Roboguru

Diketahui titik-titik ,  dan . wakil dari , wakil dari . Kosinus sudut antara vektor dan vektor  adalah ...

Pertanyaan

Diketahui titik-titik straight A left parenthesis 2 comma space 4 comma space 1 right parenthesisstraight B left parenthesis 4 comma space 6 comma space 1 right parenthesis dan straight C left parenthesis 3 comma space 5 comma space 5 right parenthesis. AB with rightwards arrow on top wakil dari straight u with rightwards arrow on top, AC with rightwards arrow on top wakil dari straight v with rightwards arrow on top. Kosinus sudut antara vektor straight u with rightwards arrow on top dan vektor straight v with rightwards arrow on top adalah ...

  1. 1 half

  2. 1 third

  3. 1 fourth

  4. 1 fifth

  5. 1 over 6

Pembahasan Soal:

Misalkan terdapat dua buah titik straight A open parentheses x subscript 1 comma space y subscript 1 comma space z subscript 1 close parentheses dan straight B open parentheses x subscript 2 comma space y subscript 2 comma space z subscript 2 close parentheses. Vektor AB dapat didefinisikan sebagai:

AB with bar on top equals straight B minus straight A equals open parentheses table row cell straight x subscript 2 minus straight x subscript 1 end cell row cell straight y subscript 2 minus straight y subscript 1 end cell row cell straight z subscript 2 minus straight z subscript 1 end cell end table close parentheses

Sedangkan panjang vektor AB with rightwards arrow on top adalah:

open vertical bar AB with rightwards arrow on top close vertical bar equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared plus open parentheses z subscript 2 minus z subscript 1 close parentheses squared end root.

Lalu ingat juga rumus dot product antara dua vektor yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar end fraction end cell end table.

Pada soal diketahui straight A left parenthesis 2 comma space 4 comma space 1 right parenthesisstraight B left parenthesis 4 comma space 6 comma space 1 right parenthesis dan straight C left parenthesis 3 comma space 5 comma space 5 right parenthesis. AB with rightwards arrow on top wakil dari straight u with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B minus straight A end cell row cell straight u with rightwards arrow on top end cell equals cell open parentheses table row 4 row 6 row 1 end table close parentheses minus open parentheses table row 2 row 4 row 1 end table close parentheses end cell row cell straight u with rightwards arrow on top end cell equals cell open parentheses table row 2 row 2 row 0 end table close parentheses end cell end table

Panjang vektor straight u with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight u with rightwards arrow on top close vertical bar end cell equals cell square root of 2 squared plus 2 squared plus 0 squared end root end cell row blank equals cell square root of 4 plus 4 plus 0 end root end cell row blank equals cell square root of 8 end cell row blank equals cell 2 square root of 2 end cell end table

Sedangkan jika AC with rightwards arrow on top wakil dari straight v with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top end cell equals cell straight C minus straight A end cell row cell straight v with rightwards arrow on top end cell equals cell open parentheses table row 3 row 5 row 5 end table close parentheses minus open parentheses table row 2 row 4 row 1 end table close parentheses end cell row cell straight v with rightwards arrow on top end cell equals cell open parentheses table row 1 row 1 row 4 end table close parentheses end cell end table

Panjang vektor straight v with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight v with rightwards arrow on top close vertical bar end cell equals cell square root of 1 squared plus 1 squared plus 4 squared end root end cell row blank equals cell square root of 1 plus 1 plus 16 end root end cell row blank equals cell square root of 18 end cell row blank equals cell 3 square root of 2 end cell end table

Selanjutnya, kosinus sudut antara vektor straight u with rightwards arrow on top dan vektor straight v with rightwards arrow on top dapat dicari sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell fraction numerator straight u with rightwards arrow on top times straight v with rightwards arrow on top over denominator open vertical bar straight u with rightwards arrow on top close vertical bar open vertical bar straight v with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses 2 comma space 2 comma space 0 close parentheses times open parentheses 1 comma space 1 comma space 4 close parentheses over denominator 2 square root of 2 open parentheses 3 square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 2 open parentheses 1 close parentheses plus 2 open parentheses 1 close parentheses plus 0 open parentheses 4 close parentheses over denominator 6 open parentheses 2 close parentheses end fraction end cell row blank equals cell 4 over 12 end cell row blank equals cell 1 third end cell end table

Dengan demikian, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

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.

Roboguru

Diketahui segitiga ABC dengan koordinat dan maka nilai Cosinus sudut antara sisi AB dan AC adalah ...

Pembahasan Soal:

Vektor AB dapat ditentukan dengan cara berikut.

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 space space space space space end cell equals cell open parentheses table row 0 row cell negative 1 end cell row cell negative 1 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses end cell row cell space space space space space end cell equals cell open parentheses table row cell negative 2 end cell row 0 row cell negative 1 end cell end table close parentheses end cell end table 

Panjang vektor AB adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar stack A B with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 2 close parentheses squared plus 0 squared plus open parentheses negative 1 close parentheses squared end root end cell row blank equals cell square root of 4 plus 0 plus 1 end root end cell row blank equals cell square root of 5 end cell end table 

Vektor AC dapat ditentukan dengan cara berikut.

stack A C with rightwards arrow on top equals c with rightwards arrow on top minus a with rightwards arrow on top space space space space space equals open parentheses table row 1 row 1 row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses space space space space space equals open parentheses table row cell negative 1 end cell row 2 row cell negative 2 end cell end table close parentheses 

Panjang vektor AC adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar stack A C with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 1 close parentheses squared plus 2 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 4 plus 4 end root end cell row blank equals 3 end table 

Nilai cos antara sisi AB dan AC ditentukan dengan cara berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator stack A B space with rightwards arrow on top times stack A C with rightwards arrow on top over denominator open vertical bar stack A B space 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 cell negative 2 end cell row 0 row cell negative 1 end cell end table close parentheses open parentheses table row cell negative 1 end cell row 2 row cell negative 2 end cell end table close parentheses over denominator square root of 5 times 3 end fraction end cell row blank equals cell fraction numerator 2 plus 0 plus 2 over denominator 3 square root of 5 end fraction end cell row blank equals cell fraction numerator 4 over denominator 3 square root of 5 end fraction times fraction numerator square root of 5 over denominator square root of 5 end fraction end cell row blank equals cell fraction numerator 4 square root of 5 over denominator 15 end fraction end cell end table 

Dengan demikian, nilai Cosinus sudut antara sisi AB dan AC adalah  fraction numerator 4 square root of 5 over denominator 15 end fraction.

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.

 

Roboguru

Diketahui vektor  dan vektor . Kosinus sudut antara vektor  dan vektor  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

Asumsi soal: vektor b with rightwards arrow on top equals 2 i with rightwards arrow on top plus 8 j with rightwards arrow on top minus 2 k with rightwards arrow on top

Berdasarkan konsep di atas, dapat ditentukan panjang vektor a with rightwards arrow on top dan b with rightwards arrow on top sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 4 close parentheses squared plus 0 squared plus 4 squared end root end cell row blank equals cell square root of 16 plus 16 end root end cell row blank equals cell square root of 32 end cell row blank equals cell 4 square root of 2 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 2 squared plus 8 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 4 plus 64 plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 72 end cell row blank equals cell 6 square root of 2 end cell end table

Hasil operasi perkalian kedua vektor, yaitu

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 negative 4 end cell row 0 row 4 end table close parentheses open parentheses table row 2 row 8 row cell negative 2 end cell end table close parentheses end cell row blank equals cell negative 4 times 2 plus 0 times 8 plus 4 times open parentheses negative 2 close parentheses end cell row blank equals cell negative 8 plus 0 minus 8 end cell row blank equals cell negative 16 end cell end table

Kosinus sudut antara vektor stack text a end text with rightwards arrow on top dan vektor stack text b end text with rightwards arrow on top 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 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 cos space alpha end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row cell cos space alpha end cell equals cell fraction numerator negative 16 over denominator 4 square root of 2 times 6 square root of 2 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator negative 16 over denominator 48 end fraction end cell row cell cos space alpha end cell equals cell negative 1 third end cell end table

Oleh karena itu, jawaban yang tepat adalah D.

Roboguru

Jika sudut antara vektor  dan vektor  adalah , maka  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

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

Panjang vektor a with rightwards arrow on top dan b with rightwards arrow on top pada soal di atas adalah sebagai berikut.

open vertical bar a with rightwards arrow on top close vertical bar equals square root of 7 squared plus open parentheses square root of p close parentheses squared plus 4 squared end root equals square root of p plus 65 end root

open vertical bar b with rightwards arrow on top close vertical bar equals square root of 4 squared plus open parentheses square root of p close parentheses squared plus open parentheses negative 2 close parentheses squared end root equals square root of p plus 20 end root

Hasil perkalian vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah 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 cell open parentheses table row 7 row cell square root of p end cell row 4 end table close parentheses open parentheses table row 4 row cell square root of p end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell 7 times 4 plus square root of p times square root of p plus 4 times open parentheses negative 2 close parentheses end cell row blank equals cell 28 plus p minus 8 end cell row blank equals cell p plus 20 end cell end table

Nilai p 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 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 p plus 20 end cell equals cell square root of p plus 65 end root times square root of p plus 20 end root times cos space 45 degree end cell row cell p plus 20 end cell equals cell square root of open parentheses p plus 65 close parentheses open parentheses p plus 20 close parentheses end root times 1 half square root of 2 end cell row cell 2 p plus 40 end cell equals cell square root of p squared plus 85 p plus 1300 end root times square root of 2 end cell row cell 4 p squared plus 160 p plus 1600 end cell equals cell 2 open parentheses p squared plus 85 p plus 1300 close parentheses end cell row cell 2 p squared plus 80 p plus 800 end cell equals cell p squared plus 85 p plus 1300 end cell row cell p squared minus 5 p minus 500 end cell equals 0 row cell open parentheses p plus 20 close parentheses open parentheses p minus 25 close parentheses end cell equals 0 end table

p equals negative 20 space text atau end text space p equals 25

Oleh karena itu, jawaban yang tepat adalah C.

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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