Given: measure 1, measure 2, measure 3, and measure 4 formed by two intersecting segments Prove: measure 1 and measure 3 are congruent. NEED HELP ASAP!!

Step-by-step explanation:

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Given that ∠1, ∠2, ∠3, and ∠4 are formed by two intersecting segments, and that:

∠1 and  ∠2 form a linear pair  

∠2 and ∠3 form a linear pair  

These linear pair relationships suggest that by the Linear Pair Postulate:

∠1 and ∠2 are supplementary.    ⇒  Linear Pair Postulate

∠2 and ∠3 are supplementary.    ⇒  Linear Pair Postulate

m∠2 + m∠ 3 = 180°  ⇒  Definition of supplementary angles.

m∠1 + m∠3  = m∠2 + m∠ 3  ⇒  Substitution Property of Equality.          

m∠1  =  m∠3   ⇒  Subtraction Property of Equality.  

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

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