B. A certain chocolate bar costs $50.00 per piece. However, if you buy more than 5 pieces they will mark down the price to P48.00 per piece. Use a piecewise
function to represent the cost in terms of the number of chocolate bars bought.

P(x) = $50.00*x     if    0 ≤ x ≤ 5

P(x) = $48.00*x     if     5 < x

Step-by-step explanation:

A piecewise function is a function that acts differently in different regions of the domain.

In this case we know that:

P = $50.00 per piece if the number of pieces is less or equal than 5.

P = $48.00 per piece if the number of pieces is more than 5.

Then if we define our variable x as the number of pieces of chocolate, we can write the price function as:

P(x) = $50.00*x     if    0 ≤ x ≤ 5

P(x) = $48.00*x     if     5 < x

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

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step-by-step explanation: