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|x+2|<8
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x<6
:)
are u sure J?
-10 < x < 6
I'm pretty sure I think. Well I see it as you can not do anything in the absolute value. It becomes what is in there x+2<8 treat the < sign as if it were an = sign suntract 2 And then x<6
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Or am I thinking of a different function?
Yes I looked it up on maythway it is the answer :)
|x+2|<8 -8<x+2<8 -10<x<6 interval notation (-10,6)
I dont understand how I am wrong? Why do you
the interpretation of |x+2|<8 is all the x-values that are within 8 units of -2. So -2-8=-10 and -2+8=6. See the drawing|dw:1317599561866:dw|
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