What pattern do you notice in the numbers below? 1 = 1 × 1, and 1 × 1 × 1, and 1 × 1 × 1 64 = 8 × 8, and 4 × 4 × 4 729 = 27 × 27, and 9 × 9 × 9 4,096 = 64 × 64, and 16 × 16 × 16 a. they are perfect squares. b. they are perfect cubes. c. they are both perfect squares and perfect cubes. d. they are neither perfect squares nor perfect cubes.

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

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.

52 \43 i don't know this i tried