+0  
 
0
740
7
avatar+2498 

 Jan 8, 2016

Best Answer 

 #2
avatar+128079 
+10

ax +by + c = 0   subtract   ax, c from both sides

 

by  = -ax - c       divide both sides by b

 

y  = (-a/b) x - (c/b)     or, in another form

 

y =  -[a/b]x +  (-[c/b] )

 

Note,  the line has a negative slope.....so   -[a/b] < 0   implies  that a/b itself must be positive

 

Also, the y intercept, (-[c/b] ),  is positive........thus, -[c/b]  > 0     which implies that c/b < 0

 

So.....as guest has said, I and III are correct  and (E) is the answer.....

 

 

 

cool cool cool

 Jan 8, 2016
 #1
avatar
+10

OK Solveit..... put it into standard   y = mx + b form

 

y = -a/b x - c/b        From this you can see the slope (and by looking at the graph) is NEGATIVE   so  m = -a/b is negative which means a/b is POSITIVE     so I is an answer

b is the y intercept of the line    or  -c/b  from the graph we can see that -c/b is POSITIVE or > 0 so c/b must be NEGATIVE which means III is true also

 

So I would say answer   'E'          if I did this all right!

 Jan 8, 2016
 #2
avatar+128079 
+10
Best Answer

ax +by + c = 0   subtract   ax, c from both sides

 

by  = -ax - c       divide both sides by b

 

y  = (-a/b) x - (c/b)     or, in another form

 

y =  -[a/b]x +  (-[c/b] )

 

Note,  the line has a negative slope.....so   -[a/b] < 0   implies  that a/b itself must be positive

 

Also, the y intercept, (-[c/b] ),  is positive........thus, -[c/b]  > 0     which implies that c/b < 0

 

So.....as guest has said, I and III are correct  and (E) is the answer.....

 

 

 

cool cool cool

CPhill Jan 8, 2016
 #3
avatar+2498 
+5

Thanks Guest and CPhill !

 Jan 8, 2016
 #4
avatar
0

that finger of yours is so weird

 Jan 8, 2016
 #5
avatar+128079 
0

OK, Solveit........"guest" actually arrived at the correct answer, first.....I just kinda' confirmed it....!!!

 

 

cool cool cool

 Jan 8, 2016
 #6
avatar
0

That finger is very weirdsmiley

 Jan 8, 2016
 #7
avatar+2498 
0

my finger look like a good guy

 Jan 9, 2016

9 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar