The growth of a city is described by the population function p(t) = P0e^kt where P0 is the initial population of the city, t is the time in years, and k is a constant. If the population of the city atis 19,000 and the population of the city atis 23,000, which is the nearest approximation to the population of the city at

Answer:

27,800

Step-by-step explanation:

We need to obtain the initial population(P0) and constant value (k)

Population function : p(t) = P0e^kt

At t = 0, population = 19,000

19,000 = P0e^(k*0)

19,000 = P0 * e^0

19000 = P0 * 1

19000 = P0

Hence, initial population = 19,000

At t = 3; population = 23,000

23,000 = 19000e^(k*3)

23000 = 19000 * e^3k

e^3k = 23000/ 19000

e^3k = 1.2105263

Take the ln

3k = ln(1.2105263)

k = 0.1910552 / 3

k = 0.0636850

At t = 6

p(t) = P0e^kt

p(6) = 19000 * e^(0.0636850 * 6)

P(6) = 19000 * e^0.3821104

P(6) = 19000 * 1.4653739

P(6) = 27842.104

27,800 ( nearest whole number)

3 0

2 years ago

The population of two different villages is modeled by the equations shown

vlada-n [284]

Answer:

Year = 1995 and population = 315

Year = 2015 and population = 715

Step-by-step explanation:

It is given that the population of two different villages is modeled by the given equations:

$y=x^2-30x+540$

$y=20x+15$

The population of both villages are same after x years after 1980 if

$x^2-30x+540=20x+15$

$x^2-30x+540-20x-15=0$

$x^2-50x+525=0$

Splitting the middle term, we get

$x^2-15x-35x+525=0$

$x(x-15)-35(x-15)=0$

$(x-15)(x-35)=0$

$x=15,35$

It means, after 15 or 35 years, the population will same.

For x=15, years is $1980+15=1995$ and population is

$y=20(15)+15=315$

For x=35, years is $1980+35=2015$ and population is

$y=20(35)+15=715$ .

Therefore, population are equation in Year = 1995 and population = 315 or Year = 2015 and population = 715.

7 0

3 years ago

Need help badly don’t have answer to this problem.

NikAS [45]

There are a couple of ways to tackle this one, using the 45-45-90 rule or just using the pythagorean theore, let's use the pythagorean theorem.

the angle at A is 45°, and its opposite side is BC, the angle at C is 45° as well, and its opposite side is AB, well, the angles are the same, thus BC = AB.

hmmm le'ts call hmmm ohh hmmm say z, thus BC = AB = z.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2 \implies (6\sqrt{2})^2=z^2+z^2
\qquad 
\begin{cases}
c=\stackrel{6\sqrt{2}}{hypotenuse}\\
a=\stackrel{z}{adjacent}\\
b=\stackrel{z}{opposite}\\
\end{cases}\\\\\\ 6^2(\sqrt{2})^2=2z^2\implies 36(2)=2z^2\implies \cfrac{36(2)}{2}=z^2\implies 36=z^2
\\\\\\
\sqrt{36}=z\implies 6=z$

4 0

3 years ago

Read 2 more answers

Find the equation of the line that is perpendicular to the given line and passes through the given point.

liberstina [14]

Y= -1/5x-3
Y= -1/5x0-3
Y=-3

4 0

2 years ago