heureka

If f(1)=3 and f(n)=-2f(n-1)+1, then f(5) is equal to?

$\small{ \begin{array}{rclcl} f(1)&& &=&3\\ f(2)&& &=& (-2)^1\cdot f(1)+1 \\ f(3)&=& (-2)\cdot [ f(2)]+1 \\ &=& (-2)\cdot[ (-2)\cdot f(1)+1 ]+1 &=& (-2)^2\cdot f(1) + (-2)^1 + 1\\ f(4)&=& (-2)\cdot [ f(3)]+1 \\ &=& (-2)\cdot[ (-2)^2\cdot f(1) + (-2) + 1 ]+1 &=& (-2)^3\cdot f(1) + (-2)^2 + (-2)^1 + 1\\ f(5)&=& (-2)\cdot [ f(4)]+1 \\ &=& (-2)\cdot[ (-2)^3\cdot f(1) + (-2)^2 + (-2) + 1 ]+1 &=& (-2)^4\cdot f(1) + (-2)^3 + (-2)^2 + (-2)^1 + 1\\ f(5)&& &=&16\cdot 3 -8 + 4 -2 + 1\\ f(5)&& &=&48 -8 + 4 -2 + 1\\ f(5)&& &=&43\\ \end{array} }$

$\begin{array}{rclcl} f(n)&=& (-2)^{n-1}\cdot f(1) + (-2)^{n-2} + (-2)^{n-3} + (-2)^{n-4}+\dots + (-2)^1 + 1\\\\ && \text{sum of a geometric sequence } ~r=-2 \quad a_1 = 1\\ && (-2)^{n-2} + (-2)^{n-3} + (-2)^{n-4}+\dots + (-2)^1 + 1 = \frac{1-(-2)^{n-1}}{1-(-2)}\\\\ f(n)&=& (-2)^{n-1}\cdot f(1) + \frac{1-(-2)^{n-1}}{3}\\ f(n)&=& (-2)^{n-1}\cdot f(1) + \frac13 - \frac{(-2)^{n-1}}{3}\\ f(n)&=& (-2)^{n-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \\ \end{array}$

$\boxed{ \begin{array}{rcll} f(n)&=& (-2)^{n-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \end{array} }$

$\begin{array}{rcll} f(5)&=& (-2)^{5-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \qquad | \qquad f(1) = 3 \\ f(5)&=& (-2)^{4}\cdot \left[ 3 -\frac13 \right] + \frac13 \\ f(5)&=& 16\cdot \left[ 3 -\frac13 \right] + \frac13 \\ f(5)&=& 16\cdot \left[ \frac83 \right] + \frac13 \\ f(5)&=& \frac{16\cdot 8 + 1}{3} \\ f(5)&=& \frac{129}{3} \\ \mathbf{f(5)}&\mathbf{=}& \mathbf{43} \end{array}$

A circle of radius 9 cm is divided into three equal sectors.

Calculate

a) The length of the arc of each sector

$\boxed{ \begin{array}{rcll} c&=&2\cdot\pi\cdot r \end{array} }$

$\begin{array}{rcll} \frac{c}{3}&=& \frac23 \cdot\pi\cdot r \\ \frac{c}{3}&=& \frac23 \cdot\pi\cdot 9\ \text{cm}\\ \frac{c}{3}&=& 6 \cdot\pi \ \text{cm} \qquad | \qquad \pi = 3.14159265359\dots\\ \frac{c}{3}&=& 18.8495559215 \ \text{cm}\\ \end{array}$

b) the area of each sector

$\boxed{ \begin{array}{rcll} A&=& \pi\cdot r^2 \end{array} }$

$\begin{array}{rcll} \frac{A}{3}&=& \frac{\pi}{3}\cdot r^2 \\ \frac{A}{3}&=& \frac{\pi}{3}\cdot 9^2 \ \text{cm}^2 \\ \frac{A}{3}&=& \frac{\pi}{3}\cdot 81\ \text{cm}^2\\ \frac{A}{3}&=& 27\cdot \pi \ \text{cm}^2 \qquad | \qquad \pi = 3.14159265359\dots\\ \frac{A}{3}&=& 84.8230016469\ \text{cm}^2\\ \end{array}$

Find the area of a regular hexagon with apothem length of 5 centimeters

a = apothem

a = 5 cm

$\begin{array}{rcll} \mathbf{ A_n } & \mathbf{=} & \mathbf{ n\cdot a^2 \cdot \tan( \frac{180^{\circ}}{n} ) } \end{array}$

hexagon: n = 6

$\begin{array}{rcll} A_6 &=& 6\cdot a^2 \cdot \tan( \frac{180^{\circ}}{6} )\\ A_6 &=& 6\cdot 5^2 \cdot \tan( 30^{\circ} ) \qquad &| \qquad \tan( 30^{\circ} )=\frac13 \cdot \sqrt{3} \\ A_6 &=& 6\cdot 5^2 \cdot \frac13 \cdot \sqrt{3} \\ A_6 &=& 2\cdot 5^2 \cdot \sqrt{3} \\ A_6 &=& 2\cdot 25 \cdot \sqrt{3} \\ A_6 &=& 50 \cdot \sqrt{3} \\ A_6 &=& 86.6025403784\ \text{cm}^2 \\ \end{array}$

The area of a regular hexagon with apothem length of 5 centimeters is $86.6\ \text{cm}^2$

solve for x: cos(x)=0.342/0.798

$\begin{array}{rcll} \cos(x) &=& \frac{0.342}{0.798} \\ \cos(x) &=& 0.42857142857 \qquad & | \qquad \pm\arccos{()}\\ x &=& \pm\arccos{( 0.42857142857 )}\\ x &=& \pm64.6230664748^{\circ}\\ \mathbf{ x_1 } & \mathbf{=} & \mathbf{64.6230664748^{\circ} }\\\\ x_2 &=& -64.6230664748^{\circ} \\ x_2 &=& -64.6230664748^{\circ}+ 360^{\circ} \\ \mathbf{ x_2 } & \mathbf{=} & \mathbf{295.376933525^{\circ}} \end{array}$

this is not a arcus function "arc cosh" is wrong, this is a area - function area cosh(x) or arcosh(x) is right!

If 4^x5^3x+1 = 10^2x+1, prove that x = log2 / log5

$\small{ \begin{array}{rcll} 4^x\cdot 5^{3x+1} &=& 10^{2x+1} \quad &| \quad \log_{10} \\ \log_{10} ( 4^x\cdot 5^{3x+1} ) &=& \log_{10} ( 10^{2x+1} ) \quad &| \quad \log_{10} (10^{2x+1}) = 2x+1 \\ \log_{10} ( 4^x\cdot 5^{3x+1} ) &=& 2x+1 \quad &| \quad \log_{10} (a\cdot b) = \log_{10} (a) + \log_{10} (b)\\ \log_{10} ( 4^x ) + \log_{10} ( 5^{3x+1} ) &=& 2x+1 \quad &| \quad \log_{10} (a^b) = b\cdot \log_{10} (a) \\ x\cdot \log_{10} ( 4 ) + (3x+1) \cdot \log_{10} ( 5 ) &=& 2x+1 \\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) + \log_{10} ( 5 ) &=& 2x+1 \quad &| \quad -2x\\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) -2x + \log_{10}(5) &=& 1 \quad &| \quad - \log_{10}(5)\\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) -2x &=& 1 - \log_{10}(5)\\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& 1 - \log_{10}(5) \quad &| \quad 1 = \log_{10}(10) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(10) - \log_{10}(5)\quad &| \quad \log_{10} (a) - \log_{10} (b) = \log_{10} ( \frac{a}{b} ) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(\frac{10}{5}) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(2) \quad &| \quad 2 = \log_{10}(10^2) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -\log_{10}(10^2) ] &=& \log_{10}(2) \quad &| \quad b\cdot \log_{10} (a) = \log_{10} (a^b)\\ x\cdot [ \log_{10}(4) + \log_{10}(5^3) -\log_{10}(10^2) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(\frac{ 4\cdot 5^3 } {10^2} ) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(\frac{ 500 } { 100 } ) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(5) ] &=& \log_{10}(2) \quad &| \quad : \log_{10}(5) \\ \mathbf{ x } & \mathbf{ = } & \mathbf{ \frac{ \log_{10}(2) } { \log_{10}(5) } } \end{array} }$

$nodes: \begin{array}{ccccccccccc} & & & & 27 & \rightarrow & 28 & \rightarrow & 29 & \rightarrow & 30 & \rightarrow & B(31)\\ & & & & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\ & & & & 22 & \rightarrow & 23 & \rightarrow & 24 & \rightarrow & 25 & \rightarrow & 26 \\ & & & & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\ & & & & 17 & \rightarrow & 18 & \rightarrow & 19 & \rightarrow & 20 & \rightarrow & 21 \\ & & & & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\ 10 & \rightarrow & 11 &\rightarrow & 12 & \rightarrow & 13 & \rightarrow & 14 & \rightarrow & 15 & \rightarrow & 16 \\ \uparrow & & \uparrow & & \uparrow \\ 7 & \rightarrow & 8 &\rightarrow & 9 \\ \uparrow & & \uparrow & & \uparrow \\ 4 & \rightarrow & 5 &\rightarrow & 6 \\ \uparrow & & \uparrow & & \uparrow \\ A(1)& \rightarrow & 2 &\rightarrow & 3 \\ \end{array}$

$\small{\text{ entrys $ \qquad 1 = $ one way $\quad 0 = $no way from node x to node y}}$

Matrix A =

Marix A*A =

...

$\text{Matrix }\ A^{12} =A\cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A =$

$A^{12} : \text{ Matrix element[A][B] } =350 \quad\text{ (12-station-way)}$

integration of 1/(x^2 + 4x + 6) equals what

How can you make 10 out of the numbers 1,5,6,8, and 12 using any or all operations?

 $\begin{array}{rcll} \frac{8\cdot 12}{6} - 5 - 1 &=& 10\\\\ \frac{8\cdot 12}{1+5} - 6 &=& 10\\\\ \frac{5\cdot 6}{12-8-1} &=& 10\\\\ 8- \frac{12}{6} + 5 - 1&=& 10\\\\ 8+ \frac{(5-1)\cdot 6}{12} &=& 10\\\\ 6\cdot (\frac{12}{8}+1)-5&=& 10\\\\ 12-\frac{5+8-1}{6} &=& 10\\\\ \frac{6\cdot 12}{(5-1)} - 8 &=& 10 \end{array}$

So, not entire sure how to go about doing this one using the substitution method, so i'd appreciate any help :\

$\int_{1}^{e^3} \frac{dx}{x \cdot [1+\ln{(x)} ] }$

$\begin{array}{rcll} \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } \\\\ && \text{substitute }~z = 1+\ln{(x)} \qquad dz = \frac{dx}{x} \qquad dx = x\cdot dz \\\\ \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } &=& \int_{1}^{e^3} \dfrac{x\cdot dz}{x \cdot z } \\\\ &=& \int_{1}^{e^3} \dfrac{dz}{ z } \\\\ &=& \left[~ \ln{(z)} ~\right]_{1}^{e^3} \\\\ &=& \left[~ \ln{(~ 1+\ln{(x)} ~)} ~\right]_{1}^{e^3} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{(~ 1+\ln{(1)}~)} \qquad | \qquad \ln{(1)} = 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{( 1+0)} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{(1)} \qquad | \qquad \ln{(1)} = 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} \qquad | \qquad \ln{(e^3)} = 3\\\\ &=& \ln{( 1+3 )} \\\\ \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } &=& \ln{( 4 )} \\\\ \mathbf{\int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } }&\mathbf{=} & \mathbf{1.38629436112} \\\\ \end{array}$