Part 1 attempt all questions 1 given that a = {3, 5, 7,9}, b = {2,4,6,8} and c = {x: 3 < x < 9} are subsets of a universal set, u = {2, 3, 4, 5, 6, 7, 8, 9}. find: a) an (b'n c') b) (aub) n (buc)

a)  A n  (B' n C') =  {3, 9}.

b) (AUB) n (BUC) = {2,4,5,6,7,8}

Step-by-step explanation:

a)  Rewriting C ; C = {4,5,6,7,8}

B' = {3,5,7,9} and C' = {2, 3, 9}

so (B' n C') = {3, 9}

and A n  (B' n C') =  {3, 9}.

b)  {A U B} = { 2,3,4,5,6,7,8,9} and {B U C} = {2,4,5,6,7,8}

so (AUB) n (BUC) = {2,4,5,6,7,8}

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400

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