Maya runs 100m in 13.4s and amy can run 100m in 14.1s. if amy were to finish a 100m race at the same time as maya, how much of a head start, in meters, will amy need? what information is known and unknown? and what proportion can you write to solve this problem?

In this case we will use the general equation:

 d = s t (distance = speed*time)

Therefore

Distance Amy can run in 13.4s:

d = (100m/14.1s)*(13.4s)

d = 95.04meters

Headstart needed = total distance - distance run in 13.4s (time Maya runs 100m)

headstart needed = 100-95.04 = 4.96meters

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry