Mitglied: heureka

$\dfrac { 650\ \text{Euro} } { 17,5\ \text{Std.} } \\\\ = 37\ \dfrac{\text{Euro}}{\text{Std.} }\\\\ =\dfrac { 37\ \text{Euro} } { 1\ \text{Std.} } \\\\$

37 Euro pro 1 Stunde

heureka18.02.2015

+26396

Wie kann man geradem und ungeraden dreisatz unterscheiden ?

Siehe gerader Dreisatz: $\frac{a}{b} = \frac{c}{x}$

http://lehrerfortbildung-bw.de/faecher/mathematik/bs/bej/dreisatz/anl_gerade/anleitung_gerader_dreisatz.pdf

Siehe ungerader Dreisatz: $a*b = c*x$

http://lehrerfortbildung-bw.de/faecher/mathematik/bs/bej/dreisatz/anl_ungerade/anleitung_ungerader_dreisatz.pdf

Beispiel Mischrechnung Dreisatz:

http://lehrerfortbildung-bw.de/faecher/mathematik/bs/bej/dreisatz/anl_misch/anleitung_mischungsrechnung.pdf

heureka18.02.2015

+26396

+10

What is the LCM of 8847 and 2532 ?

$\small{\text{The following formula reduces the problem of computing the least common multiple}}\\ \small{\text{to the problem of computing the greatest common divisor (GCD), }}\\ \small{\text{also known as the greatest common factor:}}\\\\ \small{\text{ $ \boxed{lcm$(a,b)=\dfrac{|a\cdot b|}{\text{gcd}(a,b)} } $ . }}$

There is the Euclidean algorithm for computing the GCD.

$\small{\text{gcd$(8847,2532)$}} \\ \small{\text{$ \begin{array}{rrrr} & & q & r\\ 8847 & 2532 & 3 & 1251 \\ 2532 & 1251 & 2 & 30\\ 1251 & 30 & 41 & 21\\ 30 & 21 & 1 & 9 \\ 21 & 9 & 2 & \textcolor[rgb]{1,0,0}{3} \\ 9 & 3 & 3 & $ algorithm Ends $ 0 \end{array} $}}\\\\ \small{\text{gcd$(8847,2532)$}} = 3 \\$

$\small{\text{lcm$(8847,2532)=\dfrac{|8847\cdot 2532|}{3} = 7466868 $}}$

heureka18.02.2015

+26396

Zwei Orte A und B möchten an einem in der Nähe befindlichen, geradlinigen Abschnitt CD eines Flussufers eine gemeinsame Trinkwasseraufbereitungsanlage E errichten. Wo ist diese anzulegen, damit die Gesamtlänge der geradlinigen Rohrleitungen AE und BE möglichst niedrig ist ?

$AE=l_1: \quad l_1^2=6^2+x^2\qquad BE=l_2: \quad l_2^2=12^2+(30-x)^2$

$\\\small{\text{Gesamtl$\ddot{a}$nge Rohrleitungen L = }}\ l_1+l_2 =\sqrt{6^2+x^2}+\sqrt{12^2+(30-x)^2}\\\\ \small{\text{$L(x)$ soll zum Minimum werden: }} \qquad L(x) = \sqrt{36+x^2}+\sqrt{144+(30-x)^2}\\\\ L'(x) = \frac{\not{2}x} {\not{2} \sqrt{36+x^2} } +\frac{\not{2}(30-x)(-1)} {\not{2} \sqrt{144+(30-x)^2} }\\ \small{\text{$L'(x)$ wird auf 0 gesetzt: }} \qquad \frac{x} {\sqrt{36+x^2} } = \frac{(30-x)} {\sqrt{144+(30-x)^2} } \quad | \quad ()^2 \\\\ \small{\text{$\frac{x^2} {36+x^2} = \frac{(30-x)^2} {144+(30-x)^2 }$}} \\\\ \small{\text{$x^2*(144+(30-x)^2) = (36+x^2)(30-x)^2 $}}\\ \small{\text{$144x^2+x^2(30-x)^2 = (36+x^2)(30-x)^2 $}}\\ \small{\text{$144x^2 = (36+\not{x^2}-\not{x^2})(30-x)^2 $}}\\ \small{\text{$144x^2 = 36(30-x)^2 \quad | \quad : 36 $}} \\ \small{\text{$4x^2 = (30-x)^2 $}} \\ \small{\text{$4x^2 = 900-60x+x^2$}} \\ \small{\text{$3x^2 = 900-60x \quad | \quad : 3$}} \\ \small{\text{$x^2 = 300-20x $}}$

$\\\small{\text{$ x^2 +20x- 300 = 0 $}} \\ \small{\text{$ x_{1,2}= -10 \pm \sqrt{100+300} $}} \\ \small{\text{$ x_{1,2}= -10 \pm 20 $}} \\ \\ \small{\text{$ x_1= -10 + 20 $}} \\ \small{\text{$x_1 = 10 $}} \\ \\ \small{\text{$x_2 = -10-20 $}} \\ \small{\text{$x_2 = -30 \quad $ keine L$\ddot{o}$sung! } }$

heureka18.02.2015

+26396

Wieviel gramm eis von 0 grad können geschmolzen und anschließend als dampf von 100 grad verwendet werden, wenn eine wärmemenge von 60212,4 J zur verfügung stehen ?

$\\Q_1 = m \cdot \lambda_S \quad \small{\text{ W$\ddot{a}$rmemenge zum Schmelzen des Eises mit $\lambda_S = 333,5 \frac{kJ}{ kg}$ }} \\ Q_2 = m \cdot c_{\text{Wasser}} \cdot \Delta T \quad \small{\text{ Erw$\ddot{a}$rmen auf $100 \ensurement{^{\circ}}C,\quad c_{Wasser} = 4,18\ \frac{kJ}{ kg\ K}$ }} \\ Q_3= m \cdot \lambda_D \quad \small{\text{ W$\ddot{a}$rmemenge zum Verdampfen des Wassers mit $\lambda_D = 2257\ \frac{kJ}{ kg}$ }} \\\\ Q_1+Q_2+Q_3 = Q = m ( \lambda_S + c_{\text{Wasser}} \cdot \Delta T+\lambda_D ) \\\\ \small{\text{ $ m = \dfrac{Q} { ( \lambda_S + c_{\text{Wasser}} \cdot \Delta T+\lambda_D) } = \dfrac{60212,4\ J\cdot \frac{1\ kJ}{1000\ J}} { (\ 333,5 + 4,18 \cdot 100+2257 \ )\frac{kJ}{ kg} } =0,02001409340\ kg $}}\\\\ m=20\ g$

heureka18.02.2015

+26396

siehe: "Eis zu Wasserdampf machen" Seite 6 ff.

und "Schmelz- und Verdampfungswärme" Seite 5

http://www.hochschule-bochum.de/fileadmin/media/izk/Mensch%20und%20Technik%2003Berechnungen%20und%20Beispiele.pdf

heureka17.02.2015

+26396

Two trains, each traveling at 60 kph [1 kilometer/minute] are on parallel tracks heading toward one another. One train is one-half kilometer long and the other train is two-thirds kilometer long. The fronts of the two trains pass the same point A at the same time. How many seconds does it take for the two trains to completely pass each other ?

$v= 1\cdot \dfrac{km}{\text{min.}}\\\\ 1.\ \text{ train speed*time: } \quad s_1 = vt \\ 2.\ \text{ train speed*time: } \quad s_2 = vt \\\\ 1.\ \text{ train long: } \quad l_1 = \frac{1}{2}\ km \\ 2.\ \text{ train long: } \quad l_2 = \frac{2}{3}\ km \\\\ \boxed{s_1+s_2=l_1+l_2}\\\\ l_1+l_2 = vt+vt=2vt\\\\ \boxed{t=\dfrac{l_1+l_2}{2v}}\\\\ t = \dfrac{1}{2\cdot \dfrac{1\ km}{\ \text{min.}} }\cdot \left( \dfrac{1}{2} + \dfrac{2}{3} \right)\ km$

$\\t = \dfrac{1}{2}\cdot \dfrac{7}{6} \ \text{min.} \\\\ t = 0.58 \overline{3} \ \text{min.}\\ \\ t = \dfrac{1}{2}\cdot \dfrac{7}{6} * 60 \ s \\\\ t = \dfrac{7*10}{2}\ s = 7*5 \ s \\ \\ t = 35 \ s$

heureka17.02.2015

+26396

x²-2x- 8 = 0 solve by factorising quadratic equation

$\\(-2)*(+4) = -8 \\ (-2) +4 = 2\\ \boxed{(x-4)(x+2) = x^2-2x-8=0 } \\ x_1 = -2\\ x_2 = 4$

