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aksik [14]
3 years ago
14

SIMPLE QUESTION (multiple choice) Justin is looking for a recreational facility to join so he can stay active after school in th

e early evenings. Which of the following details will help him stay safe and healthy?
A) Good lighting and maintenance
B) Friendly people and staff
C) Inexpensive membership fees
D) Remote location with trees
Advanced Placement (AP)
2 answers:
dmitriy555 [2]3 years ago
4 0
B. friendly people and staff<span />
Ket [755]3 years ago
3 0

Answer:

A) Good lighting and maintenance

Explanation:

As the question is talking about the detail that will help Justin stay safe and healthy, from the options given the one that adjust to that is good lighting and maintenance as good lighting will allow him to be able to see anything that may cause an accident like a stone in his way and maintenance will provide a facility that is in excellent conditions which can prevent an accident and allow him to perform his activities to be active.

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Answer:

A

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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

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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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