2. Which is larger?
The common ratio, r , in a geometric sequence whose second term is
24 and whose fifth term is 1536
or
The common difference, d , in an arithmetic sequence whose fourth
term is 16 and whose seventh term is 31.

The common difference d is larger than the common ratio r

Step-by-step explanation:

Geometric sequence

∵ The second term is 24

∴ = 24

∵

- Equate it by its value

∴ ar = 24 ⇒ (1)

∵ The fifth term is 1536

∴   = 1536

∵

- Equate it by its value

∴ = 1536 ⇒ (2)

Divide (2) by (1)

∴

- Divide up and down by ar

∴ r³ = 64

- Take ∛  for both sides

∴ r = 4

Arithmetic sequence

∵ The fourth term is 16

∴ = 16

∵ = a + (4 - 1)d

∴ = a + 3 d

- Equate it by its value

∴ a + 3d = 16 ⇒ (1)

∵ The seventh term is 31

∴ = 31

∵ = a + (7 - 1)d

∴ = a + 6 d

- Equate it by its value

∴ a + 6 d = 31 ⇒ (2)

Subtract equation (1) from equation (2) to eliminate a and find d

∵ (a - a) + (6 d - 3 d) = (31 - 16)

∴ 3 d = 15

- Divide both sides by 3

∴ d = 5

∵ r = 4 and d = 5

∴ d > r

∴ The common difference d is larger than the common ratio r

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