Michael M. answered 10/17/19
Experienced Teacher Very Good at Doing and Teaching Mathematics
3 days = 2 additional days and costs $13 - $5 = $8 extra.
Cost per additional day = $8/2 = $4.
Let d = days and c = charge.
Alternative 1:
cost structure = $5 plus $4 for each additional day
c = 5 + 4(d -1) where d >= 1
check:
d = 1: 5 + 4(1 - 1) = 5
d = 3: 5 = 4(3 - 1) = 13
Alternative 2:
simplify equation:
c = 5 + 4(d -1)
c = 5 + 4d - 4
c = 1 + 4d where d >= 1
cost structure = $1 fee plus $4 per day including first day
check:
d = 1: 1 + 4x1 = 5
d = 3: 1 + 4x3 = 13