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If the sides of a triangle have measures 3x + 4, 6x – 1 and 8x + 2, find all possible values of x.

Question 12 options:

a)

x > -4/3


b)

0 < x < 1/6


c)

1/6 < x < 3/11


d)

x > 0

User RobertR
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2 Answers

0 votes

Answer:

Answer: x>-4/3 is correct.

Explanation:

User Ryan Roemer
by
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2 votes

Answer: x>-4/3

Step-by-step explanation:All sides must be greater of zero:

3x+4>0

3x> -4 Divide with 3

x> -4/3

6x-1>0

6x>1 Divide with 6

x>1/6

8x+2>0

8x> -2 Divide with 8

x> -2/8

x> -1/4

Least of that numbers is -4/3= -1.3333

x> -4/3 is solution

User KallDrexx
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