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=

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