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wolverine [178]
3 years ago
10

19. The teaching of standards and

Advanced Placement (AP)
1 answer:
babymother [125]3 years ago
8 0
I think the answer you’re looking for is
C.
- I hope this helps, I got it right. Enjoy the rest of your day/night, please make me brainliest!
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Differences between the U.S/Canada border and U.S/Mexico border. Why do you think those borders are controlled so differently? D
Alexxx [7]

Answer:

Since I cannot see the data,  am trying my hardest. Check below for comments.

  • The Canadian border is 3 times larger, at around 5500 miles long.
  • The Southern border is more discussed. Possibly for immigration reasons. We do not have a whole lot of Canadians wanting to come here.
  • The Mexico border is more talked about in terms of guarding.
  • The Mexico bored is "highly defended"

Explanation:

5 0
3 years ago
How many vowels are there in what you think you become<br>​
Wewaii [24]

Answer:

there are 9 vowels in the sentence

what you think you become

hope this helped you

please mark as the brainliest (ㆁωㆁ)

6 0
3 years ago
Read 2 more answers
Select two skills from the list in the article that you feel you have to offer in the workplace.
saw5 [17]

Answer:

I need the list please

Explanation:

5 0
2 years ago
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

8 0
3 years ago
A person displays a set of rare behaviors that
ElenaW [278]
A case study
I did that already
4 0
3 years ago
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