Question

Alguém poderia me ajudar pfvr nessas questões pfvr

Answer 1

Resposta:

7)a)

$a + b + c = 180 \\ \frac{a}{1} = \frac{b}{2} = \frac{c}{6} = \frac{a + b + c}{1 + 2 + 6} = \frac{180}{9} = 20 \\ \\ \frac{a}{1} = 20 == > a = 1 \times 20 = = > a = 20 \\ \frac{b}{2} = 20 = = > b = 2 \times 20 = = > b = 40 \\ \frac{c}{6} = 20 = = > c = 6 \times 20 = = > c = 120$

b)

$a + b + c = 900 \\ \frac{a}{4} = \frac{b}{5} = \frac{c}{9} = \frac{a + b + c}{4 + 5 + 9} = \frac{900}{18} = 50 \\ \\ \frac{a}{4} = 50 == > a = 4 \times 50 = = > a = 200 \\ \frac{b}{5} = 50 = = > b = 5 \times 50 = = > b =250 \\ \frac{c}{9} = 50 = = > c = 9\times 50 = = > c = 450$

8)A=x

B=y

C=z

$\frac{x}{1000}+\frac{y}{1500}+\frac{z}{2500}=\frac{x + y + z}{1000+1500 + 2500}=\frac{20.000}{5000}= 4 \\ \\ \frac{x}{1000}= 4 \\ x =1000\times4\\ x = 4000\\ \\\frac{y}{1500}= 4 \\ y = 1500 \times 4 \\ y = 6000 \\ \frac{z}{2500} = 4 \\ z = 2500 \times 4 \\ z = 10.000$