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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 ✨

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📈 Analytical overview of Telegram channel AddMath Spmnetic!™⚡️

Channel AddMath Spmnetic!™⚡️ (@addmathspmnotes) in the English language segment is an active participant. Currently, the community unites 38 064 subscribers, ranking 4 957 in the Education category and 10 234 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 38 064 subscribers.

According to the latest data from 26 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 1 022 over the last 30 days and by 31 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 12.29%. Within the first 24 hours after publication, content typically collects 6.25% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 4 675 views. Within the first day, a publication typically gains 2 376 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 11.
  • Thematic interests: Content is focused on key topics such as addmath, untuk, 629/4, math, eqn.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻 For any enquiries, please directly ask in our discussion group ✨”

Thanks to the high frequency of updates (latest data received on 27 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

38 064
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Posts Archive
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+2

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...🔥🔥

photo content

photo content

photo content

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
+1
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..