answer:To solve for the range of omega given the function f(x)=2sin(omega x+frac{pi}{3})-sqrt{3} that has exactly three zeros in the interval [0,frac{pi}{2}], we follow these steps: 1. **Understand the condition for zeros**: The zeros of f(x) correspond to the values of x for which 2sin(omega x+frac{pi}{3})-sqrt{3} = 0. Simplifying this gives sin (omega x+frac{pi}{3}) = frac{sqrt{3}}{2}. 2. **Identify the sine values**: The equation sin (omega x+frac{pi}{3}) = frac{sqrt{3}}{2} has solutions when omega x+frac{pi}{3} is equal to the angles that give a sine value of frac{sqrt{3}}{2}. 3. **Determine the interval for omega x + frac{pi}{3}**: Since there are exactly three zeros in [0,frac{pi}{2}], the angle omega x+frac{pi}{3} must span across a specific interval to hit three such values. This interval is from frac{pi}{3} to frac{omegapi}{2}+frac{pi}{3}. 4. **Set the inequalities**: The first zero occurs at frac{pi}{3}, and the last zero must occur before 2pi +frac{2pi}{3} for there to be exactly three zeros. Therefore, we have the inequalities 2pi +frac{pi}{3}leq frac{omegapi}{2}+frac{pi}{3} < 2pi +frac{2pi}{3}. 5. **Solve for omega**: Subtracting frac{pi}{3} from each part of the inequality and then multiplying by frac{2}{pi} gives 4leq omega < frac{14}{3}. Therefore, the range of omega for which the function f(x)=2sin(omega x+frac{pi}{3})-sqrt{3} has exactly three zeros in the interval [0,frac{pi}{2}] is [4,frac{14}{3}). So, the correct answer is boxed{D}.

answer:First, combine like terms in the expression: [ (8b^3 + 104b^2 - 9) - (-9b^3 + b^2 - 9) = 8b^3 + 104b^2 - 9 + 9b^3 - b^2 + 9 = 17b^3 + 103b^2. ] Now, factor out the greatest common factor from the resultant polynomial: [ 17b^3 + 103b^2 = b^2(17b + 103). ] Thus, the completely factored form of the expression is: [ boxed{b^2(17b + 103)}. ]

answer:Solution: (1) (2 frac {7}{9})^{ frac {1}{2}}-(2 sqrt {3}-pi)^0-(2 frac {10}{27})^{- frac {2}{3}}+0.25^{- frac {3}{2}}; The original expression equals sqrt { frac {25}{9}}-1-( frac {64}{27})^{- frac {2}{3}}+( frac {1}{4})^{- frac {3}{2}} = frac {5}{3}-1-( frac {27}{64})^{ frac {2}{3}}+4^{ frac {3}{2}} = frac {2}{3}- frac {9}{16}+8 = 8 frac {5}{48} So, the answer for the first part is boxed{8 frac {5}{48}}. (2) Given the condition: 0<x<1, Therefore, x^{ frac {1}{2}}-x^{- frac {1}{2}}<0 Thus, (x^{ frac {1}{2}}-x^{- frac {1}{2}})^2=x+x^{-1}-2. Since x+x^{-1}=3, (x^{ frac {1}{2}}-x^{- frac {1}{2}})^2=1 Hence, x^{ frac {1}{2}}-x^{- frac {1}{2}}=-1, which gives us the final answer boxed{-1}.

answer:First, we solve the system of equations to find the intersection point of the two given lines: begin{cases} 2x+y+2=0 2x-y+2=0 end{cases} By adding both equations, we obtain: 4x+0y+4=0 Rightarrow x=-1 Substitute x=-1 into the first equation: 2(-1)+y+2=0 Rightarrow y=0 Thus, the intersection point of 2x+y+2=0 and 2x-y+2=0 is (-1,0). Next, since the line we are looking for is perpendicular to the line x+y=0, it will have a slope that is the negative reciprocal of the slope of x+y=0. The slope of x+y=0 is -1, hence the perpendicular line will have a slope of 1. Using the point-slope form (y-y_1)=m(x-x_1), where m is the slope, and (x_1,y_1) is the point (-1,0), we can write the equation of the desired line: y-0=1(x+1) Rightarrow y=x+1 Rearranging the equation to the general form, we get: x-y+1=0 Thus, the equation of the line that passes through the intersection of 2x+y+2=0 and 2x-y+2=0 and is perpendicular to x+y=0 is: boxed{x-y+1=0} In this problem, we used the intersection of lines and the concept of perpendicularity to find the slope and eventually the equation of the desired line. This type of problem is a fundamental one in the study of lines in coordinate geometry.