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juin [17]
2 years ago
5

Students are asked to memorize a list of 100 words. The students are given periodic quizzes to see how many words they have memo

rized. The function L gives the number of words memorized at time t. The rate of change of the number of words memorized is proportional to the number of words left to be memorized. Which of the following differential equations could be used to model this situation, where k is a positive constant?
A. dL/dt=kL
B. dL/dt=100−kL
C. dL/dt=k(100−L)
D. dL/dt=kL−100
Mathematics
1 answer:
Yuri [45]2 years ago
6 0

Answer:

C. dL/dt=k(100−L)

Step-by-step explanation:

The rate of change of the number of words memorized is proportional to the number of words left to be memorized.

The rate of change is dL/dt.

The rate is a constant k. Since the rate is proportional to the number of words left to be memorized, and the initial number of words is 100, k has to be multiplied by (100-L).

So the correct answer is:

C. dL/dt=k(100−L)

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In the problem it is already given that Weston Laundry washed 285.38 pounds of towel and 353.47 pounds of sheets from local hotels in 1 day. Firstly we have to find the total pounds of linens that Weston Laundry has washed in 7 days. Then only will it be possible to find the amount of linen washed in a day.
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4 0
2 years ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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Answer:

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