Roboguru

Nyatakan dalam bentuk pangkat negatif. h.

Pertanyaan

Nyatakan dalam bentuk pangkat negatif.

h. begin mathsize 14px style open parentheses 1 over 25 close parentheses squared end style 

Pembahasan Soal:

Ingat!

Sifat bilangan berpangkat

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over a to the power of m end cell equals cell a to the power of negative m end exponent end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of m close parentheses to the power of n end cell equals cell a to the power of m n end exponent end cell end table

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 over 25 close parentheses squared end cell equals cell open parentheses 1 over 5 squared close parentheses squared end cell row blank equals cell open parentheses 5 to the power of negative 2 end exponent close parentheses squared end cell row blank equals cell 5 to the power of open parentheses negative 2 cross times 2 close parentheses end exponent end cell row blank equals cell 5 to the power of negative 4 end exponent end cell end table end style

Dengan demikian, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 1 over 25 close parentheses squared end cell end table dinyatakan dalam pangkat negatif menjadi table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 5 to the power of negative 4 end exponent end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Hermawan

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Tentukan hasil operasi bilangan berpangkat berikut! g. h.

Pembahasan Soal:

Ingatlah sifat-sifat bilangan berpangkat berikut.

left parenthesis straight i right parenthesis space a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent left parenthesis ii right parenthesis space a to the power of negative m end exponent equals 1 over a to the power of m

Berdasarkan sifat-sifat di atas, maka:

g)

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of negative 9 end exponent divided by 4 cubed end cell equals cell 4 to the power of negative 9 minus 3 end exponent end cell row blank equals cell 4 to the power of negative 12 end exponent end cell row blank equals cell 1 over 4 to the power of 12 end cell row blank equals cell fraction numerator 1 over denominator 16.777.216 end fraction end cell end table

h)

table attributes columnalign right center left columnspacing 0px end attributes row cell y to the power of 14 divided by y to the power of negative 2 end exponent end cell equals cell y to the power of 14 minus open parentheses negative 2 close parentheses end exponent end cell row blank equals cell y to the power of 14 plus 2 end exponent end cell row blank equals cell y to the power of 16 end cell end table

Dengan demikian, hasil dari  4 to the power of negative 9 end exponent divided by 4 cubed equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator 16.777.216 end fraction end cell end table dan Error converting from MathML to accessible text..

1

Roboguru

Nyatakan bilangan berikut dengan meggunakan pangkat positif!

Pembahasan Soal:

Ingat kembali sifat bilangan berpangkat untuk pangkat negatif berikut.

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

Berdasarkan sifat di atas, maka diperoleh perhitungan berikut ini.

negative 12 to the power of negative 10 end exponent equals fraction numerator 1 over denominator negative 12 to the power of 10 end fraction

Jadi, bilangan negative 12 to the power of negative 10 end exponent dinyatakan dengan pangkat positif adalah fraction numerator 1 over denominator negative 12 to the power of 10 end fraction.

1

Roboguru

Hitunglah hasil dari bilangan pecahan berpangkat berikut. a.

Pembahasan Soal:

Ingat!

Definisi bilangan berpangkat:

  •  a to the power of n equals stack stack a cross times a cross times... cross times a with underbrace below with n below 

Sifat bilangan berpangkat:

  • a to the power of negative n end exponent equals 1 over a to the power of n 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 1 fifth close parentheses to the power of negative 3 end exponent end cell equals cell 1 over open parentheses negative begin display style 1 fifth end style close parentheses cubed end cell row blank equals cell fraction numerator 1 over denominator open parentheses negative begin display style 1 fifth end style close parentheses cross times open parentheses negative begin display style 1 fifth end style close parentheses cross times open parentheses negative begin display style 1 fifth end style close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator negative begin display style 1 over 125 end style end fraction end cell row blank equals cell 1 cross times open parentheses negative 125 over 1 close parentheses end cell row blank equals cell negative 125 end cell end table 

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 1 fifth close parentheses to the power of negative 3 end exponent end cell end table adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative 125 end cell end table.

0

Roboguru

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Ingat kembali:

straight a to the power of straight m space colon space straight a to the power of straight n equals straight a to the power of straight m minus straight n end exponent straight a to the power of negative straight n end exponent equals 1 over straight a to the power of straight n open parentheses straight a to the power of straight m close parentheses to the power of straight n equals straight a to the power of straight m cross times straight n end exponent  

Maka:

open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals open parentheses straight a squared close parentheses cubed times straight b cubed space colon space straight a squared times straight b to the power of 4 open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals straight a to the power of 6 times straight b cubed space colon space straight a squared times straight b to the power of 4 open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals straight a to the power of 6 minus 2 end exponent times straight b to the power of 3 minus 4 end exponent open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals straight a to the power of 4 times straight b to the power of negative 1 end exponent open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals straight a to the power of 4 times 1 over straight b open parentheses straight a squared straight b close parentheses cubed space colon space open parentheses straight a squared straight b to the power of 4 close parentheses equals straight a to the power of 4 over straight b 

Jadi, jawaban yang benar adalah B.

0

Roboguru

Nilai dari

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of negative 3 end exponent cross times 4 to the power of negative 2 end exponent end cell equals cell 1 over 2 cubed cross times 1 over 4 squared end cell row blank equals cell 1 over 2 cubed cross times fraction numerator 1 over denominator 2 squared cross times 2 squared end fraction end cell row blank equals cell 1 over 2 to the power of 3 plus 2 plus 2 end exponent end cell row blank equals cell 1 over 2 to the power of 7 end cell row blank equals cell 2 to the power of negative 7 end exponent end cell end table  

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved