Math first year problem first aid good has a heavy reward

In the first question, let the truck speed be x kilometers per hour.

Because a truck travels the same distance as a car, there is an equation:

2x + 1*x + 15/60*x + 2x = (x+30)*1 + x+30)*(1 - 1/3)*2

Solve the equation to get x = ..

In the second question, let's say B catches up with A x hours after departure.

A completed the journey in 11 hours at a speed of 1 11 B completed the journey in 5 hours at a speed of 1 5

A had been walking for 4 hours when B departed.

4 * 1 11 + x * 1 11 = x * 1 5 Solve the equation to get x = .

Question 1: If the speed of the truck is x, then there is.

2x+ to solve the equation, you can find x

Question 2: You can't calculate it with a one-dimensional equation, because you don't give a distance, so you can only set two unknowns, and the method is the same as the above problem.

1.The speed of the vehicle is set to x kilometers per hour.

Equal column equivalence according to the distance traveled.

2x+x+15/60*x+2x=x+30+4*（x+30）÷3x=242.It can be seen that A took 11 hours to complete the journey, and B took 5 hours to complete the journey.

If the total distance is considered as a unit of 1, then V A = 1 11 B = 1 5 B has already walked for 4 hours when B sets off. According to the equation of the equal series of distances.

Set B walked for x hours.

1/11*4+1/11*x=1/5*x

x = 3 and 1/3 = 200 points.

Answer: 1:20 chase.

The hard plane can be the anti-Kashgar probably hard to spill through Kashgar feel that the Kashgar electrical processing Caesar machine.

1.In the triangular EDA, there is.

Angle EAC (acute angle) = angle D + angle DEA · 1) In the same way, angle EAC (acute angle) = angle B + angle BCA · 2) Angle F + Angle FEA + Angle FCA + Angle EAC (obtuse angle) = 360 degrees, so angle F + Angle FEA + Angle FCA = Angle EAC (acute angle) ·· 3) Obtained from the known.

Angular BCA = 2 angular FCA, angular DEA = 2 angular FEA 4).

From (1), (2), (3), (4) obtain:

2 angle f = angle b + angle d

2.From the answer of 1, we get: 2x=2+4=6, and x=3

1.When the number of people traveling to the hidden tank is 25, travel agency A pays a fee of 25*660*yuan; B travel agency pays 23*660* yuan.

When the number of tourists is 40, travel agency A pays 19,800 yuan; B travel agency paid 20,064 yuan.

2. Let the number of people be x

660*x* solves the equation x=32

3. When the number of people is less than 32, it is more suitable to choose travel agency B; When the number of people is greater than 32, it is more cost-effective to choose travel agency A; When the number of people is equal to 32, the two travel agencies are equally cost-effective.

There are x people in class A and y people in class B.

According to the title, 300 < 6+9(x-1) <400

300 < 13+8(y-1) <4006+9(x-1)=13+8(y-1)

Solve 33 2 3 < x < 44 7 9< y <

9x-3=8y+5

x=8y+8/9

Because the number of people must be an integer.

So 34 is less than or equal to x less than or equal to 44

37 is less than or equal to x less than or equal to 49

and x is a multiple of 8.

8y+8 is a multiple of 9.

So x=40, y=44

x+y=40+44=84

The number of students in Class A and Class B is 84.

Set the number of donors in class A as x, the number of donors in class B as y, and the donation in class A as 6 + (x-1) * 9 that is, 9x-3

The donation for Class B is 13+(y-1)*8, i.e. 8y+5

Then 9x-3=8y+5, i.e. (x-8) (y-x)=8 and 300<9x-3<400

So x is 34 to 44, and x-8 is an integer from 26 to 36, and y-x is an integer, so can be a multiple of 8, only x-8=32, that is, x=40, so y-x=5, so y=45

So the sum of the numbers is 85

Question 1: The first pair.

12, the second pair 11, the third pair 10, and so on, the answer is 12+11+10+·· 1+1 (first pair) =

Question 2: When the distance difference between A and B is less than or equal to 400,400 (50-46)=100 minutes.

Tested. It is found that A is at the midpoint of the DC edge and B is at the midpoint of the ed edge, that is, when A travels to the D point, that is, A runs (50*100+200) 50=

Question 3: Let the swimmer's speed be x and the water velocity be y, and there is a problem that can be achieved from the time the kettle is dropped to the time it is chased, and the movement time of the kettle and the swimmer is equal, that is:

x/6+2)/(x+y)=2/y

The solution is y=1 12 km/h.