Greetings:
If a and b each correspond to a point on the real line, then the distance from one point to the other is equal to |a - b| (or, equivalently, |b - a|). Thus the distance from x to 0 is written |x - 0| or simply |x|, and the distance from x to 5 is |x - 5|. Thus x is farther from 0 than it is from 5 implies |x| > |x - 5| (or, |x - 5| < |x|).
b) x is more than b units from a implies |x - a| > b. x is less than c units from a implies |x - a| < c. Since b < |x - a| and |x - a| < c, we conclude b < |x - a| < c.
Regards,
Rich B.
--- In Math4u@yahoogroups.com, dghbosa alias <redgate3000@...> wrote:
>
> Hi everyone can you help me to solve these questions :
>
> 1.Calculate [ ( -2 ) ^ 0 y ^ -1 ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1 y ^ -4 ]
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> 2. Use the absolute value notation " | | " to discribe :
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> a. x is farther from 0 than it is from 5 .
>
> b. x is more than b units from a but less than c units from a .
>
>
> ---------------------------------
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