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??

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