Respuesta :

kudzordzifrancis

Answer:

[tex]d \le -3[/tex]

Step-by-step explanation:

The given inequality is;

[tex]\frac{d-3}{10} \ge \frac{2d+3}{5}+\frac{d+3}{3}[/tex]

Clear the fraction by multiply through by 30;

[tex]3(d-3) \ge 6(2d+3)+10(d+3)[/tex]

Expand;

[tex]3d-9\ge 12d+18+10d+30[/tex]

Group like terms;

[tex]3d-12d-10d \ge 18+30+9[/tex]

[tex]-19d \ge 57[/tex]

Divide through by -19 and reverse the sign.

[tex]d \le -3[/tex]

zainebamir540

Answer:

Choice A is correct. d ≤ -3 is the answer.

Step-by-step explanation:

 We have given an inequality:

[tex]\frac{d-3}{10}\geq \frac{2d+3}{5}+\frac{d+3}{5}[/tex]

We have to solve the inequality.

[tex]\frac{d-3}{10}\geq \frac{3(2d+3)+5(d+3)}{15}=\frac{6d+9+5d+15}{15}=\frac{11d+24}{15}[/tex]

[tex]\frac{d-3}{10}\geq \frac{11d+24}{15}[/tex]

[tex]15(d-3)\geq10(11d+24)[/tex]

[tex]15d-45\geq110d+240[/tex]

15d-110d ≥ 240+45

-95d ≥ 285

d ≤ -3 which is the answer.

Otras preguntas