Roger can run one mile in 14minutes. jeff can run one mile in 12minutes. if jeff gives roger a 1minute head start, how long will it take before jeff catches up to roger? how far will each have run?

v1 = 1/14 miles per min the average speed of Roger

v2 = 1/12 miles per min the average speed of Jeff

t = the time in minutes that Jeff runs before catches up to Roger

t + 1 = the time in minutes that Roger runs before Jess catches up with him

since the both run the same distance and distance = speed*time we have

v1*(t + 1) = v2*t

(1/14)(t + 1) = (1/12)*t

by solving we find

t = 6 min

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

i would be 2,000 just get the square root of 50 which is 2500 and the divid 4/5 and last thing multiply the answer you get for that by the square root of 50

step-by-step explanation: