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: