Z+6/3=2z/4 Please solve for z❤️

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

Carly has 15 stuffed animals. How many ways are there to organize her top 3 favorites?

Ivanshal [37]

Well, the answer will depend on whether the order will count or not (based on Permutations and Combinations). If the order counts, then we would use the formula for Permutations, which is:
$\frac{n!}{(n-r)!}$
Where n is the number of items you have, and r is the number of times you choose from the items.
$\frac{15!}{(15-3)!}$
Which simplifies to
$\frac{15!}{(12)!}$
Which simplifies to 15*14*13 (because all the numbers 1-12 in the factorial canceled out), which gets us the answer 2730.

Now, if you wanted to find the number of ways to order the toys without replacement (order doesn't count), you would use the formula:
$\frac{n!}{r!(n-r)!}$
The variables are still the same, but you are now multiplying by r!.
$\frac{15!}{3!(15-3)!}$
Simplifies to
$\frac{15!}{3!(12)!}$
Which simplifies to (using the same cancellation method above)
$\frac{2730}{3!}$
Dividing 2730 by 3! will get us an answer of 455.

Really, it depends on whether they are ordered or not. In this case (since you didn't specify whether the order mattered), the answer would be 455 or 2730.

:)

4 0

3 years ago

To join a gym, you have to pay an initial fee of $50 and $30 per month. Write an equation representing total cost (t).

Vladimir [108]

50+30m= t

m: number of months
t: total

6 0

4 years ago

Read 2 more answers

A farmer owns about 960 acres of land. If his farm is square, what would be the length of each side in miles, to the nearest ten

OLEGan [10]

It is given in the question that

A farmer owns about 960 acres of land. If his farm is square.