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Tentukan bayangan titik  oleh transformasi: g. Rotasi

Pertanyaan

Tentukan bayangan titik straight R open parentheses negative 1 comma space 6 close parentheses oleh transformasi:

g. Rotasi straight R open square brackets straight O comma space 120 degree close square brackets

Pembahasan Soal:

Rotasi titik straight A open parentheses x comma space y close parentheses sebesar theta searah jarum jam terhadap titik pusat straight O open parentheses 0 comma space 0 close parentheses diperoleh bayangannya straight A apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses.

 open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space theta end cell cell negative sin space theta end cell row cell sin space theta end cell cell cos space theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Akan ditentukan bayangan titik straight R open parentheses negative 1 comma space 6 close parentheses, oleh transformasi Rotasi straight R open square brackets straight O comma space 120 degree close square brackets,

table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space 120 degree end cell cell negative sin space 120 degree end cell row cell sin space 120 degree end cell cell cos space 120 degree end cell end table close parentheses open parentheses table row cell negative 1 end cell row 6 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 1 half end cell cell negative 1 half square root of 3 end cell row cell 1 half square root of 3 end cell cell negative 1 half end cell end table close parentheses open parentheses table row cell negative 1 end cell row 6 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 half minus 6 over 2 square root of 3 end cell row cell negative 1 half square root of 3 minus 6 over 2 end cell end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half minus 3 square root of 3 end cell row cell negative 3 minus 1 half square root of 3 end cell end table close parentheses end cell end table 

Jadi, diperoleh bayangannya adalah straight R apostrophe open parentheses 1 half minus 3 square root of 3 comma space minus 3 minus 1 half square root of 3 close parentheses.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Koordinat bayangan titik  setelah ditranslasi  dan dilanjutkan dengan rotasi berpusat di  sejauh  adalah ...

Pembahasan Soal:

Ingat konsep translasi titik begin mathsize 14px style open parentheses x comma space y close parentheses end style sejauh begin mathsize 14px style straight T equals open parentheses table row a row b end table close parentheses end style dan rotasi berpusat di begin mathsize 14px style straight O open parentheses 0 comma space 0 close parentheses end style sejauh begin mathsize 14px style 180 degree end style:

begin mathsize 14px style table row cell straight P open parentheses x comma space y close parentheses end cell cell rightwards arrow with T open parentheses table row a row b end table close parentheses on top end cell cell straight P apostrophe open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell x plus a end cell row cell y plus b end cell end table close parentheses end cell row cell straight P open parentheses x comma space y close parentheses end cell cell rightwards arrow with open square brackets R open parentheses 0 comma 0 close parentheses comma space 180 degree close square brackets on top end cell cell straight P apostrophe open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell end table end style  

 Koordinat bayangan titik begin mathsize 14px style straight P open parentheses 8 comma space 4 close parentheses end style ditranslasi sejauh begin mathsize 14px style open parentheses table row 3 row 1 end table close parentheses end style adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 8 row 4 end table close parentheses plus open parentheses table row 3 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 8 plus 3 end cell row cell 4 plus 1 end cell end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table end style 

Diperoleh koordinat hasil translasi begin mathsize 14px style straight P apostrophe open parentheses 11 comma space 5 close parentheses end style, kemudian dilanjutkan rotasi berpusat di begin mathsize 14px style straight O open parentheses 0 comma space 0 close parentheses end style sejauh begin mathsize 14px style 180 degree end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close parentheses open parentheses table row 11 row 5 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 11 end cell row cell negative 5 end cell end table close parentheses end cell end table end style 

Dengan demikian bayangan titik P adalah begin mathsize 14px style P apostrophe apostrophe open parentheses negative 11 comma space minus 5 close parentheses end style

Oleh karena itu jawaban yang benar adalah B.

0

Roboguru

Garis  dirotasi oleh . Persamaan bayangan garis  yang terjadi berbentuk .... (Modifikasi UN 2014)

Pembahasan Soal:

Diketahui,

  • straight l identical to 3 x minus y minus 1 equals 0
  • Dirotasikan oleh  begin mathsize 14px style open square brackets straight O open parentheses 0 comma space 0 close parentheses comma space straight R open parentheses straight pi over 4 close parentheses close square brackets end style 

Ditanyakan,

  • Bayangan garis

Gunakan persamaan matriks berikut.

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space straight theta end cell cell negative sin space straight theta end cell row cell sin space straight theta end cell cell cos space straight theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space 45 end cell cell negative sin space 45 end cell row cell sin space 45 end cell cell cos space 45 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row blank blank blank row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator open parentheses begin display style 1 half square root of 2 space end root times 1 half square root of 2 end style close parentheses minus open parentheses begin display style 1 half square root of 2 times negative 1 half square root of 2 end style close parentheses end fraction open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 2 over 4 minus open parentheses negative 2 over 4 close parentheses end style end fraction open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 1 end style end fraction open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell row cell negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell end table close parentheses end cell end table

Maka di dapat x equals 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe dan y equals negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe

Masukan ke persamaan garis straight l identical to 3 x minus y minus 1 equals 0,

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus y minus 1 end cell equals 0 row cell 3 open parentheses 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe close parentheses minus open parentheses negative 1 half x apostrophe plus 1 half square root of 2 y apostrophe close parentheses minus 1 end cell equals 0 row cell 3 over 2 square root of 2 x apostrophe plus 3 over 2 square root of 2 y apostrophe plus 1 half x apostrophe minus 1 half square root of 2 y apostrophe minus 1 end cell equals 0 row cell 4 over 2 square root of 2 x apostrophe plus 2 over 2 square root of 2 y apostrophe minus 1 end cell equals 0 row cell 2 square root of 2 x apostrophe plus square root of 2 y apostrophe minus 1 end cell equals 0 row cell 4 x apostrophe plus 2 y apostrophe minus square root of 2 end cell equals 0 row blank blank blank end table

Jadi, bayangan garisnya adalah .table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank apostrophe end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank apostrophe end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Bayangan titik  yang direfleksikan terhadap garis , dilanjutkan refleksi terhadap garis  dan kemudian dilanjutkan rotasi pusat O bersudut  radian adalah , maka koordinat titik A adalah ....

Pembahasan Soal:

Jika diketahui sebarang titik dengan koordinat open parentheses x comma space y close parentheses, maka koordinat bayangan hasil pencerminan x equals a, dinotasikan:

open parentheses 2 a minus x comma space y close parentheses 

Jika diketahui sebarang titik dengan koordinat open parentheses x comma space y close parentheses, maka koordinat bayangan hasil pencerminan y equals b, dinotasikan:

open parentheses x comma space 2 b minus y close parentheses  

Persamaan transformasi dengan pusat straight O left parenthesis 0 comma space 0 right parenthesis dan sudut rotasi theta berlawanan arah jarum jam, ditulis straight R left square bracket 0 comma space theta right square bracket , untuk pemetaan left parenthesis x comma space y right parenthesis ke left parenthesis x apostrophe comma space y apostrophe right parenthesis dapat dinyatakan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos theta end cell cell negative sin theta end cell row cell sin theta end cell cell cos theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row blank blank blank end table 

Diketahui: bayangan titik A adalah left parenthesis negative 4 comma space 6 right parenthesis

Transformasi refleksi terhadap garis x equals negative 2

 table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 a minus x comma space y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell row cell open parentheses 2 times open parentheses negative 2 close parentheses minus x comma space y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell row cell open parentheses negative 4 minus x comma space y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell end table 

Kemudian, refleksi terhadap garis y equals 3

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x comma space 2 b minus y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell row cell open parentheses negative 4 minus x comma space 2 times 3 minus y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell row cell open parentheses negative 4 minus x comma space 6 minus y close parentheses end cell equals cell open parentheses 4 comma space 6 close parentheses end cell row blank blank blank end table  

Kemudian dirotasi straight R open square brackets 0 comma space straight pi over 2 close square brackets adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos 90 degree end cell cell negative sin 90 degree end cell row cell sin 90 degree end cell cell cos 90 degree end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell negative 4 end cell row 6 end table close parentheses end cell equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row cell negative 4 minus x end cell row cell 6 minus y end cell end table close parentheses end cell row cell open parentheses table row cell negative 4 end cell row 6 end table close parentheses end cell equals cell open parentheses table row cell 0 times open parentheses negative 4 minus x close parentheses minus 1 times open parentheses 6 minus y close parentheses end cell row cell 1 times open parentheses negative 4 minus x close parentheses plus 0 times open parentheses 6 minus y close parentheses end cell end table close parentheses end cell row cell open parentheses table row cell negative 4 end cell row 6 end table close parentheses end cell equals cell open parentheses table row cell negative 6 plus y end cell row cell negative 4 minus x end cell end table close parentheses end cell end table       

Dari kesamaan matriks di atas, diperoleh:

  table attributes columnalign right center left columnspacing 0px end attributes row cell negative 6 plus y end cell equals cell negative 4 end cell row cell negative 6 plus y plus 6 end cell equals cell negative 4 plus 6 end cell row cell therefore space y end cell equals 2 end table 

 table attributes columnalign right center left columnspacing 0px end attributes row cell negative 4 minus x end cell equals 6 row cell negative 4 minus x plus 4 end cell equals cell 6 plus 4 end cell row cell negative x end cell equals 10 row cell space therefore space x end cell equals cell negative 10 end cell end table  

Sehingga diperoleh koordinat titik A adalah open parentheses negative 10 comma space 2 close parentheses.

