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Wie kann ich cos(sin^-1(x)) umschreiben ??

Find the reciprocal of

\(\frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \)    

( The letter "i" refers to the complex number)

Formula:

\(\begin{array}{|rcll|} \hline z &=& x+i\cdot y \\ \frac{1}{z} &=& \frac{x}{x^2+y^2} - i\cdot \frac{y}{x^2+y^2} \\ \hline \end{array}\)

We have:

\(x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2}\)

so the reciprocal is:

\(\begin{array}{|rcll|} \hline z &=& x+i\cdot y \quad & | \quad x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2} \\ z &=& \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ (\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{(\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ \frac24+\frac24} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{\frac24+\frac24} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ 1} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{1} \\ \mathbf{\dfrac{1}{z}} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}} {2} + i\cdot \dfrac{\sqrt{2}} {2} } \\ \hline \end{array}\)

7log(3x)+3log(5x)=15 what is x

i assume log is the natural logarithm

\(\begin{array}{|rcll|} \hline 7\cdot log(3x)+3 \cdot log(5x) &=& 15 \\ 7\cdot (~log(3)+log(x) ~)+3 \cdot (~log(5)+log(x) ~) &=& 15 \\ 7\cdot log(3) + 7\cdot log(x) +3\cdot log(5) + 3\cdot log(x)&=& 15 \\ 10\cdot log(x) + 7\cdot log(3) +3\cdot log(5) &=& 15 \\ 10\cdot log(x) + log(3^7) + log(5^3) &=& 15 \\ 10\cdot log(x) + log(3^7\cdot 5^3) &=& 15 \quad & | \quad -log(3^7\cdot 5^3) \\ 10\cdot log(x) &=& 15 -log(3^7\cdot 5^3) \quad & | \quad : 10 \\ log(x) &=& \frac{ 15 -log(3^7\cdot 5^3) } {10 } \\ log(x) &=& \frac{ 15}{10} - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \log(~(3^7\cdot 5^3)^{\frac{1}{10}}~) \\ log(x) &=& 1.5 - \log(3^\frac{7}{10}\cdot 5^\frac{3}{10}) \\ log(x) &=& 1.5 - \log(3^{0.7}\cdot 5^{0.3}) \quad & | \quad e^{()} \\ x &=& e^{1.5 - \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot e^ {- \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot \frac{ 1 } { e^{\log(3^{0.7}\cdot 5^{0.3}) } } \\ x &=& e^{1.5} \cdot \frac{ 1 } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ e^{1.5} } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ 4.48168907034 } {2.15766927997\cdot 1.62065659669 } \\ x &=& \frac{ 4.48168907034 } {3.49684095207 } \\ \mathbf{x} & \mathbf{=} & \mathbf{1.28163938016178...} \\ \hline \end{array}\)

Ich fahre jeden Tag 100 km mit dem Auto. Der Sprit kostet zurzeit 1.40 pro Liter.

Mein Durschnitsverbrauch liegt bei 9 Liter.

Wie teuer ist eine fahrt?

\(\begin{array}{|rcll|} \hline && 100\ km \cdot \frac{1.40€}{1\ l} \cdot \frac{9\ l}{100\ km } \\ &=& 100\ \not{km} \cdot \frac{1.40€}{1\ \not{l} } \cdot \frac{9\ \not{l}}{100\ \not{km} } \\ &=& 100\ \cdot 1.40 \cdot \frac{9}{100}\ € \\ &=& 1.40 \cdot 9 \ € \\ &=& 12.6 \ € \\ \hline \end{array}\)

How to calculate CubeRoot in excel?? Example 6nth√1,10*1,15*1,02 HELP??

=(1,10*1,15*1,02 )^(1/6)

= 1.04339427520

square root   =(1,10*1,15*1,02 )^(1/2)

cube root      =(1,10*1,15*1,02 )^(1/3)

6th root         =(1,10*1,15*1,02 )^(1/6)

...

A car travelling at 90 km/h is 500 m behind another car travelling at 70 km/h in the same direction.

How long will it take the first car to catch the second?

