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Akzente setzen in Latex:

Compute.

\(i+i^2+i^3+\cdots+i^{258}+i^{259}.\)

geometric series:
\(\begin{array}{|rcll|} \hline a &=& i\\ r &=& i \\ n &=& 259 \\\\ s &=& i\cdot \dfrac{(1-i^{259}) }{1-i}\\\\ s &=& \dfrac{i-i^{260} }{1-i} \qquad & | \qquad i^{4m} &=& 1 \\\\ s &=& \dfrac{i-i^{4\cdot 65} }{1-i} \qquad & | \qquad i^{4\cdot 65} &=& 1 \\\\ s &=& \dfrac{i-1}{1-i} \\\\ s &=& -\dfrac{1-i}{1-i} \\\\ s &=& -1 \\ \hline \end{array} \)

Why when I draw the graph for sqrt(x) it appears to have values for x<0?

Given A(-5,-4) and B(7, 18), find the point two-fifths of the way from A to B.

\(\begin{array}{|rcll|} \hline \vec{A} &=& \binom{-5}{-4} \\ \vec{B} &=& \binom{7}{18} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{A}+(\vec{B}-\vec{A})\cdot \lambda \qquad & | \qquad 0\le \lambda\le 1 \\ && & | \qquad \lambda = 0 \Rightarrow \vec{x} = \vec{A}\\ && & | \qquad \lambda = 1 \Rightarrow \vec{x} = \vec{B}\\ \vec{x} &=& \vec{A} + \lambda\cdot \vec{B} + \lambda\cdot\vec{A} \\ \vec{x} &=& (1-\lambda)\cdot\vec{A}+ \lambda\cdot \vec{B} \qquad & | \qquad \lambda = \dfrac25 \\\\ \vec{x} &=& (1-\frac25) \cdot\vec{A}+ \frac25 \cdot \vec{B} \\ \vec{x} &=& \frac35 \cdot\vec{A} + \frac25 \cdot \vec{B} \\ \vec{x} &=& 0.6 \cdot\vec{A} + 0.4 \cdot \vec{B} \\ \vec{x} &=& 0.6 \cdot\binom{-5}{-4} + 0.4 \cdot \binom{7}{18} \\ \vec{x} &=& \binom{-3}{-2.4} + \binom{2.8}{7.2} \\ \vec{x} &=& \binom{-3+2.8}{-2.4+7.2} \\\\ \mathbf{\vec{x}} &\mathbf{=}& \mathbf{\dbinom{-0.2}{4.8} }\\ \hline \end{array}\)

The Point is (-0.2, 4.8)

If the point (x,3) is equidistant from (3,-2) and (7,4), find the value of x.

\(\begin{array}{|rcll|} \hline \vec{X} &=& \binom{x}{3} \\ \vec{P_1} &=& \binom{3}{-2} \\ \vec{P_2} &=& \binom{7}{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline |\vec{X}-\vec{P_1}| &=& |\vec{P_2}-\vec{X}| \\ |\binom{x}{3}-\binom{3}{-2}| &=& |\binom{7}{4}-\binom{x}{3}| \\ |\binom{x-3}{3-(-2)}| &=& |\binom{7-x}{4-3}| \\ |\binom{x-3}{5}| &=& |\binom{7-x}{1}| \\ (x-3)^2+5^2 &=& (7-x)^2 + 1^2 \\ (x-3)^2+25 &=& (7-x)^2 + 1 \\ x^2 -6x + 9 +25 &=& 49 - 14 x + x^2 + 1 \\ x^2 -6x + 34 &=& 50 - 14 x + x^2 \\ -6x + 34 &=& 50 - 14 x \qquad & | \qquad +14x -34\\ 8x &=& 16 \qquad & | \qquad :8 \\ x &=& 2 \\ \hline \end{array}\)

The equidistant is \(\sqrt{1+25}=\sqrt{26}=5.09901951359\)

Calculate 10hrs into degrees

\(\begin{array}{|rcll|} \hline && \dfrac{360^{\circ}} {24~ h} \cdot 10~ h \\ &=& \left( \dfrac{3600}{24} \right)^{\circ}\\ &=& 150^{\circ} \\ \hline \end{array}\)

1) Betrachten Sie den Markt für ein Produkt (x=menge, P=Preis), für das aufgrund von Erfahrungen die beiden Verkaufsmengen- und Preiskombinationen (x1,p1) = 100,30 und x2,p2 = 150,25 vorliegen. 

