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Q.[tex].[/tex]Diketahui suatu fungsi :f(p) = 2p × 200p + 2p

Berikut ini adalah pertanyaan dari Neuil12 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Q..
Diketahui suatu fungsi :
f(p) = 2p × 200p + 2p + 2p³ - 20
Ditanya ⬇
f(2) = ...
f(3) = ...
f(4) = ...
f(5) = ...
f(6) = ...
f(7) = ...
f(8) = ...
f(9) = ...
f(10) = ...
f(11) = ...
f(12) = ...


-^- ​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

➩ Diketahui :

\sf \bf \tt f(p) = 2p × 200p + 2p + 2p³ - 20

➩ Ditanya :

f(2) = ...

f(3) = ...

f(4) = ...

f(5) = ...

f(6) = ...

f(7) = ...

f(8) = ...

f(9) = ...

f(10) = ...

f(11) = ...

f(12) = ...

➩ Jawab :

  • Menentukan nilai f(p)

\sf \bf \tt f(p) = 2p × 200p + 2p + 2p³ - 20

\bold{\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20}

  • Menentukan nilai f(2)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(2) = 400(2²) + 2(2) + 2(2³) - 20

\sf \bf \tt f(2) = 400(2 × 2) + 2(2) + 2(2×2×2) - 20

\sf \bf \tt f(2) = 400(4) + 2(2) + 2(2×2×2) - 20

\sf \bf \tt f(2) = 1.600+ 2(2) + 2(2×2×2) - 20

\sf \bf \tt f(2) = 1.600+ 4 + 2(2×2×2) - 20

\sf \bf \tt f(2) = 1.600+ 4 + 2(4×2) - 20

\sf \bf \tt f(2) = 1.600+ 4 + 2(8)- 20

\sf \bf \tt f(2) = 1.600+ 4 + 16- 20

\sf \bf \tt f(2) = 1.604 + 16- 20

\sf \bf \tt f(2) = 1.620- 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(2) = 1.600}}}}

  • Menentukan nilai f(3)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(3) = 400(3²) + 2(3) + 2(3³) - 20

\sf \bf \tt f(3) = 400(3×3) + 2(3) + 2(3×3×3) - 20

\sf \bf \tt f(3) = 400(9) + 2(3) + 2(3×3×3) - 20

\sf \bf \tt f(3) = 3.600+ 2(3) + 2(3×3×3) - 20

\sf \bf \tt f(3) = 3.600+ 6 + 2(3×3×3) - 20

\sf \bf \tt f(3) = 3.600+ 6 + 2(9×3) - 20

\sf \bf \tt f(3) = 3.600+ 6 + 2(27)- 20

\sf \bf \tt f(3) = 3.600+ 6 + 54- 20

\sf \bf \tt f(3) = 3.606 + 54- 20

\sf \bf \tt f(3) = 3.660 - 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(3) = 3.640}}}}

  • Menentukan nilai f(4)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(4) = 400(4²) + 2(4) + 2(4³) - 20

\sf \bf \tt f(4) = 400(4×4) + 2(4) + 2(4×4×4) - 20

\sf \bf \tt f(4) = 400(16) + 2(4) + 2(4×4×4) - 20

\sf \bf \tt f(4) = 6.400+ 2(4) + 2(4×4×4) - 20

\sf \bf \tt f(4) = 6.400+ 8 + 2(4×4×4) - 20

\sf \bf \tt f(4) = 6.400+ 8 + 2(16×4) - 20

\sf \bf \tt f(4) = 6.400+ 8 + 2(64)- 20

\sf \bf \tt f(4) = 6.400+ 8 + 128- 20

\sf \bf \tt f(4) = 6.408+ 128- 20

\sf \bf \tt f(4) = 6.536- 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(4) = 6.516}}}}

  • Menentukan nilai f(5)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(5) = 400(5²) + 2(5) + 2(5³) - 20

\sf \bf \tt f(5) = 400(5×5) + 2(5) + 2(5×5×5) - 20

\sf \bf \tt f(5) = 400(25) + 2(5) + 2(5×5×5) - 20

\sf \bf \tt f(5) = 10.000+ 2(5) + 2(5×5×5) - 20

\sf \bf \tt f(5) = 10.000+ 10 + 2(5×5×5) - 20

\sf \bf \tt f(5) = 10.000+ 10 + 2(25×5) - 20

\sf \bf \tt f(5) = 10.000+ 10 + 2(125)- 20

\sf \bf \tt f(5) = 10.000+ 10 + 250- 20

\sf \bf \tt f(5) = 10.010+ 250- 20

\sf \bf \tt f(5) = 10.260- 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(5) = 10.240}}}}

  • Menentukan nilai f(6)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(6) = 400(6²) + 2(6) + 2(6³) - 20

\sf \bf \tt f(6) = 400(6×6) + 2(6) + 2(6×6×6) - 20

\sf \bf \tt f(6) = 400(36) + 2(6) + 2(6×6×6) - 20

\sf \bf \tt f(6) = 14.400+ 2(6) + 2(6×6×6) - 20

\sf \bf \tt f(6) = 14.400+ 12 + 2(6×6×6) - 20

\sf \bf \tt f(6) = 14.400+ 12 + 2(36×6) - 20

\sf \bf \tt f(6) = 14.400+ 12 + 2(216)- 20

\sf \bf \tt f(6) = 14.400+ 12 + 432- 20

\sf \bf \tt f(6) = 14.412 + 128- 20

\sf \bf \tt f(6) = 14.540 - 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(6) = 14.520}}}}

  • Menentukan nilai f(7)

\sf \bf \tt f(p) = 400p² + 2p + 2p³ - 20

\sf \bf \tt f(7) = 400(7²) + 2(7) + 2(7³) - 20

\sf \bf \tt f(7) = 400(7×7) + 2(7) + 2(7×7×7) - 20

\sf \bf \tt f(7) = 400(49) + 2(7) + 2(7×7×7) - 20

\sf \bf \tt f(7) = 19.600+ 2(7) + 2(7×7×7) - 20

\sf \bf \tt f(7) = 19.600+ 14 + 2(7×7×7) - 20

\sf \bf \tt f(7) = 19.600+ 14 + 2(49×7) - 20

\sf \bf \tt f(7) = 19.600+ 14 + 2(343)- 20

\sf \bf \tt f(7) = 19.600+ 14 + 686- 20

\sf \bf \tt f(7) = 19.614+ 686- 20

\sf \bf \tt f(7) = 20.300 - 20

\bold{\underline{\boxed{\pink{\sf \bf \tt f(7) = 20.280}}}}

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Last Update: Thu, 20 Oct 22
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