heureka

4• 7, 5, 8, 4, 9, 3, ..............

$$\textcolor[rgb]{1,0,0}{7}, 5, \textcolor[rgb]{1,0,0}{8}, 4, \textcolor[rgb]{1,0,0}{9}, 3, \textcolor[rgb]{1,0,0}{10}$$

3• 23, 45, 89, 177, .............

$$23, 45, 89, 177, \textcolor[rgb]{1,0,0}{353} \qquad T_n = 2^{n-1}\cdot 22 +1 \quad n = 1,2, \cdots$$

What's the distance from Barnard's star from Earth ?

Distance 5.980 ± 0.003 ly
              (1.834 ± 0.001 pc)

1 ly =   9.4607×10¹⁵ m

1 ly =  6.3241×10⁴ au (astronomical units)

1• 8723, 3872, 2387, ..........

$$8723, 3872, 2387, \textcolor[rgb]{1,0,0}{7238}$$

Let triangle ABC be A(2,-1,0), B(1,2,3), C(-1,0,1), with its orthocenter being H(a,b,c), what is a, b, and c ?

$$\\\small{\text{
$
\vec{u}=\vec{a}-\vec{c}=
\left(
\begin{array}{c}2-(-1)\\-1-0 \\0-1 \end{array}
\right)=\left(
\begin{array}{c}3\\-1 \\-1 \end{array}
\right)\qquad
\vec{v}=\vec{b}-\vec{c}=
\left(
\begin{array}{c}1-(-1)\\2-0 \\3-1 \end{array}
\right)=\left(
\begin{array}{c}2\\2 \\2\end{array}
\right)
$
}}\\\\
\small{\text{
$
u=|\vec{u}|=\sqrt{3^2+(-1)^2+(-1)^2}=\sqrt{11} \qquad
v=|\vec{v}|=\sqrt{2^2+2^2+2^2}=\sqrt{12}
$
}}\\\\
\small{\text{
$
\vec{u_0}=\dfrac{\vec{u}}{u}=
\frac{1}{\sqrt{11}}
\left(
\begin{array}{c}3\\-1 \\-1 \end{array}
\right)\qquad
\vec{v_0}=\dfrac{\vec{v}}{v}=
\frac{1}{\sqrt{12}}
\left(
\begin{array}{c}2\\2 \\2 \end{array}
\right)\qquad
$
}}\\\\
\boxed{
\small{\text{
$
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}
}\\\\
\boxed{
\small{\text{
$
\vec{H}=(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}+ \lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ] + \vec{c}
$
}}
}\\\\
\boxed{
\small{\text{
$
\vec{H}-\vec{c}=
(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]
=
(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}+ \lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]
$
}}
}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]
-\lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]
=(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}
-(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} \quad | \quad \cdot \vec{v_0}
$
}}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]\cdot \vec{v_0}
-\underbrace{\lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]\cdot \vec{v_0}}_{=0}
=(\vec{v_0}\cdot \vec{u}) \cdot (\vec{v_0}\cdot \vec{v_0} )
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]\cdot \vec{v_0}
=(\vec{v_0}\cdot \vec{u}) \cdot \underbrace{(\vec{v_0}\cdot \vec{v_0} )}_{=1}
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [\underbrace{(\vec{v}\cdot \vec{v_0})}_{=v} - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) ]
=(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [v - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) ]
=(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}$$

$$\boxed{
\small{\text{
$
\mu=
\dfrac{
(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
}
{v - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) }
\qquad
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}
}\\\\
\small{\text{
$
\vec{v}\cdot \vec{u} = \left(\begin{array}{c} 2\\2 \\2 \end{array}\right)
\cdot \left(\begin{array}{c} 3\\-1 \\-1 \end{array}\right) =6-2-2 = 2
$
}}\\\\
\small{\text{
$
\vec{v_0}\cdot \vec{u} = \dfrac{\vec{v}\cdot \vec{u}}{v} = \dfrac{2}{\sqrt{12}}
\qquad \vec{u_0}\cdot \vec{v} = \dfrac{\vec{v}\cdot \vec{u}}{u} = \dfrac{2}{\sqrt{11}}
\qquad \vec{u_0}\cdot \vec{v_0} = \dfrac{\vec{v}\cdot \vec{u}}{v\cdot u} = \dfrac{2}{\sqrt{11}\cdot \sqrt{12}}
$
}}\\\\
\small{\text{
$
\mu=
\dfrac{
\dfrac{2}{\sqrt{12}}
-\dfrac{2}{\sqrt{11}} \cdot \dfrac{2}{\sqrt{11}\cdot \sqrt{12}}
}
{\sqrt{12} - \dfrac{2}{\sqrt{11}} \cdot \dfrac{2}{\sqrt{11}\cdot \sqrt{12}} }
=\dfrac{
\dfrac{1}{\sqrt{12}}\cdot \left(2-\dfrac{4}{11} \right) }
{\dfrac{1}{\sqrt{12}}\cdot \left( 12 -\dfrac{4}{11} \right) }
= \dfrac{18}{128} = \dfrac{9}{64}
$
}}$$

$$\\\small{\text{
$
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}\\
\small{\text{
$
\vec{H}=(1-\mu )[(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]+\mu\vec{v}+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(1-\dfrac{9}{64} \right)\cdot
\left[ \dfrac{2} { \sqrt{11} } \cdot \vec{u_0} \right]
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(\dfrac{55}{64} \right)\cdot \left( \dfrac{2} { \sqrt{11} }\right) \cdot \dfrac{\vec{u}}{u}
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(\dfrac{55}{64} \right)\cdot \left( \dfrac{2} { \sqrt{11} }\right) \cdot \dfrac{\vec{u}}{\sqrt{11}}
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\dfrac{10}{64}\cdot \vec{u}+ \dfrac{9}{64}\cdot \vec{v} + \vec{c}
$
}}$$

$$\\ \small{\text{
$
\vec{H}=
\dfrac{10}{64}\cdot
\left(\begin{array}{c}3 \\ -1\\ -1\end{array}\right)
+ \dfrac{9}{64}\cdot
\left(\begin{array}{c}2 \\ 2\\ 2\end{array} \right)
+
\left( \begin{array}{c}-1 \\ 0\\ 1\end{array} \right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c}\dfrac{30}{64}+\dfrac{18}{64}-1 \\ \\ -\dfrac{10}{64}+\dfrac{18}{64}+0\\ \\ -\dfrac{10}{64}+\dfrac{18}{64}+1\end{array}\right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c} -\dfrac{1}{4} \\\\ \dfrac{1}{8} \\\\ 1 \dfrac{1}{8}
\end{array}
\right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c} -0.25 \\ 0.125 \\ 1.125
\end{array}
\right)
$
}}$$

Berechnen Sie alle komplexen Lösungen der folgenden Gleichung: z^2=-1+j {z=-sqrt(j-1), z=sqrt(j-1)} ist als Lösung nicht richtig meint der Lehrer es ist zu einfach. Wie mache ich das sonst noch ?

   k = 0,1,2, ... , n-1

$$\boxed{\ z = a+b\cdot i \qquad r = \sqrt{a^2+b^2}\ }\\\\
z^2= -1 +j \qquad a = -1 \text{ und } b = 1 \qquad r = \sqrt{(-1^2)+1^2} \quad r = \sqrt{2}\\\\
\small{\text{Wir erhalten f\"ur den Betrag $ r = \sqrt{2} $}}\\\\
\boxed{
\sqrt{z}=\sqrt{\sqrt{2}}\cdot \left(
\cos( \frac{\varphi+2k\pi } {2}) + i \cdot \sin( \frac{\varphi+2k\pi }{2} )
\right) \quad k = 0,1
}$$

 $$\boxed{\ \varphi = arg(z) = \arctan{(\frac{b}{a})}\ } \qquad \varphi = \arctan{(\frac{1}{-1})} = -\frac{\pi}{4} = \frac{3}{4}\cdot \pi\\\\
\small{\text{1. L\"osung ( k=0 ):}}\\
\sqrt{z} = \sqrt{r}\left[
\cos( \frac{\varphi}{2}) + i\cdot \sin ({\frac{\varphi}{2}) }
\right]\\\\
\sqrt{z} = \sqrt{\sqrt{2} }\left[
\cos( \frac{\frac{3}{4}\cdot \pi}{2}) + i\cdot \sin ({\frac{\frac{3}{4}\cdot \pi}{2}) }
\right]\\\\
\sqrt{z} = \sqrt{\sqrt{2} }\left[
\cos( \frac{3}{8}\cdot \pi}) + i\cdot \sin ({\frac{3}{8}\cdot \pi}) }
\right]\\\\
\boxed{\sqrt{z} = 0.45508986 + i\cdot 1.098684113}$$ 

$$\small{\text{2. L\"osung ( k=1 ):}}\\
\sqrt{z} = \sqrt{r}\left[
\cos( \frac{\varphi+2\pi}{2}) + i\cdot \sin ({\frac{\varphi+2\pi}{2}) }
\right]\\\\
\sqrt{z} = \sqrt{\sqrt{2} }\left[
\cos( \frac{\frac{3}{4}\cdot \pi+2\pi}{2}) + i\cdot \sin ({\frac{\frac{3}{4}\cdot \pi+2\pi}{2}) }
\right]\\\\
\sqrt{z} = \sqrt{\sqrt{2} }\left[
\cos( \frac{11}{8}\cdot \pi}) + i\cdot \sin ({\frac{11}{8}\cdot \pi}) }
\right]\\\\
\boxed{\sqrt{z} = -0.45508986 - i\cdot 1.098684113}$$

pq+12=3p+pr, and q-r=7, what is p ?

$$pq+12=3p+pr \quad | \quad :p \\\\
q + \frac{12}{p} = 3 + r \quad | \quad r = q-7 \\\\
\not{q} + \frac{12}{p} = 3 + \not{q}-7 \\\\
\frac{12}{p} = -4\\\\
p = -\frac{12}{4}\\\\
p = - 3$$

twice the sum of a number and 3 is equal to three times the difference of the number and 6. find the number

The number is x:

2(x+3) = 3(x-6)

2x + 6 = 3x -18

x = 18 + 6

x = 24

Your Question for all:

You can place weights on both side of weighing balance and you need
to measure all weights between 1 and 1000.
For example if you have weights 1 and 3,now you can measure 1,3 and 4
like earlier case, and also you can measure 2,by placing 3 on one side
and 1 on the side which contain the substance to be weighed.
So question again is how many minimum weights and of
what denominations you need to measure all weights from 1kg to 1000kg.

What diameter should you make a copper wire so its resistance is 0.015 $$\Omega$$ per meter of length ? 

The resistance R of a copper wire with the length l can be calculated with the following formula:

 

where

R = resistance (ohms, Ω)

ρ = resistivity (ohm meter, Ω m)  Resistivity Copper: 1.7 x 10^-8 Ω m

l = length of conductor (m)

A = cross-sectional area of conductor (m²)

d  = the nominal diameter of the wire

 

We have:

$$\frac{R}{l}= 0.015\ \frac{\Omega}{m}$$

$$\boxed{\dfrac{R}{l}= \dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
\dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
d^2 = \dfrac{4\cdot \rho}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \rho_{copper} = 1.7 \cdot 10^{-8}\ \Omega\ m\\\\\\
d^2 = \dfrac{4\cdot 1.7 \cdot 10^{-8}\ \Omega\ m}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \sqrt{} \\\\\\
d = 2 \cdot 10^{-4} \sqrt{ \dfrac{1.7}{\pi\cdot 0.015} }\ m \\\\
d = 0.00120125135\ \mathrm{m} \\
d = 1.2\ \mathrm{mm}$$