answer:To solve this, we first find the derivative of the function f(x)=x- frac {1}{3}sin 2x+asin x, which is f'(x)=1- frac {2}{3}cos 2x+acos x. Given that the function is monotonically increasing on mathbb{R}, we have f'(x)geqslant 0 always, which translates to 1- frac {2}{3}cos 2x+acos xgeqslant 0, or equivalently, frac {5}{3}- frac {4}{3}cos ^{2}x+acos xgeqslant 0. Let t=cos x where -1leqslant tleqslant 1, we then have 5-4t^{2}+3atgeqslant 0. When t=0, the inequality is obviously true; For 0 < tleqslant 1, we have 3ageqslant 4t- frac {5}{t}. Since 4t- frac {5}{t} is increasing on (0,1], at t=1, it reaches its maximum value of -1, which gives 3ageqslant -1, or ageqslant - frac {1}{3}; For -1leqslant t < 0, we have 3aleqslant 4t- frac {5}{t}. Since 4t- frac {5}{t} is increasing on [-1,0), at t=-1, it reaches its minimum value of 1, which gives 3aleqslant 1, or aleqslant frac {1}{3}. Combining the above, we find that the range of a is [- frac {1}{3}, frac {1}{3}], Therefore, the answer is: boxed{[- frac {1}{3}, frac {1}{3}]}. By finding the derivative of f(x) and discussing the value of t=cos x where -1leqslant tleqslant 1, we can determine the range of a by discussing different cases for t and utilizing the monotonicity of functions to find the maximum and minimum values, thus solving the inequality. This problem tests the application of derivatives to determine monotonicity, the method of solving problems where an inequality always holds, the use of parameter separation and substitution, and the application of function monotonicity, making it a medium-level question.

answer:Start with the given equation: 2sin ^{2}α+sin αcos α-3cos ^{2}α= frac {7}{5} Divide both sides by sin ^{2}α+cos ^{2}α (which equals to 1): frac{2sin ^{2}α+sin αcos α-3cos ^{2}α}{sin ^{2}α+cos ^{2}α}= frac {7}{5} This simplifies to: frac{2tan ^{2}α+tan α-3}{tan ^{2}α+1}= frac {7}{5} Solving the quadratic equation for tan α, we get: tan α=2 quad text{or} quad tan α=-frac {11}{3} Thus, the possible values for tan α are boxed{2} and boxed{-frac {11}{3}}.

answer:We start by drawing triangle PQR with the given side lengths. Point M, as the midpoint of overline{QR}, splits QR into two segments of length QR/2 = 13 each. Applying Apollonius's theorem, which states that in any triangle, the sum of the squares of any two medians equals frac{1}{2} times the sum of the squares of the three sides minus the square of the median, to median PM, we have: Let PM be the median to side QR, then: [ PM^2 + left(frac{QR}{2}right)^2 = frac{1}{2}(PQ^2 + PR^2 + QR^2) ] [ PM^2 + 13^2 = frac{1}{2}(24^2 + 32^2 + 26^2) ] [ PM^2 + 169 = frac{1}{2}(576 + 1024 + 676) ] [ PM^2 + 169 = frac{1}{2}(2276) ] [ PM^2 + 169 = 1138 ] [ PM^2 = 1138 - 169 ] [ PM^2 = 969 ] [ PM = sqrt{969} ] [ PM = boxed{31} ]

answer:Solution: Due to symmetry, the region corresponding to |x|+|y|≤n is four times the region corresponding to the system of inequalities: begin{cases} x+y≤n x≥0 y≥0end{cases} The coordinates of the line x+y=n with the axes are (0, n) and (n, 0). Thus, the area of the corresponding region S = 4 × frac {1}{2}×n×n = 2n². Hence, the answer is: boxed{2n^2} The key to solving this problem is to find the area of the region in the first quadrant using symmetry. This problem primarily tests the application of inductive reasoning.