1.grafik fungsi R(x) = c + bx + ax^2 simetris terhadap sumbu y jika a. a= 0 b. b= 0 c. c= 0 d. b^2 - 4ac = 0 2. DIberikan K(x-2) = [tex] \frac{K(x)+3}{K(x)-1} [
Matematika
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Pertanyaan
1.grafik fungsi R(x) = c + bx + ax^2 simetris terhadap sumbu y jika
a. a= 0
b. b= 0
c. c= 0
d. b^2 - 4ac = 0
2. DIberikan K(x-2) = [tex] \frac{K(x)+3}{K(x)-1} [/tex] .
jika K(6)= 2 maka nilai dari [k(10)x K(8)].
= a.2
b.5
c.8
d.10
Jawab pake CARA
a. a= 0
b. b= 0
c. c= 0
d. b^2 - 4ac = 0
2. DIberikan K(x-2) = [tex] \frac{K(x)+3}{K(x)-1} [/tex] .
jika K(6)= 2 maka nilai dari [k(10)x K(8)].
= a.2
b.5
c.8
d.10
Jawab pake CARA
1 Jawaban
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1. Jawaban arsetpopeye
1) R(x) = c + bx + ax^2 akan simetris dengan sumbu y jika
Sumbu simetri : x = 0
=> b/(-2a) = 0
=> b = 0(-2a)
=> b = 0
2) k(6) = 2
k(x - 2) = [k(x) + 3] / [k(x) - 1]
Jika x = 8
k(8 - 2) = [k(8) + 3] / [k(8) - 1]
k(6) = [k(8) + 3] / [k(8) - 1]
2 = [k(8) + 3] / [k(8) - 1]
2[k(8) - 1] = [k(8) + 3]
2k(8) - 2 = k(8) + 3
2k(8) - k(8) = 3 + 2
k(8) = 5
x = 10
k(10 - 2) = [k(10) + 3] / [k(10) - 1]
k(8) = [k(10) + 3] / [k(10) - 1]
5 = [k(10) + 3] / [k(10) - 1]
5[k(10) - 1] = [k(10) + 3]
5k(10) - 5 = k(10) + 3
5k(10) - k(10) = 3 + 5
4k(10) = 8
k(10) = 8/4 = 2
k(10) x k(8) = 2 x 5 = 10