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Natali [406]
3 years ago
10

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

QUESTION 1

The given inequality is

|m|-2>0

We group like terms to get,

|m|>2


This implies that,

-m>2 or m>2.

We simplify the inequality to get,

m or m>2.

We can write this interval notation to get,

(-\infty,-2)\cup (2,+\infty).


QUESTION 2

|x-4|-3\:>\:5.

We group like terms to get,


|x-4|\:>\:5+3.


|x-4|\:>\:8

We split the absolute value sign to get,

-(x-4)\:>\:8 or x-4\:>\:8


This implies that,


x-4\: or x-4\:>\:8


x\: or x\:>\:8+4


x\: or x\:>\:12


We can write this interval notation to get,

(-\infty,-4)\cup (12,+\infty).


QUESTION 3

The given inequality is

|6+9x|\leq 24


We split the absolute value sign to obtain,

-(6+9x)\leq 24 or (6+9x)\leq 24


This simplifies to

6+9x\ge -24 and 6+9x\leq 24


9x\ge -24-6 and 9x\leq 24-6


9x\ge -30 and 9x\leq 18


x\ge -\frac{10}{3} and x\leq 2

-\frac{10}{3}\leq x\leq2

We write this in interval form  to get,

[-\frac{10}{3},2]


QUESTION 4

The given inequality is

|1-5a|>29

We split the absolute value sign to get,

-(1-5a)>29 or 1-5a>29

This simplifies to,

1-5a\: or 1-5a\:>\:29


This implies that,

-5a\: or -5a\:>\:29-1


-5a\: or -5a\:>\:28


a\:>\:6 or a\:

We write this in interval notation to get,

(-\infty,-\frac{28}{5})\cup (6,+\infty)















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