heureka

I have this formula that works well to convert wire gauge to mm:

mm dia = 0.127 * 92 ^ ((36-AWG)/39)

Following this, the same source offers the following as an inverse formula(mm to gauge):

AWG = -(ln (dia. mm/0.127) / (ln (92) * 39) – 36)

\(\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \qquad \text{AWG} =\ ? \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad : 0.127 \\\\ \dfrac{ dia_{mm} }{0.127} &=& 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad \ln{()}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \ln{ \left( 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \right)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \left( \frac{36-\text{AWG}}{39} \right) \cdot \ln{(92)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& ( 36-\text{AWG} ) \cdot \frac{ \ln{(92)} } {39} \quad &|\quad \cdot \frac{39}{ \ln{(92)} } \\\\ \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 - \text{AWG} \quad &|\quad +\text{AWG} \\\\ \text{AWG} + \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 \quad &|\quad -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} \\\\ \mathbf{\text{AWG} } &\mathbf{=}& \mathbf{36 -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} }\\ \hline \end{array} \)

121^x=11

\(\begin{array}{|rcll|} \hline 121^x &=& 11 \quad &|\quad 121 = 11^2 \\ (11^2)^x &=& 11 \quad &|\quad (11^2)^x = 11^{2x}\\ 11^{2x} &=& 11 \\ 11^{2x} &=& 11^1 \\ 2x &=& 1 \\ \mathbf{x} & \mathbf{=} & \mathbf{ \frac12 }\\ \hline \end{array}\)

 Help

\(\begin{array}{|rcll|} \hline 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ \sqrt{16x+8} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \quad & | \quad512 = 8^3 \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{3\cdot 2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{6x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > 8^{6x-2\cdot \sqrt{4x+2} } \\ \sqrt{4x+2} > 6x-2\cdot \sqrt{4x+2} \\ 3\cdot \sqrt{4x+2} > 6x \quad & | \quad : 3\\ \sqrt{4x+2} > 2x \\ \hline \end{array}\)

Let's find the domain:

\( \begin{array}{|rcll|} \hline 4x+2 &\ge & 0 \quad & | \quad -2\\ 4x &\ge & -2 \quad & | \quad : 4 \\ x &\ge & -\frac12 \\ \hline \end{array} \)

The domain is \([-\frac12, \infty)\)

case differentiation

1. \( x\ge 0\)

\(\begin{array}{|rcll|} \hline \sqrt{4x+2} &>& 2x \quad & | \quad \text{square both sides}\\ 4x+2 &>& 4x^2 \quad & | \quad -4x \\ 2 &>& 4x^2-4x \quad & | \quad :4 \\ \frac12 &>& x^2-x \\ \frac12 &>& \left(x-\frac12 \right)^2-\frac14 \quad & | \quad + \frac14\\ \frac12+ \frac14 &>& \left(x-\frac12 \right)^2 \\ \frac34 &>& \left(x-\frac12 \right)^2 \quad & | \quad \sqrt{ } \\ \frac{\sqrt{3}}{2} &>& | x-\frac12 | \\ | x-\frac12 | &<& \frac{\sqrt{3}}{2} \\ \Rightarrow \quad - \frac{\sqrt{3}}{2} &<& x-\frac12 < \frac{\sqrt{3}}{2} \quad & | \quad +\frac12 \\ - \frac{\sqrt{3}}{2} +\frac12 &<& x-\frac12 +\frac12 < \frac{\sqrt{3}}{2} +\frac12 \\ \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ \hline \end{array}\)

case differentiation
2. x < 0 , because the domain is \(x \ge -\frac12\)

\(\begin{array}{|rcll|} \hline x &<& 0 \\ \text{the domain is } x &\ge& -\frac12 \text{ or } \\ -\frac12 & \le& x \\ \Rightarrow \quad -\frac12 & \le & x < 0 \\\\ \underbrace{\sqrt{4x+2}}_{\ge 0} &>& \underbrace{2x}_{<0} \quad & | \quad \text{always true}\\ \hline \end{array} \)

together:

\(\begin{array}{|rcll|} \hline \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ -\frac12 & \le & x < 0 \\\\ \mathbf{ -\frac12 } & \mathbf{ \le } & \mathbf{ x < \frac {1+\sqrt{3}}{2} } \\ \hline \end{array}\)

Please Help!!

