Ask question

Ask question

All categories

AleksandrR [38]

3 years ago

14

Does anyone know what S= Please Help

1 answer:

fomenos3 years ago

3 0

$\frac{-s + 7}{4} = -10 \\4(\frac{-s + 7}{4}) = 4(-10) \\-s + 7 = -40 \\\frac{-s}{-1} = \frac{-40}{-1} \\s = 40$

You might be interested in

(3,-8), (-9,9)<br> What is the slope

Mandarinka [93]

The answer will be m= -17/12

8 0

2 years ago

Read 2 more answers

Write an inequality to model the situation.<br> The temperature must be kept below 32 degrees.

Nikitich [7]

Answer:

x<32

Step-by-step explanation:

3 0

3 years ago

Find the circumference of a circle with a diameter 1 1/2 inches

loris [4]

Answer:

C≅17.28

Step-by-step explanation:

Step one:If you divide 11 by 2 it would give you 5.5

Step two:Then use π which is 3.14

Step three:Multiply 5.5*3.14 which gives you 17.27 round to the nearest tenth would be 17.28

Hope it helps!

3 0

2 years ago

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

Plz help I'll give brainliest

Ad libitum [116K]

Answer: The answer is 144 I believe

8 0

2 years ago