Consider the function represented by 9x+3y= 12 with x as the independent variable. how can this function be written using
function notation?
o av=-=x+
o 0) = -3x+4
o px) =-x+
o f) = - 3y+ 4

f(x)=-3x+4

(can't see some of your choices)

Step-by-step explanation:

We want x to be independent means we want to write it so when we plug in numbers we can just choose what we want to plug in for x but y's value will depend on our choosing of x.

So we need to solve for y.

9x+3y=12

Subtract 9x on both sides

     3y=-9x+12

Divide both sides by 3:

     y=-3x+4

Replace y with f(x).

    f(x)=-3x+4

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

i have no idea just need points sorry

step-by-step explanation: