heureka

If the sqare root of -1 is "i", then what is i^2, 3, and 4. Do you see the pattern? Now, following this knowledge, what is i^10? (you can do it!)

See the imaginary coordinate system. The real axis in x (-1,+1) and the imaginary axis in y ( -i, + 1).

Set a Point on ( 1,0) for $i^0$ so $i^0 = 1$

Move the Point counter clockwise 90 degrees you have the Point (0,i) for $i^1$ so $i^1 = i$

Move the Point again counter clockwise 90 degrees you have the Point (-1,0) for $i^2$ so $i^2 = -1$

Move the Point again counter clockwise 90 degrees you have the Point (0,-i) for $i^3$ so $i^3 = -i$

Move the Point again counter clockwise 90 degrees you have the Point (1,0) for $i^4$ so $i^4 = 1$

etc.

$i^{10}=-1$

f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term

$\small{ \begin{array}{lrcl} & f_n &=& 3\cdot f_{n-1} - 60 \qquad a=3 \qquad c=-60\\ \text{or general}: & f_n &=& a\cdot f_{n-1} + c \\ \hline \\ & f_1 &=& a\cdot f_0 + c \\\\ & f_2 &=& a\cdot f_1 + c \\ & f_2 &=& a\cdot [a\cdot f_0 + c] + c \\ & f_2 &=& a^2\cdot f_0 + a\cdot c + c \\ & f_2 &=& a^2\cdot f_0 + c\cdot(1+a) \\\\ & f_3 &=& a\cdot f_2 + c \\ & f_3 &=& a\cdot [a^2\cdot f_0 + c\cdot(1+a)] + c \\ & f_3 &=& a^3\cdot f_0 + a\cdot c\cdot(1+a) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(a+a^2) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(1+a+a^2) \\\\ & f_4 &=& a\cdot f_3 + c \\ & f_4 &=& a\cdot [a^3\cdot f_0 + c\cdot(1+a+a^2)] + c \\ & f_4 &=& a^4\cdot f_0 + a\cdot c\cdot(1+a+a^2) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(a+a^2+a^3) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(1+a+a^2+a^3) \\\\ \hline \text{general solution}: & f_n &=& a^n\cdot f_0 + c\cdot (1+a+a^2+a^3+...+ a^{n-1}) \\ & && 1+a+a^2+a^3+...+ a^{n-1} = \frac{1-a^n}{1-a}\\ & f_n &=& a^n\cdot f_0 + c\cdot \frac{1-a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 + c\cdot \left( \frac{1}{1-a} - \frac{a^n}{1-a} \right) \\ & f_n &=& a^n\cdot f_0 + c\cdot\frac{1}{1-a} - c\cdot\frac{a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 - c\cdot\frac{a^n}{1-a} + c\cdot\frac{1}{1-a} \\ & && \boxed{~ \begin{array}{lrcl} & f_n &=& a^n\cdot \left( f_0 - c\cdot\frac{1}{1-a} \right) + c\cdot\frac{1}{1-a} \\ & f_n &=& x + \beta \cdot a^n \\ &&& x = c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - x \end{array} ~}\\ \hline \\ \text{general}: & f_n &=& a\cdot f_{n-1} + c \\ \text{general solution}: & f_n &=& x + \beta \cdot a^n \qquad x = \frac{c}{1-a} \qquad \beta = f_0 - x \\\\ & f_n &=& 3\cdot f_{n-1} -60 \qquad a=3 \qquad c=-60\\\\ & f_1 &=& 3 \cdot f_0 - 60 \\ & 3 \cdot f_0 &=& f_1 + 60 \\ & f_0 &=& \frac{ f_1 + 60 } {3} \qquad f_1 = 20\\ & f_0 &=& \frac{ 20 + 60 } {3} \\ & f_0 &=& \frac{ 80 } {3} \\\\ & x &=& \frac{c}{1-a} \\ & x &=& \frac{-60}{1-3} \\ & x &=& \frac{-60}{-2} \\ & x &=& 30 \\ \\ & \beta &=& f_0 - x \\ & \beta &=& \frac{ 80 } {3} - 30 \\ & \beta &=& -\frac{ 10 } {3} \\\\ & f_n &=& 30 -\frac{ 10 } {3} \cdot a^n \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 3^{10} \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 59049 \\ & f_{10} &=& 30 -196830\\ & \mathbf{f_{10} }& \mathbf{=} &\mathbf { -196800 } \end{array} }$

Guten Morgen :) {nl} ich brauche mal ganz dringend eure Hilfe bei der Berechnung einer Stabkraft in einem Fachwerk.

