heureka

find using first principles the derivative of cosx

$$\boxed{ e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} }$$

$$e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} \quad | \quad \frac {d}{dx}$$

$$\left( e^{i\phi} \right)' = ie^{i\phi} = \left( \cos{(\phi)}+i\sin{(\phi)} \right) ' = (\cos{(\phi)})'+i(\sin{(\phi)})'$$

$$ie^{i\phi} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\left( \cos{(\phi)}+i\sin{(\phi)} \right) = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\cos{(\phi)}+\underbrace{i^2}_{i^2=-1}\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\cos{(\phi)}-\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ \textcolor[rgb]{1,0,0}{ -\sin{(\phi)} } + i\textcolor[rgb]{0,0,1}{\cos{(\phi)} }= \textcolor[rgb]{1,0,0}{ (\cos{(\phi)})'}+i\textcolor[rgb]{0,0,1} {(\sin{(\phi)})'}\\ \\ \textcolor[rgb]{1,0,0}{\boxed{ (\cos{(\phi)})'= -\sin{(\phi)} }} \\ \textcolor[rgb]{0,0,1}{\boxed{ (\sin{(\phi)})'= \cos{(\phi)} }} \\$$

(12^4)*(3^-2)*(4^-2)

$$(12^4)*(3^{-2})*(4^{-2}) \\\\ = \dfrac{12^4}{(3^2)(4^2)} \\\\ = \dfrac{12^4}{(3*4)^2} \\\\ = \dfrac{12^4}{12^2} \\\\ = 12^{4-2} \\\\ = 12^{2} \\\\ = 144$$

was ist 555 zum quadrat

$$555^2 = 555*555 = 308025$$  

555 zum Quadrat ist 308025

Is 46^78+89^67 dividable with 5  ?

yes !

$$\boxed{46^{78} + 89^{67} \stackrel{?}{=} 0 \mod 5} \\\\ \small{ \underbrace{(46^{78}\mod 5)}_{ 1^{78} \mod 5} + \underbrace{(89^{67}\mod 5)}_{-1^{67}\mod 5} \stackrel{?}{=} 0 \quad | \quad (46 \mod 5 = 1) \ and \ (89 \mod 5 = 4 = -1) } \\\\\\ \small{ 1^{78} = 1 \ and \ (-1)^{67} = -1}\\\\ \small{ ( 1 \mod 5) + (-1 \mod 5) = [ (1-1) \mod 5 ]=( 0 \mod 5) = 0 \quad q.e.d. }$$

A plus b equals 1. A squared plus b squared is 2. What is a cubed plus b cubed

$$a+b=1 \\ a^2+b^2 = 2\\ a^3+b^3 = ?$$

I.

$$a^3+b^3 = (a+b)(a^2-ab+b^2) \quad | \quad a+b=1 \ and \ a^2 +b^2 = 2 \\ a^3+b^3 = 1*(2-ab) \\ a^3 + b^3 = 2 -ab$$

II.

$$(a+b)^2 = a^2+ 2ab + b^2 \quad | \quad a+b=1 \ and \ a^2+b^2 = 2\\ 1^2 = 2 + 2ab \\ -1 = 2ab \\ ab = -\frac{1}{2}$$

III.

$$a^3+b^3 = 2-ab \quad | \quad ab = -\frac{1}{2} \\ a^3+b^3 = 2 - ( -\frac{1}{2} ) \\ a^3+b^3 = 2 + \frac{1}{2} \\ \boxed{a^3+b^3 = 2.5}$$

target number 150

5,6,4,12,19

$$150 = 12 *( 19 - 5 - \dfrac{6}{4} )$$

$$150 = 6+ 4*(5+12+19)$$

The count of bacteria in a culture was 800 after 10 minutes and 1800 after 40 minutes. What was the initial size of the culture?

$$\boxed{b = b_0 * a^{t} } \\\\ b = bacteria \\ b_0=bacteria_{Initial size} \\ t= time\\ \begin{array}{lrcl} \hline (1) & 800 & = & b_0 * a^{10} \quad | \quad t = 10 \quad b = 800\\ (2)& 1800 & = & b_0 * a^{40} \quad | \quad t = 40 \quad b = 1800 \\ \hline \end{array}$$

$$(2):(1) \quad \frac{1800}{800} = \frac{b_0*a^{40} } {b_0*a^{10} }\\\\ \frac{18}{8} = \frac{a^{40} } {a^{10} } = a^{40-10}=a^{30} \quad| \quad \sqrt[30]{}\\\\ \boxed{a=\sqrt[30]{ \frac{18}{8} }} \\\\ (1) \quad b_0 = \dfrac{800}{a^{10}} = \dfrac{800} { \left(\sqrt[30]{ \frac{18}{8} }\right)^{10} } } = \dfrac{800} { \left( \frac{18}{8} }\right)^{\frac{10}{30}} } } = \dfrac{800} { \left( \frac{18}{8} }\right)^{\frac{1}{3}} } } = \dfrac{800} { \frac{ \sqrt[3]{18} } { \sqrt[3]{8} } } = \dfrac{800} { \frac{ \sqrt[3]{18} } { \sqrt[3]{2^3} } } \\\\ = \dfrac{800} { \frac{ \sqrt[3]{18} } { 2 } } = \dfrac{1600}{ \sqrt[3]{18} } = 610.514262695$$

the initial size of the culture is 610.514262695

Nullstellen von x2-1,6+0,2=0

Gesucht sind die Nullstellen von

$$x^2-1.6+0.2 = 0 \\ x^2 - 1.4 = 0 \quad | \quad +1.4\\ x^2 = 1.4 \quad | \quad \pm\sqrt{} \\ x_{1,2}= \pm\sqrt{1.4} \\ x_{1,2}= \pm 1.18321595662 \\ \boxed{x_1 = 1.18321595662 \ oder \ x_2 = -1.18321595662}$$

Die beiden Nullstellen von $$x^2-1.6+0.2 = 0$$  lauten $$x_1 = 1.18321595662$$  und $$x_2 = -1.18321595662$$

what is the greatest common factor of 1134218 and 1135620

Use the euklid algorithmn:

$$a= 1135620 \ and \ b= 1134218 \ and \ a> b\\ \\ \begin{array}{c|c|c|c|c|c} \hline a & b & q & r \\ \hline 1135620 & 1134218 & \frac{a}{b}=\textcolor[rgb]{0,1,0}{1} & \textcolor[rgb]{0,0,1}{1402} & 1135620 = \textcolor[rgb]{0,1,0}{1} * 1134218 + \textcolor[rgb]{0,0,1}{1402} \\ 1134218 (old\ b) & 1402(old\ r) & \frac{a}{b}=\textcolor[rgb]{0,1,0}{809} & \textcolor[rgb]{0,0,1}{0} &1134218 = \textcolor[rgb]{0,1,0}{809} * 1134218 + \textcolor[rgb]{0,0,1}{0}& r= 0 \ \textcolor[rgb]{1,0,0}{Stop }\\ \hline \end{array}$$

if  r = 0 then in the line before r = 1402 is the greatest common factor

What would  x^6-y^6 be when completely factored?

$$\textcolor[rgb]{1,0,0}{ a^3 + b^3 = (a + b)(a^2-ab + b^2) }\\ \textcolor[rgb]{1,0,0}{ a^3 - b^3 = (a - b)(a^2 + ab + b^2) } \\ (a^6-b^6)=(a^3 + b^3)*(a^3 - b^3 )=(a + b)(a^2-ab + b^2)(a - b)(a^2 + ab + b^2)$$

$$a=x \text{ and } b=y$$ $$(x^6-y^6)=(x + y)(x^2-xy + y^2)(x - y)(x^2 + xy + y^2)$$