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rishi2377
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yes I get E too

x >-2 then
x+2<6
x<4

if x<-2

-(x+2)<6
-x-2<6
-x<8>-8

E it is..
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FN
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yes I get E too

x >-2 then
x+2<6
x<4

if x<-2

-(x+2)<6
-x-2<6
-x<8

x>-8

E it is..
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rishi2377
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fresinha12
yes I get E too

x >-2 then
x+2<6
x<4

if x<-2

-(x+2)<6
-x-2<6
-x<8>-8

E it is..



my mistake just edited it. OA is E. Thx
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rishi2377
fresinha12
yes I get E too

x >-2 then
x+2<6
x<4

if x<-2

-(x+2)<6
-x-2<6
-x<8>-8

E it is..


thx OA is A thx guys


funny, what is the source?
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rishi2377
--(-9)--((-8)--(-7)--(-6)--(-5)--(-4)--(-3)--(-2)--(1)--0--1--2--3--4--5

X on the number line, the shaded interval (-8 to 4) is the graph of which of the following inequalities?

A. |x| ≤ 4
B. |x| ≤ 8
C.|x-2| ≤ 4
D.|x-2| ≤ 6
E. |x+2| ≤ 6


This one isn't particularly difficult. Were u suppose to give us this bit of info: (-8 to 4).

Anyway. E. |x+2|---> x+2 ≤6 ---> x ≤4

|x+2| ---> x+2>=-6 x>=-8 so -8 ≤x ≤4
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ashkrs
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rishi2377
fresinha12
yes I get E too

x >-2 then
x+2<6
x<4

if x<-2

-(x+2)<6
-x-2<6
-x<8>-8

E it is..


thx OA is A thx guys


Then OA is wrong . Change it!
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Note that the centre of the interval is -2, and the endpoints are 6 units from the centre- Remember that |x-y| is the distance between x and y

Thus |x-(-2)| ≤ 6



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