heureka

$$\small{ \text{sum $ \ s = \underbrace{(7.9*10^{-1})}_{=a} + (7.9* 10^{-1} ) * 10^{-1} + (7.9* 10^{-1} ) * 10^{-2} + (7.9* 10^{-1} ) * 10^{-3} + \dots + (7.9* 10^{-1} ) * 10^{-(n-1)} $ }}$\\\\\\$ \small{\text{ $ r = \dfrac{ (7.9* 10^{-1} ) * 10^{-1} }{ (7.9* 10^{-1} )} = 10^{-1} $ }}$\\\\$ \small{\text{ $ s = \dfrac{ a }{ 1-r } = \dfrac{(7.9* 10^{-1} ) }{1-10^{-1} } $ }}$\\\\$ \small{\text{ $ s = \dfrac{(7.9* 10^{-1} ) }{1- 10^{-1} } * \dfrac{ 10^{1}}{10^{1} } $ }}$\\\\$ \small{\text{ $ s = \dfrac{ 7.9 } { 10^{1} -1} = \dfrac{7.9}{9} $ }}$\\\\$ \small{\text{ $ s = \dfrac{7.9}{9} = 0.87777777777777777777777777\overline{7} = 0.8\overline{7} $ }}$$

The age of a man 3 years ago was 11 times the age of his son. The ratio of their ages after 15 years will be 2:1

$$\small{\text{ s = son and m = man }$\\\\$ \begin{array}{lrcllrcl} (1) & (m-3) &=& 11*(s-3) & | \quad (m-3) &=& 11s-33\\ (2) & \frac{m+15}{s+15} &=& \frac{2}{1} & | \quad m+15 &=& 2s+30\\ \hline (2)-(1):& 18 &=& -9s+63 & | :9\\ &2 &=& -s+7\\ & s&=&5 \\ \hline (1): & m-3 &=& 55-33\\ & m &=& 3+22\\ & m &=& 25 \\ \hline \end{array}$$

The age of the man is 25 and the age of his son is 5.

  $$\small{\text{Arithmetical sequence : } }\boxed{\ a_n=a_1+(n-1)*d\ } \\\\ a_1=7 \text{ and } d = 4 \text{ , so we have } a_n = 7 + (n-1) * 4 = 3+4n \\ 7+0*4,\ 7+1*4,\ 7+2*4,\ 7+3*4, \ +\dots+,\ 7+n*4$$

$$=\dfrac{z^{k-1}-2z^k+z^{k+1}}{z^{k+1}-z^k} =\dfrac{z^k*z^{-1}-2z^k+z^k*z}{z^k*z-z^k}\\\\ =\dfrac{z^k(z^{-1}-2+z)}{z^k(z-1)} =\dfrac{z-2+z^{-1}}{z-1}\\\\ =\dfrac{(z-1)-1+z^{-1}}{z-1} =\dfrac{z-1}{z-1}-\dfrac{1-z^{-1}}{z-1}\\\\ =1-\dfrac{1-\frac{1}{z}}{z-1} =1-\dfrac{z-1}{z*(z-1)}\\\\ =1-\dfrac{1}{z}$$

Die Schnittpunkte P und Q ?

-2x²-8x-6=4x²+16x+12

$$\small{\text{ Gleichung 1: $ y=4x^2+16x+12$ }}\\ \small{\text{ Gleichung 2: $ y=-2x^2-8x-6$ }}\\ \small{\text{ Schnittpunkte finden durch gleichsetzen der beiden Gleichungen: $ y_s=4x_s^2+16x_s+12 = -2x_s^2-8x_s-6$ }}\\ \small{\text{ $4x_s^2+16x_s+12 = -2x_s^2-8x_s-6 \quad | \quad +2x_s^2 $ }}\\ \small{\text{ $ 6x_s^2+16x_s+12 = -8x_s-6$ }}\\ \small{\text{ $ 6x_s^2+16x_s+12 = - 8x_s-6 \quad | \quad +8x_s $ }}\\ \small{\text{ $ 6x_s^2+24x_s+12 = -6 $ }}\\ \small{\text{ $ 6x_s^2+24x_s+12 = -6 \quad | \quad +6 $ }}\\ \small{\text{ $ 6x_s^2+24x_s+18= 0 $ }}\\ \small{\text{ $ 6x_s^2+24x_s+18= 0 \quad | \quad :6 $ }}\\ \small{\text{ $ x_s^2+4x_s+3= 0 $ }}\\ \small{\text{ $ x_{s_{1,2}} = -2\pm\sqrt{4-3 } $ }}\\ \small{\text{ $ x_{s_{1}} = -2+1 = -1 $ }}\\ \small{\text{ $ x_{s_{2}} = -2-1 = -3 $ }}\\ \small{\text{ $ y_{s_{1}} = 4*(-1)^2+16*(-1)+12 = 4-16+12 = 0 $ }}\\ \small{\text{ $ y_{s_{2}} = 4*(-3)^2+16*(-3)+12 = 36 - 48 + 12 = 0 $ }}\\$$

Schnittpunkt 1 hat die Werte x = -1 und y = 0

Schnittpunkt 2 hat die Werte x = -3 und y = 0

f(x)=-(4x^2-x)^(-1/2)         g(x)=2x

$$\small{\text{ $f(x)= - [ (4x^2-x)^{ (-\frac{1}{2}) } ] $ and $g(x)=2x $ }}\\ \\ \small{\text{ $ f'(x) = - \left[ (-\frac{1}{2}) (4x^2-x)^{ (-\frac{1}{2}-1) }*(8x-1) \right] $ }} \\ \\ \small{\text{ $ f'(x) = (\frac{1}{2}) (4x^2-x)^{ (-\frac{3}{2}) }*(8x-1) $ }} \\ \\ \small{\text{ $ f'(x) = \dfrac{(8x-1)} {2*(4x^2-x)^{ (1.5)} }$ and $g'(x)=2 $ }} \\ \\ \small{\text{ $ \dfrac{f'(x)}{g'(x)} = \dfrac{(8x-1)} {4*(4x^2-x)^{ (1.5)} } = \dfrac{(8-\frac{1}{x})} {4*x^{(1.5-1)}(4x-1)^{ (1.5)} } = \dfrac{(8-\frac{1}{x})} {4*x^{(0.5)}(4x-1)^{ (1.5)} } $ }} \\ \\ \small{\text{ $ \lim \limits_{x\longrightarrow \infty} \dfrac{(8-\frac{1}{x})} {4*x^{(0.5)}(4x-1)^{ (1.5)} } = 0 $ }}$$

$$\small{\text{The best solutions: $< 0.001$ \$ difference }}\\\\ \small{\text{ $ \begin{array}{rcccr} \# & & & & \small{\text{difference in \$ }} \\ 1. & \$ 4.22 & \$ 2.08 & \$ 0.81 & 0.000144 \\ 2. & \$ 3.21 & \$ 3.21 & \$ 0.69 & 0.000171 \\ 3. & \$ 4.62 & \$ 1.35 & \$ 1.14 & -0.000180 \\ 4. & \$ 3.22 & \$ 3.20 & \$ 0.69 & 0.000240 \\ 5. & \$ 4.37 & \$ 1.87 & \$ 0.87 & 0.000447 \\ 6. & \$ 3.23 & \$ 3.19 & \$ 0.69 & 0.000447 \\ 7. & \$ 3.85 & \$ 2.53 & \$ 0.73 & -0.000565 \\ 8. & \$ 4.25 & \$ 2.04 & \$ 0.82 & 0.000600 \\ 9. & \$ 3.24 & \$ 3.18 & \$ 0.69 & 0.000792 \end{array} $ }}$$

(by  Computing)

......what are the possible prices of the three items : $$\\ \$\ 4.62 \qquad \$\ 1.35 \qquad \$\ 1.14 \\ \dots$$

Why does -4^2 equal -16 and not +16 ?

$$-4^2 = (-1)[(4)^2]=(-1)[16]=-16\\\\ (-4)^2 = [(-1)*(4)^2]=(-1)^2*(4)^2=1*16=16\\$$

If the RR Lyrae stars in a globular cluster have apparent magnitudes of 18, how far away is the cluster ?

You need the absolute magitude too.

$$d = 10^{0.2(m-M+5)} \quad \small{\text {where d is in pc }}\\ m = \small{\text { Apparent magnitudes }} \\ M = \small{\text { Absolute magnitudes }} \\ \small{\text{ 1 pc = 3.26 light years }}$$