Given: ΔABC with Prove: Statement Reason 1. given 2. Addition Property of Equality 3. using common denominators 4. AB = AD + DB CB = CE + EB segment addition 5. Substitution Property of Equality 6. ∠ABC ≅ ∠DBE Reflexive Property of Congruence 7. ΔABC ~ ΔDBE SAS similarity criterion 8. Corresponding angles of similar triangles are congruent. 9. If the corresponding angles formed by two lines cut by a transversal are congruent, then the lines are parallel. 26 What is the missing step in this proof? A. ∠BAC ≅ ∠ACB B. ∠BDE ≅ ∠DEB C. ∠ADE ≅ ∠DEC D. ∠BAC ≅ ∠BDE E. ∠ABC ≅ ∠DBE

E no.is missing

Step-by-step explanation:

I hope I am right

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

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: