Question

sendo J={32÷[(9-3)÷2+5]+3}×2 e R={[(42÷6+1)÷2+1]×3-7}÷4×3+1 qual valor de R ÷J​

Answer 1

Resposta:

1/2 ou 0,5

Explicação passo-a-passo:

J={32÷[(9-3)÷2+5]+3}×2

J={32÷[(6)÷2+5]+3}×2

J={32÷[3+5]+3}×2

J={32÷[8]+3}×2

J={4+3}×2

J={7}×2

J=14

R={[(42÷6+1)÷2+1]×3-7}÷4×3+1

R={[(7+1)÷2+1]×3-7}÷4×3+1

R={[(8)÷2+1]×3-7}÷4×3+1

R={[4+1]×3-7}÷4×3+1

R={[5]×3-7}÷4×3+1

R={15-7}÷4×3+1

R={8}÷4×3+1

R=2×3+1

R=6+1

R=7

R/J = 7/14 = 1/2 = 0,5