heureka

2, 11, 20, 29...
Which term in this sequence is 227?

\(\begin{array}{|rcll|} \hline & \text{difference} \\ \hline a_1=2 \\ & \color{red}9 \\ a_2=11 \\ & \color{red}9 \\ a_3=20 \\ & \color{red}9 \\ a_4=29 \\ \ldots \\ a_n =227 \\ \hline \end{array}\)

This is a Arithmetic Sequence.

\(\text{Formula:}~ \boxed{a_n=a_1 + (n-1)d,\text{ with } a_1=2\text{ and }d = 9 }\)

\(\begin{array}{ll} \text{where:}\\ & a_1~ \text{ is the first term, and} \\ & d~ \text{ is the difference between the terms (called the "common difference")} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline 227 &=& 2 + (n-1)9 \\ 227 &=& 2 + 9n-9 \\ 227 &=& 9n-7 \\ 227+7 &=& 9n \\ 234 &=& 9n \\ n &=& \dfrac{234}{9} \\ \mathbf{n} & \mathbf{=} & \mathbf{26} \\ \hline \end{array}\)

Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.

\(\text{p in the form $4n+1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n+1)^2 - 1 \\ &=& 16n^2+8n+1-1 \\ &=& 8n(n+1) \quad\Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)

\(\text{p in the form $4n-1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n-1)^2 - 1 \\ &=& 16n^2-8n+1-1 \\ &=& 8n(n-1)\quad \Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)

\(\text{Euler's theorem}\\ \text{This states that if $a$ and $n$ are relatively prime $gcd(a,n) = 1$} \\ \text{then} \\ {\displaystyle a^{\varphi (n)}\equiv 1\mod n.} \)

\(\begin{array}{|rcll|} \hline a^{\varphi (3)} &\equiv& 1\pmod{3} \quad | \quad \varphi (3) = 2 \\ a^2 &\equiv& 1\pmod{3} \quad | \quad \text{if } gcd(a,3) = 1, \text{ so $a$ is a prime number and not } 3 \\ p^2 &\equiv& 1\pmod{3} \quad | \quad p\ne 3! \\ \text{so } p^2 -1 &\equiv& 0 \pmod{3} \quad\Rightarrow \quad 3 | p^2-1 \\ \hline \end{array}\)

\(\text{If $8 | p^2-1$ and $3 | p^2-1$ so also divides $24|p^2-1$}\)

Hallo CarlNowi,

du kannst beliebige Zahlen wählen, doch für eine einfache Rechnung habe ich die kleinsten

Zahlen eben die 1 und die -1 genommen.

exponent problem
solve
\(x\)  for  \(\large 3^{(2x+2)}+27^{(x+1)}=36\)

\(\text{Formula:}\\ \large{\boxed{\left(a^b\right)^c = a^{bc}=a^{cb}=\left(a^c\right)^b}}\)

\(\begin{array}{|rcll|} \hline 3^{(2x+2)}+27^{(x+1)} &=& 36 \quad | \quad 27 = 3^3 \\ 3^{2( x+1)}+\left(3^3\right)^{(x+1)} &=& 36 \quad | \quad \left(3^3\right)^{(x+1)} = 3^{3(x+1)}\\ 3^{2( x+1)}+ 3^{3(x+1)} &=& 36 \\ \left(3^{(x+1)}\right)^{2} + \left(3^{(x+1)}\right)^{3} &=& 36 \\ && \text{we substitute:} ~ \boxed{u = 3^{(x+1)}} \\\\ u^2+u^3&=& 36 \\ u^3+u^2-36 &=& 0 \\\\ u_1 &=& 3 \\ u_2 &=& -2-2i\sqrt{2} \\ u_3 &=& -2+2i\sqrt{2} \\\\ u &=& 3^{(x+1)} \quad | \quad u = 3 \\ 3 &=& 3^{(x+1)} \\ 3^1 &=& 3^{(x+1)} \\ 1 &=& x+1 \\ \mathbf{ x } & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

how do you expand, then express in a single trig ratio and/or constant?
\(4 \sin(x) \cos(x) \Big(\sec^3(x) \cos(x) + \csc(x) \sec(x)\Big)\)

\(\text{Formula:}\\ \text{$\csc(x) = \dfrac{1}{\sin(x)} $} \\ \text{$\sec(x) = \dfrac{1}{\cos(x)} $} \\\)

\(\begin{array}{|rcll|} \hline && 4 \cdot\sin(x) \cos(x) \Big(\sec^3(x) \cos(x) + \csc(x) \sec(x)\Big) \\\\ &=& 4 \cdot\sin(x) \cos(x) \left( \dfrac{\cos(x)}{\cos^3(x)} + \dfrac{1}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\sin(x) \cos(x) \left( \dfrac{1}{\cos^2(x)} + \dfrac{1}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\left( \dfrac{\sin(x) \cos(x)}{\cos^2(x)} + \dfrac{\sin(x) \cos(x)}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\left( \dfrac{\sin(x)}{\cos(x)} + 1 \right) \quad | \quad \dfrac{\sin(x)}{\cos(x)}=\tan(x) \\\\ &\mathbf{=}& \mathbf{ 4 \cdot \Big( \tan(x) + 1 \Big )} \\ \hline \end{array}\)

Kann mir jemand helfen , wie ich diesen Term durch Partialbruchzerlegung vereinfache?
Bitte mit Rechenweg
\(\displaystyle \dfrac{x^2+3x-4} { (x-4)*(x-2)^3 }\)

\(\small{ \begin{array}{|lrcll|} \hline & \dfrac{x^2+3x-4} { (x-4)(x-2)^3 } &=& \dfrac{A}{x-4} + \dfrac{B}{x-2}+ \dfrac{C}{(x-2)^2}+ \dfrac{D}{(x-2)^3} \quad | \quad \cdot (x-4)(x-2)^3 \\\\ & x^2+3x-4 &=& A(x-2)^3 + B(x-4)*(x-2)^2+ C(x-4)*(x-2) \\ &&& + D(x-4) \\\\ x=2 & 2^2+3\cdot 2-4 &=& A(2-2)^3 + B(2-4)*(2-2)^2+ C(2-4)*(2-2) \\ &&& + D(2-4) \\ & 6 &=& A\cdot 0 + B\cdot 0+ C\cdot 0 -2D \\ & 2D &=& -6 \\ & \mathbf{ D } & \mathbf{=} & \mathbf{-3} \\\\ x=4 & 4^2+3\cdot 4-4 &=& A(4-2)^3 + B(4-4)*(4-2)^2+ C(4-4)*(4-2) \\ &&& + D(4-4) \\ &24 &=& 8A+B\cdot 0 + C\cdot 0 +D\cdot 0 \\ & 8A &=& 24 \\ & \mathbf{ A } & \mathbf{=} & \mathbf{3} \\\\ x=1 & 1^2+3-4 &=& A(1-2)^3 + B(1-4)*(1-2)^2+ C(1-4)*(1-2) \\ &&& + D(1-4) \\ & 0 &=& -A-3B+3C-3D \quad | \quad A = 3, \quad D=-3\\ & 0 &=& -3-3B+3C-3(-3) \\ & 0 &=& -3-3B+3C+9 \quad | \quad : 3 \\ & 0 &=& -1-B+C+3 \\ & B &=& -1+C+3 \\ & \mathbf{ B } & \mathbf{=} & \mathbf{C+2} \\\\ x=-1 & (-1)^2+3\cdot(-1)-4 &=& A(-1-2)^3 + B(-1-4)*(-1-2)^2+ C(-1-4)*(-1-2) \\ &&& + D(-1-4) \\ & -6 &=& A(-3)^3 + B(-5)*(-3)^2+ C(-5)*(-3) + D(-5) \\ & -6 &=& -27A -45B+ 15C + -5D \quad | \quad A = 3, \quad D=-3\\ & -6 &=& -27\cdot 3 -45B+ 15C -5\cdot(-3) \\ & -6 &=& -27\cdot 3 -45B+ 15C+ 15 \quad | \quad B=C+2 \\ & -6 &=& -27\cdot 3 -45(C+2)+ 15C+ 15 \quad | \quad :3 \\ & -2 &=& -27-15(C+2)+ 5C+ 5 \\ & -2 &=& -27-15C-30+ 5C+ 5 \\ & -2 &=& -27-10C-30+ 5 \\ & 10C &=& 2 -27 -30+ 5 \\ & 10C &=& -50 \\ & \mathbf{ C } & \mathbf{=} & \mathbf{-5} \\\\ & B &=& C+2 \\ & B &=& -5+2 \\ & \mathbf{ C } & \mathbf{=} & \mathbf{-3} \\\\ & \mathbf{\dfrac{x^2+3x-4} { (x-4)*(x-2)^3 }} &\mathbf{=}& \mathbf{ \dfrac{3}{x-4} - \dfrac{3}{x-2}- \dfrac{5}{(x-2)^2}- \dfrac{3}{(x-2)^3} } \\ \hline \end{array} }\)

