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If y=−m^2, which of the following must be true?
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01 Jun 2012, 01:55
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If y=−m^2, which of the following must be true? I. y is negative. II. m is nonnegative. III. If m is negative then y is negative. A. I only B. II only C. III only D. I and II only E. II and III only
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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 02:20
Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative. II. m is nonnegative. III. If m is negative then y is negative.
A. I only B. II only C. III only D. I and II only E. II and III only Tricky question. +1. If y=−m^2, which of the following must be true?First of all notice m^2 is always nonnegative, so m^2 is nonpositive (zero or negative), which means that y is zero when m=0 and y is negative for ANY other value of m. I. y is negative > not necessarily true, if m=0 then y=m^2=0; II. m is nonnegative. m can take ANY value: positive, negative, zero. We don't have any restrictions on its value; III. If m is negative then y is negative. m is negative means that m is not zero. As discussed above if m is other than zero (positive or negative) then y is negative: y=(negative^2)=positive=negative (y=(positive^2)=positive=negative). So, this option is always true. Answer: C (III only). Hope it's clear.
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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 03:14



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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 03:28
Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative. II. m is nonnegative. III. If m is negative then y is negative.
A. I only B. II only C. III only D. I and II only E. II and III only Condition I If m=0 then y=0, hence not always true COndition II m can be anything, +ve, 0 or ve. m^2 will be either 0 (for m=0) or +ve Condition III if m is ve (or to say m is less than 0) than m^2 is +ve. Hence m^2 is always ve. if m is ve than y IS ve Always True



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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 04:03
Bunuel wrote: Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative. II. m is nonnegative. III. If m is negative then y is negative.
A. I only B. II only C. III only D. I and II only E. II and III only Tricky question. +1. If y=−m^2, which of the following must be true?First of all notice m^2 is always nonnegative, so m^2 is nonpositive (zero or negative), which means that y is zero when m=0 and y is negative for ANY other value of m. I. y is negative > not necessarily true, if m=0 then y=m^2=0; II. m is nonnegative. m can take ANY value: positive, negative, zero. We don't have any restrictions on its value; III. If m is negative then y is negative. m is negative means that m is not zero. As discussed above if m is other than zero (positive or negative) then y is negative: y=(negative^2)=positive=negative (y=(positive^2)=positive=negative). So, this option is always true. Answer: C (III only). Hope it's clear. If m is positive then too y is negative so statement iii is contradicted ?



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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 04:42
Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is ve" That means it is saying that "y=m^2" is always ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is nonnegative" That means it is saying that in "y=m^2" m is either 0 or +ve But it is not true, m can be ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always ve The only time when y is not ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is ve (less than 0) y is veAlways True



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Re: If y=−m^2, which of the following must be true?
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01 Jun 2012, 14:33
manulath wrote: Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is ve" That means it is saying that "y=m^2" is always ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is nonnegative" That means it is saying that in "y=m^2" m is either 0 or +ve But it is not true, m can be ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always ve The only time when y is not ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is ve (less than 0) y is veAlways TrueI think both of you are correct, I was thinking the question to be " Which of the following is ALWAYS true " Please correct me if I am wrong . I) which of the following MUST be true II) which of the following is ALWAYS true , I think for both of these the answer to this question will be different ?



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Re: If y=−m^2, which of the following must be true?
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02 Jun 2012, 07:26
Joy111 wrote: manulath wrote: Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is ve" That means it is saying that "y=m^2" is always ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is nonnegative" That means it is saying that in "y=m^2" m is either 0 or +ve But it is not true, m can be ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always ve The only time when y is not ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is ve (less than 0) y is veAlways TrueI think both of you are correct, I was thinking the question to be " Which of the following is ALWAYS true " Please correct me if I am wrong . I) which of the following MUST be true II) which of the following is ALWAYS true , I think for both of these the answer to this question will be different ? No, they are the same. Option III must be true because it's always true.
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Re: If y=−m^2, which of the following must be true?
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02 Jun 2012, 07:55
hm, I ignored m=0, since 0 cant have a sign. is it ok to write y=0!?
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Re: If y=−m^2, which of the following must be true?
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Re: If y=−m^2, which of the following must be true?
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02 Jun 2012, 08:13
Bunuel wrote: LalaB wrote: hm,
I ignored m=0, since 0 cant have a sign. is it ok to write y=0!? y=0 simply means that y=0, so there is nothing wrong with that. aha, I found in wiki  http://en.wikipedia.org/wiki/Sign_%28mathematics%29The number zero is neither positive nor negative, and therefore has no sign. In arithmetic, +0 and −0 both denote the same number 0, and the negation of zero is zero itself.I thought it is not allowed to write 0 ,since 0 is neither negative, nor positive. but wiki says 0 is ok and it means just 0. ok, Bunuel, now I agree with u
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Re: If y=−m^2, which of the following must be true?
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Re: If y=−m^2, which of the following must be true?
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19 Jun 2013, 08:00
Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative. II. m is nonnegative. III. If m is negative then y is negative.
A. I only B. II only C. III only D. I and II only E. II and III only Given that \(y = m^2\) If m=0, then y =0 > This eliminates option I. Again, m can be a nonnegative value as there is no restriction on that, thus statement II gets eliminated. Multiply both sides by m assuming \(m\neq{0}\). Thus, \(my = m^3\). Now, if m<0,then \(m^3<0\)and \(m^3>0\). Hence, as my>0, m and y must have the same sign. C.
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Re: If y=−m^2, which of the following must be true?
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30 May 2014, 13:26
The biggest issue here is obviously noticing that m could be 0, thus making y=0 as well. The first option is a tricky one, but normally the other options will at least offer a hint and/or insight into the trick. In this case, option II mentions "nonnegative," which suggests that the reader considers that m could be 0. Summary: Read all of the option choices! They're trying to give you hints



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If y=−m2, which of the following must be true? I) y is negative.
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14 Jun 2015, 12:07
If y=−m^2, which of the following must be true? I) y is negative. II) m is nonnegative. III) If m is negative then y is negative. I only II only III only I and II only II and III only Please help with this question. Thanks



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Re: If y=−m^2, which of the following must be true?
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Re: If y=−m^2, which of the following must be true?
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25 Nov 2016, 09:34
yep, that m=0 does the trick. tricky one!



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