heureka

How many numbers are in the list 2008,2003,1998,...8,3

arithmetic sequence: 

$$\small{\text{ $ a_1,\, a_2,\, a_3,\, a_4,\, a_5,\, a_6,\, \cdots ,\, a_n $}} \qquad \small{\text{ $\boxed{ a_n=a_1+(n-1)\cdot d } \qquad $ or $\qquad \boxed{ n=1+\dfrac{a_n-a_1}{d} } $ }}$$

If the distance between the numbers   $$\small{\text{$d = -5$}}$$ and $$\small{\text{$a_1=2008$}}$$  and $$\small{\text{$a_n=3$}}$$  then we find n:

$$\small{\text{$ \begin{array}{rcl} n &=& 1+\dfrac{a_n-a_1}{d} \\\\ n &=& 1 + \dfrac{3-2008}{(-5)}\\\\ n &=& 1 + \dfrac{-2005}{(-5)}\\\\ n &=& 1 + \dfrac{2005}{5}\\\\ n &=& 1 + 401\\\\ n &=& 402 \end{array} $}}$$

There are 402 numbers in the list.

Find the coefficient of x^6 in the expansion of (2x+1)^12 

$$\\\small{\text{$(2x+1)^{12} $}}\\\\ \small{\text{ $ = \binom{12}{0}\cdot(2x)^{12}\cdot 1^0 + \binom{12}{1}\cdot(2x)^{11}\cdot 1^1 + \binom{12}{2}\cdot(2x)^{10}\cdot 1^2 + \binom{12}{3}\cdot(2x)^{9}\cdot 1^3 + \binom{12}{4}\cdot(2x)^{8}\cdot 1^4+ $}}\\\\ \small{\text{ $ + \binom{12}{5}\cdot(2x)^{7}\cdot 1^5 + \textcolor[rgb]{1,0,0}{ \binom{12}{6}\cdot(2x)^{6}\cdot 1^6 } + \binom{12}{7}\cdot(2x)^{5}\cdot 1^7 + \binom{12}{8}\cdot(2x)^{4}\cdot 1^8+ $}}\\\\ \small{\text{ $ + \binom{12}{9}\cdot(2x)^{3}\cdot 1^9 + \binom{12}{10}\cdot(2x)^{2}\cdot 1^{10} + \binom{12}{11}\cdot(2x)^{1}\cdot 1^{11} + \binom{12}{12}\cdot(2x)^{0}\cdot 1^{12} $}}$$

Coefficient of $$\small{\text{$x^6$}}$$  :

$$\small{\text{ $ \begin{array}{rl} &\textcolor[rgb]{1,0,0}{\binom{12}{6}\cdot(2x)^{6}\cdot 1^6} \\\\ =&\binom{12}{6}\cdot(2x)^{6}\\\\ =&\binom{12}{6}\cdot2^6\cdot x^6\\\\ =&\binom{12}{6}\cdot 64 \cdot x^6\\\\ =&924\cdot 64 \cdot x^6\\\\ =&59136\cdot x^6\\\\ \end{array} $ }}$$

The coefficient of $$\small{\text{$x^6$}}$$  is 59136

find the area of the triangle having the indicated sides and angle:

C=84.5°, a=32, b=40

$$\boxed{ 2\cdot A = a\cdot b \cdot \sin{( C ) } }\\\\ \small{\text{$2\cdot A = 32 \cdot 40 \cdot \sin{(84.5\ensurement{^{\circ}}}) $}} \\ \small{\text{$ 2\cdot A = 1280 \cdot 0.99539619837 $}}\\ \small{\text{$ 2\cdot A = 1274.10713391 $}}\\ \small{\text{$ A = 637.053566955 $}}$$

Line RS passes through points R (5, 3) and S (-1, 0). 

Line PQ is parallel to line RS and passes through points P (3, -1) and Q (-2, y). 

What are the coordinates of point Q ?

$$\small{\text{Line PQ is parallel to line RS - cross product }} \boxed{|(\vec{S}-\vec{R}) \times (\vec{Q}-\vec{P})| = 0}\\ \\ \left| \left[ \binom{-1}{0}-\binom{5}{3} \right] \times \left[ \binom{-2}{y}-\binom{3}{-1} \right] \right| = 0\\\\ \left| \binom{-6}{-3} \times \binom{-5}{y+1} \right| = 0\\\\ (-6)\cdot(y+1)-(-3)\cdot(-5) = 0\\ (-6)\cdot(y+1)-15 = 0\\ (-6)\cdot(y+1)=15\\\\ y+1=-\frac{15}{6}\\\\ y=-\frac{15}{6}-1\\\\ y=-\frac{21}{6}\\\\ y= -\frac{7}{2}\\\\ y=-3.5\\\\ \boxed{\vec{Q}=\binom{-2}{-3.5}}$$

In einem Gefäß befinden sich eine weiße(w), vier rote(r) und fünf blaue(b) Kugeln. Es werden gleichzeitig zwei Kugeln gezogen.

Baum:

1w,4r,5b ( Anzahl Kugeln = 10 )

1. Ziehung  w gezogen ( 1/10 )

                  r gezogen  ( 4/10 )

                  b gezogen ( 5/10 )

2. Ziehung w und  r gezogen ( 1 / 10 ) * ( 4 /9 )

                 w und b gezogen ( 1 / 10 ) * ( 5 / 9 )

                 r und r gezogen ( 4 / 10) * ( 3 / 9 )

                 r und b gezogen ( 4 / 10 ) * ( 5 / 9 )

                 r und w gezogen ( 4 / 10 ) * ( 1 / 9 )

                 b und r gezogen ( 5 / 10 ) * ( 4 / 9 )

                 b und b gezogen ( 5 / 10 ) * ( 4 / 9 )

                 b und w gezogen ( 5 / 10 ) * ( 1 / 9 )

a) Mit welcher Wahrscheinlichkeit werden zwei verschiedenfarbige Kugeln gezogen?

$$\small{\text{ Farben wr, wb, rb,rw, br,bw: }}\\ \small{\text{ $ \frac{1}{10}\cdot \frac{4}{9} +\frac{1}{10}\cdot \frac{5}{9} +\frac{4}{10}\cdot \frac{5}{9} +\frac{4}{10}\cdot \frac{1}{9} +\frac{5}{10}\cdot \frac{4}{9} +\frac{5}{10}\cdot \frac{1}{9} =\frac{1\cdot4+1\cdot 5+ 4\cdot 5 +4\cdot 1 + 5\cdot 4 + 5\cdot 1}{10\cdot 9} =\dfrac{58}{90}=64,\overline{4}\% $ }}$$

$$=\dfrac{ \binom11\cdot\binom41\cdot\binom50 +\binom11\cdot\binom40\cdot\binom51 + \binom10\cdot \binom41\cdot \binom51 } {\binom{10}{2}}$$

b) Wie groß ist die Wahrscheinlichkeit, dass höchstens eine der gezogenen Kugeln rot ist?

$$\small{\text{ Farben wr,wb, rb, rw, br, bb, bw: }}\\ \small{\text{ $ \frac{1}{10} \cdot \frac{4}{9} +\frac{1}{10}\cdot \frac{5}{9} +\frac{4}{10}\cdot \frac{5}{9} +\frac{4}{10}\cdot \frac{1}{9} +\frac{5}{10}\cdot \frac{4}{9} +\frac{5}{10}\cdot \frac{4}{9} +\frac{5}{10}\cdot \frac{1}{9} =\frac {1\cdot 4+ 1\cdot 5+ 4\cdot 5 +4\cdot 1 + 5\cdot 4 +5\cdot 4 + 5\cdot 1 }{10\cdot 9} =\dfrac{78}{90}=86,\overline{6}\% $ }}$$

$$=\dfrac{ \binom41\cdot\binom61 + \binom40\cdot\binom62 } {\binom{10}{2}}$$

If the density of methane gas is 0.66kg/m3, then what is the density of methane in cm3 ?

$$0.66\ \dfrac{ \rm{kg} } { \rm{m^3} } \\\\ =0.66\ \dfrac{ \rm{kg} } { \rm{m}\cdot\rm{m}\cdot\rm{m} } \\\\ =0.66\ \dfrac{ \rm{kg} } { \rm{m}\cdot\rm{m}\cdot\rm{m} } \cdot \dfrac{ 1\ \rm{m} }{ 100\ \rm{cm }} \cdot \dfrac{ 1\ \rm{m} }{ 100\ \rm{cm }} \cdot \dfrac{ 1\ \rm{m} }{ 100\ \rm{cm }}\\\\ =0.66\ \dfrac{ \rm{kg} } { \rm{\not{m}}\cdot\rm{\not{m}}\cdot\rm{\not{m}} } \cdot \dfrac{ 1\ \rm{\not{m}} }{ 100\ \rm{cm}} \cdot \dfrac{ 1\ \rm{\not{m}} }{ 100\ \rm{cm}} \cdot \dfrac{ 1\ \rm{\not{m}} }{ 100\ \rm{cm}}\\\\ =0.66\ \dfrac{ \rm{kg} } { 100\ \rm{cm } \cdot 100\ \rm{cm} \cdot 100\ \rm{cm} } \\\\ =0.66\ \dfrac{ \rm{kg} } { 100\cdot 100\cdot 100\cdot \rm{cm}\cdot \rm{cm}\cdot \rm{cm}} \\\\ =0.66\ \dfrac{ \rm{kg} } { 100\cdot 100\cdot 100\cdot \rm{cm^3}} \\\\ =\dfrac{0.66\ }{100\cdot 100\cdot 100}\cdot \dfrac{ \rm{kg} } { \rm{cm^3}} \\\\ =\dfrac{0.66\ }{10^2\cdot 10^2\cdot 10^2}\cdot \dfrac{ \rm{kg} } { \rm{cm^3}} \\\\ =\dfrac{0.66\ }{ 10^6 } \cdot \dfrac{ \rm{kg} } { \rm{cm^3}} \\\\ =0.66\cdot 10^{-6} \cdot \dfrac{ \rm{kg} } { \rm{cm^3}}$$

$$\\=6.6 \cdot 10^{-1} \cdot 10^{-6} \cdot \dfrac{ \rm{kg} } { \rm{cm^3} }\\\\ =6.6 \cdot 10^{-7}\cdot \dfrac{ \rm{kg} } { \rm{cm^3} }\\\\$$

3 1/2 - 2 2/3 = ?

$$3 \frac12 - 2 \frac23 \\\\ =3 +\frac12 - 2 -\frac23 \\\\ =3-2 +\frac12-\frac23 \\\\ =1+\frac12-\frac23 \\\\ =1\cdot \frac22 \cdot \frac33 +\frac12\cdot\frac33- \frac23\cdot \frac22\\\\ =\frac{1\cdot 2 \cdot 3+ 1\cdot 3 - 2 \cdot 2}{2\cdot 3}\\\\ =\frac{6+3-4}{6}\\\\ =\frac{5}{6}$$

or

$$3 \frac12 - 2 \frac23 \\\\ =\frac{3\cdot2+1} {2}-\frac{2\cdot 3+2}{3}\\\\ =\frac{7} {2}-\frac{8}{3}\\\\ =\frac{3\cdot 7-2\cdot 8}{2\cdot 3}\\\\ =\frac{21-16}{6}\\\\ =\frac{5}{6}$$

$$\lim \limits_{n\rightarrow +\infty} {\sqrt{(n^2-n)}-n } =\ ?$$

Wir formen um:

$$=\lim \limits_{n\rightarrow +\infty} {\sqrt{n(n-1)}-n }\\ =\lim \limits_{n\rightarrow +\infty} {\sqrt{\frac{n}{n}\cdot n(n-1)}-n }\\ =\lim \limits_{n\rightarrow +\infty} {\sqrt{\frac{1}{n}\cdot n^2(n-1)}-n }\\ =\lim \limits_{n\rightarrow +\infty} { n \sqrt{\frac{n-1}{n}}-n }\\ =\lim \limits_{n\rightarrow +\infty} { n \sqrt{1-\frac{1}{n}}-n }\\ =\lim \limits_{n\rightarrow +\infty} { n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }\\$$

Hallo Melody,

my english is not so good, but i start.

We have a Point P, the coordinate is ( x,y ). So we have P(x,y).

You can also use atan2, see examples below, to get the angle in the quadrant  (I, II, III, and IV):

$$\\\text{Formula:}\\ \alpha = \mathrm{atan2}\ {(\Delta y, \Delta x)}$$

Examples: