heureka

5x cubed divided by 10x to the power of 7

$\begin{array}{rcl} \dfrac{(5x)^3}{(10x)^7} \\ &=&\dfrac{(5x)^3}{(2\cdot 5\cdot x)^7} \\ &=&\dfrac{(5x)^3}{2^7\cdot (5\cdot x)^7} \\ &=&\dfrac{1} {2^7\cdot (5x)^{7-3}} \\ &=&\dfrac{1} {2^7\cdot (5x)^{4}} \\ \dfrac{(5x)^3}{(10x)^7} &=&\dfrac{1} {2^7\cdot (5x)^{4}} \\ \end{array}$

Folgende Angabe . √10x-√40x BRUCHSTRICH √10x Wie mache ich die √10x weg , also weg vom Nenner kürzen geht ja nicht aufgrund der Differenz

$\begin{array}{rcl} \dfrac{ \sqrt{10x} - \sqrt{40x} } {\sqrt{10x} } \\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x}) } {\sqrt{10x} } \cdot \dfrac{\sqrt{10x}}{\sqrt{10x}}\\\\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x})\cdot\sqrt{10x} } {\sqrt{10x}\sqrt{10x} }\\\\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x})\cdot\sqrt{10x} } { \sqrt{10x}^{\frac{1}{2}} }\\\\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x})\cdot\sqrt{10x} } { 10x^{\frac{2}{2}} }\\\\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x})\cdot\sqrt{10x} } { 10x^1 }\\\\ &=& \dfrac{ (\sqrt{10x} - \sqrt{40x})\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ \sqrt{10x}\cdot\sqrt{10x} - \sqrt{40x}\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ \sqrt{10x}^{\frac{1}{2}} - \sqrt{40x}\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ 10x^{\frac{2}{2}} - \sqrt{40x}\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ 10x^1 - \sqrt{40x}\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ 10x - \sqrt{40x}\cdot\sqrt{10x} } { 10x }\\\\ &=& \dfrac{ 10x - \sqrt{40x\cdot 10x } } { 10x }\\\\ &=& \dfrac{ 10x - \sqrt{400x^2 } } { 10x }\\\\ &=& \dfrac{ 10x - 20x } { 10x }\\\\ &=& \dfrac{ -10x } { 10x }\\\\ \mathbf{\dfrac{ \sqrt{10x} - \sqrt{40x} } {\sqrt{10x} }} &\mathbf{=}& \mathbf{-1} \\\\ \end{array}$

Lösen sie die folgenden Gleichungen und Ungleichungen in der Grundmenge N!

Schreiben sie alle Umformungsschritte mit auf.

Geben sie die Lösungen in form einer Lösungsmenge an!

d.) 3x-7< 2+x

$\begin{array}{rcl} 3x-7 &\le& 2+x \qquad | \qquad -x \\ 2x-7 &\le& 2 \qquad | \qquad +7 \\ 2x &\le& 9 \qquad | \qquad :2 \\ x &\le& \frac{9}{2} \\ \mathbf{x} &\mathbf{\le}& \mathbf{4,5} \end{array}$

$L = \{ x ~| ~x \le4.5 \}$

wie sieht hier der gesamte lösungsweg aus?:

(((2*x+1))/((x-1)))+(((2*x+4))/((1-x)))+(((x-9))/((x**2-1)))=(((3-8*x))/((1-x**2)))

$\begin{array}{rcl} \dfrac{ 2 x+1 }{ x-1 } +\dfrac{ 2x+4 }{ 1-x } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 }{ x-1 } -\dfrac{ 2x+4 }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 -(2x+4) }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 - 2x - 4) }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3 }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 } \qquad | \qquad x^2-1 = (x-1)(x+1)\\\\ \dfrac{ ( -3 ) }{( x-1 )}\cdot \dfrac{(x+1)}{(x+1)} +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ ( -3 )(x+1) }{ x^2-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3x-3 }{ x^2-1 }+\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3x-3 +x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -2x-12 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -2x-12 }{ x^2-1 } &=&-\dfrac{ 3-8x }{ x^2-1 }\\\\ -\dfrac{ 2x+12 }{ x^2-1 } &=&-\dfrac{ 3-8x }{ x^2-1 }\\\\ \dfrac{ 2x+12 }{ x^2-1 } &=&\dfrac{ 3-8x }{ x^2-1 }\\\\ \dfrac{ 2x+12 }{ x^2-1 }-\dfrac{ 3-8x }{ x^2-1 } &=&0\\\\ \dfrac{ 2x+12-( 3-8x ) }{ x^2-1 } &=&0\\\\ \dfrac{ 2x+12-3+8x }{ x^2-1 } &=&0\\\\ \dfrac{ 10x+9 }{ x^2-1 } &=&0\\\\ 10x+9 &=&0\\ 10x &=&-9\\ x &=& \dfrac{-9}{10}\\ \mathbf{x} &\mathbf{=}& \mathbf{-0,9} \end{array}$

Guten Abend ,

Ich hätte eine frage wie rechne ich die Textaufgabe aus ?

Die frage lautet : Eine 5,50m Leiter wird an einen Baum gelehent. Der Fuß der Leiter steht dabei 1,80m vor dem Baum.

