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Diketahui a = 2 p ​ + q ​ , b = p ​ − 4 q ​ dan c = m p ​ + ( m + n − 1 ) q ​ dengan m dan n konstanta serta vektor p ​ dan q ​ dua vektor tidak saling sejajar. Jika c = 2 a − 2 b . Nilai ( m + n ) adalah ...

Diketahui  dan  dengan  dan  konstanta serta vektor  dan  dua vektor tidak saling sejajar. Jika . Nilai  adalah ...

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N. Puspita

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nilai adalah .

nilai open parentheses m plus n close parentheses adalah 9

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Diketahui dan , maka Diketahui pula dan , maka Dari kesamaan vektor tersebut diperoleh Substitusi ke , diperoleh Sehingga Jadi, nilai adalah .

Diketahui a with rightwards arrow on top equals 2 p with rightwards arrow on top plus q with rightwards arrow on top dan b with rightwards arrow on top equals p with rightwards arrow on top minus 4 q with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards arrow on top minus 2 b with rightwards arrow on top end cell equals cell 2 open parentheses 2 p with rightwards arrow on top plus q with rightwards arrow on top close parentheses minus 2 open parentheses p with rightwards arrow on top minus 4 q with rightwards arrow on top close parentheses end cell row blank equals cell open parentheses 4 p with rightwards arrow on top plus 2 q with rightwards arrow on top close parentheses minus open parentheses 2 p with rightwards arrow on top minus 8 q with rightwards arrow on top close parentheses end cell row blank equals cell 4 p with rightwards arrow on top minus 2 p with rightwards arrow on top plus 2 q with rightwards arrow on top plus 8 q with rightwards arrow on top end cell row blank equals cell 2 p with rightwards arrow on top plus 10 q with rightwards arrow on top space end cell end table

Diketahui pula c with rightwards arrow on top equals m p with rightwards arrow on top plus open parentheses m plus n minus 1 close parentheses q with rightwards arrow on top dan c with rightwards arrow on top equals 2 a with rightwards arrow on top minus 2 b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell 2 a with rightwards arrow on top minus 2 b with rightwards arrow on top end cell row cell m p with rightwards arrow on top plus open parentheses m plus n minus 1 close parentheses q with rightwards arrow on top end cell equals cell 2 p with rightwards arrow on top plus 10 q with rightwards arrow on top space end cell row blank blank blank end table 

Dari kesamaan vektor tersebut diperoleh 

m equals 2 space dan space m plus n minus 1 equals 10 

Substitusi m equals 2 ke m plus n minus 1 equals 10, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell m plus n minus 1 end cell equals 10 row cell 2 plus n minus 1 end cell equals 10 row cell 1 plus n end cell equals 10 row n equals 9 end table 

Sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row cell m plus n end cell equals cell 2 plus 9 end cell row blank equals 11 end table 

Jadi, nilai open parentheses m plus n close parentheses adalah 9

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