asinus

What is the probability that at least one of them picks pineapple, and at least one of them picks pumpkin pie?

The probability that at least one of them picks pineapple is \(\color{blue}\frac{1}{4}\cdot 3 =\frac{3}{4}\ ,\)
the probability that at least one of them picks pumpkin pie is \(\color{blue}\frac{1}{3}\cdot 3=1\ .\)

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Given a right triangle CAB with the right angle at B and angle bisector AD, let AD = AC.

Solution:

The question is nonsensical.

The length of the angle bisector of an acute angle in a right triangle is not equal to the length of the hypotenuse.

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With AB = AC = 1 and angle BAC = 60°, it is an equilateral triangle with side length 1. D lies on AB, E lies on the arc BC. The condition BD = DE is only met if E is identical to C and D is identical to A. F lies on DB, B lies on the arc DE. The condition DF = FG is only met if F is identical to B and G is identical to C.

Thus, the length DF = AB = 1.

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The maximum possible distance between their endpoints is the diameter of the circle, namely 2r.

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What is DE?

The lines extended from BD and CE form triangle ABC.

AB=13, AC=19, BC=3.

The following holds: |AC-AB| < BC.

Therefore, there is no triangle with side lengths {13, 19, 3}.

The question does not yield a sine answer.

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Find the set of all possible values of a/b.

\(a+b=1\\ a=1-b\\ b=1-a\\ \dfrac{a}{b}=\dfrac{a}{1-a}\)

\(\color{blue}Possibility_{\frac{a}{b}}= \{0<\mathbb R <∞\}\)

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Let k be a positive real number and find  k, so that the line is tangent to the circle.

\(y=\sqrt{k-x^2}\\ y=-x+3+k\)

\(\color{blue}k\notin \{+\mathbb{R}\}\)

\(y=\sqrt{k-x^2}\\ y=-x+3+k \)

\(y=\sqrt{{\color{blue}1.365}-x^2}\\ y=-x+3\color{blue}-1.365\)

\(\color{blue}k\in \{ \approx 1.365,\ \approx-1.365\}\)   determined with graphics. 

If +k is replaced by -k in the problem, it is solvable. 

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Where is the circle sector shown??

What is the radius of the circle?

\(A_{sector}=\dfrac{\pi r^2\alpha}{360°}\\ A_{sector}=\dfrac{60}{\pi}\\ r_{sector}=\sqrt{\dfrac{60\cdot 360°}{\pi \alpha}\cdot \dfrac{2\pi}{360°}}\\ \color{blue}r_{sector}=\sqrt{\dfrac{120}{\alpha }}\)

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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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