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yanalaym [24]
3 years ago
11

Describe ONE difference between the level of industrialization typically seen in more developed countries and the level seen in

less developed countries.
Advanced Placement (AP)
2 answers:
RideAnS [48]3 years ago
8 0

Industrialized countries have a greater capacity to produce goods and services than other countries.

Industrialization is a term that refers to the production of goods and services. Industrialization is related to the development of new technologies and machines to optimize production processes.

According to the above, more developed countries have a greater capacity to produce goods and services, such as the United States, Russia, Japan, and China, because much of its economy is based on the production of services and goods. On the other hand, less developed countries (also known as developing or underdeveloped countries) do not yet have a high capacity to produce goods and services. On the contrary, their economies are based on other fields such as agriculture or resource extraction.

Learn more in: brainly.com/question/7721379

PSYCHO15rus [73]3 years ago
6 0

I'll use the countries.. America and Pakistan, Americas industries is more advancely and more largely developed then Pakistan's industries

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My fellow math brodas, help
DiKsa [7]

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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