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

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

3

step-by-step explanation: