answer:Let the common ratio of the arithmetic sequence {a_n} be q, then a_1q=2, a_1q^{5}=32, So a_1=1, q=2, So S_{100}= frac {1times(1-2^{100})}{1-2}=boxed{2^{100}-1}. Thus the answer is: D. Based on a_2=2, a_6=32, we find a_1 and q, then use the formula for the sum of the first n terms of a geometric sequence to solve the problem. This problem mainly tests the application of geometric sequences, and establishing condition relations based on geometric sequences to find the first term and common ratio is the key to solving this problem.

answer:Given the power function f(x)=(m-2)x^{m^2-2m}, we need to find the value of m that makes this a valid power function. Given that the coefficient of x in a power function is a constant, we compare the coefficient of x in the given function to the standard form of a power function. In the standard form, the coefficient of x is constant, and the exponent is a degree of the polynomial or a power of x. Here, the coefficient before x^{m^2-2m} is (m-2). Since this must be a constant for f(x) to be a power function, and the standard form of a power function has a coefficient of x as 1 (if it’s not specified otherwise), we equate m-2 to 1 to find the value of m. Starting from the equation m-2=1, we can solve for m: [m-2=1] [m=1+2] [m=3] Therefore, the value of m that makes f(x)=(m-2)x^{m^2-2m} a power function is m=3. Thus, the final answer is boxed{3}.

answer:1. Let D be the point of intersection of the tangents to the circumcircle Gamma of triangle ABC at points B and C. 2. Our goal is to prove that points A, S, and D are collinear. 3. First, observe that the center S of the direct similarity mapping B to A and A to C implies: [ angle BSA = angle ASC. ] 4. We need to demonstrate the alignment of A, S, and D. This requires us to examine the angles involved. 5. By angle chasing using the properties of similarities and cyclic quadrilaterals, we note: [ widehat{BSC} = 2 widehat{BAC}. ] 6. For the tangents B and C intersecting at D, using properties of the tangents from a point to a circle, we have: [ widehat{BDC} = 180^circ - 2 widehat{BAC}. ] 7. From the above, it follows that points B, C, D, and S form a cyclic quadrilateral since: [ widehat{BSC} + widehat{BDC} = 2 widehat{BAC} + (180^circ - 2 widehat{BAC}) = 180^circ. ] 8. Consequently, the angles subtended by the segments from S and D give: [ widehat{BSD} = widehat{BCD} = widehat{BAC}. ] 9. Simultaneously, note that: [ widehat{BSA} = 180^circ - widehat{BAC}. ] 10. Hence, it follows that A, S, and D are collinear since: begin{align*} widehat{BSA} + widehat{BSD} &= (180^circ - widehat{BAC}) + widehat{BAC} &= 180^circ. end{align*} 11. Therefore, the line AS serves as the angle bisector of triangle ABC. # Conclusion: [ boxed{(AS) text{ est la symédiane issue de } A text{ dans } triangle ABC}. ]

answer:To solve the given problem, we start by manipulating the given equation for z as follows: [iz = 2 - i] To find z, we isolate it on one side: [z = frac{2 - i}{i}] To simplify the fraction, we multiply both numerator and denominator by i: [z = frac{(2 - i) cdot i}{i cdot i} = frac{2i - i^2}{i^2}] We know that i^2 = -1, substituting this in gives: [z = frac{2i - (-1)}{-1} = frac{2i + 1}{-1} = -1 - 2i] Therefore, z = -1 - 2i. Now, we need to find the conjugate of z, denoted as overline{z}. The conjugate of a complex number a + bi is a - bi. Applying this to z: [overline{z} = -1 - (-2i) = -1 + 2i] Thus, the answer is encapsulated as: [boxed{-1 + 2i}] Referring to the given options, this corresponds to: [boxed{C}]