**Thanks so much**

Guest Feb 21, 2018

edited by
Guest
Feb 21, 2018

edited by Guest Feb 21, 2018

edited by Guest Feb 21, 2018

edited by Guest Feb 21, 2018

edited by Guest Feb 21, 2018

#1**+1 **

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?

ElectricPavlov
Feb 21, 2018

#3**+1 **

Is it ? A= 5 +.4N A-5 = .4N N=(A-5)/.4 hours

I think that is what is meant....then d would be

N= (10-5)/.4 hours

N = 12.5 hours to reach 10 kgals.

ElectricPavlov
Feb 21, 2018

#4**0 **

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Gloriawoods
Feb 21, 2018

#5**+1 **

1)

a) A = f(n) = 5 + .4 ( n )

b) f^{-1}A = 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

CPhill
Feb 21, 2018