Question

Question.
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f(x) = x³ + x!! × 2^x
f(5) = ...
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[tex]\mathbb\color{hotpink}{{{RULES : }}}[/tex]
↬Pakai cara!
↬ngasal → report!
↬Rapi!
↬Salah → koreksi 30 menit!
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Answer 1

SOAL 1

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f(x) = x³ + x!! × 2^x

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f(5) = 5³ + 5!! × 2^5

f(5) = (5.5.5) + (5.3.1) × (2.2.2.2.2)

f(5) = (5.5.5) + (5.3.1) × (4.2.2.2)

f(5) = (5.5.5) + (5.3.1) × (8.2.2)

f(5) = (25.5) + (15.1) × (16.2)

f(5) = 125 + 15 × 32

f(5) = 125 + 480

f(5) = 605

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Answer 2

« Penyelesaian Soal »

[ Soal ]

f(x) = x³ + x!! × 2^x

f(5) = ...

____________

• Menentukan Niali f(5)

• f(x) = x³ + x!! × 2^x
• f(5) = x³ + x!! × 2^x
• f(5) = 5³ + 5!! × 2^5
• f(5) = ( 5 × 5 × 5 ) + ( 5 × 3 × 1 ) × ( 2 × 2 × 2 × 2 × 2 )
• f(5) = ( 25 × 5 ) + ( 15¹) × ( 4 × 4 × 2 )
• f(5) = ( 125 ) + ( 15 × ( 16 × 2 )
• f(5) = ( 125 + ( 15 × 32 )
• f(5) = ( 125 + 480 )
• f(5) = 605

Kesimpulan ↓

• Maka, Hasil dari f(5) = 605

Semoga Bisa Bermanfaat-!