Which ordered pairs are solutions to the inequality y−2x≤−3?

select each correct answer.

(5, −3)(5, −3)

(0, −2)(0, −2)

(−6, −3)(−6, −3)

(1, −1)(1, −1)

(7, 12)

we will proceed to resolve each case to determine the solution

we have

we know that

If an ordered pair is the solution of the inequality, then it must satisfy the inequality.

case a)

Substitute the value of x and y in the inequality

-------> is true

so

The ordered pair is a solution

case b)

Substitute the value of x and y in the inequality

-------> is False

so

The ordered pair is not a solution

case c)

Substitute the value of x and y in the inequality

-------> is False

so

The ordered pair is not a solution

case d)

Substitute the value of x and y in the inequality

-------> is True

so

The ordered pair is a solution

case e)

Substitute the value of x and y in the inequality

-------> is False

so

The ordered pair is not a solution

Verify

using a graphing tool

see the attached figure

the solution is the shaded  area below the line

The points A and D lies on the shaded area, therefore the ordered pairs A and D are solution of the inequality

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

it has the same slope and a different y-intercept

hope this : )