Everyone, help me solve this math problem! 30

Suppose A and B are starting points A and B, the first meeting is at C, and the second meeting is at D. Make the diameter EF so that the distance between d and f is 60 and the distance between b and f is 20. Because when A walked 60 during the first meeting, and A and B walked a semicircle together, it can be seen that before the second meeting, the semicircle EDF was the last part of the road A and B took, and the distance A walked was 60.

In the same way, on the other semicircle ECF, CF is the road traveled by A, and the distance is 60; EC is the path that B has walked. Since the bf is 20, the black obvious ae is also 20, according to the title, it is known that ac is the road traveled by a is 60, so ec=ac-ae=40, so the semicircle ecf is ec+cf=100, so the circumference of the circle is 200, that is, the runway is 200 meters long!!

Let the length of the runway be x meters, the velocity of A is y, the velocity of B is z, and when x 2>80 is x >160, then there is a system of equations.

60/y=(x/2-60)/z

x/2+80)/y=(x-80)/z

Divide the upper and lower equations to give equation 60 (x 2+80)=(x 2-60) (x-80).

Solve the equation to get x=200

When x 2<80, i.e. x <160 then there is a system of equations.

60/y=(x/2-60)/z

80-x/2+x)=(x-80)/z

The solution of the equation is x=200, which is inconsistent with the condition x<160, so the solution is invalid.

So x=200, i.e. the circular runway is 200 meters long.

When x 2<80, i.e. x <160 then there is a system of equations.

60/y=(80-60)/z

160/y=80/z

There is no solution to this system of equations.

So x=200, i.e. the circular runway is 200 meters long.

Let the runway length be 2x

x-60) 60=(2x-80) (x+80) so x=100

So the runway is 200 meters long.

Let the three sides be a, b, and c

Know a+b+c=10

and the sum of the two sides is greater than the third side.

So the longest side should be less than 5

Let the longest side be a

So a<5

and a+b+c=10

So a>10 3

So a=4 when a=4, b+c=6

Let c be the shortest side of the oak block, then c 3

When c = 1, b = 5 does not form a triangle at this time.

When c=2 and b=4 next to the simple chaos, it is an isosceles triangle.

When c = 3 and b = 3, it is also an isosceles triangle at this time.

So the possibility of flavor 3, 3, 4 or 2, 4, 4

1. If the first paragraph is 1 long, the remaining 9 will be divided into 2 sections. Apparently 1,8;2，7；3,6；4, 5 can't do it, let's go!

2. Let the first paragraph be 2 long, and the remaining 8 will be divided into 2 sections. 1,7；2,6；3,5；No, don't do it! But 4, 4 can.

3, set the first section of the length of 3, and the remaining 7 of the wild excitement is divided into 2 sections. 1,6；2,5 No, let's go! But 3,4 can.

4. Let the first paragraph be 4 long, and the remaining 6 will be divided into 2 sections. 1, 5 no, She! But Song leather socks 2, 4 can.

5. Set the first paragraph to be 5 long, then the remaining 5 is divided into 2 sections. Neither is advisable.

None of them can be taken after that, because the sum of the two segments is less than the length of the first segment.

To sum up, there are two types: 2, 4, 4 and 3, 3, 4.

Vector a=(sinx,cosx),vector b=(cosx,cosx)b+a=(sinx+cosx,2cosx)f(x)=a· (b+a）=sin x+cosxsinx+2cos x=1+cos x+1 2sin2x=1 2cos2x+1 2sin2x+3 2

2/2sin（2x+π/4）+3/2

f(x) maximum is 2 2+3 2 and the minimum positive period is when 2 2sin(2x+ 4) 0 holds the inequality +2k 2x+ 4 0+2k (k z)x [3 8 +k ,-1 8 +k ].

f(x)=sinx(sinx+cosx)+2cos2 x=3 2+2 root number2 sin(2x+4 parts) maximum value(3+root number2) 2

Minimum positive period,

2x+4 belongs to [2k, 2k + k belongs to an integer, so x belongs to [k - 8, k +3 8] The second question on the first floor is wrong, and it is reversed.

Master Wang walked x kilometers per hour.

Arrive at the unit 10 minutes in advance, Master Wang will save 5 minutes for the car to walk the distance from the platform to the unit, and walk 50*5 60=25 6 kilometers from the platform to the unit.

A total of 55 + 5 = 60 minutes, x = 25 6 1 = 25 6 (km h).