Mathematics, 25.09.2020 14:01 skatingflower

Determine whether the point (3,3) is in the solution set of the system of inequalities below. y ≤ –x – 2

y ≥ –5x + 2

Answers: 2

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Answer from: aminamuhammad11

The ordered pair is not a solution of the system D

explanation:

y < -3

y ≤ 2/3x - 4

y > -3

y ≥ 2/3x - 4

y < -3

y ≥ 2/3x - 4

y > -2

y ≤ 2/3x - 4

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each case

Case A) we have

> inequality A

> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

> is not true

therefore

The ordered pair is not a solution of the system A

Case B) we have

> inequality A

> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

> is true

Inequality B

> is true

therefore

The ordered pair is a solution of the system B

Case C) we have

> inequality A

> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

> is not true

therefore

The ordered pair is not a solution of the system C

Case D) we have

y > -2 > inequality A

> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

Inequality A

> is not true

therefore

The ordered pair is not a solution of the system D

Answer from: Quest

add the table and i will answer correctly

step-by-step explanation:

Answer from: Quest

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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