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
-10 ≥ 2x + 14
-24 ≥ 2x [subtract 14 from both sides]
2x ≤ -24 [read right to left]
x ≤ -12 [divide both sides by 2]
Check:
Let x = -13 (It is ≥ -12)
Is this true: -10 ≥ 2(-13) + 14 ?
-10 ≥ -26 + 14 ?
-10 ≥ -14 ?yes
Let x = -11 (it is not ≥ -12)
Is this false:: -10 ≥ 2(-11) + 14 ?
-10 ≥ -22 + 14 ?
-10 ≥ -8 ?yes
³√81 × 3^⅔
= 81^⅓ × 3^⅔
= 3^⁴(⅓) × 3^⅔
= 3^1⅓ × 3^⅔
= 3²
= 9 inches²
