You are not logged in.
Graph the set on the real number line
number line goes form -7 to 7
Last edited by Payari (2006-10-16 02:38:55)
Live Love Life
Offline
The tricky part of that question is the modulus function. What that does is it works out the value of the expression inside the two lines, and then gets rid of the minus sign if there is one.
|3| = 3, |-2| = 2, etc.
So when you have an inequality with a modulus sign, you need to split it up into 2 normal equalities.
In this case, that would be x-3 < 4 and -(x-3) < 4.
The first one becomes x < 7, and the second becomes -x < 1, and so x > -1.
Combining these two gives your final answer as -1 < x < 7.
Why did the vector cross the road?
It wanted to be normal.
Offline
thats a tough one
but she is asking for to graph the set on the number line?
so put a dot on -1 or 7 ?
Desi
Raat Key Rani !
Offline
Put a circle on -1 and 7 and draw a line between them. Make sure the circles aren't filled in, to indicate that it's < instead of ≤.
Why did the vector cross the road?
It wanted to be normal.
Offline
The tricky part of that question is the modulus function.
Modulus? You mean absolute value?
Draw a line on a piece of paper, and draw two ticks on it. Label one -4 and one 7. Draw an open circle on each, and highlight the line inbetwee.
That's the graph.
This is because |(-4 - 3)| = |-7| = 7, and thus, any value greater than -4 will be on your line. Also, |7 - 3| = |4| = 4, and thus, any value less than 7 will be on your line.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
The tricky part of that question is the modulus function.
Modulus? You mean absolute value?
Draw a line on a piece of paper, and draw two ticks on it. Label one -4 and one 7. Draw an open circle on each, and highlight the line inbetwee.
That's the graph.
This is because |(-4 - 3)| = |-7| = 7, and thus, any value greater than -4 will be on your line. Also, |7 - 3| = |4| = 4, and thus, any value less than 7 will be on your line.
...but we're looking for |x-3|<4, not |x-3|<7.
-1 < x < 7 is correct, me thinks.
Bad speling makes me [sic]
Offline
The tricky part of that question is the modulus function.
Modulus? You mean absolute value?
They're just the same thing with different names, really. I usually call it modulus when it's written as |x| and absolute value when it's written as abs(x). But it doesn't matter, it's the same thing.
Why did the vector cross the road?
It wanted to be normal.
Offline
Apparently, according to wikipedia, modulus is absolute value in Britian. I find that extremely horrible terminology since the "other" modulus (remainder) is entirely different.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
Apparently, according to wikipedia, modulus is absolute value in Britian. I find that extremely horrible terminology since the "other" modulus (remainder) is entirely different.
We don't use "modulus" for "remainder" in britain, so there isn't a problem over here. I'm pretty sure we don't use it for anything else, either, so all is sound as long as you pick a system and stick to it
Bad speling makes me [sic]
Offline
Is the modulus of a complex number |z| an american or british term ? or both ?
Offline
Is the modulus of a complex number |z| an american or british term ? or both ?
It is certainly used in Britain... don't know about the USA.
Bad speling makes me [sic]
Offline