asinus

$BC = 3, AC = 19\ and\ AB = 13.\\ BC<(AC-AB)\\ \color{blue }The\ requirements\ are\ pointless.$

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f(x)=12(x+2)/(x+1) grafic

At which integer-x is f(x) a integer?

There, f(x)=b.

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No, that's not helpful. It would be helpful if you explained your solution to us.

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Find the number of ordered pairs (a,b).

$\dfrac{{\color{blue}a} + 2}{a + 1} = \dfrac{\color{blue}b}{12}\\ $

$\dfrac{{\color{blue}1}+2}{1+1}=\dfrac{\color{blue}18}{12}\\ \dfrac{{\color{blue}3}+2}{3+1}=\dfrac{\color{blue}15}{12}\\ \dfrac{{\color{blue}5}+2}{5+1}=\dfrac{\color{blue}14}{12}\\ $

$\dfrac{{\color{blue}11}+2}{11+1}=\dfrac{\color{blue}13}{12}$

$\color{blue}(a,b)\in\{(1,18)\ ; (3,15)\ ; (5,14)\ ; (11,13)\}$

The number of solutions is 4.

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$Let\ \omega\ be\ a\ nonreal\ root\ of\ x^3 = 1.\\ Compute\ (1 - \omega + \omega^2)^6 + (1 + \omega - \omega^2)^6.$

$x^3=1\\ x=\sqrt[3]{1}\\ \color{blue}\sqrt[3]{1}\ is\ not\ a\ transcendental\ number.\ \sqrt[3]{1}=1 \\ \omega=1\\ (1 - 1 + 1^2)^6 + (1 + 1 - 1^2)^6=2$

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$0$

$Find\ the\ maximum\ value\ of\ \ 3x + 4y + 5z + x^3 + \dfrac{4x^2 y}{z} + \dfrac{z^5}{xy^2}.$

Plane geometry

$z^2=x^2+y^2\ \color{green }Pythagorean\ theorem\\ \color{blue}y=x\\ z^2=2x^2\\ \color{blue}z=x\sqrt{2}\\ x^2+y^2+z^2=1\\ x^2+x^2+2x^2=1\\$

$4x^2=1\\ \color{blue}x=\dfrac{1}{2}$

$f(x)=3x + 4y + 5z + x^3 + \dfrac{4x^2 y}{z} + \dfrac{z^5}{xy^2}\\ f(x)=3x+4x+5x\sqrt{2}+x^3+\dfrac{4x^3}{x\sqrt{2}}+\dfrac{(x\sqrt{2})^5}{x^3}\\ f(x)=x^3+6\sqrt{2}x^2+5\sqrt{2}x+7x\ \color{green}according\ to\ WolframAlpha\\ f(x_{0.5})=0.5^3+6\sqrt{2}(0.5)^2+5\sqrt{2}(0.5)+7(0.5)=9.28185\\ \dfrac{d}{dx}=3x^2+3⋅2^{\frac{5}{2}}x+5√2+7=0\\ {\color{blue}x\in \{-4.6477};-1.009\}$

$f(x)_{max}=(-4.6477)^3+6\sqrt{2}(-4.6477)^2+5\sqrt{2}(-4.6477)+7(-4.6477)\\ \color{blue}f(x)_{max}=17.4979$

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Find AB.

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$AB=\sqrt{409+(\frac{15}{sin(120°-\arctan{\frac{20}{3}})})^2-2*\sqrt{409}*\frac{15}{sin(120^°-\arctan{\frac{20}{3}})}*\cos{60°}}=22.4$

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The path to the farthest corner is \sqrt (2^2+1^2)= \sqrt 5.

Franklin reaches 100% of the cube surface.

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If K is the area of ​​the triangle, what should be determined if K is the area of ​​the triangle?

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Find the area of the shaded region.

Where can I find the shaded region?

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Let cos A = x

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$\sin 4A = \sin A + \sin 2A$     Find acute angle A.

$\sin 4A=8\sin A\cos^3 A-4\sin A\cos A\\ \cos^3 A=\\ \sin 2A=2\sin A\cos A\\ 8\sin A\cos^3 A-4\sin A\cos A=\sin A+2\sin A\cos A$

will be continued

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Find length PS.

PS is half the base of an equilateral triangle OSA with side length 5.

$\color{blue}\overline{PS}=2.5$

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$\color{blac}Find\ \angle CAD\ in\ degrees.\\$

Thanks webbinsight for the useful suggestions to solve this tricky task!

$\angle ADB =\angle ADE=\alpha\\ \angle ACB=\angle ACE=\beta$

The angles intersection AD/EC are twice:

$(120°-\alpha )\ and\ (60°+\alpha )$

The angles intersection AC/BD are twice:

$(30°+\beta)\ and\ (150°-\beta)\\ \angle CAD=60°+\alpha -\beta$

In the triangle (intersection AC/BD) - A-D the following applies:

$\alpha =180°-(150°-\beta )-(60°+\alpha -\beta)\\ \color{blue}\beta = 15°$

The angles intersection BD/CE are twice:

$(180°-30°-2\beta=120°)\ and\ (60°)$

In the triangle (intersection AD/CE) -  (intersection CE/BD) - D the following applies:

$(120°)+(60°+\alpha )+(\alpha )=180°\\ \color{blue} \alpha =0\ ?\ ?\ ?$

I ask everyone for help.

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Find WX

$U(-3,0), V(3,0), Y(-2,2), Z(2,2)\\ UY(-2.5,1)\\ m=\frac{y_Y-y_U}{x_Y-x_U}=\frac{2-0}{(-2)-(-3)}\\ m=2$

$f(x)=m(x-x_U)+y_U=2(x-(-3))+0\\ f(x)=2x+6\\ f(x_M)=-\frac{1}{m}(x-x_{UY})+y_{UY}=-\frac{1}{2}(x-(-2.5))+1\\ f(x_M)=-0.5x-0.25\\ x=0\\ \color{blue}M(0,-0.25)\ ,\ Center\ of\ the\ enclosing\ circle$

$r=\sqrt{x_Z\ ^2+(y_Z-y_M)^2}=\sqrt{2^2+(2-(-0.25))^2}\\ \color{blue}r=3.0104\ ,\ Radius\ of\ the\ enclosing\ circle$

$f_{circ}(x)=y_M\pm \sqrt{r^2-x^2}\\ 1=-0.25+\sqrt{3.0104^2-x^2}\\ (1+0.25)^2=3.0104^2-x^2\\ x_{WX}=\pm \sqrt{3.0104^2-1.25^2}\\ x_{WX}=\pm \sqrt{7.5}=\pm 2.7386 \\ \color{blue}\overline{WX}=5.4772$

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$-\sqrt{3}sec(\theta)=2\\ -\sqrt{3}\cdot \dfrac{1}{cos(\theta)}=2\\ cos(\theta)=\dfrac{-\sqrt{3}}{2}\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (\arccos\dfrac{-\sqrt{3}}{2})+2\pi n\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (150^{\circ})+2\pi n\\ \color{blue}\theta=\dfrac{5\pi}{6}+2\pi n$

Multiply the DEG-result of  $\arccos \dfrac{-\sqrt{3}}{2} \ by\ \dfrac{\pi}{180^{\circ}}(=1)$ to express it as a fraction. To express the multiple result of $\arccos \dfrac{-\sqrt{3}}{2}$  add $2\pi n.$

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What is sin C in an acute triangle?

$A=\arcsin \dfrac{2}{5}\\ A=23.5782^{\circ}\\ B=\arcsin \dfrac{2}{\sqrt{7}}\\ B=49.1066^\circ \\ C=180^{\circ }-23.5782^{\circ}-49.1066^\circ\\ \color{red }C=107.3152^{\circ}\\ The\ triangle\ is\ obtuse-angled!\\ \color{blue}\sin C=0.9547$

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 What is the area of the triangle with side lengths tan A, tan B, and tan C?

$f_{AC}(x)=\sqrt{9^2-x^2}\\ f_{BC}(x)=\sqrt{6^2-x^2+10x-5^2}\\ 81-x^2=36-x^2+10x-25\\ 10x=81-36+25\\ x_C=\dfrac{81-36+25}{10}\\ x_C=7\\ y_C=\sqrt{9^2-7^2}\\ C\ (7,5.6569)$

$\angle B=180^\circ -atan\ \frac{y_c}{x_C-x_B}=180^\circ-atan\ \frac{5.6569}{7-5}\\ \angle B=109.4712^\circ\\ \color{red}tan\ \angle B=-2.8284\\ The\ sides\ of\ a\ real\ triangle\ are\ greater\ than\ zero.$

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My question to CPhill:

How can one formulate the equation AB / BD = AC / DC without knowing the angles of triangle ABC?

Please describe how I can arrive at this final solution.

Thanks in advance for your answer!

I will not be available until April 27, 2025.

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