A cable car begins its trip by moving up a hill. As it moves up, it gains elevation at a constant rate of 50 feet/minute until it reaches the peak at 2,000 feet. Then, as the car moves down to the hill's base, its elevation drops at the same rate. The equation that models the cable car's elevation, e, after t minutes is e = |t? | +. The cable car's elevation will be 750 feet after minutes.

Question

Patsy

Mathematics

29 August, 05:48

A cable car begins its trip by moving up a hill. As it moves up, it gains elevation at a constant rate of 50 feet/minute until it reaches the peak at 2,000 feet. Then, as the car moves down to the hill's base, its elevation drops at the same rate. The equation that models the cable car's elevation, e, after t minutes is e = |t? | +. The cable car's elevation will be 750 feet after minutes.

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Brennan Schmidt · Answer 1

he cable car's elevation will be 750 feet after 15 minutes

Step-by-step explanation:

Given:

elevation, e = 2,000 feet

rate, I = 50 feet/minute

The equation that models the car's elevation, e, after t minutes is e = |t

The cable car's elevation will be 750 feet after how many minutes?

since, e = It

for 2,000 feet = 50 feet/minute * t

t = 2000/50 = 40minutes

using the same constant rate, I for 750ft, T =

since, e = It

for 750 feet = 50 feet/minute * t

t = 750/50 = 15minutes