3/4 plus 1/4
Step-by-step explanation:
Time taken for the amount deposited to grow to $5000 will be found as follows:
A=P(1+r/100*n)^nt
where:
A=amount
P=principle
r=rate
n=number of terms
t=time
thus plugging in our values we have:
5000=3587.39=(1+4.2/200)^2t
solving for t we have:
1.393771=1.021^2t
2t=ln1.393771/ln1.021
2t=16.111
t=8.055685 years~8 years
Answer:

Step-by-step explanation:
Volume of water in the tank = 1000 L
Let y(t) denote the amount of salt in the tank at any time t.
Initially, the tank contains 60 kg of salt, therefore:
y(0)=60 kg
<u />
<u>Rate In</u>
A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.
=(concentration of salt in inflow)(input rate of solution)

<u>Rate Out</u>
The solution is mixed and drains from the tank at the same rate.
Concentration, 
=(concentration of salt in outflow)(output rate of solution)

Therefore, the differential equation for the amount of Salt in the Tank at any time t:

Amphitrite1040Answer:
Amphitrite1040
Step-by-step explanation: