Astore has two types of animal feed available. type a contains 9 pounds of oats and 3 pounds of corn per bag. type b contains 2 pounds of oats and 10 pounds of corn per bag. a farmer wants to combine the two types so that the resulting mixture has at least 57 pounds of oats and at least 75 pounds of corn. the store only has 15 bags of type a feed and 15 bags of type b feed in stock. type a costs $4 per bag, and type b costs $4 per bag. how many bags of each type should the farmer buy to minimize his cost?

Farmer will buy 5 bags of type A and 6 bags of Type B.

Step-by-step explanation:

Type A contains : 9 pounds of oats and 3 pounds of corn per bag

Type B contains : 2 pounds of oats and 10 pounds of corn per bag

Farmer wants the mixture that contains : At least 57 pounds of oats and 75 pounds of corn.

Let the farmer mixes the type A feed = x pounds

and type B feed = y pounds

Amount of oats when mixed : (9x + 2y) pounds

Amount of corn when mixed : (3x + 10y) pounds

Equation for oats : 9x + 2y = 57 --------(1)

Equation for corn : 3x + 10y = 75 ---------(2)

Multiply equation (2) by 3 and subtract it from equation (1)

3(3x + 10y) - (9x + 2y) = 75×3 - 57

9x + 30y - 9x - 2y = 225 - 57

28y = 168

y =

y = 6 bags

from equation (1)

9x + 2×6 = 57

9x + 12 = 57

9x = 57 - 12

9x = 45

x = 5 bags

Since the store has 15 bags of each type, costing the same as $4 per bag.

Therefore, cost of 5 bags of type A and 6 bags of type B will cost = (5×4) + (6×4)

= 20 + 24

= $44

Farmer will buy 5 bags of type A and 6 bags of type B.

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