Formula:

\(\begin{array}{|rcll|} \hline x^{\log_x(y)} &=& y \quad &|\quad \log_y() \\ \log_y(x^{\log_x(y)}) &=& \log_y(y) \quad &|\quad \log_y(y)=1 \\ \log_y(x^{\log_x(y)}) &=& 1 \\ \mathbf{\log_x(y)\cdot \log_y(x) } &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

We substitute:

\(\begin{array}{|rcll|} \hline \log_{a+b}(m) &=& \dfrac{1 }{\log_{m}(a+b)} \\ \log_{a-b}(m) &=& \dfrac{1 }{\log_{m}(a-b)} \\ \hline \end{array}\)

So:

\(\begin{array}{|lcll|} \hline \log_{a+b}(m) + \log_{a-b}(m) -2\cdot \log_{a+b}(m) \cdot \log_{a-b}(m) \\ & \log_{a+b}(m) = \dfrac{1 }{\log_{m}(a+b)} \\ & \log_{a-b}(m) = \dfrac{1 }{\log_{m}(a-b)} \\ = \dfrac{1}{\log_{m}(a+b)} + \dfrac{1}{\log_{m}(a-b)} -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}(a-b)+\log_{m}(a+b)}{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}[(a-b)\cdot(a+b)] }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}[(a^2-b^2)] }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \quad &|\quad a^2-b^2 = m^2\\\\ = \dfrac{\log_{m}(m^2) }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{2\cdot \log_{m}(m) }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \quad &|\quad \log_{m}(m) = 1\\\\ = \dfrac{2 }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = 0 \\ \hline \end{array} \)

Ich benötige Hilfe bei der Berechnung des Goldenen Schnittes. Ich weiß das a sich zu b verhält wie a+b zu a, also a/b = (a+b)/a. Jetzt ist nur die Frage wie berechne ich denn b, wenn ich nur a habe. Auflösen der Gleichung ist mir nicht gelungen.

\(\begin{array}{|lr|} \hline \begin{array}{|rcll|} \hline \dfrac{ a }{b} &=& \dfrac{a+b}{a}\\ \hline \end{array} &\text{Gegeben: a} &\text{Gesucht: b}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{ a }{b} &=& \dfrac{a+b}{a} \\ \dfrac{ a }{b} &=& \dfrac{a}{a}+\dfrac{b}{a} \\ \dfrac{ a }{b} &=& 1+\dfrac{b}{a} \\ \dfrac{ a }{b} &=& 1+ \frac{1}{\frac{a}{b}} \quad & | \quad \text{Goldene Schnitt } = \frac{a}{b} = \varphi\\ \varphi &=& 1+ \dfrac{1}{\varphi} \quad & | \quad -1\\ \mathbf{\varphi -1} &\mathbf{=}&\mathbf{ {\dfrac{1}{\varphi}} }\quad & | \quad \mathbf{\color{red}!!!}\\ \hline \end{array} \)

Die Berechnung des goldenen Schnittes: \(\varphi\)

\(\begin{array}{|rcll|} \hline \varphi -1 &=& \dfrac{1}{\varphi} \quad & | \quad \cdot \varphi \\ \varphi^2 -\varphi &=& 1 \quad & | \quad -1 \\ \varphi^2 -\varphi -1 &=& 0 \\\\ \varphi_{1,2} &=& \dfrac{1\pm \sqrt{1-4\cdot(-1)} }{2} \\ \varphi_{1,2} &=& \dfrac{1\pm \sqrt{5} }{2} \\\\ \mathbf{\varphi_1}=\mathbf{\varphi} & \mathbf{=} & \mathbf{\dfrac{1+ \sqrt{5} }{2} = 1.61803398875\dots} \\ \mathbf{\varphi_2} & \mathbf{=} & \mathbf{\dfrac{1- \sqrt{5} }{2} = 1-\varphi = -0.61803398875\dots} \\ \hline \end{array} \)

Die Berechnung von: b

\(\begin{array}{|rcll|} \hline \dfrac{a}{b} &= \varphi \\ \varphi &=& \dfrac{a}{b} \quad & | \quad \cdot b \\ b\cdot \varphi &=& a \quad & | \quad : \varphi \\ b &=& \dfrac{a}{\varphi} \\\\ b_1 &=& \dfrac{a}{\varphi_1} \quad & | \quad \varphi_1 = \varphi \\ b_1 &=& \dfrac{a}{\varphi} \\ b_1 &=& a\cdot \dfrac{1}{\varphi} \quad & | \quad \dfrac{1}{\varphi} = \varphi -1 \\ \mathbf{b_1} &\mathbf{=}& \mathbf{(\varphi -1 )\cdot a} \\\\ b_2 &=& \dfrac{a}{\varphi_2} \quad & | \quad \varphi_2 = 1-\varphi \\ b_2 &=& \dfrac{a}{1-\varphi } \quad & | \quad -\dfrac{1}{\varphi} = -(\varphi -1) = 1-\varphi \\ b_2 &=& \dfrac{a}{\dfrac{-1}{\varphi} } \\ \mathbf{b_2} &\mathbf{=}& \mathbf{ -\varphi\cdot a }\\ \hline \end{array}\)

Mir liegt folgende Formel vor:

(x^3+3x^2+3x)^-1=-1 Kann ich hier dierekt mit Quadratischer Ergänzung arbeiten sprich meinen Therm umformen in (x+1)^3 und und dann das ^-1 anwenden? sprich ich komme dann auf (x+1)^2=0 ???

1. Aus der Formelsammlung:

\(\begin{array}{|ccll|} \hline (x+1)^3 = x^3+3x^2+3x+1 \\ \text{oder}\\ x^3+3x^2+3x=(x+1)^3-1\\ \hline \end{array} \)

2.

\(\begin{array}{|rcll|} \hline (x^3+3x^2+3x)^{-1} &=& -1 \\ \dfrac{1}{x^3+3x^2+3x}&=& -1 \quad &|\quad x^3+3x^2+3x=(x+1)^3-1\\ \dfrac{1}{(x+1)^3-1}&=& -1 \quad &|\quad \cdot [(x+1)^3-1]\\ 1&=& -1 \cdot [(x+1)^3-1] \quad &|\quad \cdot -1\\ -1&=& 1 \cdot [(x+1)^3-1] \\ -1&=& (x+1)^3-1 \quad &|\quad +1\\ 0&=& (x+1)^3 \\ (x+1)^3 &=& 0 \quad &|\quad \sqrt[3]{}\\ x+1 &=& \sqrt[3]{0}\\ x+1 &=& 0 \quad &|\quad -1 \\ \mathbf{x} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)

(2k+1) ³ ausmultiplizieren

\(\begin{array}{|rcll|} \hline && (2k+1)^3 \\ &=& (2k+1)^2\cdot (2k+1) \\ &=& [ (2k)^2 + 2\cdot 2k \cdot 1 + 1^2 ]\cdot (2k+1) \\ &=& ( 4k^2 + 4k + 1 )\cdot (2k+1) \\ &=& ( 4k^2 + 4k + 1 )\cdot 2k + ( 4k^2 + 4k + 1 )\cdot 1 \\ &=& ( 4k^2 + 4k + 1 )\cdot 2k + 4k^2 + 4k + 1 \\ &=& 4k^2\cdot 2k + 4k\cdot 2k + 1\cdot 2k + 4k^2 + 4k + 1 \\ &=& 8k^3 + 8k^2 + 2k + 4k^2 + 4k + 1 \\ &=& 8k^3 + 12k^2 + 6k + 1 \\\\ \mathbf{(2k+1)^3} & \mathbf{=} & \mathbf{ 8 k^3+12 k^2+6 k+1 }\\ \hline \end{array} \)

Sei a eine Folge, die folgendermaßen rekursiv definiert ist:

a₁ = 2

a_n+1 = 2 −  1/a_n 

Zeigen Sie mittels vollst. Induktion: an = (n+1)/ n .

