Answer:
Step-by-step explanation:6*3 7*3 8*3 9*3
Answer:
x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I
Step-by-step explanation:
1−2sin2 x≤−sin x ⇒ (2sin x+1)(sin x−1)≥0
sin x≤−1/2 or sin x≥1
−5π/6+2nπ≤x≤−π/6+2nπ or , n ϵ I x=(4n+1)π/2, n ϵ I⇒ -5π6+2nπ≤x≤-π6+2nπ or , n ϵ I x=4n+1π2, n ϵ I (as sin x = 1 is valid only)
In general⇒ In general x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I
Answer:
-20
Step-by-step explanation:
Follow the PEDMAS order (from top to bottom):
Parentheses
Exponents
Division and Multiplication
Addition and Subtraction
(-5 × 9 - 1) ÷ 2 + (5)2 - 7
(-45 - 1) ÷ 2 + 10 - 7
-46 ÷ 2 + 10 - 7
-23 + 10 - 7
-13 - 7
-20
Subtract 1/9 - 9 to isolate the x term on the left side
J = 16 solve for j by simplifying both sides of the equation, then isolating the variable