\(\begin{array}{cll} \text{Let } v_1 &=& 90\ \frac{km}{h} \\ \text{Let } v_2 &=& 70\ \frac{km}{h} \\ \text{Let } d = 500\ m &=& 0.5\ km \\ \text{Let } t &=& time \\ \end{array}\)

if d is the distance:

\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& v_1\cdot t - 0.5 \\ d_{\text{car}_2} &=& v_2\cdot t \\ \hline \end{array}\)

The first car catch the second if \(d_{\text{car}_1} = d_{\text{car}_2}\)

\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& d_{\text{car}_2} \\ v_1\cdot t -0.5\ km &=& v_2\cdot t \quad & | \quad -v_2\cdot t +0.5 \\ v_1\cdot t - v_2\cdot t &=& 0.5\ km \\ t\cdot ( v_1 - v_2 ) &=& 0.5\ km \quad & | \quad :( v_1 - v_2 ) \\ t &=& \frac {0.5\ km} { v_1 - v_2 } \quad & | \quad v_1 = 90\ \frac{km}{h} \qquad v_2 = 70\ \frac{km}{h} \\ t &=& \frac {0.5\ km} { 90\ \frac{km}{h} - 70\ \frac{km}{h} } \\ t &=& \frac {0.5\ km} { 20\ \frac{km}{h} } \\ t &=& \frac {0.5} { 20 }\ h \\ t &=& 0.025\ h \quad & | \quad 1\ h = 60\min.\\ t &=& 0.025\cdot 60\min.\\ \mathbf{t} & \mathbf{=} & \mathbf{1.5\min.} \\ \hline \end{array}\)

(Korrektur)

Der minimale Abstand beträgt:
\(\begin{array}{|rcll|} \hline && \sqrt{\left( 0-\frac{\sqrt{2} }{2} \right)^2+ \left( 1-\frac12 \right)^2} \\ &=& \sqrt{\frac24+\frac14} \\ &=& \sqrt{\frac{3}{4}} \\ &=& \dfrac{\sqrt{3}}{2} \\ &=& 0,86602540378 \\ \hline \end{array} \)

Bestimme den minimalen Abstand zwischen dem Punkt P = (0,1) ∈ R^2 und der Parabel E = {(x,y) ∈ R^2 : y = x^2}.

Finde einen Punkt Q = (x₀,y₀) in der Parabel (y₀ = x^2, x₀), in dem dieser minimale Abstand angenommen wird.

Gegeben:

Punkt  \(P = (0,1) \qquad x_p = 0\quad y_p = 1\)

Parabel \(y = x^2\)

Gesucht:

Der mimimale Abstand von P zur Parabel am Punkt Q

Abstand Punkt und Parabel zum Quadrat:

\(\begin{array}{|rcll|} \hline d^2 = D &=& (y-y_p)^2 + (x-x_p)^2 \qquad & | \qquad y = x^2\\ D &=& (x^2-y_p)^2 + (x-x_p)^2 \\ D &=& x^4-2x^2y_p+y_p^2+x^2-2xx_p + x_p^2 \\ D &=& x^4+x^2(1-2y_p)-x\cdot 2x_p +x_p^2 +y_p^2 \\ \hline \end{array} \)

Mimimale Abstand (Extrempunkt):

1. Ableitung null setzen, und 2. Ableitung muss > 0 sein.

1. Ableitung von D:

\(\begin{array}{|rcll|} \hline D &=& x^4+x^2(1-2y_p)-x\cdot 2x_p +x_p^2 +y_p^2 \\ D' &=& 4x^3 +2x(1-2y_p)-2x_p \\ \hline \end{array}\)

1. Ableitung von D gleich null setzen:

\(\begin{array}{|rcll|} \hline D' = 4x^3 +2x(1-2y_p)-2x_p &=& 0 \quad & | \quad x_p = 0 \qquad y_p = 1 \\ 4x^3 +2x(1-2\cdot 1)-2\cdot 0 &=& 0 \\ 4x^3 +2x(-1) &=& 0 \\ 4x^3 -2x &=& 0 \quad & | \quad : 2 \\ 2x^3 -x &=& 0 \\ x\cdot (2x^2 -1) &=& 0 \\ \hline \end{array} \)

Überprüfung der 3 Lösungen ob ein relatives Mimimum (D'' >  0) vorliegt.

