Solve using proper methods. Show work. (25 POINTS)Initially a tank contains 10,000 liters of liquid at the time t = 0 minutes a tap is opened, a liquid then follows out of the tank. The volume of the liquid V liters, which remains in the tank after t minutes is given by V = 10,000(0.933)^ta) Find the value of V after 5 minutes.b) Find how long, to the nearest second, it takes for half of the initial amount of liquid to follow out of the tank.c) The tank is regarded as effectively empty when 95% of the liquid has flowed out. Show that it takes almost three quarters of an hour for this to happen.d) (i) Find the value of 10,000 - V when t = 0.001 minutes(ii) Hence or otherwise, estimate the initial flow rate of the liquid. Give your answer in liters per minute, correct to two significant figures.

Answer:a) 7069.82 Litersb) 600 secondsc) Shown belowd) (i) 0.6935 liters   (ii)  Since 0.6935 liters in 0.001 minute, so 693.5 liters per minute is as estimate (in liters per minute)Step-by-step explanation:a)We simply put 5 into t of the equation and get the value of V. So:[tex]V=10,000(0.933)^t\\V=10,000(0.933)^5\\V=7069.82[/tex]So after 5 minutes the amount remaining is 7069.82 Litersb)half of the initial amount is half of 10,000 which is 5000. So we substitute 5000 into V and solve for t using logarithms.Note: [tex]ln(a^b)=blna[/tex]Thus, we have:[tex]V=10,000(0.933)^t\\5000=10,000(0.933)^t\\0.5=(0.933)^t\\ln(0.5)=ln((0.933)^t)\\ln(0.5)=tln(0.933)\\t=\frac{ln(0.5)}{ln(0.933)}\\t=9.99[/tex]Thus, t = 9.9949 minutes.To get answer in seconds, we multiply by 60. Thus 9.9949*60= 600 secondsc)95% empty means 5% remaining. 5% of 10,000 = 0.05 * 10,000 = 500. We plug in 500 into V and solve for t as the previous step. Shown below:[tex]V=10,000(0.933)^t\\500=10,000(0.933)^t\\0.05=0.933^t\\ln(0.05)=ln(0.933^t)\\ln(0.05)=tln(0.933)\\t=\frac{ln(0.05)}{ln(0.933)}\\t=43.1972[/tex]So it takes around 43.1972 minutes to empty 95%. Since three-quarters of an hour is  [tex](\frac{3}{4})(60)=45[/tex] minutes, we have shown that the time it takes (43.1972 minutes) is very close to three-quarters of an hour.d)We plug in 0.001 into t and find V. Then we subtract that value from 10,000. This is just finding how much water has been removed in 0.001 minutes. Let's do this:[tex]V=10,000(0.933)^t\\V=10,000(0.933)^{0.001}\\V=9999.3065\\Now\\10,000 - 9999.3065 = 0.6935[/tex] So, 0.6935 liters