0}{1} \\
\end{array}$$

$$2015_2~=~\textcolor[rgb]{1,0,0}{1}~1~1~1~1~0~1~1~1~1~1$$

i need help this this question. in a geometric sequence the second term is 28 and the fifth term is 1792. find the 8th term.

$$Formula: \boxed{~ a_n=a\cdot r^{n-1} ~}$$

$$a_2=28=a\cdot{r}^1 \qquad a_5 = 1792=a\cdot r^4\qquad a_8 = a\cdot r^7$$

$$\dfrac{a_5}{a_2}=\dfrac{a\cdot r^4}{a\cdot r^1} = r^3\\\\\\
\small{\text{$
\begin{array}{rcl}
r^3 &=& \dfrac {1792}{28}\\\\
r &=& \sqrt[3]{\dfrac {1792}{28}}\\\\
r&=&\sqrt[3]{64}\\\\
r &=&4
\end{array}
$}}$$

$$a=\dfrac{28}{r}=\dfrac{28}{4}=7$$

$$\\a_8 = a\cdot r^7\\
a_8= 7\cdot 4^7\\
a_8 = 7\cdot 16384\\
a_8 = 114688$$

The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

$$\\A_{circle}=\pi r^2 \\
A_{square}=(r \sqrt{2})^2=2r^2\\
A_{goblet}=\frac{ A_{circle} - A_{square} }{4} = \frac{\pi r^2 - 2r^2}{4}\\\\
A = 4\cdot
\left[
A_{quadrant}-2\cdot A_{goblet}- A_{triangle}
\right]\\\\
A = 4\cdot
\left[
\frac{\pi r^2 }{4}
-2\cdot \left( \frac{\pi r^2 - 2r^2}{4} \right)
- \frac{ \left( r\sqrt{2}-r \right)^2 }{2}
\right]\\\\
A = \pi r^2 -2\cdot \left(\pi r^2 - 2r^2\right)
- 2\cdot \left( r\sqrt{2}-r \right)^2\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 2\cdot \left( r\sqrt{2}-r \right)^2\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 2\cdot \left(2r^2-2\sqrt{2}r^2 +r^2 \right)\\\\
A = \pi r^2 -2\pi r^2 + 4r^2
- 4r^2 +4\sqrt{2}r^2-2r^2\\\\
A = -\pi r^2 +4\sqrt{2}r^2-2r^2\\\\
A=r^2\cdot (4\sqrt{2}-2-\pi)$$

Sin(3x)=-0.6         x = ?

$$\small{\text{$
\begin{array}{|lrcl|lrcl|}
\hline
I.& &&& II. &\\
&&&& \boxed{\sin{(\alpha)} = \sin{(180\ensurement{^{\circ}}-\alpha)} }\\
\hline
& && && && \\
&\sin(3x) &=& -0.6 && \sin(180\ensurement{^{\circ}} -3x) &=& -0.6 \\
& && && && \\
&3x &=& \arcsin{(-0.6)} && 180\ensurement{^{\circ}} -3x &=& \arcsin{(-0.6)} \\
& && && && \\
&x &=& \dfrac{ \arcsin{(-0.6)}}{3} && 3x &=& 180\ensurement{^{\circ}} -\arcsin{(-0.6)} \\
& && && && \\
&x &=& -12.2899658819\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}} && x &=& \dfrac{ 180\ensurement{^{\circ}} -\arcsin{(-0.6)} }{3} \\
& && && && \\
& && && x &=& 72.2899658819\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}} \\
& && && && \\
\hline
\end{array}
$}}$$

k = 1,2, 3...

y - 8 = 4/3*(x - 8) in standard form

$$\small{\text{$
\begin{array}{rcll}
y - 8 &=& \frac{4}{3}\cdot (x - 8) \quad &| \quad +8\\\\
y &=& \frac{4}{3}\cdot (x - 8) +8\\\\
y &=& \frac{4}{3}\cdot x - \frac{4}{3}\cdot 8 +8\\\\
y &=& \frac{4}{3}\cdot x - \frac{4}{3}\cdot 8 +\frac{3}{3}\cdot8\\\\
y &=& \frac{4}{3}\cdot x - 4\cdot \frac{8}{3} +3\cdot\frac{8}{3}\\\\
y &=& \frac{4}{3}\cdot x - \frac{8}{3}\\\\
\end{array}
$}}$$

$$\begin{array}{rrr}
\mathbf{9} & \mathbf{4} & \mathbf{20} \\
\mathbf{8} & \mathbf{5} & \mathbf{12} \\
\mathbf{7} & \mathbf{6} & \mathbf{\textcolor[rgb]{150,0,0}{?}} \\
\end{array}$$

