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Tugas matik i
- 1. Kelompok F
Ketua : ERVIANATUN NASIHAH (03)
Anggota : EKA DEVI WIDYANTARI (07)
SHERLY FELIANA L (09)
WINDA KARCANA YANTI (26)
1. p = 60 –x
l = 30 - x
Penyelesaian : (p – x) (l – x) = L
(60 – x) (30 – x) = 1000
1800 – 60x – 30x + x2 = 1000
1800 – 90x + x2 = 1000
1800 – 1000 – 90x + x2 = 0
800 – 90x + x2 = 0
x2 – 90x + 800 = 0
Jadi persamaan kuadratnya adalah x2 – 90x + 800 = 0
2.
50 cm
x
x
x
x
40 cm
200 cm
x
p = 50 – 2x
l = 40 – 2x
L=pxl
200 = (50 – 2x) (40 – 2x)
200 = 2000 – 100x – 80x + 4x2
- 2. 200 = 2000 – 180x +4x2
0 = 2000 – 200 – 180x +4x2
0 = 1800 – 180x +4x2
0 = 4x2 – 180x +1800
0 = x2 – 45x +450
0 = x2 – 30x – 15x + 450
0 = x (x – 30) – 15 (x – 30)
0 = (x – 15) (x – 30)
(x – 15) = 0 atau( x – 30) = 0
x – 15 = 0
x = 15
x – 30 = 0
x = 30
p = 50 – 2x
p = 50 – 2x
= 50 – 2.15
= 50 – 30
=50 – 2.30
atau
= 20
l = 40cm – 2x
=50 - 60
=-10
l = 40cm – 2x
= 40cm – 2.15 atau
=40 – 2.30
= 40cm – 30
=40 - 60
= 10cm
=-20
t = 50cm – 20cm : 2 atau t = 40cm – 10cm : 2
=15cm
=15cm
- 3. V=pxlxt
V = 20cm x 10cm x 15cm
V = 3000 cm3
Jadi Volume tempat air adalah 3000cm3
3. Volume mula – mula
V=
=
Volume karena penambahan jari – jari sebanyak 24
2
V1 =
V1 =
.3
(r + 24)2
Volume karena penambahan tinggi 24
V2 =
V2 =
V2 = 9 r2
V1 = V2
2
= 9
r2 + 48r + 576 = 9r2
0 = 9r2 – r2 – 48r – 576
0 = 8r2 – 48r – 576
0 = r2 – 6r – 72
0 = r2 + 6r – 12r + 72
0 = r (r + 6) – 12 (r + 6)
0 = (r – 12) (r + 6)
- 4. (r – 12) = 0 atau ( r + 6) = 0
r – 12 = 0
r+6=0
r = 12
r = -6
Jadi jari – jari kerucut semula 12cm
4. P1 = x – 1 jam
P2 = x jam
P1 + P2 = 1,2 jam
Penyelesaian : 1 : (
1:(
1:
1x
) = 1,2
+
) = 1,2
= 1,2
= 1,2
x2 – x = 2x – 1 . 1,2
x2 – x = 2,4x – 1,2
x2 = 2,4 x + x – 1,2
x2 = 3,4x – 1,2
0 = -x2 + 3,4 – 1,2
x2 – 3,4 + 1,2 = 0
10x2 – 34x + 12 = 0
5x (2x – 6) + 2(2x – 6) = 0
(5x + 2) (2x – 6) = 0
(5x + 2) = 0 atau (2x – 6) = 0
5x = -2 atau
2x = 6
x1 =
x2 =
x2 = 3
Jadi waktu yang dibutuhkan printer jenis kedua adalah 3 jam