heureka

e = one diagonal

f = the other diagonal

h ist the height of the rhombus

\(\boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\\)

or

\(\begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }}\\ h &=&\frac{ 1 }{ \frac { \sqrt{ e^2 + f^2 }}{ e\cdot f}} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{e^2}{e^2\cdot f^2} + \frac{f^2}{e^2\cdot f^2} }} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{1}{f^2} + \frac{1}{e^2} }} \\ \frac{1}{h} &=&\sqrt{ \frac{1}{f^2} + \frac{1}{e^2} } \\ \frac{1}{h^2} &=&\frac{1}{f^2} + \frac{1}{e^2} \\ \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \end{array} ~}\)

how high is rhombus if diagonals are 36 cm and 12cm

e = one diagonal

f = the other diagonal

\(e = 12\ cm\) and \(f = 36\ cm\)

The four sides all have the same length \( a\)

\(h\) ist the height of the rhombus

cos-rule

\(\begin{array}{rcll} e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 - 2a^2\cdot \cos{(B)} \\\\ 2A+2B &=& 360^\circ \\ A+B &=& 180^\circ \\ B &=& 180^\circ - A \\ \cos{(B)} &=& \cos{(180^\circ - A)} = -\cos{(A)} \\\\ e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 + 2a^2\cdot \cos{(A)} \\\\ \cos{(A)} = \frac{2a^2-e^2}{2a^2} &=& \frac{f^2-2a^2}{2a^2}\\ 2a^2-e^2 &=& f^2-2a^2 \\ 4a^2 &=& e^2 + f^2\\ 2a &=& \sqrt{ e^2 + f^2 } \\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{ \sqrt{ e^2 + f^2 } } {2} }\\ \end{array}\)

\(\begin{array}{rcll} \cos{(A)} &=& \frac{2a^2-e^2}{2a^2} \\ \cos^2{(A)} &=& \frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& 1-\frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - (2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - 4a^2 + 4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot (4a^2-e^2) }{4a^4} \qquad | \qquad 4a^2 = e^2 + f^2\\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot ( e^2 + f^2-e^2) }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \qquad | \qquad \sin^2{(A)}=1-\cos^2{(A)}\\ \sin^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \\ \mathbf{\sin{(A)}} &\mathbf{=}&\mathbf{\frac{ e\cdot f }{2a^2} } \\ \end{array}\)

\(\begin{array}{rcll} h &=& a\cdot \sin{(A)} \\ h &=& a\cdot \frac{ e\cdot f }{2a^2} \\ h &=&\frac{ e\cdot f }{2a} \qquad & | \qquad 2a = \sqrt{ e^2 + f^2 }\\ \end{array} \\ \boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\\)

\(\begin{array}{rcll} h &=&\frac{ 12\cdot 36 }{\sqrt{ 12^2 + 36^2 }} \\ h &=&\frac{ 432 }{\sqrt{ 1440 }} \\ h &=&\frac{ 432 }{37.9473319220} \\\\ \mathbf{h} &\mathbf{=}& \mathbf{11.3841995766\ cm }\\ \end{array}\)

666+666+666+6+6+6 = 999+999+9+9

800/x - 800/6x = 8

sorry i posted this again. i didnt understand the steps given. i was told to multiply it by 2000/3x or something similar. i need it either in a more basic more detailed procedure.

I asume \(\begin{array}{rcl} \frac{800}{x} - \frac{800}{6x} &=& 8 \end{array}\)

\(\begin{array}{rcll} \frac{800}{x} - \frac{800}{6x} &=& 8 \\\\ \frac{800}{x} - \frac{800}{x}\cdot \frac16 &=& 8 \\\\ \frac{800}{x} \cdot \left( 1 - \frac16 \right) &=& 8 \\\\ \frac{800}{x} \cdot \left( \frac66 - \frac16 \right) &=& 8 \\\\ \frac{800}{x} \cdot \left( \frac56 \right) &=& 8 \qquad & |\qquad \cdot x \\\\ 800 \cdot \left( \frac56 \right) &=& 8\cdot x \qquad & |\qquad :8 \\\\ \frac{800}{8} \cdot \left( \frac56 \right) &=& x \\\\ 100 \cdot \left( \frac56 \right) &=& x \\\\ \frac{500}{6} &=& x \\\\ 83.3333333333\dots &=& x \\\\ \mathbf{x} & \mathbf{=} & \mathbf{83\frac13} \\\\ \end{array}\)

Please give the next 4 terms of the following sequence. Consider it an intellectual amusement and thanks!:

1, 4, 18, 48, 100, 180, 294, 448, 648...........

