INTEGER REAL-WORLD PROBLEMS I DONT KNOW HOW TO DO THIS PLEASE I WILL GIVE BRAINILEST

Integer | Absolute Value | Opposite

1. -45 |45| 45

2. 60 |60| -60

3. -20 |20| 20

4. -15 |15| 15

5. -125 |125| 125

6. 0 0 0

7. 37 |37| -37

8. 1287 |1287| -1287

9. -5394 |5394| 5394

10. 10 |10| -10

11. 0 0 0

12. -15 |15| 15

13. -47 |47| 47

14. 5000 |5000| -5000

15. -8 |8| 8

16. 243 |243| -243

17. -13 |13| 13

18. -571 |571| 571

19. 33 |33| -33

20. 84 |84| -84

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

10 you would take 50/ .2

step-by-step explanation: