3 years ago

A number D is less than or equal to 2 and greater than or equal to negative 2

Mathematics

1 answer:

Alenkinab [10]3 years ago

4 0

$D \le 2$

$D \ge -2$

$-2 \le D$

$-2 \le D \le 2$

Answer: -2 ≤ D ≤ 2

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$\bold{\text{Answer:}\quad \dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}$

<u>Step-by-step explanation:</u>

$.\quad \dfrac{-5x}{8x+7}-\dfrac{6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}+\dfrac{-6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}\bigg(\dfrac{3x+1}{3x+1}\bigg)+\dfrac{-6x^3}{3x+1}\bigg(\dfrac{8x+7}{8x+7}\bigg)\\\\\\=\dfrac{-15x^2-5x}{(8x+7)(3x+1)}+\dfrac{-48x^4-42x^3}{(8x+7)(3x+1)}\\\\\\=\large\boxed{\dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}$