Captain Ashley has a ship, the H. M.S Crimson Lynx. The ship is two furlongs from the dread pirate Umaima and her merciless band of thieves. If her ship hasn't already been hit, Captain Ashley has probability \dfrac{1}{2}
2
1

start fraction, 1, divided by, 2, end fraction of hitting the pirate ship. If her ship has been hit, Captain Ashley will always miss.
If her ship hasn't already been hit, dread pirate Umaima has probability \dfrac{1}{3}
3
1

start fraction, 1, divided by, 3, end fraction of hitting the Captain's ship. If her ship has been hit, dread pirate Umaima will always miss.
If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that both the pirate and the Captain hit each other's ships?

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

-4, -2, 1, 4

step-by-step explanation:

domain = x values

range = y values

step-by-step explanation: