At a store, 3 pieces of cake and 4 pieces of bread together cost $18.60. The total cost of 2 pieces of cake and 2 glasses of
juice is $12.34. If one piece of cake costs 2 times as much as
one piece of bread, what is the total cost of 5 pieces of cake
and 4 glasses of juice? Include two decimal places in your
answer. For example if the answer is $12 enter it as $12.00.

9514 1404 393

  $28.40

Step-by-step explanation:

Using the given information, we can write three equations relating the three item prices. Let c, b, and j represent the cost of 1 item each of cake, bread, and juice.

  3c +4b = 18.60

  2c +2j = 12.34

  c = 2b

Multiplying the third equation by 2 gives an expression for 4b that we can substitute into the first equation.

  2c = 4b

  3c +2c = 18.60

  c = 18.60/5 = 3.72

Doubling the second equation almost gives us the purchase we want.

  2(2c +2j) = 2(12.34)

  4c +4j = 24.68

Adding one more cake to this give the desired purchase:

  (4c +4j) +c = 24.68 +3.72

  5c +4j = 28.40

The total cost of 5 pieces of cake and 4 glasses of juice is $28.40.

what is the question here

step-by-step explanation:

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