answer this quadratic word problem, 20 points and also i will mark brainliest,
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average speed = Total distance / Total time
Time = 2 hours
Speed = S1 of train 1
Speed = S2 of train 2
Distance = 50km
t = d/s
2 = 50 / (1+5)S
12S= 50
S = 50/12 = 4.1 KMPH
Time = 2 hours
Speed = S1 of train 1
Speed = S2 of train 2
Distance = 50km
t = d/s
2 = 50 / (1+5)S
12S= 50
S = 50/12 = 4.1 KMPH
AryanK10:
10 km, 15 km and 20 km
Let the speed of the second train = x km/hr
The speed of the first train = x+5 km/hr
Distance covered after two hours by the first train is 2(x+5) km.
Distance covered by the second train after two hours is 2x km.
(2x)^2 + 4(x+5)^2=(50)^2 (Using pythagoras theorem)
Solve
We get x=-20 and x=15
The speed of the first train = 15+5=20km/hr
The speed of the first train = x+5 km/hr
Distance covered after two hours by the first train is 2(x+5) km.
Distance covered by the second train after two hours is 2x km.
(2x)^2 + 4(x+5)^2=(50)^2 (Using pythagoras theorem)
Solve
We get x=-20 and x=15
The speed of the first train = 15+5=20km/hr
this is right then
yes
points please
and comments
didnt you get it?
i put it when i asked
u didnt mark brailiest
brainliest
Answered by
1
Let s = the speed of the northbound trainThen(s+5) = the speed of the westbound train:This is a right triangle problem: a^2 + b^2 = c^2The distance between the trains is the hypotenusedist = speed * timeThe time is 2 hrs, so we havea = 2s; northbound train distanceb = 2(s+5) = (2s+10); westbound distancec = 50; distance between the two trains:(2s)^2 + (2s+10)^2 = 50^24s^2 + 4s^2 + 40s + 100 = 2500 Arrange as a quadratic equation4s^2 + 4s^2 + 40s + 100 - 2500 = 08s^2 + 40s - 2400 = 0:Simplify, divide by 8:s^2 + 5s - 300 = 0:Factors to(s - 15)(s + 20) = 0:The positive solution is what we want heres = 15 mph is the speed of the northbound trainthen5 + 15 = 20 mph is the speed of the westbound train::Check this; find the distance (d) between the trains using these distancesNorthbound traveled 2(15) = 30 miWestbound traveled 2(20) = 40 mid = d = 50
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