Solve the equation on the interval 0 less than or equals theta less than 2 pi 0≤θ<2π. 2 cosine theta plus 1 equals 0 2cosθ+1=0

θ = 2π/3 or 5π/3

Step-by-step explanation:

2 cosθ + 1= 0

when  + 1 goes to R H S Side than it becomes negative by transposition   method

2 cos = -1

now by dividing by 2 both sides we get

cosθ  =  -

since cosθ is negative

Cosθ = - cos 60°

it means either cosθ lies in 2nd or third quadrant

θ = π - π/3

or

θ = 2π- π/3

so θ = 2π/3 or 5π/3

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9

answer: 12,000

step-by-step explanation: first you ar e supposed to add. then multiply then subtract