Lkm= 7x-5 nkm= 3x +9 find the angle to lkm

The measure of the angle m<LKM is 156degrees

The point where two lines meet is called angle.

From the diagram shown:

m<LKN + m<NKM = m<LKM and:

m<LKN = m<NKM

Hence m<NKM + m<NKM = m<LKM

2m<NKM = m<LKM

2(3x+9) = 7x-5

6x+18 = 7x - 5

6x-7x = -5-18

-x = -23

x = 23

Get the angle m<LKM

m<LKM = 2m<NKM

m<LKM = 2 (3x+9)

m<LKM = 6x+18

m<LKM = 6(23) + 18

m<LKM =  138+18

m<LKM = 156degrees

Hence the measure of m<LKM is 156 degrees

Learn more here: link

Given =

∠lkm = 7x - 5

∠nkm = 3x +9

find = value of∠ lkm.

To proof -  let us assume that we takeapoint say n and draw a line nk.

                nk is angle bisector it divided angle in two equal parts.

( as shown in the diagram given below)

thus we have

∠lkm = ∠lkn + ∠nkm

7x-5 = 3x+9 +3x +9

x = 23

thus the value of∠ lkm

∠lkm = 7x-5  

(put the value of x)

thus

∠lkm = 7×23 - 5

∠lkm = 156°

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