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alexdok [17]
3 years ago
5

The logistic equation for the population​ (in thousands) of a certain species is given by:

Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

\frac{dp}{dt} =3p-2p^2

Divide both sides by 3p-2p^2  and multiply both sides by dt:

\frac{dp}{3p-2p^2}=dt

Integrate both sides:

\int\ \frac{1}{3p-2p^2}  dp =\int\ dt

Evaluate the integrals and simplify:

p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=-1

Now, let's evaluate the limit:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 }  \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5

c. As we did before, let's find the constant C1 for the initial condition given:

p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=1.75

Now, let's evaluate the limit:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 }  \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5

d. To figure out that, we need to do the same procedure as we did before. So,  let's find the constant C1 for the initial condition given:

p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=-\frac{1}{2} =-0.5

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 }  \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5

Therefore, a population of 2000 never will decline to 800.

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Determine the astm grain size number if 33 grains per square inch are measured at a magnification of 270×
strojnjashka [21]
8.9

The equation for the grain size is expressed as the equality:
Nm(M/100)^2 = 2^(n-1)
where
Nm = number of grains per square inch at magnification M.
M = Magnification
n = ASTM grain size number

Let's solve for n, then substitute the known values and calculate.
Nm(M/100)^2 = 2^(n-1)

log(Nm(M/100)^2) = log(2^(n-1))
log(Nm) + 2*log(M/100) = (n-1) * log(2)
(log(Nm) + 2*log(M/100))/log(2) = n-1
(log(Nm) + 2*log(M/100))/log(2) + 1 = n

(log(33) + 2*log(270/100))/log(2) + 1 = n
(1.51851394 + 2*0.431363764)/0.301029996 + 1 = n
(1.51851394 + 0.862727528)/0.301029996 + 1 = n
2.381241468/0.301029996 + 1 = n
7.910312934 + 1 = n
8.910312934 = n

So the ASTM grain size number is 8.9

If you want to calculate the number of grains per square inch, you'd use the
same formula with M equal to 1. So:
Nm(M/100)^2 = 2^(n-1)
Nm(1/100)^2 = 2^(8.9-1)
Nm(1/10000) = 2^7.9
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Nm = 2388564.458

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7 0
3 years ago
SOMEONE PLEASE HELP ME WITH THESE 3 MATH QUESTIONS!!!!!
Len [333]

Look at the pictures (examples)

m∠1 = m∠6 → answer: none

m∠2 = m∠3 → answer: none

m∠2 = m∠C → answer: none



7 0
3 years ago
The point-slope form of the equation of a line that passes through points (8,4) and (0, 2) is y-
Oksi-84 [34.3K]

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

y = mx + b

Where:

m: It's the slope

b: It is the cut-off point with the y axis

While the point-slope equation of a line is given by:

y-y_ {0} = m (x-x_ {0})

Where:

m: It's the slope

(x_ {0}, y_ {0}):It is a point through which the line passes

In this case we have a line through:

(8,4) and (0,2)

Therefore, its slope is:

m = \frac {2-4} {0-8} = \frac {-2} {- 8} = \frac {1} {4}

Its point-slope equation is:

y-4 = \frac {1} {4} (x-8)

Then, we manipulate the expression to find the equation of the slope-intersection form:

y-4 = \frac {1} {4} x- \frac {8} {4}\\y-4 = \frac {1} {4} x-2\\y = \frac {1} {4} x-2 + 4\\y = \frac {1} {4} x + 2

Therefore, the cut-off point with the y-axis is b = 2

ANswer:

y = \frac {1} {4} x + 2

7 0
3 years ago
1 2/3*J=15 What is J?
Wewaii [24]

Answer:

Do you mean:

(1+(2/3))j = 15

((3/3)+(2/3))j = 15

((3+2)/3)j = 15

(5/3)j = 15

j = 15 / (5/3)

j = 15 * (3/5)

j = 9

Step-by-step explanation:

5 0
3 years ago
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