asinus

Find the radius of the table.

\(2r^2=( \sqrt{2\cdot 6^2}+r)^2\\ 2r^2=(6\sqrt{2}+r)^2 \ |^\sqrt{}\\ r\sqrt{2}=6\sqrt{2}+r\\ r(\sqrt{2}-1)=6\sqrt{2}\\ r=\dfrac{6\sqrt{2}}{\sqrt{2}-1}\\ \color{blue}r=20.485\)

The radius of the table is 20.485

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Find the area of trapezoid EFGH.

The height of the trapezoid EFGH is h. 

\(A_{EFP}=\frac{\overline{EF}}{2} \cdot \frac{h}{3}=6\\ \overline{EF}=\frac{36}{h}\\ A_{GHP}=\frac{\overline{GH}}{2} \cdot \frac{2h}{3}=6\\ \overline{GH}=\frac{18}{h}\)

\(A_{EFGH}=\frac{\frac{36}{h}+\frac{18}{h}}{2}\cdot h\\ \color{blue}A_{EFGH}=27\)

The area of trapezoid EFGH is 27.

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\( \dfrac{1}{\sqrt2+\sqrt 3}+\dfrac{1}{\sqrt 2-\sqrt 3}\\ =\dfrac{\sqrt 2-\sqrt 3+\sqrt 2+\sqrt 3}{2-3}\\ =\dfrac{2\sqrt{2}}{-1}\\ \color{blue}=-2\sqrt{2}\ or =-\sqrt{8}\)

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Hallo mellojello!

Beweise mit vollständiger Induktion:

Es gilt:    \(\bigwedge _{n\in\mathbb N,n\ge 2}:\ \color{blue}n^2>n+1\)

Induktionsanfang:

\(n=2\\ linke Seite:2^2=4\\ rechte Seite:2+1=3\\ F\ddot ur\ n=2\ ist\ die\ linke\ Seite\ gr\ddot osser\ als\ die\ rechte\ Seite.\\ Die Aussage\ ist\ richtig. \)

Die Induktionsannahme (IA) lautet:

\(\bigwedge _{n\in\mathbb N,n\ge 2}:\ \color{blue}n^2>n+1\)

Der Induktionsschluss von n nach n+1 ist:

\(\bigwedge _{(n+1)\in\mathbb N,n\ge 2}:\ \color{blue}(n+1)^2>(n+1)+1\)

linke Seite:

\((2+1)^2=\color{blue}9\)

rechte Seite:

\((2+1)+1=\color{blue}4\)

Für \(\bigwedge _{(n+1)\in\mathbb N,n\ge 2}:\ \color{blue}(n+1)^2>(n+1)+1\) ist die linke Seite (9) grösser als die rechte Seite (4), q.e.d.

Die Induktionsannahme ist bewiesen.

Find the area of the rectangle.

\(2(a+b)=40\\ a=\frac{40}{2}-b\\ 10^2\cdot \sqrt{2}^2=a^2+b^2\\ 200=400-40b+2b^2\\ b^2-20b+100=0\\ b=10\pm \sqrt{100-100}\\ b=10\\ a=10\\ A=ab=10\cdot 10\\ \color{blue} A=100\)

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If I cut the wire in half and pass $175$ volts through it, how many milliamps of current will I measure?

\(R=\frac{V}{I}=\frac{500V}{25\cdot 10^{-3}A }\\ R=20 k\Omega\)

The resistance of the two halves of the wire in combination is \(R_2=\frac{20k\Omega}{4}=5k\Omega .\)

\(I=\frac{V}{R_2}=\frac{175V}{5\cdot 10^3 \Omega}\cdot \frac{10^3mA}{A}\\ \color{blue}I=35mA\)

The current of the two halves of the wire in combination is 35mA.

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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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