Q-The general solution of inequality cos 2 x≤- sin x is

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

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

the answer is: the length is 5/6 yards.

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step-by-step explanation: