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?

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

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