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Genrish500 [490]

2 years ago

15

What is cloes to 0.0009999

2 answers:

Nastasia [14]2 years ago

8 0

<span>.0008888 or 1
hope this helps</span>

denis23 [38]2 years ago

6 0

Is this a joke?
Then here:
.0008888 or 1

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The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

a

$y(t) = y_o e^{\beta t}$

b

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

c

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

So

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

So

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

So

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

So

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

so

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

How do you solve the equation:

Molodets [167]

Copy and paste the answer on quizlet

5 0

2 years ago

mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for

frez [133]

mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .

<u>Step-by-step explanation:</u>

Here we have , mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for the same length of time, We need to find how long must he leave it to gain 5600 in interest . Let's find out:

Let mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8% for time x months , So total interest gain is 5600 i.e.

⇒ $\frac{8000(12)}{100}(x) +\frac{2000(8)}{100}(x) =5600$

⇒ $\frac{8000(12)+2000(8)}{100}(x) =5600$

⇒ $(80(12)+20(8))(x) =5600$

⇒ $(960+160)(x) =5600$

⇒ $(1020)(x) =5600$

⇒ $x =\frac{5600}{1020}$

⇒ $x =5.5$

Therefore , mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .

6 0

2 years ago

7 - 5n = 6 + 6(n + 2)

postnew [5]

7- 5n=6+6(n+2)
7- 5n=6+6n+12
7 -5n=18+6n
5n-6n=18-7
-11n=11
n=-1

4 0

2 years ago

Unit 5 lesson 4.3 quiz answers

son4ous [18]

Answer:

There is nothing here to answer

Step-by-step explanation:

6 0

2 years ago