heureka

Gesucht ist eine Natürliche Zahl 0 <= k <=6 sodass 22222^55555+55555^22222 kongruent zu k mod 7 ist

$$\small{ \text{
$
22222^{55555}+55555^{22222} \quad | \quad 22222 = 4 \mod 7 \qquad 55555= 3 \mod 7
$
}}\\
\small{ \text{
$
\equiv 4^{55555}+3^{22222} \\
$
}}\\
\small{ \text{
$
\equiv ( 4^{5} )^{11111}+( 3^{2} )^{11111 } \quad | \quad 4^5 = 2 \mod 7 \qquad 3^2= 2 \mod 7
\\
$
}}\\
\small{ \text{
$
\equiv 2^{11111}+ 2^{11111 }
$
}}\\
\small{ \text{
$
\equiv 2*2^{11111}
$
}}\\
\small{ \text{
$
\equiv 2^{11112 }
$
}}\\
\small{ \text{
$
\equiv (2^{24})^{463} \quad | \quad 2^{24} = 1 \mod 7
$
}}\\
\small{ \text{
$
\equiv 1^{463} \mod 7 \quad | \quad 1^{463} = 1
$
}}\\
\small{ \text{
$
\equiv 1 \mod 7
$
}}$$

k=1

9*(ln(x) / x)^2.

 $$\small{
\text{
Je kleiner x wird, desto mehr strebt der Quotient
$\frac{\ln(x)}{x} $ gegen$ - \infty $. }$\\$ \text{ Da wir aber noch quadrieren $\left(\frac{\ln(x)}{x} \right)^2 $ wird daraus ein $+\infty!$
}}$$

Gesucht ist ein Natürliche Zahl 0<= k <= 6 sodass die summe = 1²+2²+3²+....+10000² kongruent zu k mod 7 ist.

$$\small{
$
\boxed{1^2+2^2+3^2+....+n^2 = \frac{n(n+1)(2n+1)}{6}} \\\\
\text{$
1^2+2^2+3^2+....+10000^2 = \frac{10000(10000+1)(2*10000+1)}{6} =\frac{10000*10001*20001}{6}=333383335000 $}
$
}$$

333383335000 mod 7 = 2

k= 2

(x-(i-1))*(x-(i+1))

$$\small{\text{
$
(x-(i-1))*(x-(i+1)) =[(x-i)+1] [(x-i)-1]= (x-i)^2-1
$
}}\\
\small{\text{
$
=x^2-2xi+i^2-1 \quad | \quad i^2=-1$
}}\\
\small{\text{
$
=x^2-2xi-2 = (x^2-2) - 2xi $
}}$$

ln ( 10.75 + 6.98i )

$$\small{\text{
$z=10.75+6.98\ i \quad \ln{(z)} = ?$
}}\\
\small{
Solution:
\boxed{ln{(z)} = \ln{(|z|)} + i*(arg(z)+2\pi k ) }
}
\\\\
\small{\text{
I. $ |z| = \sqrt{10.75^2+6.98^2} = 12.8172891050 $
}}\\
\small{\text{
II. $ \phi\ensurement{^{\circ}} = \tan^{-1}{(\frac{6.98}{10.75} )} = 32.9957575073\ensurement{^{\circ}} $
}}\\
\small{\text{
III. $ arg(z) = \phi \ensurement{^{\circ}} *
\frac{ \pi }{ 180\ensurement{^{\circ}} }+2\pi k = 0.57588460769 + 2\pi k$
}}\\\\
\small{\text{
$
\ln{(z)} = \ln{(12.8172891050 )}+i*(0.57588460769+2\pi k)
$
}}\\
\small{\text{
$
\boxed{ \ln{(10.75+6.08\ i)} = 2.55079497086+(0.57588460769+2\pi k) \ i } \quad k=0,1,2\dots
$
}}$$

