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Matematika
tia848
Pertanyaan
tolong dijawab ya... sudah mau diperiksa hari ini!
1 Jawaban
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1. Jawaban hendrisyafa
31) y = 2 sin (3x- π/2)
m = y'
= 2 cos (3x-π/2). 3
= 6 cos (3x-π/2) --> x = π/6
= 6. cos (3.π/6 - π/2)
= 6 cos (0)
= 6
pers GS
y - b = m (x -a)
y - 0 = 6 (x - π/6)
y = 6 (x - π/6)
32) y = cos (π/2) + 2 / sin (π/2)
= 2/1
= 2 --> (π/2, 2)
m = y' --> u = cos x +2 --> u' = - sin x
v = sin x ---> v' = cos x
= u'v - v'u/ v²
= -sin x sin x - cosx (cos x+2) / sin²x
= -sin²x-cos²x-2 cos x / sin²x
= -1-2 cos x / sin²x
= -1-2 cos π/2 / sin² π/2
= -1 / 1
= -1
Pers GS
y - b = m (x - a)
y - 2 = -1 (x - π/2)
= -x + π/2
x+y - 2 - π/2 = 0
------------------------ x 2
2x+2y- 4 -π = 0
2x+2y- (4+π) = 0