answer:According to the problem, the ellipse dfrac {x^{2}}{25}+ dfrac {y^{2}}{m^{2}}=1(m > 0) and the hyperbola dfrac {x^{2}}{7}- dfrac {y^{2}}{n^{2}}=1(n > 0) have the same foci, then we have 25-m^{2}=7+n^{2}, rearranging gives: m^{2}+n^{2}=18, and by the inequality dfrac {m^{2}+n^{2}}{2}geqslant left( dfrac {m+n}{2}right)^{2}, we get left( dfrac {m+n}{2}right)^{2}leqslant 9, which implies m+nleqslant 6, thus, the maximum value of m+n is 6; Therefore, the answer is: boxed{B}. Based on the problem, by analyzing the geometric properties of the ellipse and hyperbola, we obtain 25-m^{2}=7+n^{2}, rearranging gives: m^{2}+n^{2}=18, and then by applying the properties of basic inequalities, we can derive the answer. This question examines the geometric properties of ellipses and hyperbolas, involving the properties of basic inequalities, with the key being to establish the relationship between m^{2} and n^{2}.

answer:Since M={x|(x+2)(x-1)<0}, we have M={x|-2<x<1}, Since N={x|x+1<0}, we have N={x|x<-1}, Therefore, Mcap N={x|-2<x<-1} Hence, the correct choice is boxed{C}. **Analysis:** According to the problem, M={x|(x+2)(x-1)<0} and N={x|x+1<0}. Solve for M and N, and then calculate their intersection according to the definition and operation rules of set intersection.

answer:Given 4^{x^2+6x+9}=16^{x+3}, we know 16 can be written as 4^2. Therefore, the equation transforms to: [ 4^{x^2+6x+9} = 4^{2(x+3)} ] Equating the exponents since the bases are the same, we have: [ x^2 + 6x + 9 = 2(x + 3) ] [ x^2 + 6x + 9 = 2x + 6 ] [ x^2 + 4x + 3 = 0 ] Solving this quadratic equation by factoring gives: [ (x+3)(x+1) = 0 ] [ x = -3, -1 ] Check for both solutions in the original equation: - For x = -3, 4^{0} = 16^{-3}, both sides equal 1. - For x = -1, 4^4 = 16^{-1}, both sides equal 1/16. Sum of all possible values of x is -3 + (-1) = boxed{-4}.

answer:Let's denote the rate at which pipe A fills the tank as A and the rate at which pipe B fills the tank as B. Pipe A can fill the tank in 12 hours, so its rate is 1/12 of the tank per hour. Pipe B can fill the tank in 15 hours, so its rate is 1/15 of the tank per hour. When both pipes are open together, their combined rate is the sum of their individual rates: Combined rate = A + B = 1/12 + 1/15 To add these fractions, we need a common denominator, which is 60 (the least common multiple of 12 and 15): Combined rate = (5/60) + (4/60) = 9/60 = 3/20 So, the combined rate of both pipes is 3/20 of the tank per hour. To find out how long it will take for both pipes to fill the tank together, we take the reciprocal of the combined rate: Time = 1 / (Combined rate) = 1 / (3/20) = 20/3 = 6.67 hours Therefore, it will take approximately boxed{6.67} hours for both pipes A and B to fill the tank when they are open together.