there is no pic here love
I had this same problem also, with adding a picture. when you add a pic, come back to me and I'll help.
I believe that it’s the first one
Answer:
12<x<16
Step-by-step explanation:
if c^2 > a^2 + b^2
then the triangle is obtuse
since 20 is the longest side
20^2 > x^2 + (x+4)^2
simplify
400 > x^2 + (x+4)(x+4)
FOIL
400 > (x^2 ) + (x^2 + 4x+4x+16)
combine like terms
400 > (2x^2 + 8x+16)
divide by 2
400/2 > 2x^2 /2 + 8x/2 + 16/2
200 > x^2 + 4x +8
subtract 200 from each side
0> x^2 + 4x +8-200
0> x^2 +4x-192
Factor
0 > (x-12) ( x+16)
using the zero product property
0> x-12 0 > x+16
12>x -16>x
x must be greater than 12
we know the longest side is 20
x+4 < 20
subtract 4
x< 16
x>12 and x < 16
12<x<16
EDIT:
Order of operations doesn't matter in addition. So you can add them in any order.
