If ef = 5w + 20, fg = 7w + 28, and eg = 72, find the value of w. find ef and fg

 w = 2, EF = 30 , FG = 42

Explanation:

We have EF = 5w + 20  and FG = 7w + 28

 Since these points E, F and G are col-linear points we have.

    EG = EF + FG

  And given that EG = 72

So, EF+EG = 72

     5w + 20 + 7w + 28 = 72

     12w + 48 = 72

      12 w = 24

       w = 2

So EF = 5w + 20 = 5x2 + 20 = 30

     FG = 7w + 28 = 7x2 + 28 = 42

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