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Diketahui vektor a=3i+21​j​−21​k; b=i+5j​−k dan c=23​i. Hasil dari a+21​b−c adalah ...

Pertanyaan

Diketahui vektor straight a with rightwards arrow on top equals 3 straight i with rightwards arrow on top plus 1 half straight j with rightwards arrow on top minus 1 half straight k with rightwards arrow on top; straight b with rightwards arrow on top equals straight i with rightwards arrow on top plus 5 straight j with rightwards arrow on top minus straight k with rightwards arrow on top dan straight c with rightwards arrow on top equals 3 over 2 straight i with rightwards arrow on top. Hasil dari straight a with rightwards arrow on top plus 1 half straight b with rightwards arrow on top minus straight c with rightwards arrow on top adalah ...

  1. 4 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus straight k with rightwards arrow on top

  2. 3 1 half straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus straight k with rightwards arrow on top

  3. 4 straight i with rightwards arrow on top plus 5 1 half straight j with rightwards arrow on top minus straight k with rightwards arrow on top

  4. 2 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top plus straight k with rightwards arrow on top

  5. 2 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus straight k with rightwards arrow on top

Pembahasan Soal:

Misalkan terdapat vektor straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top, straight b with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top dan straight z adalah skalar. maka berlaku:

open parentheses straight i close parentheses space straight a with rightwards arrow on top plus-or-minus straight b with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 plus-or-minus straight b subscript 1 end cell row cell straight a subscript 2 plus-or-minus straight b subscript 2 straight a subscript 3 plus-or-minus straight b subscript 3 end cell end table close parentheses equals open parentheses straight a subscript 1 plus-or-minus straight b subscript 1 close parentheses straight i with rightwards arrow on top plus open parentheses straight a subscript 2 plus-or-minus straight b subscript 2 close parentheses straight j with rightwards arrow on top plus open parentheses straight a subscript 3 plus-or-minus straight b subscript 3 close parentheses straight k with rightwards arrow on top left parenthesis ii right parenthesis space straight z straight a with rightwards arrow on top equals open parentheses table row cell za subscript 1 end cell row cell za subscript 2 end cell row cell za subscript 3 end cell end table close parentheses equals za subscript 1 straight i with rightwards arrow on top plus za subscript 2 straight j with rightwards arrow on top plus za subscript 3 straight k with rightwards arrow on top

Berdasarkan sifat-sifat operasi vektor tersebut, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with rightwards arrow on top plus 1 half straight b with rightwards arrow on top minus straight c with rightwards arrow on top end cell equals cell open parentheses table row 3 row cell begin inline style 1 half end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses plus 1 half open parentheses table row 1 row 5 row cell negative 1 end cell end table close parentheses minus open parentheses table row cell begin inline style 3 over 2 end style end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell begin inline style 1 half end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses plus open parentheses table row cell begin inline style 1 half end style end cell row cell begin inline style 5 over 2 end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses minus open parentheses table row cell begin inline style 3 over 2 end style end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus begin inline style 1 half end style minus begin inline style 3 over 2 end style end cell row cell begin inline style 1 half end style plus begin inline style 5 over 2 end style end cell row cell negative begin inline style 1 half end style minus begin inline style 1 half end style end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row 3 row cell negative 1 end cell end table close parentheses end cell row blank equals cell 2 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus straight k with rightwards arrow on top end cell end table

Dengan demikian, hasil operasi vektor pada soal 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 cell straight i with rightwards arrow on top end cell 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 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight j with rightwards arrow on top end cell 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 straight k with rightwards arrow on top end cell end table.

Oleh karena itu, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui vektor a=3i+21​j​−21​k, b=i+5j​−k, dan  c=23​i. Hasil dari a+21​b−c adalah ...

Pembahasan Soal:

Diketahui bahwa vektor a with rightwards arrow on top,b with rightwards arrow on top, dan c with rightwards arrow on top merupakan vektor basis (i,j,k) yang berada pada dimensi tiga yang dapat diubah ke vektor baris berada pada dimensi tiga sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell 3 i with rightwards arrow on top plus 1 half j with rightwards arrow on top minus 1 half k with rightwards arrow on top ⟶ a with rightwards arrow on top equals open parentheses 3 comma 1 half comma negative 1 half close parentheses end cell row blank blank blank row cell b with rightwards arrow on top end cell equals cell i with rightwards arrow on top plus 5 j with rightwards arrow on top minus k with rightwards arrow on top ⟶ b with rightwards arrow on top equals left parenthesis 1 , 5 comma negative 1 right parenthesis end cell row blank blank blank row cell c with rightwards arrow on top end cell equals cell 3 over 2 i with rightwards arrow on top ⟶ c with rightwards arrow on top equals open parentheses 3 over 2 comma 0 , 0 close parentheses end cell end table

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus 1 half b with rightwards arrow on top minus c with rightwards arrow on top end cell equals cell open parentheses 3 comma 1 half comma negative 1 half close parentheses plus 1 half open parentheses 1 , 5 comma negative 1 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 3 comma 1 half comma negative 1 half close parentheses plus open parentheses 1 half comma 5 over 2 comma negative 1 half close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 3 1 half comma 6 over 2 comma negative 2 over 2 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 7 over 2 comma 6 over 2 comma negative 2 over 2 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 4 over 2 comma 6 over 2 comma negative 2 over 2 close parentheses end cell row blank equals cell open parentheses 2 , 3 comma negative 1 close parentheses end cell end table

Jadi, hasil dari a with rightwards arrow on top plus 1 half b with rightwards arrow on top minus c with rightwards arrow on top adalah open parentheses 2 , 3 comma negative 1 close parentheses.

1

Roboguru

Jika a=3i+4j​−5k, b=i+2j​−3k dan c=−i+2j​+32​k, maka vektor satuan d=a−2b+3c adalah ...

Pembahasan Soal:

Vektor satuan dari vektor tiga dimensi straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses adalah sebagai berikut.

fraction numerator straight a with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar end fraction equals fraction numerator open parentheses straight a subscript 1 comma space straight a subscript 2 comma space straight a subscript 3 close parentheses over denominator square root of straight a subscript 1 squared plus straight a subscript 2 squared plus straight a subscript 3 squared end root end fraction.

Sebelum mencari vektor satuan dari straight d with rightwards arrow on top, hitunglah operasi vektornya terlebih dahulu, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight d with rightwards arrow on top end cell equals cell straight a with rightwards arrow on top minus 2 straight b with rightwards arrow on top plus 3 straight c with rightwards arrow on top end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row 3 row 4 row cell negative 5 end cell end table close parentheses minus 2 open parentheses table row 1 row 2 row cell negative 3 end cell end table close parentheses plus 3 open parentheses table row cell negative 1 end cell row 2 row cell begin inline style 2 over 3 end style end cell end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row 3 row 4 row cell negative 5 end cell end table close parentheses minus open parentheses table row 2 row 4 row cell negative 6 end cell end table close parentheses plus open parentheses table row cell negative 3 end cell row 6 row 2 end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row cell 3 minus 2 minus 3 end cell row cell 4 minus 4 plus 6 end cell row cell negative 5 plus 6 plus 2 end cell end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row cell negative 2 end cell row 6 row 3 end table close parentheses end cell end table 

Selanjutnya, vektor satuan straight d with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight d with rightwards arrow on top over denominator open vertical bar straight d with rightwards arrow on top close vertical bar end fraction end cell equals cell fraction numerator open parentheses negative 2 comma space 6 comma space 3 close parentheses over denominator square root of open parentheses negative 2 close parentheses squared plus 6 squared plus 3 squared end root end fraction end cell row blank equals cell fraction numerator open parentheses negative 2 comma space 6 comma space 3 close parentheses over denominator square root of 4 plus 36 plus 9 end root end fraction end cell row blank equals cell fraction numerator open parentheses negative 2 comma space 6 comma space 3 close parentheses over denominator square root of 49 end fraction end cell row blank equals cell fraction numerator open parentheses negative 2 comma space 6 comma space 3 close parentheses over denominator 7 end fraction end cell row blank equals cell open parentheses negative 2 over 7 comma space 6 over 7 comma space 3 over 7 close parentheses end cell row blank equals cell negative 2 over 7 straight i with rightwards arrow on top plus 6 over 7 straight j with rightwards arrow on top plus 3 over 7 straight k with rightwards arrow on top end cell end table

Dengan demikian, jawaban yang benar adalah B.

0

Roboguru

Diketahui vektor a=4i−23​j​+2k; b=2i+5j​−3k dan c=2j​−3k. Hasil dari a−21​b+21​c adalah ...