heureka16.02.2015

+26396

+11

see: https://en.wikipedia.org/wiki/Derangement

$\small{\text{Suppose that a professor has had 4 of his students }} \\ \small{\text{ - student A, student B, student C, and student D - }}\\ \small{\text{take a test and wants to let his students grade each other's tests. }}\\ \small{\text{Of course, no student should grade his or her own test. }}\\ \small{\text{How many ways could the professor hand the tests back }}\\ \small{\text{to the students for grading, }}\\ \small{\text{such that no student received his or her own test back? }}\\ \small{\text{Out of 24 possible permutations (4!) for handing back the tests, }}\\ \small{\text{there are only 9 derangements: }}\\ \boxed{BADC, BCDA, BDAC,CADB, CDAB, CDBA,DABC, DCAB, DCBA.}$

$\begin{array}{cccccr} 1& 2& 3& 4& &\small{\text{ natural Position}} \\ A& B& C& D \\ A& & & & &A \small{\text{ natural Position }} 1 \\ & B& & & &B \small{\text{ natural Position }} 2 \\ & & C& & &C \small{\text{ natural Position }} 3 \\ & & & D & &D \small{\text{ natural Position }} 4 \\ \end{array}\\\\ \small{\text{Fixed-Point-Free:}}$

$\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 1. \small{\text{ Derangement }} & B& A& D& C \\ B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & & & C & \end{array} $ }} \\\\\\\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 2. \small{\text{ Derangement }} & B& C& D& A \\ B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & & & A & \end{array} $ }}$

$\\\small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 3. \small{\text{ Derangement }} & B& D& A& C \\ B \small{\text{ not at his own(natural position) Position }} 2 & B & & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & & & C & \end{array} $ }} \\\\\\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 4. \small{\text{ Derangement }} & C& A& D& B \\ C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & &D& & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & & & B & \end{array} $ }}$

$\\\small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 5. \small{\text{ Derangement }} & C& D& A& B \\ C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & & & B & \end{array} $ }} \\\\\\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 6. \small{\text{ Derangement }} & C& D& B& A \\ C \small{\text{ not at his own(natural position) Position }} 3 & C & & & & \\ D \small{\text{ not at his own(natural position) Position }} 4 & & D & & & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & & & A & \end{array} $ }}$

$\\\small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 7. \small{\text{ Derangement }} & D& A& B& C \\ D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & A & & & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & & & C & \end{array} $ }} \\\\\\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 8. \small{\text{ Derangement }} & D& C& A& B \\ D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & &A& & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & & & B & \end{array} $ }} \\\\\\\ \small{\text{ $ \begin{array}{lcccr} & 1& 2& 3& 4 \\ 9. \small{\text{ Derangement }} & D& C& B& A \\ D \small{\text{ not at his own(natural position) Position }} 4 & D & & & & \\ C \small{\text{ not at his own(natural position) Position }} 3 & & C & & & \\ B \small{\text{ not at his own(natural position) Position }} 2 & & &B& & \\ A \small{\text{ not at his own(natural position) Position }} 1 & & & & A & \end{array} $ }}$

$\small{\text{In every other permutation of this 4-member set, }}\\ \small{\text{at least one student gets his or her own test back.}}$

heureka16.02.2015

+26396

+15

What does the ! mean ?

The subfactorial or left factorial, written, is the number of ways that n objects can be arranged where no object appears in its natural position (known as "derangements.")

The formula:

$\begin{array} $ !n = n! \cdot \displaystyle\sum^{n}_{i = 0} \frac{(-1)^i}{i!} =n!\cdot\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots +(-1)^n \ \frac{1}{n!} \right)$ \end{array}$

Example !6:

$\begin{array} $ !6 = 6!\cdot\left(1-\dfrac{1}{1!}+\dfrac{1}{2!} -\dfrac{1}{3!} +\dfrac{1}{4!} -\dfrac{1}{5!} + \dfrac{1}{6!} \right)= 265 $ \end{array}$

Example !7:

$\small{\text{$ \begin{array}{rcl} !7 &=& 7! * \left(1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{6!}-\dfrac{1}{7!}\right)\\\\ &=& 5040*\left(\not{1}-\not{1}+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720}-\frac{1}{5040}\right) \\\\ &=& \frac{5040}{2}-\frac{5040}{6}+\frac{5040}{24}-\frac{5040}{120}+\frac{5040}{720}-\frac{5040}{5040}\\\\ &=& 2520-840+210-42+7-1\\\\ &=&1854 \end{array}$}}$

heureka16.02.2015

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