What must be added to the last term of the polynomial expression below to make it a perfect square trinomial? a. 4
b. 8
c. 16
d. 64

It's gonna be the click thanks

Given:
vi=80 gal
cin=2 lb/gal
qin=2 gal/min
qout=2 gal/min
(since qin=qout, the volume is constant)
assume xo(original amount)=0
cout=x/v=x/80

dx/dt=cinqin-coutqout

dx/dt=(2)(/80)(2)

dx/dt=2[(160-x)/80]

dx/dt=(160-x)/40 (separable de)

dx/(160-x)=dt/40 (integrate)

-ln(160-x)=(1/40)t+c (general solution)

initial condition: xo=0,t=0
-ln(160-0)=(1/40)(0)+c
c=-ln(160)=-5.0752

-ln(160-x)=(1/40)t-5.0752 (particular solution)

a.) x at any time t x(t)

-ln(160-x)=(1/40)t-5.0752

ln[(160-x)^-1]=(0.025)t-5.0752

e^{ln[(160-x)^-1]}=e^(0.025t-5.0752)

(160-x)^-1=e^(0.025t-5.0752)

1/(160-x)=e^(0.025t-5.0752) (isolate x)

1/[e^(0.025t-5.0752)]=160-x

x(t)=160-{1/[e^(0.025t-5.0752)]}

b. if c=1 lb/gal, t=?

from c=x/v

1=x/80
x=80

-ln(160-x)=0.025t-5.0752

-ln(160-80)=0.025t-5.0752
-ln(80)=0.025t-5.0752

0.025t=5.0752-ln(80)

0.025t=0.6932

t=27.728 mins