9^x=27 lösungsweg

ich habe herausgefunden das 27 = 3^3

\(\begin{array}{|rcll|} \hline 9^x &=& 27 \quad & | \quad 27 = 3^3 \\ 9^x &=& 3^3 \quad & | \quad 9 = 3^2 \\ \left(3^2 \right)^x &=& 3^3 \quad & | \quad \text{Formel: }~ \left(a^b \right)^c = a^{bc} \\ 3^{2x} &=& 3^3 \\ 2x &=& 3 \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{3}{2}} \\ \hline \end{array}\)

Find the area enclosed by the graph of the parametric equations
\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

\(\text{Second attempt:} \\ \begin{array}{|rcll|} \hline x &=& 6 \cos( t) \sin( t) \\ y &=& 6 \cos^2( t) \\ \hline x^2+y^2 &=& \Big( 6\cos( t) \sin( t) \Big)^2 + \Big( 6 \cos^2( t) \Big)^2 \\ x^2+y^2 &=& 6^2 \cos^2( t) \sin^2( t) + 6^2 \cos^2( t)\cos^2( t) \\ x^2+y^2 &=& 6^2 \cos^2( t)\Big( \underbrace{\sin^2( t) + \cos^2( t)}_{=1} \Big) \\ x^2+y^2 &=& 6^2 \cos^2( t) \\ x^2+y^2 &=& 6\cdot 6 \cos^2( t) \quad & | \quad y = 6 \cos^2( t) \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{6y} \\\\ y^2-6y+x^2 &=& 0 \\ (y-3)^2-9+x^2 &=& 0 \\ (y-3)^2+x^2 &=& 9 \\ \mathbf{(y-3)^2+(x+0)^2} & \mathbf{=} & \mathbf{3^2} \\ \hline \end{array}\\ \begin{array}{|l|} \hline \text{This is a circle with center $(0,3)$ and radius = $3$ } \\ \text{The area enclosed is $ \pi r^2 = \pi \cdot 3^2 = \mathbf{9 \pi} $ } \\ \hline \end{array}\)

Find the area enclosed by the graph of the parametric equations

\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

\(\text{Formula 1:} \\ \begin{array}{|rcll|} \hline 2 \cos( t) \sin( t) &=& \sin(2t) \quad & | \quad \cdot 3 \\ \mathbf{6 \cos( t) \sin( t) } & \mathbf{=} & \mathbf{3\sin(2t)} \\ \hline \end{array} \)

\(\text{Formula 2:} \\ \begin{array}{|rcll|} \hline 2 \cos^2( t) &=& 1+\cos(2t) \quad & | \quad \cdot 3 \\ \mathbf{6 \cos^2( t) } & \mathbf{=} & \mathbf{3+3\cos(2t)} \\ \hline \end{array}\)

