answer:Since for x geq 0, we have f(x+2) = f(x), therefore the period of the function is T = 2. Because f(x) is an even function on (-infty, +infty), and for x in [0, 2), f(x) = log_{2}(x+1), therefore f(-2010) + f(2011) = f(2010) + f(2011) = f(0) + f(1) = log_{2}1 + log_{2}(1+1) = 1. Thus, the answer is boxed{1}.

answer:Solution: (1) From 3rho^{2}=12rhocos theta-10 (rho > 0), we get 3x^{2}+3y^{2}=12x-10, which simplifies to (x-2)^{2}+y^{2}= frac {2}{3}. Therefore, the Cartesian equation of curve C_1 is: (x-2)^{2}+y^{2}= frac {2}{3}. (2) According to the problem, we can assume Q(4cos theta,2sin theta). From (1), we know the center of circle C_1 is at (2,0), then |QC|= sqrt {(4cos theta-2)^{2}+4sin ^{2}theta}= sqrt {12cos ^{2}theta-16cos theta+8} =2 sqrt {3(cos theta- frac {2}{3})^{2}+ frac {2}{3}}. Therefore, when cos theta= frac {2}{3}, |QC|_{min}= frac {2 sqrt {6}}{3}. Therefore, |PQ|_{min}= boxed{frac { sqrt {6}}{3}}.

answer:: 1. **Understanding the Problem Setup**: We are given a circle with center (O) and radius (r) tangent to the sides (BA) and (BC) of angle (ABC) at points (M) and (N) respectively. A line passing through point (M) parallel to (BC) intersects the ray (BO) at point (K). A point (T) is chosen on the ray (MN) such that (angle MTK = frac{1}{2} angle ABC). We need to find the length of segment (BO) given (KT = a). 2. **Identifying Key Points and Relationships**: Let (L) be the point where (MN) intersects the ray (BO). Notice that triangle (MBN) is isosceles with (BM = BN) because both lines from (B) to (M) and (N) are tangent to the circle and tangent segments to the same circle from a common external point are equal. Since (BL) is the bisector of angle (angle B) in isosceles triangle (MBN), it also serves as the height and median of triangle (MBN). 3. **Right-Angled Triangles and Similarity**: Since (MK parallel BC) and (BL) is the height, (angle KLM = 90^circ), making triangles (KLM) and (BLN) congruent. Therefore, (MK = BN). Given that (angle MKL = angle LBC) and (angle MTK = frac{1}{2} angle ABC), triangles (MKT) and (MLK) are similar because they share two angles. Hence, [ angle MKT = 90^circ, ] which makes triangle (MKT) similar to triangle (ONB). 4. **Proportions and Equation**: Using the similarity of triangles (MKT) and (ONB), [ frac{KT}{MK} = frac{BN}{ON} Longleftrightarrow frac{a}{MK} = frac{BN}{r}. ] Since (MK = BN), this simplifies to: [ frac{a}{BN} = frac{BN}{r} Longleftrightarrow BN^2 = ar. ] 5. **Applying the Pythagorean Theorem**: We now use the Pythagorean theorem to find (BO): [ BO = sqrt{BN^2 + ON^2}. ] Substituting (BN^2 = ar): [ BO = sqrt{ar + r^2} = sqrt{r(a + r)}. ] # Conclusion: [ boxed{sqrt{r(a+r)}} ]

answer:1. **Isolate C and B in terms of A:** - From 501C - 2004A = 3006: [ 501C = 2004A + 3006 implies C = 4A + 6 ] - From 2502B + 6006A = 10010: [ 2502B = 10010 - 6006A implies B = -2.4A + 4 ] 2. **Introduce D and solve for D in terms of A:** - Given D = A + 2. 3. **Calculate the average of A, B, C, and D:** - Substitute expressions for B, C, D into the average formula: [ frac{A + B + C + D}{4} = frac{A + (-2.4A + 4) + (4A + 6) + (A + 2)}{4} ] - Simplify the expression: [ A - 2.4A + 4A + A + 4 + 6 + 2 = 3.6A + 12 ] - Now, compute the average: [ frac{3.6A + 12}{4} = 0.9A + 3 ] - Since A is not uniquely determined from the equations, and without additional constraints, we assume A = 0 for simplicity (can be any real number if not restricted): [ 0.9 times 0 = 0 + 3 = 3 ] 4. **Conclusion:** - Assuming A=0 for simplicity, the average of A, B, C, and D is 3. The final answer is boxed{D) 3}