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3 years ago

8

Can anybody help plzzzz I’m making brainliest IF correct!!!!

2 answers:

Paul [167]3 years ago

7 0

Answer: 12xy + 32

Step-by-step explanation: IM SO SORRY IF IM WRONG SKJKAJD i did this really quickly so you may want to wait for another answer but im pretty sure im right!!

iren [92.7K]3 years ago

7 0

12xy + 32

Sorry if I'm stu pid.

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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

4 years ago

Please help me simplify 9x-3(2+4x)=

denis23 [38]

solution

1) <span>Expand
</span><span>9x−6−12x

2) </span><span>Gather like terms
</span><span>(9x−12x)−6

3) </span> <span>Simplify
</span><span><span>−3x−6</span></span>

4 0

4 years ago

Read 2 more answers

What is it called when you have two burgers but one of them is half?

antiseptic1488 [7]

1 1/2? I pretty sure this is the answer.

4 0

3 years ago

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Hi ik no one would have any hope in my friend basically he got super sick .. and he said nothing would make him feel happier if

nalin [4]

Answer:

Give your friend our best and hope they get better.

Step-by-step explanation:

3 0

3 years ago

What is greater one seventh or one nineth

Anton [14]

Method 1:

Look at the picture.

$\dfrac{1}{7}$ is greater than $\dfrac{1}{9}$

$\boxed{\dfrac{1}{7}>\dfrac{1}{9}}$

Method 2:

Find the common denominator:

multiples of 7: 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...

multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ...

63 = 7 · 9

$\dfrac{1}{7}=\dfrac{1\cdot9}{7\cdot9}=\dfrac{9}{63}\\\\\dfrac{1}{9}=\dfrac{1\cdot7}{9\cdot7}=\dfrac{7}{63}$

$\dfrac{9}{63} > \dfrac{7}{63}$

therefore

$\dfrac{1}{7} > \dfrac{1}{9}$

3 0

3 years ago

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