I need help solving this

One more then im done, ty ^^

OleMash [197]

Answer:

5

Step-by-step explanation:

5 0

2 years ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

4 years ago

A dog walker charges a flat rate of $6 per walk plus an hourly rate of $30. How much does the dog walker charge for a 45 minute

zvonat [6]

Step 1

Find the equation in function notation

Let

h-------> the number of hours

y-------> the function for the walker fee in dollars

we know that

the hourly rate is $30\frac{\$}{hour}$

$y=6+30h$

in this linear equation

the independent variable is the variable h

the dependent variable is the variable y

Convert to function notation

Let

$f(h)=y$

$f(h)=6+30h$

Step 2

Find how much does the dog walker charge for a 45 minute walk

Convert the time in hours

$1\ hour=60\ minutes$

$45\ minutes=45/60=0.75\ hours$

substitute in the equation

For $h=0.75\ hours$

$f(0.75)=6+30*(0.75)=\$28.5$

therefore

the answer is

the dog walker charge for a 45 minute walk $\$28.5$

Statements

Step 3

The dependent variable is the number of hours true or false

The statement is false,

because the independent variable is the number of hours and the dependent variable is the walker fee in dollars

Step 4

The function for the walker fee is f(h)= 30h+6 true or false

The statement is True --------> see the Step 1

Step 5

The dog walker charges $22.5 for a 45 min walk true or false

The statement is False

Because the dog walker charges for a 45 minute walk $\$28.5$

5 0

3 years ago

the ratio of girls to boys in the sixth grade is 5:3. If there are 400 sixth graders, how many boys are there?

PilotLPTM [1.2K]

Answer:

boys = 250

girls = 150

Step-by-step explanation:

5g = 3b eq. 1

g + b = 400 eq. 2

g = girls

b = boys

From the eq. 2

g = 400 - b

Replacing this last eq. on eq. 1:

5(400-b) = 3b

5*400 + 5*-b = 3b

2000 - 5b = 3b

2000 = 3b + 5b

2000 = 8b

2000/8 = b

250 = b

From eq. 2

g + 250 = 400

g = 400 - 250

g = 150

Check:

from eq. 1

5*150 = 3*250 = 750

3 0

2 years ago

Camille had a very fun birthday party with lots of friends and family attending. The party lasted for 3 hours. She and her frien

Elena-2011 [213]

Camille spent 22 minutes and 30 seconds opening presents.

Since Camille had a very fun birthday party with lots of friends and family attending, and the party lasted for 3 hours, and she and her friends played games for 3/8 of the time, ate pizza and cake for 50% of the time , and spent the remainder of the time opening presents, to determine the amount of time spent opening presents the following calculation must be performed, posing the following linear equation:

Total time - time spent on other activities = time spent opening presents

3 - (3 x 3/8) - (3 x 0.5) = X
3 - 1.125 - 1.5 = X
3 - 2.625 = X
0.375 = X

1 = 60
0.375 = X
0.375 x 60 = X
22.5 = X

Therefore, Camille spent 22 minutes and 30 seconds opening presents.

Learn more in brainly.com/question/11897796

7 0

3 years ago

I need help solving this ​

I need help solving this