Roboguru

Diketahui titik ,  dan . Agar vektor posisi dari  tegak lurus pada vektor posisi dari  dan vektor posisi dari , maka  adalah ...

Pertanyaan

Diketahui titik text A end text open parentheses negative 1 comma space 2 comma space 1 close parenthesestext B end text open parentheses 2 comma space minus 2 comma space 2 close parentheses dan text C end text open parentheses x comma space y comma space z close parentheses. Agar vektor posisi dari text C end text tegak lurus pada vektor posisi dari text A end text dan vektor posisi dari text B end text, maka text C end text adalah ...

  1. open parentheses 4 comma space 3 comma space 1 close parentheses 

  2. open parentheses 0 comma space 1 comma space minus 1 close parentheses 

  3. open parentheses 4 comma space 3 comma space minus 2 close parentheses 

  4. open parentheses 1 comma space 2 comma space 0 close parentheses 

  5. open parentheses 3 comma space 0 comma space minus 2 close parentheses 

Pembahasan Soal:

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Jika vektor a with rightwards arrow on top tegak lurus dengan vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space 90 degree end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Misal vektor posisi dari text C end text open parentheses x comma space y comma space z close parentheses.

Vektor posisi c with rightwards arrow on top tegak lurus a with rightwards arrow on top sehingga diperoleh persamaan (1) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times a with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals 0 row cell negative x plus 2 y plus z end cell equals 0 end table

Vektor posisi c with rightwards arrow on top tegak lurus b with rightwards arrow on top sehingga diperoleh persamaan (2) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row 2 row cell negative 2 end cell row 2 end table close parentheses end cell equals 0 row cell 2 x minus 2 y plus 2 z end cell equals 0 end table

Dari persamaan (1) dan (2) diperoleh persamaan (3) berikut.

table row cell negative x plus 2 y plus z end cell equals cell 0 space space space space end cell row cell 2 x minus 2 y plus 2 z end cell equals cell 0 space plus end cell row cell x plus 3 z end cell equals cell 0 space space space space end cell end table

Diperoleh x equals negative 3 z sehingga dengan metode sbstitusi dapat ditentukan nilai y berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 2 y plus z end cell equals 0 row cell negative open parentheses negative 3 z close parentheses plus 2 y plus z end cell equals 0 row cell 4 z plus 2 y end cell equals 0 row cell 2 y end cell equals cell negative 4 z end cell row y equals cell negative 2 z end cell end table

Dapat ditentukan perbandingan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x colon y colon z end cell equals cell negative 3 z colon negative 2 z colon z end cell row blank equals cell 3 colon 2 colon negative 1 end cell end table

Diperoleh text C end text open parentheses 3 comma space 2 comma space 1 close parentheses 

Oleh karena itu, tidak ada jawaban yang tepat.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui segitiga  dengan ,  dan . Besar sudut  adalah ...

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Jika vektor a with rightwards arrow on top dan b with rightwards arrow on top membentuk sudut alpha, maka 

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha

Komponen dari vektor stack A B with rightwards arrow on top dan stack A C with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B end text minus text A end text end cell row blank equals cell open parentheses table row 7 row 6 row 5 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row 5 row 7 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell text C-A end text end cell row blank equals cell open parentheses table row 1 row 6 row 2 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses end cell end table

Berdasarkan konsep di atas, dapat ditentukan besar sudut text BAC end text sebagai berikut.

Misal: alpha equals angle text BAC end text

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top times stack A C with rightwards arrow on top end cell equals cell open vertical bar stack A B with rightwards arrow on top close vertical bar open vertical bar stack A C with rightwards arrow on top close vertical bar space cos space alpha end cell row cell cos space alpha end cell equals cell fraction numerator stack A B with rightwards arrow on top times stack A C with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar times open vertical bar stack A C with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 4 row 5 row 7 end table close parentheses open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses over denominator square root of 4 squared plus 5 squared plus 7 squared end root times square root of open parentheses negative 2 close parentheses squared plus 5 squared plus 4 squared end root end fraction end cell row blank equals cell fraction numerator 4 times open parentheses negative 2 close parentheses plus 5 times 5 plus 7 times 4 over denominator square root of 16 plus 25 plus 49 end root times square root of 4 plus 25 plus 16 end root end fraction end cell row blank equals cell fraction numerator negative 8 plus 25 plus 28 over denominator square root of 90 times square root of 45 end fraction end cell row blank equals cell fraction numerator 45 over denominator 3 square root of 10 times 3 square root of 5 end fraction end cell row blank equals cell fraction numerator 45 over denominator 45 square root of 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell 1 half square root of 2 end cell end table

