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Hunter-Best [27]

3 years ago

5

You are riding the bus to school and you realize it is taking longer because of all the stops you are making. The time it takes

to get to school, measured in minutes, is modeled using the function g(x) = x4 − 3x2 + 4x − 5, where x is the number of stops the bus makes. If the bus makes 2 stops after you board, how long does it take you to get to school?

2 answers:

Sedaia [141]3 years ago

8 0

Answer:

7 minutes

Step-by-step explanation:

If x represents the number of stops, and there are 2 stops after you board, solving g(2) will give you the time it takes to get to school allowing time for x = 2 stops:

$g(2)=(2)^4-3(2)^2+4(2)-5$

This gives you, in minutes, that

g(2) = 7

nikitadnepr [17]3 years ago

6 0

Answer:

7 minutes

Step-by-step explanation:

start with formula

g(x) = x^4 - 3x^2 + 4x - 5

substitute x with number of stops (2)

g(2) = 2^4 - 3(2^2) + 4(2) - 5

simplify using p.e.m.d.a.s: start with exponents

g(2) = 16 - 3(4) + 4(2) - 5

multiply

g(2) = 16 - 12 + 8 - 5

subtract/add

16 - 12 = 4

4 + 8 = 12

12 - 5 = <u>7</u>

<u></u>

<u>input: 2</u>

<u>output: 7</u>

<u>ordered pair: (2,7)</u>

<u></u>

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

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

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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:

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Solving for C1:

$C_1=-1$

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

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

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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:

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

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

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.

6 0

3 years ago