heureka

Try your hand on this problem and see how to goes:

(Computer simulation)

The EXACT probability is:  $\frac{129\ 230\ 640}{6^{11}} = \frac{129\ 230\ 640}{362\ 797\ 056} = 0.356206418609968 = 35.621\ \%$

Auf wie viele Arten kann man 9 verschiedene Geschenke an 4 Kinder verteilen, wenn das jüngsteKind 3 und alle anderen 2 Geschenke erhalten?

Sind die Geschenke alle verschieden, so ist die Anzahl aller Möglichkeiten, $r_1$ Geschenke an das erste Kind, $r_2$ Geschenke an das zweite Kind, usw. zu verteilen, gleich dem Multinomialkoeffizienten

$\dfrac{n!}{r_1!r_2!\dots r_k!}$ .

$n=9$

$\begin{array}{rcll} \dfrac{n!}{2!2!2!3!} &=& \dfrac{9!}{2!2!2!3!} \\\\ &=& \dfrac{ 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9 } {2!2!2!3!} \\\\ &=& \dfrac{ 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9 } {2\cdot 2\cdot 2 \cdot 6} \\\\ &=& \dfrac{ 362880 } {48} \\\\ &=& 7560 \end{array}$

Auf 7560 verschiedene Arten können 9 verschiedene Geschenke an 4 Kinder verteilt werden, wenn 3 Kinder 2 Geschenke erhalten und ein Kind 3 Geschenke.

100th power

6300/6600 to the 100th power

$\begin{array}{rcll} \left( \frac{ 6300 } { 6600 } \right) ^{100} &=& \left( \frac{ 21 } { 22 } \right) ^{100} \\\\ &=& \left( 0.9\overline{54} \right) ^{100} \\\\ \mathbf{ \left( \frac{ 6300 } { 6600 } \right) ^{100} } & \mathbf{=} & \mathbf{0.00954248292} \end{array}$

(sin x * tan x) / (1 - cos x) - 1 = sec x
Show that left equals right.

$\begin{array}{rcll} \frac{ \sin{( x )} \cdot \tan{( x )} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \sec {( x )} \qquad & | \qquad \sec {( x )} = \frac{1}{\cos {( x )}}\\\\ & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin{( x )} \cdot \tan{( x )} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \qquad & | \qquad \tan {( x )} = \frac{\sin {( x )} }{\cos {( x )}} \\\\ \frac{ \sin{( x )} \cdot \frac{\sin {( x )} }{\cos {( x )}} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} - \cos {( x )}\cdot [ 1 - \cos{( x )} ] } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} - \cos {( x )} + \cos^2{( x )} } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \qquad & | \qquad \sin{( x )}^2 + \cos^2 {( x )} = 1 \\\\ \frac{ [ 1 - \cos {( x )} ] } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ 1 } { \cos {( x )} } & = & \frac{1}{\cos {( x )}} \end{array}$

(1 + sin x)/(cos x) + (cos x)/(1 + sin x) = (4 sin x)/(sin 2x) {nl} Show that left equals right, and sin 2x is a double-angle identity. I despise Trig, haha.

$\begin{array}{rcll} \frac{ 1 + \sin{( x )} } { \cos {( x )} } + \frac{ \cos {( x )} } { 1 + \sin {( x )} } &\overset{?}{=}& \frac{ 4 \sin {( x )} } { \sin {( 2x )} } \qquad & | \qquad \sin {( 2x )} = 2\cdot \sin {( x )}\cos {( x )} \\\\ &\overset{?}{=}& \frac{ 4 \sin {( x )} } { 2\cdot \sin {( x )}\cos {( x )} } \\\\ &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 1 + \sin{( x )} } { \cos {( x )} } + \frac{ \cos {( x )} } { 1 + \sin {( x )} } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ [1 + \sin{( x )}]^2 + [\cos^2 {( x )}] } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 1+2\sin{( x )} + \sin{( x )}^2 + \cos^2 {( x )} } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \qquad & | \qquad \sin{( x )}^2 + \cos^2 {( x )} = 1 \\\\ \frac{ 1+2\sin{( x )} + 1} { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2+2\sin{( x )} } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2\cdot[ 1+\sin{( x )} ] } { \cos {( x )}\cdot [1 + \sin {( x )} ] } &\overset{?}{=}& \frac{ 2 } { \cos {( x )} } \\\\ \frac{ 2 } { \cos {( x )} } &=& \frac{ 2 } { \cos {( x )} } \end{array}$

What is the first term of the geometric sequence presented in the table below?
 

n      3      6

a_n  12   −324

$\small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} }$

$\begin{array}{rcll} a_3 &=& a_i = 12 \qquad & i = 3\\ a_6 &=& a_j =-324 \qquad & j = 6\\ a_1 &=& a_k = \ ? \qquad & k = 1 \end{array}$

$\begin{array}{rcll} a_1 &=& 12 ^{ \left( \frac{6-1}{6-3} \right)} \cdot (-324)^{ \left( \frac{1-3}{6-3} \right) } \\ a_1 &=& 12 ^{\left( \frac{5}{3} \right) } \cdot (-324)^{ \left( \frac{-2}{3} \right) } \\ a_1 &=& \frac{ 12 ^{\left( \frac{5}{3} \right) } } { (-324)^{ \left( \frac{2}{3} \right) } }\\ a_1 &=& \sqrt[3]{ \frac{ 12^5 } { (-324)^{2} } } \\ a_1 &=& \sqrt[3]{ \frac{ 248832 } { 104976 } } \\ a_1 &=& \sqrt[3]{ 2.37037037037 } \\ a_1 &=& 1.33333333333 \\ a_1 &=& \frac43 \\ \end{array}$

partially differentiate $F = \frac{4xy}{x^2+y^2}$

$F_x = F \cdot ( \frac{1}{x} - \frac{2x}{x^2+y^2} ) \\ F_y = F \cdot ( \frac{1}{y} - \frac{2y}{x^2+y^2} )$

circle inversion (unit circle (R=1) )  and axis-rotation 90 degrees.

circle inversion: $x' = \frac{R^2\cdot x}{x^2+y^2}\\ y' = \frac{R^2\cdot y}{x^2+y^2}\\ r'\cdot r = R^2$

What is $a_{30}$

$\small{ \boxed{~ \text{arithmetric sequence: }\quad a_k = a_i \frac{j-k}{j-i} + a_j\frac{k-i}{j-i} ~} }\\ \small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} }$

$a_6 = a_i = 50 \qquad i = 6\\ a_{11}= a_j =35 \qquad j = 11\\ a_{30}= a_k = \ ? \qquad k = 30$

$a_{30} = 50 \cdot ( \frac{11-30}{11-6} )+35\cdot ( \frac{30-6}{11-6} ) \\ a_{30}=50 \cdot ( \frac{-19}{5} ) +35\cdot ( \frac{24}{5} ) \\ a_{30}=-19\cdot 10 + 7 \cdot 24\\ a_{30}=-190+168\\ a_{30}=-22$

qHow many terms are in the arithmetic sequence $7,\ 0,\ -7,\ \dots \ , \ -175$

$\begin{array}{rcl} a_n &=& a_1 +(n-1)\cdot d\\ n-1 &=& \frac{ a_n-a_1 } {d}\\ n &=& 1+\frac{ a_n- a_1 }{d} \end{array}$

$a_n = -175 \qquad a_1 = 7 \qquad d=-7\\ n = 1 + \frac{-175-7}{-7}\\ n= 1+ 25+1\\ \mathbf{n=27}$

The answer is A.