Ace manufacturing has determined that the cost of labor for producing x transmissions is

L(x) = 0.3x2 + 400x + 550 dollars,

While the cost of materials is

M(x) = 0.1x2 + 50x + 800 dollars.

Write a polynomial T(x) that represents the total cost of materials and labor for producing x transmissions.

Evaluate the total cost polynomial for x =500.

Find the cost of labor for 500 transmissions and the cost of materials for 500 transmissions.

Any and all help would be greatly appreciated! I am totally confused.

To find the polynomial T(x) that represents the total cost of materials and labor for producing x transmissions, you add the cost of labor (L(x)) and the cost of materials (M(x)):

T(x) = L(x) + M(x)

Substituting the given expressions for L(x) and M(x), we have:

T(x) = (0.3x^2 + 400x + 550) + (0.1x^2 + 50x + 800)

Simplifying T(x), we combine like terms:

T(x) = 0.3x^2 + 0.1x^2 + 400x + 50x + 550 + 800

T(x) = 0.4x^2 + 450x + 1350

So, the polynomial T(x) representing the total cost of materials and labor for producing x transmissions is 0.4x^2 + 450x + 1350.

To evaluate the total cost polynomial for x = 500, we substitute x = 500 into T(x):

T(500) = 0.4(500^2) + 450(500) + 1350

T(500) = 0.4(250000) + 225000 + 1350

T(500) = 100000 + 225000 + 1350

T(500) = 326350

The total cost of producing 500 transmissions, according to the given polynomial, is $326350.

To find the cost of labor for 500 transmissions, we substitute x = 500 into the labor cost polynomial, L(x):

L(500) = 0.3(500^2) + 400(500) + 550

L(500) = 0.3(250000) + 200000 + 550

L(500) = 75000 + 200000 + 550

L(500) = 276550

The cost of labor for producing 500 transmissions, according to the given polynomial, is $276550.

To find the cost of materials for 500 transmissions, we substitute x = 500 into the materials cost polynomial, M(x):

M(500) = 0.1(500^2) + 50(500) + 800

M(500) = 0.1(250000) + 25000 + 800

M(500) = 25000 + 25000 + 800

M(500) = 50800

The cost of materials for producing 500 transmissions, according to the given polynomial, is $50800.