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Angelina_Jolie [31]
2 years ago
13

The park service that administers a state park estimates that there are 495 deer in the park. They decide to remove deer accordi

ng to the differential equation dP/dt = −0.1P.
a. Show that the solution to the differential equation dP/dt = −0.1P is P = 495e^−0.1t , where t is measured in years and P is the population of deer. Use it to find the deer population in 5 years to the nearest deer.

b. After this 5-year period, no human intervention is taken and the deer population grows again. From that time, the deer population increases directly proportional to 650−P, where the constant of proportionality is k. Find an equation for the deer population P(t) in terms of t and k for this 5-year period.

c. Using the growth model from part b), 1 year later the deer population is 350. Find k.

d. Using the growth model from part b) and the value of k from part c), find lim t→∞ P(t)
Mathematics
1 answer:
Vlada [557]2 years ago
6 0

Answer:

your answer is (A.)

Step-by-step explanation:

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The coordinate grid shows points A through K. What point is a solution to the system of inequalities? y &lt; −2x + 10 y &lt; 1 o
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The graph that represents the system of inequalities will be y ≤ −2x + 3; y ≤ x + 3

<h3>How to depict the graph?</h3>

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Enter the equation of the circle described below.<br><br> Center (3,-2), radius = 5
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3 0
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Read 2 more answers
Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a
SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

Area_{A}=0.25*Area_{B} (Equation\ 1)\\

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}

put this value into equation 1

Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\  Area_{B}=80\ cm^{2}

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)}  \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm

Now, put the value of length(2B) into equation 3 to know length (1B)

length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm

I really hope this helps you, have a great day.

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