H(n)= -2n^2+4 answer

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

Write the equation of a circle with center (4, -5) and radius = square root of 13

Anon25 [30]

<h2>Hello!</h2>

The answer is:

The equation of the given circle is:

$(x-4)^{2}+(y+5)^{2}=13$

The equation of a circle is given by the following equation:

$(x-h)^{2} +(y-k)^{2}=r^{2}$

We are given the center point (4,-5) and the radius.

So,

Where:

$h=x=4\\k=y=-5\\r=\sqrt{13}$

Then, substituting into the circle equation, we have:

$(x-4)^{2} +(y-(-5))^{2}=(\sqrt{13})^{2}$

Hence, the simplified equation of the circle is:

$(x-4)^{2}+(y+5)^{2}=13$

Have a nice day!

3 0

3 years ago

Read 2 more answers

A.4<br> -3<br> B. 3<br> - 4<br> C. 3<br> 4<br> D. 4<br> 3

Paul [167]

The answer should be A

8 0

3 years ago

Write an equation of a line in slope intercept form (y = mx + b) going through the points

Alexeev081 [22]

Answer:

y = - $\frac{1}{2}$ x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and b the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 2, 0) and (x₂, y₂ ) = (4, - 3)

m = $\frac{-3-0}{4+2}$ = $\frac{-3}{6}$ = - $\frac{1}{2}$ , thus

y = - $\frac{1}{2}$ x + b ← is the partial equation

To find b substitute either of the 2 points into the partial equation

Using (- 2, 0 ), then

0 = 1 + b ⇒ b = 0 - 1 = - 1

y = - $\frac{1}{2}$ x - 1 ← in slope- intercept form

5 0

4 years ago

The chance of having either a boy or a girl is 50%. What is the probability of the Smiths having two girls?

PilotLPTM [1.2K]

Based on the fact that the chance of having a boy or a girl is 50%, the probability of the Smiths having two girls is 25%.

<h3>What is the probability of having two girls?</h3>

When you are faced with two independent events happening one after the other, the way to find their probability is:

= Probability of first event x Probability of second event

The probability of the Smiths getting two girls is therefore:

= 50% x 50%

= 25%

Find out more on independent probability at brainly.com/question/1374659

#SPJ1

5 0

2 years ago