Jadi, jawaban yang tepat adalah A.

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Roboguru

Titik P (-3, 1) dipetakan oleh rotasi dengan pusat O sejauh , dilanjutkan dengan translasi . Peta titik P adalah...

Pembahasan Soal:

B a y a n g a n space d a r i space r o t a s i space 90 degree space a d a l a h  open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row x row y end table close parentheses  R o t a s i space t i t i k space P left parenthesis negative 3 comma 1 right parenthesis space d e n g a n space r o t a s i space 90 degree space a d a l a h  open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row 3 row cell negative 1 end cell end table close parentheses  open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row 1 row 3 end table close parentheses    B a y a n g a n space d a r i space t r a n s l a s i space t e r h a d a p space m a t r i k s space left parenthesis a comma b right parenthesis space a d a l a h  open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row a row b end table close parentheses plus open parentheses table row x row y end table close parentheses    B a y a n g a n space d a r i space t i t i k space left parenthesis 1 comma 3 right parenthesis space o l e h space t r a n s l a s i space left parenthesis 3 comma 4 right parenthesis space a d a l a h  open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses equals open parentheses table row 3 row 4 end table close parentheses plus open parentheses table row 1 row 3 end table close parentheses  open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses equals open parentheses table row 4 row 7 end table close parentheses

0

Roboguru

Bayangan garis , jika diputar sejauh  dengan pusat  adalah ....

Pembahasan Soal:

Rotasi sejauh straight pi over 4 dengan pusat O akan menghasilkan bayangan:


table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space straight pi over 4 end cell cell negative sin space straight pi over 4 end cell row cell sin space straight pi over 4 end cell cell cos space straight pi over 4 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell end table


Bentuk persamaan di atas dapat diubah ke bentuk persamaan matriks elimiasi Gauss-Jordan di bawah ini.


open parentheses table row cell table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table end cell cell negative table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table end cell row cell table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table end cell cell table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table end cell end table table row vertical ellipsis row vertical ellipsis end table table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses table row cell square root of 2 B subscript 1 rightwards arrow B subscript 1 end cell row cell square root of 2 B subscript 2 rightwards arrow B subscript 2 end cell end table open parentheses table row 1 cell negative 1 end cell row 1 1 end table table row vertical ellipsis row vertical ellipsis end table table row cell square root of 2 x apostrophe end cell row cell square root of 2 y apostrophe end cell end table close parentheses table row blank row cell B subscript 2 minus B subscript 1 rightwards arrow B subscript 2 end cell end table open parentheses table row 1 cell negative 1 end cell row 0 2 end table table row vertical ellipsis row vertical ellipsis end table table row cell square root of 2 x apostrophe end cell row cell square root of 2 y apostrophe minus square root of 2 x apostrophe end cell end table close parentheses table row blank row cell 1 half B subscript 2 rightwards arrow B subscript 2 end cell end table open parentheses table row 1 cell negative 1 end cell vertical ellipsis cell square root of 2 x apostrophe end cell row 0 1 vertical ellipsis cell 1 half square root of 2 y apostrophe minus 1 half square root of 2 x apostrophe end cell end table close parentheses table row cell B subscript 1 plus B subscript 2 rightwards arrow B subscript 1 end cell row blank end table open parentheses table row 1 0 vertical ellipsis cell 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell row 0 1 vertical ellipsis cell 1 half square root of 2 y apostrophe minus 1 half square root of 2 x apostrophe end cell end table close parentheses


Sehingga didapatkan table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell end table dan table attributes columnalign right center left columnspacing 0px end attributes row y equals cell negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell end table.


Substitusikan kedua persamaan tersebut ke dalam persamaan garis 3 x minus y plus 2 equals 0, sehingga didapatkan persamaan bayangan garis tersebut adalah:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus y plus 2 end cell equals 0 row cell 3 open parentheses 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe close parentheses minus open parentheses negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe close parentheses plus 2 end cell equals 0 row cell 2 square root of 2 x apostrophe minus square root of 2 y apostrophe plus 2 end cell equals 0 row blank blank blank row cell 2 x apostrophe minus y apostrophe plus fraction numerator 2 over denominator square root of 2 end fraction end cell equals 0 row cell 2 x apostrophe minus y apostrophe plus square root of 2 end cell equals 0 end table end style


Jadi, persamaan garis bayangannya adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table.

Oleh karena itu, jawaban yang benar adalah B.

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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