The two trains above are docked at St. Mark's Train Station and are assigned to parallel tracks. At 7am, Train #941 leaves the station traveling 48 mph. Three hours later, Train #940 leaves
the station traveling in the same direction as Train #941. At 5pm, Train #940 catches Train
#941.
Find the speed of Train #940.
Hel

The speed of Train #940 is 68. mph

Step-by-step explanation:

The given parameters of the w=question are;

The time at which train #941 leaves the station = 7 a.m.

The speed with which train #941 travels, v = 48 mph

The time at which train #940 leaves the station = 3 hours after train #941

Therefore;

The time at which train #940 leaves the station = 3 hours + 7 a.m. = 10 a. m.

The direction of train #940 = The direction of train #941

The time at which train #940 catches train #941 = 5 p.m.

We have;

The time, 't', (after departure of train #941) it takes train #940 to catch train #941, is given as follows;

t = 5 p.m. - 7 a.m.

∴ t = 17:00 hours - 07:00 hours = 10 hours

The time (after departure of train #941) it takes train #940 to catch train #941, t = 10 hours

The distance from St. Mark's Train Station train #940 catches Train #941, 'd', is given as follows;

d =  The speed of travel of train #941 × t

∴ d = 48 mph × 10 h = 480 miles

The distance from St. Mark's Train Station train #940 catches Train #941, d = 480 miles

The time train

The time, t₂, it takes Train #940 to catch up with train #941 is given as follows;

t₂ = 5 p.m. - 10 a.m. = 17:00 hours - 10:00 hours = 7 hours

The speed of Train #940, v₂ = d/t₂

∴ v₂ = (480 miles)/(7 hours) = 68. mph

The speed of Train #940, v₂ = 68. mph

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

the area of the rectangle is 35units^2.

step-by-step explanation:

graph it and find the distance.

it is a 5*7 rectangle.

therefore the area is 35units^2.