answer:Let's denote the length of the train as ( L ) (in meters). When the train passes the tree, it only needs to cover its own length, which it does in 120 seconds. When it passes the platform, it needs to cover its own length plus the length of the platform (500 meters) in 170 seconds. The speed of the train should be the same in both cases since the speed is not mentioned to change. Therefore, we can set up two equations based on the formula for speed (( speed = frac{distance}{time} )): 1. When passing the tree: [ speed = frac{L}{120} ] 2. When passing the platform: [ speed = frac{L + 500}{170} ] Since the speed is the same in both cases, we can set the two expressions equal to each other: [ frac{L}{120} = frac{L + 500}{170} ] Now, we can solve for ( L ): [ 170L = 120(L + 500) ] [ 170L = 120L + 60000 ] [ 170L - 120L = 60000 ] [ 50L = 60000 ] [ L = frac{60000}{50} ] [ L = 1200 ] So, the length of the train is boxed{1200} meters.

answer:First, let's find out the total time of the concert without the intermission. Since the intermission was 10 minutes, we subtract that from the total concert time. The total concert time is 1 hour and 20 minutes. 1 hour is equal to 60 minutes, so the concert was 60 + 20 = 80 minutes long. Now, subtract the 10-minute intermission: 80 minutes - 10 minutes = 70 minutes of actual performance time. We know that all songs were 5 minutes long except for one that was 10 minutes long. Let's first account for the 10-minute song. That leaves us with 70 minutes - 10 minutes = 60 minutes for the remaining songs. Since each of the remaining songs is 5 minutes long, we can divide the remaining 60 minutes by 5 minutes per song to find out how many songs were performed. 60 minutes / 5 minutes per song = 12 songs Now, we add back the one song that was 10 minutes long. 12 songs (5 minutes each) + 1 song (10 minutes) = 13 songs in total. Therefore, the singer sang boxed{13} songs during the concert.

answer:(I) According to the question, f(x)=frac{sqrt{3}}{2}cos (2x+ frac{pi }{6})+{sin }^{2}x = frac{1}{4}cos 2x - frac{sqrt{3}}{4}sin 2x + frac{1}{2} =frac{1}{2}cos (2x+ frac{pi }{3})+frac{1}{2}. From 2kpi -pi leqslant 2x+ frac{pi }{3}leqslant 2kpi ,kin mathbb{Z}, we get kpi -frac{2pi }{3}leqslant xleqslant kpi -frac{pi }{6},kin mathbb{Z}. Thus, the interval where f(x) is monotonically increasing is [kpi -frac{2pi }{3},kpi -frac{pi }{6}],kin mathbb{Z}. (II) According to the question, f(x)=(frac{sqrt{3}}{2}cos varphi -frac{1}{2})cos 2x-frac{sqrt{3}}{2}sin varphi sin 2x+frac{1}{2}. Since the maximum value of the function f(x) is frac{3}{2}, then (frac{sqrt{3}}{2}cos varphi -frac{1}{2}{)}^{2}+(frac{sqrt{3}}{2}sin varphi {)}^{2}=1. From this, we get cos varphi =0, and given that 0 leqslant varphi < pi, We have varphi = boxed{frac{pi }{2}}.

answer:(Ⅰ) When x=2, g(x) = a - |x-2| reaches its maximum value of a, Since f(x) = |x+1| + |x-3| geq 4, and only when -1 leq x leq 3, f(x) reaches its minimum value of 4, Since the inequality f(x) < g(x) has solutions, Therefore, a > 4, which means the range of the real number a is (4, +infty). (Ⅱ) When x= frac{7}{2}, f(x) = 5, Then g(frac{7}{2}) = -frac{7}{2} + a + 2 = 5, solving this gives a = frac{13}{2}, Therefore, when x < 2, g(x) = x + frac{9}{2}, Setting g(x) = x + frac{9}{2} = 4, we get x = -frac{1}{2} which is in the interval (-1, 3), Therefore, b = -frac{1}{2}, and thus a+b = boxed{6}.