heureka

wie stelle ich folgende Formel nach r um?

0=4pi-2*16/r^3

$$\begin{array}{lcl} 0 & = & 4\pi-2*\dfrac{16}{r^3 } \quad | \quad :4 \\ \\ 0 & = & \pi -2* \dfrac{4}{r^3} \\ \\ 0 & = & \pi -\dfrac{8}{r^3} \quad | \quad +\dfrac{8}{r^3} \\ \\ \dfrac{8}{r^3} & = & \pi \quad | \quad \updownarrow \\\\ \dfrac{r^3}{8} & = & \dfrac{1}{\pi} \quad | \quad *8 \\ \\ r^3 & = & \dfrac{8}{\pi} \\\\ r & = & \sqrt[3]{ \dfrac{8}{\pi} } \\\\ r & = & \sqrt[3]{ \dfrac{2^3}{\pi} } \\\\ r & = & \dfrac{2}{\sqrt[3]{ \pi }} } \\\\ \end{array}$$

Was ist die Hauptstadt der Slovakei : Bratislava 

$$\small{\text{ $ \frac{10 \sqrt40}{\sqrt5} =10 \sqrt{ \frac {40} {5} }= 10\sqrt{8} = 10 \sqrt{4*2}= 10*2*\sqrt{2} = 20\sqrt{2} = 28.2842712475 $ }}$$

what is 1190 divisible by

$$\small{\text{ prime factorization: $ 1190 = 2^{\textcolor[rgb]{1,0,0}{1}}*5^{\textcolor[rgb]{0,1,0}{1}}*7^{\textcolor[rgb]{0,0,1}{1}}*17^{\textcolor[rgb]{0,1,1}{1}} $ }}$$

$$\small{\text{ The number of the divisors $ = (\textcolor[rgb]{1,0,0}{1} + 1) *(\textcolor[rgb]{0,1,0}{1} + 1) *(\textcolor[rgb]{0,0,1}{1} + 1) *(\textcolor[rgb]{0,1,1}{1} + 1) = 2*2*2*2 = 16$ }}$$

$$\begin{array}{|r|c|c|c|c|c|l|l|} \hline n & 1 & 2 & 5 & 7 & 17 & & devisor\\ \hline 1 & x & & & & & &= 1 \\ 2 & & x & & & & &= 2 \\ 3 & & & x & & & &= 5 \\ 4 & & & & x & & &= 7 \\ 5 & & & & & x & &=17 \\ 6 & & x & x & & & 2*5 &= 10 \\ 7 & & x & & x & & 2*7 &= 14 \\ 8 & & x & & & x & 2*17 &= 34 \\ 9 & & & x & x & & 5*7 &= 35 \\ 10 & & & x & & x & 5*17 &= 85 \\ 11 & & & & x & x & 7*17 &= 119 \\ 12 & & x & x & x & & 2*5*7 &= 70 \\ 13 & & x & x & & x & 2*5*17 &= 170 \\ 14 & & x & & x & x & 2*7*17 &= 238 \\ 15 & & & x & x & x & 5*7*17 &= 70 \\ 16 & & x & x & x & x & 2*5*7*17 &= 1190 \\ \hline \end{array}$$

A sprinkler system on a farm is set to spray water over a distance of 35 meters and to rotate through an angle of 140degrees. find the area of the region that can be watered

$$\\ \small{\text{ $A=\pi r^2 $ is the area of the circle. $ A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) $ is the area of the Region can be watered. }} \\ \small{\text{ $ r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2$ }}\\ \small{\text{ The area of the Region can be watered is $ \textcolor[rgb]{1,0,0}{1496.62\;m^2 } $ }}$$

Find an equation of the line containing the given pair of points (-5,-8) and (-9,-2)

$$\small{\text{ $\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1} \Rightarrow \boxed{y=\left( \dfrac{y_2-y_1}{x_2-x_1}\right)x+\dfrac{y_1x_2-x_1y_2}{x_2-x_1}} $ }} \\ \\ \small{\text{ $P_1 (x_1=-5, y_1=-8) \qquad P_2 (x_2=-9, y_2=-2) $ }} \\ \\ \small{\text{ y=\left( \dfrac{ (-2)-(-8)}{(-9)-(-5) }\right)x +\dfrac{(-8)(-9)-(-5)(-2)} { (-9)-(-5) } $ }} \\ \\ \small{\text{ $ y=- \dfrac{ 6 }{ 4 }x +\dfrac{ 72-10 } { -4 } $ }} \\ \\ \small{\text{ $ y=-1.5x -15.5$ }}$$

Check: -8 = -1.5(-5) - 15.5  ok

            -2 = -1.5(-9) - 15.5  ok

the equation c2=p can have zero, one, or two solutions. what can you say about the number of solutions of the equation c3=p?

$$\small{\text{ $c^n$ = p has always n solutions ( real and imaginary solutions together ) }}$$

what is the remainder of 2663 divided by 259?

$$\dfrac{2663}{259} = 10 \small{\text{ remainder }} 73 = 10*259 + 73$$

Solve 3x^3 - 21x^2 - 50x + 220 = 0

$$\begin{array}{c|l} \quad & \quad 3x^3-21x^2-50x+220=0 \quad |\; :3 \\ \quad & \quad \\ x^3+ax^2+bx+c=0 \quad & \quad x^3-7x^2-\frac{50}{3}x+\frac{220}{3}=0 \\ \quad & \quad \\ \quad & \quad a=-7; \; b=-\frac{50}{3}; \;c=\frac{220}{3} \\ \quad & \quad \\ B=\frac{3b-a^2}{3} \quad C=\frac{2a^3-9ab+27c}{27} \quad & \quad B=\frac{3(-\frac{50}{3})-(-7)^2}{3}=-\frac{99}{3}=-33 \quad \\ \quad & \quad \\ \quad & \quad C=\frac{2(-7)^3-9(-7)(-\frac{50}{3})+27(\frac{220}{3})}{27} =-\frac{-686-1050+1980}{27}=\frac{244}{27} \quad \\ \quad & \quad \\ d=\sqrt{-\frac{4}{3}B} \quad & \quad d=\sqrt{-\frac{4}{3}(-33)}=\sqrt{4*11}=2\sqrt{11}=6.633249581\\ \quad & \quad \\ q=\dfrac{C}{d^3}} \quad & \quad q=\frac{244}{27}\frac{1}{8(\sqrt{11})^3}}=\frac{61}{54}\frac{1}{(\sqrt{11})^3}=0.030963286\\ \quad & \quad \\ 4q <1 \;! \small{\text{ 3 real solutions }}\quad & \quad 4q = 0.123853144 < 1\;!\\ \quad & \quad \\ \phi\ensurement{^{\circ}}= \sin^{-1}{(4q)} \quad & \quad \phi\ensurement{^{\circ}}= \sin^{-1}{(0.123853144)} \\ \quad & \quad \\ \quad & \quad \phi\ensurement{^{\circ}}=7.114531178\ensurement{^{\circ}}\\ \quad & \quad \\ x_1=d*sin{(\frac{1}{3}\phi\ensurement{^{\circ}})}-\frac{a}{3} \quad & \quad x_1=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*7.11451178\ensurement{^{\circ}} ) }=\frac{7}{3}+2\sqrt{11}*0.041378847 \\ \quad & \quad x_1=2.607809554\\ \quad & \quad \\ x_2=d*sin{(\frac{1}{3} (\phi\ensurement{^{\circ}} +360\ensurement{^{\circ}} ) )}-\frac{a}{3} \quad & \quad x_2=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}* (7.11451178\ensurement{^{\circ}}+360\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*0.844594254 \\ \quad & \quad x_2=7.935737816\\ \quad & \quad \\ x_3=d*sin{(\frac{1}{3} (\phi\ensurement{^{\circ}} +720\ensurement{^{\circ}} ) )}-\frac{a}{3} \quad & \quad x_3=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}* (7.11451178\ensurement{^{\circ}}+720\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*(-0.885973101) \\ \quad & \quad x_3=-3.543547371\\ \end{array}$$

4^x+3=10^2-x

$$\\\small{\text{ $ 4^{x+3}=10^{2-x} \quad 4^x*4^3=10^2*10^{-x} \quad (4*10)^x=\frac{5^2}{4^2} \quad 40^x = 1.25^2 \; | \;\ln $ }} \\ \small{\text{ $ x\ln{(40)}= 2*\ln{(1.25)} \quad x = 2*\frac{\ln{(1.25)}}{\ln{(40)}} \quad x=0.12098175291$ }}$$