Need answer fast please!!!

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

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;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 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, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

A bacterial culture starts with 500 bacteria and doubles in size every half-hour.

tekilochka [14]

Answer:

a). 32000

b). $T_{t}=500\times 4^{t}$

c). 1259

Step-by-step explanation:

Growth of a bacteria is always exponential. Therefore, population of the bacteria is represented by the the geometric sequence.

Sum of the bacterial population after t hours will be represented by

$T_{n}=ar^{n}$

Where a = population at the start

r = ratio with the population is growing

n = time or duration of the growth in one hour

a). Population of 500 bacteria gets doubled after half an hour.

Or gets 4 times after an hour

This sequence will have a common ratio r = 4

and initial population a = 500

Therefore, population of the bacteria after 3 hours will be

$T_{3}=500\times 4^{3}$

$T_{3}=32000$

b). After t hours number of bacteria will be represented by

$T_{t}=500\times 4^{t}$

c). We have to calculate the population after 40 minutes.

That means duration 't' = 40 minutes of $\frac{2}{3}$ hours

By the formula,

$T_{\frac{2}{3}}=500\times 4^{\frac{2}{3}}$

$T_{\frac{2}{3}}=1259.92$ ≈ 1259

Therefore, number of bacteria after 40 minutes will be 1259.

8 0

3 years ago

Read 2 more answers

What is -5/16 as a decimal

Lelechka [254]

Answer:

-0.3125

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A bottle of soda had 2 2/4 of the daily recommended sugar if you were to drink 5/9 of the bottle how much of the daily recommend

Blizzard [7]

Hey friend,

the answer, you divide 4 2/7 by 2.

half of 4 is 2 and half of 2/7 is 1/7

so you get 2 1/7 of the daily recommended sugar.

6 0

3 years ago

To help plan its nursing staff schedule, a large hospital uses simple exponential smoothing to forecast the daily number of hosp

Morgarella [4.7K]

Answer:386

Step-by-step explanation:

We have given

Smoothing parameter $\left ( \alpha \right )=0.56$

Forecasted demand $\left ( F_t\right )=385$

Actual demand $\left ( D_t\right )=386$

And Forecast is given by

$F_{t+1}=\alpha D_t+\left ( 1-\alpha \right )F_t$

$F_{t+1}=0.56\cdot 386+\left ( 1-0.56\right )385=385.56\approx 386$

$F_{t+2}=0.56\cdot 386+\left ( 1-0.56\right )385.56=385.806\approx 386$

$F_{t+3}=0.56\cdot 386+\left ( 1-0.56\right )385.806=385.914\approx 386$

$F_{t+4}=0.56\cdot 386+\left ( 1-0.56\right )385.914=385.962\approx 386$

3 0

3 years ago