Question 1161706
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Here is the distance-rate-time table for both the car and plane
<table border = "1" cellpadding = "5"><tr><td></td><td>Distance</td><td>Rate or Speed</td><td>Time</td></tr><tr><td>Car</td><td>200</td><td>200/t</td><td>t</td></tr><tr><td>Plane</td><td>920</td><td>920/t</td><td>t</td></tr></table>
You start off filling out the distances for each. Those are the given numeric values. The time t is the same for both the car and plane. The rate or speed is found using the formula r = d/t, which comes from d = r*t. The table is optional, but may be handy to keep track of everything.


We know that the plane goes 180 mph faster than the car, so this means
plane's speed = (car's speed) + 180
920/t = (200/t) + 180
920 = 200 + 180t
on the last step, I multiplied every term by t to clear out the fractions


From here, let's solve for t
920 = 200 + 180t
920-200 = 180t
720 = 180t
180t = 720
t = 720/180
t = 4


Both car and plane travel for t = 4 hours.
Doing so, the car's speed is r = d/t = 200/4 = 50 mph
and the plane's speed is r = d/t = 920/4 = 230 mph
The plane's speed is 230-50 = 180 mph greater than that of the car


Answer: plane's speed = <font color=red size=4>230 mph</font>
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