\(\small{ \begin{array}{rcrcrcrcrcrcrcrcrcr} 1 &,& 4 &,& 18&,& 48&,& 100&,& 180&,& 294&,& 448&,& 648&,& \cdots\\ 1 &,& 2\cdot 2 &,&3\cdot 6&,&4\cdot 12&,&5\cdot 20&,&6\cdot 30&,&7\cdot 42&,&8\cdot 56&,&9\cdot 72&,& \cdots\\ 1 &,& 2\cdot 2\cdot 1 &,&3\cdot 3\cdot 2&,&4\cdot 4\cdot 3 &,&5\cdot 5\cdot 4&,&6\cdot 6\cdot 5&,&7\cdot 7\cdot 6&,&8\cdot 8\cdot 7&,&9\cdot 9\cdot 8&,& \cdots \end{array}\\ \begin{array}{lcl} \\ \mathbf{a_1} &\mathbf{=}& \mathbf{1}\\ \mathbf{a_n }& \mathbf{=}& \mathbf{ n^2\cdot (n-1) \qquad n> 1} \\\\ a_{10} &=& 10^2\cdot 9 = 900 \\ a_{11} &=& 11^2\cdot 10 = 1210 \\ a_{12} &=& 12^2\cdot 11 = 1584 \\ a_{13} &=& 13^2\cdot 12 = 2028 \\ \end{array} }\)

wir haben 12 blumen . 5 Tulpen, 2 Rosen, 2 Hyazinthen und 3 weitere verschiedene Blumen handelt. Wie viele Möglichkeiten zur Anordnung der verschiedene Blumenarten haben wir ?

Permutation:

\(\begin{array}{rcl} \frac{12!}{5!~\cdot 2!~\cdot 2!~\cdot 1!~\cdot 1!~\cdot 1!} &=& \frac{12!}{5!~\cdot 2!~\cdot 2!} \\\\ &=& \frac{479001600}{120~\cdot 2~\cdot 2} \\\\ &=& \frac{479001600}{4802} \\\\ &=& 997920 \end{array}\)

1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?

\(\small{ \begin{array}{rcr} a_1 &=& 1\\ a_2 &=& 6\\ a_3 &=& 20\\ a_4 &=& 51\\ a_5 &=& 189\\ a_6 &=& 517\\ a_7 &=& 2197\\ a_8 &=& 4823\\ a_9 &=& 14496\\ \end{array} }\)

\(\begin{array}{|cr|r|r|r|r|} \hline &1&20&189&2197&14496 \\ +&6&+51&+517&+4823&+\color{red}55555 \\ \hline =&7&7\underbrace{1}_{a_1}&70\underbrace{6}_{a_2}&70\underbrace{20}_{a_3}&700\underbrace{51}_{a_4} \\ \hline \end{array}\)

next comes \(\color{red}55555\)

1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?

\(\small{ \begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1} && 6 && 20 && 51 && 189 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 14 && 31 && 138 && 328 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 9} && 17 && 107 && 190 && 1352 && \cdots \\ \text{3. Difference } {\color{red}d_3 = 8} \\ \text{4. Difference } {\color{red}d_4 = 82} \\ \text{5. Difference } {\color{red}d_5 = -89} \\ \text{6. Difference } {\color{red}d_6 = 1175} \\ \text{7. Difference } {\color{red}d_7 = -4908} \\ \text{8. Difference } {\color{red}d_8 = 19363} \\ \end{array} }\)

\(\small{ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ &+& \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } + \binom{n-1}{5}\cdot {\color{red}d_5 }\\ &+& \binom{n-1}{6}\cdot {\color{red}d_6 } + \binom{n-1}{7}\cdot {\color{red}d_7 } + \binom{n-1}{8}\cdot {\color{red}d_8 }\\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 1 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 9 } \\ &+& \binom{n-1}{3}\cdot {\color{red} 8 } + \binom{n-1}{4}\cdot {\color{red} 82 } + \binom{n-1}{5}\cdot {\color{red} (-89) } \\ &+& \binom{n-1}{6}\cdot {\color{red} 1175 } + \binom{n-1}{7}\cdot {\color{red} (-4908) } + \binom{n-1}{8}\cdot {\color{red} 19363 } \\\\ a_{10} &=& \binom{9}{0}\cdot {\color{red} 1 } + \binom{9}{1}\cdot {\color{red} 5 } + \binom{9}{2}\cdot {\color{red} 9 } \\ &+& \binom{9}{3}\cdot {\color{red} 8 } + \binom{9}{4}\cdot {\color{red} 82 } + \binom{9}{5}\cdot {\color{red} (-89) } \\ &+& \binom{9}{6}\cdot {\color{red} 1175 } + \binom{9}{7}\cdot {\color{red} (-4908) } + \binom{9}{8}\cdot {\color{red} 19363 } \\ a_{10} &=& 1\cdot {\color{red} 1 } + 9\cdot {\color{red} 5 } + 36\cdot {\color{red} 9 } \\ &+& 84\cdot {\color{red} 8 } + 126\cdot {\color{red} 82 } + 126\cdot {\color{red} (-89) } \\ &+& 84\cdot {\color{red} 1175 } + 36\cdot {\color{red} (-4908) } + 9\cdot {\color{red} 19363 } \\ a_{10} &=& 1 + 45 + 324 +672 + 10332 -11214 +98700 - 176688+174267 \\ \mathbf{a_{10}} & \mathbf{=}& \mathbf{96439} \end{array} }\)

kann mir jemand bei den nullstellen zu  \(f(x)=x^2+ln(x^2)\)  helfen?

