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AddMath Spmnetic!™⚡️
If f(x) = ax+ b, if f(1) = 4 and f(2) = 7. Find f(3)
guys, may I know your working for this question?
AddMath Spmnetic!™⚡️
If f(x) = ax+ b, if f(1) = 4 and f(2) = 7. Find f(3)
y=ax+b tu linear function Yang nj ikut pola je
f(1) = 4
f(2) = 7
So 7-4=3
So .. f(3) = 7+3
f(1) = 4
f(2) = 7
So 7-4=3
So .. f(3) = 7+3
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Jason (5 Radium)
y=ax+b tu linear function Yang nj ikut pola je f(1) = 4 f(2) = 7 So 7-4=3 So .. f(3) = 7+3
congrats to amni madiha...
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Jason (5 Radium)
y=ax+b tu linear function Yang nj ikut pola je f(1) = 4 f(2) = 7 So 7-4=3 So .. f(3) = 7+3
remember not all linear functions are arithemic sequences, but all arithemic sequences are linear functions..
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AddMath Spmnetic!™⚡️
If f(x) = ax+ b, if f(1) = 4 and f(2) = 7. Find f(3)
congrats to Nisha also...👍🏽👍🏽👍🏽
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Jason (5 Radium)
haha, a very nice question...
btw JEE advanced is one of the most toughest exams for undergraduate students..
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Forwarded from Mathematics Spmnetic!™ 🔢 (Afiq)
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🤲❤️
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Forwarded from Mathematics Spmnetic!™ 🔢 (Afiq)
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
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Jason (5 Radium)
Photo
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
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