Pembahasan Soal:

Misalkan terdapat vektor straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top, straight b with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top dan straight z adalah skalar. maka berlaku:

open parentheses straight i close parentheses space straight a with rightwards arrow on top plus-or-minus straight b with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 plus-or-minus straight b subscript 1 end cell row cell straight a subscript 2 plus-or-minus straight b subscript 2 straight a subscript 3 plus-or-minus straight b subscript 3 end cell end table close parentheses equals open parentheses straight a subscript 1 plus-or-minus straight b subscript 1 close parentheses straight i with rightwards arrow on top plus open parentheses straight a subscript 2 plus-or-minus straight b subscript 2 close parentheses straight j with rightwards arrow on top plus open parentheses straight a subscript 3 plus-or-minus straight b subscript 3 close parentheses straight k with rightwards arrow on top left parenthesis ii right parenthesis space straight z straight a with rightwards arrow on top equals open parentheses table row cell za subscript 1 end cell row cell za subscript 2 end cell row cell za subscript 3 end cell end table close parentheses equals za subscript 1 straight i with rightwards arrow on top plus za subscript 2 straight j with rightwards arrow on top plus za subscript 3 straight k with rightwards arrow on top

Berdasarkan sifat-sifat operasi vektor tersebut, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with rightwards arrow on top minus 1 half straight b with rightwards arrow on top plus 1 half straight c with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative begin inline style 3 over 2 end style end cell row 2 end table close parentheses minus 1 half open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses plus 1 half open parentheses table row 0 row 2 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row cell negative begin inline style 3 over 2 end style end cell row 2 end table close parentheses minus open parentheses table row 1 row cell begin inline style 5 over 2 end style end cell row cell negative begin inline style 3 over 2 end style end cell end table close parentheses plus open parentheses table row 0 row 1 row cell negative begin inline style 3 over 2 end style end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus 1 end cell row cell negative begin inline style 3 over 2 end style minus begin inline style 5 over 2 end style plus 1 end cell row cell 2 plus begin inline style 3 over 2 end style minus begin inline style 3 over 2 end style end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell 3 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top end cell end table 

Dengan demikian, hasil operasi vektor pada soal adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight i with rightwards arrow on top end cell 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 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight j with rightwards arrow on top end cell 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 cell straight k with rightwards arrow on top end cell end table.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Diketahui vektor-vektor u=2i+3j​+k, v=2i+4j​+k, dan w=2i−3j​+4k. Hasil dari 4u−(3v−w) adalah ...

Pembahasan Soal:

Diketahui bahwa vektor u with rightwards arrow on top, v with rightwards arrow on top, dan w with rightwards arrow on top merupakan vektor basis left parenthesis i comma j comma k right parenthesis spaceyang berada pada dimensi tiga yang dapat diubah ke vektor baris berada pada dimensi tiga sehingga diperoleh

u with rightwards arrow on top equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top plus k with rightwards arrow on top ⟶ u with rightwards arrow on top equals open parentheses 2 , 3 comma 1 close parentheses  v with rightwards arrow on top equals 2 i with rightwards arrow on top plus 4 j with rightwards arrow on top plus k with rightwards arrow on top ⟶ v with rightwards arrow on top equals left parenthesis 2 , 4 comma 1 right parenthesis  w with rightwards arrow on top equals 2 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus 4 k with rightwards arrow on top ⟶ w with rightwards arrow on top equals open parentheses 2 comma negative 3 , 4 close parentheses

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 u with rightwards arrow on top minus open parentheses 3 v with rightwards arrow on top minus w with rightwards arrow on top close parentheses end cell equals cell 4 open parentheses 2 , 3 comma 1 close parentheses minus left parenthesis 3 left parenthesis 2 , 4 comma 1 right parenthesis minus open parentheses 2 comma negative 3 , 4 close parentheses right parenthesis end cell row blank equals cell open parentheses 8 , 12 , 4 close parentheses plus left parenthesis open parentheses 6 , 12 , 3 close parentheses minus open parentheses 2 comma negative 3 , 4 close parentheses right parenthesis end cell row blank equals cell open parentheses 8 , 12 , 4 close parentheses plus open parentheses 4 , 15 comma negative 1 close parentheses end cell row blank equals cell open parentheses 4 comma negative 3 , 5 close parentheses end cell row blank equals cell 4 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus 5 k with rightwards arrow on top end cell end table

Jadi, hasil dari a with rightwards arrow on top plus 1 half b with rightwards arrow on top minus c with rightwards arrow on top adalah 4 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus 5 k with rightwards arrow on top.

Oleh karena itu, jawaban yang benar adalah C.

1

Roboguru

Diketahui p​=3i+2j​−2k; q​=i−4j​+k dan r=13i+4j​−7k. Jika r=mp​+nq​ maka m−n=...

Pembahasan Soal:

Misalkan terdapat vektor straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top, straight b with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top dan straight z adalah skalar. maka berlaku:

open parentheses straight i close parentheses space straight a with rightwards arrow on top plus-or-minus straight b with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 plus-or-minus straight b subscript 1 end cell row cell straight a subscript 2 plus-or-minus straight b subscript 2 straight a subscript 3 plus-or-minus straight b subscript 3 end cell end table close parentheses equals open parentheses straight a subscript 1 plus-or-minus straight b subscript 1 close parentheses straight i with rightwards arrow on top plus open parentheses straight a subscript 2 plus-or-minus straight b subscript 2 close parentheses straight j with rightwards arrow on top plus open parentheses straight a subscript 3 plus-or-minus straight b subscript 3 close parentheses straight k with rightwards arrow on top left parenthesis ii right parenthesis space straight z straight a with rightwards arrow on top equals open parentheses table row cell za subscript 1 end cell row cell za subscript 2 end cell row cell za subscript 3 end cell end table close parentheses equals za subscript 1 straight i with rightwards arrow on top plus za subscript 2 straight j with rightwards arrow on top plus za subscript 3 straight k with rightwards arrow on top

Pada soal diketahui straight p with rightwards arrow on top equals 3 straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top minus 2 straight k with rightwards arrow on top equals open parentheses table row 3 row 2 row cell negative 2 end cell end table close parentheses; straight q with rightwards arrow on top equals straight i with rightwards arrow on top minus 4 straight j with rightwards arrow on top plus straight k with rightwards arrow on top equals open parentheses table row 1 row cell negative 4 end cell row 1 end table close parentheses dan straight r with rightwards arrow on top equals 13 straight i with rightwards arrow on top plus 4 straight j with rightwards arrow on top minus 7 straight k with rightwards arrow on top equals open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses. maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight r with rightwards arrow on top end cell equals cell straight m straight p with rightwards arrow on top plus straight n straight q with rightwards arrow on top end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell straight m open parentheses table row 3 row 2 row cell negative 2 end cell end table close parentheses plus straight n open parentheses table row 1 row cell negative 4 end cell row 1 end table close parentheses end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell open parentheses table row cell 3 straight m end cell row cell 2 straight m end cell row cell negative 2 straight m end cell end table close parentheses plus open parentheses table row straight n row cell negative 4 straight n end cell row straight n end table close parentheses end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell open parentheses table row cell 3 straight m plus straight n end cell row cell 2 straight m minus 4 straight n end cell row cell negative 2 straight m plus straight n end cell end table close parentheses end cell end table

Diperoleh SPLDV yaitu:

open curly brackets table attributes columnalign left end attributes row cell 3 straight m plus straight n equals 13 space... left parenthesis straight i right parenthesis end cell row cell 2 straight m minus 4 straight n equals 4 space... left parenthesis ii right parenthesis end cell row cell negative 2 straight m plus straight n equals negative 7 space... left parenthesis iii right parenthesis end cell end table close

Ambil persamaan (ii) dan (iii) sehingga didapat nilai straight n yaitu:

bottom enclose table attributes columnalign right center left columnspacing 2px end attributes row cell 2 straight m minus 4 straight n end cell equals 4 row cell negative 2 straight m plus straight n end cell equals cell negative 7 space plus end cell end table end enclose table attributes columnalign right center left columnspacing 2px end attributes row cell space space space space space space minus 3 straight n end cell equals cell negative 3 end cell row straight n equals 1 end table

Substitusi persamaan nilai straight n ke persamaan (i) sehingga didapat nilai straight m yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 straight m plus straight n end cell equals 13 row cell 3 straight m plus 1 end cell equals 13 row cell 3 straight m end cell equals cell 13 minus 1 end cell row cell 3 straight m end cell equals 12 row straight m equals 4 end table

Dengan demikian, didapat nilai straight n equals 1 dan straight m equals 4, maka pengurangan m dengan n adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight m minus straight n end cell equals cell 4 minus 1 end cell row cell straight m minus straight n end cell equals 3 end table

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

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