H(n) = n^2+ 4n <br> g(n)= 2n -3 <br> Find h(g(-2))

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

Calculate 6 P 6 note n p r equals n Over N - r

Murrr4er [49]

Answer:

7

Step-by-step explanation:

n! Might be 7, and n might be 6. r might also be 6. So 6 - 6= 0! = 1

So I think it might be 7 ÷ 1, which is 7.

7 0

3 years ago

If the can of dogfood holds 8 ounces and the bag of dogfood holds 30 times the volume , how many pounds of dogfood does the bag

Maru [420]

Answer:

15 pound in a bag

Step-by-step explanation:

one can= 8 oz

one bag= 30 times the amount in a can

8 oz x 30 = 240 oz in a bag

Question asked how many pound is in one bag of dog food:

We know that 16 oz = 1 pound

240 oz = x pound

To find x pound we divide 240 oz by 16 oz-->

240 oz/ 16 oz = 15 pound

7 0

4 years ago

Working conditions in the United States in the early 1900's can best be described as?

Dmitry_Shevchenko [17]

Working conditions in the United States in the early 1900’s can best be described as atrocious.

6 0

3 years ago

Read 2 more answers

Please help me !!!!!!

cestrela7 [59]

Given:

The set of pair and graphs.

To find:

The domain and range.

Solution:

We know that,

Domain is the set of x-values or input values.

Range is the set of y-values or output values.

(a)

The given set of ordered pairs is

{(-3,3),(5,5),(-3,2),(5,3)}

Here, the x-coordinates are -3, 5, -3, 5.

A set contains distinct values.

Therefore, the domain is {-3,5}.

(b)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,1),(-1,0.5),(-1,3),(0,0),(0,2),(1,0.5),(1,3)(2,1)}

Here, the y-values are 1, 0.5, 3, 0, 2, 0.5, 3, 1.

Therefore, the range is {0, 0.5, 1, 2, 3}.

(c)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,3),(-1,3),(0,1),(2,4)}

Here, the x-values are -2, -1,0, 2.

Therefore, the domain is {-2,-1,0,2}.

4 0

3 years ago

H(n) = n^2+ 4n g(n)= 2n -3 Find h(g(-2))