In may you used 700 kilowatt-hours of energy for electricity.
a. calculate the total electrical energy used in joules
b. calculate the average power use in watts
c. assume that your power supplier generates electricity by burning oil. note that 1 liter of oil releases 12 million joules of energy. how much oil is needed to generate the electricity you use? give answer in both liters and joules.

i am really having a hard time with this..

1 kilowatt-hours = 3.6x10^6 joules

Converting 900 kilowatt-hours to joules, we have:

900*(3.6*10^6)=3.24*10^9 joules, or

Total Electrical Energy use = 3,240,000,000 joules

"The total electrical energy use was 3,240,000,000 joules."

b)

The month of May has 31 days and 1 day  is 24 hours. So May has:

31*24=744 hours

Now, we divide 900 kW-hr by the number of hours in the month (744 hrs) to get average power use:

kW. Since 1000 Watts = 1 kW, we multiply this by 1000 to get the answer in Watts:

1.21 * 1000 = 1210 Watts

c)

Part 1:

The conversion efficiency of most generating stations is 33%. So we need to multiply the total electrical energy use (in joules) by a factor of 3 to get the amount of joules required. So:

(3.24 * 10^9)*3 = 9.72 * 10^9 joules

We divide this by 12 million joules to get the number of liters:

So, 810 liters is needed

Part 2:

We know that 0.2642 gallons is equal to 1 liters. To get the number of gallons needed, we multiply 810 by 0.2642. So:

810 * 0.2642 = 214.00 gallons

"To generate the electricity you used, 214.00 gal is needed."

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)