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Hasil operasi 2(−43​)−5(32​) adalah ....

Pertanyaan

Hasil operasi 2 open parentheses table row cell negative 4 end cell row 3 end table close parentheses minus 5 open parentheses table row 3 row 2 end table close parentheses adalah ....

Pembahasan Soal:

Dengan memperhatikan aturan dan sifat operasi pada vektor, maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 open parentheses table row cell negative 4 end cell row 3 end table close parentheses minus 5 open parentheses table row 3 row 2 end table close parentheses end cell equals cell open parentheses table row cell negative 8 end cell row 6 end table close parentheses minus open parentheses table row 15 row 10 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 23 end cell row cell negative 4 end cell end table close parentheses end cell end table

Jadi hasil operasi table attributes columnalign right center left columnspacing 0px end attributes row cell 2 open parentheses table row cell negative 4 end cell row 3 end table close parentheses minus 5 open parentheses table row 3 row 2 end table close parentheses end cell equals cell open parentheses table row cell negative 23 end cell row cell negative 4 end cell end table close parentheses end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kurnia

Mahasiswa/Alumni Universitas Jember

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diketahui koordinat titik A(4,−3),B(−1,−5) dan C(−2,3). Jika a,bdanc berturut-turut vektor posisi titik A, B, dan C, hasil 2a−3b+c adalah....

Pembahasan Soal:

Diketahui titik-titik koordinat straight A left parenthesis 4 comma space minus 3 right parenthesis comma space straight B left parenthesis negative 1 comma space minus 5 right parenthesis dan straight C open parentheses negative 2 comma space 3 close parentheses. karena a with rightwards arrow on top comma space b with rightwards arrow on top space dan space c with rightwards arrow on top berturut-turut vektor posisi titik A, B, dan C, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 3 end cell end table close parentheses semicolon space b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses semicolon space c with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank blank blank row blank rightwards arrow cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top plus c with rightwards arrow on top equals 2 open parentheses table row 4 row cell negative 3 end cell end table close parentheses minus 3 open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 cross times 4 end cell row cell 2 cross times open parentheses negative 3 close parentheses end cell end table close parentheses minus open parentheses table row cell 3 cross times open parentheses negative 1 close parentheses end cell row cell 3 cross times open parentheses negative 5 close parentheses end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 8 row cell negative 6 end cell end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 15 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table    

Jadi, hasil table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table.

0

Roboguru

Diketahui koordinat titik A(4,−3),B(−1,−5),danC(−2,3). Jika a,b,danc berturut-turut vektor posisi titik A, B, dan C, hasil 2a−3b+c adalah ....

Pembahasan Soal:

Diketahui A left parenthesis 4 comma negative 3 right parenthesis comma space B left parenthesis negative 1 comma negative 5 right parenthesis comma space dan space C left parenthesis negative 2 comma 3 right parenthesis, serta a with rightwards arrow on top comma space b with rightwards arrow on top comma space dan space c with rightwards arrow on top berturut-turut vektor posisi titik A, B, dan C, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell 2 open parentheses table row 4 row cell negative 3 end cell end table close parentheses minus 3 open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 8 row cell negative 6 end cell end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 15 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Diketahui vektor d=(21​),e=(−42​), dan f​=(05​). Tentukan hasil operasi-operasi vektor berikut. 4) −3d+4e−2f​

Pembahasan Soal:

Gunakan konsep penjumlahan dan pengurangan vektor serta perkalian vektor dengan skalar.

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell negative 3 d with rightwards arrow on top plus 4 e with rightwards arrow on top minus 2 f with rightwards arrow on top end cell equals cell negative 3 open parentheses table row 2 row 1 end table close parentheses plus 4 open parentheses table row cell negative 4 end cell row 2 end table close parentheses minus 2 open parentheses table row 0 row 5 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 6 end cell row cell negative 3 end cell end table close parentheses plus open parentheses table row cell negative 16 end cell row 8 end table close parentheses minus open parentheses table row 0 row 10 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 6 plus open parentheses negative 16 close parentheses minus 0 end cell row cell negative 3 plus 8 minus 10 end cell end table close parentheses end cell row cell negative 3 d with rightwards arrow on top plus 4 e with rightwards arrow on top minus 2 f with rightwards arrow on top end cell equals cell open parentheses table row cell negative 22 end cell row cell negative 5 end cell end table close parentheses end cell end table end style 

Jadi, diperoleh hasil operasinya adalah begin mathsize 14px style negative 3 d with rightwards arrow on top plus 4 e with rightwards arrow on top minus 2 f with rightwards arrow on top equals open parentheses table row cell negative 22 end cell row cell negative 5 end cell end table close parentheses end style.

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Roboguru

Jika p​=⎝⎛​−3612​⎠⎞​,q​=⎝⎛​23−1​⎠⎞​,r=⎝⎛​4−2−6​⎠⎞​, hasil 32​p​+2q​−21​r adalah ....

Pembahasan Soal:

Diketahui, p with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 6 row 12 end table close parentheses comma space q with rightwards arrow on top equals open parentheses table row 2 row 3 row cell negative 1 end cell end table close parentheses comma space r with rightwards arrow on top equals open parentheses table row 4 row cell negative 2 end cell row cell negative 6 end cell end table close parentheses, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 over 3 p with rightwards arrow on top plus 2 q with rightwards arrow on top minus 1 half r with rightwards arrow on top end cell equals cell 2 over 3 open parentheses table row cell negative 3 end cell row 6 row 12 end table close parentheses plus 2 open parentheses table row 2 row 3 row cell negative 1 end cell end table close parentheses minus 1 half open parentheses table row 4 row cell negative 2 end cell row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 4 row 8 end table close parentheses plus open parentheses table row 4 row 6 row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row 11 row 9 end table close parentheses end cell end table

Jadi, jawaban yang tepat adalah D.

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Roboguru

Diketahui vektor c=4i+2j dan vektor e=6i−4j. Tentukan vektor 3c−e

Pembahasan Soal:

Dengan vektor kolom,

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 c with rightwards arrow on top minus e with rightwards arrow on top end cell equals cell 3 open parentheses table row 4 row 2 end table close parentheses minus open parentheses table row 6 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 12 row 6 end table close parentheses minus open parentheses table row 6 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row 10 end table close parentheses end cell end table

Jadi, diperoleh vektor 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 c 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 e 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 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank i 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 10 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank j end table.

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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