Wie lautet die Nachfragefunktionm, wenn eine lineaere Beziehung unterstellt wird, und wieviel kann dann voraussichtlich verkauft werden, wenn der Preis auf 20 gesenkt wird?

x=Menge, P=Preis

Gesuchte Gleichung: P = x*a + b

1. Preiskombination: 30 = 100*a + b

2. Preiskombination: 25 = 150*a + b

\(\begin{array}{lrcl} (1): & 30 &=& 100\cdot a + b \\ (2): & 25 &=& 150\cdot a + b \\ \hline \\ (1) - (2): &30-25 &=& 100\cdot a + b - (150\cdot a + b)\\ & 5 &=& 100\cdot a + b - 150\cdot a - b \\ & 5 &=& 100\cdot a - 150\cdot a \\ & 5 &=& -50 \cdot a \\ & a &=& -\dfrac{5}{50} \\ & \mathbf{a} &\mathbf{=} & \mathbf{-0,1} \\ \\ \hline \\ & 30 &=& 100\cdot a + b \qquad & | \qquad a = -0,1 \\ & 30 &=& 100\cdot (-0,1) + b \\ & 30 &=& -10 + b \\ & b &=& 30+10 \\ & \mathbf{b} &\mathbf{=} & \mathbf{40} \\ \\ \hline \\ & \mathbf{P} &\mathbf{=}& \mathbf{-0,1\cdot x+ 40} \\ & 20 &=& -0,1\cdot x+ 40 \\ & 0,1\cdot x &=& 40-20\\ & 0,1\cdot x &=& 20\\ & x &=& \dfrac{20}{0,1}\\ & \mathbf{x} &\mathbf{=} & \mathbf{200} \\ \\ \hline \end{array}\)

Es können 200 Produkte verkauft werden, wenn der Preis auf 20 sinkt.

Gesuchte Gleichung: P =  -0,1 * x + 40

oder:

x=Menge, P=Preis
Gesuchte Gleichung: x = P*a + b
1. Preiskombination: 100 = 30*a + b
2. Preiskombination: 150 = 25*a + b

\(\begin{array}{lrcl} (1): & 100 &=& 30\cdot a + b \\ (2): & 150 &=& 25\cdot a + b \\ \hline \\ (1) - (2): &100-150 &=& 30\cdot a + b - (25\cdot a + b)\\ & -50 &=& 30\cdot a + b - 25\cdot a - b \\ & -50 &=& 30\cdot a - 25\cdot a \\ & -50 &=& 5 \cdot a \\ & a &=& -\dfrac{50}{5} \\ & \mathbf{a} &\mathbf{=} & \mathbf{-10} \\ \\ \hline \\ & 100 &=& 30\cdot a + b \qquad & | \qquad a = -10 \\ & 100 &=& 30\cdot (-10) + b \\ & 100 &=& -300 + b \\ & b &=& 300+100 \\ & \mathbf{b} &\mathbf{=} & \mathbf{400} \\ \\ \hline \\ & \mathbf{x} &\mathbf{=}& \mathbf{-10\cdot P+ 400} \\ & x &=& -10\cdot 20+ 400 \\ & x &=& -200+400\\ & \mathbf{x} &\mathbf{=} & \mathbf{200} \\ \\ \hline \end{array}\)

Es können 200 Produkte verkauft werden, wenn der Preis auf 20 sinkt.

Gesuchte Gleichung: x =  -10 * P + 400

An equation

\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x} \qquad \qquad x\ne 0\)

How many real roots does it have:

a)Exactly 3 real roots

b)Exactly 2 real roots

c)Exactly 1 real root

d)It doesn't have any real roots

\(\begin{array}{rcll} |\frac{1}{x}-\frac{2}{x+1}| &=& \frac{2}{x} \qquad & | \qquad \text{square both sides} \\ \left(\frac{1}{x}-\frac{2}{x+1} \right)^2 &=& \left(\frac{2}{x} \right)^2 \\ \left(\frac{x+1-2x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \left(\frac{ 1- x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \frac{ (1- x)^2}{x^2(x+1)^2} &=& \frac{4}{x^2} \qquad & | \qquad \cdot x^2 \qquad \text{solution }~ x = 0 ~ \text{impossible}\\ \frac{ (1- x)^2}{ (x+1)^2} &=& 4 \\ 4\cdot(x+1)^2 &=& (1- x)^2 \\ 4\cdot( x^2+2x+1) &=& (1-2x+x^2) \\ 4x^2 + 8x+4 &=& 1-2x+x^2 \\ 3x^2 + 10x+3 &=& 0\\ \end{array} \)

