AddMath Spmnetic!™⚡️
前往频道在 Telegram
This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻 For any enquiries, please directly ask in our discussion group ✨
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网络:📈 Telegram 频道 AddMath Spmnetic!™⚡️ 的分析概览
频道 AddMath Spmnetic!™⚡️ (@addmathspmnotes) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 36 598 名订阅者，在 教育 类别中位列第 5 096，并在 印度 地区排名第 10 645 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 36 598 名订阅者。
根据 15 九月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 1 128，过去 24 小时变化为 20，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 10.63%。内容发布后 24 小时内通常能获得 6.02% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 3 890 次浏览，首日通常累积 2 205 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 11。
- 主题关注点： 内容集中在 addmath, untuk, 629/4, math, eqn 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻
For any enquiries, please directly ask in our discussion group ✨”
凭借高频更新（最新数据采集于 16 九月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
36 598
订阅者
+2024 小时
+2137 天
+1 12830 天
帖子存档
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🇲🇾🇲🇾Malaysia Day 16/9 Special Class❤️❤️
We will discuss actual SPM 2025 MATHS Paper 2 Section B& Section C together. Feel free to join the class and share the link around to your friends tonight. Sharing is caring🫶
meet.google.com/zzt-ygrr-ikr
📅Time: 8PM-10PM
Questions discussed: *8PM-10PM*
‼️‼️Limited to first 100 people only
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Repost from Thespmnetic™ 🕵🏻🕵🏻♀
Countdown SPM:
68 days left (9 minggu 5 hari)
Biar usaha mencecah langit, supaya result nanti boleh mencecah awan - by dee
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SOALAN PERCUBAAN MATEMATIK TAMBAHAN (2023-2026)
DILENGKAPI DENGAN LANGKAH PENYELESAIAN & PENJELASAN
TINGKATAN 4 BAB 1 FUNGSI:
https://addmaths-atlas.pages.dev/guide/f4-functions/
TINGKATAN 4 BAB 2 FUNGSI QUADRATIK
https://addmaths-atlas.pages.dev/guide/f4-quadratic-functions
TINGKATAN 5 BAB 1 SUKATAN MEMBULAT
https://addmaths-atlas.pages.dev/guide/f5-circular-measure
TINGKATAN 5 BAB 2 PEMBEZAAN
https://addmaths-atlas.pages.dev/guide/f5-differentiation
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KERTAS PERCUBAAN MATEMATIK TAMABHAN 2026
DILENGKAPI DENGAN LANGKAH PENYELESAIAN & PENJELASAN
JOHOR BAHRU
K1: https://addmaths-atlas.pages.dev/trials/johor-bahru/paper-1.html?lang=bm
K2: https://addmaths-atlas.pages.dev/trials/johor-bahru/paper-2.html?lang=bm
JOHOR BATU PAHAT
K1: https://addmaths-atlas.pages.dev/trials/johor-batu-pahat/paper-1.html?lang=bm
K2: https://addmaths-atlas.pages.dev/trials/johor-batu-pahat/paper-2.html?lang=bm
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Repost from Thespmnetic™ 🕵🏻🕵🏻♀
Countdown SPM:
69 days left (9 minggu 6 hari)
Either study hard for a better future or do nothing and ruin your future.
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Since f(x) = ax² + bx + c = 0 for all real numbers x, by letting x = 0, we get f(0) = a(0²) + b(0) + c = 0, which further implies that c = 0.
This directly simplifies f(x) to:
f(x) = ax² + bx = 0.
By letting x = 1, f(1) = a(1²) + b(1) = 0 implies that a + b = 0.
By letting x = -1, f(-1) = a(-1)² + b(-1) = 0 implies that a - b = 0.
Adding the two equations gives 2a = 0, which implies that a = 0.
Substitute a = 0 into either equation yields b = 0.
So it's proven that if f(x) = ax² + bx + c = 0 for all real number x, then a = b = c = 0.
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how to really know ur potential and improve?
1. time yourself 2 hours *PAPER 1) or 2 hours 30 minutes (PAPER 2), stop writting when timer ends.
2. mark using answer scheme, K0 , N0 most of the times, dont simply put N1 just bcz ur answer correct.
3. Know why K0
4. know why and where P1 is given
5. differentiate between the steps that can be skipped and cannot be skipped
6. Know how to recheck ur answer and ur working.
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Suppose we are given f(x) = ax^2 + bx + c. It is easy to see that if a = b = c = 0, then f(x) = 0 for all (real) numbers x.
The challenge is to prove the other direction. Can you prove that if f(x) = 0 for all (real) numbers x, then a = b = c = 0?
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1st step in this chapter is to know how to convert the equation , this exercise would be helpful.
