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prohojiy [21]
4 years ago
10

A carmen no le gustan mis zapatos

Advanced Placement (AP)
1 answer:
damaskus [11]4 years ago
5 0
Carmen does not like my shoes
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A presidential candidate emerges from a new minor party that supports radical changes to the national government. This candidate
Zielflug [23.3K]

Answer:

the third one

Explanation:

ones with popular voted get the electoral votes. the fight is always between Democrat or republican because they have the wider audience. third parties have no way of breaking that cycle since they don't have a wide range audience. this ensures that it is a two party system and third parties can't crash it unless they find a way to reach a wider audience

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3 years ago
True or False: If you ride a bicycle in Florida, your bicycle is legally defined as a vehicle.
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True. Bicycles are considered as vehicles in Florida and the bicyclists, drivers
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Students should ve aware of all the fallowing before asking for letters of recommendatchion except
MArishka [77]
D. Which of his or her friends should write a letter
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4 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
AP Computer Science Principles QUESTIONS (MAX POINTS)
marishachu [46]

Answer:

bmkglvmrpfgvrnrgrgjr]pg

regrgnvaer

;fgrgpf

rn

fgr

Explanation:vner;iofgjrfgjrh'rm[o

RjfwGO

NEG

ewg\vn

rfgwr

[gr

gr

7 0
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