Atrain covered the distance of 400 km between a and b at a certain speed. on the way back it covered 2/5 of the distance at that same speed and then it decreased its speed by 20 km/hour. find the speed of the train at the end of its journey from b back to a, if the entire trip took 11 hours.

s - 20 = 60 km/hr

Step-by-step explanation:

The distances are the same going there and coming back.

So

Givens

d = 400

2/5 * d = 2/5 * 400 = 160

The rest of the trip back = 400 - 160 = 240

Equation

400/(s) + 160/s + 240/(s - 20) = 11 hours.

Solution

distance / speed = time or

d/s = time

Multiply through by s*(s - 20)

400*(s - 20) + 160*(s - 20) + 240s = 11*s*(s - 20)       Remove the brackets.

400s - 8000 + 160s - 3200 + 240s= 11(s^2 - 20s)   Collect like terms on the left.

560s - 11200 + 240s = 11(s^2 - 20s)                          Remove the brackets on the rt.

800s - 11200 = 11s^2 - 220s                                   Switch and add 220s

11s^2 - 220s + 220s = 800s +220s - 11200          Combine

11s^2 = 1020s - 11200                                            Put everything on the left.

11s^2 - 1020s + 11200 = 0

this quadratic factors into

(11s - 140 )(s - 80) = 0

You have 2 solutions

11s - 140 = 0

11s = 140

s = 140/11

s = 12.727272... This solution won't work because the speed has no room to drop down 20 km / hour

s - 80 = 0

s = 80 km hour

But that is not the answer

The answer you want is s - 20 which is 60 km/hr.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

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step-by-step explanation:

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    -7                           -7

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