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aleksandr82 [10.1K]
2 years ago
7

Which of the following situations would a local government address?

Advanced Placement (AP)
2 answers:
pickupchik [31]2 years ago
5 0

Answer:

D. Local government deals with what goes on within a given city. The other types of government would be national.

Tatiana [17]2 years ago
4 0

Answer:

direct street lights - D

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Sidana [21]

Help me too

Explanation:

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How did tea change history in India?
Lina20 [59]
It was, and still is, a staple import of the united kingdom, which at one time had possession of the 13 colonies of America, India and Hong Kong. As such, these three countries maintain to this day a strong trade network with the UK and her former colonies.
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3 years ago
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How many solutions exist for the given equation? 3x + 13 = 3(x + 6) + 1
Deffense [45]

Answer:

There is no solutions to this problem.

Because when you solve it you get an untrue statement. When this happens, it means that there is nothing that can make it true.

Explanation:

3x + 13 = 3(x + 6) + 1  use distribute property

3x + 13 = 3x + 18 + 1 combine like terms

3x + 13 = 3x + 19 subtract 3x from both sides

13 = 19

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2 years ago
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Laurie is trying to stay within 10 feet of her current diving depth of –30 feet (with regard to sea level) so that the light is
Hitman42 [59]

Answer: x + 30 = 10 and x + 30 = –10

Explanation: The two equations can be used to find the minimum and maximum depths Laurie wants to stay between is the below:

x + 30 = 10 and x + 30 = -10

Hope this helps!

8 0
3 years ago
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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

8 0
3 years ago
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