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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 ✨

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Tarmoq: Sejarah spmnetic!™Hindiston9 910Taʼlim4 922...

📈 Telegram kanali AddMath Spmnetic!™⚡️ analitikasi

AddMath Spmnetic!™⚡️ (@addmathspmnotes) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 38 215 obunachidan iborat bo'lib, Taʼlim toifasida 4 922-o'rinni va Hindiston mintaqasida 9 910-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 38 215 obunachiga ega bo‘ldi.

01 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 939 ga, so‘nggi 24 soatda esa 28 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 12.94% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 6.07% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 4 942 marta ko‘riladi; birinchi sutkada odatda 2 318 ta ko‘rish yig‘iladi.
Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 14 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 02 Noyabr, 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.

38 215

Obunachilar

+2824 soatlar

+1907 kun

+93930 kun

4 942

Post ko'rishlar

~ 2 31824 soatlar

~ 3 24348 soatlar

12.94%

Muloqot nisbati

~ 5

Kuniga postlar

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38 217

saya akan hantar lagi 50 soalan yg bukan susah sangat guyss ( cuba buat semua soalan ni kalau u guys aim credit shj)

38 217

ok guys 10 soalan ady, pls react a bit to make me continue posting

38 217

guys since tmr u guys have malay exam, lemme send 10 problems as i said, but u guys can try it after exam

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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

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Thanks to TonyCWX

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Circular measure+ differentiation

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Repost from 9a+ 🤝

addmath f4 ch 5 progressions