Suppose that a large mixing tank initially holds 900 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 4 lb/gal, determine a differential equation (in lb/min) for the amount of salt A(t) (in lb) in the tank at time t > 0. (Use A for A(t).)

dA/dt= 12-2A(t)/900+t

Step by step Explanation:

Given:

rate of 3 gal/min

concentration of the solution entering is 4 lb/gal,

The equation of the solution can be expressed as

dA/dt = R(1)- R(2)

Where R(1)=( concentration of the salt at inflow* rate of input of given brine)

R(2)= ( concentration of the salt at inflow* rate of output of given brine)

R(1)= 3 gal/min× 4 lb/gal=12Ib/min

But the solution is been pumped slowly, the rate of the accumulation can calculated as;

(3 gal/min- 2 gal/min)=1gal/min

But after t min we will be having 900+t gallons present in the tank

CHECK THE ATTACHMENT FOR COMPLETE EXPLANATION

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.