Two trains simultaneously left points m and n and headed towards each other. the distance between point m and point n is 380 mi. the speed of the train, which started, from point n was 5 mph faster than the speed of the other train. in two hours after the departure the distance b

not complete question pls finish it

Step-by-step explanation:

Lets denote position M as our 0 point. Thus point N is at 380 mi

Let m = speed of train starting from point M. Let xm be distance train has traveled.

Let n = speed of train starting from point N. Let xn be distance train has traveled.

Since speed = distance/time we have

xm = 0+m*t (remember we are taking point M as 0 point)

xn = 380-n*t (point N is at 380

We are given speed of train starting at point N was 5 mph faster than one starting at point M. Thus

n = m+5

We are told after t = 2 hrs that the distance between the trains was 30 mi. Thus

|xm-xn| = |(0+m*t)-(380-n*t)| = |m*t-380+n*t| = |m*t-380+(m+5)*t| = |2m*t+5t-380|

Plug in t = 2

|2m*2+5*2-380| = |4m+10-380|=|40-370| = 30

So we have two equations

4m-370 = 30 which means m = 400/4 = 100 mph

or

-(4m-370) = 30 which means 4m-370 = -30 which means m = 340/4 = 85 mph

From this n = 105 mph or n = 90 mph

Option 1:

If m = 100 mph then xm = 100*2 = 200 miles

This means n = 105 mph and xn = 380-105*2 = 170 miles

I think we can discard this because this means the trains have passed each other. I'm thinking the problems wants before they pass each other.

Option 2:

If m = 85 mph then xm = 85*2 = 170 miles

This means n = 90 mph and xn = 380-90*2 = 200 miles

This is probably the likely case.

So your answer is the train leaving from M was at 85 mph and the train leaving from point N was at 90 mph.

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)