If cos 65 degree +cos 55 degree = cos θ, find θ in radians
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<span>The rate of the express train is 15 mph faster than the freight train.
Let's grasp onto this piece of information.
The rate at which the freight train goes will be labeled "x" (keep in mind rate is just the speed)
If the freight train has a speed of "x" then the express train which is going 15mph faster than the freight train can be represented as "x + 15mph."
Our result is in HOURS, but our speed is in Miles per Hour (Miles/Hour). Meaning we need to multiply our speeds by some time (t).
Let's set up two equations:
Freight Train: 18 hours * x mph = distance
Express Train: 15 hours * (x mph + 15mph) = distance
They are both the same distance so we can set the two equations equal to each other like this:
18 hours * x mph = 15 hours * (x mph + 15mph)
That right side is a bit of a thorn in our side, so let's simplify it by multiplying all the terms in the parenthesis by 15 hours (keep in mind the hours in the units cancel out).
18 hours * x mph = 15x miles + 15 * 15 miles
18 hours * x mph = 15x miles + 225 miles
Let's simplify the left side (</span>
)
18x miles = 15x miles + 225 miles
Get all the "x" terms on one side by subtracting 15x on both sides.
18x - 15x = 225 miles
Simplify
3x = 225 miles
Divide by 3 on both sides to solve for x.
225/3
225 = 25*9
(25*9)/3 = 25*3 = 75mph
So "x" which we chose was the rate of the freight train is 75mph.
Adding 15mph to the freight train rate gives you the express trains rate which is 90mph.
Hope that helped.
Let's grasp onto this piece of information.
The rate at which the freight train goes will be labeled "x" (keep in mind rate is just the speed)
If the freight train has a speed of "x" then the express train which is going 15mph faster than the freight train can be represented as "x + 15mph."
Our result is in HOURS, but our speed is in Miles per Hour (Miles/Hour). Meaning we need to multiply our speeds by some time (t).
Let's set up two equations:
Freight Train: 18 hours * x mph = distance
Express Train: 15 hours * (x mph + 15mph) = distance
They are both the same distance so we can set the two equations equal to each other like this:
18 hours * x mph = 15 hours * (x mph + 15mph)
That right side is a bit of a thorn in our side, so let's simplify it by multiplying all the terms in the parenthesis by 15 hours (keep in mind the hours in the units cancel out).
18 hours * x mph = 15x miles + 15 * 15 miles
18 hours * x mph = 15x miles + 225 miles
Let's simplify the left side (</span>
18x miles = 15x miles + 225 miles
Get all the "x" terms on one side by subtracting 15x on both sides.
18x - 15x = 225 miles
Simplify
3x = 225 miles
Divide by 3 on both sides to solve for x.
225/3
225 = 25*9
(25*9)/3 = 25*3 = 75mph
So "x" which we chose was the rate of the freight train is 75mph.
Adding 15mph to the freight train rate gives you the express trains rate which is 90mph.
Hope that helped.