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Juliette [100K]
4 years ago
11

Jill decides to study for her biology final exam by studying for 20 minutes each day for the next three weeks, as opposed to stu

dying all the material at once. Which study technique is she using?
Advanced Placement (AP)
1 answer:
NemiM [27]4 years ago
3 0
Jill is using a technique known as Spaced Repetition. In this technique, the learner studies in short periods with great frequency over a long period of time, frequently testing herself along the way. This aids in long-term retention for the learner. 

The opposite of this technique would be cramming, also known as block learning. Spaced Repetition is widely accepted as the superior study technique. 
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Solve and reduce to lowest terms: 6/11 x 1/3=<br><br> 6/33<br> 7/14<br> 3/11<br> 2/11
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2/11 I think if I’m wrong sorry
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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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