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erastovalidia [21]
3 years ago
14

.

Mathematics
2 answers:
kondor19780726 [428]3 years ago
8 0
ख.दुध
चिडिया
There is no underlined word in question no.3
Tomtit [17]3 years ago
5 0

Answer:

I dont know how to read hindi can u pls translate it into English

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Geometry Help? Plz I Beg Of You!! *pic inside*
Natalija [7]
1a) if all the angles are perpendicular, they all =90°
7 0
2 years ago
Ryder Industries is considering a project that will produce cash inflows of $92,000 a year for five years. What is the internal
RSB [31]

Answer:

20.02%

Step-by-step explanation:

Formula : NVP = 0 =-P_0 + \frac{P_1}{(1+IRR)} + \frac{P_2}{(1+IRR)^2} + . . . +\frac{P_n}{(1+IRR)^n}

P_0 = 275000

n = 1,2,3,4,5

Substitute the values in the formula :

0 =-275000 + \frac{92000}{(1+IRR)} + \frac{92000}{(1+IRR)^2} + \frac{92000}{(1+IRR)^3}+\frac{92000}{(1+IRR)^4}+\frac{92000}{(1+IRR)^5}

275000 = \frac{92000}{(1+IRR)} + \frac{92000}{(1+IRR)^2} + \frac{92000}{(1+IRR)^3}+\frac{92000}{(1+IRR)^4}+\frac{92000}{(1+IRR)^5}

Solving for IRR using calculator

IRR = 20.02

Hence the internal rate of return if the initial cost of the project is $275,000 is 20.02%

4 0
3 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
3 years ago
What is the formal for surface area for a rectangular prism​
tester [92]

Answer I guess:

A = Length times width

8 0
3 years ago
Read 2 more answers
I need help I don't get it new concepts
koban [17]
C IS THE RIGHT thing
4 0
3 years ago
Read 2 more answers
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