Determinar la suma de los ángulos exteriores de un poligono de 14 lados:​

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )

hello,

as   long as they add up to 180 its correct.

the answer to 1/4x+5/3x-4=2-1/12x is x=3

hope this if needed hmu