, 02.02.202009389706948

How to solve the rational algebraic expression.

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
Mulitiplying by 5

solution:

let x= any real no.

0.2 = 1/5 (in fraction form)

x/ 0.2 = x/1/5

x/1/5 = x (5/1)

x(5/1) = 5x

medyo magulo sorry
In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, is an algebraic expression.

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