Car a and car B start from two places 395km apart at the same time. Car a stops for one hour due to fault and then goes on. As a result, car B meets car a on the way 4.5 hours later. It is known that the speed of car B is 45km per hour. The speed of car a is calculated

Car a and car B start from two places 395km apart at the same time. Car a stops for one hour due to fault and then goes on. As a result, car B meets car a on the way 4.5 hours later. It is known that the speed of car B is 45km per hour. The speed of car a is calculated

Set the speed of car a as X km / h
From the meaning of the title, it is concluded that:
3.5x+4.5*45=395
3.5x+202.5=395
3.5x=192.5
x=55
A: the speed of car a is 55 km / h
Suppose the speed of car a is x kilometers per hour.
3.5x+4.5×45=395
x=55
A: car a travels 55 kilometers per hour.
Let the speed of car a be X.
3.5x+45*4.5=395
Er ~ ~ ~ what grade are you in? Question: Grade Four
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6/5-4/2=\x09\x093/2-2/1=\x094/1+3/2=
5/3-3/1= 8/4-6/1= 3/5-4/3=
6/2+3/1=\x099/5-2/1=\x097/2+3/1=
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9/3-3/1=\x0912/7-8/3=\x095/2+3/1=
1-9/4=\x099/5+9/8=\x09\x09 8/1+8/7=
24/19-24/13=\x0936/19+36/3=
7/3+7/4=\x098/11-8/1=\x094/1-9/1=
13/12-13/3=\x099/8-9/1=\x091-6/1=
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8/2-13/2=\x095/1+7/1=\x093/1+2/1=
4/1+2/1=\x092/1-5/1=\x097/5+8/1=
5/2-3/1=\x096/1+5/2=\x099/5+8/1=
4/1-6/1=\x095/2+7/1=\x096/1+2/1=
9/2-12/3=\x09\x093/2+8/1=\x095/2-3/1=
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1/4+1/8=\x09\x092/3-1/6=\x09\x093/7+5/14=
2/5-2/15=\x09\x097/8-9/16=\x09\x091/32+1/4=
2/9+7/9=\x09\x093/4-9/16=\x09\x0925/32-3/8=
3/5+13/15=\x09\x0923/25+1/5=
21/25-1/5=
3/14+2/7=
2/3+1/9=
5/7-3/14=
7/9+1/18=
19/21+3/7=
23/35-1/5=
3/4+1/16=
3/8-1/16=
5/9+4/9=
You must choose me
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I really don't have that many. 1) 12 / 13-8 / 13 = 2) 8 / 9-2 / 9 = 3) 14 / 15-10 = 4) 11 / 13-7 / 13 = 5) 8 / 10-1 / 10 = 6) 13 / 14-7 / 14 = 7) 5 / 9 + 1 / 9 = 8) 5 / 13-4 / 13 = 9) 4 / 13-1 / 13 = 10) 4 / 5-2 / 5 = 11) 9 / 12 + 3 / 12 = 12) 7 / 13-3 / 13 = 13) 1 / 12 + 11 / 12 = 1
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