Mérida Open Akron: Saville comes back to beat Kalinina
Daria Saville won against Anhelina Kalinina 4-6, 6-2, 6-2 on Wednesday and will face the winner of the match between Romanian Jaqueline Cristian and Spaniard Paula Badosa, the second seed, in the next round
Image Credit: AI/ Reuters/ Panoramic
Australian qualifier Daria Saville won against Ukrainian Anhelina Kalinina 4-6, 6-2, 6-2 to reach the last eight of the Mérida Open Akron on Wednesday evening.
Saville, ranked No 121, will face the winner of the match between Romanian Jaqueline Cristian and Spaniard Paula Badosa, the second seed, next.
In the previous round, the Australian beat Ukrainian Marta Kostyuk, the No 6 seed (6-4, 7-6 (6)).
Kalinina, ranked No 52, won against French qualifier Leolia Jeanjean (6-4, 1-6, 6-1) during the 32.
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