Tugas Matematika.docx

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Tugas Matematika

Nama : Ida Ayu Ardia Peramesti Ningrat No

: 05

Kelas

: XI IPA 4

SMA Negeri 6 Denpasar

1. Susi berbelanja bulanan dengan membeli sabun cuci muka. Pada bulan 1 susi membeli 2 pond’s, 3 clean&clear, 4 citra seharga Rp. 125.000. pada bulan ke 2 susi membeli 3 pond’s, 4 clean&clear, 2 citra seharga Rp. 140.000. dan pada bulan ke 3 susi membeli 4 pond’s, 2 clean&clear, 4 citra seharga Rp.150.000. jika pada bulan ke 4 susi ingin membeli 1 pond’s, 1clean&clear, 1 citra berapa susi harus membayar? Selesaikan dengan matrik.

 dokumentasi

 jawaban

2

3

4

X

3

4

2

Y

2

4

z

4

Det A =

2

3

=

125.000 140.000 150.000

4

3

4

2

4

2

4

= 2 (16 - 4) – 3 ( 12 – 8 ) + 4 (6-16) = 2 ( 12 ) – 3 ( 4 ) + 4 ( -10 ) = 24 – 12 + (-40) = -28 Det X =

125

3

4

140

4

2

150

2

4

= 125 ( 16 - 4 ) – 3 ( 560 – 300 ) + 4 ( 280 – 600) = 125 ( 12 ) – 3 ( 260 ) + 4 ( -320 ) = 1500 – 780 + (-1280) = -560

Det Y =

2

125

4

3

140

2

4

150

4

= 2 ( 560 – 300 ) – 125 ( 12 – 8 ) + 4 ( 450 – 560 ) = 2 ( 260 ) – 125 ( 4 ) + 4 ( -110 ) = 520 – 500 + (-440) = -420 Det X =

2

3

125

3

4

140

4

2

150

= 2 ( 600 – 280 ) – 3 ( 450 – 560 ) + 125 ( 6- 16 ) = 2 ( 320 ) – 3 (-110) + 125 (-10) = 640 – (-330) + (-1250) = -280

X = -560. 000 = 20.000 -28 Y = -420.000 = 15.000 -28

Z = -280.000

= 10.000

-28 Jadi, 1 pond’s + 1 clean&clear + 1 citra = 20.000 + 15.000 + 10.000 = 45.000 1. Nenek mempunyai 4 orang cucu yang sedang berbelanja di super market. Cucu pertama membeli 3 tanggo, 2 nabati, 4 selamat dengan harga Rp. 94.000. cucu kedua membeli 2 tanggo, 1 nabati, 3 selamat dengan harga Rp. 63.000. dan cucu ketiga membeli 1 tanggo, 3 nabati, 5 selamat dengan harga Rp. 89.000. jika cucu keempat ingin membeli 1 tanggo, 1 nabati, 1 selamat berapa nenek harus membayar? Selesaikan dengan matrik.  Dokumentasi

 Jawaban

3

2

4

X

2

1

3

Y

1

3

5

Z

Det A =

3

2

4

2

1

3

1

3

5

94. 000 =

63.000 89. 000

=

= 3 ( 5 – 9) – 2 ( 10 – 3 ) + 4 ( 6 – 1 ) = 3 ( -4 ) – 2 ( 7 ) + 4 ( 5 ) = -12 – 14 + 20 = -6

Det X =

94

2

4

63

1

3

89

3

5

= 94 ( 5 – 9 ) – 2 ( 315 – 267 ) + 4 ( 189 – 89 ) = 94 ( -4 ) – 2 ( 48 ) + 4 ( 100 ) = -376 – 96 + 400 = -72

Det Y =

3

94

4

2

63

3

1

89

5

= 3 ( 315 – 267 ) – 94 ( 10 – 3 ) + 4 ( 178 – 63 ) = 3 ( 48 ) – 94 ( 7 ) + 4 ( 115 ) = 144 – 658 + 460 = -54

Det Z =

3

2

94

2

1

63

1

3

89

= 3 ( 89 – 189 ) – 2 ( 178 – 63 ) + 94 ( 6 – 1 ) = 3 ( -100 ) – 2 ( 115 ) + 94 ( 5 ) = -300 – 230 + 470 = -60

X = -72.000 -6

= 12.000

Y = -54.000

= 9.000

-6 Z = -60.000

= 10.000

-6

Jadi, 1 tanggo + 1 nabati + 1 selamet = 12.000 + 9.000 + 10. 000 = 31.000

2. Sebuah swalayan menawarkan 4 parcel, dimana parcel 1 berisi 2 redeo, 3 oreo, 2 slai o’lai dengan harga Rp. 54.000. Parcel 2 berisi 2 redeo, 1 oreo, 4 slai o’lai dengan harga Rp. 50.000, dan parcel 3 berisi 1 redeo, 4 oreo, 3 slai o’lai dengan harga Rp. 59.000. jika parcel ke 4 yang berisi 2 redeo, 2 oreo, 2 slai o’lai berapa harga parcel teresebut? Jelaskan dengan matrik.  Dokumentasi

 Jawaban

2 3 2

X

2 1 4

Y

50. 000

1 4 3

Z

59. 000

Det A =

2

3

2

2

1

4

1

4

3

=

54. 0000

= 2 ( 3 – 16 ) – 3 ( 6 – 4 ) + 2 ( 8 – 1 ) = 2 ( -13 ) – 3 ( 2 ) + 2 ( 7 ) =-26 – 6 + 14 = -18

Det X =

54 3

2

50 1

4

59 4

3

= 54 ( 3 – 16 ) – 3 ( 150 – 236 ) + 2 ( 200 – 59 ) = 54 ( -13 ) – 3 ( -86 ) + 2 ( 141 ) = -702 – ( -258 ) + 282 = -162

Det Y =

2 54 2 2 50 4 1 59 3

= 2 ( 150 – 236 ) – 54 ( 6 – 4 ) + 2 ( 118 – 50 ) = 2 ( -86 ) – 54 ( 2 ) + 2 ( 68 ) = -172 – 108 + 136 = -144

Det Z =

2 3

54

2 1

50

1 4

59

= 2 ( 59 – 200 ) – 3 ( 118 – 50 ) + 54 ( 8 – 1 ) = 2 ( -141 ) – 3 ( 68 ) + 54 ( 7 ) = -282 – 204 + 378 = -108

X = -162.000

= 9.000

-18 Y = -144.000

= 8.000

-18 Z = -108.000

= 6.000

-18

Jadi 2 redeo, 2 oreo, 2 slai o’lai = 2 ( 9.000 ) + 2 ( 8.000 ) + 2 ( 6.000 ) = 18.000 + 16.000 + 12.000 = 46.000

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