Question

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Answer 1

The answer would be: 
d. 0, $12.40, $24.79, $37.19, $49.58; The practical range of this function is 0≤f(g) ≤49.58

The number of gallons needed is the value of g variable, so you need to put the variable value into the equation. The calculation would be:
f(g)= 2.479g
f(0)= 2.479(0)=0
f(5)= 2.479(5)=12.395= 12.40
f(10)= 2.479(10)=24.79
f(15)= 2.479(15)=37.185= 37.19
f(20)= 2.479(20)=49.58

The range of the function would be more than(≥) 0 but less than(≤) 49.58. If you put it together, the sign would look like 0≤f(g)≤49.58