The sides of a triangle have lengths of x, x + 4, and 20. if the longest side is 20, which of the following values of x would make the triangle obtuse?

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

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}

step-by-step explanation: