heureka

-9x - 3 > 51  | +3

-9x      > 54  | : -9 !

   x      < -6 

Hi Melody,

Convert (4, 300°) to rectangular form

$$\\x= 4 * \cos{(300\ensuremath{^\circ}} )}=4*\cos{(60\ensuremath{^\circ}} )}=4*\frac{1}{2}=2\\ y=4*\sin{(300\ensuremath{^\circ}} )}=4*( -\sin{(60\ensuremath{^\circ}} )} )=-4*\frac{1}{2}*\sqrt{3}=-2\sqrt{3}\\ (2,-2\sqrt{3})$$

$$\\ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } =\lambda p_1\\\\ \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\lambda p_2$$

$$\dfrac{ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } { \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\dfrac{\lambda p_1}{\lambda p_2}$$

$$\Rightarrow (1) \quad \dfrac {p_1}{p_2}=\dfrac{3}{2}\dfrac{x_2}{x_1}$$

$$\begin{array}{rcccc} &I.&&II.&\\\\ (2)\qquad p_1x_1+p_2x_2=m \quad | \quad &\times&\dfrac{1}{p_2} \quad | \quad &\times&\dfrac{1}{p_1} \end{array}$$

$$I.\quad \dfrac{p_1}{p_2}x_1+x_2 = \dfrac{m}{p_2} \quad \Rightarrow \quad \dfrac{3}{2}\dfrac{x_2}{x_1}x_1+x_2=\dfrac{m}{p_2}\quad \Rightarrow \quad \dfrac{5}{2}x_2=\dfrac{m}{p_2}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_2=\dfrac{2}{5}\dfrac{m}{p_2}}}$$

$$II.\quad x_1+\dfrac{p_2}{p_1}x_2 = \dfrac{m}{p_1}\quad\Rightarrow \quad x_1+ \dfrac{2}{3}\dfrac{x_1}{x_2}x_2=\dfrac{m}{p_1} \quad \Rightarrow \quad \dfrac{5}{3}x_1=\dfrac{m}{p_1}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_1=\dfrac{3}{5}\dfrac{m}{p_1}} }$$

5 cm : 1.3 m  ?

$$\\\dfrac{5{cm} }{1.3m}=\dfrac{5\not{c}\not{m}*\dfrac{1m}{100\not{c}\not{m}}} {1.3m}=\dfrac{5\not{m}}{100*1.3\not{m}}=\dfrac{5}{130}\\\\ =\dfrac{1}{26}= 0.03846153846$$

I remember the question is:

solve for all values of "x" using whatever method

Die Polynomdivision liefert die Faktorisierung der Ausgangsgleichung:   $$-x^3+2x^2-x+2=0$$

$$-x^3+2x^2-x+2=\textcolor[rgb]{1,0,0}{(-x^2-1)(x-2)}=0$$

Hieraus ergeben sich die Lösungen der qubischen Gleichung, wenn man die Faktoren einzelnt null setzt:

$$(\underbrace{ -x^2-1}_{=0}) (\underbrace{ x-2 }_{=0}) =0$$

1.)  $$x-2=0 \quad \Rightarrow \textcolor[rgb]{0,0,1}{x = 2}$$

2.) $$-x^2-1=0 \quad \Rightarrow x^2+1=0$$

$$\\x^2=-1\\x_{1,2}=\pm\sqrt{-1}\\ \textcolor[rgb]{0,0,1}{x = x_1=i} \quad \text(imagin\ddot are L\ddot osung) \\ \textcolor[rgb]{0,0,1}{x=x_2=-i} \quad \text(imagin\ddot are L\ddot osung)$$

Es gibt nur eine reelle Lösung x = 2

Thank you, Melody

$$\left({\sqrt{5}}\right)^6 \quad ?$$

Für die Wurzel aus 5 kann man auch schreiben: $$\sqrt{5}=\left(5\right)^ {\frac{1}{2}}$$

Oben eingesetzt haben wir nun:   $$\left( 5^{\frac{1}{2}} \right)^{6}$$

$$\left({\sqrt{5}}\right)^6 =\left( 5^{\frac{1}{2}} \right)^{6} =5^{\left(\frac{1}{2}*6\right)} =5^{\left(\frac{6}{2}\right)} =5^{\left(\frac{3}{1}\right)} =5^{3} =5*5*5 =125$$

$$\dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$

$$\dfrac{1}{x}=\dfrac{20-14}{14*20}=\dfrac{6}{14*20}$$

$$x=\dfrac{14*20}{6}=46.6666666667$$  

for the wife alone 46.6666666667 days

P.S.

$$\begin{array}{lcr} \frac{1\;V}{20\;days} & \mbox{rate for 1 day} & \mbox{man}\\\\ \frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{woman}\\\\ \frac{1\;V}{20\;days} + \frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{man and woman}\\\\ \left(\frac{1\;V}{20\;days} + \frac{1\;V}{x\;days}\right) & \times \mbox{ 14 days} &=1\mbox{ V} \end{array}\\\\\\ \Rightarrow \dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$