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\(a^3b^4c^7>0\) Which of the following statements must be true? Indicate all such statements. A. \(ab\) is negative B. \(abc\) is positive C. \(ac\) is positive
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Carcass wrote: \(a^3b^4c^7>0\)
Which of the following statements must be true?
Indicate all such statements.
A. \(ab\) is negative
B. \(abc\) is positive
C. \(ac\) is positive the above statement will be true if any of the following statement are true, 1. a , b , c are all positive numbers 2. if a is +ve, b is ve and c is +ve 3. if a is ve, b is +ve and c is ve Therefore only option C is always true
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Carcass wrote: \(a^3b^4c^7>0\)
Which of the following statements must be true?
Indicate all such statements.
A. \(ab\) is negative
B. \(abc\) is positive
C. \(ac\) is positive Key Concept: (any number)^(EVEN INTEGER) ≥ 0 First, since \(a^3b^4c^7>0\), we know that a ≠ 0, b ≠ 0, and c ≠ 0 Next, since \(b^4\) must be POSITIVE, we can safely divide both sides of the inequality by \(b^4\) to get: \(a^3c^7>0\) Also, since \(c^6\) must be POSITIVE, we can safely divide both sides of the inequality by \(c^6\) to get: \(a^3c>0\) Finally, since \(a^2\) must be POSITIVE, we can safely divide both sides of the inequality by \(a^2\) to get: \(ac>0\) So, the ONLY relevant conclusion we can make is that \(ac>0\) Answer: C Cheers, Brent
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Answer: C
A: if ab <0 then: a<0 and b>0 : then c should be negative to have a^3*b^4*c^7 >0 a>0 and b<0 : then it’s ok because b’s power is even. So A is possible but not obligatory. B: it’s possible, But it’s not obligatory. C: ac is positive: True Because power of a and c are odd, they both should have same sign, both negative or positibve to have the whole expression positive.
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the above statement will be true if any of the following statement are true,
1. a , b , c are all positive numbers
2. if a is +ve, b is ve and c is +ve
3. if a is ve, b is +ve and c is ve
Therefore only option C is always true



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GreenlightTestPrep wrote: Carcass wrote: \(a^3b^4c^7>0\)
Which of the following statements must be true?
Indicate all such statements.
A. \(ab\) is negative
B. \(abc\) is positive
C. \(ac\) is positive Key Concept: (any number)^(EVEN INTEGER) ≥ 0 First, since \(a^3b^4c^7>0\), we know that a ≠ 0, b ≠ 0, and c ≠ 0 Next, since \(b^4\) must be POSITIVE, we can safely divide both sides of the inequality by \(b^4\) to get: \(a^3c^7>0\) Also, since \(c^6\) must be POSITIVE, we can safely divide both sides of the inequality by \(c^6\) to get: \(a^3c>0\) Finally, since \(a^2\) must be POSITIVE, we can safely divide both sides of the inequality by \(a^2\) to get: \(ac>0\) So, the ONLY relevant conclusion we can make is that \(ac>0\) Answer: C Cheers, Brent



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Fatemeh wrote: Answer: C A: if ab <0 then: a<0 and b>0 : then c should be negative to have a^3*b^4*c^7 >0 a>0 and b<0 : then it’s ok because b’s power is even. So A is possible but not obligatory.
B: it’s possible, But it’s not obligatory.
C: ac is positive: True Because power of a and c are odd, they both should have same sign, both negative or positibve to have the whole expression positive. I'm confused with this questionanswer at this point, why is B not a "must be true" statement?nvm got it, got tricked abc:   + > + 347:  + + > 



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how come B isn't an answer?



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shadowmr20 wrote: how come B isn't an answer? We're told that \(a^3b^4c^7>0\) So, it COULD be the case that \(a=1\), \(b=1\) and \(c=1\) Statement B says \(abc\) must be positive However, if \(a=1\), \(b=1\) and \(c=1\) then \(abc\) is negative So, we can eliminate B Cheers, Brent
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GreenlightTestPrep wrote: shadowmr20 wrote: how come B isn't an answer? We're told that \(a^3b^4c^7>0\) So, it COULD be the case that \(a=1\), \(b=1\) and \(c=1\) Statement B says \(abc\) must be positive However, if \(a=1\), \(b=1\) and \(c=1\) then \(abc\) is negative So, we can eliminate B Cheers, Brent if b is equal 1 then it must be positive as rules coz \(b^4\). If a or c is equal 1 not the b then we can easily eliminate the answer choice B.
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Brent's answer is a really clever one, and a reminder that when you see inequalities like this on the GRE, you should look to manipulate it so that it's still true. You'll likely find a cool shortcut.









