heureka

what is 0.4 recurring as a fraction

$\begin{array}{|rcll|} \hline 10\cdot 0.\bar{4} &=& 4.\bar{4} \\ 1\cdot 0.\bar{4} &=& 0.\bar{4} \\\\ 10\cdot 0.\bar{4} - 1\cdot 0.\bar{4} &=& 4.\bar{4}-0.\bar{4} \\ 9\cdot 0.\bar{4} &=& 4 \\ 0.\bar{4} &=& \frac{4}{9} \\ \mathbf{ 0.4\ recurring } &\mathbf{=}& \mathbf{ \frac{4}{9} }\\ \hline \end{array}$

11 17 23 29 find, in terms of n, an expression for the nth term of this arithmatic sequence

$\begin{array}{|rcll|} \hline a_1 = 11+0\cdot 6 &=& 11 \\ a_2 = 11+1\cdot 6 &=& 17 \\ a_3 = 11+2\cdot 6 &=& 23 \\ a_4 = 11+3\cdot 6 &=& 29 \\ \dots \\ a_n = 11+(n-1)\cdot 6 &=& 5+6n \\ \hline \end{array}$

What is the Fibonacci Sequence?

What is the mean? 0.91 0.91 0.91 0.91 0.54 0.68 0.93 0.91 0.68

$\begin{array}{|rcll|} \hline && \frac{0.91+ 0.91+ 0.91+ 0.91+ 0.54+ 0.68+ 0.93+ 0.91+ 0.68}{9} \\ &=& \frac{5\cdot 0.91 +2\cdot 0.68+ 0.54+ 0.93 }{9} \\ &=& \frac{7.38}{9} \\ &=& 0.82 \\ \hline \end{array}$

-4x+y=4

2x+2y=13

$\begin{array}{|lrcll|} \hline (1) & -4x+y &=& 4 \\ & y = 4+4x \\ (2) & 2x+2y &=& 13 \\ & 2x+2[4+4x]&=& 13 \\ & 2x+8+8x &=& 13 \\ & 10x+8&=& 13 \\ & 10x&=& 13 -8 \\ & 10x&=& 5 \\ & x&=& \frac{5}{10} \\ & x&=& \frac12 \\ & \mathbf{x} & \mathbf{=} & \mathbf{0.5} \\\\ & y &=& 4+4x \\ & y &=& 4+4\cdot {\frac12} \\ & y &=& 4+2 \\ & \mathbf{y} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}$

hi Melody,

typesetting polynom division can easy be done with the polynom package:

evaluate cosec(10)+cosec(50)-cosec(70)

Formula:

