• Matéria: Matemática
  • Autor: Navy123566
  • Perguntado 8 anos atrás

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Anexos:

Respostas

respondido por: Anônimo
1


→BÔC
→BÔA
→AÔC
→BÔE
→BÔC e CÔE
→BÔD e DÔE



A)
x = 3(180-x)
x = 540-3x
4x = 540
x=  \frac{540}{4}
x=135°

B)

 \left \{ {{x+y=90} \atop {x=2y}} \right.
2y+y=90
3y=90
y=30
⇒x+y=90
x+30=90
x=90-30
x=30

C)
180°- 72° 15' 18''
179° 60' 60'' - 72° 15' 18''
=107° 45' 42''

D)
2(90°- 42° 27')
2 · 47° 37'
=95° 14'
 
E)
 \left \{ {{x+y=90} \atop {x-y=50}} \right.
APLICA MÉTODO DA ADIÇÃO
2x=140
x= \frac{140}{2}
x=70°
⇒x+y=90
   70+y=90
    y=20°

F)
2x-40+ \frac{x}{2} =180
 \frac{4x-80+x+35=360}{2}
4x-80+x+35=360
4x + x = 360 + 35
5x=395
x= \frac{395}{5}
x=79

1°ANGULO
2x-40
2·79 - 40
158-40
118°

2° ANGULO 
 \frac{79}{2} + 35
39° 30' + 35°
74° 30'
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PELO AMOR DE DEUS DE VALOR EU DEMOREI A RESOLVER TUDO
ABRAÇO




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