\(\text{Substitute:} \\ \begin{array}{|rcll|} \hline x &=& 6 \cos( t) \sin( t) \quad & | \quad \mathbf{6 \cos( t) \sin( t) =3\sin(2t)} \\ &=& 3\sin(2t) \\ &=& 0+ 3\sin(2t) \\\\ y &=& 6 \cos^2( t) \quad & | \quad \mathbf{6 \cos^2( t)=3+3\cos(2t)} \\ &=& 3+3\cos(2t) \\\\ \hline x &=& 0+3\sin(2t) \\ y &=& 3+3\cos(2t) \\ \hline \end{array}\\ \begin{array}{|l|} \hline \text{This is a circle with center $(0,3)$ and radius = $3$ } \\ \text{The area enclosed is $ \pi r^2 = \pi \cdot 3^2 = \mathbf{9 \pi} $ } \\ \hline \end{array}\)

The graph of the parametric equations
\(\begin{align*} x&=\cos t,\\ y&=\sin t, \end{align*}\)
meets the graph of the parametric equations
\(\begin{align*} x &= 2+ 4\cos s,\\ y &= 3+4\sin s, \end{align*}\)
at two points.
Find the slope of the line between these two points.

\( \text{circle 1:} \\ \begin{array}{|rcll|} \hline x &=& \cos(t) \\ y &=& \sin(t) \\ \hline x^2+y^2 &=& \cos^2(t) + \sin^2(t) \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)

\(\text{circle 2:} \\ \begin{array}{|rcll|} \hline x &=& 2+4\cos(t) \quad \text{ or } \quad 4\cos(t) = x-2 \\ y &=& 3+4\sin(t) \quad \text{ or } \quad 4\sin(t) = y-3 \\ \hline x^2+y^2 &=& \Big( 2+4\cos(t) \Big)^2 + \Big( 3+4\sin(t) \Big)^2 \\ x^2+y^2 &=& 4+16\cos(t)+16\cos^2(t) + 9 + 24\sin(t) + 16\sin^2(t) \\ x^2+y^2 &=& 13+16\cos(t)+ 24\sin(t) + 16\Big(\cos^2(t) + \sin^2(t) \Big) \\ x^2+y^2 &=& 13+16\cos(t)+ 24\sin(t) + 16 \\ x^2+y^2 &=& 29+16\cos(t)+ 24\sin(t) \\ x^2+y^2 &=& 29+4\cdot4\cos(t)+ 6\cdot 4\sin(t) \\ x^2+y^2 &=& 29+4\cdot (x-2)+ 6\cdot (y-3) \\ x^2+y^2 &=& 29+4x-8+6y-18 \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{3+4x +6y} \\ \hline \end{array} \)

\(\text{intersection between the two circles:} \\ \begin{array}{|rcll|} \hline \text{circle 2: }~\mathbf{x^2+y^2} & \mathbf{=} & \mathbf{3+4x +6y} \quad &| \quad \text{circle 1: }~ \mathbf{x^2+y^2= 1} \\ 1 &=& 3+4x +6y \\ 6y+4x+3 &=& 1 \\ 6y &=& -4x-3 +1 \\ 6y &=& -4x-2 \\\\ y &=& \dfrac{-4x-2}{6} \\\\ y &=& -\dfrac{4}{6}x -\dfrac{2}{6} \\\\ y &=& -\dfrac{2}{3}x -\dfrac{1}{3} \\ \hline \end{array} \)

\(\text{the line between these two points:} \\ \begin{array}{|rcll|} \hline \mathbf{ y } & \mathbf{=} & \mathbf{ \underbrace{-\dfrac{2}{3}}_{\text{slope}}x -\dfrac{1}{3} } \\ \hline \end{array} \)

The slope of the line between these two points is \(\mathbf{-\dfrac{2}{3}}\).