On the Shanghai Nanjing Expressway, a car was driving at a constant speed of 108km / h. The driver suddenly found that there was an emergency ahead. After 0.6s, he began to brake, and after 4.4S, he taxied 52m, and the car stopped. The average speed of the driver from finding the situation to stopping was______ ．

∵ v = 108km / h = 30m / s, ∵ 0.6s the distance of the car is s = VT = 30m / s × 0.6s = 18m; the distance of the car from finding the situation to stopping is s' = S + s sliding = 18m + 52m = 70m; the time used is t '= t + T sliding = 0.6s + 4.4S = 5S; the average speed from finding the situation to stopping is v' = s' = 70m5s = 14m / S; so the answer is: 14m / s

As shown in the figure, given that angle 1 = angle 2, angle 3 = 80 degrees, calculate the degree of angle 4

The total length of Luning expressway is 274.08 km. A bridge car and a bus leave from Shanghai and Nanjing at the same time. The average speed of cars is 118.4 km per hour, and that of buses is 110 km per hour. After a few hours, the two cars meet on the way?

Setting: after X hours, two cars meet on the way
118.4x+110x=274.08
x=1.2
A: after 1.2 hours, the two cars met on the way

Sum Sn = 1 / (1 * 4) + 1 / (2 * 7) +. + 1 / N * (3N + 1)

1 / N (3N + 1) = 3 [(3N + 1-3n) / 3N (3N + 1)] = 3 [1 / 3n-1 / (3N + 1)] Sn = 3 [(1 / 3 + 1 / 6 + 1 / 9 +... + 1 / 3n) - (1 / 4 + 1 / 7 + 1 / 10 +... 1 / (3N + 1))] define ψ (k) = Lim [n →∞] (lnn - (1 / K + 1 / (K + 1) +... + 1 / N)), then Sn = ψ (K + 1) - ψ (K + 4 / 3) + 3 - π√ 3 / 6-3ln3 / 2

The railway between a and B is 40km longer than the highway. The car starts from a with a speed of 40km / h. After 0.5h, the train starts from B with a speed of 60km / h. as a result, the car arrives at B one hour later than the train?

Let the highway be x, then the railway be x + 40. From a to B, the time taken by the car is x / 40, and the time taken by the train is (x + 40) / 60. From the title, we know that the time taken by the car is 0.5 + 1 = 1.5h longer than that of the train. The equation can be formulated. X / 40 - (x + 40) / 60 = 1.5

() - (X & sup2; + 3xy) = - XY-1 / 3Y & sup2; if x + y = 5, then 3-x-y + () if X-Y = 3 / 4, then 4 (Y-X) = ()

（x^2+2xy-1/3y^2）-（x²+3xy）= -xy—1/3y²
If x + y = 5, then 3-x-y = (- 2)
If X-Y = 3 / 4, then 4 (Y-X) = (- 3)

A batch of cement consumed more than half a ton in the first day, and the remaining one-third less than two tons in the second day, leaving 16 tons of raw materials

With the solution of the equation, we first assume that there is x tons of cement
[x - (2 / 1X + 1) times 3 / 1-2] + 2 / 1 + 1 = x-16
[(2 / 1x-1) times 3 / 1-2] + 2 / 1X + 1 = x-16
6/1X+2/1X-3/1-2+1=X-16
3/2X-3/4=X-16
X=44
I can't use the multiply sign. Please forgive me
Another three-thirds is one-third

Let f (x) = m · n, where vector M = (sin (x + π / 4), a), n = (2cos (x + π / 4), 1), X ∈ R, and function y = f (x) image passes through point (π / 3,1)
(1) Finding the value of real number a
(2) Finding monotone interval and symmetric center of function f (x)
(3) If f (a) = 7 / 4 and a ∈ (π / 2, π), find the value of F (A / 2)

First determine f (x)
f(x)=m·n=(sin(x+π/4)*(2cos(x+π/4)+a*1=cos(2x)+a
(1) Because the function y = f (x) passes through the point (π / 3,1)
Then 1 = cos (2 * π / 3) + A, a = 3 / 2
That is, f (x) = cos (2x) + 3 / 2
(2) Because 2K π - π

When 2 tons of yellow sand are transported to the construction site, 25% of it will be used on the first day, and 14% on the second day. How much is left?

1-25-14 = 720

Let the image of the function y = f (x) be symmetric with respect to the line x = 1. When x is less than or equal to 1, the square of F (x) = (x + 2) minus 1, then when x is greater than 1, f (x) =?