\(\begin{array}{|rcll|} \hline a_{n+1}:= 2 - \frac{1}{a_n}\qquad a_1 = 2 \qquad \text{ Vermutung } a_n = \frac{n+1}{n} \quad n \in N \\ \hline \end{array} \)

Induktionsanfang:

\(\begin{array}{|rcll|} \hline n=1 \qquad a_1 = \frac{1+1}{1} = 2~ \checkmark \\ \hline \end{array}\)

Induktionsvoraussetzung:

\(\begin{array}{|rcll|} \hline a_n = \frac{n+1}{n} \quad n \in N \\ \hline \end{array}\)

Induktionsschritt: \(n\rightarrow n+1\)

\(\begin{array}{|rcll|} \hline a_{n+1}= 2 - \frac{1}{a_n} &\overset{IV}{=}& 2-\frac{1}{\frac{n+1}{n}} \\ &=& 2-\frac{n}{n+1} \\ &=& \frac{2\cdot (n+1)-n}{n+1} \\ &=& \frac{(n+1)+(n+1)-n}{(n+1)} \\ &=& \frac{(n+1)+1}{(n+1)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a_n = \frac{n+1}{n} \text { oder }\\ a_{n+1} = \frac{(n+1)+1}{(n+1)}\\ \hline \end{array}\)

Express

\(P(z)=z^4-z^3+z^2+2a\)

as a product of two real quadratic factors given that 1+i is a root of P(z) and a is a real number.

\(\begin{array}{|lcll|} \hline \text{Let } z_1 = 1+i\\\\ \text{If } z_1 \text{ is a root of } P(z) \text{ then } (z-z_1)=(z-(1+i)) \text{ divides } P(z) \\ \text{If } (1+i) \text{ is a root of } P(z) \text{ then the complex conjugate } (1-i)=z_2 \text{ is a root of } P(z) \\ \text{If } z_2 \text{ is a root of } P(z) \text{ then } (z-z_2)=(z-(1-i)) \text{ divides } P(z) \\ \text{The first real quadratic factor is }\\ (z-(1+i))\cdot (z-(1-i)) \\ = [(z-1)-i]\cdot [(z-1)+i]\\ = (z-1)^2 -i^2 \\ = (z-1)^2 +1 \\ = z^2-2z+1+1 \\ = z^2-2z+2 \\\\ P(1+i)=0\\ (1+i)^4-(1+i)^3+(1+i)^2+2a = 0 \\ (1+i)^2[ (1+i)^2-(1+i)+1]+2a = 0 \\ (1+i)^2[ (1+i)^2-i]+2a = 0 \\ (1+2i+i^2)( 1+2i+i^2 -i)+2a = 0 \\ (1+2i-1)( 1+2i-1-i )+2a = 0 \\ ( 2i )( i )+2a = 0 \\ 2i^2 +2a = 0 \\ -2 +2a = 0 \\ 2a = 2 \\ a = 1 \\\\ \text{also } P(1-i)=0 \Rightarrow a = 1 \\\\ \text{The second real quadratic factor is } z^4-z^3+z^2+2 : (z^2-2z+2) = z^2+z+1 \\ \text{so } z^4-z^3+z^2+2 = (z^2-2z+2)\cdot ( z^2+z+1 ) \\ \hline \end{array}\)

wie bestimme ich den Bruch von 0,0262626262626(periodisch)

Wir bezeichnen 0,0262626262626(periodisch) mit a,

also a = 0,0262626262626(periodisch)

also sind 10a = 0,262626262626(periodisch)

und es sind 1000a = 26,2626262626(periodisch)

somit sind

\(\begin{array}{|rcll|} \hline 1000a - 10a &=& 26,2626262626(\text{periodisch}) - 0,262626262626(\text{periodisch}) \\ 990a &=& 26,2626262626(\text{periodisch}) - 0,262626262626(\text{periodisch}) \\ 990a &=& 26 \quad &| \quad :990\\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{26}{990}}\\ \hline \end{array} \)

Der Bruch

\(\begin{array}{|rcll|} \hline \frac{26}{990} = 0,0262626262626(\text{periodisch})\\ \hline \end{array} \)