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Firibgarlarga uchmang! Telemetrio bunday kanallarni topadi va belgilaydi 👉 Belgini ko‘rmoqchi bo‘lsangiz, obuna bo‘ling

AddMath Spmnetic!™⚡️

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

📈 Telegram kanali AddMath Spmnetic!™⚡️ analitikasi

AddMath Spmnetic!™⚡️ (@addmathspmnotes) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 38 041 obunachidan iborat bo'lib, Taʼlim toifasida 4 957-o'rinni va Hindiston mintaqasida 10 234-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 38 041 obunachiga ega bo‘ldi.

26 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 1 022 ga, so‘nggi 24 soatda esa 31 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 12.29% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 6.25% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 4 675 marta ko‘riladi; birinchi sutkada odatda 2 376 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 27 Oktabr, 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.

38 041
Obunachilar
+3124 soatlar
+1707 kun
+1 02230 kun
Postlar arxiv
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Day 02 of studying addmaths everyday

Try this guys. This should be easy
Try this guys. This should be easy

Here is the solution.
+1
Here is the solution.

It is given that the diameter of the circle is divided into 4 equal parts, find the area of the shaded region if the radius o
It is given that the diameter of the circle is divided into 4 equal parts, find the area of the shaded region if the radius of the circle = 4.

Recently I saw a few videos talking about ambiguous case of the law of sine for solutions of triangle. Question: When you are learning this topic, do you find it difficult to understand it? And why?
Anonymous voting

Repost from Sembang Maths
Percayalah cakap saya, mula lah sekarang. Jangan tunggu saat akhir. Pesanan yang sama dari saya, sebab saya dah tengok pelbag
Percayalah cakap saya, mula lah sekarang. Jangan tunggu saat akhir. Pesanan yang sama dari saya, sebab saya dah tengok pelbagai situasi pelajar saat2 last nak spm

wahh

day 1 of studying addmaths everyday...🔥🔥

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Let x=length of the rectangle h = height of the rectangle Black line = h/tan(60°) = (h√3)/3 Using Pythagoras' Theorem, [(h√3)/3 + x]² + h² = 36 h²/3 + (2xh√3)/3 + x² + h² - 36 = 0 h²(4/3) + [(2x√3)/3]h + (x²-36) = 0 h = [-(2x√3)/3 + √[(4x²/3)-(16/3)(x²-36)]]/(8/3) h = [-x√3 + 3√(48-x²)]/4 Area of the rectangle, A = xh = x([-x√3 + 3√(48-x²)]/4) dA/dx = (72-3x²-x√(144-3x²))/(2√(48-x²)) At Maximum, dA/dx = 0 (72-3x²-x√(144-3x²))/(2√(48-x²)) = 0 72-3x²-x√(144-3x²) = 0 => x=-6 or x=2√3 Max A = 6√3

Kepada batch 09 juga, good luck for your upcoming exams, especially spm 2026. Be ready with every subject, conquer the chapter, jangan belajar last minit, dengar cakap cikgu dan jangan skip sekolah/kelas tambahan

Kepada batch 08 yang still ada dalam channel ni, mintak semua untuk bawa bertenang dan ada perasaan redha dengan keputusan spm anda. Saya doa dan pasti anda semua calon spm 2025 akan dapat straight A!! Amin🤲❤️

JANJANG/PROGRESSIONS •gp C5F4
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JANJANG/PROGRESSIONS •gp C5F4

JANJANG/PROGRESSIONS •gp(geometric progression) C5F4
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JANJANG/PROGRESSIONS •gp(geometric progression) C5F4

JANJANG/PROGRESSIONS •gp (geometric progression) C5F4
+1
JANJANG/PROGRESSIONS •gp (geometric progression) C5F4

btw JEE advanced is one of the most toughest exams for undergraduate students..