heureka

x + 2y - 6 = 0

$$x + 2y - 6 = 0\\ 2y=-x+6\\ \boxed{y=f(x)=-\frac{1}{2}x+3}\\\\ \small{\text{ r = radius of the circle }}\\ \small{\text{ $ x=f(x)=r $ }}\\ \small{\text{ $ x=-\frac{1}{2}x+3 $ }}\\ \small{\text{$\frac{1}{2}x+x=3 $}}\\ \small{\text{$\frac{3}{2}x=3 $}}\\ \small{\text{$x=\frac{2}{3}*3=2$}}\\ \small{\text{$r=2$. The center of the circle is $(x_m,\ y_m) =\ ( 2,\ 2)$ }}\\ \small{\text{The equation of the circle is: }}\\ \small{\text{ $ (x-x_m)^2+(y-y_m)^2=r^2 \qquad (x-2)^2+(y-2)^2=2^2 $ }}$$

hello i am 11 yer alt wat is 123+123+420? please giff me answer

Die Antwort ist:  $${\mathtt{123}}{\mathtt{\,\small\textbf+\,}}{\mathtt{123}}{\mathtt{\,\small\textbf+\,}}{\mathtt{420}} = {\mathtt{666}}$$

x-2y=-2       x^2-y^2=7

$$x-2y=-2 \qquad x=2y-2 \\\\ x^2-y^2=7 \qquad (2y-2)^2-y^2=7\\ 4y^2-8y+4-y^2=7\\ 3y^2-8y+4=7\\ 3y^2-8y-3=0\\\\ \small{\text{ $ y_{1,2}=\dfrac{8\pm\sqrt{64-4*3*(-3)} } {2*3} =\dfrac{8\pm\sqrt{64-36} } {6} =\dfrac{8\pm\sqrt{100} } {6} =\dfrac{8\pm10 } {6} $ }}\\\\ \small{\text{ $ y_1=\dfrac{8+10 } {6}=\dfrac{18 } {6}=3 $ }}\\\\ \small{\text{ $ y_2=\dfrac{8-10}{6}=-\dfrac{2}{6}=-\dfrac{1}{3} $ }}\\\\ \small{\text{ $ x_1=2*3-2 = 4 $ }}\\ \small{\text{ $ x_2=2*(-\frac{1}{3})-2 = -2\frac{2}{3} $ }}$$

Zwischen x-Achse, y-Achse und der Funktion f(x)= $${{\mathtt{e}}}^{{\mathtt{\,-\,}}{\mathtt{x}}}$$  soll ein Rechteck möglichst großer Fläche gelegt werden.

$$\small{\text{ $ \boxed{A=x*f(x) \qquad \frac{dA}{dx}=1*f(x)+x*f'(x)\qquad \frac{dA}{dx}=0 \text{ setzen} } $ }}\\\\ \small{\text{ $ f(x) + x * f'(x)=0 \qquad | \quad f(x) = e^{-x} \quad f'(x) = e^{-x}*(-1) $ }}\\ \small{\text{ $ e^{-x}+x*(-1)e^{-x}=0 $ }}\\ \small{\text{ $ e^{-x}-xe^{-x}=0 $ }}\\ \small{\text{ $ xe^{-x}=e^{-x} $ }}\\ \small{\text{ $ x=\frac{e^{-x} }{ e^{-x} } $ }}\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{x=1} $ }}\\\\ \small{\text{ $ A=x*e^{-x} \quad | \quad x = 1 $ }}\\ \small{\text{ $ A_{max}=1*e^{-1} $ }}\\\\ \small{\text{ $ \textcolor[rgb]{1,0,0}{A_{max}=\dfrac{1}{e}} = 0.36787944117 $ }}\\$$

Berechnen Sie mit den Grenzwertsätzen von l´Hospital :  lim _x->0    (1-cos(x))/x² ----> =0/0

$$\small{\text{ $ \lim\limits_{x\to 0} \dfrac{1-\cos{(x)} }{x^2} \quad \stackrel{l'Hospital}{=} \quad \lim\limits_{x\to 0} \dfrac{\sin{(x)} }{2x} \quad \stackrel{l'Hospital}{=} \qaud \lim\limits_{x\to 0} \dfrac{\cos{(x)} }{2}=\frac{1}{2} \quad | \quad \lim\limits_{x\to 0} (\cos{(x)}) =\cos(0)=1 $ }}$$

(m+3)2-(m+3)(m-3)

$$(m+3)2-(m+3)(m-3)\\ (m+3)(2-(m-3))\\ (m+3)(2-m+3)\\ (m+3)(5-m)\\$$

solve the given equation for the variable:

6x - 3(x - 1) = 4(2x + 1)

$$6x - 3(x - 1) = 4(2x + 1)\\ 6x-3x-3(-1) = 4*(2x)+4*1\\ 6x-3x+3=8x+4\\ 3x+3=8x+4 \quad | \quad -3x\\ 3 = 8x-3x+4\\ 3=5x+4 \quad | \quad -4\\ 3-4=5x\\ -1=5x\\ 5x=-1\quad | \quad :5\\ x=-\frac{1}{5}$$

For triangle HJK, j = 31, m∠H = 132 degrees, m∠J = 21 degrees, and m∠K = 27 degrees. Find h to the nearest whole number.

$$\small{\text{ $ \tan{(132)}=\frac{h}{p} \qquad \tan{(27)}=\frac{h}{q} \qquad p+q=31\ m $ }}\\ \small{\text{ $ p=\frac{h}{\tan{(132)}} \qquad q=\frac{h}{\tan{(27)}} \qquad \frac{h}{\tan{(132)}} +\frac{h}{\tan{(27)}}=31\ m $ }}\\ \small{\text{ $ h* \left( \frac{1}{\tan{(132)}} + \frac{1}{\tan{(27)}} \right)=31\ m $ }}\\ \small{\text{ $ h* \left( -0.90040404430 + 1.96261050551 \right)=31\ m $ }}\\ \small{\text{ $ h*1.06220646121 =31\ m $ }}\\ \small{\text{ $ h =\frac{31}{1.06220646121}\ m = 29.1845334520\ m $ }}\\ \small{\text{ h to the nearest whole number: $ h = 29\ m $ }}$$

$$\sqrt{x+9}=9 \quad | \quad ()^2\\ x+9=81 \\ x=81-9\\ x=72$$

find three consecutive integers so that three times the middle integer is five more than the sum of the first and third.

$$3a_2=5+a_1+a_3 \quad | \quad a_1 = a_2-1 \qquad a_3=a_2+1\\ 3a_2=5+(a_2-1)+(a_2+1)\\ 3a_2=5+a_2+a_2\\ 3a_2=5+2a_2\\ a_2=5\\ a_1 = a_2-1 = 5-1=4\\ a_3=a_2+1 = 5+1=6$$

4, 5, 6