heureka

$$\small{\text{Factorize $(81b^4 - 18b^2 + 1)$. }}\\\\$$

$$\small{\text{$
\begin{array}{rcl}
81b^4 - 18b^2 + 1 \\
&=& [9b^2-1]^2 \\
&=& [(3b-1)(3b+1) ]^2\\
&=& (3b-1)^2(3b+1)^2 \\
&=& (3b-1)(3b-1) (3b+1)(3b+1) \\
d. &=& (3b+1)(3b+1) (3b-1)(3b-1) \\
\end{array}
$}}$$

$$x^{\frac23} \cdot y^{\frac13}\\\\
=(x^2)^{\frac13}\cdot y^{\frac13}\\\\
=(x^2\cdot y)^{\frac13}\\\\
=\sqrt[3]{x^2\cdot y}$$

tan A = 2/3 and tan b= -3/5 . what exact value of cot (A-B)

$$\small{\text{$
\boxed{~\cot{(A-B)}
=\dfrac{ 1+\tan{(A)} \cdot \tan{(B)} }{ \tan{(A)} - \tan{(B)} }
~}
$}}$$

$$\small{\text{$
\begin{array}{lcl}
\cot{(A-B)} &=&\dfrac
{ 1+\dfrac23 \cdot \left(\dfrac{-3}{5}\right) }
{ \dfrac23 - \left(\dfrac{-3}{5}\right)}\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac23 \cdot \left(\dfrac{3}{5}\right) }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac25 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac{19}{15} }\\\\
\cot{(A-B)} &=& \left(\dfrac35 \right) \cdot \left(\dfrac{15}{19}\right)\\\\
\mathbf{\cot{(A-B)}} & \mathbf{=} & \mathbf{\dfrac{9}{19}} \\\\
\end{array}
$}}\\\\$$

$$\\\small{\text{
In mathematics, Euler's identity (also known as Euler's equation) is the equality $e^{i \pi} + 1 = 0$
}}$$

$$\small{\text{$
\begin{array}{lcl}
(4x^{2n-2}y^{4}-6x^{n}y^{n+3}+9x^{2}y^{2n+2})(2x^{n-1}y^{2}+3xy^{n+1})\\\\
=(4x^{2n-2}y^{4}-6x^{n}y^{n+3}+9x^{2}y^{2n+2})(2x^{n-1}y^{2})
+(4x^{2n-2}y^{4}-6x^{n}y^{n+3}+9x^{2}y^{2n+2})(3xy^{n+1})\\\\
=4x^{2n-2}y^{4}2x^{n-1}y^{2}
-6x^{n}y^{n+3}2x^{n-1}y^{2}
+9x^{2}y^{2n+2}2x^{n-1}y^{2}
+4x^{2n-2}y^{4}3xy^{n+1}
-6x^{n}y^{n+3}3xy^{n+1}
+9x^{2}y^{2n+2}3xy^{n+1}\\\\
=8x^{3n-3}y^{6}\textcolor[rgb]{1,0,0}{-12x^{2n-1}y^{n+5}}\textcolor[rgb]{0,0,1}{+18x^{n+1}y^{2n+4}}\textcolor[rgb]{1,0,0}{+12x^{2n-1}y^{n+5}}\textcolor[rgb]{0,0,1}{-18x^{n+1}y^{2n+4}}
+27x^{3}y^{3n+3}\\\\
=8x^{3n-3}y^{6}+27x^{3}y^{3n+3}\\\\
=y^3( 8x^{3n-3}y^{3}+27x^{3}y^{3n})\\\\
\end{array}
$}}$$

x^x=2.  How to solve this equation?

$$\small{\text{$
\begin{array}{rcl}
x^x &=&2 \qquad | \qquad \ln{()}\\
x\ln(x) &=& \ln{(2)} \qquad | \qquad x = e^{ln{(x)}}\\
e^{ln{(x)}}\cdot \ln{(x)} & =& \ln{(2)} \qquad | \qquad z = \ln{(x)} \\
e^z\cdot z &=& \ln{(2)}\\
\end{array}
$}}\\\\
\small{\text{
If we have $\boxed{~a=z\cdot e^z~}$,
the solution is $\boxed{~z=W(a)~}$. and $\boxed{~x=e^z~} $}}\\
\small{\text{
In mathematics, the Lambert W function, is also called the omega function or product logarithm
}}$$

