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2 years ago

12

Cents 140 3 Minutes 3 1 1

2 answers:

coldgirl [10]2 years ago

5 0

Answer:

25

Step-by-step explanation:

PLS GIVE BRAINLIEST

Lerok [7]2 years ago

4 0

Answer:

25

Step-by-step explanation:

PLEASE GIVE ME BRAINLEIST

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n/5+4=5 what is n? nwfowjvksncknknssnpxjvspdojsovjdsjsdbosdjoisdjosdjosdjgsodgjsodgjsdogjsdogjsdgosdjgosdjgosd

yaroslaw [1]

N = 45

Hope this helps!!!

7 0

3 years ago

Read 2 more answers

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

ivanzaharov [21]

Answer: 18w^2 - 6w + 4

Explanation:

(4 + 7w^2) - (6w - 11w^2)
= 4 + 7w^2 - 6w + 11w^2
= 18w^2 - 6w + 4

5 0

3 years ago

How does the function g(x)= - (x - 1) compare to the parent function f(x) = x ?

grandymaker [24]

Reflection across x axis and move 1 unit to the right

3 0

3 years ago

Four times a number

BlackZzzverrR [31]

Four times a number would equal 4n

6 0

3 years ago

Read 2 more answers