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Mamont248 [21]
3 years ago
13

What is connection between crime and unequal access to basic services? Explain at least , four connecting factors.

Advanced Placement (AP)
1 answer:
tatyana61 [14]3 years ago
4 0
Define basic services.

1) first connection between crime & unequal access is lack of recurring income (money)

2) homelessness?

3) lack of or lower education

4) this one may sound strange but feeling of entitlement; "I have to watch out for myself since society isn't/didn't; I deserve it; it's not fair others have it & I don't. "
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Your anger will put you in danger cause your mind will not be calm. You will not be patient and you might take risks of accident in your anger. You might crash with a car or have another accident. You would run your car in high speed or break the traffic signal and get multiple fines . You would be in a high risk
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Primates have been able to inhabit much of the earth due to their ability to
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All have binocular vision with fields of view that significantly overlap, resulting in true three dimensional (3-D) depth perception or stereoscopic vision. ... Depth perception is an invaluable tool for animals that need to move quickly. It allows them to judge the distances to important objects in their environment.

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My fellow math brodas, help
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A. Depending on which variable you choose to integrate with, you can capture the total bounded region with either -2 ≤ x ≤ (-1 + √5)/2 or 1 ≤ y ≤ (5 + √5)/2, where the upper endpoints correspond to the coordinates of the appropriate intersections:

y = x² + 1

⇒   x = (x² - 2)² - 2

⇒   x⁴ - 4x² - x + 2 = 0

⇒   (x - 2) (x + 1) (x² + x - 1) = 0

⇒   x = 2, x = -1, x = -1/2 ± √5/2

⇒   y = 5, y = 2, y = (5 ± √5)/2

On the other hand, we can compute the areas of A and B separately, then sum those integrals. Area A is easier to compute by integrating with respect to y over 2 ≤ y ≤ (5 + √5)/2, while area B is easier to find by integrating x over -1 ≤ x ≤ (-1 + √5)/2.

B. I'll stick to the split-region approach.

First, we find equations for the appropriates halves of either parabola:

• y = x² + 1   ⇒   x = ± √(y - 1)

and x = -√(y - 1) describes the left half of the blue parabola;

• x = (y - 3)² - 2   ⇒   y = 3 ± √(x + 2)

and y = 3 - √(x + 2) describe the bottom half of the red parabola.

Now we can set up the integrals.

Area of A:

\displaystyle \int_2^{(5+\sqrt5)/2} \left(\left(-\sqrt{y-1}\right) - \left((y-3)^2-2\right)\right) \, dy \\ ~~~~~~~~ = -\int_2^{(5+\sqrt5)/2} \left((y-3)^2 - 2 + \sqrt{y-1}\right) \, dy

Area of B:

\displaystyle \int_{-1}^{(-1+\sqrt5)/2} \left(\left(3-\sqrt{x+2}\right) - \left(x^2+1\right) \right) \, dx \\ ~~~~~~~~ = - \int_{-1}^{(-1+\sqrt5)/2} \left(x^2 - 2 + \sqrt{x+2}\right) \, dx

Alternatively, one can prove that the regions A and B are symmetric across the line y = x + 3, so we can simply pick one of these integrals and double it.

C. Computing the integrals, we find

area of A = 2/3

area of B = 2/3

and so the total area is 2/3 + 2/3 = 4/3.

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