$$\small{\text{$
\begin{array}{lrcl}
\text{We have }& z &=& \ln{(x)}\\
\text{and } & e^z\cdot z &=& \ln{(2)}\\
\text{so } & z &=& W(\ln{(2)})\\
\text{and } & \ln{(x)}&=& W(\ln{(2)})\\\\
& e^{\ln{(x)}} &=& e^{W(\ln{(2)})}\\
\text{finally }& x &=& e^{W(\ln{(2)})}\\
\text{also } & \mathbf{ x }& \mathbf{=} & \mathbf{\dfrac{\ln{(2)}}{W(\ln{(2)})} }
\end{array}
$}}\\\\$$

$$\small{\text{ Numerical Evaluation:}}\\ \small{\text{ The $W$ function may be approximated using Newton's method, with successive approximations to}}\\ \small{\text{ $z=W(a)$ (so $a=ze^z$) being $z_{j+1}=z_j-\dfrac{z_j e^{z_j}-a}{e^{z_j}+z_j e^{z_j}}$. }}\\
\small{\text{We start the Iteration with $z_0 = 1$}}\\$$

$$\small{\text{$
\begin{array}{rclcrcl}
z_1 &=& 1 - \dfrac{ 1\cdot e^1 - \ln{(2)} }{e^1+ 1\cdot e^1 } = 0.62749729872 &\quad & x_1 &=& \dfrac{ \ln{(2)} } {z_1} = 1.10462177603\\\\
z_2 &=& z_1-\dfrac{z_1 e^{z_1}- \ln{(2)} }{e^{z_1}+z_1 e^{z_1}} = 0.24045491796 &\quad & x_2 &=& \dfrac{ \ln{(2)} } {z_2} = 2.88264921519\\\\
z_3 &=& z_2-\dfrac{z_2 e^{z_2}- \ln{(2)} }{e^{z_2}+z_2 e^{z_2}} = 0.48596644327 &\quad & x_3 &=& \dfrac{ \ln{(2)} } {z_3} = 1.42632724987\\\\
z_4 &=& z_3-\dfrac{z_3 e^{z_3}- \ln{(2)} }{e^{z_3}+z_3 e^{z_3}} = 0.44585119296 &\quad & x_4 &=& \dfrac{ \ln{(2)} } {z_4} = 1.55466036989\\\\
z_5 &=& z_4-\dfrac{z_4 e^{z_4}- \ln{(2)} }{e^{z_4}+z_4 e^{z_4}} = 0.44443778365 &\quad & x_5 &=& \dfrac{ \ln{(2)} } {z_5} = 1.55960452971\\\\
z_6 &=& z_5-\dfrac{z_5 e^{z_5}- \ln{(2)} }{e^{z_5}+z_5 e^{z_5}} = 0.44443609102 &\quad & x_6 &=& \dfrac{ \ln{(2)} } {z_6} = 1.55961046945\\\\
z_7 &=& z_6-\dfrac{z_6 e^{z_6}- \ln{(2)} }{e^{z_6}+z_6 e^{z_6}} = 0.44443609102 &\quad & x_7 &=& \dfrac{ \ln{(2)} } {z_7} = 1.55961046946\\\\
\end{array}
$}}$$

x = 1.55961046946

24% of the people at a shopping mall are men. The number of boys is 3/8 of the number of men. The number of women is 4 times the number of boys. The number of girls is 504 more than the number of men. How many males are there in the shopping mall?

x are the number of all people.

m = men,  b = boys, w = women, g = girls

$$\small{
\begin{array}{rcl}
m&=& 0,24x\\\\
b&=& \dfrac{3}{8}m \\
w &=& 4\cdot b = 4\cdot \dfrac{3}{8}m = \dfrac{3}{2}m\\
g &=& 504 + m\\\\
x &=& m+b+w+g\\\\
x &=& m + \dfrac{3}{8}m+\dfrac{3}{2}m+504 + m\\\\
x &=& 2m+ \dfrac{3}{8}m+\dfrac{3}{2}m+504 \\\\
x &=& \dfrac{16+3+12}{8}m+504\\\\
x &=& \dfrac{31}{8}m+504\\\\
x &=& \dfrac{31}{8}\cdot 0.24x+504\\\\
x &=& 31\cdot 0.03x+504\\\\
x &=& 0.93x+504\\\\
x- 0.93x &=&504\\\\
0.07x &=&504\\\\
x &=& \dfrac{504\cdot 100}{7}\\\\
x &=& 7200\\\\
m &=& 0.24x\\
m &=& 0.24\cdot 7200 \\
\mathbf{m} & \mathbf{=} & \mathbf{1728}
\end{array}
}$$