\(\begin{array}{|rcll|} \hline 3x^2 + 10x+3 &=& 0 \qquad \qquad a = 3 \qquad b=10 \qquad c = 3 \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{10^2-4\cdot 3 \cdot 3} } { 2\cdot 3 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{100-36} } { 6 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{64} } {6 } \\\\ x_{1,2} &=& \frac{ -10 \pm 8 } {6 } \\\\ x_{1} &=& \frac{ -10 + 8 } { 6 } \\ x_{1} &=& \frac{ -2 } { 6 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ - \frac13 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ x_{2} &=& \frac{ -10 - 8 } { 6 } \\ x_{2} &=& \frac{ -18 } { 6 } \\ \mathbf{ x_{2} }&\mathbf{=}& \mathbf{ - 3 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ \hline \end{array}\)

d) It doesn't have any real roots

let f(x) = 6x^2 +x. Use the definition of the derivative as the limit of the difference quoetient to find f '(x).

\(\begin{array}{rcll} f(x) &=& 6x^2+x\\ f(x+h)&=&6(x+h)^2+(x+h)\\\\ \text{The difference quotient is:}\\ \dfrac{\Delta x}{\Delta y} &=& \dfrac{f(x+h)-f(x)}{h}\\ &=& \dfrac{6(x+h)^2+(x+h)-(6x^2+x)}{h}\\ &=& \dfrac{6(x+h)^2+ x +h-6x^2 - x }{h}\\ &=& \dfrac{6(x+h)^2 +h-6x^2 }{h}\\ &=& \dfrac{6(x^2+2xh+h^2) +h-6x^2 }{h}\\ &=& \dfrac{6x^2+12xh+6h^2 +h-6x^2 }{h}\\ &=& \dfrac{ 12xh+6h^2 +h }{h}\\ &=& \dfrac{ h( 12x+6h +1) }{h}\\ \dfrac{\Delta x}{\Delta y} &=& 12x+6h +1 \end{array}\)

\(\begin{array}{rcll} f'(x) &=& \dfrac{dy}{dx} \\ \dfrac{dy}{dx} &=& \lim \limits_{h\to 0} { \left( 12x+6h +1 \right) } \\ \dfrac{dy}{dx} &=& \left( 12x+6\cdot 0 +1 \right) \\ \dfrac{dy}{dx} &=& \left( 12x+1 \right) \\\\ \mathbf{f'(x)} &\mathbf{=}& \mathbf{12x+1} \end{array}\)

 Simplify 

\(\dfrac{\log\sqrt8+\log\sqrt{125}-\log\sqrt{27}}{\log2+\log\sqrt[3]{5}-\log\sqrt[3]{12}}\)

\(\begin{array}{rcll} && \dfrac{\log\sqrt8+\log\sqrt{125}-\log\sqrt{27}}{\log2+\log\sqrt[3]{5}-\log\sqrt[3]{12}}\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} - \log{\sqrt[3]{3\cdot2^2}} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} - \log(\sqrt[3]{3}\cdot \sqrt[3]{2^2}) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\log\sqrt[3]{2^2} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\log(2^\frac23 ) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\frac23\cdot \log(2) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)-\frac23\cdot \log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac33\cdot\log(2)-\frac23\cdot \log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac13\cdot\log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac13\cdot\log(2) +\frac13\cdot\log(5) -\frac13\cdot\log(3) }\\\\ &=& \dfrac{ \frac32\cdot\log(2)+\frac32\cdot\log(5)-\frac32\cdot\log(3) }{\frac13\cdot\log(2) +\frac13\cdot\log(5) -\frac13\cdot\log(3) }\\\\ &=& \dfrac{\frac32}{\frac13} \left( \dfrac{ \log(2)+\log(5)-\log(3) }{\log(2) +\log(5) -\log(3) } \right) \\\\ &=& \dfrac{\frac32}{\frac13} \\\\ &=& \frac32 \cdot \frac31 \\\\ &=& \frac92 \\\\ &=& 4.5 \end{array}\)