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The first time they met 16 meters away from B and the second time they met 18 meters away from a Solving linear equation with one variable
Let AB distance be x, the speed ratio of a and B is fixed, and the time for each encounter is equal. According to the formula: time = distance / speed
First meeting:
(x-16) / speed a = 16 / speed B
Second meeting:
(2x-18) / speed a = (x + 18) / speed B
(x-16) / 16 = (2x-18) / (x + 18) multiply both sides by 16 (x + 18)
(X-16)(X+18)=16(2X-18)
X^2+2X-18*16=32X-16*18
X^2-30X=0
X(X-30)=0
We get: X1 = 0, X2 = 30, because the distance of AB cannot be zero, so we take x = 30
A and B two cars, from AB two places at the same time, meet two hours, then continue to move on, a arrived at B, B is 60 kilometers away from a
Given that the speed ratio of two cars is 2:3, what is the speed of a and B cars?
A 54, B 36
Party A and Party B walk from ab at the same time. The distance between the first meeting place and a accounts for two fifths of the whole journey. After meeting, the two cars continue to move forward and reach a respectively
After arriving at AB, they turn around and return. The second time they meet is two fifths of a kilometer away from B. how many kilometers are the two places apart
The distance between the first meeting place and a place accounts for two fifths of the whole journey, which means that a has done two fifths of the whole journey. When the second meeting, a and B have done three whole journey. At this time, a should have done 2 / 5 × 3 = 6 / 5 of the whole journey
The two cars leave from AB and meet at 2 o'clock. Then they go on and pass by car a at 1.5 o'clock. Car B is 35 kilometers away from A. what's the distance between the two places?
Please tell me in detail and use the proportion method to solve the problem
Let X be the distance
x/2=(2x-35)/3.5
x=140
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