asinus

I = V/R

$R_1=\frac{V_1}{I_1}=\frac{500V}{0.025A}=\color{blue}20k\Omega$

I halve the length of the wire, hence the resistance $R_1$.

$I_2=\frac{V_2}{R_2}=\frac{175V}{10k\Omega}=\color{blue}17.5mA$

I apply the voltage$V_2$ to both halves of the wire, thus halving the resistance $R_1$.

$I_3=\frac{V_2}{R_3}=\frac{175V}{5k\Omega}=\color{blue}35mA$ in both wire halves together.

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$\dfrac{x+4}{x+5} = \dfrac{x-3}{2} + \dfrac{x + 7}{5}\\ \dfrac{10(x+4)}{10(x+5)}=\dfrac{(x+5)(x-3)+(x+5)(x+7)}{10(x+5)}\\ 10x+40=x^2+2x-15+x^2+12x+35\\ 2x^2+4x-20=0\\ x^2-2x-10=0\\ x_{1,2}=1\pm \sqrt{1+10}\\ \color{blue}x\in \{-2.3166, 4.3166\}$

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What is DE ?

$\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2\cdot \overline{AB}\cdot \overline{AC}\cdot cos\ \alpha\\ \alpha =acos\ ( \dfrac{\overline{AB}^2+\overline{AC}^2-\overline{BC}^2}{2\cdot \overline{AB}\cdot \overline{AC}})\\ \alpha =acos\ (\dfrac{13^2+19^2-3^2}{2\cdot 13\cdot 19})=acos\ (1.0546\ ...)\\ \alpha =\ unreal$

The segments 3, 13 and 19 do not form a closed triangle.

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Die Dichte des Holzes sei $\rho$. Dann gilt

$20^3cm^3\cdot \rho\cdot g=20^2\cdot 17cm^3\cdot \dfrac{1g}{cm^3}\cdot g\\ \rho =\dfrac{20^2\cdot 17g}{20^3cm^3}\\ \color{blue}\rho=0,85g/cm^3$

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Der Preis der Ware sei x. Dann gilt

$0,96x-0,95x=2,5\\ 0,01x=2,5\\ x=\dfrac{2,5}{0,01}=250$

Die Ware kostet 250€.

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 Wie ist hier die Eingabelogik:  tan ß = 1  | tan^-1     ß =  45° ?

Hallo calypso1, betätige die Felder wie folgt:

$\circ Deg\\ atan\\ 1\\ =\\ \color{blue}Ergebnis:tan^{-1}(1)=45^{\circ}$

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$\dfrac{12,5km\cdot t}{h}+\dfrac{15,5km\cdot t}{h}=140km\\ \dfrac{28km\cdot t}{h}=140km\\ t=\dfrac{140km\cdot h}{28km}\\ \color{blue}t=5h$

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Find WX.

$h_{yz}=\sqrt{r^2-2^2}\\ h_{uv}=\sqrt{r^2-3^2}\\ h_{yz}-h_{uv}=2\\ \sqrt{r^2-2^2}-\sqrt{r^2-3^2}=2\\ r=\frac{\sqrt{145}}{4}=3.0104\ (WolframAlpha)$

$h_{yz}=2.25\\ h_{uv}=0.2500\\ h_{wx}=(2.25+0.250)/2=1.25\\ \overline{WX}=2 \sqrt{r^2-h_{wx}^2}=2\sqrt{3.0104^2-1.25^2}\\ \color{blue}\overline{WX}=5.4772$

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May 18 2024

https://web2.0calc.com/questions/geometry_18743#r1

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Find a + b.

${\color{blue}m_{AB}=2}=\frac{y_b-y_a}{x_b-x_a}=\frac{b^2-a^2}{b-a}=\frac{(b+a)(b-a)}{b-a}\color{blue}=b+a\\ \color{blue}a+b=2$

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