SOLUTION: −5≥−11
−3≤−9
f 3x−10=17, then x=9
(−1)50=−1 or (−1)99=−1
If a≠3 then |a|≠3.
The square of any real number is a positive number.
7 < 5 or 3 > 1
Algebra ->
Conjunction
-> SOLUTION: −5≥−11
−3≤−9
f 3x−10=17, then x=9
(−1)50=−1 or (−1)99=−1
If a≠3 then |a|≠3.
The square of any real number is a positive number.
7 < 5 or 3 > 1
Log On
Question 1203911: −5≥−11
−3≤−9
f 3x−10=17, then x=9
(−1)50=−1 or (−1)99=−1
If a≠3 then |a|≠3.
The square of any real number is a positive number.
7 < 5 or 3 > 1 Answer by greenestamps(13200) (Show Source):
