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The water usage at a car wash is modeled by the equation [tex]$W(x)=4x^3+6x^2-11x+7$[/tex], where [tex]$W$[/tex] is the amount of water in cubic feet and [tex][tex]$x$[/tex][/tex] is the number of hours the car wash is open. The owners of the car wash want to cut back their winter usage during a drought and decide to close the car wash early two days a week. The amount of decrease in water used is modeled by [tex]$D(x)=x^3+2x^2+15$[/tex], where [tex]$D$[/tex] is the amount of water in cubic feet and [tex][tex]$x$[/tex][/tex] is time in hours.

Write a function, [tex]$C(x)$[/tex], to model the water used by the car wash on a shorter day.

A. [tex]$C(x)=4x^3+4x^2-11x+8$[/tex]
B. [tex][tex]$C(x)=3x^3+4x^2-11x-8$[/tex][/tex]
C. [tex]$C(x)=3x^3+4x^2-11x+8$[/tex]
D. [tex]$C(x)=4x^3+4x^2-11x-8$[/tex]

Sagot :

GinnyAnsweridn
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]
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