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Korolek [52]
3 years ago
11

HELP!!!!

Advanced Placement (AP)
2 answers:
Tatiana [17]3 years ago
7 0

Answer:

true

Explanation:

i think im sorry if i wrong pls vote me brainlest

vredina [299]3 years ago
3 0
True because I just took it
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I need help, can you try to help with all the questions because they are too hard for me
maw [93]

Explanation:

1. Employed as she was working before the holidays.

2.No, as people are considered employed if they did any work at all for pay or profit during the survey reference week.

3.Yes, as she won't be getting any pay or profit.

4.No, as to be counted as 'unemployed' you will have to be looking for work, or currently available to work. But, she is working as a substitute.

5.Yes, because she will have to look for a job.

6. I'm not sure about that one sorry

7.Yes, because he won't be getting pay or profit.

8.He will be considered in the labour force. As he won't have a job or not looking for one.

9.Focusing on school, have no time. etc.

10.That question is for you.

6 0
3 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
How did the California gold rush best represent the new american identity​
harina [27]

Answer:

The gold rush was 1949

Explanation:

3 0
3 years ago
What is 1000000000000000000 divided by 17 +27045
VLD [36.1K]

Answer:

5.88235294E16

Explanation:

7 0
3 years ago
PLZ HELP MEH!!!!!!!!!!!!!!!!!!
Murljashka [212]

What do you need help with?

5 0
4 years ago
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