5th grade math problems

The speed of the train does not change, so divide the two journeys by the two times.

Let the length of the car be x468+x) 35=x 9

9(468+x)=35x

26x=4212

x=162A: The length of the train is 162 meters.

Let the commander be x

468+x）/35=x/9

9(468+x)=35x

26x=4212

x=162A: The length of the train is 162 meters.

Equations learned in the 5th grade of elementary school... I forgot whether I had studied at that time.

If you don't use the equation, then consider that the time taken to pass through the bridge is the length of the bridge + the length of the train, then the time taken by the train to pass the distance of 468 meters is (35-9) seconds, then the train speed is 468 (35-9), then the train length is.

468 (35-9)*9=162 meters.

Solution: Let the speed be V and the train length L

From the title: the first time T1 is 9 seconds, the second time T2 is 35 seconds, and the length of the shovel is 468 meters.

l=v* t1①

l+s=v* t2②

Lianli:

l = 162 meters.

A type of travel problem is a type of travel problem from the front of the car to the rear of the car off the bridge. The pole is regarded as a line or a point, its length is not counted, it takes 9 seconds, that is, the train is exactly the length of the car, over the 468-meter iron bridge, the train is 468 meters more than just now, it took 35-9 26 seconds, the speed is: 468 26 18 meters per second, the length of the car:

18*9 162 meters.

468 (35-9) = 18 meters.

18*9 162 meters.

Use the correspondence to find the velocity. (The length of the signal light is considered 0).

Let the length of the car be x, then the speed of the car is x 9, and the length of the whole car should be added to the length of the bridge to cross the bridge, 468 + x = (x 9) * 35

4212+9x=35x

26x=4212

x=162A: The length of the train is 162 meters.

d The area cannot be determined because the upper bottom cannot be determined.

a plus units.

Bottom = x High = y

Trapezoid=1 2*xy

Parallelogram = xy

xy-1/2xy=24