Complete the statements. 1.the solution of 3tan2x − 1 = 0 if x is in the first quadrant is °. 2. the solution of 3tan2x − 1 = 0 if x is in the second quadrant is. 1. a.30 b.60 c.150 d. 240 2. a.30 b.60 c.150 d.240

1. 30

2. 150

Step-by-step explanation:

Lets assume tan(x) = u

Now we solve for 'u'

add 1 on both sides

, divide both sides by 3

Take square root on both sides

We replace tan(x) for 'u'

x = 30 because  in first quadrant

      x = 30      (tan is positive in first quadrant)

  x = 150 because  in second quadrant

     tan is negative in second quadrant

it is b)

hope this : )

step-by-step explanation:

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