Let the slower train's velocity be x-21
Let the faster train's velocity be x
We know that the approach speed is the sum of both speeds, so x+x -21= 2x-21.
The approach rate is given by Distance/time = 471/3 = 157mpH
x+x-21=157
2x=157+21
2x=178
x=89mph
The slower train is travelling 89-21 = 68mph
The faster train is travelling 89mph.
<span>x+y=30.....(1) and x-y=2......(2) so(1)+(2)......x+y+x-y=30+2 so 2x=32...x=16 now apply x=16 in (1) so we get......16+y=30 so .......y=14</span>
Answer: C
The x row goes up every time by two and the y decreases by 4.
6-4=2 ect...
MARK THE BEST PLZ!!
Left column. Right column
6m. 6x10^0m
120,000m. 12x10^4m OR 1.2x10^5m
0.0012m. 12×10^4m OR 1.2×10^-3m
300m. 3×10^2m
0.6m. 6x10^-1m
0.0009m 9x10^-4m
6000m. 6x10^3m