Wie hoch reicht die Leiter ?

$\text{Pythagoras}\\$

$\begin{array}{rcl} h^2 + 1,80^2 &=& 5,50^2\\ h^2 &=& 5,50^2 - 1,80^2\\ h^2 &=& 30,25 - 3,24\\ h^2 &=& 27,01 \qquad | \qquad \sqrt{}\\ h &=& \sqrt{27,01} \\ \mathbf{h} &\mathbf{=} & \mathbf{5,20\ m} \end{array}$

Die Leiter reicht 5,20 m hoch.

14868=14250(1+r/12)^36 r=?

$\begin{array}{rcl} 14868 &=& 14250\cdot (1+ \frac{r}{12} )^{36} \\ 14250\cdot(1+ \frac{r}{12} )^{36} &=&14868 \\ (1+ \frac{r}{12} )^{36} &=& \frac{ 14868 }{ 14250 } \qquad | \qquad \sqrt[36]{}\\ 1+ \frac{r}{12} &=&\sqrt[36]{\frac{ 14868 }{ 14250 } }\\ \frac{r}{12} &=&\sqrt[36]{\frac{ 14868 }{ 14250 } }-1 \\ r &=& 12 \cdot \left( \sqrt[36]{\frac{ 14868 }{ 14250 } }-1 \right) \\ r &=& 12 \cdot \left( \sqrt[36]{ 1.04336842105 }-1 \right) \\ r &=& 12 \cdot \left( 1.00117998301-1 \right) \\ r &=& 12 \cdot 0.00117998301 \\ \mathbf{r} &\mathbf{=}& \mathbf{0.01415979617} \end{array}$

There were 480 members in a computer lab last year. This year, the number of male members has been reduced by 1/7 and the number of female members has been reduced by 40%. There are twice as many male members as female members now. (a) how many male members are there in the club now? (b) what is the percentage decrease in the membership of the club?

m = male, f = female last year

M = male, F = female this year

$m + f = 480 \rightarrow m = 480-f$

$M = \frac67 \cdot m\\ F = 0.6\cdot f \\ F = \frac12 \cdot M = \frac12 \cdot \frac67 \cdot m = \frac{3}{7}\cdot m\\ \begin{array}{rcl} \hline F = 0.6\cdot f &=& \frac{3}{7}\cdot m\\ 0.6\cdot f &=& \frac{3}{7}\cdot m \qquad | \qquad m=480-f\\ 0.6\cdot f &=& \frac{3}{7}\cdot (480-f)\\ 1.4\cdot f &=& 480-f\\ 2.4\cdot f &=& 480\\ f &=& \frac{480}{2.4}\\ \mathbf{f} &\mathbf{=}& \mathbf{200}\\\\ F&=& 0.6\cdot f \\ \mathbf{F} &\mathbf{=}& \mathbf{120}\\\\ M&=& 2F\\ M&=& 2\cdot 120\\ \mathbf{M} &\mathbf{=}& \mathbf{240}\\\\ m&=&\frac76 \cdot M\\ m&=& \frac76 \cdot 240\\ \mathbf{m} &\mathbf{=}& \mathbf{280}\\\\ \end{array}$

(a) how many male members are there in the club now?

There are 240 male members in the club now

(b) what is the percentage decrease in the membership of the club?

$\frac{M+F}{m+f} = \frac{360}{480}=0.75$

The percentage decrease in the membership of the club is - 25%

In an exhibition hall, the ratio of the number of females to the number of males was 5:6. The ratio of the number of girls to the number of women was 2:3. The ratio of the number of boys to the number of men was 3:4. If there were 60 more men than women, how many people were there altogether?

f = female, m = male, G = girls, W = women, B = Boy, M = men

$\frac{f}{m} = \frac56 \rightarrow \frac{G+W}{B+M}=\frac56 \\ \frac{G}{W} = \frac23 \rightarrow W=\frac32 G \\ \frac{B}{M} = \frac34 \rightarrow B=\frac34 M \rightarrow B= \frac34 (60+W) = \frac34 (60+\frac32 G) = 45 + \frac{9}{8} G\\ M=60+W \rightarrow M = 60 + \frac32 G$

$G+W = G+\frac32 G = \frac52 G\\ B+M = 45 + \frac98 G + 60 + \frac32 G = 105 + \frac{21}{8} G\\ \begin{array}{rcl} \hline \frac{G+W}{B+M}&=&\frac56\\ \frac{ \frac52 G}{105 + \frac{21}{8} G} &=& \frac56 \\ 15G &=& 5\cdot(105 + \frac{21}{8} G ) \\ 15G &=& 525 + 13.125 G\\ 1.875G &=& 525\\ G&=& \frac{525}{1.875} \\ \mathbf{G}& \mathbf{=} & \mathbf{280}\\ \end{array}$

$W=\frac32 G =\frac32 \cdot 280 \\ \mathbf{W} \mathbf{=} \mathbf{420}\\ $

M = 60 + W

M = 60 + 420

M = 480

$B=\frac34 M \\ B=\frac34 480\\ \mathbf{B=360}$

280 + 420 + 480 + 360 = 1540 people were there altogether