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Y_Kistochka [10]
3 years ago
6

A Koch snowflake begins with an equilateral triangle with side length of 1 (stage 0). Trisect each side and attach a smaller equ

ilateral triangle to the center of each side (stage 1). Repeat for each triangle to obtain stage 2. Find the perimeter of the stage 2 snowflake.
Mathematics
1 answer:
nata0808 [166]3 years ago
7 0
Each triangle will have a side length of 1/3.
Since you do this for the 3 sides, each equilateral triangle has a perimeter of 1, so the 3 triangles will have a total perimeter of the stage 2 snowflake = 3
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Its not clearly given that whether EBF = 2x + 9 or 2x - 9.

I have written the solution for both.

If EBF = 2x + 9,

then ABF = 6x + 26 and ABE = 11x - 31.

Now, ABE = ABF + EBF

11x - 31 = (6x + 26) + (2x + 9)

           = (6x + 2x) + (26 + 9)

           = 8x + 35

11x - 8x = 35 + 31

3x = 66

x = 22

Therefore, ABF = 6x + 26 = 6(22) + 26 = 132 + 26 = 158°

If EBF = 2x - 9,

then ABF = 6x + 26 and ABE = 11x - 31.

Now, ABE = ABF + EBF

11x - 31 = (6x + 26) + (2x - 9)

           = (6x + 2x) + (26 - 9)

           = 8x + 17

11x - 8x = 17 + 31

3x = 48

x = 16

Therefore, ABF = 6x + 26 = 6(16) + 26 = 96 + 26 = 122°

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which means the value of the line integral is 3 times the area of the circle, or 27\pi.

b. The closed sphere has no boundary, so by Stokes' theorem the integral is 0.

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Answer:

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