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16 Sep 2014, 01:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
53% (01:09) correct
47%
(01:15)
wrong
based on 330
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If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
16 Sep 2014, 01:25
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Bookmarks
Official Solution:
If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
Note that \(\sqrt{a^2}=|a|\).
Next, since both \(x < 0\) and \(y < 0\), then \(|x| = -x\) and \(|y| = -y\).
Hence, we can express the given expression as follows:
\(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}=\)
\(=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\)
\(=\frac{-x}{x}-\sqrt{y^2}=\)
\(=-1-|y|=\)
\(=-1-(-y)=\)
\(=-1+y\)
Answer: D
If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
Note that \(\sqrt{a^2}=|a|\).
Next, since both \(x < 0\) and \(y < 0\), then \(|x| = -x\) and \(|y| = -y\).
Hence, we can express the given expression as follows:
\(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}=\)
\(=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\)
\(=\frac{-x}{x}-\sqrt{y^2}=\)
\(=-1-|y|=\)
\(=-1-(-y)=\)
\(=-1+y\)
Answer: D
31 Aug 2016, 07:48
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xnthic
having a hard time understanding this basic rule: \(\sqrt{x^2}=|x|\)
Assuming it's an integer, shouldn't this be the positive integer of x? why |x|
Assuming it's an integer, shouldn't this be the positive integer of x? why |x|
MUST KNOW: \(\sqrt{x^2}=|x|\):
The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).
So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).
What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
For example if x = -5, then \(\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|\)
Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html
DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58
Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
Absolute value tips: absolute-value-tips-and-hints-175002.html
DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58
Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
