answer:Solution: (I) Since the foci of the hyperbola dfrac {x^{2}}{5}-y^{2}=1 are the vertices of the ellipse C: dfrac {x^{2}}{a^{2}}+ dfrac {y^{2}}{b^{2}}=1(a > b > 0), and the eccentricities of the ellipse and the hyperbola are reciprocal to each other, therefore a= sqrt {6}, e_{text{hyperbola}}= dfrac { sqrt {6}}{ sqrt {5}}, e_{text{ellipse}}= dfrac { sqrt {5}}{ sqrt {6}}= dfrac {c}{a}, therefore c= sqrt {5}, b= sqrt {6-5}=1, therefore The equation of the ellipse C is dfrac {x^{2}}{6}+y^{2}=1. (II) When the slope of line MN is 0, by |MN|= dfrac {4 sqrt {3}}{3}, then M( dfrac {2 sqrt {3}}{3},y), then y= dfrac { sqrt {7}}{3}, then the y-intercept of line MN is dfrac { sqrt {7}}{3}, When the slope of line MN does not exist, it has no intersection with the y-axis, Let MN be: y=kx+m, (kneq 0) Combining begin{cases} y=kx+m dfrac {x^{2}}{6}+y^{2}=1end{cases}, we get (1+6k^{2})x^{2}+12kmx+6m^{2}-6=0, x_{1}+x_{2}=- dfrac {12km}{1+6k^{2}}, x_{1}x_{2}= dfrac {6m^{2}-6}{1+6k^{2}}, triangle =(12km)^{2}-4(1+6k^{2})(6m^{2}-6) > 0, triangle =144k^{2}-24m^{2}+24 > 0, therefore m^{2} < 6k^{2}+1, |MN|= sqrt {(1+k^{2})[(x_{1}+x_{2})^{2}-4x_{1}x_{2}]}= dfrac {4 sqrt {3}}{3}, therefore sqrt {(1+k^{2})[(- dfrac {12km}{1+6k^{2}})^{2}-4times dfrac {6m^{2}-6}{1+6k^{2}}]}= dfrac {4 sqrt {3}}{3}, After simplification, we get m^{2}= dfrac {39k^{2}-18k^{4}+7}{9k^{2}+9}, therefore m^{2}= dfrac {39k^{2}-18k^{4}+7}{9k^{2}+9} < 6k^{2}+1, After rearrangement: 36k^{4}+12k^{2}+1 > 0, which means 6k^{2}+1 > 0, kin(-infty,0)cup(0,+infty), Then m^{2}= dfrac {39k^{2}-18k^{4}+7}{9k^{2}+9}= dfrac {-18(k^{2}+1)^{2}+75(k^{2}+1)-50}{9(k^{2}+1)}, let k^{2}+1=t, (t > 1), Then f(t)=-2t- dfrac {50}{9t}+ dfrac {25}{3}, (t > 1), differentiate f′(t)=-2+ dfrac {50}{9t^{2}}, Setting f′(t) > 0, we find 1 < t < dfrac {5}{3}, Setting f′(t) < 0, we find t > dfrac {5}{3}, Thus, f(t) is increasing in (1, dfrac {5}{3}) and decreasing in ( dfrac {5}{3},+infty), therefore When t= dfrac {5}{3}, f(t) reaches its maximum value, which is dfrac {5}{3}, therefore The maximum value of m is dfrac {5}{3}, In summary, the maximum value of m is boxed{dfrac {5}{3}}.

answer:First, let's denote the number of oranges received by Alice, Becky, and Chris as a, b, and c respectively, where each person must receive at least three oranges. We can define new variables: a = a' + 3, quad b = b' + 3, quad c = c' + 3, where a', b', and c' are non-negative integers representing the additional oranges each person receives beyond the minimum of three. The total number of oranges is then expressed as: (a' + 3) + (b' + 3) + (c' + 3) = 30, which simplifies to: a' + b' + c' + 9 = 30, a' + b' + c' = 21. The problem of finding non-negative integer solutions to the equation a' + b' + c' = 21 can be approached using the stars and bars theorem. The number of ways to distribute n indistinguishable objects (oranges) into k distinguishable boxes (people) is given by: binom{n+k-1}{k-1}, where n = 21 and k = 3: binom{21+3-1}{3-1} = binom{23}{2}. Calculating binom{23}{2}: binom{23}{2} = frac{23 times 22}{2} = 253. Thus, the number of ways Alice, Becky, and Chris can each receive at least three oranges is 253. The final answer is boxed{C}.

answer:**Answer:** First, determine the period of the function based on the given conditions, then adjust -2007.5 into the interval [0,1] using the periodicity and evenness of the function to solve. Analysis: Since f(x) is an even function defined on mathbb{R} and f(x+1)+f(x)=3 (1) Therefore, f(-x+1)+f(-x)=3, which means f(x-1)+f(x)=3 (2) From (1) and (2), we get f(x+1)=f(x-1), therefore, the period T of f(x) is 2, Therefore, f(-2007.5)=f(-2008+0.5)=f(0.5)=2-0.5=boxed{1.5}. Hence, the correct option is B.

answer:Given the sum of the first n terms of a geometric sequence, we can determine the first few terms as follows: 1. The first term a_1 is equal to the sum of the first term itself, i.e., a_1=S_1=frac{1}{2}3^{1+1}-a=frac{9}{2}-a. 2. The second term a_2 can be found by subtracting the sum of the first term from the sum of the first two terms, i.e., a_2=S_2-S_1=(frac{1}{2}3^{2+1}-a)-(frac{9}{2}-a)=9. 3. Similarly, the third term a_3 can be found by subtracting the sum of the first two terms from the sum of the first three terms, i.e., a_3=S_3-S_2=(frac{1}{2}3^{3+1}-a)-(frac{27}{2}-a)=27. Since this is a geometric sequence, the square of the second term should be equal to the product of the first and third terms, i.e., a_2^2=a_1a_3. Substituting the values we found, we get: 9^2=(frac{9}{2}-a) times 27. Solving for a, we find a=boxed{frac{3}{2}}.