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Limit -x2/1-cosx x→0
limit 0/0 L'Hopital turunkan pembilang dan penyebutnya lim x→0 -x² / (1 - cos x) = lim x→0 -2x/(0 - (-sin x)) = -2 lim x→0 x/sin x = -2 . 1 = -2
limit 0/0
L’Hopital
turunkan pembilang dan penyebutnya
lim x→0 -x² / (1 – cos x)
= lim x→0 -2x/(0 – (-sin x))
= -2 lim x→0 x/sin x
= -2 . 1
= -2
See lessTentukan jumlah dari semua bilangan ganjil 2 digit
barisan aritmatika bilangan ganjil 2 digit 11, 13, 15, 17, ... , 99 a = 11 b = 13 - 11 = 2 n = (Un - a)/b + 1 n = (99 - 11)/2 + 1 n = 44 + 1 n = 45 Sn = n/2 (a + Un) S45 = 45/2 (11 + 99) S45 = 45/2 × 110 S45 = 45 × 55 S45 = 45(50 + 5) S45 = 2250 + 225 S45 = 2475 jumlah smRead more
barisan aritmatika
bilangan ganjil 2 digit
11, 13, 15, 17, … , 99
a = 11
b = 13 – 11 = 2
n = (Un – a)/b + 1
n = (99 – 11)/2 + 1
n = 44 + 1
n = 45
Sn = n/2 (a + Un)
S45 = 45/2 (11 + 99)
S45 = 45/2 × 110
S45 = 45 × 55
S45 = 45(50 + 5)
S45 = 2250 + 225
S45 = 2475
jumlah smua bil ganjil 2 digit = 2475
See lessIni berapa ya kak hasilnya?
eksponen ((2a⁵b) / (16a²b⁴))^-1 = (16a²b⁴) / (2a⁵b) = 16/2 . a²/a⁵ . b⁴/b = 8 . 1/a^(5 - 2) . b^(4 - 1) = 8 . 1/a³ . b³ = 8b³/a³
eksponen
((2a⁵b) / (16a²b⁴))^-1
= (16a²b⁴) / (2a⁵b)
= 16/2 . a²/a⁵ . b⁴/b
= 8 . 1/a^(5 – 2) . b^(4 – 1)
= 8 . 1/a³ . b³
= 8b³/a³
See lessBantuin kak jangan asal ya
eksponen (2x^-3 y²)³ = 2³ . (x^-3)³ . (y²)³ = 8 x^-9 . y⁶ = 8y⁶/x⁹
eksponen
(2x^-3 y²)³
= 2³ . (x^-3)³ . (y²)³
= 8 x^-9 . y⁶
= 8y⁶/x⁹
See lessKak bantuin 1 soal aja
eksponen (a/b)^-1 = b/a (-2/3k)³ × (-2/3k)^-4 = (-2/3k)^(3 + (-4)) = (-2/3k)^-1 = -3k/2
eksponen
(a/b)^-1 = b/a
(-2/3k)³ × (-2/3k)^-4
= (-2/3k)^(3 + (-4))
= (-2/3k)^-1
= -3k/2
See lessQuizzzz Match
dimensi tiga soal satu kubus r = 6 cm O tengah BE P tengah BF ∆BEG sama sisi Jarak G ke BE = jarak G ke tengah BE = GO = √(GF² + FP² + PO²) = √(6² + 3² + 3²) = √54 = 3√6 cm soal dua kubus r = 3 cm ∆AEG X pada AG EX ⊥ AG Jarak E ke AG = EX = AE × EG / AG = rRead more
dimensi tiga
soal satu
kubus
r = 6 cm
O tengah BE
P tengah BF
∆BEG sama sisi
Jarak G ke BE
= jarak G ke tengah BE
= GO
= √(GF² + FP² + PO²)
= √(6² + 3² + 3²)
= √54
= 3√6 cm
soal dua
kubus
r = 3 cm
∆AEG
X pada AG
EX ⊥ AG
Jarak E ke AG
= EX
= AE × EG / AG
= r × r√2 / r√3
= 1/3 r√6
= 1/3 × 3√6
= √6 cm
See lessTentukan nilai x yang memenuhi persamaan berikut: |x+1|²-5|x+1|=-6.
nilai mutlak |x + 1|² - 5|x + 1| = -6 a² - 5a + 6 = 0 (a - 2)(a - 3) = 0 a - 2 = 0 → a = 2 atau a - 3 = 0 → a = 3 a = 2 |x + 1| = 2 x + 1 = 2 x = 1 atau x + 1 = -2 x = -3 a = 3 |x + 1| = 3 x + 1 = 3 x = 2 x + 1 = -3 x = -4 Nilai x yang memenuhi : -4 , -3 , 1Read more
nilai mutlak
|x + 1|² – 5|x + 1| = -6
a² – 5a + 6 = 0
(a – 2)(a – 3) = 0
a – 2 = 0 → a = 2
atau
a – 3 = 0 → a = 3
a = 2
|x + 1| = 2
x + 1 = 2
x = 1
atau
x + 1 = -2
x = -3
a = 3
|x + 1| = 3
x + 1 = 3
x = 2
x + 1 = -3
x = -4
Nilai x yang memenuhi :
-4 , -3 , 1 , 2
See less