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enot [183]
3 years ago
14

Thayer sometimes finds it difficult to understand what his instructor is saying. He raises his hand in class and asks his instru

ctor to elaborate on what was just said, but it rarely helps. What should Thayer do to overcome this obstacle?
A.
during class, ask other students what the instructor means
B.
raise more questions in class
C.
join a study group of his peers
D.
try to do the best he can with the information he has
Advanced Placement (AP)
2 answers:
Ksju [112]3 years ago
8 0

Option A would be disruptive. Option B may would likely not help and may also be disruptive. Option D would likely not work because he can't get through not understanding. The answer would be Option C.

Sholpan [36]3 years ago
6 0

The correct answer is C. Join a study group of his peers

Explanation:

During classes in some cases, students find difficulties in the process of understanding content. The first step to address this is to directly ask the teacher about it during class; however, if this does not help the student, he or she will need to try other methods to solve it. This includes trying to get a personal class with the same teacher or another teacher, try to learn about the topic of the class by yourself or looking for help in peers by joining or creating study groups.

According to this, the best option for Thayer is to join a study group of his peers because through a study group he can clarify concepts by discussing, and revising the content of the class. Also, his peers can explain to him aspects he does not understand, which might help him to understand better.

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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

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