$$\small{
\begin{array}{rrrrrrr}
1.& 85g =& & 4 \times 15g &+ 1 \times 25g& \\
2.& 85g =& 1 \times 5g &+ 2 \times 15g & & + 1 \times 50g\\
3.& 85g =& 1 \times 5g &+ 2 \times 15g & + 2 \times 25g \\
4.& 85g =& 2 \times 5g & & + 1 \times 25g & + 1 \times 50g\\
5.& 85g =& 2 \times 5g & & + 3 \times 25g \\
6.& 85g =& 2 \times 5g &+ 5 \times 15g \\
7.& 85g =& 3 \times 5g &+ 3 \times 15g & + 1 \times 25g \\
8.& 85g =& 4 \times 5g &+ 1 \times 15g & & + 1 \times 50g\\
9.& 85g =& 4 \times 5g &+ 1 \times 15g & + 2 \times 25g \\
10.& 85g =& 5 \times 5g &+ 4 \times 15g \\
11.& 85g =& 6 \times 5g &+ 2 \times 15g & + 1 \times 25g \\
12.& 85g =& 7 \times 5g & & & + 1 \times 50g \\
13.& 85g =& 7 \times 5g & & + 2 \times 25g \\
14.& 85g =& 8 \times 5g &+ 3 \times 15g \\
15.& 85g =& 9 \times 5g &+ 1 \times 15g & + 1 \times 25g \\
16.& 85g =& 11 \times 5g &+ 2 \times 15g \\
17.& 85g =& 12 \times 5g & & + 1 \times 25g \\
18.& 85g =& 14 \times 5g &+ 1 \times 15g \\
19.& 85g =& 17 \times 5g \\
\end{array}
}$$

See the question was: Height of a right triangle without area :

$$\\(1) \quad Area = \dfrac{h_c*c}{2} \\\\
(2) \quad Area = \dfrac{a*b}{2}\\\\
\dfrac{h_c*c}{2} = \dfrac{a*b}{2} \\\\
h_c*c = a*b \\\\
h_c= \dfrac{a*b}{c} \\\\$$

Whats the pattern for 1 -3 9 ? 81 ? -243. ? ?

$$\begin{array}{lrcl}
n& & & \\
1& 1 & = & 3^0 \\
2& -3 &=& (-3)^1 \\
3& 9 &=& 3^2\\
4& -27 &=& (-3)^3 \\
5& 81 &=& 3^4\\
6&-243 &=& (-3)^5\\
7&729 &=& 3^6\\
8&-2187 &=& (-3)^7\\
\dots
\end{array}
\\
\small{\text{
The formula is: $(-3)^{n-1}$
}}$$

the coordinates of the points A and B are (-4,5) and (-5,-4) respectively.A' is reflection image of A with respect to y-axis.B is rotated anticlockwise about the origin O through 90^。     to B'

isA'B perpendicular to AB'?explain your answer.

$$\small{\text{
The vector dot product $\vec{A'B} *\vec{AB'}=0$, if $\vec{A'B}$ perpendicular $\vec{AB'}$ .
}}\\$
\small{\text{
We calculate: $\vec{A'B} = \vec{A'}-\vec{B}=\left(\begin{array}{c}4\\5\end{array}\right)-\left(\begin{array}{c}-5\\-4\end{array}\right)=\left(\begin{array}{c}4-(-5)\\5-(-4)\end{array}\right)}=\left(\begin{array}{c}9\\9\end{array}\right)$
}}$\\$
\small{\text{
and calculate: $\vec{AB'} = \vec{A}-\vec{B'}=\left(\begin{array}{c}-4\\5\end{array}\right)-\left(\begin{array}{c}4\\-5\end{array}\right)=\left(\begin{array}{c}-4-4)\\5-(-5)\end{array}\right)}=
\left(\begin{array}{c}-8\\10\end{array}\right)
$
}}$\\$
\small{\text{
$\vec{A'B} *\vec{AB'} =\left(\begin{array}{c}9\\9\end{array}\right)
*\left(\begin{array}{c}-8\\10\end{array}\right) =9*(-8)+9*10=-72+90=18
$
}}$\\$
\small{\text{
$18 \ne 0 $ so $ \vec{A'B} $ not perpendicular to $ \vec{AB'}$
}}$$

How do I find the height of a triangle without the area??

$$\boxed{h_c=\dfrac{a*b}{c}}$$