Mit der Lambertschen W-Funktion (Omegafunktion) könnte man sofort auf das Ergebnis schließen.

W(x) ist eine inverse Funktion zu\( f(x) = xe^x .\)
\(xe^x = y\) hat damit die Lösung \( x=W(y).\)

Ansonsten bleiben nur die Iterationsmethoden.

Beispiel Lambertsche W-Funktion:

Wir ersetzen \(x^2\) mit \(u\) und erhalten \(y = u+\ln{(u)} = 0\).

oder

\(\begin{array}{rcll} u &=& -\ln{(u)} \\ u &=& 0-\ln{(u)} \qquad &| \qquad \ln{(1)} = 0 \\ u &=& \ln{1}-\ln(u) \qquad &| \qquad \ln{(1)}-\ln{(u)} = \ln{(\frac{1}{u})} \\ u &=& \ln{(\frac{1}{u})}\qquad &| \qquad e^{()} \\ e^u &=& e^{\ln{(\frac{1}{u})} } \\ e^u &=& \frac{1}{u} \\ u\cdot e^u &=& 1 \\ \end{array}\)

Wir können jetzt die Lambertsche W-Funktion anwenden. \(u=W(1)\)

\(W(1) \) ist aber gerade die  Omega-Konstante\(\small{ \begin{array}{l} \Omega = 0.5671432904097838729999686622103555497538157871865125081351310792230457930866\dots \end{array} }\)

und erhalten als Ergebnis für \(x=\pm\sqrt{\Omega}\)

Die zwei Nullstellen lauten also:

\(x_1 = 0.7530891649796748157965833628063502497431317507325243408136858446\dots\)

und

\(x_2 = -0.7530891649796748157965833628063502497431317507325243408136858446\dots\)

I cant find the angle between the vertices c (1,1,1) a (1,-4,2) b (-5,2,7) Can you help me?

I set:

\(\small{ \begin{array}{lcr} a_x &=& 1\\ a_y&=&-4\\ a_z&=&2 \end{array} \qquad \begin{array}{lcr} b_x &=& -5\\ b_y&=&2\\ b_z&=&7 \end{array} \qquad \begin{array}{lcr} c_x &=& 1\\ c_y &=& 1\\ c_z &=& 1\\ \end{array} }\)

Differences of the vectors:
\(\small{ \begin{array}{lcr} \vec{b}-\vec{a} &=& \bigl(\begin{smallmatrix} b_x-a_x\\b_y-a_y\\b_z-a_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr)\\ \vec{a}-\vec{b} &=& -\bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{c}-\vec{b} &=& \bigl(\begin{smallmatrix} c_x-b_x\\c_y-b_y\\c_z-b_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr)\\ \vec{b}-\vec{c} &=& -\bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{a}-\vec{c} &=& \bigl(\begin{smallmatrix} a_x-c_x\\a_y-c_y\\a_z-c_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)\\ \vec{c}-\vec{a} &=& -\bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \\ \end{array} }\)

In the plane of the triangle (ABC) the angles are:

\(\small{ \begin{array}{rcl} \tan{(A)} &=& \frac{ \vert (\vec{c}-\vec{a}) \times (\vec{b}-\vec{a}) \vert} {(\vec{c}-\vec{a})\cdot (\vec{b}-\vec{a}) } = \frac{ \left| \bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {25 } \qquad (I.)\\ A &=& \arctan{(1.74218253923)}\\ A &=& 60.1444921468^\circ \\ \tan{(B)} &=& \frac{ \vert (\vec{a}-\vec{b}) \times (\vec{c}-\vec{b}) \vert} {(\vec{a}-\vec{b})\cdot (\vec{c}-\vec{b}) } = \frac{ \left| \bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {72 } \qquad (I.)\\ B &=& \arctan{(0.60492449279)}\\ B &=& 31.1707710617^\circ \\ \tan{(C)} &=& \frac{ \vert (\vec{b}-\vec{c}) \times (\vec{a}-\vec{c}) \vert} {(\vec{b}-\vec{c})\cdot (\vec{a}-\vec{c}) } = \frac{ \left| \bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)} = \frac{ 43.5545634808 } {1 } \qquad (I.)\\ C &=& \arctan{(43.5545634808)}\\ C &=& 88.6847367915^\circ \\ \end{array} }\)

\(A+B+C = 60.1444921468^\circ + 31.1707710617^\circ + 88.6847367915^\circ = 180^\circ \)