Im pretty sure it’s A,B,D
Y = 8x + 12
Now, plug in the 60.
60 = 8x + 12
Subtract 12 from both sides.
48 = 8x
Divide both sides by 8.
6 = x
Swap sides.
x = 6
It would take 6 hours for the pool to have 60 inches of water inside it.
<h2>T
he car is about 6.6 years old.</h2>
Step-by-step explanation:
Given : An equation for the depreciation of a car is given by
, where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years. The value of a car is half what it originally cost. The rate of depreciation is 10%.
To find : Approximately how old is the car?
Solution :
The value of a car is half what it originally cost i.e. 
The rate of depreciation is 10% i.e. r=10%=0.1
Substitute in the equation, 


Taking log both side,




Therefore, the car is about 6.6 years old.