Question

There are 20 parrots at the animal sanctuary. Their population is increasing at a rate of 15% per year. There are also 24 snakes at the sanctuary. Each year 4 more snakes are born.

Part A: Write functions to represent the numbers of parrots and snakes at the sanctuary throughout the years.
Part B: How many parrots are at the sanctuary after 10 years? How many snakes are at the sanctuary after the same number of years? Assume there are no deaths to the animals during this time.
Part C: After approximately how many years is the number of parrots and snakes the same?

Answer 1

\[s _{n}=\frac{ n }{ 2 }\left\{ 2a+\left( n-1 \right)d \right\} \]
for parrots a=20
\[d=\frac{ 15 }{ 100 }\]
for snakes
a=24
d=4

Answer 2

Its so cunfussing

Answer 3

for parrots
\[s _{n}=\frac{ n }{ 2 }\left\{ 2*20+\left( n-1 \right)\frac{ 15 }{ 100 } \right\}\]
\[=\frac{ n }{ 2 }\left\{ 20+\left( n-1 \right)\frac{ 3 }{ 20 } \right\}=\frac{ n }{ 2 }\left\{ \frac{ 400+3n-3 }{ 20 } \right \}=\frac{ n }{ 2 }\left\{ \frac{ 3n+397 }{ 20 } \right\}\]

Answer 4

for C
\[\frac{ n }{ 2 }\left\{ 2*20+\left( n-1 \right)\frac{ 15 }{ 100 } \right\}=\frac{ n }{ 2 }\left\{ 2*24+\left( n-1 \right)4 \right\}\]
\[40+\left( n-1 \right)\frac{ 3 }{ 20 }=48+\left( n-1 \right)4\]
find n

Answer 5

so the answer for 1 is 20?

Answer 6

how do i foind n

Answer 7

for B n=10

Answer 8

Let x be the year.
x = 0 is the starting year when there are 20 parrots and 24 snakes.

Let function \(p(x)\) represent the parrot population.
Let function \(s(x)\) represent the snake population.
You notice that both functions are functions of x, the year.

Answer 9

In year 0, the beginning point, there are 20 parrots.
Each year, there are 15% more.
Let x = 0 when there are 20 parrots.

Start:
\(p(0) = 20\) parrots

At year 1, there is a 15% increase in the parrot population, so you have
\(p(1) = 20 \times 1.15\) parrots

At year 2, there are
\(p(2) = (20 \times 1.15) \times 1.15 = 20 \times (1.15 \times 1.15) = 20 \times 1.15 ^2\) parrots

At year 3, there are:
\(p(3) = 20 \times 1.15^2 \times 1.15 = 20 \times 1.15^3\) parrots

The above suggests that for any year x, the parrot population is:
\(f(x) = 20(1.15)^x\)

Answer 10

In year 0, there are 24 snakes.
Each year, there are 4 more snakes.
Let x = 0 when there are 24 snakes.

Start:
\(s(0) = 24\)

At year 1, there is an increase of 4 in the snake population, so you have
\(s(1) = 24 + 4\) snakes

At year 2, there are
\(s(2) = 24 + 4 + 4 = 24 + 4 \times 2\) snakes

At year 3, there are:
\(s(3) = 24 + 4 \times 2 + 4 = 24 + 4 \times 3\) snakes

The above suggests that for any year x, the snake population is:
\(s(x) = 24 + 4x\)

Answer 11

Now you have two functions, p(x) and s(x) that tell you what the parrot and snake populations are for any year starting at year 0.

Answer 12

That was part A.
For part B, find (p(10) and s(10).
For part C, set the functions equal and solve for x, the year.

Answer 13

ty

Answer 14

@surjithayer
The growth of the parrot population is exponential growth, like compound interest.
It grows 15% from the last year total.
p(0) = 20
p(1) = 20(1.15)^1 = 23
p(2) = 20(1.15)^2 = 26.45
p(3) = 20(1.15)^3 = 30.42
p(4) = 20(1.15)^4 = 34.98
As you can see the growth in the first year is only 3, but from year 3 to year 4 the growth is approximately 4.5

With the snake population, the population starts at 24 at year 0. It increases by 4 each year.
There is a year after year 0 in which the parrot population will catch up to the snake population.

To find that year, we equate the two functions and solve for x, the year.

\(20(1.15)^x = 24 + 4x\)

This is not an easily solved equation by algebra, so we can use a table of values to find when the parrot population reaches the snake population.

x p(x) s(x)
0 20 24
1 23 28
2 26.4 32
3 30.4 36
4 35.0 40
5 40.2 44
6 46.3 48
7 53.2 52

You can see that the parrot population is already larger by the 7th year.
By doing a linear interpolation between years 6 and 7, you get an answer of approximately 6.5 years for the two populations to be equal.

Answer 15

you are correct.

Answer 16

Thanks.