Write the slop form of an equation of the line thet passes throughtl (4-3)and(2,1

y+2x-5= 0

Step-by-step explanation:

Applying,

(y-y₁) = m(x-x₁) Equation 1

Where m = slope

But,

m = (y₂-y₁)/(x₂-x₁)

Given: y₁ = -3, y₂ = 1, x₁ = 4, x₂ = 2

m = (1+3)/(2-4)

m = 4/-2

m = -2

Substituting into equation 1

(y+3) = -2(x-4)

y+3 = -2x+8

y = -2x+8-3

y = -2x+5

y = 5-2x

y+2x-5 = 0

step-by-step explanation:

if u want da anser kill your teeser ifu hav wun

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