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ad-work [718]
3 years ago
12

Is the Mandelbrot set locally connected?

Mathematics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

Step-by-step explanation:

It is conjectured that the Mandelbrot set is locally connected. This famous conjecture is known as MLC (for Mandelbrot locally connected). By the work of Adrien Douady and John H. Hubbard, this conjecture would result in a simple abstract "pinched disk" model of the Mandelbrot set. In particular, it would imply the important hyperbolicity conjecture mentioned above.

The work of Jean-Christophe Yoccoz established local connectivity of the Mandelbrot set at all finitely renormalizable parameters; that is, roughly speaking those contained only in finitely many small Mandelbrot copies.[19] Since then, local connectivity has been proved at many other points of {\displaystyle M}M, but the full conjecture is still open.

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Lucy borrowed $425 from Colleen. Colleen charges Lucy 5.3% interest if Lucy pays Colleen back in 3 years how much does she pay C
Sonja [21]

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7 0
2 years ago
Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.
galben [10]

Answer:

∠EFD ≅ ∠EGD ⇒ A

Arc ED ≅ arc FD ⇒ C

m arc FD = 120° ⇒ E

Step-by-step explanation:

Let us revise some facts

  1. Equal chords subtended equal arcs
  2. The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
  3. The measure of a central angle is equal to the measure of its subtended arc
  4. If one angle of an isosceles triangle measure 60° then the triangle is equilateral
  5. The sum of the measures of the interior angles of any quadrilateral is 360°

In the quadrilateral  CDGE

∵ m∠G = 60°

∵ m∠GDC = m∠GEC = 90°

- By using the 5th rule above

∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°

∴ 60 + 90 + m∠DCE + 90 = 360

∴ 240 + m∠DCE = 360

- Subtract 240 from both sides

∵ m∠DCE = 120°

In circle C

∵ ∠DCE is a central angle subtended by arc DE

∵ ∠DFE is an inscribed angle subtended by arc DE

- By using the 2nd rule above

∴ m∠DFE = \frac{1}{2} m∠∠DCE

∵ m∠DCE = 120°

∴ m∠DFE = \frac{1}{2} (120)

∴ m∠DFE = 60°

- That means ∠EFD ≅ ∠EGD because their measure is 60°

∴ ∠EFD ≅ ∠EGD

In Δ EFD

∵ EF = FD

∵ m∠DFE = 60°

- By using the 4th rule above

∴ Δ EFD is an equilateral triangle

∴ ED = FD = FE

In circle C

∵ Side ED subtended by arc ED

∵ Side FD subtended by FD

∵ Side ED ≅ side FD ⇒ proved

- By using the 1st rule above

∴ Arc ED ≅ arc FD

∵ m∠ECD = 120°

∵ ∠ECD is a central angle subtended by arc ED

- By using the 3rd rule above

∴ m∠ECD = m arc ED

∴ m of arc ED = 120°

∵ Arc ED ≅ arc FD

∴ m arc ED = m arc FD

∴ m arc FD = 120°

5 0
2 years ago
Read 2 more answers
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