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Answered by
1
let the speed of the first var be x
a/q
5(x)+5(x+5)=425
5x+5x+25=425
10x=425-25
10x=400
x=400/10
x=40
speed of first car= 40km/hr
speed of second car=45km/hr
a/q
5(x)+5(x+5)=425
5x+5x+25=425
10x=425-25
10x=400
x=400/10
x=40
speed of first car= 40km/hr
speed of second car=45km/hr
mathematics10:
thanks
Answered by
3
Let the speed of the 1st car be x.
Then the speed of the other car will be 5 km more than that of the other = x + 5.
Given that distance = 425.
Time = 5hrs.
We know that Time t = distance/rate
Then r = distance/time
= 425/5
= 85.
x + x + 5 = 85
2x + 5 = 85
2x = 85 - 5
2x = 80
x = 80/2
x = 40.
The average speed of the 1st car = 40km/hr.
average speed of the 2nd car = x + 5 = 40 + 5 = 45km/hr.
Hope this helps!
Then the speed of the other car will be 5 km more than that of the other = x + 5.
Given that distance = 425.
Time = 5hrs.
We know that Time t = distance/rate
Then r = distance/time
= 425/5
= 85.
x + x + 5 = 85
2x + 5 = 85
2x = 85 - 5
2x = 80
x = 80/2
x = 40.
The average speed of the 1st car = 40km/hr.
average speed of the 2nd car = x + 5 = 40 + 5 = 45km/hr.
Hope this helps!
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