Diperoleh nilai cos space alpha equals 1 half square root of 2 sehingga alpha equals 45 degree 

Oleh karena itu, jawaban yang tepat adalah B.

0

Roboguru

Diketahui , , dan .  dan  berturut-turut mewakili vektor  dan . Tentukan: d.

Pembahasan Soal:

Jika a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka a with rightwards arrow on top times b with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses times open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses end cell row blank equals cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell end table 

Vektor Posisi adalah vektor yang berpangkal di pusat koordinat open parentheses 0 comma space 0 close parentheses dan berujung di suatu titik open parentheses x comma space y close parentheses.

Diketahui straight A open parentheses 2 comma space 6 comma space minus 5 close parentheses, straight B open parentheses negative 1 comma space 3 comma space 7 close parentheses, dan straight C open parentheses 4 comma space minus 1 comma space 8 close parentheses.

Nilai vektor posisi akan sama dengan koordinat titik ujungnya, maka tentukan AB with rightwards arrow on top dan BC with rightwards arrow on top.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell b with rightwards arrow on top minus a with rightwards arrow on top end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses minus open parentheses table row 2 row 6 row cell negative 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 minus 2 end cell row cell 3 minus 6 end cell row cell 7 plus 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses end cell row cell BC with rightwards arrow on top end cell equals cell c with rightwards arrow on top minus b with rightwards arrow on top end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 1 end cell row 8 end table close parentheses minus open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row cell 4 plus 1 end cell row cell negative 1 minus 3 end cell row cell 8 minus 7 end cell end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses end cell end table   

d. open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses u with rightwards arrow on top close parentheses squared minus 12 open parentheses u with rightwards arrow on top times v with rightwards arrow on top close parentheses plus 9 open parentheses v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses squared minus 12 open parentheses open parentheses table row cell negative 3 end cell row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses close parentheses plus 9 open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses squared end cell row blank equals cell 4 open parentheses open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared plus 12 squared close parentheses minus 12 open parentheses negative 15 plus 12 plus 12 close parentheses plus 9 open parentheses 5 squared plus open parentheses negative 4 close parentheses squared plus 1 squared close parentheses end cell row blank equals cell 4 times 162 minus 12 times 9 plus 9 times 42 end cell row blank equals cell 648 minus 108 plus 378 end cell row blank equals 918 end table end style 

Jadi, Error converting from MathML to accessible text..

0

Roboguru

Jika  dan  maka besar sudut yang dibentuk oleh vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali  penjumlahan vektor, panjang vektor dan besar sudut pada vektor berikut.

  • Jika top enclose a equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan top enclose b equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka

top enclose a plus top enclose b equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis k

top enclose a minus top enclose b equals left parenthesis a subscript 1 minus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 minus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 minus b subscript 3 right parenthesis k 

  • Jika theta adalah sudut antara vektor top enclose a space dan space top enclose b, maka cos space theta equals fraction numerator a bullet b over denominator open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar end fraction.

