Solve each equation for all values of x over the interval [0,360°]. 1) 2 sin x + 1 = 0 2) 2 sin2x – 1 = 0 3) √3 tan x + 1 = 0 4) 2 sin2x – 5 sin x + 2 = 0 5) sin2x – 2 sin x – 3 = 0 6) sin2x – 3 sin x + 2 = 0
7) 4 cos2x – 3 = 0 8) sin x = tan x 9) sin x + cos x = 0
10) sin x + sin x cos x = 0 11) tan2x – √3 tan x = 0 12) cos x = 3 cos x – 2 13) 4 sin2x – 4 sin x + 1 = 0 14) cos 2x = cos x 15) cos x sin 2x = 0
i need help

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation:

answer: 1 box of pens at the store ? ? pens in the box

i'm guessing it might be 40.