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The changing rate of the amount of water in the watering can is a constant rate of 0.15 gallon per minute.
The initial volume of water in the can at time = 0 is 5 gallons
The function will be a linear function in the form of
where
is the volume of water in the can after 
minutes
is the changing rate
is the initial volume when time is 0
Substitute all the values into the form we have
The initial volume of water in the can at time = 0 is 5 gallons
The function will be a linear function in the form of
Substitute all the values into the form we have
Answer:
D. 5
Step-by-step explanation:
From the information given:
The derivatives of functions and functions have a different average rate. The average rate expanded over an interval of two distinct point (a f(a)) and (b f(b)) is represented by the equation:
where; the two points are:
(-1 , f( -1) ) & ( 1, f(1) )
For f(1); we have:
f(1) = 1 +3(1)³ - 5(1)² + 2(1) - 2
f(1) = 1 + 3 - 5 + 2 -2
f(1) = - 1
For f( -1); we have:
(-1)⁴ + 3(-1)³ - 5(-1)²+2(-1)-2
= 1 - 3 - 5 -2 - 2
= -11
∴
= 5