If n denotes a numner to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be..
1) less than 10
2) between 1 and 1/10
3) between 1/10 and 0
4) between 0 and 1/10
5) greater than 10
my way was...
n^2 , 1/100
1/10 < n < 1/10
since n is negative
1/10 < n
n needs to be reciprocal
10 < n
but which is greater than 10 and which is not answer (not even an option)
Could anyone pls take a look at my procedure and give me advise where I got it wrong...
inequity 54
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 gabriel
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Everything until here is finemagical cook wrote:If n denotes a numner to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be..
1) less than 10
2) between 1 and 1/10
3) between 1/10 and 0
4) between 0 and 1/10
5) greater than 10
my way was...
n^2 , 1/100
1/10 < n < 1/10
since n is negative
1/10 < n
When you take the reciprocal u need to take it on both the sides (this applies to all mathematical operations for example if you multiply one side with 2 then you have to do the same with the other side if you dont then you are changing the relation between the two sides, the same goes for addition, subtraction, division, reciprocals etc.), so the last step should have been 10<1/n.. which basically means n < 10 .. so the answer should be A ..magical cook wrote: n needs to be reciprocal
10 < n

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magical cook, can you confirm that A is the right answer? i'm not seeing how 10<1/n is actually n<10i'm missing something here? if n is actually negative shouldn't taking it over to the other side yield n>10 making the answer E? if you can confirm would appreciate it  thanks

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n^2 < 1/100
i.e n <1> 1/10
since n is to the left of 0, hence n > 1/10
i.e n > 1/10
i.e 10 > 1/n
or 1/n (reciprocal of n)< 10
A
i.e n <1> 1/10
since n is to the left of 0, hence n > 1/10
i.e n > 1/10
i.e 10 > 1/n
or 1/n (reciprocal of n)< 10
A
Regards
Samir
Samir