answer:If the painter has 32 hours of work left and each room takes 8 hours to paint, then he has 32 / 8 = 4 rooms left to paint. Since there are a total of 9 rooms and he has 4 rooms left to paint, he must have already painted 9 - 4 = boxed{5} rooms.

answer:**Analysis** This problem tests the application of arithmetic sequences and the relationship between roots and coefficients. It is a basic problem. According to the relationship between roots and coefficients in arithmetic sequences, find the value of a_{5}+a_{7}, and then find the sum of the first 11 terms of {a_{n}}. **Solution** Given that a_{5} and a_{7} are the roots of the equation x^{2}{-}2x{-}6{=}0, we have: a_{5} + a_{7} = 2 Since {a_{n}} is an arithmetic sequence, the average of two terms is equal to the arithmetic mean of these terms. Hence, a_{6} = frac{1}{2}(a_{5}+a_{7}) = 1 Now, let's find the sum of the first 11 terms, denoted by S_{11}: S_{11} = frac{11 times (a_{1}+a_{11})}{2} As a_{6} is the middle term of this arithmetic sequence with an odd number of terms, we can rewrite the sum as: S_{11} = 11a_{6} = 11 times 1 = boxed{11}

answer:Solution: (Ⅰ) Let f(x)=ax^{2}+bx+c, From f(0)=1, we get c=1, Thus f(x)=ax^{2}+bx+1. Since f(x+1)-f(x)=2x, We have a(x+1)^{2}+b(x+1)+1-(ax^{2}+bx+1)=2x. That is 2ax+a+b=2x, So begin{cases} 2a=2 a+b=0end{cases}, Therefore begin{cases} a=1 b=-1end{cases}, Thus f(x)=x^{2}-x+1. (Ⅱ) f(x)=x^{2}-x+1=(x- dfrac {1}{2})^{2}+ dfrac {3}{4}, So when xin[-1,1], y_{min}=f( dfrac {1}{2})= dfrac {3}{4}, y_{max}=f(-1)=3, Therefore, the range of the function is boxed{left[ dfrac {3}{4},3right]}. (Ⅲ) According to the problem, x^{2}-x+1 > 2x+m always holds in [-1,1]. That is x^{2}-3x+1-m > 0 always holds in [-1,1]. Let g(x)=x^{2}-3x+1-m, its axis of symmetry is the line x= dfrac {3}{2}, So g(x) is decreasing in [-1,1]. Therefore, it is sufficient that g(1) > 0, That is 1^{2}-3times1+1-m > 0, Solving this, we get boxed{m < -1}.

answer:(1) Since overrightarrow{a}//overrightarrow{b}, it follows that frac{1}{2}f(x)= frac{1}{2}sin x+ frac{ sqrt{3}}{2}cos x, which implies f(x)=2sin (x+ frac{π}{3}). Thus, the minimum positive period of the function f(x) is 2π, and the maximum value is 2. (2) From f(2A- frac{π}{6})=1, we get sin (2A+ frac{π}{6})= frac{1}{2}. Given 0 < A < π, we have frac{π}{6} < 2A+ frac{π}{6} < frac{13π}{6}, which implies 2A+ frac{π}{6}= frac{5π}{6}, hence A= frac{π}{3}. By the sine law, we have frac{BC}{sin A}= frac{AC}{sin B}. Substituting the given values, we obtain AC= frac{BCcdot sin B}{sin A}=boxed{2}.