What is the answer of K

The population of a small rural town in the year 2006 was 2,459. The population can be modeled by the function below, where f(x)

Zepler [3.9K]

f(x)=2,459(0.92)^t

Using the information given above, complete the following statements. The percent change is %. The percent change represents .

f(x)=2,459(0.92)^t

Given equation is an exponential function and in the form of f(x) = a(1-r)^x

Where 'a' is the initial population

'r' is the decay rate

'x' is the time

f(x) = a(1-r)^x

Given function is f(x)=2,459(0.92)^t

1 - r = 0.92

r = 1 - 0.92

r = 0.08

r= 0.08 %

Percentage change = 0.08%

Percentage change represents the decay rate.

8 0

2 years ago

Read 2 more answers

What is the measure of an exterior angle in a regular 19-gon?

Mashutka [201]

The measure is about 19!

7 0

3 years ago

HELP!!!!!!!!!!!!!

Misha Larkins [42]

Answer:

P'(7,0)Q'(2,1) R'(12,-6)

Step-by-step explanation:

P(3,5)------>P'(3+4,5-5)

P'(7,0)

Q(-2,6)----->Q'(-2+4,6-5)

Q'(2,1)

R(8,-1)------>R'(8+4,-1-5)

R'(12,-6)

8 0

3 years ago

What’s 433 rounded to nearest hundreds

MAXImum [283]

Answer:

400

Step-by-step explanation:

Looking at the number 433, we have to look at the number in the tens place (in this case, 3) to determine if we will round down to 400, or up to 500.

Because the number 3 is less than 5, we will round down to 400.

6 0

3 years ago

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Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

2 years ago

What is the answer of K < -5 ?