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Virza Azzahra
Student XII IPS
8. Tentukan titik beloknya fungsi y = 8Sin(2x – 60) + 12 pada selang (0°. 90°) =
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Putra Gema Nusa
Rockstar Teacher GURU SMA
= y = 8 s10 (2x – 60°) Il 1.09 – x2) se 91 -> X 9 - 8 số (2x - 6ơ) + 2 (0 , 90°) Ditanya: title belok # 2x – 60° = 0° tk. 360° # 2X-60° = 180° + 1.360° = Sin 2x 60° tl.360° 2x = 240° + k. 360° y' - 2.8 cos (2x–600) x = 30° +le. 180° X = 120* t l. 180° y's .sk = 0,X= 30° Ф y" 2.16 (- sin (2x-60*)) # y =8 sin (2x - 60°) + 12 X = 30° y" – 32 sin (2x – 60") = 8 sin (2-36-60°) + 12 # titile belole terjadi ketika y" =0 8 sin (60%-60°) + 12 – 32 Sin (2x – 60°) = 0 8 sin 0° +12 sin (2x -60%) = 0 = 8.0 +12 & titile belole sin (2x - 60°) = sin 0° °) = = x s = 12 (x,y)= (30", 12) Dahamify
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