$\begin{array}{|rcll|} \hline \sin(10^{\circ}) = \cos(90^{\circ}-10^{\circ}) = \cos(80^{\circ}) \\ \sin(50^{\circ}) = \cos(90^{\circ}-50^{\circ}) = \cos(40^{\circ}) \\ \sin(70^{\circ}) = \cos(90^{\circ}-70^{\circ}) = \cos(20^{\circ}) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && csc(10^{\circ})+csc(50^{\circ})-csc(70^{\circ}) \\ &=& \frac{1}{\sin(10^{\circ})} + \frac{1}{\sin(50^{\circ})} - \frac{1}{\sin(70^{\circ})} \\ &=& \frac{1}{\cos(80^{\circ})} + \frac{1}{\cos(40^{\circ})} - \frac{1}{\cos(20^{\circ})} \\ &=& \frac{\cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ})} {\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\\\ && \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ}) \\ &&= \frac{ 2\cdot\sin(20^{\circ}) \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(40^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(80^{\circ})\cos(80^{\circ})} {4\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(160^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(20^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ 1 } {8} \\\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\ &=& 8\cdot \{~ \cos(20^{\circ})[\cos(40^{\circ})+\cos(80^{\circ})]-\cos(40^{\circ})\cos(80^{\circ}) ~\} \\\\ && \cos(40^{\circ}) = \cos(60^{\circ}-20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})+\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(80^{\circ}) = \cos(60^{\circ}+20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})-\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ})+\cos(80^{\circ}) = 2 \cos(60^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot [~ \cos(20^{\circ})2 \cos(60^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \quad | \quad \cos(60^{\circ}) = \frac12\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\\\ && \cos(40^{\circ}) = \cos(80^{\circ}-40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})+\sin(80^{\circ})\sin(40^{\circ}) \\ && -\cos(60^{\circ})=\cos(120^{\circ}) = \cos(80^{\circ}+40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})-\sin(80^{\circ})\sin(40^{\circ}) \\ && \cos(40^{\circ})-\cos(60^{\circ})= 2\cos(80^{\circ})\cos(40^{\circ}) \\\\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot [\cos(40^{\circ})-\cos(60^{\circ})] ~\} \\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac12 \cos(60^{\circ}) ~\} \quad | \quad \cos(60^{\circ}) = \frac12 \\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\\\ && 1 = \cos(20^{\circ}-20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ}) = \cos(20^{\circ}+20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})-\sin(20^{\circ})\sin(20^{\circ}) \\ && 1+\cos(40^{\circ}) = 2 \cos(20^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot \{~ \frac12[1+\cos(40^{\circ})] -\frac12 \cdot \cos(40^{\circ})+\frac14 ~ \} \\ &=& 8\cdot [~ \frac12 + \frac12 \cdot \cos(40^{\circ}) -\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\ &=& 8\cdot (~ \frac12 +\frac14 ~) \\ &=& 8\cdot (~ \frac34 ~) \\ &=& \mathbf{6} \\ \hline \end{array}$

compute 

1 - 2 + 3 - 4 + $\dots$+ 2005 - 2006 + 2007 = ?

Geometric series:

$\begin{array}{|rcll|} \hline 1+x+x^2+x^3+x^4+x^5+\dots + x^n &=& \dfrac{1-x^{n+1}}{1-x} \qquad -1 < x < 1 \\ \hline \end{array}$

Derivative:

$\begin{array}{|rcll|} \hline 1+2x+3x^2+4x^3+5x^4+6x^5+\dots + nx^{n-1} &=& \dfrac{1-x^n[~n(1-x)+1~]}{(1-x)^2} \\ \hline \end{array}$

x $\rightarrow$ -x

$\begin{array}{|rcll|} \hline 1-2x+3x^2-4x^3+5x^4-6x^5+-\dots + n(-x)^{n-1} &=& \dfrac{1-(-x)^n[~n(1+x)+1~]}{(1+x)^2} \\ \hline \end{array}$

x = 1:

$\begin{array}{|rcll|} \hline 1-2+3-4+5-6+-\dots + n(-1)^{n-1} &=& \dfrac{1-(-1)^n[~2n+1~]}{4} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^{2007}n (-1)^{n - 1} = 1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007 \\\\ &=& \dfrac{1-(-1)^{2007}[~2\cdot2007+1~]}{4} \\\\ &=& \dfrac{1+4015}{4} \\\\ &=& \dfrac{4016}{4} \\\\ &\mathbf{=}& \mathbf{1004} \\ \hline \end{array}$

6x−y+3z = 28

5x−2y+2z = 28

−3x−y+4z = -17

We are given a regular heptagon of side length 1. Let S be the set of points that are within a distance of 1 from some point on or inside the heptagon, but not including the heptagon itself. Find the area of S

$\begin{array}{|rcll|} \hline A_1 &=& 1\cdot 1 \\ &=& 1 \\\\ A_2 &=& \pi \cdot 1^2 \cdot \frac{ \frac{360^{\circ}}{7} } {360^{\circ}} \\ &=& \pi \cdot \frac{1}{7} \\\\ S &=& 7\cdot A_1 + 7 \cdot A_2 \\ &=& 7\cdot 1 + 7 \cdot \pi \cdot \frac{1}{7} \\ &=& 7+\pi \\ &=& 10.1415926536 \\ \hline \end{array}$