Help me, please!!!

Question

Data were collected on the number of kilowatt-hour (kWh) of electricity customers used each month. The data were fit into a function. The function f(x) = 650 sin(0.6x - 2.8) + 962 gives the number of kWh of electricity used during month x. Month 1 corresponds to January. The electric company charges $0.0715 per kWh up to 500kWh. Then the company charges S0.1042 per kWh used above 500. The company also charges $8.00 in fees. How many kWh would a customer use in April, and what would the total cost be?

kWh used in April =

Total cost for April =

Answer 1

f(4) = 650sin(24-2.8)+962 = 1424.255

kWh used in April = 1424.255

If over 500kWh then the cost is:

C(x) = 500•0.0715 + (f(x) - 500)•0.1042 + 8

C(4) = 35.75 + (924.255)(0.1042) + 8 = 140.05

Total cost for April = $140.05

Answer 2

So January corresponds to month 1, which means April corresponds to month 4.

The function f(x) takes in a month x and spits out the kWh, so plug in x = 4 and find f(4) to get the number of kWh in April.

Next we find the total cost. Let's say that the number of kWh in April is less than 500, for example 200kWh.

In this case, the cost would just be $0.0715 x 200kWh.

But now let's say that the number of kWh in April is greater than 500, for example 1200kWh.

In this case for the first 500kWh the cost becomes $0.0715 x 500kWh, and for the remaining 700kWh the cost is $0.1042 x 700kWh. Add the two costs to get the total. So $0.0715 x 500kWh + $0.1042 x 700kWh