ten more than 3 times a real number "j" is greater than negative 31. what are all possible values of "j"?!
2007-02-23 13:01:02 · 6 answers · asked by almeda_94 1 in Science & Mathematics ➔ Mathematics
In other words, solve 3j + 10 > -31. Thus, subtract 10 and divide by 3.
2007-02-23 13:04:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Any number greater than -13 and 2/3, including -13 and 2/3.
2007-02-23 21:08:28 · answer #2 · answered by vegtabletarian 2 · 0⤊ 0⤋
If I'm reading this correctly we have
10 + (3 * j) > -31.
so 3 * J > -41
J > - 13 2/3
2007-02-23 21:07:41 · answer #3 · answered by davidbgreensmith 4 · 0⤊ 0⤋
3j+10>-31
3j>-41
j>-13 2/3
2007-02-23 21:06:45 · answer #4 · answered by leo 6 · 0⤊ 0⤋
3j + 10 > -31
3j > -41
j > -41/3
2007-02-23 21:04:18 · answer #5 · answered by Tom :: Athier than Thou 6 · 0⤊ 0⤋
3j + 10 > -31
3j > -41
j> -41/3
2007-02-23 21:04:17 · answer #6 · answered by 7 · 0⤊ 1⤋