There must have been a fixed cost involved,
cost = a(HCD) + b
case 1:
38.10 = 19(HCD) + b
case 2:
82.65 = 46(HCD) + B
Subtract them:
44.55 = 27HCD
HCD = 1.65
sub back into 38.10 = 19(HCD) + b to find b
now you have the whole equation, plug in HCD = 21
The monthly cost (in dollars) of water use is a linear function of the amount of water used (in hundreds of cubic feet, HCF). The cost for using 19 HCF of water is $38.10, and the cost for using 46 HCF is $82.65. What is the cost for using 21 HCF of water?
3 answers
so would the equation look like 38.10=19(21)+b ?
No
go back and find b
from
38.10 = 19(HCD) + b with HCD = 1.65
Then you will have
c = 19 * (HCD) + whatever b is
go back and find b
from
38.10 = 19(HCD) + b with HCD = 1.65
Then you will have
c = 19 * (HCD) + whatever b is