answer:1. Let ( triangle ABC ) be a triangle with ( BC = 80 ) and ( angle ABC = 60^circ ). We are given that ( AB + AC = 90 ).2. Draw the altitude ( AH ) from ( A ) to ( BC ), where ( H in BC ). In ( triangle ABH ), since ( angle ABH = 60^circ ), we have: [ BH = frac{1}{2} AB quad text{and} quad AH = frac{sqrt{3}}{2} AB ]3. In ( triangle AHC ), we have: [ HC = BC - BH = 80 - frac{AB}{2} ] and [ AC = 90 - AB ]4. By the Pythagorean theorem in ( triangle AHC ), we have: [ AH^2 + HC^2 = AC^2 ]5. Substituting the expressions for ( AH ), ( HC ), and ( AC ), we get: [ left( frac{sqrt{3}}{2} AB right)^2 + left( 80 - frac{AB}{2} right)^2 = left( 90 - AB right)^2 ]6. Simplifying each term: [ left( frac{sqrt{3}}{2} AB right)^2 = frac{3}{4} AB^2 ] [ left( 80 - frac{AB}{2} right)^2 = 80^2 - 80 cdot AB + frac{AB^2}{4} ] [ left( 90 - AB right)^2 = 90^2 - 180 cdot AB + AB^2 ]7. Combining these, we get: [ frac{3}{4} AB^2 + 80^2 - 80 AB + frac{AB^2}{4} = 90^2 - 180 AB + AB^2 ]8. Simplifying further: [ frac{3}{4} AB^2 + frac{1}{4} AB^2 + 6400 - 80 AB = 8100 - 180 AB + AB^2 ] [ AB^2 + 6400 - 80 AB = 8100 - 180 AB + AB^2 ]9. Canceling ( AB^2 ) from both sides: [ 6400 - 80 AB = 8100 - 180 AB ]10. Solving for ( AB ): [ 100 AB = 1700 ] [ AB = 17 ]11. Since ( AB = 17 ), the shortest side is ( boxed{17} ).

answer:From the given information, we know that theta is in the third quadrant.Hence,cos theta = -sqrt{1 - sin^2 theta} = -sqrt{1 - left(-frac{4}{5}right)^2} = -frac{3}{5}.Therefore, the answer is boxed{-frac{3}{5}}.This can be obtained using the fundamental trigonometric identity sin^2 theta + cos^2 theta = 1.This question primarily tests basic knowledge and operations of simple trigonometric functions.

answer:AnalysisThis problem examines the application of the monotonicity and even-odd properties of functions: solving inequalities involves the use of derivatives to determine monotonicity and the method of solving logarithmic inequalities. It is considered a medium difficulty and commonly mistaken problem. By finding the derivative of the function and determining the interval of increase, and then considering the even-odd property of the function, the inequality f(ln x)+f(ln frac{1}{x}) 0, f′(x) > 0, f(x) is increasing, and f(-x)=xsin x+cos (-x)+(-x)^{2}=f(x), indicating it is an even function, i.e., f(x)=f(|x|), thus the inequality f(ln x)+f(ln frac{1}{x}) < 2f(1),is equivalent to f(ln x) < f(1) which is f|ln x| < f(1), then |ln x| < 1, i.e., -1 < ln x < 1,solving this gives frac{1}{e} < x < e. Therefore, the correct choice is boxed{D}.

answer:Solve: From the recurrence relation, when m geqslant 2, we have x_{2 m+1}=x_{2 m}+x_{2 m-1}=2 x_{2 m-1}+2 x_{2 m-2},and x_{2 m-2}=x_{2 m-1}-x_{2 m-3},thus x_{2 m+1}=4 x_{2 m-1}-2 x_{2 m-3}.Let a_{m}=x_{2 m-1}, then a_{m+1}=4 a_{m}-2 a_{m-1}, m geqslant 2.Similarly, let b_{m}=x_{2 m}, we can also get b_{m+1}=4 b_{m}-2 b_{m-1}, m geqslant 2.The sequences left{a_{m}right} and left{b_{m}right} have the same recurrence relation, whose characteristic equation is lambda^{2}-4 lambda+2=0,solving which gives the characteristic roots lambda=2 pm sqrt{2}. Therefore,begin{array}{l}a_{m}=alpha_{1}(2+sqrt{2})^{m}+beta_{1}(2-sqrt{2})^{m}, m=1,2, cdots b_{m}=alpha_{2}(2+sqrt{2})^{m}+beta_{2}(2-sqrt{2})^{m}, m=1,2, cdotsend{array}where alpha_{1}, beta_{1}, alpha_{2}, beta_{2} are undetermined constants. Furthermore, bybegin{array}{l}a_{1}=x_{1}=2, a_{2}=x_{3}=x_{2}+x_{1}=5, b_{1}=x_{2}=3, b_{2}=x_{4}=x_{3}+2 x_{2}=11,end{array}we can calculate quad alpha_{1}=frac{1}{4}(3-sqrt{2}), beta_{1}=frac{1}{4}(3+sqrt{2}),alpha_{2}=frac{1}{4}(1+2 sqrt{2}), beta_{2}=frac{1}{4}(1-2 sqrt{2}) .Therefore, when n=2 m-1, x_{n}=frac{1}{4}(3-sqrt{2})(2+sqrt{2})^{m}+frac{1}{4}(3+sqrt{2})(2-sqrt{2})^{m}; when n=2 m, x_{n}=frac{1}{4}(1+2 sqrt{2})(2+sqrt{2})^{m}+frac{1}{4}(1-2 sqrt{2})(2-sqrt{2})^{m}.

answer:Analysis: We can directly use the formula given in the problem to calculate. When n=2, 2^{n-1}(2^n - 1) = 6. When n=3, 2^{n-1} - 1 = 3, which is a prime number, so 2^{n-1}(2^n - 1) = 4 times 7 = 28. Therefore, the next perfect number after 6 is boxed{28}.

answer:4. Let S=left{x_{1}, x_{2}, cdots, x_{n}right}, and x_{1}<x_{2}<cdots<x_{n}, then frac{x_{1}+x_{2}}{2}<frac{x_{1}+x_{3}}{2}<cdots<frac{x_{1}+x_{n}}{2}< frac{x_{2}+x_{n}}{2}<frac{x_{3}+x_{n}}{2}<cdots<frac{x_{n-1}+x_{n}}{2}, therefore, A_{s} contains at least 2 n-3 elements. On the other hand, if we take S={2,4,6, cdots, 2 n}, then A_{s}={3,4,5, cdots, 2 n-1}, which has only 2 n-3 elements. Thus, the minimum number of elements in A_{S} is 2 n-3.