-1 <=y<=7/3 ................(1)
=> 0<=|y|<=7/3...........(2)
how to derive (2) from (1)
can anybody explain in simpler way in steps ?
thanks
2007-06-17 06:36:14 · 4 answers · asked by calculus 1 in Science & Mathematics ➔ Mathematics
As y varies from -1 to 0 and then to 7/3, the value of |y| varies from 1 to 0 and then back up to 7/3. Since 7/3 > 1 then
0 ≤ |y| ≤ 7/3
Doug
2007-06-17 06:41:34 · answer #1 · answered by doug_donaghue 7 · 1⤊ 0⤋
If (1) is true, then (2) is true. This is because 7/3 = 2.33333 is greater than 1.000. If (1) is true, there are two possibilities: (a) y is negative, and is between -1 and 0.0, or (b) y is positive, and is between 0 and 7/3. If (a) is true, then |y| <= 1, which is included as a possibility in (2).
2007-06-17 13:43:47 · answer #2 · answered by cosmo 7 · 0⤊ 0⤋
y= a possible negative # just with the - taken away so that it's a positive
2007-06-17 13:39:38 · answer #3 · answered by kazekage786 1 · 0⤊ 0⤋
http://regentsprep.org/regents/math/absvalue/Labsolute.htm
2007-06-17 13:40:22 · answer #4 · answered by Anonymous · 0⤊ 0⤋