Thanks for pointing out my mistake... yes i cannot divide it by xy...
But are you suggesting that x2 and y2 are different variables? If yes then the question does not make sense at all...
@nuttyvaru: Can you throw some light on this?
well.. I am convinced that I was right to begin with and the answer has to be option E only.. but for ur help I have updated the original question..
I believe I copied and pasted it, and that what made the superscript look like a subscript in here.. and I didnt notice that... anyway.. hope this helps clarify things for you.. 😉
Is sqrt ((x-3)^2) = 3 - x? A) x not equal to 3 B) -x x > 0
I know the answer but no clue?
I feel sqrt can give value of x-3 or 3-x and for x-3 = 3-x gives x = 3 so x not equal to 3 rules out that but if sqrt value is 3-x then its valid for all.
Is sqrt ((x-3)^2) = 3 - x? A) x not equal to 3 B) -x x| > 0
I know the answer but no clue?
I feel sqrt can give value of x-3 or 3-x and for x-3 = 3-x gives x = 3 so x not equal to 3 rules out that but if sqrt value is 3-x then its valid for all.
Hey buddy ...Ans is B ..
statement in bold is incorrect .. square root of a positive number is always non negative..
(x-3)^2 is positive and root of is non negative only... Hence, root of is x-3 and not just x-3 or 3-x...
so, the question asked rephrases to , is x-3 = 3-x ??
for this to hold true, 3-x > 0 i.e x So, finally question boils down to is x
St 1 : x is not equal to 3 ...this does not tell us if x is less than 3 ...not sufficient ..
St 2 : -x x >0... |x is always positive, so -x should be positive or x should be negative ...i.e xHence always, x Hence, Ans B
square root of a positive number is always non negative..
(x-3)^2 is positive and root of is non negative only... Hence, root of is x-3| and not just x-3 or 3-x...
Hi,
Can you explain me how can "square root of a positive number be always non negative"... sqrt(x) takes positive and negative values. I think you meant something else here.
Also, |x-3 is equivallent to saying +/-(x-3) which would be either (x-3) or -(x-3) that is 3-x...
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105? (1) x is a multiple of 9. (2) y is a multiple of 25.
1 is not sufficient to answer the question. From 2, we can conclude that y is a multiple of 5. So xy is multiple of 3, 7 and 5 and hence 105 So the answer is B
Is sqrt ((x-3)^2) = 3 - x? A) x not equal to 3 B) -x x > 0
I know the answer but no clue?
I feel sqrt can give value of x-3 or 3-x and for x-3 = 3-x gives x = 3 so x not equal to 3 rules out that but if sqrt value is 3-x then its valid for all.
Sqrt ((x-3)^2) = 3- x only if x -3 From A, we cannot say if x From B, x If x So B is sufficient to answer the question. Hence choice B.
Can you explain me how can "square root of a positive number be always non negative"... sqrt(x) takes positive and negative values. I think you meant something else here.
Also, x-3| is equivallent to saying +/-(x-3) which would be either (x-3) or -(x-3) that is 3-x...
Hey ...conventionally square root of positive no takes non negative values only ..its the same way as we say 0! =1
eg root 9 is only +3 and not -3 ... And By x-3 i meant the positive absolute value of diff between x and 3 ...it is not the same as (x-3) or (3-x), infact one of the realtion would hold true depending on value of x, but not both
For instance if xand if x>3 then sqrt ((x-3)^2) is x-3 ,
So to sum up , sqrt ((x-3)^2) = |x-3...
can post up few more links regarding sign convention of roots if u need..
There are approximately 2.2 pounds in one kilogram. To the nearest eighteenth, how many eighteenths of a kilogram are in one pound?
1. 7 2. 8 3. 9 4. 39 5. 40
I am not sure what this problem means even..please help.
he he ..this is only tying to play with words .. ok...there would be 18 eighteenths in 1 kilogram ( simply speaking its like 2 halves, 4 quarters etc) i.e in other words 18 eighteenths of a kilogram in 2.2 pounds so in 1 pound there would be 18/2.2 = 8
xy can never simultaneously be lesser than zero and greater than 1 ..only your 1st argument has to hold true..
question says its x square =3/14 ...it has 2 poss roots ...+ve n -ve even if u say x^2 >3/14 it does not imply x is greater than root 3/14 only ..
eg ..x^2> 25 it implies x>5 or x similarly for statement 2 also
Detailed approaches have been discussed 2 pages back ...everybody concurs Ans is E ..
First of all my sincire apologies for severely overlooking the subtle rule that: a^2 > b^2 => A> B; A Still, i've come up with following X> SQRT (3/14) ; x Y > 1 ; Y Now big Question is that by drawing the 4 possible combinations here on number line....overlapping area results that... XY(1-XY) > 0.