There are 1728 males in the shopping mall

Boys = 648

Woman = 2592

Girls = 2232

16% of the pens in a stationery shop are red. The number of black pens is 3/4 the number of red pens. The number of green pens is 5/7 the total number if red and black pens. The rest of the pens in the shop are blue. The number of blue pens is 180 more than the number of red, black and green pens. Find the total number of pens in the shop

x are the number of the pens.

$$\\\small{
\begin{array}{rcl}
\textcolor[rgb]{150,0,0}{\text{red}} &=& 0,16x\\\\
\text{black}& =& \dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}} \\\\
\textcolor[rgb]{0,150,0}{green} &=& \dfrac{5}{7}( \textcolor[rgb]{150,0,0}{\text{red}} +\text{black} ) = \dfrac{5}{7}\textcolor[rgb]{150,0,0}{\text{red}} + \dfrac{5}{7}\cdot \dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}} = \dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}} \\\\
\textcolor[rgb]{0,0,150}{blue}& =& 180 + ( \textcolor[rgb]{150,0,0}{\text{red}} +\text{black} + \textcolor[rgb]{0,150,0}{green} )
= 180 + \textcolor[rgb]{150,0,0}{\text{red}} +\dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}}+\dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}} = 180 + 3\cdot \textcolor[rgb]{150,0,0}{\text{red}} \\\\\\
x &=& \textcolor[rgb]{150,0,0}{\text{red}} +\text{Black}
+\textcolor[rgb]{0,150,0}{green}+\textcolor[rgb]{0,0,150}{blue}\\\\
x &=& \textcolor[rgb]{150,0,0}{\text{red}}
+\dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}}
+\dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}}
+180 + 3\cdot \textcolor[rgb]{150,0,0}{\text{red}} \\\\
x &=& 6\cdot \textcolor[rgb]{150,0,0}{\text{red}} +180\\\\
x &=& 6\cdot 0.16x+180\\\\
x &=& 0.96x+180\\\\
x-0.96x &=& 180\\\\
0.04x &=& 180\\\\
x &=& \dfrac{180\cdot 100}{4}\\\\
\mathbf{x}& \mathbf{=}& \mathbf{4500}
\end{array}
}$$

The total number of pens in the shop is 4500

red = 720

black = 540

green = 900

blue = 2340

(n+5    über

 n+3)

$$\small{\text{Formel: $ \boxed{~\binom{n}{k}=\dfrac{n!}{k!(n-k)!} ~} $
$
\begin{array}{rcl}
\dbinom{n+5}{n+3} &=& \dfrac{(n+5)!}{(n+3)![n+5-(n+3)]!} \\\\
\dbinom{n+5}{n+3} &=& \dfrac{(n+5)!}{(n+3)!(n+5-n-3)!} \\\\
\dbinom{n+5}{n+3} &=& \dfrac{(n+5)!}{(n+3)!\cdot 2!} \qquad | \qquad 2!=2\cdot 1 = 2 \\\\
\dbinom{n+5}{n+3} &=& \dfrac{(n+5)!}{(n+3)!\cdot 2} \\\\
\dbinom{n+5}{n+3} &=& \dfrac{(n+5)!}{(n+3)!\cdot 2}\qquad | \qquad (n+5)! = (n+3)! \cdot(n+4) \cdot (n+5) \\\\
\dbinom{n+5}{n+3} &=& \dfrac{ (n+3)! \cdot(n+4) \cdot (n+5)}
{(n+3)!\cdot 2}\\\\
\dbinom{n+5}{n+3} &=& \dfrac{ (n+4) \cdot (n+5)}
{2}\\\\
\mathbf{\dbinom{n+5}{n+3}} &\mathbf{=} & \mathbf{\dfrac{n^2+9n+20}
{2} }\\\\
\end{array}
$}}$$