answer:First, we calculate ( n^2 = (2^{40}5^{15})^2 = 2^{80}5^{30} ). - The total number of divisors of ( n^2 ) is given by ( (80 + 1)(30 + 1) = 81 times 31 = 2511 ). - To find divisors of ( n^2 ) less than ( n ), we pair each divisor with its complement regarding ( n^2 ). Half of these divisors, excluding ( n ) itself, are less than ( n ): ( frac{2511 - 1}{2} = 1255 ). - The divisors of ( n ) are ( (40 + 1)(15 + 1) = 41 times 16 = 656 ). All of these divisors are less than ( n ) and included in the 1255. To find the divisors of ( n^2 ) that are less than ( n ) but do not divide ( n ), we subtract the number of divisors of ( n ) from the divisors of ( n^2 ) less than ( n ): [ 1255 - 656 = 599 ] Thus, the number of positive integer divisors of ( n^2 ) that are less than ( n ) but do not divide ( n ) is ( boxed{599} ).

answer:Since (e^{x}+ frac {1}{2}x^{2})^{′}=e^{x}+x, we have ∫_{ 0 }^{ 1 }(e^{x}+x)dx=(e^{x}+ frac {1}{2}x^{2} )|_{ 0 }^{ 1 }=(e^{1}+ frac {1}{2}×1^{2})-(e^{0}+ frac {1}{2}×0^{2})=e- frac {1}{2}． Therefore, the answer is boxed{e- frac {1}{2}}． The solution involves finding the antiderivative of the integrand and then directly applying the Fundamental Theorem of Calculus. This question tests the Fundamental Theorem of Calculus. The key to solving this problem is to memorize the derivative formulas of basic elementary functions, thereby finding the antiderivative of the integrand. This is a basic question.

answer:Given a convex quadrilateral and a point M inside it, we are to prove that the sum of the distances from point M to the vertices of the quadrilateral is less than the sum of the pairwise distances between the vertices of the quadrilateral. 1. Let diagonals AC and BD intersect at point O. Without loss of generality, assume that M lies inside the triangle BOC. 2. Let the line CM intersect side AB at point Q. Considering Triangle Inequality: 3. In triangle BQC, by the triangle inequality, we have: [ MB + MC leq MQ + QC ] 4. Now, consider another application of the triangle inequality in triangle BCO: [ MQ + QC leq MQ + QO + OC ] 5. Recognize that the segment QO is part of diagonal AC. Therefore, we have: [ MB + MC leq BQ + QO + OC = BO + OC ] 6. This inequality implies that the sum of the distances from M to points B and C is less than or equal to the sum of the distances from point O to points B and C. 7. Similarly, consider the triangle inequality for the remaining pairs. For instance, for point A and point D: [ MA + MD leq AQ + QM + MC + CD ] 8. By breaking it down, we get: [ MA + MD leq AQ + QB + BA ] Summing Up Inequalities: 9. Summing up all such inequalities for all pairs (A, D) and (B, C), we have: [ MA + MB + MC + MD < AB + BC + CD + DA ] Conclusion: 10. Therefore, the sum of the distances from point M to the vertices of the quadrilateral is less than the sum of the pairwise distances between the vertices of the quadrilateral, i.e., [ MA + MB + MC + MD < AB + BC + CD + DA ] blacksquare

answer:To solve the equation frac{1}{2}x^3 + 4 = 0, we follow these steps: 1. Start with the given equation: frac{1}{2}x^3 + 4 = 0 2. Isolate the cubic term: frac{1}{2}x^3 = -4 3. Multiply both sides by 2 to eliminate the fraction: x^3 = -4 times 2 x^3 = -8 4. Find the cube root of both sides to solve for x: x = sqrt[3]{-8} x = -2 Therefore, the root of the equation is boxed{x = -2}.