asinus

Hi aw007!

\(Point\ A\ is\ the\ origin.\  \overline{AB}=1\\\)

\( \color{blue}Point\ C:\\f_{AC}(x)=tan 20°\cdot x\\ f_{BC}(x)=tan 43°\cdot (x-1)\\ tan 20°\cdot x=tan 43°\cdot (x-1)\\ (tan 43°-tan 20°)\cdot x-tan 43°=0\\ {\color{blue}x_C}=\frac{tan 43°}{tan 43°-tan 20°}=\color{blue}1.64018\\ {\color{blue}y_C}=\frac{tan 43°\cdot tan 20°}{tan 43°-tan 20°}=\color{blue}0.59698\)

\(\color{blue}Point\ D:\\m_{AD}=tan\ \frac{20°}{3}\\ f_{AD}(x)=tan\ \frac{20^\circ}{3}\cdot x\\ y_D=tan\ \frac{20^\circ}{3}\cdot x_D\\ f_{CD}(x)=m(x-x_C)+y_C\\ m_{CD}=tan\ (180°-(180-20-23)°-\frac{23°}{3})=tan\ 35.\overline 3°\\ y_D=m_{CD}(x_D-x_C)+y_C=tan\ 35.\overline 3(x_D-1.64018)+0.59698\)

\(tan\ \frac{20^\circ}{3}\cdot x_D=tan\ 35.\overline 3°\cdot x_D-tan\ 35.\overline 3°\cdot 1.64018+0.59698\\ x_D=\frac{tan\ 35.\overline 3°\cdot 1.64018-0.59698}{tan\ 35.3\overline 3°-tan\ \frac{20°}{3}}\\ \color{blue}x_D=0.95564\\ y_D=tan\ \frac{20^\circ}{3}\cdot x_D\\ \color{blue}y_D=0.11170\)

\(\color{blue}Point\ E:\\ f_{AE}(x)=tan\ \frac{2\cdot 20^\circ}{3}\cdot x\\ y_E=tan\ \frac{40^\circ}{3}\cdot x_E\\ f_{CE}(x)=m(x-x_C)+y_C\\ m_{CE}=tan\ (180°-(180-20-23)°-\frac{2\cdot 23°}{3})=tan\ 27.6\overline 6°\\ y_E=m_{CE}(x_E-x_C)+y_C=tan\ 27.6\overline 6(x_E-1.64018)+0.59698\)

\(tan\ \frac{40^\circ}{3}\cdot x_E=tan\ 27.6\overline 6(x_E-1.64018)+0.59698\\ tan\ \frac{40^\circ}{3}\cdot x_E=tan\ 27.6\overline 6\cdot x_E-tan\ 27.6\overline 6\cdot 1.64018+0.59698\\ .\\ x_E=\frac{tan\ 27.6\overline 6\cdot 1.64018-0.59698}{tan\ 27.6\overline 6-tan\ 13.3\overline3°}\\ \color{blue}x_E=0.91524\)

\(y_E=tan\ \frac{40^\circ}{3}\cdot x_E=tan\ \frac{40^\circ}{3}\cdot 0.91524\\ \color{blue}y_E=0.21692\)

\(m_{DE}=\frac{y_E-y_D}{x_E-x_D}=\frac{0.21692-0.11170}{0.91524-0.95564}\\ m_{DE}=-2.60446\)

\(\angle ADF=\angle DAB+atan \ m_{DE}=\frac{20°}{3}-atan\ (-2.60446)\\ \angle ADF=75\frac{2}{3}^\circ\\ \color{blue}The\ degree\ measure\ of\ angle\ AFD\ is\ 75\frac{2}{3}^\circ.\)

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Hallo Wickilif3a!

2.a) zwei Drittel von 162 €

\(162\ €\cdot\frac{2}{3}=\color{blue}108\ €\)

2.b) 24 % des Drittels von 162 €

\(\frac{162\ €}{3}\cdot \frac{24}{100}=\color{blue}12.96\ €\)

3. Wie hoch ist der Preis pro kg für dieses Fleisches?

\(2.8\ kg:18.06\ €=1\ kg:x\\ 2.8\ kg\cdot x=18,06\ €\cdot 1\ kg\ |\ /2.8\ kg\\ x=\frac{18,06\ €\cdot 1\ kg}{2.8\ kg}\\ x=\color{blue}6.45\ €\)

Der Preis pro kg für dieses Fleisch ist 6.45 €.

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Hallo Wickilif3a!

\(g=27-5/3*4+22/46-35\\ g=27-\frac{5}{3}\cdot 4+\frac{22}{46}-35\\ g=\frac{27\cdot 3\cdot 46-5\cdot 46\cdot 4+22\cdot 3-35\cdot 3\cdot 46}{3\cdot 46}=-\frac{1958}{138}\\ g=-\frac{979}{69}\)

\(h=7-1/7+16/21\\ h=\frac{7\cdot 7\cdot 21-1\cdot 21+16\cdot 7}{7\cdot 21}\\ h=\frac{1029-21+112}{147}\\ h=\frac{1120}{147}\)

\(i=a+b/c=\frac{ac+b}{c}\)

\(j=p/3-5/2p\\ j=\frac{p}{3}-\frac{5}{2}\cdot p\\ j= \frac{p\cdot 2-5\cdot 3\cdot p}{3\cdot 2}\\ j=-\frac{13p}{6}\)

Eine Anleitung zur Fraktionenrechnung findest du unter dem Link.

Find all points (x, y) that are 5 units away from the point (2, 7) and that lie on the line 2x + y = 13

Helo ABJelly!

\(f_1(x)=\pm\sqrt{5^2-(x-2)^2}+7\\ f_2(x)=-2x+13\\ f_1(x)=f_2(x)\\ \pm\sqrt{5^2-(x-2)^2}+7=-2x+13\\ \sqrt{5^2-(x-2)^2}\ ^2=(-2x+6)^2\\ 5^2-(x-2)^2=4x^2-24x+36\)

\(25-x^2+4x-4=4x^2-24x+36\\ 5x^2-28x+15=0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \color{blue}x\in \{\frac{3}{5},5\}\\ \color{blue}f_1(x)=11.8\\ \color{blue}f_2(x)=3\)

The points we are looking for are (0.6,11.8) and (5,3).

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When Edith finishes the race, how far behind is Foley, in km?

Hello Lilliam0216!

\(v_d=24km/t_d\\ v_e=(24-5)km/t_d=19km/t_d\\ v_f=(24-6)km/t_d=18km/t_d\\ t_e=\frac{24km}{v_e}=\frac{24km\cdot t_d}{19km}\\ s_f=v_f\cdot t_e=\frac{18km}{t_d}\cdot \frac{24km\cdot t_d}{19km}=22.737km\)

Foley was (24-22.737)km=1.263km behind when Edith finished the race.

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How many additional workers should be hired?

Hello Lilliam0216!

I assume the 5 workers can complete the job in around 120 hours.

\(\psi =\frac{J}{5\cdot 120h}\\ J=5\cdot 120h\cdot \psi =(5+1)\cdot (120-8)\cdot \psi=(5+x)\cdot (120-28)\cdot \psi\\ 6\cdot 112=(5+x)\cdot 92\\ 672=460+92x\\ x=\frac{672-460}{92}=2.3\color{blue}\to 3\\ 6\cdot t=(5+\color{blue}3)\cdot 92\)

\(t=\frac{8\cdot 92}{6}=\color{blue}122\frac{2}{3}h\)

If the 5 workers completed the work in 122 2/3 hours, 3 additional workers should be hired to save 28 hours.

An additional time specification is absolutely necessary for your question.

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Hello tomtom!

Under whom are the cone and cylinder located?

What is the difference between the surface area of the sphere and the lateral surface area of the cylinder?

Hello tomtom!

\(V_c =\pi r^2\cdot 2r=2\pi r^3=144\\ r=(\dfrac{72}{\pi })^{\frac{1}{3}}\)

\(LS_c=2\pi r^2+2\pi r\cdot 2r=2\pi r^2(1+2)=6\pi r^2\\ SA_s=4\pi r^2\\ LS_c-SA_s= 2\pi r^2=2\pi \cdot (\dfrac{72}{\pi })^{\frac{2}{3}} \\ \color{blue}LS_c-SA_s=50.6954\)

The surface area of the sphere and the lateral surface area of the cylinder is 50.6954.

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No triangle can be constructed with the given data.

AB must be greater than BC.

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Find the smallest angle, in degrees.

Hello blackpanther!

\(\alpha +\alpha +d +\alpha +2d=180^\circ\\ (\alpha +2d)-\alpha =56^\circ\\ d=28^\circ\\ 3\alpha +3\cdot 28^\circ =180^\circ\\ \alpha+28^\circ =60^\circ\\ \color{blue}\alpha =32^\circ\\ \beta =60^\circ\\ \gamma =88^\circ \)

\(\color{blue}\alpha ,the\ smallest\ angle\ is\ 32^\circ .\) 

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