asinus

What is the area of rectangle ZWXY?

Let ZW be the baseline and WX be the height of the rectangle ZWXY, and ZW be x.

Then:

\(\overline{WX}=\dfrac{x-2}{tan\ 60°}\\ \overline{AX}=\dfrac{tan\ 30°(x-2)}{tan\ 60°}=\dfrac{x-2}{3}\\ \)

\(\triangle ZAY\\ (x-\overline{AX})^2+\overline{WX}^2=4^2 \\ (x-\dfrac{x-2}{3})^2+(\dfrac{x-2}{tan\ 60°})^2=16\\ (\dfrac{2x}{3}+\dfrac{2}{3})^2+(\dfrac{x}{tan\ 60°}-\dfrac{2}{\tan\ 60°})^2=16\\ x\in \{-4,\dfrac{32}{7}\}\ \color{blue}WolframAlpha\)

\(A_{ZWXY}=x\cdot \dfrac{x-2}{tan\ 60°}=\dfrac{32}{7}\cdot \dfrac{\frac{32}{7}-2}{tan\ 60°}\\ \color{blue}A_{ZWXY}=6.7868\)

The area of rectangle ZWXY is 6.7868.

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The number of solutions to
\(x_1 + x_2 + x_3 + x_4 + x_5 \le 1\)

in nonnegative integers:

\(Each\ x_n\ can\ become\ =1, the\ remainings\ x_n\ become =0,\\ or\ all\ x_n\ become =0.\)

\(Example:\\ x_1\in \{1\}\\ x_{2}, x_{3},x_{4},x_{5} \in \{0\}\\ \color{blue}or\\ x_1,x_2, x_3,x_4,x_5 \in \{0\}\)

\(\color{blue}The\ number\ of\ solutions\ for\ nonnegative\ integers\ is\ 6.\)

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

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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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