At 2:45 p.m., a jet is located 56 mi due east of a city. A second jet is located 30 mi due north of the city. To the nearest tenth of a mile, what is the distance between the two jets?

hoiuu May 8, 2019

#1**+2 **

Let the coordinates of the first jet be (56, 0) and the second be ( 0, 30)

The distance is

sqrt [ ( 56 - 0)^2 + ( 0 - 30)^2 ] =

sqrt [ 56^2 + 30^2 ] =

sqrt [ 3136 + 900 ] =

sqrt (4036) ≈ 63.5 miles

CPhill May 8, 2019

#2**+2 **

Same answer asn Chris'

Use Pythagorean theorem one leg is from city to 1st jet = 56 mi

other leg is from city to 2nd jet = 30 mi

Hypotenuse is distance between the jets d^2 = 56^2 + 30^2 d = 63.53 miles

ElectricPavlov May 8, 2019