? = Was ist die Antwort!

$$\begin{array}{rrrrlr}
\mathbf{9} & & \qquad \mathbf{4} & & \qquad \mathbf{20} & \\
& -1 & & +1 & & -8 \\
\mathbf{8} & & \qquad \mathbf{5} & & \qquad \mathbf{12} \\
& -1 & & +1 & & -8 \\
\mathbf{7} & & \qquad \mathbf{6} & & \qquad \mathbf{\textcolor[rgb]{150,0,0}{12-8=4}} \\
\end{array}$$

40x+10x^2=o ?

$$\small{\text{$
\begin{array}{rcl}
40x+10x^2 &=& 0 \\\\
\underbrace{ \textcolor[rgb]{150,0,0}{\mathbf{10x}} }_{=0}\cdot ( \underbrace{ \textcolor[rgb]{0,0,150}{\mathbf{4+x}} }_{=0} ) &=& \mathbf{ 0 } \\\\
\end{array}
$}}\\\\
\textcolor[rgb]{150,0,0}{\mathbf{10x}} = 0 \qquad | \qquad : 10 \\
\boxed{ \textcolor[rgb]{150,0,0}{\mathbf{x}} = 0}\\\\
\textcolor[rgb]{0,0,150}{\mathbf{4+x}} = 0 \qquad | \qquad -4 \\
\boxed{ \textcolor[rgb]{0,0,150}{\mathbf{x}} = -4}$$

$$\\\small{\text{check: }}\\
\small{\text{
$
\begin{array}{rcl}
40\cdot \textcolor[rgb]{150,0,0}{0} + 10\cdot \textcolor[rgb]{150,0,0}{0}^2 &=& 0 \\
0+0 &=& 0 \small{\text{ okay}}\\
\end{array}
$}}\\\\
\small{\text{
$
\begin{array}{rcl}
40\cdot (\textcolor[rgb]{0,0,150}{-4}) + 10\cdot (\textcolor[rgb]{0,0,150}{-4})^2 &=& 0 \\
-160 + 10\cdot 16 &=& 0\\
-160 + 160 &=& 0 \\
0 &=& 0 \small{\text{ okay}}
\end{array}
$}}$$

change 1/3 into an equivalent fraction with a denominator of 24

$$\mathbf{ \dfrac{1} {3} \cdot \dfrac{ \textcolor[rgb]{1,0,0}{8} } { \textcolor[rgb]{1,0,0}{8} } = \dfrac{1 \cdot 8} {3\cdot 8} =\dfrac{8} {24} }$$

What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2) ?

$$\small{\text{$
\begin{array}{rcl}
1-5(x^3+x^2) &=& -5x^2-4(x^3+2) \\\\
1-5(x^3+x^2) + 5x^2 +4(x^3+2) &=& 0\\\\
1 - 5x^3 - 5x^2 + 5x^2 + 4x^3 + 8 &=& 0 \\\\
-5x^3 - 5x^2 + 5x^2 + 4x^3 + 9 &=& 0 \\\\
-5x^3 \underbrace{- 5x^2 + 5x^2}_{=0} + 4x^3 + 9 &=& 0 \\\\
-5x^3 + 4x^3 + 9 &=& 0 \\\\
-x^3 + 9 &=& 0 \\\\
9 &=& x^3 \\\\
x^3 &=& 9 \qquad | \qquad \sqrt[3]{}\\\\
x &=& \sqrt[3]{9}\\\\
x &=& 2.0800838231
\end{array}
$}}\\\\
\mathbf{ x = 2.0800838231 }$$

4.) sin^4 alpha−cos^4 alpha=

$$\small{\text{$
\begin{array}{rcl}
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) } \\
&=& - (~ \cos^4{ ( \alpha ) } - \sin^4{ (\alpha) } ~)\\
&=& - (~
( \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } ) \cdot
( \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } )
~)\\\\
&&$Formeln:\\\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } = \cos{(2\cdot\alpha)} }}\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } = 1 }}\\\\
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) }&=&-\cos{(2\cdot\alpha)}
\end{array}
$}}$$