answer:(1) Since the point (2,0) lies on the ellipse C, we have frac{4}{a^{2}}+ frac{0}{b^{2}}=1, which implies a^{2}=4. As the eccentricity of the ellipse C is frac{1}{2}, we have frac{c}{a}= frac{1}{2}, which is equivalent to frac{a^{2}-b^{2}}{a^{2}}= frac{1}{4}. Solving for b^{2}, we get b^{2}=3. Therefore, the equation of the ellipse C is frac{x^{2}}{4}+ frac{y^{2}}{3}=1. (2) Let P(-1,y_{0}), where y_{0}in(-frac{sqrt{3}}{2}, frac{sqrt{3}}{2}). quadquad (i) When the slope of line MN exists, let the equation of line MN be y-y_{0}=k(x+1). Let M(x_{1},y_{1}) and N(x_{2},y_{2}). From the system of equations begin{cases} 3x^{2}+4y^{2}=12 y-y_{0}=k(x+1)end{cases}, we obtain (3+4k^{2})x^{2}+(8ky_{0}+8k^{2})x+(4y_{0}^{2}+8ky_{0}+4k^{2}-12)=0. So, x_{1}+x_{2}=-frac{8ky_{0}+8k^{2}}{3+4k^{2}}. Since P is the midpoint of MN, we have frac{x_{1}+x_{2}}{2}=-1, which implies -frac{8ky_{0}+8k^{2}}{3+4k^{2}}=-2. Thus, the slope of line MN is k_{MN}= frac{3}{4y_{0}} (y_{0}neq 0). As line l is perpendicular to line MN, we have k_{l}=- frac{4y_{0}}{3}. So, the equation of line l is y-y_{0}=- frac{4y_{0}}{3}(x+1), which simplifies to y=- frac{4y_{0}}{3}(x+ frac{1}{4}). Clearly, line l always passes through the fixed point (-frac{1}{4}, 0). quadquad (ii) When the slope of line MN does not exist, the equation of line MN is x=-1. In this case, line l is the x-axis, which still passes through the point (-frac{1}{4}, 0). In conclusion, line l always passes through the fixed point boxed{(-frac{1}{4}, 0)}.

answer:Let's denote the number of new boarders that joined the school as ( x ). Originally, the ratio of boarders to day students was 5 to 12. This means that for every 5 boarders, there were 12 day students. We know there were initially 150 boarders, so we can set up a proportion to find the initial number of day students: [ frac{5}{12} = frac{150}{text{day students}} ] Solving for the number of day students, we get: [ text{day students} = frac{150 times 12}{5} = frac{150 times 12}{5} = 30 times 12 = 360 ] So, there were initially 360 day students. After the new boarders joined, the ratio changed to 1 to 2. This means that for every boarder, there are 2 day students. The total number of boarders after the new ones joined is ( 150 + x ), and the number of day students remains the same (360), as no boarders became day students and vice versa, and no students left the school. So we can set up the following equation: [ frac{150 + x}{360} = frac{1}{2} ] Cross-multiplying to solve for ( x ), we get: [ 2(150 + x) = 1 times 360 ] [ 300 + 2x = 360 ] Subtracting 300 from both sides, we get: [ 2x = 60 ] Dividing both sides by 2, we find: [ x = 30 ] So, boxed{30} new boarders joined the school.

answer:1. Identify zeros of the numerator and denominator. The numerator (x) is zero at (x = 0), and the denominator (x + 2) is zero at (x = -2). 2. Create a sign chart for (frac{x}{x+2}): - When (x < -2), both numerator (x) and denominator (x+2) are negative, making the fraction positive. - When (-2 < x < 0), (x) is negative and (x+2) is positive, making the fraction negative. - When (0 < x), both (x) and (x+2) are positive, making the fraction positive. 3. Consider the inequality (frac{x}{x+2} geq 0) and the additional constraint (x < 5): - The solution includes intervals where (frac{x}{x+2} geq 0). - Valid intervals from the sign chart are ((-∞, -2) cup [0, ∞)). - Since (x) must also be less than (5), we intersect this with ((-∞, 5)) to get the final solution. Conclusion: (boxed{(-infty, -2)cup[0, 5)})

answer:First, let's calculate f( dfrac {1}{8}). Since dfrac {1}{8} > 0, we use the first part of the function definition, f(x) = log_{2}x. So, f( dfrac {1}{8}) = log_{2} dfrac {1}{8} = log_{2} 2^{-3} = -3. Next, we need to find f(f( dfrac {1}{8})) = f(-3). Since -3 leqslant 0, we use the second part of the function definition, f(x) = 3^{x}. So, f(-3) = 3^{-3} = dfrac {1}{27}. Therefore, f(f( dfrac {1}{8})) = boxed{dfrac {1}{27}}.