Probably like 5 ft of carpet
Answer:
E. 800%
Step-by-step explanation:
Since,
u = 8d/h² __________ eqn (1)
Now, density (d) is doubled and usage (h) is halved.
Hence the new life (u'), becomes:
u' = 8(2d)/(0.5h)²
u' = 8(8d/h²)
using eqn (1), we get:
u' = 8u
In percentage,
u' = 800% of u
In words, the percentage increase in useful life of the equipment is <u>800%</u>.
im pretty sure its the fourth choice but its been a while since ive done graphing