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Pachacha [2.7K]
3 years ago
5

becky visited the great pyramid of Giza she learned that the pyramid was originally 481 feet tall and it’s square base has size

measuring 751 feet in link Becky wants to know the volume of the pyramid what is the volume of the pyramid?
Mathematics
1 answer:
qaws [65]3 years ago
7 0

Answer:

90428160.3 ft³

Step-by-step explanation:

The question is on volume of a pyramid with a square base

The formulae for volume of a pyramid is V=( l×w×h)/3

where l is length of pyramid base, w is width of pyramid base and h is the height of the pyramid

The pyramid has a square base thus l=w= 751 feet

The height of the pyramid is h=481 feet

v= (751×751×481)/3

v=90428160.3 ft³

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Well it's a one out of six chance
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Read 2 more answers
A tank contains 100 L of water. A solution with a salt con- centration of 0.4 kg/L is added at a rate of 5 L/min. The solution i
Fantom [35]

Answer:

a) (dy/dt) = 2 - [3y/(100 + 2t)]

b) The solved differential equation gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time

The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)

The rate of change of the volume of solution = dV/dt

Rate of flow into the tank = Fᵢ = 5 L/min

Rate of flow out of the tank = F = 3 L/min

(dV/dt) = Fᵢ - F

(dV/dt) = (Fᵢ - F)

dV = (Fᵢ - F) dt

∫ dV = ∫ (Fᵢ - F) dt

Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t

V - 100 = (Fᵢ - F)t

V = 100 + (5 - 3)t

V = 100 + (2) t

V = (100 + 2t) L

Component balance for the amount of salt in the tank.

Let the initial amount of salt in the tank be y₀ = 0 kg

Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min

Amount of salt in the tank, at any time = y kg

Concentration of salt in the tank at any time = (y/V) kg/L

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Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)

(dy/dt) = 2 - (3y/V)

(dy/dt) = 2 - [3y/(100 + 2t)]

To solve this differential equation, it is done in the attached image to this question.

The solution of the differential equation is

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration after 20 minutes.

After 20 minutes, volume of water in tank will be

V(t) = 100 + 2t

V(20) = 100 + 2(20) = 140 L

Amount of salt in the tank after 20 minutes gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵

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y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵

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Hope this Helps!!!

8 0
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Answer:

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we know that

To Round a number

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c) But increase it by  1 if the next digit is  5 or more (this is called rounding up)

In this problem we have

0.0825

We want to keep the digit 2

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Answer:

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x = 9000 / 225

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maw [93]
The answer is going to be A.) -2

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