AddMath Spmnetic!™⚡️
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This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻 For any enquiries, please directly ask in our discussion group ✨
Ko'proq ko'rsatish
Tarmoq:📈 Telegram kanali AddMath Spmnetic!™⚡️ analitikasi
AddMath Spmnetic!™⚡️ (@addmathspmnotes) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 36 598 obunachidan iborat bo'lib, Taʼlim toifasida 5 096-o'rinni va Hindiston mintaqasida 10 645-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 36 598 obunachiga ega bo‘ldi.
15 Sentabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 1 128 ga, so‘nggi 24 soatda esa 20 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 10.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 6.02% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 3 890 marta ko‘riladi; birinchi sutkada odatda 2 205 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 11 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent addmath, untuk, 629/4, math, eqn kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻
For any enquiries, please directly ask in our discussion group ✨”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 16 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
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| Sana | Obunachilarni jalb qilish | Esdaliklar | Kanallar | |
| 15 Sentabr | +21 | |||
| 14 Sentabr | +22 | |||
| 13 Sentabr | +40 | |||
| 12 Sentabr | +50 | |||
| 11 Sentabr | +33 | |||
| 10 Sentabr | +25 | |||
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| 08 Sentabr | +40 | |||
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| 05 Sentabr | +37 | |||
| 04 Sentabr | +72 | |||
| 03 Sentabr | +42 | |||
| 02 Sentabr | +62 | |||
| 01 Sentabr | +67 |
Kanal postlari
🇲🇾🇲🇾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
| 2 | Countdown SPM:
68 days left (9 minggu 5 hari)
Biar usaha mencecah langit, supaya result nanti boleh mencecah awan - by dee | 903 |
| 3 | 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 | 2 201 |
| 4 | Identify the mathematical flaw | 1 753 |
| 5 |  Matn yo'q... | 1 751 |
| 6 | 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 | 1 795 |
| 7 | Guys do anyone have mrsm addmx paper with skema? | 1 592 |
| 8 | Countdown SPM:
69 days left (9 minggu 6 hari)
Either study hard for a better future or do nothing and ruin your future. | 1 378 |
| 9 | solution by tonyXWX08 | 2 130 |
| 10 | 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. | 2 097 |
| 11 | idk when , but maybe one day I will share how to recheck answer | 1 776 |
| 12 | 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. | 1 812 |
| 13 | 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? | 2 270 |
| 14 |  2026 kedah paper 1 | 1 972 |
| 15 |  2026 Johor Batu Pahat Paper 2 | 1 895 |
| 16 |  2026 melaka paper 2 | 1 598 |
| 17 |  2026 melaka paper 1 | 1 395 |
| 18 | 1st step in this chapter is to know how to convert the equation , this exercise would be helpful. | 1 345 |
| 19 |  +1form 4 chapter 6 question | 1 416 |
| 20 | https://www.youtube.com/watch?v=_90pLf3ma5s | 1 618 |
