Jawaban:
1.
2. 1024
3.
4.
5. -1/2
Penjelasan dengan langkah-langkah:
= g(f(x))
= 2 - (2x/(x-1))²
= 2 - 4x²/(x²-2x+1)
= (2x²-4x+2-4x²))/(x²-2x+1)
= (-2x²-4x+2)/(x²-2x+1)
2.
T(t)= 2(5)-1
T(5)= 10-1
T(5)= 9
T= 9
Masukkan lagi:
= 2 × 2⁹
= 2 × 512
= 1024
sin²A+cos²A= 1
(2/3)²+cos²A= 1
4/9+cos²A= 1
cos²A= 5/9
Ambil negatif:
cos A= -akar 5/3.
tan A= sin A/cos A
= 2/3/-akar 5/3
= -2/akar 5
= -2/5 akar 5.
sin A= 1/5
(1/5)²+cos²A= 1
1/25+cos²A= 1
cos²A= 24/25
cos A= akar 24/5
sin 2A
= 2 sin A cos A
= 2 × 1/5 × akar 24/5
= 2 × akar 24/25
=
5.
sin a cos b - cos a sin b = sin(a-b)
sin 15° cos 45° - cos 15° sin 45°= sin (15-45)
= sin (-30°)
= sin (180-210°)
= sin 210°
= sin (180°+30°)
= -sin 30°
= -1/2
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Jawaban:
1.
2. 1024
3.
4.
5. -1/2
Penjelasan dengan langkah-langkah:
1.
= g(f(x))
= 2 - (2x/(x-1))²
= 2 - 4x²/(x²-2x+1)
= (2x²-4x+2-4x²))/(x²-2x+1)
= (-2x²-4x+2)/(x²-2x+1)
2.
T(t)= 2(5)-1
T(5)= 10-1
T(5)= 9
T= 9
Masukkan lagi:
= 2 × 2⁹
= 2 × 512
= 1024
3.
sin²A+cos²A= 1
(2/3)²+cos²A= 1
4/9+cos²A= 1
cos²A= 5/9
Ambil negatif:
cos A= -akar 5/3.
tan A= sin A/cos A
= 2/3/-akar 5/3
= -2/akar 5
= -2/5 akar 5.
4.
sin A= 1/5
sin²A+cos²A= 1
(1/5)²+cos²A= 1
1/25+cos²A= 1
cos²A= 24/25
cos A= akar 24/5
sin 2A
= 2 sin A cos A
= 2 × 1/5 × akar 24/5
= 2 × akar 24/25
=
5.
sin a cos b - cos a sin b = sin(a-b)
sin 15° cos 45° - cos 15° sin 45°= sin (15-45)
= sin (-30°)
= sin (180-210°)
= sin 210°
= sin (180°+30°)
= -sin 30°
= -1/2