<span><span> <span>Akar akar persamaan kuadrat 2x² - 3x -1 = 0 adalah x1 dan x2. Persamaan kuadrat baru yang akar akarnya satu lebih kecil dari dua kali akar akar persamaan kuadrat di atas adalah ........</span></span><span><span><span>A.x² - x - 4 = 0</span><span>B.x² + 5x - 4 = 0</span><span>C.x² - x + 4 = 0</span></span><span><span>D.x² + x + 4 = 0</span><span>E.x² - 5x - 4 = 0</span></span></span><span>Jawaban : A
Penyelesaian :
Akar-akar persamaan lama : x1 dan x2
Akar-akar persamaan baru : xA dan xB
xA = 2x1 - 1
xB = 2x2 - 1
xA + xB = (2x1 - 1) + (2x2 - 1)
= 2 (x1 + x2) - 2
= 2 () - 2
= 3 - 2
xA + xB = 1
xA . xB = (2x1 - 1) (2x2 - 1)
= 4 x1.x2 - 2(x1 + x2) + 1
= 4.(-) - 2() + 1
= -2 - 3 + 1
xA . xB = -4
Jadi persamaan kuadrat baru : x² - (xA + xB)x + xA . xB = 0
x² - x - 4 = 0
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The area would be 35 cm squared.
Lets break it down:
Triangle- The area of a triangle is base×height÷2
so if we plug in the numbers, the base would be 5 (the length) and the height would be 4 (the height). 5×4= 20÷2= 10.
Square- The area of a square is base×height so if we pluged in the numbers, the base is 5 (the length) and the height is 5 (the height). 5×5= 25.
10+25= 35 cm squared
Hope this helps!
4sin²(x) = 5 - 4cos(x)
4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
2 - 2cos(2x) = 5 - 4cos(x)
- 2 - 2
-2cos(2x) = 3 - 4cos(x)
-2[2cos²(x) - 1] = 3 - 4cos(x)
-4cos²(x) + 2 = 3 - 4cos(x)
- 2 - 2
-4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
4cos²(x) - 4cos(x) + 1 = 0
[2cos(x) - 1]² = 0
2cos(x) - 1 = 0
+ 1 + 1
2cos(x) = 1
2 2
cos(x) = ¹/₂
cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
x = 60, 300
x = π/3, 5π/3
[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π
