2. El cociente de 15 y un número.

The population of a community, p(x), is modeled by this exponential function, where x represents the number of years since the p

NeX [460]

An exponential model is used to represent growth of a population

The population after 3 years is 2585

The function is given as:

$\mathbf{P(x) = 2400(1.025)^x}$

3 years after, means that x = 3

So, we have:

$\mathbf{P(3) = 2400(1.025)^3}$

Evaluate exponents

$\mathbf{P(3) = 2400 \times 1.076890625}$

$\mathbf{P(3) = 2584.5375}$

Approximate

$\mathbf{P(3) = 2585}$

Hence, the population after 3 years is 2585

Read more about population growth at:

brainly.com/question/3160736

5 0

3 years ago

Amd the third is the last one i wont ask again

Musya8 [376]

Answer:

9.3

Step-by-step explanation:

2/.214

3 0

2 years ago

Simplify completely the quantity x squared plus x minus 12 over quantity x squared minus x minus 20 divided by the quantity 3 x

posledela

X² + x - 12 / x² - x - 20 ÷ 3x² - 24x + 45 / 12x² - 48x - 60

x² + x - 12 / x² - x - 20 * 12x² - 48x - 60 / 3x² - 24x + 45

(x² + x - 12)(12x² - 48x - 60)
(x² - x - 20)(3x² - 24x + 45)

12x^4 - 48x³ - 60x² + 12x³ - 48x² - 60x - 144x² + 576x + 720
3x^4 - 24x³ + 45x² - 3x³ + 24x² - 45x - 60x² + 480x - 900

12x^4 - 48x³ + 12x³ - 60x² - 48x² - 144x² - 60x + 576x + 720
3x^4 - 24x³ - 3x³ + 45x² + 24x² - 60x² - 45x + 480x - 900

12x^4 - 36x³ - 252x² + 516x + 720
3x^4 - 27x³ + 9x² + 435x - 900

12(x^4 - 3x³ - 21x² + 43x + 60) 
3(x^4 - 9x³ + 3x² + 145x + 300)

4(x^4 - 3x³ - 21x² + 43x + 60) 
 (x^4 - 9x³ + 3x² + 145x + 300)

4 0

3 years ago

Jonathan parents told him that for every 5 hours of homework or reading he completed

Murrr4er [49]

Finish your statement... I'm anxious to help!

4 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago