Work out (7×10^5) ÷ (2×10^2)
1 answer:
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Answer:
The differential equation for the amount of salt A(t) in the tank at a time t > 0 is .
Step-by-step explanation:
We are given that a large mixing tank initially holds 500 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.
The concentration of the solution entering is 4 lb/gal.
Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;
where, = concentration of salt in the inflow
input rate of brine solution
and = concentration of salt in the outflow
outflow rate of brine solution
So, = 4 lb/gal
3 gal/min = 12 lb/gal
Now, the rate of accumulation = Rate of input of solution - Rate of output of solution
= 3 gal/min - 2 gal/min
= 1 gal/min.
It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.
So, = concentration of salt in the outflow
outflow rate of brine solution
= =
Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;
=
or .
Answer:
[See Below]
Step-by-step explanation:
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✦ Solve for :
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✦ Solve Parenthesis first:
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✦ Add all the values together:
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So Mikayla wrote the decimal correctly.
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