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Sistem persamaan linier dan kuadrat y= 5x -4 y=x²+6-16 tentukan himpunan penyelesaian (hp) dari splk di atas dan kemudian gambarlah grafiknya
SPLK y = y x² + 6x - 16 = 5x - 4 x² + 6x - 16 - 5x + 4 = 0 x² + x - 12 = 0 (x + 4)(x - 3) = 0 x = -4 atau x = 3 x = -4 y = 5x - 4 = 5(-4) - 4 = -24 (-4 , -24) x = 3 y = 5.3 - 4 = 11 (3,11) HP = {(-4,-24) , (3,11)
SPLK
y = y
x² + 6x – 16 = 5x – 4
x² + 6x – 16 – 5x + 4 = 0
x² + x – 12 = 0
(x + 4)(x – 3) = 0
x = -4 atau x = 3
x = –4
y = 5x – 4 = 5(-4) – 4 = -24
(–4 , –24)
x = 3
y = 5.3 – 4 = 11
(3,11)
HP = {(–4,–24) , (3,11)
See lessSederhana dari 3/√3-6√2
aljabar = 3(√3 + 6√2) / ((√3)² - (6√2)²) = 3(√3 + 6√2) / (3 - 72) = -3/69 (√3 + 6√2) = -(√3 + 6√2)/23
aljabar
= 3(√3 + 6√2) / ((√3)² – (6√2)²)
= 3(√3 + 6√2) / (3 – 72)
= –3/69 (√3 + 6√2)
= –(√3 + 6√2)/23
See lessTentukan daerah himpunan penyelesaian pertidaksamaan berikut: 3x+6y<_12
pertidaksamaan 3x + 6y ≤ 12 kedua ruas bagi (3) x + 2y ≤ 4 x + 2y = 4 • titik potong dg sumbu x → y = 0 x + 2.0 = 4 x = 4 tipot A(4,0) • titik potong dg sumbu y → x = 0 0 + 2y = 4 y = 2 tipot B(0,2) buat garis melalui A dan B arah arsiran → uji titik (0,0) x + 2y .Read more
pertidaksamaan
3x + 6y ≤ 12
kedua ruas bagi (3)
x + 2y ≤ 4
x + 2y = 4
• titik potong dg sumbu x → y = 0
x + 2.0 = 4
x = 4
tipot A(4,0)
• titik potong dg sumbu y → x = 0
0 + 2y = 4
y = 2
tipot B(0,2)
buat garis melalui A dan B
arah arsiran → uji titik (0,0)
x + 2y … 4
0 + 2.0 … 4
0 < 4 memenuhi
arah arsiran memuat titik (0,0)
Lihat gambar
daerah arsiran = daerah himpunan penyelesaian
See lessSederhanakanlah bentuk logaritme berikut( gunakan sifat sifat logaritme) log3+log5
logaritma log a + log b = log (a × b) log 3 + log 5 = log (3 × 5) = log 15
logaritma
log a + log b = log (a × b)
log 3 + log 5
= log (3 × 5)
= log 15
See lessTentukan himpunan penyelesaian dari persamaan nilaik mutlak -3|5-x|= -6
nilai mutlak -3 |5 - x| = -6 |5 - x| = -6/-3 = 2 5 - x = 2 → x = 3 atau 5 - x = -2 → x = 7 HP = {3 , 7}
nilai mutlak
-3 |5 – x| = -6
|5 – x| = -6/-3 = 2
5 – x = 2 → x = 3
atau
5 – x = -2 → x = 7
HP = {3 , 7}
See lessMohon bantuannya beserta diisi cara terima kasih
Dimensi Tiga Kubus r = 12 cm P tengah DH O tengah AC AC diwakili bidang ACP OP sejajar BH X pada BH PX ⊥ BH ∆BPH PH = 1/2 r = 6 cm BH = r√3 = 12√3 cm DB = r√2 = 12√2 cm Jarak AC dan BH = jarak OP dan BH = PX = HP × DB / HB = 6 × 12√2 / 12√3 = 2√6 cm
Dimensi Tiga
Kubus
r = 12 cm
P tengah DH
O tengah AC
AC diwakili bidang ACP
OP sejajar BH
X pada BH
PX ⊥ BH
∆BPH
PH = 1/2 r = 6 cm
BH = r√3 = 12√3 cm
DB = r√2 = 12√2 cm
Jarak AC dan BH
= jarak OP dan BH
= PX
= HP × DB / HB
= 6 × 12√2 / 12√3
= 2√6 cm
See lessMohon bantuannya karena besok akan dikumpul. terima kasih
Dimensi Tiga Kubus r = 12 cm BG = √(BC² + CG²) = r√2 = 12√2 cm P tengah BG DF diwakili bidang CDEF AB diwakili bidang ABGH ABGH ⊥ CDEF BP ⊥ CF Jarak AB ke DG = BP = 1/2 BG = 1/2 × 12√2 = 6√2 cm
Dimensi Tiga
Kubus
r = 12 cm
BG = √(BC² + CG²) = r√2 = 12√2 cm
P tengah BG
DF diwakili bidang CDEF
AB diwakili bidang ABGH
ABGH ⊥ CDEF
BP ⊥ CF
Jarak AB ke DG
= BP
= 1/2 BG
= 1/2 × 12√2
= 6√2 cm
See less