A university spent $1.8 million to install solar panels atop a parking garage. These panels will have a capacity of 400 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 20%, that electricity can be purchased at $0.10 per kilowatt-hour (kWh), and that the marginal cost of electricity production using the solar panels is zero.

9241.08 Hrs

Explanation:

Calculation to determine Approximately how many hours per year will the solar panels need to operate to enable this project to break

Let n represent hour of operation per year for breakeven

Breakeven annual cost = Annual savings

$1,800,000 * (A/P, 20%,20) = 400 * n * 0.1

$1,800,000 * 0.205357 = 400 * n * 0.1

n = 369,643/ 40

n= 9241.075

n=9241.08 Hrs (Approximately)

Therefore Approximately how many hours per year will the solar panels need to operate to enable this project to break will be 9241.08 Hrs

start by isolating the variable on one side by adding 43w^2 to both sides.   the equation will read:

25w^2+5=0

*subtract 5 from both sides*

25w^2=-5

*divide both sides by 25*

w^2=-1/5

*take the square root of both sides*

w=√(-1/5)

the square root of -1 is i.   the square root of 5 remains √5

the answer is i/√5.

hope this

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