Thanks so much
Starts with 5000 then adds 400gal/hr (the Q asks for f(n) to give thanswer in THOUSANDS of gallons)
A= 5+ .400N kgal where kgal is kilo-gallon (1000 gal)
b: Not sure if that is entered correctly...I am unclear on the notation N=f−1(A).
c: f(10) = 5 + .400(10) = 9. kgal
d: Like b above...I do not understand your notation....can you clarify for me?
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a) A = f(n) = 5 + .4 ( n )
b) f-1A = f-1(n)
Write y for f(n)....and we have that
y = 5 + .4(n) subtract 5 from both sides
y - 5 = .400(n) divide both sides by .4
[ y - 5] /. 4= n "swap" y and n
[ n - 5] / .4= y for y, write f-1 (n)
[ n - 5] / .4 = f-1(n)
c) f(10) is the amount of water present in the tank at 6PM =
5 + .4(10) = 5 + 4 = 9 ⇒ 9000 gallons
d) f-1(10) = [ 10 - 5 ] / .4 = 12.5 this is the number of hours after 8AM when the tank will contain 10,000 gallons of water ⇒ 8:30 PM