Prove the following: let a , a, b , b, and c c be sets. if a ⊆ b a⊆b and b ⊆ c , b⊆c, then a ⊆ c . a⊆c. let a a and b b be sets. then a − b = a ∩ b c a−b=a∩bc . let a , b , and c a, b, and c be sets. if ( a ⊆ b a⊆b and a ⊆ c a⊆c) then a ⊆ b ∩ c . a⊆b∩c. let a and b a and b be sets. a ⊆ b a⊆b if and only if b c ⊆ a c bc⊆ac . let a , b , and c a, b, and c be sets. if a ⊆ b a⊆b then a × c ⊆ b × c .

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

hit or miss i guess they never miss huh

step-by-step explanation:

you got a boyfriend i bet he doesnt kiss ya

0.038

step-by-step explanation: