Prep Club for GRE
  • Close

Kudos owned by the user


GreenlightTestPrep received 3 Kudos for post Re: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=.

GiverPostDate
SivhHarishRe: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=16-Jul-2022
CarcassRe: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=14-Jul-2022
YashZRe: If jkjk, jk0jk0, and j2x=k2y=jkj2x=k2y=jk, then x=14-Jul-2022

Page 1 of 1

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact