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4 years ago

Help on horizontal asymptote!

Mathematics

2 answers:

user100 [1]4 years ago

7 0

Answer:

A. None

Step-by-step explanation:

They are no horizontal asymptotes. I get helped from another website when comes to graphs and looking for asymptotes. Good luck on your exam!

Westkost [7]4 years ago

5 0

Answer: A. none

<u>Step-by-step explanation:</u>

There are three rules when finding the horizontal asymptote (H.A.)

Consider "m" as the degree of the numerator and "n" as the degree of the denominator.

Rules:

m > n: no H. A (use long division to find the slant asymptote)
m = n: H.A. is y = leading coefficient of m ÷ leading coefficient of n
m < n: H.A. is y = 0

In the equation given, the numerator is x⁵ + ... and the denominator is x⁴ + ...

m = 5, n = 4

m > n so there is no H.A

<em>but you can use long division to find the Slant Asymptote.</em>

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Step-by-step explanation:

According to the given table, the variables have a linear relation, because there's a constant change involved.

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Step-by-step explanation:

For this case we have the following info:

$P(0) = 660$ represent the initial amount of bacteria

$P(6) = 3960$ represent the population of bacteria after 6 hours

Part a

For this case we use the proportional model given by:

$\frac{dP}{dt}= kP$

Where P represent the population at time t and k is a constant of proportionality.

We can rewrite the model like this:

$\frac{dP}{P} = k dt$

And if we integrate both sides we got:

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$P = e^{kt +c}= e^{kt} e^C$

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$660 = P_o e^{0k}, P_o = 660$

We have the following model:

$P(t) = 660 e^{kt}$

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$3960 = 660e^{6k}$

We can divide both sides by 660 and we got:

$6= e^{6k}$

Now we can apply natural log on both sides and we got:

$ln (6)= 6k$

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Part b

For this case we just need to replace t=3 into the model and we got:

$P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617$

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