# Aufgabe 5.1 restart; b := (11-(2*sqrt(2)-sqrt(3))^2)/sqrt(48);

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simplify(b);

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c := abs(Pi^exp(1) - exp(Pi));

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evalf[25](c);

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z := (1+I)/((2+I)*(4+I));

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Re(z);

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Im(z);

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abs(z);

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s := sum(1/k, k=1..100);

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evalf(s);

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# Aufgabe 5.2 restart; f := (1/x - x^3)/((1+x)*(1/x+x));

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simplify(f);

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p := x^4+5*x^3-7*x^2-29*x+30;

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factor(p);

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g := x*(a-x)^2 + x^2*(3*a-x);

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ge := expand(g);

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eval(ge, [a=2, x=1/2]);

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eval(g, [a=2, x=1/2]);

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pk := binomial(S,k)*binomial(N-S,n-k)/binomial(N,n);

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mu := sum(k*pk,k=0..n);

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v := sum((k-mu)^2*pk,k=0..n);

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# Aufgabe 5.3 restart; S := solve(z^9+z=0, z);

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evalf(S);

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Sn := sum(k^2, k=1..n);

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Sn := simplify(Sn);

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solve(Sn = 2745429470, n);

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# das gesuchte n ist 2019 restart; gl1 := (1/10)*p + (4/10)*r = (15/100)*2; # Gleichung fuer Alkoholmenge

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gl2 := p + r = 2; # Gleichung fuer Fluessigkeitsmenge

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solve({gl1, gl2}, {p, r});

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# Man braucht 1/3 Liter Rum und 5/3 Liter Prosecco