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# Help with problem

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287
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Thanks so much

Feb 21, 2018
edited by Guest  Feb 21, 2018
edited by Guest  Feb 21, 2018
edited by Guest  Feb 21, 2018

#1
+18055
+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?

Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018
#2
0

Oh, my mistake, the formatting got messed up. I'll update the main post in sec

Guest Feb 21, 2018
#3
+18055
+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
+2
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Feb 21, 2018
#5
+99586
+1

1)

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

Feb 21, 2018