1994 Torneo Internazional Femmin di Palermo – Doubles
Karin Kschwendt and Natalia Medvedeva were the defending champions, but none competed this year.
Doubles | |
---|---|
1994 Torneo Internazional Femmin di Palermo | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 6–1, 6–0 |
Draw | 16 (1WC/1Q) |
Seeds | 4 |
Ruxandra Dragomir and Laura Garrone won the title by defeating qualifiers Alice Canepa and Giulia Casoni 6–1, 6–0 in the final.
Seeds
Barbara Rittner / Patricia Tarabini (First round) Silvia Farina / Irina Spîrlea (First round) Dominique Monami / Petra Ritter (Quarterfinals) Bettina Fulco-Villella / Olga Lugina (Semifinals)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 2 | 3 | ||||||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 2 | 6 | 3 | |||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 2 | 6 | |||||||||||||||||||||
![]() ![]() | 1 | 1 | ![]() ![]() | 3 | 6 | 6 | |||||||||||||||||||||
3 | ![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 4 | 2 | ||||||||||||||||||||
![]() ![]() | 4 | 3 | 3 | ![]() ![]() | 6 | 3 | 4 | ||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 4 | 6 | 6 | |||||||||||||||||||||
![]() ![]() | 4 | 1 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
![]() ![]() | 6 | 77 | Q | ![]() ![]() | 1 | 0 | |||||||||||||||||||||
![]() ![]() | 2 | 64 | ![]() ![]() | 0 | 2 | ||||||||||||||||||||||
WC | ![]() ![]() | 4 | 7 | 2 | 4 | ![]() ![]() | 6 | 6 | |||||||||||||||||||
4 | ![]() ![]() | 6 | 5 | 6 | 4 | ![]() ![]() | 3 | 5 | |||||||||||||||||||
![]() ![]() | 6 | 77 | Q | ![]() ![]() | 6 | 7 | |||||||||||||||||||||
![]() ![]() | 4 | 61 | ![]() ![]() | 4 | 7 | 1 | |||||||||||||||||||||
Q | ![]() ![]() | 77 | 3 | 6 | Q | ![]() ![]() | 6 | 5 | 6 | ||||||||||||||||||
2 | ![]() ![]() | 65 | 6 | 3 |
gollark: x³ + x² + x + 1 = 0 has 3 solutions because it has a degree of 3 for instance.
gollark: It's not "raised to" anything. For this the degree is just the maximum power it contains.
gollark: With reals it's n or less which is bad and nobody likes it.
gollark: It means that for a polynomial P(x) with degree n, P(x) = 0 has exactly n solutions.
gollark: … no.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.