2. Ableitung von D:

\(\begin{array}{|rcll|} \hline D' &=& 4x^3 +2x(1-2y_p)-2x_p \\ D'' &=& 12x^2 + 2(1-2y_p)-2x_p \quad & | \quad x_p = 0 \qquad y_p = 1 \\ D'' &=& 12x^2 + 2(1-2\cdot 1)-2\cdot 0 \\ D'' &=& 12x^2 + 2(-1) \\ D'' &=& 12x^2 - 2 \\ \hline \end{array}\)

Die 3 Lösungen finden:

\(\begin{array}{|rcll|} \hline x\cdot (2x^2 -1) &=& 0 \\ x &=& 0 \Rightarrow x_1 = 0 \text{ keine Lösung, da } D'' = 12\cdot 0 - 2 < 0 \\\\ 2x^2 -1 &=& 0 \quad & | \quad +1 \\ 2x^2 &=& 1 \quad & | \quad : 2 \\ x^2 &=& \frac12 \\ x &=& \pm\sqrt{\frac12} \\ x_2 &=& +\sqrt{\frac12} \\ x_2 &=& \frac{ \sqrt{2} }{2} \text{ eine Lösung, da } D'' = 12\cdot (\frac{\sqrt{2} }{2})^2 - 2 > 0 \\\\ x_3 &=& -\sqrt{\frac12} \\ x_3 &=& -\frac{ \sqrt{2} }{2} \text{ eine Lösung, da } D'' = 12\cdot \frac{-\sqrt{2} }{2})^2 - 2 > 0 \\ \hline \end{array}\)

Die beiden Punkte Q sind:

\(\begin{array}{|rcll|} \hline x_{q_1} &=& \frac{ \sqrt{2} }{2} \\ y_{q_1} &=& ( \frac{ \sqrt{2} }{2} )^2 = \frac12 \\\\ x_{q_2} &=& - \frac{ \sqrt{2} }{2} \\ y_{q_2} &=& ( -\frac{ \sqrt{2} }{2} )^2 = \frac12 \\\\ \hline \end{array} \)

\(Q_1 (\frac{ \sqrt{2} }{2}, \frac12 )\)

\(Q_2 (\frac{ -\sqrt{2} }{2}, \frac12 )\)

Der minimale Abstand beträgt:
\(\begin{array}{|rcll|} \hline && \sqrt{(0-\frac{\sqrt{2} }{2})^2+(1-\frac12)^2} \\ &=& \sqrt{\frac14} \\ &=& \frac{1}{\sqrt{2}} \\ &=& \frac{\sqrt{2}}{2} \\ \hline \end{array} \)

A^B=C^D

Find D

\(\begin{array}{|rcll|} \hline C^D &=& A^B \quad &| \quad \ln() \text{ both sides }\\ \ln(C^D) &=& \ln(A^B) \quad &| \quad \ln(a^b) = b\cdot \ln(a) \\ D\cdot \ln(C) &=& B\cdot ln(A) \quad &| \quad :\ln(C) \\ \mathbf{D} & \mathbf{=} & \mathbf{B\cdot \frac{ln(A)}{ln(C)}} \\ \hline \end{array}\)

Converting from slope-intercept to standard form

how do I turn this y= 29.95x+99

\(\begin{array}{|c|c|} \hline \text{Slope-intercept form } y = mx+b & \text{Standard form } Ax+By = C \\ \hline \begin{array}{rcll} y &=& 29.95x+99 \qquad &| -29.95x \\ -29.95x + y &=& 99 \qquad &| \cdot 20 \\ -599x + 20y &=& 1980\\ \end{array} & \begin{array}{rcll} \\ \\ -599x + 20y &=& 1980\\ \end{array} \\ \hline \end{array}\)