 

  • open vertical bar top enclose a close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis top enclose a plus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 plus 1 right parenthesis i plus left parenthesis 1 plus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 plus 2 right parenthesis k end cell row blank equals cell 0 i plus 0 j plus 4 k end cell row cell left parenthesis top enclose a minus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 minus 1 right parenthesis i plus left parenthesis 1 minus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 minus 2 right parenthesis k end cell row blank equals cell negative 2 i plus 2 j plus 0 k end cell row blank blank blank end table 

Sehingga besar sudut  yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator left parenthesis top enclose a plus top enclose b right parenthesis bullet left parenthesis top enclose a minus top enclose b right parenthesis over denominator open vertical bar top enclose a plus top enclose b close vertical bar times open vertical bar top enclose a minus top enclose b close vertical bar end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 i plus 0 j plus 4 k right parenthesis bullet left parenthesis negative 2 i plus 2 j plus 0 k right parenthesis over denominator square root of 0 squared plus 0 squared plus 4 squared end root cross times square root of left parenthesis negative 2 right parenthesis squared plus 2 squared plus 0 squared end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 cross times 0 right parenthesis plus left parenthesis 0 cross times 2 right parenthesis plus left parenthesis 4 cross times 0 right parenthesis over denominator square root of 0 plus 0 plus 16 end root cross times square root of 4 plus 4 plus end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 0 over denominator square root of 16 cross times square root of 8 end fraction end cell row cell cos space theta end cell equals 0 row cell cos space theta end cell equals cell cos space 90 degree end cell row theta equals cell 90 degree end cell end table 

Jadi, besar sudut yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis  adalah 90 degree.

Oleh karena itu, jawaban yang benar adalah D.

 

0

Roboguru

Diketahui titik-titik ,  dan . Jika titik  merupakan proyeksi titik  pada garis  maka panjang  sama dengan ...

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Panjang proyeksi vektor stack text a end text with rightwards arrow on top pada vektor stack text b end text with rightwards arrow on top dapat ditentukan menggunakan rumus berikut.

open vertical bar c with rightwards arrow on top close vertical bar equals open vertical bar fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction close vertical bar

Komponen vektor stack A B with rightwards arrow on top dan stack A C with rightwards arrow on top pada soal di atas dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B-A end text end cell row blank equals cell open parentheses table row 10 row 1 row 3 end table close parentheses minus open parentheses table row 2 row 1 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 8 row 0 row 6 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell text C-A end text end cell row blank equals cell open parentheses table row 8 row 4 row cell negative 1 end cell end table close parentheses minus open parentheses table row 2 row 1 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row 3 row 2 end table close parentheses end cell end table

Berdasarkan konsep di atas, panjang proyeksi vektor stack A C with rightwards arrow on top pada vektor stack A B with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar text AD end text close vertical bar end cell equals cell open vertical bar fraction numerator stack A C with rightwards arrow on top times stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses table row 6 row 3 row 2 end table close parentheses open parentheses table row 8 row 0 row 6 end table close parentheses over denominator square root of 8 squared plus 0 squared plus 6 squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 6 times 8 plus 3 times 0 plus 2 times 6 over denominator square root of 100 end fraction close vertical bar end cell row blank equals cell open vertical bar 60 over 10 close vertical bar end cell row blank equals 6 end table

Diperoleh panjang text AD=6 end text 

Oleh karena itu, jawaban yang tepat adalah C.

0

Roboguru

Diketahui vektor  tegak lurus dengan vektor . Jika  maka

Pembahasan Soal:

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Jika vektor a with rightwards arrow on top tegak lurus vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Berdasarkan konsep di atas, nilai x dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row 2 row 3 row x end table close parentheses open parentheses table row 3 row cell negative 6 end cell row 2 end table close parentheses end cell equals 0 row cell 2 times 3 plus 3 open parentheses negative 6 close parentheses plus 2 x end cell equals 0 row cell 6 minus 18 plus 2 x end cell equals 0 row cell negative 12 plus 2 x end cell equals 0 row cell 2 x end cell equals 12 row x equals 6 end table

Diperoleh vektor x equals 6 sehingga a with rightwards arrow on top equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top plus 6 k with rightwards arrow on top dan c with rightwards arrow on top equals 6 i with rightwards arrow on top minus 5 j with rightwards arrow on top plus k with rightwards arrow on top. Dapat ditentukan operasi penjumlahan dan pengurangan vektor berikut.

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

Diperoleh table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a 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 b 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 cell c with rightwards arrow on top 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 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 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell k with rightwards arrow on top end cell end table

Oleh karena itu, jawaban yang tepat 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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