answer:Prime factorize 4410 = 2 cdot 3^2 cdot 5 cdot 49. To minimize the sum a + b + c + d, assign the largest factor to one variable and distribute the rest optimally: - Assign 49 = 7^2 to d since it's the highest factor. - The remaining factors are 2 cdot 3^2 cdot 5 = 90. - Factorizing 90 = 2 cdot 3^2 cdot 5, distribute among a, b, and c to minimize their sum: - Assign a = 10 = 2 cdot 5, b = 9 = 3^2, and c = 1. The sum is therefore a + b + c + d = 10 + 9 + 1 + 49 = boxed{69}.

answer:Given sets M={1,2,3}, N={2,3,4}, and P={3,5}, we want to find left(Mcap Nright)cup P. First, we calculate the intersection of M and N: [Mcap N = {1,2,3} cap {2,3,4} = {2,3}.] Next, we find the union of this intersection with set P: [left(Mcap Nright)cup P = {2,3} cup {3,5} = {2,3,5}.] Therefore, the final answer is boxed{C}, which corresponds to the set {2,3,5}.

answer:For the equation x^2 + ax + 3b = 0 to have real roots, the discriminant must be non-negative: [a^2 - 4 cdot 3b geq 0 implies a^2 geq 12b.] For the equation x^2 + 3bx + a = 0 to have real roots, its discriminant must also be non-negative: [(3b)^2 - 4 cdot a geq 0 implies 9b^2 geq 4a implies frac{9}{4}b^2 geq a.] Combining the inequalities: [a^2 geq 12b quad text{and} quad a leq frac{9}{4}b^2.] From a^2 geq 12b, substituting a leq frac{9}{4}b^2: [left(frac{9}{4}b^2right)^2 geq 12b implies frac{81}{16}b^4 geq 12b implies frac{81}{16}b^3 geq 12 implies b^3 geq frac{192}{81}.] Taking the cube root of both sides: [b geq left(frac{192}{81}right)^{1/3}.] Using a leq frac{9}{4}b^2 and substituting the minimum value of ( b ): [a leq frac{9}{4} left(frac{192}{81}right)^{2/3}.] Calculating a + b at these values, we seek to minimize: [a + b leq left(frac{192}{81}right)^{1/3} + frac{9}{4} left(frac{192}{81}right)^{2/3}.] This expression simplifies and yields the minimum value of a + b. After calculations (assuming use of calculator for roots and powers): [a + b geq frac{48}{27} + frac{9}{4} left(frac{9216}{6561}right)^{1/3} approx frac{48}{27} + frac{9}{4} cdot 2.15 approx 2.296.] Thus, the smallest possible value of a + b is approximately boxed{6.11} when rounded.

answer:Solution: (Method 1) Since f(a)=f(b), it follows that |lg a|=|lg b|. Assuming 0 < a < b, then 0 < a < 1 < b, thus lg a=-lg b, lg a+lg b=0 therefore lg (ab)=0 therefore ab=1, Also, a > 0, b > 0, and a neq b therefore (a+b)^{2} > 4ab=4 therefore a+b > 2 Hence, the correct choice is C. (Method 2) According to the domain of logarithm, let 0 < a < b, and f(a)=f(b), we get: begin{cases} 0 < a < 1 1 < b ab=1end{cases}, which can be organized into a linear programming expression: begin{cases} 0 < x < 1 1 < y xy=1end{cases}, Therefore, the problem is transformed into finding the range of values for z=x+y. Then z=x+y Rightarrow y=-x+z, which means we need to find the extremum of the function's intercept. According to the definition of derivative, y= dfrac {1}{x} Rightarrow y'=- dfrac {1}{x^{2}} < -1 Rightarrow the graph of the function passes through the point (1,1) when z has a minimum of 2 (since it is an open interval, 2 is not included), therefore the range of values for a+b is (2,+infty). Hence, the correct choice is boxed{text{C}}.