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Hunter-Best [27]

3 years ago

7

Carlos invested some money 9 years ago in a savings account that earned a

1 answer:

Ray Of Light [21]3 years ago

7 0

The answer is 769.2592592592593

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What is the solution to the following equation?

lara31 [8.8K]

C is the correcr answer

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

Divide by using long<br><br> division.<br><br> (x²-x-6)<br><br> =<br><br> (x-3)

Fudgin [204]

Answer: x + 2

Step-by-step explanation:

Given the equation to solve by long division :

Divide x² - x - 6 by x - 3

x in the divisor is used to divide the x² in the Dividend to obtain a value of x as the quotient. The quotient value is multiplied by each value in the Dividend and so on.

This steps taken obtain the quotient is clearly shown in the picture attached.

5 0

3 years ago

A store purchased a digital camera and marked it up 100% from the original cost of $863.57. A week later, the store placed the d

uysha [10]

Step-by-step explanation:

863.57 ×40/100 = 345.428

................................................

6 0

3 years ago

Read 2 more answers

By how much does -12 exceed -15

kotegsom [21]

Answer:

By 3.

Step-by-step explanation:

-12-3=-15

Hope this helps!

7 0

2 years ago

Read 2 more answers