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Raportează o problemă de traducere
+rep nice teammate
+rep fullfocus
+rep absolute
+rep 300 iq
+𝚛𝚎𝚙 𝚋𝚎𝚜𝚝 𝚙𝚕𝚊𝚢𝚎𝚛 🏆
+rep tryhard
+rep best player in the world
+rep psychopath
𝙵𝚊𝚟𝚘𝚞𝚛𝚒𝚝𝚎 𝚏𝚘𝚛 𝚏𝚊𝚟𝚘𝚞𝚛𝚒𝚝𝚎.
𝙲𝚘𝚖𝚖𝚎𝚗𝚝 𝚏𝚘𝚛 𝙲𝚘𝚖𝚖𝚎𝚗𝚝 .
𝙰𝚛𝚝𝚠𝚘𝚛𝚔
Comment for comment ❤️❤️
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Rate 4 Rate
素敵な週末を過ごしましょう
扩列
Only if you show me your boobs
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ᴄᴏᴍᴍᴇɴᴛ ꜰᴏʀ ᴄᴏᴍᴍᴇɴᴛ 扩列+
add me free
⠀⠀⠀⢀⣤⣶⠾⠿⠛⠛⠛⠛⠛⠿⠿⣶⣤⣀
⠀⣠⡾⠋⠁⠀⠀⠀⢀⣠⣤⠤⢤⣤⣄⠀⠈⠙⢿⣦
⣸⡏⠀⠀⠀⣀⣤⠾⠋⠁⠀⠀⠀⣸⠟⠀⠀⠀⠀⠹⣷
⣿⣇⠀⠀⠀⠙⠳⢦⣄⡀⠀⠀⠈⢳⣦⠀⠀⠀⠀⣰⣿
⢹⣿⣶⣤⣀⠀⠀⠀⠈⠙⠳⠶⠶⠿⠋⠀⢀⣤⣾⣿⣿⣶⣶⣦⡀
⠘⣿⣿⣿⣿⣿⣶⣶⣦⣤⣤⣤⣤⣶⣶⣿⣿⣿⣿⣿⠋⠉⠙⣿⣧
⠀⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣏⠀⠀⢠⣿⡿
⠀⠀⠻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⣿⣶⣶⣿⡿
⠀⠀⠀⠘⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠋⠀⠈⠉⠉⠁
⠀⠀⠀⠀⠀⠈⠛⠻⢿⣿⣿⣿⡿⠿⠛⠁
ᴄᴏᴍᴍᴇɴᴛ ꜰᴏʀ ᴄᴏᴍᴍᴇɴᴛ 扩列+
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀ Ya ho~
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀i'm an active user⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Can we be friends if you don't mind?⠀
⠀
⠄⡜⠸⢰⡐⠄⠄⠄⠄⠄⣇
⠄⣯⡏⣘⣎⣂⣵⢀⢾⡄⡼
⠄⠏⣎⠟⣻⣿⢻⠃⢈⡝
⠄⠄⠹⠋⢉⣵⣮⣰⡚
⠄⠄⠄⠄⠸⣿⣿⡏⣷⢹⣦
⠄⠄⠄⢀⡄⣿⣿⡇⣾⡏⣻⡄
⠄⠄⢴⣿⣿⢹⣿⡇⣿⣧⢿⣇
⠄⠸⣸⣿⣿⢸⣿⡇⣿⣿⣟⢿⣦⣀
⠄⠄⠈⠛⠛⠈⣿⣷⢻⡿⢟⣣⣭⣭⣝⡲⢶⣶⣤⣄⡀
⠄⠄⠄⠄⠄⠸⣿⢟⣤⣾⣿⣿⣿⣿⣿⣿⣷⡹⣿⣿⣿⣷⣄
⠄⠄⠄⠄⠄⢀⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⢻⣿⣿⣿⣿⣆
⠄⠄⠄⢀⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠱⡜⣿⣿⣿⣿⡿⣾⣷⠄
⠄⣠⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢛⣵⠇⡇⣿⣿⣿⢟⣵⢸⣿⡇
⣼⣿⣭⣶⣶⣶⣶⣝⡻⣿⣿⡿⠿⡛⠁⠄⠁+REP ⠄⠄⣵⣿⣿⠟
⠹⣿⣿⣿⣿⣿⣿⣿⣿⣶⣶⣴⡸⣿⣧⣀⡤⣤⠄⠄⠄⠄⠄⢷⢰⠞ add me
ᴄᴏᴍᴍᴇɴᴛ ꜰᴏʀ ᴄᴏᴍᴍᴇɴᴛ 扩列+
Here's the link to the guide with the animation: https://gtm.you1.cn/sharedfiles/filedetails/?id=3250146188
If you like it, feel free to drop some comments and likes, and share your thoughts with me. I always appreciate your feedback, praise, and if you're feeling extra generous, rewards are always welcome 😊.
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀在直角坐标系 xOy中
曲线C1的方程为y=k|x|+2.以坐标原点为极点, X 轴正半轴为极轴建立极坐标系,曲线C2的极坐标方程为p+2p cos Ꮎ -3 = 0
(1)求 C2的直角坐标方程
(2)若C1与C2有且仅有三个公共点,求C1的方程
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⢀⣤⣶⠾⠿⠛⠛⠛⠛⠛⠿⠿⣶⣤⣀
⠀⣠⡾⠋⠁⠀⠀⠀⢀⣠⣤⠤⢤⣤⣄⠀⠈⠙⢿⣦
⣸⡏⠀⠀⠀⣀⣤⠾⠋⠁⠀⠀⠀⣸⠟⠀⠀⠀⠀⠹⣷
⣿⣇⠀⠀⠀⠙⠳⢦⣄⡀⠀⠀⠈⢳⣦⠀⠀⠀⠀⣰⣿
⢹⣿⣶⣤⣀⠀⠀⠀⠈⠙⠳⠶⠶⠿⠋⠀⢀⣤⣾⣿⣿⣶⣶⣦⡀
⠘⣿⣿⣿⣿⣿⣶⣶⣦⣤⣤⣤⣤⣶⣶⣿⣿⣿⣿⣿⠋⠉⠙⣿⣧
⠀⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣏⠀⠀⢠⣿⡿
⠀⠀⠻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⣿⣶⣶⣿⡿
⠀⠀⠀⠘⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠋⠀⠈⠉⠉⠁
⠀⠀⠀⠀⠀⠈⠛⠻⢿⣿⣿⣿⡿⠿⠛⠁
🌈 Passer une très belle semaine, les ami(e)s gamers ! 🌈
Pls like and maybe give award (if you want to ofc). THANK YOU !!!!
https://gtm.you1.cn/profiles/76561199062620898/recommended/1172380/
🌈 Greetings, gamers ! Happy sunny weekend, play and dream big ! 🌈
🖤Feel free to add me🖤
🖤Accepting everyone🖤
🖤Comment for Comment🖤