Answer:
C
Step-by-step explanation:
Cost in location A: [tex]2n-4-2n^2[/tex]
Cost in location B: [tex]\dfrac{5}{4}n-\dfrac{6}{5}n^2-\dfrac{11}{3}[/tex]
Combined cost: [tex](2n-4-2n^2)+\left(\dfrac{5}{4}n-\dfrac{6}{5}n^2-\dfrac{11}{3}\right)[/tex]
Add these two polynomials. Combine the like terms:
[tex]\left(2n+\dfrac{5}{4}n\right)+\left(-2n^2-\dfrac{6}{5}n^2\right)+\left(-4-\dfrac{11}{3}\right)\\ \\=\dfrac{13}{4}n-\dfrac{16}{5}n^2-\dfrac{23}{3}\\ \\=-\dfrac{16}{5}n^2+\dfrac{13}{4}n-\dfrac{23}{3}[/tex]
