f(x)=3sin^2(x)
f'(x)=3 * (sin^2(x))' = 3 * 2sin(x) * (sinx)'=3*2sinxcosx=3sin2x
A) 20-5x≥0, 5x≤20, x≤4, x∈(-∞; 4],
Ответ: D(y)=(-∞; 4]
b)10-2x≥0, 2x≤10, x≤5, x∈(-∞;5]
x+1≥0, x≥-1, x∈[-1; +∞)
(-∞;5]∩[-1; +∞)=[-1; 5]
Ответ: D(y)=[-1; 5]
(sinα - cosα)² + 2sinαcosα = sin²α - 2sinαcosα + cos²α + 2sinαcosα =
= sin²α + cos²α = 1
<span>9(3x+7)-0,8(2x-3)=4,9+0,4x
27x+63-1,6x+2,4=4,9+0,4x
25,4х+65,4=4,9+0,4х
25,4х-0,4х=4,9-65,4
25х=-60,5
х= - 2,42</span>