asinus

What is the value of the first fraction he wrote?

\(\frac{x}{x+2}\cdot \frac{x+1}{x+3}\cdot\frac{x+2}{x+4}=10\\ \frac{x^2+x}{x^2+7x+12}=10\\x^2+x=10x^2+70x+120\\ 9x^2+69x+120=0\\ 3x^2+23x+40=0\\\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-23 \pm \sqrt{23^2-4\cdot 3\cdot 40} \over 2\cdot 3}\\ \color{blue}x\in\{-2\frac{2}{3},-5\} \)

\(\frac{x}{x+2}\cdot \frac{x+1}{x+3}\cdot\frac{x+2}{x+4}=10\\ x=-5\\ \frac{-5}{-3}\cdot \frac{-4}{-2}\cdot\frac{-3}{-1}=10\\\)

The value of the first fraction he wrote is \(\color{blue}\frac{5}{3}.\)

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\(\gamma =atan \frac{4}{3}\\ \gamma =53.1301^{\circ}\\ \alpha=\arcsin \frac{3}{5}\\ \alpha =33.8699^{\circ}\\ \color{blue}\frac{\alpha}{2}=18.4349^{\circ}\)

\(\angle ADC=(180-53.1301-18.4349)^{\circ}\\ \angle ADC=108.4349^{\circ} \)

\(tan \frac{\\alpha}{2}=\frac {\overline{BD}}{4}\\ \overline{BD}=4\cdot tan\ 18.4349^{\circ}\\ \overline{BD}=1\frac{1}{3}\\ \overline{DC}=3-1\frac{1}{3}\\ \overline{DC}=\frac{5}{3}\)

\(A_{ADC}=\frac{\overline{DC} \cdot \overline{AB}}{2}=\frac{5}{3}\cdot \frac{4}{2}=\frac{20}{6}\\ \color{blue}A_{ADC}=3\frac{1}{3}\)

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The median EC divides the triangle ABD into halves of equal area. If the area of ​​triangle ABC is 14, then the area of ​​triangle BEC is 7.

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The coordinates given the circle are not correkt. (Thanks ElectricPavlov!)

1. Determination of the slope of the chords between the points AO and BA.

\(m_{AO}=\frac{y_A}{x_A}=\frac{1}{-2}=\color{blue}-0.5\\ m_{BA}=\frac{y_B-y_a}{x_B-x_A}=\frac{2-1}{-3-(-2)}=\color{blue}-1\\\)

2. Determination of the coordinates of the centers of the chords BA and AO.

\( P_{BA}(\frac{x_B+x_A}{2},\frac{y_B+y_a}{2})\\ P_{BA}(\frac{-3+(-2)}{2},\frac{2+1}{2})\\ \color{blue}P_{BA}(-2.5,1.5) \)

\(P_{AO}(\frac{x_A+x_O}{2},\frac{y_A+y_O}{2})\\ P_{AO}(\frac{-2+0}{2},\frac{1+0}{2})\\ \color{blue}P_{AO}(-1,0.5)\)

3. Point-direction functions for the center points of the chords BA and AO. The slopes of the function graphs are the negative reciprocals of the slopes of the chords.

\(f_{BA}(x)=-\frac{1}{m_{BA}}(x-x_{PBA})+y_{PBA}\\ f_{BA}(x)=-\frac{1}{-1}(x-(-2.5))+1.5\\ {\color{blue}f_{BA}(x)}=1(x+2.5)+1.5\color{blue}=x+4\\ f_{AO}(x)=-\frac{1}{m_{AO}}(x-x_{P_{AO}})+y_{P_{AO}}\\ f_{AO}(x)=-\frac{1}{-0,5}(x-(-1))+0.5\\ {\color{blue}f_{AO}(x)}=2(x+1)+0.5\color{blue}=2x+2.5 \)

4. Determine the coordinates of point P by equating the two functions.

\(x+4=2x+2.5\\ \color{blue}x_P=1.5\\ y_P=x_P+4=1.5+4\\ \color{blue}y_P=5.5\\ .\\ \color{blue}P(1.5,5.5)\)

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 How many miles is the distance from Penthaven to Jackson?

\(s=vt\\ 2s=v\cdot 3.6h\\ 3.75h= \frac{s}{v+w}+\frac{s}{v-w}\\ 7.5h=\frac{2s}{v+w}+\frac{2s}{v-w}=\frac{v\cdot 3.6h}{v+30miles/h}+\frac{v\cdot 3.6h}{v-30miles/h}\\ 7.5=\frac{3.6v}{v+30miles/h}+\frac{3.6v}{v-30miles/h}\\ \color{blue}v=150\ miles/h\)

\(2s=v\cdot 3.6h\\ s=\frac{v\cdot 3.6h}{2}=150miles/h\cdot1.6h\\ \color{blue}s=240\ miles\)

240 miles is the distance from Penthaven to Jackson.

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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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