Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.
Answer:
Dale live 315 miles from the mountains
Step-by-step explanation:
Let y be the speed of Dale to the mountains
Time taken by Dale to the mountains=7 hrs
Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1
Time taken by Dale back home = 5hours
Since it speed increased by 18 miles per hour back home it speed = y+18
So distance traveled home =speed × time = (y+18)5 ...... eqn 2
Since distance cover is same in both the eqn 1 and eqn 2.
Eqn 1 = eqn 2
7y = (y+18)5
7y = 5y + 90
7y - 5y = 90 (collection like terms)
2y = 90
Y = 45
Substitute for y in eqn 1 to get distance away from mountain
= 7y eqn 1
= 7×45
= 315 miles.
∴ Dale leave 315 miles from the mountains
Answer: B, D, and E
Step-by-step explanation: 20% of 45 equals 9 D, D, and E all equal 9
Answer:
Step-by-step explanation:
To find : Acceleration in first 15 min . Distance between two cities Average speed of journey
Solution:
Each horizontal block is 1/8 hr = 7.5 min
Each vertical block is 10 km/hr
Time Velocity km/hr
0 Min ( 0 hr) 0
15 Min (1/4 hr) 50
45 Min (3/4 hr) 50
60 MIn ( 1 hr) 100
90 Min ( 3/2 hr) 100
120 Min ( 2hr) 0
Acceleration in first 15 min (1/4 hr) = (50 - 0)/(1/4 - 0) = 50/(1/4)
= 200 km/h²
Distance between two cities
= (1/2)(0 + 50)(1/4 - 0) + 50 * (3/4 - 1/4) + (1/2)(50 + 100)(1 - 3/4) + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)
= 25/4 + 25 + 75/4 + 50 + 25
= 125
Distance between two cities = 125 km
Average Speed of journey = 125/2 = 62.5 km/hr
Acceleration in first 15 min = 200 km/h²
Distance between two cities = 125 km
Average Speed of journey = 62.5 km/hr
Hope this helps..
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