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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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Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
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Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

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Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

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x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

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A = 2.525 -2.474+9.798 + 2.525 - 2.474

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The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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2 years ago
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Step-by-step explanation:

Given function is,

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f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

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\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

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