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Margaret [11]
3 years ago
7

Sabiendo que lo ancho de un terreno rectangular mide 24.7 m y que su perímetro es de 131.8 m. ¿Cuánto mide cada uno de sus lados

largos?
Mathematics
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

El perímetro de un terreno rectangular mide 48 m . Calcula sus dimensiones si el largo es el doble que el ancho.

Respuesta:

De largo mide 16m

De ancho mide 8m

Explicación paso a paso:

Perímetro = número de lados multiplicado por longitud del lado.

Formula: P = 2(a + b)  o    P = 2b + 2h

Donde: P = Rectángulo del perímetro

           a y b = Longitudes de lados

Asignamos las variables:

Largo = 2x

Ancho = x

Utilizamos la formula:

2(x) + 2(2x) = 48

2x + 4x = 48

6x = 48

x = 48/6

x = 8m  (ancho)

2x

2(8)

16m (largo)

Step-by-step explanation:

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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
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Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

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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

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

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3aV = πr³h

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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))]

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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/∝

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