Question

Help me!!!!! please

Answer 1

Answer:

C

Step-by-step explanation:

Cost in location A: $2n-4-2n^2$

Cost in location B: $\dfrac{5}{4}n-\dfrac{6}{5}n^2-\dfrac{11}{3}$

Combined cost: $(2n-4-2n^2)+\left(\dfrac{5}{4}n-\dfrac{6}{5}n^2-\dfrac{11}{3}\right)$

Add these two polynomials. Combine the like terms:

$\left(2n+\dfrac{5}{4}n\right)+\left(-2n^2-\dfrac{6}{5}n^2\right)+\left(-4-\dfrac{11}{3}\right)\\ \\=\dfrac{13}{4}n-\dfrac{16}{5}n^2-\dfrac{23}{3}\\ \\=-\dfrac{16}{5}n^2+\dfrac{13}{4}n-\dfrac{23}{3}$