heureka

solve the quadratic equation: (x-1)^2 = (2x+7)^2

$$(x-1)^2 = (2x+7)^2 \quad | \quad \pm\sqrt{} \\
\pm(x-1)=\pm(2x+7)\\
I. +(x-1) = +(2x+7) \quad \Rightarrow \quad x-1=2x+7\\
x_1= -8\\
II. +(x-1) = -(2x+7)\quad \Rightarrow \quad x-1=-2x-7\\
x_2 = -2$$

 $$\small{\text{
$
-a + a * ( \frac{326}{27} )^4 = a* [ (\frac{326}{27})^4 -1]
= a* [ \frac{326^4}{27^4} -1]
=a*( \frac{326^4-27^4}{27^4}) = a*21251.760280068718823
$
}}$$

Mit deiner Lösungsstrategie bist du richtig.

a=0.000523021746816

b=-0.1389327200201879

c=12.1384096902733719

Die Teiler:

           Primfaktorzerlegung

binär    2 3 5 7

1          0 0 0 1                    = 0 * 2 + 0 * 3 + 0 * 5 + 1 * 7  = 7

2          0 0 1 0                    = 0 * 2 + 0 * 3 + 1 * 5 + 0 * 7  = 5

3          0 0 1 1                    = 0 * 2 + 0 * 3 + 1 * 5 + 1 * 7  = 35

4          0 1 0 0                    = 0 * 2 + 1 * 3 + 0 * 5 + 0 * 7  = 3

5          0 1 0 1                    = 0 * 2 + 1 * 3 + 0 * 5 + 1 * 7  = 21

6          0 1 1 0                    = 0 * 2 + 1 * 3 + 1 * 5 + 0 * 7  = 15

7          0 1 1 1                    = 0 * 2 + 1 * 3 + 1 * 5 + 1 * 7  = 105

8          1 0 0 0                    = 1 * 2 + 0 * 3 + 0 * 5 + 0 * 7  = 2

9          1 0 0 1                    = 1 * 2 + 0 * 3 + 0 * 5 + 1 * 7  = 14

10        1 0 1 0                    = 1 * 2 + 0 * 3 + 1 * 5 + 0 * 7  = 10

11        1 0 1 1                    = 1 * 2 + 0 * 3 + 1 * 5 + 1 * 7  = 70

12        1 1 0 0                    = 1 * 2 + 1 * 3 + 0 * 5 + 0 * 7  = 6

13        1 1 0 1                    = 1 * 2 + 1 * 3 + 0 * 5 + 1 * 7  = 42

14        1 1 1 0                    = 1 * 2 + 1 * 3 + 1 * 5 + 0 * 7  = 30

15        1 1 1 1                    = 1 * 2 + 1 * 3 + 1 * 5 + 1 * 7  = 210

und die Zahl 1 als sechzenten Teiler

27=1.5*i1+8*i1-8*i2

27=9.5*ii1-8*i2

------------------------

-26=2*i2-8*i1+8*i2

-26=-8*i1+10*i2

10*i2=-26+8i1

i2=-2.6+0.8*i1

----------------------

27=9.5*i1-8( -2.6+0.8*i1)

27=9.5*i1+20.8-6.4*i1

3.1*i1=6.2

i1=2

------------------------

i2=-2.6+0.8*i1=-2.6+0.8*2=-2.6+1.6=-1

i1=2   and  i2=-1

$$\small{\text{
I.
$
\sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right)
\quad $
because: $ \quad b*(p_1)+b*(p_2)+b*(p_3)+... = b*(p_1+p_2+p_3+...)$
}}$\\\\$$$

$$\\\small{\text{
II.
$\prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-(m+1)+1} $\quad
because: $\ \prod \limits_{from}^{to}(a) \right)= a^{\text{to}-\text{from}+1} \quad $ example: $\ \prod \limits_{4}^{5}(a) \right)= a*a=a^{5-4+1}=a^2$
}}$\\$
\small{\text{
$\prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-m}$
}}$\\\\$$$

$$\small{\text{
III.
$
b \sum\limits_{m=0}^{n-1}a^{(n-1-m)} \right)
=b \sum\limits_{m=0}^{n-1}a^{(m)} \quad $ because: $a^{n-1}+a^{n-2}+\dots+a^1+a^0 = a^0+a^1+\dots + a^{n-2}+a^{n-1}$
}}$\\\\$$$

$$\small{\text{
IV. \ m = l:
$ \quad b \sum\limits_{m=0}^{n-1}a^{(m)} =
b \sum\limits_{l=0}^{n-1}a^{(l)} $
}}$\\\\$$$

$$\\\small{\text{
V.
$ \sum\limits_{l=0}^{n-1}a^{(l)} = a^0+a^1+a^2+\dots + a^{n-2}+a^{n-1} \quad $ this is a geometric series
}}$\\\\$
\small{\text{
The sum $s$ is:
$
a^0+a^1+a^2+\dots + a^{n-2}+a^{n-1}
$
}}$\\$
\small{\text{
$a*s$ is:
$
a^1+a^2+\dots + a^{n-1}+a^n
$
}}$\\$
\small{\text{
$s-a*s$ is:
$
a^0-a^n=1-a^n
$
}}$\\$
\small{\text{
$s(1-a)=1-a^n
$
}}$\\$
\small{\text{
$s=\frac{1-a^n}{ 1-a }$
}}$\\$
\small{\text{
$\sum\limits_{l=0}^{n-1}a^{(l)} = \dfrac{1-a^n}{ 1-a }
$
}}$$

$$\small{\text{
Result:
$
\sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right)
=b*\dfrac{1-a^n}{ 1-a }
$
}}$\\\\$$$

7 3/8 -1 1/4 = ?

$$7 \frac{3}{8} -1 \frac{1}{4}\\\\
=7 + \frac{3}{8} -1 - \frac{1}{4}\\\\
=7 -1 + \frac{3}{8} - \frac{1}{4}\\\\
=6 + \frac{3}{8} - \frac{1}{4}\\\\
=6 + \frac{3}{8} - \frac{1*2}{4*2}\\\\
=6 + \frac{3}{8} - \frac{2}{8}\\\\
=6 + \frac{1}{8}\\\\
=6\frac{1}{8}$$

Wieviele Elemente hat die Menge {k ∈ N | k teilt  210}

Primfaktorzerlegung: $$210=2^{\textcolor[rgb]{1,0,0}{1}}*3^{\textcolor[rgb]{0,1,0}{1}}*5^{\textcolor[rgb]{0,0,1}{1}}*7^{\textcolor[rgb]{0,1,1}{1}}$$

 $$\text{Anzahl Teiler }=
(1+\textcolor[rgb]{1,0,0}{1})
*(1+\textcolor[rgb]{0,1,0}{1})
*(1+\textcolor[rgb]{0,0,1}{1})
*(1+\textcolor[rgb]{0,1,1}{1})=2*2*2*2 = 16$$

if Voltage source is 12, resistance is 6800 and voltage 1 is 8 what is R2

$$\small{\text{series circuit
}}$\\$
\small{\text{$
R=6800\ \Omega \quad U = 12\ V\quad U_1 = 8\ V\quad R_2=\ ? \ \Omega
$
}}$\\$
\small{\text{
I.
$
\quad I=\frac{R_1}{U_1}=\frac{R_2}{U_2} \quad \Rightarrow \quad \boxed{R_2=\frac{U_2}{U_1}*R_1}
$
}}$\\\\$
\small{\text{
II.
$
\quad R=R_1+R_2 \quad \Rightarrow \quad \boxed{R_1=R-R_2}
$
}}$\\\\$
\small{\text{
III.
$
\quad U=U_1+U_2 \quad \Rightarrow \quad \boxed{U_2=U-U_1}
$
}}$\\\\$
\small{\text{
we substitute $U_2=U-U_1 $ and $R_1=R-R_2$
}}$\\\\$
\small{\text{
$
R_2=\left(\frac{U-U_1}{U_1}\right) *(R-R_2)\quad \Rightarrow \quad \boxed{\textcolor[rgb]{1,0,0}{ R_2 = \left( 1-\frac{U_1}{U} \right) *R } } $
}} $\\\\$
\small{\text{
$
R_2 = \left( 1-\frac{8\ V}{12\ V} \right) * 6800 \ \Omega = \left( 1-\frac{2}{3} \right) * 6800 \ \Omega =\frac{1}{3}* 6800 \ \Omega =2266.\overline{6} \ \Omega
$
}} $\\\\$
\small{\text{
$
\textcolor[rgb]{1,0,0}{R_2 = 2266.\overline{6} \ \Omega }
$
}}$$

 x=1/2, -2      what is p-q? 

$$x_1 = \frac{1}{2} \ and \ x_2=-2$$

$$2x^2-px+q=0 \quad | \quad :2 \\
x^2\underbrace{-\frac{p}{2}}_{P}x\underbrace{+\frac{q}{2}}_{Q}=0\\
x^2+ Px+Q=0 \\\\
\[[Vieta\]]\ : -(x_1+x_2) = P = -\frac{p}{2} \ and \ x_1*x_2 = Q = \frac{q}{2} \\\\
\small{\text{
$
p-q = -2P-2Q = (-2)[ -(x_1+x_2) ] - 2x_1*x_2 = 2(x_1+x_2-x_1*x_2) =2[\frac{1}{2}-2-\frac{1}{2}*(-2)]=2(\frac{1}{2}-2+1)=-1
$
}}
\\
\textcolor[rgb]{1,0,0}{p-q=-1}$$