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Genrish500 [490]
2 years ago
10

I do not need an explanation, just a simple angle number!

Mathematics
1 answer:
Murljashka [212]2 years ago
3 0

Answer:

Step-by-step explanation:

a + 143 = 180    {Linear pair}

a = 180 - 143

a = 37

37 + 93 + b = 180 {Angle sum property of triangle}

130 + b = 180

         b = 180 - 130

         b = 40

c + b = 90

c  + 40 = 90

c = 90 - 40

c = 50

d + 120 = 180   {Linear pair}

d = 180 - 120

d = 60

? = 60 + 40         {Exterior angle property of triangle}

? = 100

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Answer: 1.6\pi

Step-by-step explanation:

If a circle has a radius of 6, then the perimeter of the circle is 12pi, or about 37.68.  A circle has 360 degrees.  Thus, the arc takes up 48/360 of the circle.  Thus, simply multiply \frac{48}{360} * 12\pi to get 5.024 or 1.6pi.

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Calculus Problem
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and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

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We end up with the same integral as before except for the leading constant:

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c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

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\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
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