Question

the diagram shows the plans for the triangular park. how much longer is bc than ac?$(3x^{2}-25x-12)$ feet$(3x^{2}-5x-12)$ feet$(3x^{2}+5x-4)$ feet$(3x^{2}+25x-4)$ feetdone$bc=(10x+3x^{2}-8)$ ft$ab=(12x)$ ft$ac=(15x+4)$ ft

Explanation:

Step1: Identify lengths of BC and AC

Length of $BC = (10x + 3x^2 - 8)$ ft, Length of $AC = (15x + 4)$ ft

Step2: Subtract AC from BC

$$(10x + 3x^2 - 8) - (15x + 4)$$

Step3: Distribute the negative sign

$$10x + 3x^2 - 8 - 15x - 4$$

Step4: Combine like terms

$$3x^2 + (10x - 15x) + (-8 - 4) = 3x^2 - 5x - 12$$

Answer:

B. $(3x^2 - 5x - 12)$ feet