The first thing we are going to do is find the area of the field. To do this we are going to use the area of a square formula:

Were

is the area in square kilometers

is one of the sides of the square
We know for our problem that the side lengths of the field are 0.9 kilometers, so

. Lets replace that value in our formula to find

:

Now, to find the population density of the filed, we are going to use the population density formula:

where

is the population density in <span>in burrows per square kilometer
</span>

is the number of burrows

is the are of the field
We know that

and

, so lets replace those values in our formula:


We can conclude that the <span>density of prairie dog burrows is approximately
2444 burrws per square kilometer.</span>
Answer:
q= -1x
coordinates of p are .5, 2
Step-by-step explanation:
I apologize if I'm wrong love
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
<h3>How to solve a system of equations</h3>
In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:
x - y = 5 (1)
x² · y = 5 · x + 6 (2)
By (1):
y = x + 5
By substituting on (2):
x² · (x + 5) = 5 · x + 6
x³ + 5 · x² - 5 · x - 6 = 0
(x + 5.693) · (x - 1.430) · (x + 0.737) = 0
There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737
And the y-values are found by evaluating on (1):
y = x + 5
x₁ ≈ 5.693
y₁ ≈ 10.693
x₂ ≈ 1.430
y₂ ≈ 6.430
x₃ ≈ - 0.737
y₃ ≈ 4.263
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
To learn more on nonlinear equations: brainly.com/question/20242917
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Answer:
79.6 ft / 1 minute
Step-by-step explanation: