2019 Hardee's Pro Classic – Singles

Taylor Townsend was the defending champion,[1] but lost in the first round to Sesil Karatantcheva.

Singles
2019 Hardee's Pro Classic
Champion Kristína Kučová
Runner-up Lauren Davis
Final score3–6, 7–6(11–9), 6–2
Main article: 2019 Hardee's Pro Classic

Kristína Kučová won the title, defeating Lauren Davis in the final, 3–6, 7–6(11–9), 6–2.

Seeds
  1. Taylor Townsend (First round)
  2. Madison Brengle (First round)
  3. Anhelina Kalinina (Second round)
  4. Astra Sharma (First round)
  5. Allie Kiick (Second round)
  6. Lauren Davis (Final)
  7. Kimberly Birrell (Quarterfinals)
  8. Claire Liu (Second round)

Draw

Key

Finals
Semifinals Final
          
Francesca Di Lorenzo 63 5
Q Kristína Kučová 77 7
Q Kristína Kučová 3 711 6
6 Lauren Davis 6 69 2
Elena-Gabriela Ruse 1 63
6 Lauren Davis 6 77

Top half
First Round Second Round Quarterfinals Semifinals
1 T Townsend 4 2
  S Karatantcheva 6 6 S Karatantcheva 2 63
  D Aiava 2 2 F Di Lorenzo 6 77
  F Di Lorenzo 6 6 F Di Lorenzo 77 6
Q H Chang 4 2 PR L Hradecká 65 2
PR L Hradecká 6 6 PR L Hradecká 6 6
Q K Kawa 6 3 0r 5 A Kiick 3 4
5 A Kiick 4 6 3 F Di Lorenzo 63 5
4 A Sharma 2 6 3 Q K Kučová 77 7
Q K Kučová 6 3 6 Q K Kučová 7 5 6
  E Perez 6 6 E Perez 5 7 2
WC S Chang 1 0 Q K Kučová 77 6
Q A Myers 6 6 7 K Birrell 61 3
  L Cabrera 0 2 Q A Myers 1 5
Q O Govortsova 77 4 3 7 K Birrell 6 7
7 K Birrell 64 6 6

Bottom half
First Round Second Round Quarterfinals Semifinals
8 C Liu 7 6
Q G Talabă 5 2 8 C Liu 6 2 61
  Z Hives 2 6 2 WC UM Arconada 2 6 77
WC UM Arconada 6 4 6 WC UM Arconada 3 2
  E-G Ruse 6 6 E-G Ruse 6 6
  A Muhammad 1 4 E-G Ruse 77 6
WC L Chirico 2 1 3 A Kalinina 62 2
3 A Kalinina 6 6 E-G Ruse 1 63
6 L Davis 6 7 6 L Davis 6 77
  C Dolehide 3 5 6 L Davis 6 6
  R Zarazúa 7 6 R Zarazúa 3 2
WC A Guarachi 5 2 6 L Davis 3
  H Kuwata 3 4 S Zhuk 0r
  D Khazaniuk 6 6 D Khazaniuk 66 77 4
  S Zhuk 64 6 6 S Zhuk 78 63 6
2 M Brengle 77 4 0
gollark: servant-generic:```This package has been merged into servant 0.14.1, please use that instead if available.```
gollark: *magic*
gollark: I think that it'd basically create the following lists:0 1 1 21 1 2 3 (shifted ahead by one)and then sum them to1 2 3 5
gollark: You know, given the existence of the automatic pointfree-izer thing... we need a thing to automatically run this on entire projects.
gollark: ```haskell(....) :: ((a1 -> (a1 -> b) -> c) -> c) -> (((a1 -> b) -> a1 -> (a1 -> b) -> a1 -> (a1 -> b) -> c) -> a1 -> b) -> (a2 -> (a1 -> b) -> a1 -> (a1 -> b) -> a1 -> (a1 -> b) -> c) -> a2 -> c(....) = ((.) .)<$> (*>) (. (.)) (<*>) . ((.) >>= (. ((.) . (.)))) . ((>>) >> (<$>))```

References
  1. "W80 Dothan". www.itftennis.com.
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