A tank contains 7070 kg of salt and 20002000 L of water. Pure water enters a tank at the rate 88 L/min. The solution is mixed and drains from the tank at the rate 44 L/min. (a) What is the amount of salt in the tank initially

I'm assuming all the numbers here mistakenly got copied twice.

Let A(t) denote the amount (in kg) of salt in the tank at time t. The solution starts with 70 kg of salt, so (a) A(0) = 70.

I also assume there is more to the question, so I'll just go ahead and build the differential equation to model this situation, and solve it.

Pure water is flowing into the tank, so no salt is being introduced. The concentration of salt at time t is A(t)/2000 kg/L, and solution is drained from the tank at a rate of 4 L/min, which means salt is being removed at a rate of

(4 L/min) (A(t)/2000 kg/L) = A(t)/500 kg/min = 0.002 A(t) kg/min

Thus the amount of salt (ignoring units now) in the tank changes according to the DE,

dA(t)/dt = -0.002 A(t)

which is easy to solve because it's linear with constant coefficients.

Without going into too much detail:

dA(t)/dt + 0.002 A(t) = 0

exp(0.002t ) dA(t)/dt + 0.002 exp(0.002t ) A(t) = 0

d/dt [exp(0.002t ) A(t)] = 0

exp(0.002t ) A(t) = C

A(t) = C exp(-0.002t )

Given A(0) = 70, solve for C :

70 = C exp(0) = C

Hence

A(t) = 70 exp(-0.002t )

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation: