Answer :
To find the water used by the car wash on a shorter day, we need to subtract the decrease in water usage, [tex]\(D(x)\)[/tex], from the original water usage, [tex]\(W(x)\)[/tex]. This gives us the function [tex]\(C(x) = W(x) - D(x)\)[/tex].
Given the functions:
[tex]\[ W(x) = 4x^3 + 6x^2 - 11x + 7 \][/tex]
[tex]\[ D(x) = x^3 + 2x^2 + 15 \][/tex]
Subtract [tex]\(D(x)\)[/tex] from [tex]\(W(x)\)[/tex]:
[tex]\[ C(x) = W(x) - D(x) \][/tex]
[tex]\[ C(x) = (4x^3 + 6x^2 - 11x + 7) - (x^3 + 2x^2 + 15) \][/tex]
Now, distribute the negative sign and combine like terms:
[tex]\[ C(x) = 4x^3 + 6x^2 - 11x + 7 - x^3 - 2x^2 - 15 \][/tex]
[tex]\[ C(x) = (4x^3 - x^3) + (6x^2 - 2x^2) + (-11x) + (7 - 15) \][/tex]
Simplify each term:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]
So, the function modeling the water used by the car wash on a shorter day is:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]
Therefore, the correct option is:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]
Given the functions:
[tex]\[ W(x) = 4x^3 + 6x^2 - 11x + 7 \][/tex]
[tex]\[ D(x) = x^3 + 2x^2 + 15 \][/tex]
Subtract [tex]\(D(x)\)[/tex] from [tex]\(W(x)\)[/tex]:
[tex]\[ C(x) = W(x) - D(x) \][/tex]
[tex]\[ C(x) = (4x^3 + 6x^2 - 11x + 7) - (x^3 + 2x^2 + 15) \][/tex]
Now, distribute the negative sign and combine like terms:
[tex]\[ C(x) = 4x^3 + 6x^2 - 11x + 7 - x^3 - 2x^2 - 15 \][/tex]
[tex]\[ C(x) = (4x^3 - x^3) + (6x^2 - 2x^2) + (-11x) + (7 - 15) \][/tex]
Simplify each term:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]
So, the function modeling the water used by the car wash on a shorter day is:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]
Therefore, the correct option is:
[tex]\[ C(x) = 3x^3 + 4x^2 - 11x - 8 \][/tex]