$\small{ \begin{array}{rcll} s_6 &=& \dfrac{ \frac12 \cdot \sqrt{2}\sqrt{2}\cdot F - 2F } {\frac12 \sqrt{2}} \qquad & | \qquad \sqrt{2}\sqrt{2} =(\sqrt{2})^2= 2\\\\ s_6 &=& \dfrac{ \frac12 \cdot 2\cdot F - 2F } {\frac12 \sqrt{2}} \\\\ s_6 &=& \dfrac{ F - 2F } {\frac12 \sqrt{2}} \\\\ s_6 &=& \dfrac{ -F } {\frac12 \sqrt{2}} \\\\ s_6 &=& \dfrac{ -2F } { \sqrt{2} } \\\\ s_6 &=& \dfrac{ -2F } { \sqrt{2} } \cdot \frac{\sqrt{2}} {\sqrt{2}}\\\\ s_6 &=& \dfrac{ -2\sqrt{2}\cdot F } { \sqrt{2}\sqrt{2} }\\\\ s_6 &=& \dfrac{ -2\sqrt{2}\cdot F } { 2 } \qquad & | \qquad \sqrt{2}\sqrt{2} =(\sqrt{2})^2= 2\\\\ s_6 &=& -\sqrt{2}\cdot F \end{array} }$

For what values of c will the quadratic equation y=x2+6x+c have no real zeroes? Set up and solve an inequality for this problem. ...thanks for the help

$\begin{array}{rcl} \boxed{~ \begin{array}{rcl} ax^2+bx+c&=&0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \\ \text{discriminant D } &=& b^2-4ac\\ \text{no real zeroes } D < 0 && b^2-4ac < 0 \end{array} ~} \end{array}$

$\begin{array}{rcll} y=x^2+6x+c &=& 0 \qquad a=1 \qquad b=6 \\\\ \boxed{~ \begin{array}{rcl} b^2-4ac &<& 0 \\ \end{array} ~}\\ (-6)^2-4c &<& 0 \\ 36-4c &<& 0 \qquad & | \qquad + 4c \\ 36 &<& 4c \qquad & | \qquad :4 \\ 9 &<& c \\ c &>& 9 \\ \end{array}$

no real zeroes if c > 9

The age T, in years, of a haddock can be thought of as a function of its length L, in centimeters. One common model uses the natural logarithm, as shown in the following equation.

T = 19 − 5 ln(53 − L)

How long is a haddock that is 12 years old? (Round your answer to two decimal places.)

$\begin{array}{rcl} T &=& 19 - 5 \cdot\ln{(53 - L)} \qquad & | \qquad +5 \cdot\ln{(53 - L)}\\ T+5 \cdot\ln{(53 - L)} &=& 19 \qquad & | \qquad -T\\ 5 \cdot \ln{(53 - L)} &=& 19-T \qquad & | \qquad :5\\ \ln{(53 - L)} &=& \frac{ 19-T }{5} \qquad & | \qquad e^{()}\\ 53 - L &=& e^{\frac{ 19-T }{5}} \qquad & | \qquad \cdot(-1)\\ -53 + L &=& -e^{\frac{ 19-T }{5}} \qquad & | \qquad +53\\ L &=& -e^{\frac{ 19-T }{5}} +53\\ \boxed{~ \begin{array}{lcl} L &=& 53-e^{\frac{ 19-T }{5}} \end{array} ~}\\\\ T&=&12 \text{ years old} \\ L &=& 53-e^{\frac{ 19-12 }{5}} \\ L &=& 53-e^{\frac{ 7 }{5}} \\ L &=& 53-e^{1.4} \\ L &=& 53-4.0551999668 \\ L &=& 48.9448000332 \end{array}$

A 12 years old haddock is 48.94 cm long.

z = x * e^x

Lösung x = W(z)

Lambert W-Function oder auch Produktlogarithmus genannt!

Equation 1: x^2-y^2=0

Equation 2: x^2-4x-9=4y

(1) (x-y)(x+y) = 0

   (x-y) = 0  $\Rightarrow$  x = y

   (x+y) = 0  $\Rightarrow$ x = -y

$y = \pm x$

(2) x^2 -4x-9 = 4(-x)

     x^2  -9 = 0

     x^2 = 9

     $x = \pm 3$

    x^2 - 4x -9 = 4(x)

    x^2 - 8x - 9 = 0    factor

    (x+1)(x-9)= 0

    x+1 = 0   $\Rightarrow$ $x = -1$

    x-9 = 0  $\Rightarrow$  $x = 9$

Zeige mit vollständiger Induktion, dass für n=0,1,2,... und beliebiges x>-1 die Bernoullische Ungleichung (1+x)n >=1+nx gilt.

$\begin{array}{lcl} \mathbf{\text{Aussage: } }\\ \text{Für alle } x \in R \text{ mit } x \ge -1 \text{ und alle } n \in N \text{ gilt:}\\\\ \qquad (1 + x )^n \ge 1 + n \cdot x \quad \text{ (Bernoullische Ungleichung) } \\\\ \text{Beweis. (per Induktion nach n).}\\ \text{Sei im folgenden } x \in R \text{ mit } x \ge -1 \text{ beliebig.}\\\\ \text{Wir starten mit dem Induktionsanfang:(I.A.)} \\ \text{Für } n = 0 \text{ gilt } (1 +x)^0 = 1 \ge 1 + 0\cdot x, \text{ d.h. die Behauptung ist im Fall } n = 0 \text{ korrekt.}\\ \text{Für } n = 1 \text{ gilt } (1 +x)^1 = 1 + x \ge 1 + 1 \cdot x, \text{ d.h. die Behauptung ist im Fall } n = 1 \text{ korrekt.}\\ \text{Induktionsvoraussetzung (I.V.) Es gelte } (1 + x )^n = 1 + n \cdot x. \text{ Wir müssen zeigen, }\\ \text{dass }(1 +x)^{n+1} \ge 1 + (n + 1) \cdot x \text{ gilt.}\\\\ \end{array}\\\\ \begin{array}{lcl} \text{Es gilt: } \\\\ & (1 +x)^{n+1}& = &(1 +x)^n \cdot (1 + x )\\ &&\ge& (1 + n \cdot x ) \cdot (1 + x ) \text{ (nach I.V.)}\\ &&=& 1 +x+n \cdot x + n \cdot x^2 \\ &&=& 1 + (n+ 1) \cdot x + n \cdot x^2 \\ &&=& (1 + (n+ 1)\cdot x ) + \underbrace{ n \cdot x^2 }_{\ge 0} \\ &&\ge& 1 + ( n + 1) \cdot x\\\\ \text{also die Behauptung.} \end{array}$

wie gebe ich in den Taschenrechner 3 über 1 ein, wie berechne ich es mit dem Taschenrechner

ncr(3,1) =$\dbinom{3}{1}$= 3

shane swims 42 laps in 26 minutes and 19 seconds. how many minutes and seconds would it take shane to swim 15 laps at the same average speed

$\begin{array}{rcll} 26\ {\text{minutes}} \text{ and } 19 \ {\text{seconds}} &=& 26\ {\text{minutes}} + \frac{19}{60}\ {\text{minutes}} \\ &=& \frac{ 26 \cdot 60 + 19 } {60 }\ {\text{minutes}} \\ &=& \frac{ 1579 } {60 }\ {\text{minutes}} \\ \end{array}$

15 laps:

$\begin{array}{rcll} t \cdot \frac{ 42\ \text{laps} } { \frac{ 1579 } {60 } \ {\text{minutes} } } &=& 15\ \text{laps} \\ t &=& \frac{ 15\ \text{laps} } { 42\ \text{laps} } \cdot \frac{ 1579 } {60 } \ {\text{minutes} } \\ t &=& \frac{ 1579 } {168 } \ {\text{minutes} } \\ t &=& 9\ {\text{minutes}} \text{ and } 24 \ {\text{seconds}} \\ \end{array}$