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Diketahui titik koordinat A(−4,−1), B(−6, 2), C(3, −2), dan D(7, 4). Tentukan hasil dari: a. AB+BC  b. AB+BC+CD c. AB−BC

Pertanyaan

Diketahui titik koordinat straight A open parentheses negative 4 comma negative 1 close parenthesesstraight B open parentheses negative 6 comma space 2 close parenthesesstraight C open parentheses 3 comma space minus 2 close parentheses, dan straight D open parentheses 7 comma space 4 close parentheses. Tentukan hasil dari:

a. AB with bar on top plus BC with bar on top 

b. AB with bar on top plus BC with bar on top plus CD with bar on top

c. AB with bar on top minus BC with bar on top

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell open parentheses table row 7 row cell negative 1 end cell end table close parentheses end cell end table, table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top plus CD with rightwards arrow on top end cell equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table, dan table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top minus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 11 end cell row 7 end table close parentheses end cell end table

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah sebagai berikut.

a. table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell open parentheses table row 7 row cell negative 1 end cell end table close parentheses end cell end table

b. table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top plus CD with rightwards arrow on top end cell equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table

c. table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top minus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 11 end cell row 7 end table close parentheses end cell end table

Vektor dengan titik pangkal di text A end text open parentheses x subscript a comma space y subscript a close parentheses dan titik ujung di text B end text open parentheses x subscript b comma space y subscript b close parentheses ditentukan oleh:

stack text AB end text with rightwards arrow on top equals open parentheses table row cell x subscript b minus x subscript a end cell row cell y subscript b minus y subscript a end cell end table close parentheses

Misalkan diketahui vektor a with rightwards arrow on top equals open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell x subscript b end cell row cell y subscript b end cell end table close parentheses. Penjumlahan atau pengurangan kedua vektor tersebut dapat ditentukan oleh rumus berikut.

a with rightwards arrow on top plus-or-minus b with rightwards arrow on top equals open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses plus-or-minus open parentheses table row cell x subscript b end cell row cell y subscript b end cell end table close parentheses equals open parentheses table row cell x subscript a plus-or-minus x subscript b end cell row cell y subscript a plus-or-minus y subscript b end cell end table close parentheses

a. Hasil dari penjumlahan vektor tersebut adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 6 minus open parentheses negative 4 close parentheses end cell row cell 2 minus open parentheses negative 1 close parentheses end cell end table close parentheses plus open parentheses table row cell 3 minus open parentheses negative 6 close parentheses end cell row cell negative 2 minus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 3 end table close parentheses plus open parentheses table row 9 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 plus 9 end cell row cell 3 minus 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 7 row cell negative 1 end cell end table close parentheses end cell end table

Jadi, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell open parentheses table row 7 row cell negative 1 end cell end table close parentheses end cell end table

b. Hasil dari penjumlahan vektor tersebut adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top plus CD with rightwards arrow on top end cell equals cell open parentheses table row cell negative 6 minus open parentheses negative 4 close parentheses end cell row cell 2 minus open parentheses negative 1 close parentheses end cell end table close parentheses plus open parentheses table row cell 3 minus open parentheses negative 6 close parentheses end cell row cell negative 2 minus 2 end cell end table close parentheses plus open parentheses table row cell 7 minus 3 end cell row cell 4 minus open parentheses negative 2 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 3 end table close parentheses plus open parentheses table row 9 row cell negative 4 end cell end table close parentheses plus open parentheses table row 4 row 6 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 plus 9 plus 4 end cell row cell 3 minus 4 plus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table

Jadi, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top plus CD with rightwards arrow on top end cell equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table

c. Hasil dari pengurangan vektor tersebut adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top minus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 6 minus open parentheses negative 4 close parentheses end cell row cell 2 minus open parentheses negative 1 close parentheses end cell end table close parentheses minus open parentheses table row cell 3 minus open parentheses negative 6 close parentheses end cell row cell negative 2 minus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 3 end table close parentheses minus open parentheses table row 9 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 minus 9 end cell row cell 3 minus open parentheses negative 4 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 11 end cell row 7 end table close parentheses end cell end table

Jadi, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top minus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 11 end cell row 7 end table close parentheses end cell end table

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell open parentheses table row 7 row cell negative 1 end cell end table close parentheses end cell end table, table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top plus BC with rightwards arrow on top plus CD with rightwards arrow on top end cell equals cell open parentheses table row 11 row 5 end table close parentheses end cell end table, dan table attributes columnalign right center left columnspacing 0px end attributes row cell stack text AB end text with rightwards arrow on top minus BC with rightwards arrow on top end cell equals cell open parentheses table row cell negative 11 end cell row 7 end table close parentheses end cell end table

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Pertanyaan serupa

Diketahui titik koordinat A(−4,−1), B(−6, 2), C(3, −2), dan D(7, 4). Tentukan hasil dari: d. AB−BC+CD  e. AB−BC−CD

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