SOLUTION: Two surveyors with two-way radios leave the same point at 9:00 A.M., one walking due south at 2 mi/hr and the other due west at 4 mi/hr. How long can they communicate with one anot

Question 347630: Two surveyors with two-way radios leave the same point at 9:00 A.M., one walking due south at 2 mi/hr and the other due west at 4 mi/hr. How long can they communicate with one another if each radio has a maximum range of 2.98 miles?
Round the answer to the nearest minute.

Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
(2t)^2+(4t)^2=2.98^2
4t^2+16t^2=8.88
20t^2=8.88
t^2=8.88/20
t^2=.444
t=sqrt.444
t=.666 hours.
Proof:
(2*.666)^2+(4*.666)^2=2.98^2
1.332^2+2.664^2=8.88
1.774+7.097=8.88
8.87~8.88
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