9 – c < 2 c?
-c < -7 therefore c< 7
cannot be determined
9-c<2 Subtract 9 9-9-c<2-9 -c<-7 c>-7
nopes @miteshchvm
-c < -7 therefore if we multiply be -1 we get c > 7 i ought to correct myself ;)
Which graph shows the solution of 9 – c < 2 or –3c > 15? thats the question for my homework
lol m confused by this midnight study! i really nee sleep! :D
what would the graph look like thats basically what its asking
@TheViper and @hartnn how did you prove c > -7 !? its totally wrong as per i observed the solution, i suggest you to look at it once again
would the arrows on the graph be like <------ ------> or o-----------o
well c > 7
if -c<-7 then c>-7
and arrows would be like the second one.
o---------------o
right?
@TheViper whats the logic in that? you can't change one sign. you need to change it for whole function
when we remove negative its basically multiply by -1 and the equality signed is reversed.
@BarbieGurl yes right.. we use hollow circles when < and filled circles for <= or =>
-c < -7 -(-c) > -(-7) c > 7 thats it!! XD
\[9-c <2\] \[-c <2-9\] \[-c <-7\] \[-(c)<-7\] \[c >-7/-1\] \[c >7\]
@BarbieGurl you just go to this site and place your question ;) http://calculator.tutorvista.com/math/452/inequality-calculator.html# thats better
|dw:1348483071489:dw| so wait would it be like that or empty circles?
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