answer:Let the circle's center be O. As in the three-point case, for any triangle formed by O and two other points P and Q on the circle to be non-obtuse, the angle angle POQ at O must be acute. This means that the minor arc PQ must be less than pi/2. Now, let four points A_0, A_1, A_2, and A_3 be chosen on the circle. Consider the arc measure theta_{01} between points A_0 and A_1. The probability that theta_{01} < pi/2 is 1/2 because theta_{01} can take on any value from 0 to pi (the probability density function is uniform). Now, assuming that theta_{01} < pi/2, we need to place A_2 such that it does not create an obtuse angle at O with A_0 and A_1. The arc on which A_2 can lie to satisfy this condition spans pi - theta_{01}, which is similar to the three-point case. The probability of A_2 being in the correct position, given theta_{01}, is frac{pi - theta_{01}}{2pi} = frac{1}{2} - frac{theta_{01}}{2pi}. Lastly, A_3 must also lie in a position such that it, alongside any other two points among A_0, A_1, A_2, does not form an obtuse angle at O. This reduces the effective arc further, also symmetrically about each point, and the calculation follows similarly. If theta_{01} < pi/2, the expected value of theta_{01} is pi/4, hence the average probability that both A_2 and A_3 are suitably placed is left(frac{1}{2} - frac{pi/4}{2pi}right)^2 = left(frac{3}{8}right)^2 = frac{9}{64}. Multiplying by the probability that theta_{01} < pi/2 gives the final probability that no three points form an obtuse angle at O: [ frac{1}{2} cdot frac{9}{64} = boxed{frac{9}{128}}. ]

answer:The matrix mathbf{E} representing a dilation by a factor of 5, centered at the origin, is given by: [ mathbf{E} = begin{pmatrix} 5 & 0 0 & 5 end{pmatrix} ] To find the determinant of mathbf{E}, we calculate the product of its diagonal entries: [ det mathbf{E} = 5 times 5 = 25 ] Thus, the determinant of matrix mathbf{E} is boxed{25}.

answer:From the problem, we understand that sides BD, DC, and DA of tetrahedron B-ACD are all mutually perpendicular. This implies that the circumscribed sphere of the tetrahedron is the same as the circumscribed sphere of a rectangular box formed by extending the tetrahedron. Since the legs of the triangle ADC are AD (which is the height of the equilateral triangle and can be calculated using Pythagoras' theorem) and DC (which is half the length of side BC due to symmetry), we have: For equilateral triangle ABC, the height AD: AD = sqrt{AB^2 - left(frac{BC}{2}right)^2} = sqrt{2^2 - left(frac{2}{2}right)^2} = sqrt{4 - 1} = sqrt{3}. Thus, the lengths of the rectangle's sides that correspond to the tetrahedron's edges are 1 (half the base of the equilateral triangle), 1 (the other half), and sqrt{3} (the height of the triangle). The diagonal of the rectangular box, which is the diameter of the circumscribed sphere, can be calculated using the Pythagorean theorem in three dimensions: d = sqrt{1^2 + 1^2 + (sqrt{3})^2} = sqrt{1 + 1 + 3} = sqrt{5}. Hence, the radius r of the sphere is half of the diameter: r = frac{d}{2} = frac{sqrt{5}}{2}. The surface area A of a sphere with radius r is given by the formula: A = 4pi r^2. Substituting the value of r: A = 4pi left(frac{sqrt{5}}{2}right)^2 = 4pi cdot frac{5}{4} = 5pi. Therefore, the surface area of the circumscribed sphere of tetrahedron B-ACD is: boxed{5pi}.

answer:1. We start by applying the substitution ( t = tan frac{x}{2} ). This substitution is convenient for integrals involving trigonometric functions as it transforms them into simpler rational functions. 2. Using the identities for sine and cosine under this substitution: [ sin x = frac{2t}{1+t^2}, quad cos x = frac{1-t^2}{1+t^2}, quad dx = frac{2 , dt}{1+t^2} ] 3. Determine the new limits of integration: [ x = 0 Rightarrow t = tan 0 = 0, quad x = frac{pi}{2} Rightarrow t = tan frac{pi}{4} = 1 ] 4. Substitute these into the integral: [ int_{0}^{frac{pi}{2}} frac{cos x , dx}{1 + sin x + cos x} = int_{0}^{1} frac{left(frac{1-t^2}{1+t^2}right) left(frac{2 , dt}{1+t^2}right)}{1 + frac{2t}{1+t^2} + frac{1-t^2}{1+t^2}} ] 5. Simplify the expression inside the integral: [ int_{0}^{1} frac{left(frac{1-t^2}{1+t^2}right) left(frac{2 , dt}{1+t^2}right)}{frac{1+t^2 + 2t + 1 - t^2}{1+t^2}} = int_{0}^{1} frac{2(1-t^2) , dt}{(1+t^2)left(2+2tright)} ] 6. Further reduce: [ int_{0}^{1} frac{2(1-t^2) , dt}{2(t+1)(1+t^2)} = int_{0}^{1} frac{(1-t)(1+t) , dt}{(t+1)(1+t^2)} = int_{0}^{1} frac{(1-t) , dt}{1+t^2} ] 7. Break down the integral into two simpler integrals: [ int_{0}^{1} frac{1 , dt}{1+t^2} - int_{0}^{1} frac{t , dt}{1+t^2} ] 8. The first integral evaluates to: [ left. arctan t right|_{0}^{1} = arctan 1 - arctan 0 = frac{pi}{4} ] 9. For the second integral: [ int_{0}^{1} frac{t , dt}{1+t^2} = frac{1}{2} left. ln (1+t^2) right|_0^1 = frac{1}{2} ln 2 ] 10. Combine the results from the two integrals: [ frac{pi}{4} - frac{1}{2} ln 2 ] Hence, the final result is: [ boxed{frac{pi}{4} - frac{1}{2} ln 2} ]