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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?

 May 8, 2019
 #1
avatar+111431 
+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

 

 

 

 

cool cool cool

 May 8, 2019
 #2
avatar+25826 
0

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

 May 8, 2019

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