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3 years ago

in a country in asia, the average hourly wage in dollars from 1945 to 1995 can be modeled by the function below

Mathematics

1 answer:

ddd [48]3 years ago

4 0

Answer:

The answer is "3.03"

Step-by-step explanation:

please find the complete question in the attached file.

In 1970, the hourly wage was determined using the formula:

$\to f(x) = 0.183(x - 1970) + 3.03$

So, the hourly wage in 1970:

$=f(1970)\\\\ = 0.183(1970 -1970) + 3.03\\\\ =0.183(0) + 3.03\\\\= 0+3.03\\\\=3.03$

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3 years ago

A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L

Scorpion4ik [409]

Answer:

1. $\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

2. $y(40) = 110.873 \ kg$

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should consider y(t) to be the amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that y satisfies can be computed as follows:

$\dfrac{dy}{dt}=rate_{in} - rate_{out}$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{y(t)}{100+2t} \times3$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}$

$\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

How much salt is in the tank after 40 minutes?

So,

suppose : $e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}$

Then ,

$( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}$

$( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}$

$[y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}$

Taking the integral on both sides; we have:

$[y(t + 50)^{1.5}] = 1.6 (t + 50)^{2.5} + C$

$y = 1.6 (t+50)+C(t+50)^{-1.5}$

$y(0) = 0 = 1.6(0+50) + C ( 0 + 50)^{-1.5}$

$0 = 1.6(50) + C ( 50)^{-1.5}$

$C= -1.6(50)^{2.5}$

$y(40) = 1.6 (40 + 50)^1 - 1.6 (50)^{2.5}(50+40)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 (90)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 \times 0.001171213948$

$y(40) = 144 - 33.12693299$

$y(40) = 110.873 \ kg$

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